[{"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The Earth is now starting to get closer to being hospitable to people like us or animals like us. In the last video, we saw during the Proterozoic Eon, oxygen began to accumulate in the atmosphere. This actually caused this first snowball Earth and this mass extinction of all the anaerobic species. But it made conditions suitable for eukaryotic cells. And maybe even more important, these eukaryotic cells were able to form multicellular organisms. And we see where that starts right here on this chart, on this time clark. Multicellular life starts right over here."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it made conditions suitable for eukaryotic cells. And maybe even more important, these eukaryotic cells were able to form multicellular organisms. And we see where that starts right here on this chart, on this time clark. Multicellular life starts right over here. And I want to be clear. All of these things are a bit moving targets. As we discover more things in the geological record and we get more tools at our disposal, these numbers get tweaked."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Multicellular life starts right over here. And I want to be clear. All of these things are a bit moving targets. As we discover more things in the geological record and we get more tools at our disposal, these numbers get tweaked. But they do give you a good sense, based on our current understanding, of when these things start to appear. And coinciding with multicellular life, and this is interesting in its own right because it has its own meta-level effect on evolution, you actually start also having sexual reproduction. And what's interesting about this, why this has such a big impact on evolution, and we talk about it a lot in the biology playlist, is before evolution, variation in DNA had to be completely dependent really on mutations and just random movement around within DNA or maybe some viruses."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As we discover more things in the geological record and we get more tools at our disposal, these numbers get tweaked. But they do give you a good sense, based on our current understanding, of when these things start to appear. And coinciding with multicellular life, and this is interesting in its own right because it has its own meta-level effect on evolution, you actually start also having sexual reproduction. And what's interesting about this, why this has such a big impact on evolution, and we talk about it a lot in the biology playlist, is before evolution, variation in DNA had to be completely dependent really on mutations and just random movement around within DNA or maybe some viruses. Now with sexual reproduction, you had kind of a systematic mixing of DNA so that you got more variation in the gene pool, which allowed more selection for, or I guess you had more variants to select for. And so you kind of had an acceleration in the actual pace of evolution. So that's what we're talking."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's interesting about this, why this has such a big impact on evolution, and we talk about it a lot in the biology playlist, is before evolution, variation in DNA had to be completely dependent really on mutations and just random movement around within DNA or maybe some viruses. Now with sexual reproduction, you had kind of a systematic mixing of DNA so that you got more variation in the gene pool, which allowed more selection for, or I guess you had more variants to select for. And so you kind of had an acceleration in the actual pace of evolution. So that's what we're talking. I've looked at a bunch of sources from, they say, 1.2 billion, 1.5 billion, a little bit over a billion if you score a little bit several hundred million years ago. You start having these multicellular life forms and sexual reproduction. The other thing that we talked about in the proterozoic eon is the accumulation of oxygen allowed the ozone layer to build up."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's what we're talking. I've looked at a bunch of sources from, they say, 1.2 billion, 1.5 billion, a little bit over a billion if you score a little bit several hundred million years ago. You start having these multicellular life forms and sexual reproduction. The other thing that we talked about in the proterozoic eon is the accumulation of oxygen allowed the ozone layer to build up. Ozone is just three oxygen atoms. It is O3. And by the end of the proterozoic eon, so we're talking, I don't know, maybe 550 million years ago, give or take, tens of or maybe even 100 million years."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The other thing that we talked about in the proterozoic eon is the accumulation of oxygen allowed the ozone layer to build up. Ozone is just three oxygen atoms. It is O3. And by the end of the proterozoic eon, so we're talking, I don't know, maybe 550 million years ago, give or take, tens of or maybe even 100 million years. These are all moving targets. The ozone layer was dense enough to protect the land from UV rays. We talked about that in the last video."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And by the end of the proterozoic eon, so we're talking, I don't know, maybe 550 million years ago, give or take, tens of or maybe even 100 million years. These are all moving targets. The ozone layer was dense enough to protect the land from UV rays. We talked about that in the last video. The Earth is being bombarded with UV rays. And the ozone layer is the only thing that really keeps us from being seriously irradiated by the sun and allows land animals to actually live. And so coinciding with that time period, around 550 million years ago, you start to have life colonizing, especially significant life colonizing land."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We talked about that in the last video. The Earth is being bombarded with UV rays. And the ozone layer is the only thing that really keeps us from being seriously irradiated by the sun and allows land animals to actually live. And so coinciding with that time period, around 550 million years ago, you start to have life colonizing, especially significant life colonizing land. So life colonizes land. And this was kind of an interesting, when I first learned it, it was kind of an aha moment. You always assume that trees and grasses are kind of part of the background."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so coinciding with that time period, around 550 million years ago, you start to have life colonizing, especially significant life colonizing land. So life colonizes land. And this was kind of an interesting, when I first learned it, it was kind of an aha moment. You always assume that trees and grasses are kind of part of the background. They come part and parcel with land. But it turns out that animals colonized land before plants did. Plants didn't come into the picture until about 450 million years ago, give or take a few tens of millions of years."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You always assume that trees and grasses are kind of part of the background. They come part and parcel with land. But it turns out that animals colonized land before plants did. Plants didn't come into the picture until about 450 million years ago, give or take a few tens of millions of years. And so we're now entering the end of the Proterozoic Eon. Life has started to colonize land. We now have an ozone layer."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Plants didn't come into the picture until about 450 million years ago, give or take a few tens of millions of years. And so we're now entering the end of the Proterozoic Eon. Life has started to colonize land. We now have an ozone layer. And what happens, and actually there's another snowball glaciation or snowball Earth near the end of the Proterozoic Eon, I should say. And there's a bunch of theories about why it came about and then why it disappeared. Maybe there were volcanoes, greenhouse gases, who knows."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We now have an ozone layer. And what happens, and actually there's another snowball glaciation or snowball Earth near the end of the Proterozoic Eon, I should say. And there's a bunch of theories about why it came about and then why it disappeared. Maybe there were volcanoes, greenhouse gases, who knows. But as we enter the end of that, we start seeing life begin to flourish. And it starts to really flourish as we enter the Phanerozoic Eon. And it's not even labeled here."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe there were volcanoes, greenhouse gases, who knows. But as we enter the end of that, we start seeing life begin to flourish. And it starts to really flourish as we enter the Phanerozoic Eon. And it's not even labeled here. The Phanerozoic Eon is this chunk of time right over here. And let me write it out. This right over here is the Phanerozoic Eon."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's not even labeled here. The Phanerozoic Eon is this chunk of time right over here. And let me write it out. This right over here is the Phanerozoic Eon. And so this chart, these divisions right here are eons. And then they jump into, instead of doing eons here, they then break into eras. Eras are subsets of eons."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right over here is the Phanerozoic Eon. And so this chart, these divisions right here are eons. And then they jump into, instead of doing eons here, they then break into eras. Eras are subsets of eons. They're hundreds of millions of years. So this is the Paleozoic Era, the Mesozoic Era, and the Cenozoic Era. And that's actually our current era."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Eras are subsets of eons. They're hundreds of millions of years. So this is the Paleozoic Era, the Mesozoic Era, and the Cenozoic Era. And that's actually our current era. But perhaps the most interesting, well, I don't want to pick favorites here, but it's one of the most interesting times in the geologic era, is the first period in the Paleozoic Era, which is the first era in the Phanerozoic Eon. And that's the Cambrian Period. You might have heard of it before."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's actually our current era. But perhaps the most interesting, well, I don't want to pick favorites here, but it's one of the most interesting times in the geologic era, is the first period in the Paleozoic Era, which is the first era in the Phanerozoic Eon. And that's the Cambrian Period. You might have heard of it before. The Cambrian Period. That's about this period of time right over here. And during this period of time, the Earth experiences what we call the Cambrian Explosion."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You might have heard of it before. The Cambrian Period. That's about this period of time right over here. And during this period of time, the Earth experiences what we call the Cambrian Explosion. And that's because there's just this explosion in the number of species and genera that existed, the biodiversity, on the planet. It might just be that we had the ozone layer protecting us. Things were colonizing land."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And during this period of time, the Earth experiences what we call the Cambrian Explosion. And that's because there's just this explosion in the number of species and genera that existed, the biodiversity, on the planet. It might just be that we had the ozone layer protecting us. Things were colonizing land. It was an oxygen-rich environment. We start seeing complex, multicellular organisms. It's about that time, if you fast forward maybe a few tens of millions of years, you start seeing the first fish, the first kind of pre-amphibians or proto-amphibians."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Things were colonizing land. It was an oxygen-rich environment. We start seeing complex, multicellular organisms. It's about that time, if you fast forward maybe a few tens of millions of years, you start seeing the first fish, the first kind of pre-amphibians or proto-amphibians. You fast forward a little bit, as we get out of the Cambrian Period, we start seeing plants. So they actually draw it right over here on this land plants, or at this point right over here. And of course, these are moving targets, depending on what we discover in the fossil record."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's about that time, if you fast forward maybe a few tens of millions of years, you start seeing the first fish, the first kind of pre-amphibians or proto-amphibians. You fast forward a little bit, as we get out of the Cambrian Period, we start seeing plants. So they actually draw it right over here on this land plants, or at this point right over here. And of course, these are moving targets, depending on what we discover in the fossil record. And for me, the big aha moment here is so many of these things that you consider fundamental to what Earth is are relatively recent phenomena. Plants weren't on land until about 450 million years ago. Insects weren't on land, or did not even exist, until about 400 million years ago."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And of course, these are moving targets, depending on what we discover in the fossil record. And for me, the big aha moment here is so many of these things that you consider fundamental to what Earth is are relatively recent phenomena. Plants weren't on land until about 450 million years ago. Insects weren't on land, or did not even exist, until about 400 million years ago. Reptiles didn't exist until about 300 million years ago. So we're about right over here now. Mammals didn't exist until about 200 million years ago."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Insects weren't on land, or did not even exist, until about 400 million years ago. Reptiles didn't exist until about 300 million years ago. So we're about right over here now. Mammals didn't exist until about 200 million years ago. Birds didn't exist until about 150 million years ago. The whole dinosaur age, which we kind of consider in our distant past, that's essentially the Mesozoic Era right here. So this is the age of the dinosaurs right over here."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Mammals didn't exist until about 200 million years ago. Birds didn't exist until about 150 million years ago. The whole dinosaur age, which we kind of consider in our distant past, that's essentially the Mesozoic Era right here. So this is the age of the dinosaurs right over here. When you look at your time clock, you can see it's a relatively recent time period. And it actually ends with, we currently believe, a huge rock, a six mile in diameter rock, colliding with what is now the Yucatan Peninsula in Mexico, or right off the coast of the Yucatan Peninsula. And it destroyed all of the large land life forms, especially the dinosaurs."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the age of the dinosaurs right over here. When you look at your time clock, you can see it's a relatively recent time period. And it actually ends with, we currently believe, a huge rock, a six mile in diameter rock, colliding with what is now the Yucatan Peninsula in Mexico, or right off the coast of the Yucatan Peninsula. And it destroyed all of the large land life forms, especially the dinosaurs. And to put all of this in perspective, and actually the thing that really was an aha moment for me, plants are 450 million years ago. Grass, I kind of use this fundamental thing in nature. But grass has only been around for about, I've seen multiple estimates, 40 to 70 million years."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it destroyed all of the large land life forms, especially the dinosaurs. And to put all of this in perspective, and actually the thing that really was an aha moment for me, plants are 450 million years ago. Grass, I kind of use this fundamental thing in nature. But grass has only been around for about, I've seen multiple estimates, 40 to 70 million years. Grass is a relatively new thing on the planet. Flowers have only been around for 130 million years. So there was a time where you had dinosaurs, but you did not have flowers, and you did not have grass."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But grass has only been around for about, I've seen multiple estimates, 40 to 70 million years. Grass is a relatively new thing on the planet. Flowers have only been around for 130 million years. So there was a time where you had dinosaurs, but you did not have flowers, and you did not have grass. And so you fast forward all the way. And so when you look at this scale, it's kind of funny to look at. They say, OK, this is the time period where the dinosaurs showed up."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So there was a time where you had dinosaurs, but you did not have flowers, and you did not have grass. And so you fast forward all the way. And so when you look at this scale, it's kind of funny to look at. They say, OK, this is the time period where the dinosaurs showed up. This whole brown line is where the mammals showed up. So the dinosaurs started to show up along with the mammals. And then, of course, the dinosaurs died out here."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They say, OK, this is the time period where the dinosaurs showed up. This whole brown line is where the mammals showed up. So the dinosaurs started to show up along with the mammals. And then, of course, the dinosaurs died out here. Our ancestors, when the giant rock hit the Earth, must have been burrowed in holes or able to stash some food away or who knows what, and didn't get fully affected. I'm sure most of the large mammals were destroyed. But what's almost humbling or almost humorous or almost ridiculous when you look at this chart is they put a little dot."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then, of course, the dinosaurs died out here. Our ancestors, when the giant rock hit the Earth, must have been burrowed in holes or able to stash some food away or who knows what, and didn't get fully affected. I'm sure most of the large mammals were destroyed. But what's almost humbling or almost humorous or almost ridiculous when you look at this chart is they put a little dot. You can't even see it here. They say 2 million years ago, the first humans. And even this is being pretty generous when they say first humans."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's almost humbling or almost humorous or almost ridiculous when you look at this chart is they put a little dot. You can't even see it here. They say 2 million years ago, the first humans. And even this is being pretty generous when they say first humans. These are really the first pre-humans, the first humans that are the same as us. If you took one of those babies and you brought them up in the suburbs and gave them haircuts and stuff, they would be the same thing as we are. Those didn't exist until 200,000 years ago, give or take."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even this is being pretty generous when they say first humans. These are really the first pre-humans, the first humans that are the same as us. If you took one of those babies and you brought them up in the suburbs and gave them haircuts and stuff, they would be the same thing as we are. Those didn't exist until 200,000 years ago, give or take. 200,000 to 400,000 years ago, I've seen estimates. So this is actually a very generous period of time to say first humans. It's actually 200,000 years ago."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Those didn't exist until 200,000 years ago, give or take. 200,000 to 400,000 years ago, I've seen estimates. So this is actually a very generous period of time to say first humans. It's actually 200,000 years ago. And just to give you an idea of how new we are and how new our evolution is, it was only 5 million years ago. And I mentioned this in a previous video. It was only 5 million years ago."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's actually 200,000 years ago. And just to give you an idea of how new we are and how new our evolution is, it was only 5 million years ago. And I mentioned this in a previous video. It was only 5 million years ago. So this is just to get a sense. This is zero years. Homo sapiens sapien, only around for 200,000 years."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It was only 5 million years ago. So this is just to get a sense. This is zero years. Homo sapiens sapien, only around for 200,000 years. The Neanderthals, they were cousin species. They weren't our ancestors. Many people think they were."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Homo sapiens sapien, only around for 200,000 years. The Neanderthals, they were cousin species. They weren't our ancestors. Many people think they were. They were cousin species. We come from the same root. Although there are now theories that they might have remixed in with homo sapiens."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Many people think they were. They were cousin species. We come from the same root. Although there are now theories that they might have remixed in with homo sapiens. So maybe some of us have some Neanderthal DNA. And it shouldn't be viewed as an insult. They had big brains."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Although there are now theories that they might have remixed in with homo sapiens. So maybe some of us have some Neanderthal DNA. And it shouldn't be viewed as an insult. They had big brains. Well, they didn't necessarily have big brains. They had big heads. But that seems to imply a big brain."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They had big brains. Well, they didn't necessarily have big brains. They had big heads. But that seems to imply a big brain. But who knows? We always tend to portray them as somehow inferior. But I don't want to get into the political correctness of how to portray Neanderthals."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But that seems to imply a big brain. But who knows? We always tend to portray them as somehow inferior. But I don't want to get into the political correctness of how to portray Neanderthals. But anyway, this is a very small period of time. If you go 2 million years, then you get to the pre-human ancestors. And our family tree only diverged from the chimpanzees 5 million years ago."}, {"video_title": "Biodiversity flourishes in Phanerozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I don't want to get into the political correctness of how to portray Neanderthals. But anyway, this is a very small period of time. If you go 2 million years, then you get to the pre-human ancestors. And our family tree only diverged from the chimpanzees 5 million years ago. If you draw that on this clock right here, it would be like 2 pixels, or maybe not even 2 pixels, is when we diverged from the chimpanzees. So hopefully that gives you a sense of things. At least for me, it really puts things in perspective."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So one definition for formal charge is the hypothetical charge that would result if all bonding electrons are shared equally. So let's go down to the dot structure on the left here, which is the dot structure for methanol, and let's assign a formal charge to carbon. We need to think about the bonding electrons, or the electrons in those bonds around carbon, and we know that each bond consists of two electrons. So the bond between oxygen and carbon consists of two electrons. Let me go ahead and draw in those two electrons. Same for the bond between carbon and hydrogen. Each bond consists of two electrons, so I can go around and put in all of my bonding electrons."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So the bond between oxygen and carbon consists of two electrons. Let me go ahead and draw in those two electrons. Same for the bond between carbon and hydrogen. Each bond consists of two electrons, so I can go around and put in all of my bonding electrons. So if we want to assign a formal charge to carbon, we need to think about the number of valence electrons in the free atom, or the number of valence electrons that carbon is supposed to have. We already know that carbon is supposed to have four valence electrons. So I can put a four here, and from that four, we're going to subtract the number of valence electrons in the bonded atom, or the number of valence electrons that carbon has around it in our drawing."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Each bond consists of two electrons, so I can go around and put in all of my bonding electrons. So if we want to assign a formal charge to carbon, we need to think about the number of valence electrons in the free atom, or the number of valence electrons that carbon is supposed to have. We already know that carbon is supposed to have four valence electrons. So I can put a four here, and from that four, we're going to subtract the number of valence electrons in the bonded atom, or the number of valence electrons that carbon has around it in our drawing. And since we're doing formal charge, we need to think about all those bonding electrons being shared equally. So we think about a covalent bond. So if we have two electrons in one bond, and those two electrons are shared equally, we could split them up."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So I can put a four here, and from that four, we're going to subtract the number of valence electrons in the bonded atom, or the number of valence electrons that carbon has around it in our drawing. And since we're doing formal charge, we need to think about all those bonding electrons being shared equally. So we think about a covalent bond. So if we have two electrons in one bond, and those two electrons are shared equally, we could split them up. We could give one electron to oxygen, and one electron to carbon in that bond. We go over here to this carbon-hydrogen bond, and we could do the same thing. We have two electrons, we could split up those two electrons."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So if we have two electrons in one bond, and those two electrons are shared equally, we could split them up. We could give one electron to oxygen, and one electron to carbon in that bond. We go over here to this carbon-hydrogen bond, and we could do the same thing. We have two electrons, we could split up those two electrons. We could give one to carbon, and one to hydrogen. And we go all the way around, and we do the same thing over here, split up those electrons, and the same thing here. So how many valence electrons do we see around carbon now?"}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "We have two electrons, we could split up those two electrons. We could give one to carbon, and one to hydrogen. And we go all the way around, and we do the same thing over here, split up those electrons, and the same thing here. So how many valence electrons do we see around carbon now? So let me go ahead and highlight them. There's one, two, three, and four. So that's the number of valence electrons around carbon in our drawing."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So how many valence electrons do we see around carbon now? So let me go ahead and highlight them. There's one, two, three, and four. So that's the number of valence electrons around carbon in our drawing. So four minus four is equal to zero. So zero is the formal charge of carbon. So let me go ahead and highlight that here."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So that's the number of valence electrons around carbon in our drawing. So four minus four is equal to zero. So zero is the formal charge of carbon. So let me go ahead and highlight that here. So in this molecule, the formal charge for carbon is zero. Now let's move on to oxidation states. So you could also call these oxidation numbers."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight that here. So in this molecule, the formal charge for carbon is zero. Now let's move on to oxidation states. So you could also call these oxidation numbers. So one definition for an oxidation state is the hypothetical charge that would result if all of those bonding electrons are assigned to the more electronegative atom in the bond. So let's go to the dot structure on the right of methanol, and let's assign an oxidation state to that carbon. We need to think about our bonding electrons again, so let's go ahead and put those in."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So you could also call these oxidation numbers. So one definition for an oxidation state is the hypothetical charge that would result if all of those bonding electrons are assigned to the more electronegative atom in the bond. So let's go to the dot structure on the right of methanol, and let's assign an oxidation state to that carbon. We need to think about our bonding electrons again, so let's go ahead and put those in. So we know that each bond consists of two electrons, so I'm putting in the two electrons in each bond. And let's think about the oxidation state of that carbon. Well, first, we need to know the number of valence electrons in the free atom."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "We need to think about our bonding electrons again, so let's go ahead and put those in. So we know that each bond consists of two electrons, so I'm putting in the two electrons in each bond. And let's think about the oxidation state of that carbon. Well, first, we need to know the number of valence electrons in the free atom. So just like before, we know that carbon is supposed to have four valence electrons. So this would be a four. And from that, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons that carbon actually has in the drawing."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Well, first, we need to know the number of valence electrons in the free atom. So just like before, we know that carbon is supposed to have four valence electrons. So this would be a four. And from that, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons that carbon actually has in the drawing. This time, we need to think about an ionic bond, so we're going to pretend like a covalent bond is an ionic bond because we're going to assign all of the bonding electrons to the more electronegative atom. So there's no more sharing here. The winner takes all, the more electronegative atom is going to get all of the electrons."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And from that, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons that carbon actually has in the drawing. This time, we need to think about an ionic bond, so we're going to pretend like a covalent bond is an ionic bond because we're going to assign all of the bonding electrons to the more electronegative atom. So there's no more sharing here. The winner takes all, the more electronegative atom is going to get all of the electrons. So let's think about the electronegativities of carbon versus oxygen. We know that oxygen is more electronegative than carbon, so oxygen takes both of those electrons in that bond, so oxygen gets both of those electrons. Next, let's think about the electronegativities of carbon and hydrogen."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "The winner takes all, the more electronegative atom is going to get all of the electrons. So let's think about the electronegativities of carbon versus oxygen. We know that oxygen is more electronegative than carbon, so oxygen takes both of those electrons in that bond, so oxygen gets both of those electrons. Next, let's think about the electronegativities of carbon and hydrogen. We know that carbon is a little bit more electronegative than hydrogen. So for these two electrons, carbon's going to take both of them since carbon is more electronegative than hydrogen. And the same thing for our other carbon-hydrogen bonds."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's think about the electronegativities of carbon and hydrogen. We know that carbon is a little bit more electronegative than hydrogen. So for these two electrons, carbon's going to take both of them since carbon is more electronegative than hydrogen. And the same thing for our other carbon-hydrogen bonds. Carbon is more electronegative than hydrogen, so carbon takes those. Carbon is more electronegative than hydrogen, so carbon takes those. And so how many electrons do we have around carbon now?"}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And the same thing for our other carbon-hydrogen bonds. Carbon is more electronegative than hydrogen, so carbon takes those. Carbon is more electronegative than hydrogen, so carbon takes those. And so how many electrons do we have around carbon now? Let's count them up. That's one, two, three, four, five, and six. So now we have six electrons around carbon."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And so how many electrons do we have around carbon now? Let's count them up. That's one, two, three, four, five, and six. So now we have six electrons around carbon. So four minus six gives us negative two. So here, in this example, carbon has an oxidation state of negative two. So there's no more sharing when you're doing oxidation states, right?"}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So now we have six electrons around carbon. So four minus six gives us negative two. So here, in this example, carbon has an oxidation state of negative two. So there's no more sharing when you're doing oxidation states, right? Think about the more electronegative atom and assign both electrons to the more electronegative atom. Both formal charge and oxidation states are just really extreme methods of electron bookkeeping. They're not perfect."}, {"video_title": "Comparing formal charges to oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So there's no more sharing when you're doing oxidation states, right? Think about the more electronegative atom and assign both electrons to the more electronegative atom. Both formal charge and oxidation states are just really extreme methods of electron bookkeeping. They're not perfect. They're certainly not perfect. We're assuming that the electrons are either shared equally, perfectly, or that one atom takes both electrons, and neither of those concepts is perfect in the real world. But it works when we're drawing our dot structures and we're thinking about chemical reactions."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what happens is those two hydrogens from the hydrogen gas are added across your double bond, and they're added on the same side of where the double bond used to be. So it's a syn addition. Let's take a look at why this is a syn addition of hydrogens. So we have our metal catalyst over here. So let's go ahead and draw our flat metal catalyst. And these metals are chosen because they adsorb hydrogen really well, which means that if you bubble hydrogen gas through, the hydrogen is going to be adsorbed to the surface of that metal catalyst like that. And then your alkene comes along, and your alkene is also flat."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have our metal catalyst over here. So let's go ahead and draw our flat metal catalyst. And these metals are chosen because they adsorb hydrogen really well, which means that if you bubble hydrogen gas through, the hydrogen is going to be adsorbed to the surface of that metal catalyst like that. And then your alkene comes along, and your alkene is also flat. The portion of the molecule that contains the double bond. So these two carbons, this carbon, this carbon, are sp2 hybridized, which means that the stereochemistry around those two carbons, it's going to be flat. So this portion of the molecule is flat."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then your alkene comes along, and your alkene is also flat. The portion of the molecule that contains the double bond. So these two carbons, this carbon, this carbon, are sp2 hybridized, which means that the stereochemistry around those two carbons, it's going to be flat. So this portion of the molecule is flat. So you have one thing that's flat approaching something else that's flat. So the only way those hydrogens can add are to add them onto the same side. So if this carbon and this carbon, if you add this hydrogen to the carbon on the left and add this hydrogen to the carbon on the right, and then you go ahead and you draw the rest of the bonds, this would now be a wedge and then a dash, and then this would be a wedge and a dash."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this portion of the molecule is flat. So you have one thing that's flat approaching something else that's flat. So the only way those hydrogens can add are to add them onto the same side. So if this carbon and this carbon, if you add this hydrogen to the carbon on the left and add this hydrogen to the carbon on the right, and then you go ahead and you draw the rest of the bonds, this would now be a wedge and then a dash, and then this would be a wedge and a dash. You can see those two hydrogens have added onto the same side. So these two hydrogens are these two hydrogens for our syn addition. Notice we're also changing from sp2 hybridization to sp3 hybridization over here on the right."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if this carbon and this carbon, if you add this hydrogen to the carbon on the left and add this hydrogen to the carbon on the right, and then you go ahead and you draw the rest of the bonds, this would now be a wedge and then a dash, and then this would be a wedge and a dash. You can see those two hydrogens have added onto the same side. So these two hydrogens are these two hydrogens for our syn addition. Notice we're also changing from sp2 hybridization to sp3 hybridization over here on the right. So we have to think about stereochemistry for this reaction for your products as well. So let's take a look at an actual reaction here. And let's see if we can follow along."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Notice we're also changing from sp2 hybridization to sp3 hybridization over here on the right. So we have to think about stereochemistry for this reaction for your products as well. So let's take a look at an actual reaction here. And let's see if we can follow along. So if this was my reaction, I want to hydrogenate this alkene. So I would add some hydrogen gas, and I could choose whichever metal catalyst I wanted to. I would add two hydrogens on the same side."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's see if we can follow along. So if this was my reaction, I want to hydrogenate this alkene. So I would add some hydrogen gas, and I could choose whichever metal catalyst I wanted to. I would add two hydrogens on the same side. So I could add two hydrogens on the same side, just like I did up there. So we would get now everything changes from sp2 hybridization to sp3, so we have wedges and dashes to worry about. And usually you wouldn't see it drawn like this."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I would add two hydrogens on the same side. So I could add two hydrogens on the same side, just like I did up there. So we would get now everything changes from sp2 hybridization to sp3, so we have wedges and dashes to worry about. And usually you wouldn't see it drawn like this. That's too much work, quite frankly. It would be much easier just to say, oh, well, all I have to do is take away the double bond, and there's my product. So for some of these reactions, they're very, very, very simple."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And usually you wouldn't see it drawn like this. That's too much work, quite frankly. It would be much easier just to say, oh, well, all I have to do is take away the double bond, and there's my product. So for some of these reactions, they're very, very, very simple. Just take away the double bond, and you'll end up with your alkane-like product. Let's take a look at oxidation states for this reaction. So I'm going to redraw this reaction."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So for some of these reactions, they're very, very, very simple. Just take away the double bond, and you'll end up with your alkane-like product. Let's take a look at oxidation states for this reaction. So I'm going to redraw this reaction. And this time I'm going to draw in my atoms. And I'm also going to draw in my electrons here in a second. So I'm just drawing out all the atoms here, so I have all these methyl groups to worry about."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to redraw this reaction. And this time I'm going to draw in my atoms. And I'm also going to draw in my electrons here in a second. So I'm just drawing out all the atoms here, so I have all these methyl groups to worry about. And then I have electrons in these bonds. Each one of these bonds consists of two electrons. I'm going to go ahead and put in all of my electrons here like that."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just drawing out all the atoms here, so I have all these methyl groups to worry about. And then I have electrons in these bonds. Each one of these bonds consists of two electrons. I'm going to go ahead and put in all of my electrons here like that. Now let's assign oxidation states to those two carbons that formed our double bond. So let's look at oxidation state for the top carbon. Remember, when you're doing oxidation states, you're worried about electronegativity."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and put in all of my electrons here like that. Now let's assign oxidation states to those two carbons that formed our double bond. So let's look at oxidation state for the top carbon. Remember, when you're doing oxidation states, you're worried about electronegativity. So oxidation states are all about electronegativity. So go back and watch the earlier video on oxidation states. So we have here comparing the electronegativities of carbon and carbon."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Remember, when you're doing oxidation states, you're worried about electronegativity. So oxidation states are all about electronegativity. So go back and watch the earlier video on oxidation states. So we have here comparing the electronegativities of carbon and carbon. Well, obviously they're the exact same. So in the struggle for these electrons, it gets divided. Each one of these carbons is going to get one of these electrons."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have here comparing the electronegativities of carbon and carbon. Well, obviously they're the exact same. So in the struggle for these electrons, it gets divided. Each one of these carbons is going to get one of these electrons. So that's the case for all of these right here. So that carbon has four electrons around it. It's the exact same thing for this carbon."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Each one of these carbons is going to get one of these electrons. So that's the case for all of these right here. So that carbon has four electrons around it. It's the exact same thing for this carbon. This carbon has four electrons around it. To assign an oxidation state, we take the number of valence electrons that atom usually has, which carbon normally has four, of course. And from that, we subtract the number of electrons we just drew around it for our dot structure."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's the exact same thing for this carbon. This carbon has four electrons around it. To assign an oxidation state, we take the number of valence electrons that atom usually has, which carbon normally has four, of course. And from that, we subtract the number of electrons we just drew around it for our dot structure. So that would be four. Each one of those carbons has four. So each of these carbons has an oxidation state of zero."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And from that, we subtract the number of electrons we just drew around it for our dot structure. So that would be four. Each one of those carbons has four. So each of these carbons has an oxidation state of zero. Let's look at the product, and let's see if we can assign some oxidation states for the product. So our product over here on the right, we had a carbon, and we had some methyl groups bonded to that carbon. We added on a hydrogen."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So each of these carbons has an oxidation state of zero. Let's look at the product, and let's see if we can assign some oxidation states for the product. So our product over here on the right, we had a carbon, and we had some methyl groups bonded to that carbon. We added on a hydrogen. So each one of these carbons got a hydrogen added onto it. And let's go ahead and fill in our electrons in these bonds. So once again, each bond consists of two electrons like that."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We added on a hydrogen. So each one of these carbons got a hydrogen added onto it. And let's go ahead and fill in our electrons in these bonds. So once again, each bond consists of two electrons like that. And now we have a single bond between our carbons. And let's assign some oxidation states. So once again, we know that the two carbons have the same electronegativity."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So once again, each bond consists of two electrons like that. And now we have a single bond between our carbons. And let's assign some oxidation states. So once again, we know that the two carbons have the same electronegativity. So the tug of war for these two electrons right here, it's a tie. So it's a tie. It's a tie."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we know that the two carbons have the same electronegativity. So the tug of war for these two electrons right here, it's a tie. So it's a tie. It's a tie. What about carbon versus hydrogen? Carbon is actually more electronegative than hydrogen. So in the war over the two electrons in the carbon-hydrogen bond, carbon wins because it's a little bit more electronegative."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's a tie. What about carbon versus hydrogen? Carbon is actually more electronegative than hydrogen. So in the war over the two electrons in the carbon-hydrogen bond, carbon wins because it's a little bit more electronegative. So we're going to assign this extra electron here to carbon. And then again, carbon versus carbon, so that carbon gets that electron as well. Same thing down here."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in the war over the two electrons in the carbon-hydrogen bond, carbon wins because it's a little bit more electronegative. So we're going to assign this extra electron here to carbon. And then again, carbon versus carbon, so that carbon gets that electron as well. Same thing down here. So it's a tie. It's a tie. Carbon beats hydrogen."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Same thing down here. So it's a tie. It's a tie. Carbon beats hydrogen. And over here, it's a tie. So in the dot structure on the right, the oxidation states, the normal number of valence electrons would be 4. From that, we subtract the number of electrons in our picture here, which would be 5 electrons."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Carbon beats hydrogen. And over here, it's a tie. So in the dot structure on the right, the oxidation states, the normal number of valence electrons would be 4. From that, we subtract the number of electrons in our picture here, which would be 5 electrons. Each one of these carbons has 5 electrons around it. So it gained electron. So 4 minus 5 will give us a negative 1."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "From that, we subtract the number of electrons in our picture here, which would be 5 electrons. Each one of these carbons has 5 electrons around it. So it gained electron. So 4 minus 5 will give us a negative 1. So the oxidation states of these two carbons is negative 1. And we can look at our original oxidation states of being 0. Went from 0 to negative 1."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So 4 minus 5 will give us a negative 1. So the oxidation states of these two carbons is negative 1. And we can look at our original oxidation states of being 0. Went from 0 to negative 1. That's a decrease in the oxidation state. A decrease in the oxidation state means reduction. So this is a reduction reaction."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Went from 0 to negative 1. That's a decrease in the oxidation state. A decrease in the oxidation state means reduction. So this is a reduction reaction. So the alkene is reduced by the addition of these two hydrogens. And you'll see other definitions for oxidation states. You'll see a gain in hydrogens is reduction."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is a reduction reaction. So the alkene is reduced by the addition of these two hydrogens. And you'll see other definitions for oxidation states. You'll see a gain in hydrogens is reduction. That's another definition that's often found in organic chemistry textbooks. And while that's true, to me it makes more sense to go ahead and assign your oxidation states and watch the oxidation states change as you add those hydrogens, as your molecule gains hydrogens. So this is a reduction."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You'll see a gain in hydrogens is reduction. That's another definition that's often found in organic chemistry textbooks. And while that's true, to me it makes more sense to go ahead and assign your oxidation states and watch the oxidation states change as you add those hydrogens, as your molecule gains hydrogens. So this is a reduction. Let's look at the stereochemistry of the hydrogenation reaction. So let's do an example involving stereochemistry. So let's say your alkene, let's do that ring again."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is a reduction. Let's look at the stereochemistry of the hydrogenation reaction. So let's do an example involving stereochemistry. So let's say your alkene, let's do that ring again. Wasn't a very good one. So let's say your alkene looked something like this. And you're going to react that with hydrogen and with platinum."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say your alkene, let's do that ring again. Wasn't a very good one. So let's say your alkene looked something like this. And you're going to react that with hydrogen and with platinum. Well, your first thought might be, OK, this is simple. All I have to do is take away that double bond and I'm done. Well, sometimes that's true."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And you're going to react that with hydrogen and with platinum. Well, your first thought might be, OK, this is simple. All I have to do is take away that double bond and I'm done. Well, sometimes that's true. But in this case, we actually formed two new chirality centers. So this top carbon here is a chirality center. And this bottom carbon here is also a chirality center."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, sometimes that's true. But in this case, we actually formed two new chirality centers. So this top carbon here is a chirality center. And this bottom carbon here is also a chirality center. So sometimes it's not quite that simple. We need to think about the syn addition of those hydrogens. We need to think about the possible products that would result."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this bottom carbon here is also a chirality center. So sometimes it's not quite that simple. We need to think about the syn addition of those hydrogens. We need to think about the possible products that would result. So we're going to get two products here. Let's look at the one on the left. Well, one possibility is I could add those two hydrogens on the same side as a wedge."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We need to think about the possible products that would result. So we're going to get two products here. Let's look at the one on the left. Well, one possibility is I could add those two hydrogens on the same side as a wedge. So I have one hydrogen as a wedge, the other hydrogen as a wedge. That's our syn addition. And that means that at this top carbon here, this ethyl group must be going away from me."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, one possibility is I could add those two hydrogens on the same side as a wedge. So I have one hydrogen as a wedge, the other hydrogen as a wedge. That's our syn addition. And that means that at this top carbon here, this ethyl group must be going away from me. So there's my ethyl group going away from me. And down here at the bottom carbon, the methyl group must be going away from me. So that's one possible product."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that means that at this top carbon here, this ethyl group must be going away from me. So there's my ethyl group going away from me. And down here at the bottom carbon, the methyl group must be going away from me. So that's one possible product. The other possibility, instead of having my two hydrogens add as wedges, I can have my two hydrogens add as dashes. So there's a hydrogen, and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's one possible product. The other possibility, instead of having my two hydrogens add as wedges, I can have my two hydrogens add as dashes. So there's a hydrogen, and then here's a dash, and there's a hydrogen. So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon, now my methyl group is coming out at me like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So at this top carbon here, now my ethyl group is coming out at me. And at this bottom carbon, now my methyl group is coming out at me like this. So I have two possibilities. And if I look at these two products, I can see that they are enantiomers. They are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And if I look at these two products, I can see that they are enantiomers. They are mirror images of each other. So these two would be my enantiomers, and these would be the products of my reaction. So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So be very careful when thinking about syn additions here. Let's do one more example of a hydrogenation reaction. Let's do a bridged bicyclic compound. So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. So I'm going to, there's my bridge. And then I'm going to go like this, and then I'm going to draw that back carbon a little bit off like that."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a famous bridged bicyclic compound. Let's see if we can draw it here. So I'm going to, there's my bridge. And then I'm going to go like this, and then I'm going to draw that back carbon a little bit off like that. And my double bond is going to go right here, and then this is going to be a methyl group. And then up here are going to be two methyl groups like that. So this is alpha-pining."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then I'm going to go like this, and then I'm going to draw that back carbon a little bit off like that. And my double bond is going to go right here, and then this is going to be a methyl group. And then up here are going to be two methyl groups like that. So this is alpha-pining. This is alpha-pining right here, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pining molecule and I wanted to hydrogenate it, so I could use palladium and charcoal, or palladium and carbon."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is alpha-pining. This is alpha-pining right here, found in turpentine. And you can see there's an alkene on this. So if I took this alpha-pining molecule and I wanted to hydrogenate it, so I could use palladium and charcoal, or palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat like that. And so when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond. Think about this guy over here."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I took this alpha-pining molecule and I wanted to hydrogenate it, so I could use palladium and charcoal, or palladium and carbon. And if I think about what happens in this mechanism, I know that my metal catalyst there, my palladium, is going to be flat like that. And so when it has those hydrogens, when the palladium adsorbs those hydrogens, it's going to add those two hydrogens to my double bond. Think about this guy over here. Think about the alpha-pining as molecules like a spaceship. And the spaceship is approaching the docking station. So the spaceship is slowly going down."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Think about this guy over here. Think about the alpha-pining as molecules like a spaceship. And the spaceship is approaching the docking station. So the spaceship is slowly going down. The spaceship is going to approach the docking station. And there's only one way the spaceship can approach the docking station. And that is the way in which we have drawn it right here."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the spaceship is slowly going down. The spaceship is going to approach the docking station. And there's only one way the spaceship can approach the docking station. And that is the way in which we have drawn it right here. It could not flip upside down and approach it from the top because of the steric hindrance of these methyl groups. So this is the way that it approaches. In this part of the molecule, your alkene is the flat part."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that is the way in which we have drawn it right here. It could not flip upside down and approach it from the top because of the steric hindrance of these methyl groups. So this is the way that it approaches. In this part of the molecule, your alkene is the flat part. So it's easiest for the molecule to approach in this way. The spaceship analogy always helps my students. So there's only one product for this reaction."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "In this part of the molecule, your alkene is the flat part. So it's easiest for the molecule to approach in this way. The spaceship analogy always helps my students. So there's only one product for this reaction. And let's see if we can draw it here. So there's one product for this reaction. And let's see what it would look like."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So there's only one product for this reaction. And let's see if we can draw it here. So there's one product for this reaction. And let's see what it would look like. It would look something like this. So we have our two methyl groups right here. So the hydrogens are going to add from below."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's see what it would look like. It would look something like this. So we have our two methyl groups right here. So the hydrogens are going to add from below. So you can see the hydrogens are going to add from below. So this hydrogen, let's say it adds right here. That's going to push this methyl group up."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the hydrogens are going to add from below. So you can see the hydrogens are going to add from below. So this hydrogen, let's say it adds right here. That's going to push this methyl group up. So it's going to push that methyl group up. So that methyl group gets pushed up when that hydrogen adds right down here. And then this other hydrogen is going to add to the opposite side."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's going to push this methyl group up. So it's going to push that methyl group up. So that methyl group gets pushed up when that hydrogen adds right down here. And then this other hydrogen is going to add to the opposite side. This other hydrogen is going to add to this one right here. And so we can show the addition of that hydrogen. So there's my syn addition of these two hydrogens."}, {"video_title": "Hydrogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then this other hydrogen is going to add to the opposite side. This other hydrogen is going to add to this one right here. And so we can show the addition of that hydrogen. So there's my syn addition of these two hydrogens. And there was something else in that carbon. It was another hydrogen. So another hydrogen got pushed up right here as well."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we're going to talk a little bit about amines. And these are just organic compounds, where you have a nitrogen bonded to groups that contain carbon. So if I were to just draw some amines right here, you could have something like this, where you have a nitrogen bonded to two hydrogens, and then maybe some type of carbon chain. Maybe it's just one carbon right here. So maybe you have a carbon which is also bonded to three hydrogens like this. In this case, you would have a primary amine. If you have the nitrogen bonded to two carbons, so if you have something like this, a nitrogen bonded to two carbons, so it has one carbon chain right there."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "Maybe it's just one carbon right here. So maybe you have a carbon which is also bonded to three hydrogens like this. In this case, you would have a primary amine. If you have the nitrogen bonded to two carbons, so if you have something like this, a nitrogen bonded to two carbons, so it has one carbon chain right there. And I'm just drawing methyl groups here, because the chain could just keep on going. Maybe you have two like that, and then you have a hydrogen right there. This would be a secondary amine."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "If you have the nitrogen bonded to two carbons, so if you have something like this, a nitrogen bonded to two carbons, so it has one carbon chain right there. And I'm just drawing methyl groups here, because the chain could just keep on going. Maybe you have two like that, and then you have a hydrogen right there. This would be a secondary amine. And then finally, you could imagine if you had it bonded to three, it would be a tertiary amine. This is just to get you introduced to the terminology. Just like that."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "This would be a secondary amine. And then finally, you could imagine if you had it bonded to three, it would be a tertiary amine. This is just to get you introduced to the terminology. Just like that. Now, like all of the other new groups or new types of compounds that we've explored, what I want to do is just introduce you to the naming of it. Because one, that lets you recognize them when you hear their names, but it also, I think, on some level, familiarizes you with your structure. So let's do a couple of naming examples."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "Just like that. Now, like all of the other new groups or new types of compounds that we've explored, what I want to do is just introduce you to the naming of it. Because one, that lets you recognize them when you hear their names, but it also, I think, on some level, familiarizes you with your structure. So let's do a couple of naming examples. I drew these ahead of time. And in general, just to remember, amines are pretty high priority group. Out of all of the things that we've learned so far, the only thing that is a higher priority is actually the alcohols, or actually the thiols, which I don't even remember if I did a video on it."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So let's do a couple of naming examples. I drew these ahead of time. And in general, just to remember, amines are pretty high priority group. Out of all of the things that we've learned so far, the only thing that is a higher priority is actually the alcohols, or actually the thiols, which I don't even remember if I did a video on it. But thiols are just like alcohols, but instead of an oxygen, you have a sulfur. So this comes right after those. So let's think about this one right here."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "Out of all of the things that we've learned so far, the only thing that is a higher priority is actually the alcohols, or actually the thiols, which I don't even remember if I did a video on it. But thiols are just like alcohols, but instead of an oxygen, you have a sulfur. So this comes right after those. So let's think about this one right here. So we always want to look for our longest carbon chain. Our longest carbon chain is right here. 1, 2, 3, 4, 5, 6, 7 carbons."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about this one right here. So we always want to look for our longest carbon chain. Our longest carbon chain is right here. 1, 2, 3, 4, 5, 6, 7 carbons. You want to start numbering it closer to the functional group. 1, 2, 3, 4, 5, 6, and 7. And then the functional group is on the 2 carbon right there."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7 carbons. You want to start numbering it closer to the functional group. 1, 2, 3, 4, 5, 6, and 7. And then the functional group is on the 2 carbon right there. So first of all, 7 carbons, we would use the prefix hept. So it would be heptan. Let me write that down."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "And then the functional group is on the 2 carbon right there. So first of all, 7 carbons, we would use the prefix hept. So it would be heptan. Let me write that down. So it would be heptan. And since our functional group is an amine in this situation, it has a higher priority than the fact that this is an alkane, so this will actually define the suffix. So we would then say, well, then on the 2 carbon right over here, we have the amine group."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "Let me write that down. So it would be heptan. And since our functional group is an amine in this situation, it has a higher priority than the fact that this is an alkane, so this will actually define the suffix. So we would then say, well, then on the 2 carbon right over here, we have the amine group. So it's heptan-2-amine. And we're done. We know that we have an amine group on the 2 carbon."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So we would then say, well, then on the 2 carbon right over here, we have the amine group. So it's heptan-2-amine. And we're done. We know that we have an amine group on the 2 carbon. Now let's do this one. This one's a little bit more hairy. So first of all, we always want to figure out the longest carbon chain."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "We know that we have an amine group on the 2 carbon. Now let's do this one. This one's a little bit more hairy. So first of all, we always want to figure out the longest carbon chain. And it looks like the ring is going to be the longest carbon chain. We have 1, 2, 3, 4, 5, 6 carbons. So this thing right here is the longest chain."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So first of all, we always want to figure out the longest carbon chain. And it looks like the ring is going to be the longest carbon chain. We have 1, 2, 3, 4, 5, 6 carbons. So this thing right here is the longest chain. We only have 1, 2 carbons right there, 1 carbon, 1 carbon right over there. So our root is going to be cyclohexane. I want to make sure I have enough space."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So this thing right here is the longest chain. We only have 1, 2 carbons right there, 1 carbon, 1 carbon right over there. So our root is going to be cyclohexane. I want to make sure I have enough space. Cyclohexane. And then once again, it's an amine. So this is going to take higher priority."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "I want to make sure I have enough space. Cyclohexane. And then once again, it's an amine. So this is going to take higher priority. So we're not just going to put an e at the end and call this cyclohexane. This takes higher priority. So it'll actually define the suffix."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to take higher priority. So we're not just going to put an e at the end and call this cyclohexane. This takes higher priority. So it'll actually define the suffix. And another thing to think about is you could say, OK, this is going to be the 1 carbon. And so you could call this cyclohexane 1 amine. But in general, if this is defining it, you always assume that you're going to start numbering right there at number 1."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So it'll actually define the suffix. And another thing to think about is you could say, OK, this is going to be the 1 carbon. And so you could call this cyclohexane 1 amine. But in general, if this is defining it, you always assume that you're going to start numbering right there at number 1. So you could just call this cyclohexane amine. If it's written like this, you assume that the 1 carbon is where the amine group is attached. Now what else do we have on this thing?"}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "But in general, if this is defining it, you always assume that you're going to start numbering right there at number 1. So you could just call this cyclohexane amine. If it's written like this, you assume that the 1 carbon is where the amine group is attached. Now what else do we have on this thing? We took care of the amine. We took care of the cycle. Well, we have this ether right here."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "Now what else do we have on this thing? We took care of the amine. We took care of the cycle. Well, we have this ether right here. And this ether has 1, 2 carbons. If it was just a 2 carbon chain, it would be ethane. But since we have an ether, it's bonded to this oxygen, we call this ethoxy."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have this ether right here. And this ether has 1, 2 carbons. If it was just a 2 carbon chain, it would be ethane. But since we have an ether, it's bonded to this oxygen, we call this ethoxy. So that right there is ethoxy. And we're going to have to think about how we're going to number this. So I'll leave that alone."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "But since we have an ether, it's bonded to this oxygen, we call this ethoxy. So that right there is ethoxy. And we're going to have to think about how we're going to number this. So I'll leave that alone. And what are these over here? Well, this is a methyl group. That's a CH3 implicitly."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So I'll leave that alone. And what are these over here? Well, this is a methyl group. That's a CH3 implicitly. You don't see it drawn. There's a carbon there. And if carbon is neutral, it has to have 4 bonds."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "That's a CH3 implicitly. You don't see it drawn. There's a carbon there. And if carbon is neutral, it has to have 4 bonds. And if you only draw 1 of them, the other 3 are assumed to be the hydrogen. So this is a methyl group. And then this is also a methyl group."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "And if carbon is neutral, it has to have 4 bonds. And if you only draw 1 of them, the other 3 are assumed to be the hydrogen. So this is a methyl group. And then this is also a methyl group. So we have dimethyl. So this is also a methyl group right over there. And when we think about numbering, we could number from 1, 2, 3, or we could start numbering 1, 2, 3."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "And then this is also a methyl group. So we have dimethyl. So this is also a methyl group right over there. And when we think about numbering, we could number from 1, 2, 3, or we could start numbering 1, 2, 3. And in general, you want to go numbering in the direction where you hit the functional group first. So we want to go 1, 2, 3, 4, and 5. So this right here is 5-ethoxy."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "And when we think about numbering, we could number from 1, 2, 3, or we could start numbering 1, 2, 3. And in general, you want to go numbering in the direction where you hit the functional group first. So we want to go 1, 2, 3, 4, and 5. So this right here is 5-ethoxy. This is 5-ethoxy. And then this is 1, 2-dimethyl. These two combined, you would call this 1, 2-dimethyl."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So this right here is 5-ethoxy. This is 5-ethoxy. And then this is 1, 2-dimethyl. These two combined, you would call this 1, 2-dimethyl. And then when you want to list them in order, the ethoxy would take precedence in alphabetical, because the di you shouldn't count in the alphabetical order. This is just saying 2-methyl. So you really just want to look at the 2 of whatever you're talking about."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "These two combined, you would call this 1, 2-dimethyl. And then when you want to list them in order, the ethoxy would take precedence in alphabetical, because the di you shouldn't count in the alphabetical order. This is just saying 2-methyl. So you really just want to look at the 2 of whatever you're talking about. M comes after E in alphabetical order. So this is going to be 5-ethoxy. Actually, let me just rewrite the whole thing."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy.mp3", "Sentence": "So you really just want to look at the 2 of whatever you're talking about. M comes after E in alphabetical order. So this is going to be 5-ethoxy. Actually, let me just rewrite the whole thing. So this is going to be 5-ethoxy. 1, 2-dimethyl cyclohexanamine. And we're done."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "Let's compare the structures of ethane and ethene. Ethane is an alkane with an A-N-E ending, and it has the molecular formula C2H6. Ethene is an alkene with an E-N-E ending, and it has the molecular formula C2H4. For two carbons, six hydrogens is the maximum number that you can have, so we say that ethane is completely saturated with hydrogens. If we look at ethene, we only have four hydrogens for two carbons, so we say that ethene is unsaturated. So for two carbons, it's possible to have more, so this is unsaturated. Next, let's look at the carbons present in both molecules, and we'll start with ethane."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "For two carbons, six hydrogens is the maximum number that you can have, so we say that ethane is completely saturated with hydrogens. If we look at ethene, we only have four hydrogens for two carbons, so we say that ethene is unsaturated. So for two carbons, it's possible to have more, so this is unsaturated. Next, let's look at the carbons present in both molecules, and we'll start with ethane. This carbon is sp3 hybridized, and so is this one. So ethane contains two sp3 hybridized carbons, and we know the geometry around an sp3 hybridized carbon is tetrahedral, so we should have tetrahedral geometry around both of these carbons. For ethene, let me use a different color here."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "Next, let's look at the carbons present in both molecules, and we'll start with ethane. This carbon is sp3 hybridized, and so is this one. So ethane contains two sp3 hybridized carbons, and we know the geometry around an sp3 hybridized carbon is tetrahedral, so we should have tetrahedral geometry around both of these carbons. For ethene, let me use a different color here. So this carbon and this carbon are both sp2 hybridized, so sp2 hybridized carbons, and we know the geometry around an sp2 hybridized carbon is trigonal planar, so there's planar geometry around both of these carbons. Finally, let's look at the bonding between the two carbons. So for ethane, this sigma bond between the two carbons has some free rotation, so different conformations of the ethane molecule are possible, so we have free rotation, free rotation about the sigma bond between our two carbon atoms."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "For ethene, let me use a different color here. So this carbon and this carbon are both sp2 hybridized, so sp2 hybridized carbons, and we know the geometry around an sp2 hybridized carbon is trigonal planar, so there's planar geometry around both of these carbons. Finally, let's look at the bonding between the two carbons. So for ethane, this sigma bond between the two carbons has some free rotation, so different conformations of the ethane molecule are possible, so we have free rotation, free rotation about the sigma bond between our two carbon atoms. But for ethene, if we look at those two carbon atoms, so this one and this one, there's a double bond between those two carbons, and we know that there's no free rotation around a double bond, so there's no free rotation, so you're not gonna get different conformations for ethene, so no free rotation, so no conformations, and that affects the structure of your alkenes. On the left is ethane, and if we look at our two sp3 hybridized carbons, we can see the tetrahedral geometry around them, and for the sigma bond between those two carbons, we know there's free rotation, so here I am rotating the molecule to show different conformations of ethane. For ethene or ethylene, our two sp2 hybridized carbons have planar geometry around them, so if I rotate this molecule to the side here, you can see that it is planar, and there's no free rotation because of this double bond, because of the presence of a pi bond, so here I am trying to rotate the model set, and you can see that the molecule does not rotate."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "So for ethane, this sigma bond between the two carbons has some free rotation, so different conformations of the ethane molecule are possible, so we have free rotation, free rotation about the sigma bond between our two carbon atoms. But for ethene, if we look at those two carbon atoms, so this one and this one, there's a double bond between those two carbons, and we know that there's no free rotation around a double bond, so there's no free rotation, so you're not gonna get different conformations for ethene, so no free rotation, so no conformations, and that affects the structure of your alkenes. On the left is ethane, and if we look at our two sp3 hybridized carbons, we can see the tetrahedral geometry around them, and for the sigma bond between those two carbons, we know there's free rotation, so here I am rotating the molecule to show different conformations of ethane. For ethene or ethylene, our two sp2 hybridized carbons have planar geometry around them, so if I rotate this molecule to the side here, you can see that it is planar, and there's no free rotation because of this double bond, because of the presence of a pi bond, so here I am trying to rotate the model set, and you can see that the molecule does not rotate. You can classify alkenes according to their degrees of substitution, and if you take ethene and you take a hydrogen off and you add on an R group, you now have a monosubstituted alkene, so on the right is an example of a monosubstituted alkene, and if I put in the hydrogens, it might be a little bit more obvious. We know that this carbon has two hydrogens, and we know that this carbon has one. We have an alkyl group coming off of this carbon, a methyl group, and so this is an example of a monosubstituted alkene."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "For ethene or ethylene, our two sp2 hybridized carbons have planar geometry around them, so if I rotate this molecule to the side here, you can see that it is planar, and there's no free rotation because of this double bond, because of the presence of a pi bond, so here I am trying to rotate the model set, and you can see that the molecule does not rotate. You can classify alkenes according to their degrees of substitution, and if you take ethene and you take a hydrogen off and you add on an R group, you now have a monosubstituted alkene, so on the right is an example of a monosubstituted alkene, and if I put in the hydrogens, it might be a little bit more obvious. We know that this carbon has two hydrogens, and we know that this carbon has one. We have an alkyl group coming off of this carbon, a methyl group, and so this is an example of a monosubstituted alkene. If I want to name this alkene, we saw how to name them in an earlier video. We would make this carbon one, this is carbon two, and this is carbon three, so longest carbon chain, including our alkene, and a three-carbon alkene is called propene, so let me write that down here, so this is propene. Next, let's look at a disubstituted alkene, so now we're talking about two R groups, so here I've put in R and R prime, and R and R prime might be the same or they might be different."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "We have an alkyl group coming off of this carbon, a methyl group, and so this is an example of a monosubstituted alkene. If I want to name this alkene, we saw how to name them in an earlier video. We would make this carbon one, this is carbon two, and this is carbon three, so longest carbon chain, including our alkene, and a three-carbon alkene is called propene, so let me write that down here, so this is propene. Next, let's look at a disubstituted alkene, so now we're talking about two R groups, so here I've put in R and R prime, and R and R prime might be the same or they might be different. Here, R and R prime are on the same carbon, but it's possible to have a disubstituted alkene where R and R prime are bonded to different carbons, and then another example of a disubstituted alkene on the right here, so R and R prime are bonded to different carbons, but notice the difference. These two R groups are on opposite sides of the double bond, so this and this one are on opposite sides, whereas in this example, if I draw a line right here, both R groups are on the same side of the double bond, and we know that one of these doesn't rotate to form the other because there's no free rotation around our double bond, so here we have three examples of disubstituted alkenes. Let's look at this one down here and let's name it, so find our longest carbon chain that includes our double bond, and I wanna give the lowest number possible to our double bond, so we're gonna start right here at carbon one, and this is carbon two, and this is carbon three, so three-carbon alkene is called propene, and we have a methyl group coming off carbon two, so this would be two-methylpropene."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "Next, let's look at a disubstituted alkene, so now we're talking about two R groups, so here I've put in R and R prime, and R and R prime might be the same or they might be different. Here, R and R prime are on the same carbon, but it's possible to have a disubstituted alkene where R and R prime are bonded to different carbons, and then another example of a disubstituted alkene on the right here, so R and R prime are bonded to different carbons, but notice the difference. These two R groups are on opposite sides of the double bond, so this and this one are on opposite sides, whereas in this example, if I draw a line right here, both R groups are on the same side of the double bond, and we know that one of these doesn't rotate to form the other because there's no free rotation around our double bond, so here we have three examples of disubstituted alkenes. Let's look at this one down here and let's name it, so find our longest carbon chain that includes our double bond, and I wanna give the lowest number possible to our double bond, so we're gonna start right here at carbon one, and this is carbon two, and this is carbon three, so three-carbon alkene is called propene, and we have a methyl group coming off carbon two, so this would be two-methylpropene. In terms of which type of disubstituted alkene this is, let's go ahead and draw in our hydrogen so it's a little bit easier to see, so for this carbon, there are two hydrogens bonded to it, and then for the carbon on the left, it has a methyl group and another methyl group, so two R groups that happen to be the same, so that's this example of a disubstituted alkene where both of our R groups are bonded to one carbon. Now let's look at a trisubstituted alkene, so we have three R groups, R, R prime, and R double prime, and again, R, R prime, and R double prime might be the same or they might be different, so here's an example of a trisubstituted alkene. Let me go ahead and draw in the hydrogen on this carbon so it's easier to see that we have three R groups bonded to the double bond."}, {"video_title": "Structure and classification of alkenes.mp3", "Sentence": "Let's look at this one down here and let's name it, so find our longest carbon chain that includes our double bond, and I wanna give the lowest number possible to our double bond, so we're gonna start right here at carbon one, and this is carbon two, and this is carbon three, so three-carbon alkene is called propene, and we have a methyl group coming off carbon two, so this would be two-methylpropene. In terms of which type of disubstituted alkene this is, let's go ahead and draw in our hydrogen so it's a little bit easier to see, so for this carbon, there are two hydrogens bonded to it, and then for the carbon on the left, it has a methyl group and another methyl group, so two R groups that happen to be the same, so that's this example of a disubstituted alkene where both of our R groups are bonded to one carbon. Now let's look at a trisubstituted alkene, so we have three R groups, R, R prime, and R double prime, and again, R, R prime, and R double prime might be the same or they might be different, so here's an example of a trisubstituted alkene. Let me go ahead and draw in the hydrogen on this carbon so it's easier to see that we have three R groups bonded to the double bond. If I want to name it, I need to find the longest carbon chain that includes my double bond, so this would be carbon one, this would be carbon two, three, four, and five, so a five-carbon alkene is called pentene, so let me write that in here, and our double bond starts at carbon two, so this would be 2-pentene, and finally, I have a methyl group coming off of carbon two, so to complete the name, I need 2-methyl, so 2-methyl, 2-pentene would be the name for this trisubstituted alkene. Next, let's look at a tetrasubstituted alkene, so R, R prime, R double prime, and R triple prime, so this molecule is actually tetrasubstituted. If we find our double bond, so this carbon and this carbon, so this top carbon here, actually, let's go ahead and name it first, and then we'll look at why it's a tetrasubstituted alkene, so this would be carbon one, and then I have to follow my double bond for carbon two, so we have a cyclohexene derivative, again, we saw this in an earlier video, so let me go ahead and write cyclohexene down here, so cyclohexene, and we have a methyl group coming off of carbon one, and a methyl group coming off of carbon two, so this would be one, two dimethyl, so one, two dimethylcyclohexene would be the name, and now let's look at why it's tetrasubstituted, so let me use a different color here, so for carbon one, I have one methyl group coming off on this side, and then I have this alkyl group as part of the ring coming off of that side, so that carbon has two R groups bonded to it, and carbon two also has two R groups bonded to it, so a methyl group here, and again, I have this portion of the ring, so that's why this is a tetrasubstituted alkene."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So this is, let's see, we have a methyl group on the number two carbon, it is a pentene, and that's double bond between the number two and number three carbons. So this is two methyl, two methyl pent, pent-2-ene, pent-2-ene. So that's what we start with, we're in the presence, we're in a acidic environment, we've got, it's gonna be catalyzed by our hydronium here, and we end up with this, and how is our product different from what we started with? Well, the double bond is now gone, the number three carbon gains this hydrogen, and now the number two carbon gains a hydroxyl group. So one way to think about this is, in the presence of an acid, it's acid catalyzed, we have gained two hydrogens and an oxygen, which is, well, we've gained the built, what could be used to make a water, and this is actually called an acid-catalyzed addition of water. The water isn't sitting on one part of the molecule, but if you take the hydrogen we added and the hydroxyl we added, if you combine them, that's what you need to make a water. So let's think about how this actually happens in the presence of our hydronium."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Well, the double bond is now gone, the number three carbon gains this hydrogen, and now the number two carbon gains a hydroxyl group. So one way to think about this is, in the presence of an acid, it's acid catalyzed, we have gained two hydrogens and an oxygen, which is, well, we've gained the built, what could be used to make a water, and this is actually called an acid-catalyzed addition of water. The water isn't sitting on one part of the molecule, but if you take the hydrogen we added and the hydroxyl we added, if you combine them, that's what you need to make a water. So let's think about how this actually happens in the presence of our hydronium. So let me redraw this molecule right over here. So let me copy and paste it. So, and that's not exactly it yet, that is just with the single bond."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about how this actually happens in the presence of our hydronium. So let me redraw this molecule right over here. So let me copy and paste it. So, and that's not exactly it yet, that is just with the single bond. So let me draw, whoops, wrong tool. Let me draw the double bond there. And now let me put it in the presence of some hydronium."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So, and that's not exactly it yet, that is just with the single bond. So let me draw, whoops, wrong tool. Let me draw the double bond there. And now let me put it in the presence of some hydronium. All right, so we have an oxygen bonded to, so this would just be water, as oxygen has two lone pairs, but hydronium's a situation where oxygen is sharing one of those lone pairs with a hydrogen proton, thus making the entire molecule positive, because you could do if the hydrogen proton was positive. So there you go, this now has a positive charge. And this can be pretty reactive, because we know that oxygen is quite electronegative, it likes to keep its electrons."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And now let me put it in the presence of some hydronium. All right, so we have an oxygen bonded to, so this would just be water, as oxygen has two lone pairs, but hydronium's a situation where oxygen is sharing one of those lone pairs with a hydrogen proton, thus making the entire molecule positive, because you could do if the hydrogen proton was positive. So there you go, this now has a positive charge. And this can be pretty reactive, because we know that oxygen is quite electronegative, it likes to keep its electrons. So what if there was a way, what if there was a way for the oxygen to take back the electrons in this bond right over here, the two electrons in this bond? Well, what if one of these carbons, especially the ones that have the double bonds, what if some of the electrons from this double bond could be used to snab, to take that hydrogen proton, then oxygen can hog its electrons again? And you might say, well, that's reasonable, but which of these carbons would actually do it?"}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And this can be pretty reactive, because we know that oxygen is quite electronegative, it likes to keep its electrons. So what if there was a way, what if there was a way for the oxygen to take back the electrons in this bond right over here, the two electrons in this bond? Well, what if one of these carbons, especially the ones that have the double bonds, what if some of the electrons from this double bond could be used to snab, to take that hydrogen proton, then oxygen can hog its electrons again? And you might say, well, that's reasonable, but which of these carbons would actually do it? And to think about which of those carbons would do it, we have to turn to Markovnikov's rule. Markovnikov's rule tells us that, look, if you have a reaction like this, an alkene reaction, the carbon that's already attached to more hydrogens is more likely to gain more hydrogens. The carbon that's attached to more functional groups is more likely to gain more functional groups."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And you might say, well, that's reasonable, but which of these carbons would actually do it? And to think about which of those carbons would do it, we have to turn to Markovnikov's rule. Markovnikov's rule tells us that, look, if you have a reaction like this, an alkene reaction, the carbon that's already attached to more hydrogens is more likely to gain more hydrogens. The carbon that's attached to more functional groups is more likely to gain more functional groups. Another way to think about it is think about, well, what is the order of the carbons? Because the higher order of the carbon, the more stable it will be if it forms some type of cation. So if you look at this carbon right over here, our number two carbon, let me circle it, our number two carbon is bonded to one, two, three carbons."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "The carbon that's attached to more functional groups is more likely to gain more functional groups. Another way to think about it is think about, well, what is the order of the carbons? Because the higher order of the carbon, the more stable it will be if it forms some type of cation. So if you look at this carbon right over here, our number two carbon, let me circle it, our number two carbon is bonded to one, two, three carbons. So this is a tertiary carbon. This one right over here, this carbon, the other side of the double bond, is only bonded to one, two carbons. So this is a secondary carbon."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at this carbon right over here, our number two carbon, let me circle it, our number two carbon is bonded to one, two, three carbons. So this is a tertiary carbon. This one right over here, this carbon, the other side of the double bond, is only bonded to one, two carbons. So this is a secondary carbon. So the tertiary carbon is going to be more stable as a carbocation. You can think of it as it can spread the charge a little bit so it would be more likely to lose the electrons in one of these bonds. And so the way that we could think about this mechanism, and it might be a little bit clearer when we form the carbocation, is let's have, I'll do this in blue, let's have these two electrons that form this bond."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So this is a secondary carbon. So the tertiary carbon is going to be more stable as a carbocation. You can think of it as it can spread the charge a little bit so it would be more likely to lose the electrons in one of these bonds. And so the way that we could think about this mechanism, and it might be a little bit clearer when we form the carbocation, is let's have, I'll do this in blue, let's have these two electrons that form this bond. Well now, they form a bond with that hydrogen and now the oxygen can take back these two electrons and what is going to result? And I'm drawing an equilibrium. Remember, all of these things are going back and forth depending on how things bump into each other."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And so the way that we could think about this mechanism, and it might be a little bit clearer when we form the carbocation, is let's have, I'll do this in blue, let's have these two electrons that form this bond. Well now, they form a bond with that hydrogen and now the oxygen can take back these two electrons and what is going to result? And I'm drawing an equilibrium. Remember, all of these things are going back and forth depending on how things bump into each other. But what are we left with? So let me copy and paste this again. And I'm copying and pasting it away just so I kind of, you can view this as a backbone and I'll add what I need to add."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Remember, all of these things are going back and forth depending on how things bump into each other. But what are we left with? So let me copy and paste this again. And I'm copying and pasting it away just so I kind of, you can view this as a backbone and I'll add what I need to add. So once this happens, we have this carbon, the number three carbon, now, whoops, I keep using the wrong tool. The number three carbon now forms a bond with this hydrogen, just like that. This carbon, our number two carbon, has lost an electron."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And I'm copying and pasting it away just so I kind of, you can view this as a backbone and I'll add what I need to add. So once this happens, we have this carbon, the number three carbon, now, whoops, I keep using the wrong tool. The number three carbon now forms a bond with this hydrogen, just like that. This carbon, our number two carbon, has lost an electron. It's no longer sharing this bond. And so now, it is going to have a positive charge. It is a carbocation and once again, it is a tertiary carbocation."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "This carbon, our number two carbon, has lost an electron. It's no longer sharing this bond. And so now, it is going to have a positive charge. It is a carbocation and once again, it is a tertiary carbocation. It is bounded to one, two, three carbons. That is stable, more stable than if we did it the other way around. If this one grabbed the hydrogen somehow, then this would be a secondary carbocation."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "It is a carbocation and once again, it is a tertiary carbocation. It is bounded to one, two, three carbons. That is stable, more stable than if we did it the other way around. If this one grabbed the hydrogen somehow, then this would be a secondary carbocation. It'd be harder for it to spread that positive charge around. And what about our molecule up here? Well, let's see what it looks like now."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "If this one grabbed the hydrogen somehow, then this would be a secondary carbocation. It'd be harder for it to spread that positive charge around. And what about our molecule up here? Well, let's see what it looks like now. We have our oxygen bonded to the two hydrogens. It had one of those lone pairs and now, the electrons in this bond are now going to form another lone pair. So it took back an electron."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Well, let's see what it looks like now. We have our oxygen bonded to the two hydrogens. It had one of those lone pairs and now, the electrons in this bond are now going to form another lone pair. So it took back an electron. Or you could think of it, it gave away a hydrogen proton. And so this is now just neutral water and we see that we have a conservation of charge here. This was positively charged."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So it took back an electron. Or you could think of it, it gave away a hydrogen proton. And so this is now just neutral water and we see that we have a conservation of charge here. This was positively charged. Now, our original molecule is positively charged. And what feels good about this is we're getting close to our end product, at least on our number three carbon. We now have this hydrogen."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "This was positively charged. Now, our original molecule is positively charged. And what feels good about this is we're getting close to our end product, at least on our number three carbon. We now have this hydrogen. Now, we need to think about, well, how do we get a hydroxyl group added right over here? Well, we have all this water floating around. I could use this water molecule but the odds of it being the exact same water molecule, we don't know."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "We now have this hydrogen. Now, we need to think about, well, how do we get a hydroxyl group added right over here? Well, we have all this water floating around. I could use this water molecule but the odds of it being the exact same water molecule, we don't know. But there's all sorts of water molecules. We're in an aqueous solution. So let me draw another water molecule here."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "I could use this water molecule but the odds of it being the exact same water molecule, we don't know. But there's all sorts of water molecules. We're in an aqueous solution. So let me draw another water molecule here. So the water molecules are all equivalent but let me draw another water molecule here. And you can imagine, if they just bump into each other in just the right way, this is, water is a polar molecule. It has a partially negative end near the oxygen because the oxygen likes to hog the electrons."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw another water molecule here. So the water molecules are all equivalent but let me draw another water molecule here. And you can imagine, if they just bump into each other in just the right way, this is, water is a polar molecule. It has a partially negative end near the oxygen because the oxygen likes to hog the electrons. And then you have a partial positive end near the hydrogens because they get their electrons hogged. So you can imagine the oxygen end might be attracted to this tertiary carbocation. And so just bumping in in just the right way, it might form a bond."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "It has a partially negative end near the oxygen because the oxygen likes to hog the electrons. And then you have a partial positive end near the hydrogens because they get their electrons hogged. So you can imagine the oxygen end might be attracted to this tertiary carbocation. And so just bumping in in just the right way, it might form a bond. So let me, let's say these two electrons right over here, let's say they form a bond with this, with that number two carbon. And then what is going to result? So let me draw, so what is, what is going to result?"}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And so just bumping in in just the right way, it might form a bond. So let me, let's say these two electrons right over here, let's say they form a bond with this, with that number two carbon. And then what is going to result? So let me draw, so what is, what is going to result? Let me scroll down a little bit and let me copy and paste, whoops, let me copy and paste our original molecule again. So there we go. So what could happen?"}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw, so what is, what is going to result? Let me scroll down a little bit and let me copy and paste, whoops, let me copy and paste our original molecule again. So there we go. So what could happen? So let me construct it actually. So we have the hydrogen there. We have the hydrogen."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So what could happen? So let me construct it actually. So we have the hydrogen there. We have the hydrogen. Now this character, so we have the water molecule. So oxygen bonded to two hydrogens. You have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "We have the hydrogen. Now this character, so we have the water molecule. So oxygen bonded to two hydrogens. You have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting. And so it now forms a bond. Whoops, let me do it in that orange color. It now, it now forms an actual bond."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "You have this one lone pair that isn't reacting, but then you have the lone pair that does do the reacting. And so it now forms a bond. Whoops, let me do it in that orange color. It now, it now forms an actual bond. And we're really close to our final product. We have our hydrogen on the number three carbon. We have more than we want on our number two carbon."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "It now, it now forms an actual bond. And we're really close to our final product. We have our hydrogen on the number three carbon. We have more than we want on our number two carbon. We just want a hydroxyl group. Now we have a whole water bonded to the number two carbon. So somehow we have to get one of these other hydrogens swiped off of it."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "We have more than we want on our number two carbon. We just want a hydroxyl group. Now we have a whole water bonded to the number two carbon. So somehow we have to get one of these other hydrogens swiped off of it. Well, that could happen with just another water molecule. So let's draw that. So another water molecule someplace."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So somehow we have to get one of these other hydrogens swiped off of it. Well, that could happen with just another water molecule. So let's draw that. So another water molecule someplace. I'll do it in a different color just to differentiate. Although as we know, water, well it's hard to say even what color is water if you're looking at the molecular scale. So here we go."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So another water molecule someplace. I'll do it in a different color just to differentiate. Although as we know, water, well it's hard to say even what color is water if you're looking at the molecular scale. So here we go. We're really in the home stretch at this point. You have another water molecule. Let's say, let me pick a color."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So here we go. We're really in the home stretch at this point. You have another water molecule. Let's say, let me pick a color. So let's say these electrons right over here. Maybe they form a bond with that hydrogen proton. And then these, the electrons in that bond can go back to form a lone pair on that oxygen."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Let's say, let me pick a color. So let's say these electrons right over here. Maybe they form a bond with that hydrogen proton. And then these, the electrons in that bond can go back to form a lone pair on that oxygen. And then what are we left with? And this really is the home stretch. So we are in equilibrium with, so let me draw my five carbons."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And then these, the electrons in that bond can go back to form a lone pair on that oxygen. And then what are we left with? And this really is the home stretch. So we are in equilibrium with, so let me draw my five carbons. So let's see, I have H3C, carbon, carbon, CH2, CH3. I have a CH3. I say H3C instead of CH3."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So we are in equilibrium with, so let me draw my five carbons. So let's see, I have H3C, carbon, carbon, CH2, CH3. I have a CH3. I say H3C instead of CH3. I wrote it that way just so it's clear that the carbons are bonded to the carbons. You have the original hydrogen right over there. You have the one that we just added as part of this mechanism."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "I say H3C instead of CH3. I wrote it that way just so it's clear that the carbons are bonded to the carbons. You have the original hydrogen right over there. You have the one that we just added as part of this mechanism. You have this orange bond to now this hydroxyl group, the hydroxyl group. And it had one lone pair before. It had one lone pair before, but now it took both of the electrons from this bond to form another lone pair, to form another lone pair, which I am depicting in pink."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "You have the one that we just added as part of this mechanism. You have this orange bond to now this hydroxyl group, the hydroxyl group. And it had one lone pair before. It had one lone pair before, but now it took both of the electrons from this bond to form another lone pair, to form another lone pair, which I am depicting in pink. And then this water is now, or this water molecule is now a hydronium molecule. So let's draw that. So this is now oxygen, hydrogen, hydrogen."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "It had one lone pair before, but now it took both of the electrons from this bond to form another lone pair, to form another lone pair, which I am depicting in pink. And then this water is now, or this water molecule is now a hydronium molecule. So let's draw that. So this is now oxygen, hydrogen, hydrogen. Had one lone pair that didn't react, but it had one lone pair that I put in blue that is reacting with this hydrogen proton, with that hydrogen, just like this. And so since it got the hydrogen proton, it's giving its, or sharing its electrons now, now this has a positive, this has a positive charge just like that. Oh, I have to be very careful."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So this is now oxygen, hydrogen, hydrogen. Had one lone pair that didn't react, but it had one lone pair that I put in blue that is reacting with this hydrogen proton, with that hydrogen, just like this. And so since it got the hydrogen proton, it's giving its, or sharing its electrons now, now this has a positive, this has a positive charge just like that. Oh, I have to be very careful. In the last step, I forgot to draw the positive charge. You always wanna make sure that your charge is being conserved. We started off with a positive charge on the hydronium."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Oh, I have to be very careful. In the last step, I forgot to draw the positive charge. You always wanna make sure that your charge is being conserved. We started off with a positive charge on the hydronium. Then we have the positive charge on a tertiary carbocation right over here on a number two carbon. And now we have the positive charge would be right over here, because that oxygen, what we saw before, that oxygen, which this water molecule was neutral, but you could say, you could view it as, well, now it's going to be, it's now sharing these two electrons instead of keeping them. So you could view it as maybe it's giving it away an electron."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "We started off with a positive charge on the hydronium. Then we have the positive charge on a tertiary carbocation right over here on a number two carbon. And now we have the positive charge would be right over here, because that oxygen, what we saw before, that oxygen, which this water molecule was neutral, but you could say, you could view it as, well, now it's going to be, it's now sharing these two electrons instead of keeping them. So you could view it as maybe it's giving it away an electron. And so now it becomes positive. And then, and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium. But just like that, we are done."}, {"video_title": "Addition of water (acid-catalyzed) mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So you could view it as maybe it's giving it away an electron. And so now it becomes positive. And then, and then the positive charge finally gets transferred to that other water molecule when it becomes hydronium. But just like that, we are done. We have added a hydroxyl group and a hydrogen combined. That's a water, so that's why we call it water, addition of water, and it was catalyzed by acid. So it's acid-catalyzed addition of water."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So according to Markovnikov's rules, let's go ahead and write that down here. So you have to think about Markovnikov when you're doing this reaction. And this is an acid catalyzed reaction. So technically this reaction is at equilibrium and we will cover that at the very end of the video here. So let's look at the mechanism for the acid catalyzed addition of water across a double bond. So here I have my alkene and I have water present with sulfuric acid, right? So sulfuric acid being a strong acid will donate a proton in solution."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So technically this reaction is at equilibrium and we will cover that at the very end of the video here. So let's look at the mechanism for the acid catalyzed addition of water across a double bond. So here I have my alkene and I have water present with sulfuric acid, right? So sulfuric acid being a strong acid will donate a proton in solution. And let's say the water molecule picks up that proton. So H2O would go to H3O+. So I'm just jumping ahead to the H3O plus ion called the hydronium ion, right?"}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So sulfuric acid being a strong acid will donate a proton in solution. And let's say the water molecule picks up that proton. So H2O would go to H3O+. So I'm just jumping ahead to the H3O plus ion called the hydronium ion, right? So here is my H3O plus ion, all right? So put my lone pair of electrons on there and it's positively charged. So what's gonna happen is the pi electrons, right?"}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just jumping ahead to the H3O plus ion called the hydronium ion, right? So here is my H3O plus ion, all right? So put my lone pair of electrons on there and it's positively charged. So what's gonna happen is the pi electrons, right? The electrons in this pi bond here are going to function as a base, right? They're going to abstract a proton. They're going to accept a proton."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what's gonna happen is the pi electrons, right? The electrons in this pi bond here are going to function as a base, right? They're going to abstract a proton. They're going to accept a proton. They're going to take, let's say, this proton right here, which would cause these two electrons in this bond to kick off onto your oxygen. So acid-base equilibrium. So I'll go ahead and make this an equilibrium arrow here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "They're going to accept a proton. They're going to take, let's say, this proton right here, which would cause these two electrons in this bond to kick off onto your oxygen. So acid-base equilibrium. So I'll go ahead and make this an equilibrium arrow here. And what are we going to get if that happens, right? Well, we're going to have, let's say that the proton added to the carbon on the right, OK? So the proton added to the carbon on the right here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and make this an equilibrium arrow here. And what are we going to get if that happens, right? Well, we're going to have, let's say that the proton added to the carbon on the right, OK? So the proton added to the carbon on the right here. So I'm saying that the blue electrons on the left are going to be these electrons right here, like that. And that would mean that I took a bond away from the carbon on the left, right? This carbon over here on the left, this carbon right here, used to have four bonds to it."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the proton added to the carbon on the right here. So I'm saying that the blue electrons on the left are going to be these electrons right here, like that. And that would mean that I took a bond away from the carbon on the left, right? This carbon over here on the left, this carbon right here, used to have four bonds to it. Now it has only three bonds to it. So it ends up with a plus 1 formal charge, right? So we have a carbocation in our mechanism."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This carbon over here on the left, this carbon right here, used to have four bonds to it. Now it has only three bonds to it. So it ends up with a plus 1 formal charge, right? So we have a carbocation in our mechanism. So what's left in this acid-base reaction? We took a proton away from H3O+, which leaves us H2O, right? So here we have H2O over here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have a carbocation in our mechanism. So what's left in this acid-base reaction? We took a proton away from H3O+, which leaves us H2O, right? So here we have H2O over here. So we'll go ahead and put lone pairs of electrons in on our water molecule. And we know that water can act as a nucleophile here, right? So this lone pair of electrons is going to be attracted to something that's positively charged, so nucleophilic attack on our carbocation."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here we have H2O over here. So we'll go ahead and put lone pairs of electrons in on our water molecule. And we know that water can act as a nucleophile here, right? So this lone pair of electrons is going to be attracted to something that's positively charged, so nucleophilic attack on our carbocation. And this is technically at equilibrium as well, depending on the concentrations of your reactants. So let's go ahead and show that water molecule adding on to the carbon on the left. OK, so the carbon on the right already had a hydrogen or a proton added on to it."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this lone pair of electrons is going to be attracted to something that's positively charged, so nucleophilic attack on our carbocation. And this is technically at equilibrium as well, depending on the concentrations of your reactants. So let's go ahead and show that water molecule adding on to the carbon on the left. OK, so the carbon on the right already had a hydrogen or a proton added on to it. And the carbon on the left is going to have an oxygen now bonded to that carbon. Two hydrogens bonded to that oxygen. And there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "OK, so the carbon on the right already had a hydrogen or a proton added on to it. And the carbon on the left is going to have an oxygen now bonded to that carbon. Two hydrogens bonded to that oxygen. And there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding. This gives this oxygen right here a plus 1 formal charge. So our oxygen is now positively charged like that. And we're almost to our product, right?"}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And there was a lone pair of electrons on that oxygen that did not participate in any kind of bonding. This gives this oxygen right here a plus 1 formal charge. So our oxygen is now positively charged like that. And we're almost to our product, right? So we're almost there. We need one more acid-base reaction to get rid of that proton on our oxygen. So water can function as a base this time."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we're almost to our product, right? So we're almost there. We need one more acid-base reaction to get rid of that proton on our oxygen. So water can function as a base this time. So water comes along. And this time it's going to act as a Bronsted-Lowry base and accept a proton. So let's get those electrons in there."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So water can function as a base this time. So water comes along. And this time it's going to act as a Bronsted-Lowry base and accept a proton. So let's get those electrons in there. So this lone pair of electrons is going to, let's say it takes that proton, leaving these electrons behind on my oxygen. Once again, I'll draw my equilibrium arrows here, acid-base reaction. And I'm going to end up with an OH on the carbon on the left."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's get those electrons in there. So this lone pair of electrons is going to, let's say it takes that proton, leaving these electrons behind on my oxygen. Once again, I'll draw my equilibrium arrows here, acid-base reaction. And I'm going to end up with an OH on the carbon on the left. And the carbon on the right, there is hydrogen like that. So I added water. I ended up adding water across my double bond."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to end up with an OH on the carbon on the left. And the carbon on the right, there is hydrogen like that. So I added water. I ended up adding water across my double bond. And to be complete, this would regenerate my hydronium ion. I'd get H2O plus H plus would give me H3O plus. So hydronium is regenerated."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I ended up adding water across my double bond. And to be complete, this would regenerate my hydronium ion. I'd get H2O plus H plus would give me H3O plus. So hydronium is regenerated. And so there you go. So remember, a carbocation is present. So you have to think about Markovnikov addition."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So hydronium is regenerated. And so there you go. So remember, a carbocation is present. So you have to think about Markovnikov addition. And since a carbocation is present, you have to think about possible rearrangements. So Markovnikov and rearrangements. Let's take a look at an example where you have a rearrangement here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you have to think about Markovnikov addition. And since a carbocation is present, you have to think about possible rearrangements. So Markovnikov and rearrangements. Let's take a look at an example where you have a rearrangement here. So let's look at a reaction. So let's start off with, let's look at this as our starting alkene. And let's go ahead and think about the mechanism."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's take a look at an example where you have a rearrangement here. So let's look at a reaction. So let's start off with, let's look at this as our starting alkene. And let's go ahead and think about the mechanism. So we know H3O plus is going to be present. So H3O plus right here. So we're adding our alkene to a solution of water and sulfuric acid."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and think about the mechanism. So we know H3O plus is going to be present. So H3O plus right here. So we're adding our alkene to a solution of water and sulfuric acid. And our first step in the mechanism, the pi electrons are going to function as a base and take a proton from our hydronium ion, leaving these electrons in here, letting them kick back off onto the oxygen. So let's see what we would have from that acid-base reaction. And I realize I didn't draw an equilibrium arrow here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're adding our alkene to a solution of water and sulfuric acid. And our first step in the mechanism, the pi electrons are going to function as a base and take a proton from our hydronium ion, leaving these electrons in here, letting them kick back off onto the oxygen. So let's see what we would have from that acid-base reaction. And I realize I didn't draw an equilibrium arrow here. I'm more concerned with getting the right product. So this is our carbon skeleton. And which side do we add the hydrogen?"}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I realize I didn't draw an equilibrium arrow here. I'm more concerned with getting the right product. So this is our carbon skeleton. And which side do we add the hydrogen? Which side of the double bond? Do we add the proton to the left side, or do we add the proton to the right side? Well, we want to form the most stable carbocation we possibly can."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And which side do we add the hydrogen? Which side of the double bond? Do we add the proton to the left side, or do we add the proton to the right side? Well, we want to form the most stable carbocation we possibly can. And if we add the proton to the left side of our double bond, we end up with a secondary carbocation. This carbocation right here is secondary, because this carbon that has the positive charge is bonded to two other carbons. So this is a secondary carbocation."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, we want to form the most stable carbocation we possibly can. And if we add the proton to the left side of our double bond, we end up with a secondary carbocation. This carbocation right here is secondary, because this carbon that has the positive charge is bonded to two other carbons. So this is a secondary carbocation. If we had added the proton on to the other side of the double bond, we would have a primary carbocation. So secondary carbocation is more stable. But can we form something that's even more stable than a secondary carbocation?"}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is a secondary carbocation. If we had added the proton on to the other side of the double bond, we would have a primary carbocation. So secondary carbocation is more stable. But can we form something that's even more stable than a secondary carbocation? And of course we can. We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here, there is a hydrogen attached to that carbon."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But can we form something that's even more stable than a secondary carbocation? And of course we can. We can form a tertiary carbocation if we think about the possibility of a hydride shift. So right here, there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We move to hydrogen over here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So right here, there is a hydrogen attached to that carbon. And if the proton and these two electrons are going to move over here and form a new bond with our positively charged carbon, so we get a hydride shift at this point. So let's draw what would result from that hydride shift. We move to hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. So we added a bond to what used to be our secondary carbocation carbon."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We move to hydrogen over here. That took a bond away from this carbon. So that is the carbon that's going to end up with the positive charge now. So we added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we added a bond to what used to be our secondary carbocation carbon. And so that formal charge goes away. The formal charge moves to this carbon right here, which is now a tertiary carbocation. If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. So we have a water molecule, which is going to function as a nucleophile and attack our positively charged carbon like that."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If you look at the carbons connected to that carbon, this is a tertiary carbocation. So we know tertiary carbocations are more stable than secondary. So now we're at the step of the mechanism where a water molecule is going to come along. So we have a water molecule, which is going to function as a nucleophile and attack our positively charged carbon like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton. And we have an oxygen atom now bonded to that carbon."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have a water molecule, which is going to function as a nucleophile and attack our positively charged carbon like that. So let's go ahead and draw what the result of that nucleophilic attack would look like. So we have our carbon skeleton. And we have an oxygen atom now bonded to that carbon. So two hydrogens here. And once again, one lone pair of electrons did not participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons, off of our positively charged oxygen, and gives us our major product, with the OH adding on to this carbon right here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have an oxygen atom now bonded to that carbon. So two hydrogens here. And once again, one lone pair of electrons did not participate in that reaction, giving this oxygen a plus 1 formal charge. And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons, off of our positively charged oxygen, and gives us our major product, with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, instead of showing the last step, a water molecule comes along, takes one of the protons, off of our positively charged oxygen, and gives us our major product, with the OH adding on to this carbon right here. So this is a major product. This is our major product. And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked our secondary carbocation. And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we would get some of the alcohol that forms from the secondary carbocation. So a minor product, that's what we would get if this oxygen had attacked our secondary carbocation. And you will get some of that. But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon. So I don't have to worry about my stereochemistry here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But if your test asks for the major product, you should show the product of this rearrangement. Now, we're lucky in this instance because our product here, this carbon right here, ends up not being a chirality center. Because I have two methyl groups attached to that carbon. So I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. So let's look at this reaction right here. So we take this as our starting alkene."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I don't have to worry about my stereochemistry here. Let's do a reaction where we do have to worry about stereochemistry. So let's look at this reaction right here. So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we take this as our starting alkene. So I'll put the double bond right there. And let's make that a little bit more clear here. So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid, so H2SO4. And when we think about the mechanism, we know that we're going to add a proton to one side of the double bond. And the other side of the double bond is going to end up being our carbocation."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the double bond is between these two carbons right here. And once again, we're going to add water and sulfuric acid, so H2SO4. And when we think about the mechanism, we know that we're going to add a proton to one side of the double bond. And the other side of the double bond is going to end up being our carbocation. So the first thing to think about is, OK, which one of these two sides is going to get the proton? We want to form the most stable carbocation we can. So the proton's going to add on to the carbon on the left."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the other side of the double bond is going to end up being our carbocation. So the first thing to think about is, OK, which one of these two sides is going to get the proton? We want to form the most stable carbocation we can. So the proton's going to add on to the carbon on the left. So if I can go ahead and show the intermediate here. So for an intermediate, we don't need that arrow. We'll just go ahead and show the proton adding on to the carbon on the left."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the proton's going to add on to the carbon on the left. So if I can go ahead and show the intermediate here. So for an intermediate, we don't need that arrow. We'll just go ahead and show the proton adding on to the carbon on the left. So we get an H here. And then the carbon on the right of the double bond now ends up being our carbocation. So this is now positively charged like that."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'll just go ahead and show the proton adding on to the carbon on the left. So we get an H here. And then the carbon on the right of the double bond now ends up being our carbocation. So this is now positively charged like that. And remember, when we have a carbocation, this carbon is bonded to three other atoms. You have to think about what that looks like. So remember, carbocations, when something's bonded to three other carbons, you get this situation, where everything is in the same plane."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is now positively charged like that. And remember, when we have a carbocation, this carbon is bonded to three other atoms. You have to think about what that looks like. So remember, carbocations, when something's bonded to three other carbons, you get this situation, where everything is in the same plane. sp2 hybridized carbon exhibits trigonal planar geometry, also with your unhybridized p orbital like that. So this is your sp2 hybridized carbocation situation here. So when your water molecule comes along and acts as a nucleophile, your water molecule could end up attacking from the top here, or it could end up attacking from below here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So remember, carbocations, when something's bonded to three other carbons, you get this situation, where everything is in the same plane. sp2 hybridized carbon exhibits trigonal planar geometry, also with your unhybridized p orbital like that. So this is your sp2 hybridized carbocation situation here. So when your water molecule comes along and acts as a nucleophile, your water molecule could end up attacking from the top here, or it could end up attacking from below here. So that's where the stereochemistry comes in. So let's go ahead and take our carbocation, and let's see if we can draw the products that would result from our nucleophilic attack of water. And then we'll just go ahead and think about the proton going away in our heads for the mechanism."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when your water molecule comes along and acts as a nucleophile, your water molecule could end up attacking from the top here, or it could end up attacking from below here. So that's where the stereochemistry comes in. So let's go ahead and take our carbocation, and let's see if we can draw the products that would result from our nucleophilic attack of water. And then we'll just go ahead and think about the proton going away in our heads for the mechanism. So when you have enough practice, you can do steps in the mechanism in your head. So let's see, what would we get for our two possible products? Well, the OH could have added this way, which would push that methyl group there away from us."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll just go ahead and think about the proton going away in our heads for the mechanism. So when you have enough practice, you can do steps in the mechanism in your head. So let's see, what would we get for our two possible products? Well, the OH could have added this way, which would push that methyl group there away from us. So the methyl group would be going away from us. Or the OH could have added from the opposite side. The OH could have been the one back here, and that would have pushed the methyl group out like this."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, the OH could have added this way, which would push that methyl group there away from us. So the methyl group would be going away from us. Or the OH could have added from the opposite side. The OH could have been the one back here, and that would have pushed the methyl group out like this. And we know that these are chirality centers. We know that this is our chirality center on these guys. So we get enantiomers here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The OH could have been the one back here, and that would have pushed the methyl group out like this. And we know that these are chirality centers. We know that this is our chirality center on these guys. So we get enantiomers here. 50-50% racemic mixture for our products. And we see that the OH adds on in a Markovnikov fashion. It adds on to the side that's the most stable carbocation here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we get enantiomers here. 50-50% racemic mixture for our products. And we see that the OH adds on in a Markovnikov fashion. It adds on to the side that's the most stable carbocation here. So one more thing about this reaction. So let's do one that doesn't have any stereochemistry to worry about, and we'll try to make a different point about this reaction here. So this is our reaction."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It adds on to the side that's the most stable carbocation here. So one more thing about this reaction. So let's do one that doesn't have any stereochemistry to worry about, and we'll try to make a different point about this reaction here. So this is our reaction. So we're going to add water to this. And we'll put sulfuric acid up here. And we'll make our arrow a little bit different to illustrate the point here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is our reaction. So we're going to add water to this. And we'll put sulfuric acid up here. And we'll make our arrow a little bit different to illustrate the point here. So we could go, let's go ahead and draw an equilibrium arrow here. So let's say this whole reaction is at equilibrium. So let me get this equilibrium arrow in here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we'll make our arrow a little bit different to illustrate the point here. So we could go, let's go ahead and draw an equilibrium arrow here. So let's say this whole reaction is at equilibrium. So let me get this equilibrium arrow in here. And our product. So if we don't have to worry about stereochemistry, we think, OK, really all I have to do is think about which side of the double bond do I put my OH. And again, in Markovnikov addition, the more substituted carbon is the one that's going to get your OH."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me get this equilibrium arrow in here. And our product. So if we don't have to worry about stereochemistry, we think, OK, really all I have to do is think about which side of the double bond do I put my OH. And again, in Markovnikov addition, the more substituted carbon is the one that's going to get your OH. So the more substituted carbon would be the one on the left here. So for your product, you would say, OK, I know all I have to do is really just go and put my OH in there on the more substituted carbon, and I'm done. I don't have to worry about stereochemistry for this reaction."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And again, in Markovnikov addition, the more substituted carbon is the one that's going to get your OH. So the more substituted carbon would be the one on the left here. So for your product, you would say, OK, I know all I have to do is really just go and put my OH in there on the more substituted carbon, and I'm done. I don't have to worry about stereochemistry for this reaction. I don't have to worry about rearrangements. It's the tertiary carbocation, so that takes care of it. Now, this reaction is technically at equilibrium."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I don't have to worry about stereochemistry for this reaction. I don't have to worry about rearrangements. It's the tertiary carbocation, so that takes care of it. Now, this reaction is technically at equilibrium. And you could think about water as being one of your reactants. So if water is one of your reactants, and you think about general chemistry, Le Chatelier's principle, how do you shift in equilibrium? If you want to make more of your product, if you want to make more of this, your product, your alcohol, one way to do it would be to increase the concentration of water."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, this reaction is technically at equilibrium. And you could think about water as being one of your reactants. So if water is one of your reactants, and you think about general chemistry, Le Chatelier's principle, how do you shift in equilibrium? If you want to make more of your product, if you want to make more of this, your product, your alcohol, one way to do it would be to increase the concentration of water. So if you increase the concentration of water, the equilibrium will shift to decrease the stress that was put on the system. So you're going to get a shift to the right, and you're going to form more and more of your product here. But remember, if you have an alcohol for a product, if you react this alcohol with sulfuric acid, that's an E1 elimination reaction that we saw in earlier videos."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If you want to make more of your product, if you want to make more of this, your product, your alcohol, one way to do it would be to increase the concentration of water. So if you increase the concentration of water, the equilibrium will shift to decrease the stress that was put on the system. So you're going to get a shift to the right, and you're going to form more and more of your product here. But remember, if you have an alcohol for a product, if you react this alcohol with sulfuric acid, that's an E1 elimination reaction that we saw in earlier videos. So acid catalyzed dehydration, the addition of concentrated sulfuric acid to your alcohol can actually form your alkene. So that's a reaction that we saw earlier, an E1 elimination acid catalyzed dehydration, which your major product would be your most substituted alkene here. So you could go back the other way."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But remember, if you have an alcohol for a product, if you react this alcohol with sulfuric acid, that's an E1 elimination reaction that we saw in earlier videos. So acid catalyzed dehydration, the addition of concentrated sulfuric acid to your alcohol can actually form your alkene. So that's a reaction that we saw earlier, an E1 elimination acid catalyzed dehydration, which your major product would be your most substituted alkene here. So you could go back the other way. You could go back to the left. Let's say you decreased the concentration of water. You could shift the equilibrium to the left, and you would actually form your alkene here."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you could go back the other way. You could go back to the left. Let's say you decreased the concentration of water. You could shift the equilibrium to the left, and you would actually form your alkene here. So the way to control your equilibrium is if you want to go to the right, you just dilute your sulfuric acid. You add more water to it, which would increase this concentration. If you want to shift the equilibrium to the left, you decrease your concentration of water, which means you're using concentrated sulfuric acid."}, {"video_title": "Hydration Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You could shift the equilibrium to the left, and you would actually form your alkene here. So the way to control your equilibrium is if you want to go to the right, you just dilute your sulfuric acid. You add more water to it, which would increase this concentration. If you want to shift the equilibrium to the left, you decrease your concentration of water, which means you're using concentrated sulfuric acid. So I could just write concentrated sulfuric acid here, and that would shift your equilibrium to the left and make more of your alkene. So it all depends on what you're trying to make. And so you have to remember all that general chemistry, shifting equilibrium stuff."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Hades is also the name of the god that ran the Greek underworld, Zeus's oldest brother. And it was an appropriate name, although the idea of the ancient Greek notion of the underworld isn't exactly the more modern notion of hell. But it was a hellish environment. You had all this lava flowing around. You had things impacting the Earth from space. And as far as we can tell right now, it was completely inhospitable to life. And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You had all this lava flowing around. You had things impacting the Earth from space. And as far as we can tell right now, it was completely inhospitable to life. And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid. Maybe the collisions started to happen less and less as we started to go a few hundred million years fast forward after Thea rammed into the early Earth and formed the moon. There was something called the late heavy bombardment. And right now the consensus is that whatever we are descended from would have had to come about after the late heavy bombardment."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And to make matters worse, even though the Earth started to cool down a little bit, maybe the crust became a little bit more solid. Maybe the collisions started to happen less and less as we started to go a few hundred million years fast forward after Thea rammed into the early Earth and formed the moon. There was something called the late heavy bombardment. And right now the consensus is that whatever we are descended from would have had to come about after the late heavy bombardment. Because this was a time where so many things from outer space were hitting Earth that it was so violent that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the late heavy bombardment, but we believe that it happened because Uranus and Neptune, so this is the sun right here. That is the sun."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And right now the consensus is that whatever we are descended from would have had to come about after the late heavy bombardment. Because this was a time where so many things from outer space were hitting Earth that it was so violent that it might have killed off any kind of primitive, self-replicating organisms or molecules that might have existed before it. And I won't go into the physics of the late heavy bombardment, but we believe that it happened because Uranus and Neptune, so this is the sun right here. That is the sun. This is the asteroid belt that's outside the orbits of the inner rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics, but what that caused is gravitationally it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is the sun. This is the asteroid belt that's outside the orbits of the inner rocky planets. That Uranus and Neptune, their orbits moved outward. And I'm not going to go into the physics, but what that caused is gravitationally it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm not going to go into the physics, but what that caused is gravitationally it caused a lot of the asteroids in the asteroid belt to move inward and start impacting the inner planets. And of course, Earth was one of the inner planets. And I should make the sun like orange or something, not blue. I don't want you to think that's Earth. And it also impacted the moon. And it's more obvious on the moon because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the late heavy bombardment was Earth kind of ready for life."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't want you to think that's Earth. And it also impacted the moon. And it's more obvious on the moon because the moon does not have an atmosphere to kind of smooth over the impact. So the consensus is that only after the late heavy bombardment was Earth kind of ready for life. And we believe that the first life formed 3.8 to 4 billion years ago. Remember, G for giga, 4 billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the consensus is that only after the late heavy bombardment was Earth kind of ready for life. And we believe that the first life formed 3.8 to 4 billion years ago. Remember, G for giga, 4 billion years ago. And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, although we go into much more detail in the biology playlist, we're talking about prokaryotes."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when we talk about life at this period, we're not talking about squirrels or panda bears. We're talking about extremely simple life forms. We're talking about prokaryotes. And let me give you a little primer on that right now, although we go into much more detail in the biology playlist, we're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound, what we'd call organelles, or these little parts of the cells that perform specific functions like mitochondria."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me give you a little primer on that right now, although we go into much more detail in the biology playlist, we're talking about prokaryotes. And I'll compare them to eukaryotes. Prokaryotes are, for the most part, unicellular organisms that have no nucleuses. They also don't have any other membrane-bound, what we'd call organelles, or these little parts of the cells that perform specific functions like mitochondria. So their DNA is just kind of floating around. So let me draw this character's DNA. So it's just floating around, just like that."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They also don't have any other membrane-bound, what we'd call organelles, or these little parts of the cells that perform specific functions like mitochondria. So their DNA is just kind of floating around. So let me draw this character's DNA. So it's just floating around, just like that. And prokaryote literally means before kernel or before a nucleus. Eukaryotes do have a nucleus where all of their DNA is. So this is the nuclear membrane, and then all of its DNA is floating inside of the nucleus."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's just floating around, just like that. And prokaryote literally means before kernel or before a nucleus. Eukaryotes do have a nucleus where all of their DNA is. So this is the nuclear membrane, and then all of its DNA is floating inside of the nucleus. And then it also has other membrane-bound organelles. Mitochondria is kind of the most famous of them. So it also has things like mitochondria."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the nuclear membrane, and then all of its DNA is floating inside of the nucleus. And then it also has other membrane-bound organelles. Mitochondria is kind of the most famous of them. So it also has things like mitochondria. We'll learn more about that in future videos. Mitochondria, we believe, is essentially one prokaryote crawling inside of another prokaryote and kind of starting to become a symbiotic organism with each other. But I won't go into that right now."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it also has things like mitochondria. We'll learn more about that in future videos. Mitochondria, we believe, is essentially one prokaryote crawling inside of another prokaryote and kind of starting to become a symbiotic organism with each other. But I won't go into that right now. But when we talk about life at this period, we're talking about prokaryotes. And we still have prokaryotes on the planet. Bacteria and archaea are examples of prokaryotes."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I won't go into that right now. But when we talk about life at this period, we're talking about prokaryotes. And we still have prokaryotes on the planet. Bacteria and archaea are examples of prokaryotes. And just to give you a little bit of a tidbit right here, this kind of shows our current understanding of where we think things branched off from. So at this point of the tree is some common ancestor to prokaryotes and eukaryotes. So these are the prokaryotes right over here, the bacteria and the archaea."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Bacteria and archaea are examples of prokaryotes. And just to give you a little bit of a tidbit right here, this kind of shows our current understanding of where we think things branched off from. So at this point of the tree is some common ancestor to prokaryotes and eukaryotes. So these are the prokaryotes right over here, the bacteria and the archaea. And here is the eukaryotes. And this first living thing, or this first set of living things, we think might have just been some type of self-replicating molecules. And slowly some membrane might have come around and became a little bit more organized."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So these are the prokaryotes right over here, the bacteria and the archaea. And here is the eukaryotes. And this first living thing, or this first set of living things, we think might have just been some type of self-replicating molecules. And slowly some membrane might have come around and became a little bit more organized. DNA, RNA, maybe RNA was that original self-replicating molecule, became the method of kind of transmitting information from one generation to the next. So it's really still an open question of exactly what that first life is, or even how do you define that first life. But based on studying the genetic makeup of current organisms, this is how we think the tree of life came about."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And slowly some membrane might have come around and became a little bit more organized. DNA, RNA, maybe RNA was that original self-replicating molecule, became the method of kind of transmitting information from one generation to the next. So it's really still an open question of exactly what that first life is, or even how do you define that first life. But based on studying the genetic makeup of current organisms, this is how we think the tree of life came about. So we have one common ancestor, then they broke apart, and then the archaea and eukaryotes have a common ancestor that's different from the bacteria. And we'll talk more about that in the future. And this right here, just so you can visualize it, this is an example of bacteria."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But based on studying the genetic makeup of current organisms, this is how we think the tree of life came about. So we have one common ancestor, then they broke apart, and then the archaea and eukaryotes have a common ancestor that's different from the bacteria. And we'll talk more about that in the future. And this right here, just so you can visualize it, this is an example of bacteria. This is E. coli or Escherichia coli. It's just an example of bacteria. It comes in a bunch of shapes and forms."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this right here, just so you can visualize it, this is an example of bacteria. This is E. coli or Escherichia coli. It's just an example of bacteria. It comes in a bunch of shapes and forms. But it's a prokaryotic life form. And the earliest life forms we also think were anaerobes. These are things that did not need, one, that they did not need oxygen, and they, for the most part, found oxygen poisonous."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It comes in a bunch of shapes and forms. But it's a prokaryotic life form. And the earliest life forms we also think were anaerobes. These are things that did not need, one, that they did not need oxygen, and they, for the most part, found oxygen poisonous. And the earliest life forms also probably did not perform photosynthesis. They might have gotten their energy from other sources, chemically, from this kind of extremely volatile environment that they were in at that time. So if we fast forward a little bit, and this is actually a major event in the history of Earth."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These are things that did not need, one, that they did not need oxygen, and they, for the most part, found oxygen poisonous. And the earliest life forms also probably did not perform photosynthesis. They might have gotten their energy from other sources, chemically, from this kind of extremely volatile environment that they were in at that time. So if we fast forward a little bit, and this is actually a major event in the history of Earth. And these are huge timescales we're talking about. I mean, remember, I'm kind of just nonchalantly saying, oh, 4.6 billion years ago to 3.8 billion years ago, that's just 800 million years. Remember, and I'll talk about this, grass has only existed for 50 million years."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we fast forward a little bit, and this is actually a major event in the history of Earth. And these are huge timescales we're talking about. I mean, remember, I'm kind of just nonchalantly saying, oh, 4.6 billion years ago to 3.8 billion years ago, that's just 800 million years. Remember, and I'll talk about this, grass has only existed for 50 million years. This is 800 million years. Humans and chimpanzees only diverged 5 million years ago. This is 800 million years we're talking about."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, and I'll talk about this, grass has only existed for 50 million years. This is 800 million years. Humans and chimpanzees only diverged 5 million years ago. This is 800 million years we're talking about. From ancient Greece to now, we're only talking about 2,500 years. You multiply that times 1,000. You multiply that times 1,000, you get 2.5 million years."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is 800 million years we're talking about. From ancient Greece to now, we're only talking about 2,500 years. You multiply that times 1,000. You multiply that times 1,000, you get 2.5 million years. And this is 800 million years we're talking about. So these are extremely huge periods of time. And that's why we call them eons."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You multiply that times 1,000, you get 2.5 million years. And this is 800 million years we're talking about. So these are extremely huge periods of time. And that's why we call them eons. Eons are 500 million to a billion years. Now, the dividing line between the Hadean Eon and the Archean Eon, and it's kind of a fuzzy dividing line, but most people place it about 3.8 billion years ago, is kind of the earliest rocks that we can observe. And so we have rocks that are roughly 3.8 billion years ago, so we kind of put that as the beginning of the Archean Eon."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's why we call them eons. Eons are 500 million to a billion years. Now, the dividing line between the Hadean Eon and the Archean Eon, and it's kind of a fuzzy dividing line, but most people place it about 3.8 billion years ago, is kind of the earliest rocks that we can observe. And so we have rocks that are roughly 3.8 billion years ago, so we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean Eon, and also that's roughly when we think that the first life existed. And so we're now in the Archean Eon."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we have rocks that are roughly 3.8 billion years ago, so we kind of put that as the beginning of the Archean Eon. And so there's two things there. One, rocks have survived from the beginning of the Archean Eon, and also that's roughly when we think that the first life existed. And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we're now in the Archean Eon. And you might say, oh, maybe Earth is a more pleasant place now. But it would not be. It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It still has no to little oxygen in the environment. If you were to go to Earth at that time, it might have looked something like this. It would have been a reddish sky. You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You would have had nitrogen and methane and carbon dioxide in the atmosphere. There would have been nothing for you to breathe. There still would have been a lot of volcanic activity. This right here, these are pictures of stromatolites. And these are formed from bacteria that are bringing in sediment particles, and over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun to actually do photosynthesis."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right here, these are pictures of stromatolites. And these are formed from bacteria that are bringing in sediment particles, and over time, these things get built up. But the most significant event in the Archean period, at least in my humble opinion, was what we believe started to happen about 3.5 billion years ago. And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen. So it starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere, because you had all of this iron that was dissolved in the oceans."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is prokaryotes, or especially bacteria, evolving to actually utilize energy from the sun to actually do photosynthesis. And the real fascinating byproduct of that, other than the fact that they can now use energy directly from the sun, is that it started to produce oxygen. So it starts to produce oxygen. And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere, because you had all of this iron that was dissolved in the oceans. And let me be clear, all of the life that we're going to be talking about for really the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And at first, this oxygen, even though it was being produced by the cyanobacteria, by this blue-green bacteria, it really didn't accumulate in the atmosphere, because you had all of this iron that was dissolved in the oceans. And let me be clear, all of the life that we're going to be talking about for really the next several billion years, it all occurred in the ocean. We had no ozone layer now. The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The land was being irradiated. The land was just a completely inhospitable environment for life. So all of this was occurring in the ocean. And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see that once a lot of that iron was oxidized, and the oxygen really did start to get released in the atmosphere, it actually had, it's funny to say, a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that, because it was a catastrophe for them, but it was kind of a necessary thing that had to happen for us to happen."}, {"video_title": "Beginnings of life Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the first oxygen that actually got produced, it actually, instead of just being released into the atmosphere, it ended up bonding with the iron that was dissolved in the ocean at that time. So it actually didn't have a chance to accumulate in the atmosphere. And when we fast forward past the Archean period, we're going to see that once a lot of that iron was oxidized, and the oxygen really did start to get released in the atmosphere, it actually had, it's funny to say, a cataclysmic effect or a catastrophic effect on the other anaerobic life on the planet at the time. And it's funny to say that, because it was a catastrophe for them, but it was kind of a necessary thing that had to happen for us to happen. So for us, it was a blessing that this cyanobacteria started to pump out a lot of oxygen and eventually oxidized all of the iron and eventually released a lot of oxygen in the atmosphere and killed off all of this anaerobic bacteria so that eventually we could, us oxygen-breathing organisms could come about. But that's not going to happen for a while. We still have a few billion years before things start flopping around on the land."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with an aldehyde. Just to refresh your memory about how to recognize an aldehyde, there's a carbonyl present, a carbon double bonded to an oxygen, and there's an alkyl group on one side and a hydrogen on the other side. If we wanted to name this extremely simple molecule over here on the left, we know there are a total of four carbons and it's an alkane, so we would call this butane. If we look at a four-carbon aldehyde right over here on the right, when you number an aldehyde, you want to make your carbonyl carbon number one, so one, two, three, and four. We have a four-carbon aldehyde. What we're going to do is drop the E and add AL for aldehyde. This would be butanal."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we look at a four-carbon aldehyde right over here on the right, when you number an aldehyde, you want to make your carbonyl carbon number one, so one, two, three, and four. We have a four-carbon aldehyde. What we're going to do is drop the E and add AL for aldehyde. This would be butanal. A four-carbon aldehyde is butanal with an AL ending like that. Let's go ahead and look at some more examples. Once again, we're going to number to give our carbonyl carbon the lowest number possible."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This would be butanal. A four-carbon aldehyde is butanal with an AL ending like that. Let's go ahead and look at some more examples. Once again, we're going to number to give our carbonyl carbon the lowest number possible. The aldehyde is going to take priority over things like double bonds and alkyl groups and halogens. Let's go ahead and look at this one on the top left here. We want to number to give our carbonyl carbon the lowest number possible, so one, two, three, four, and five."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we're going to number to give our carbonyl carbon the lowest number possible. The aldehyde is going to take priority over things like double bonds and alkyl groups and halogens. Let's go ahead and look at this one on the top left here. We want to number to give our carbonyl carbon the lowest number possible, so one, two, three, four, and five. We have a five-carbon aldehyde. A five-carbon alkane would be pentane. A five-carbon aldehyde would be pentanal."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We want to number to give our carbonyl carbon the lowest number possible, so one, two, three, four, and five. We have a five-carbon aldehyde. A five-carbon alkane would be pentane. A five-carbon aldehyde would be pentanal. Let's go ahead and write pentanal here. Pentanal. Then we also have a substituent at carbon four."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "A five-carbon aldehyde would be pentanal. Let's go ahead and write pentanal here. Pentanal. Then we also have a substituent at carbon four. We have a bromine, so we have to go ahead and put that in. A four-bromo pentanal would be the IUPAC name for this molecule. Let's do this one over here on the right."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Then we also have a substituent at carbon four. We have a bromine, so we have to go ahead and put that in. A four-bromo pentanal would be the IUPAC name for this molecule. Let's do this one over here on the right. Once again, we're going to number, giving our carbonyl carbon the lowest number possible, so one, two, and three. We have a three-carbon aldehyde, but this one has a double bond in it. A three-carbon aldehyde would be called propanal, but since we have a double bond present, we're going to call it propanal."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's do this one over here on the right. Once again, we're going to number, giving our carbonyl carbon the lowest number possible, so one, two, and three. We have a three-carbon aldehyde, but this one has a double bond in it. A three-carbon aldehyde would be called propanal, but since we have a double bond present, we're going to call it propanal. We have propanal here, so let's go ahead and write that. Propanal. Then the double bond starts at carbon two right here, so we could write 2-propanal."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "A three-carbon aldehyde would be called propanal, but since we have a double bond present, we're going to call it propanal. We have propanal here, so let's go ahead and write that. Propanal. Then the double bond starts at carbon two right here, so we could write 2-propanal. We also have a phenyl group coming off of carbon three, so we could write 3-phenyl, so 3-phenyl in here. If you want to specify the stereochemistry of the double bond, you think about a hydrogen being here and a hydrogen being here, and so when you think about this double bond here, the hydrogens are on opposite sides, so you could say that they're trans, or you could use the EZ system, and your highest priority groups are on opposite sides here, so this would be E. Go ahead and write E. E-3-phenyl-2-propanal is one way to name this molecule. It's kind of the old way of doing it."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Then the double bond starts at carbon two right here, so we could write 2-propanal. We also have a phenyl group coming off of carbon three, so we could write 3-phenyl, so 3-phenyl in here. If you want to specify the stereochemistry of the double bond, you think about a hydrogen being here and a hydrogen being here, and so when you think about this double bond here, the hydrogens are on opposite sides, so you could say that they're trans, or you could use the EZ system, and your highest priority groups are on opposite sides here, so this would be E. Go ahead and write E. E-3-phenyl-2-propanal is one way to name this molecule. It's kind of the old way of doing it. If you want to do it the newer way, you would write E, and then you still have a phenyl group here, so 3-phenyl. For the new way, you would write prop, and then put the two in here, so prop-2-ene, and then L. Once again, the two specifies where the double bond is, and you still have your AL ending. This is the more modern way of doing it."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's kind of the old way of doing it. If you want to do it the newer way, you would write E, and then you still have a phenyl group here, so 3-phenyl. For the new way, you would write prop, and then put the two in here, so prop-2-ene, and then L. Once again, the two specifies where the double bond is, and you still have your AL ending. This is the more modern way of doing it. These are both IUPAC names for cinnamaldehyde. This molecule here is actually cinnamaldehyde, which is the molecule that gives cinnamon its smell. This is a fantastic smell, and you can steam distill cinnamon bark and isolate this molecule, which is a great undergraduate organic chemistry lab to do."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This is the more modern way of doing it. These are both IUPAC names for cinnamaldehyde. This molecule here is actually cinnamaldehyde, which is the molecule that gives cinnamon its smell. This is a fantastic smell, and you can steam distill cinnamon bark and isolate this molecule, which is a great undergraduate organic chemistry lab to do. That's cinnamaldehyde. Another great smelling aldehyde over here on the left is benzaldehyde. Here we have benzaldehyde, which is such a common name that it's been incorporated into IUPAC nomenclature."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This is a fantastic smell, and you can steam distill cinnamon bark and isolate this molecule, which is a great undergraduate organic chemistry lab to do. That's cinnamaldehyde. Another great smelling aldehyde over here on the left is benzaldehyde. Here we have benzaldehyde, which is such a common name that it's been incorporated into IUPAC nomenclature. Benzaldehyde smells like almonds. This is another fantastic smell. Labs with benzaldehyde are a lot of fun, too."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Here we have benzaldehyde, which is such a common name that it's been incorporated into IUPAC nomenclature. Benzaldehyde smells like almonds. This is another fantastic smell. Labs with benzaldehyde are a lot of fun, too. If you see benzaldehyde, you can use that in your IUPAC name. If we look over here on the right, we can see that benzaldehyde can form the base for the name for this molecule. Let's go ahead and write that over here."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Labs with benzaldehyde are a lot of fun, too. If you see benzaldehyde, you can use that in your IUPAC name. If we look over here on the right, we can see that benzaldehyde can form the base for the name for this molecule. Let's go ahead and write that over here. We have benzaldehyde. Then we're going to number to give this carbon on our ring number one. We have two choices."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and write that over here. We have benzaldehyde. Then we're going to number to give this carbon on our ring number one. We have two choices. We can go around the ring clockwise or counterclockwise. We know we want to give the lowest number possible to our substituents. That would be by going around clockwise."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We have two choices. We can go around the ring clockwise or counterclockwise. We know we want to give the lowest number possible to our substituents. That would be by going around clockwise. Let's do that. Two, three, and then four. We have two substituents."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "That would be by going around clockwise. Let's do that. Two, three, and then four. We have two substituents. We have a methoxy substituent here at carbon three, and we have a hydroxy substituent here at carbon four. We need to put those into alphabetical order. H comes before M. We're going to start with the hydroxy."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We have two substituents. We have a methoxy substituent here at carbon three, and we have a hydroxy substituent here at carbon four. We need to put those into alphabetical order. H comes before M. We're going to start with the hydroxy. Hopefully we'll have enough room here. Four hydroxy, and then we have three methoxy. Benzaldehyde is the IUPAC name for vanillin."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "H comes before M. We're going to start with the hydroxy. Hopefully we'll have enough room here. Four hydroxy, and then we have three methoxy. Benzaldehyde is the IUPAC name for vanillin. Vanillin, of course, is where we get our vanilla flavor from. A fantastic vanilla smell. These are white crystals which have an amazing vanilla scent to them."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Benzaldehyde is the IUPAC name for vanillin. Vanillin, of course, is where we get our vanilla flavor from. A fantastic vanilla smell. These are white crystals which have an amazing vanilla scent to them. Also a lot of fun to do labs with. A lot of aldehydes have nice smells to them. Let's look at two more aldehydes."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "These are white crystals which have an amazing vanilla scent to them. Also a lot of fun to do labs with. A lot of aldehydes have nice smells to them. Let's look at two more aldehydes. Let's look at the one on the left here. It's not benzaldehyde. We don't have a benzene ring here anymore."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at two more aldehydes. Let's look at the one on the left here. It's not benzaldehyde. We don't have a benzene ring here anymore. We have a cyclohexane. This one is going to be called cyclohexane carbaldehyde. Cyclohexane carbaldehyde is the IUPAC name for this."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We don't have a benzene ring here anymore. We have a cyclohexane. This one is going to be called cyclohexane carbaldehyde. Cyclohexane carbaldehyde is the IUPAC name for this. Carbaldehyde. Let's look at this molecule over here on the right. This is two aldehydes in the same molecule."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Cyclohexane carbaldehyde is the IUPAC name for this. Carbaldehyde. Let's look at this molecule over here on the right. This is two aldehydes in the same molecule. Let's look at this. It's going to be one, two, three, four, and five total carbons. It's two aldehydes so it's going to be diol."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This is two aldehydes in the same molecule. Let's look at this. It's going to be one, two, three, four, and five total carbons. It's two aldehydes so it's going to be diol. It's five carbons total so it's going to be pentane. We could write pentane diol here. If you wanted to put where the numbers are, it would be one, five."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's two aldehydes so it's going to be diol. It's five carbons total so it's going to be pentane. We could write pentane diol here. If you wanted to put where the numbers are, it would be one, five. You could put that in there. One, five pentane diol for two aldehydes. Let's look at ketone nomenclature next."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If you wanted to put where the numbers are, it would be one, five. You could put that in there. One, five pentane diol for two aldehydes. Let's look at ketone nomenclature next. Just to remind you of the general structure of ketones. Once again, you have a carbonyl, a carbon double bonded to an oxygen. This time you have two alkyl groups."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at ketone nomenclature next. Just to remind you of the general structure of ketones. Once again, you have a carbonyl, a carbon double bonded to an oxygen. This time you have two alkyl groups. You have two alkyl groups here. They could be the same or they could be different. I could write R prime here."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This time you have two alkyl groups. You have two alkyl groups here. They could be the same or they could be different. I could write R prime here. Either one would still be a ketone. Once again, we looked at butane earlier. We have our four carbon alkane, butane."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "I could write R prime here. Either one would still be a ketone. Once again, we looked at butane earlier. We have our four carbon alkane, butane. What about if we had a four carbon ketone? This right here is a four carbon ketone. It wouldn't be butane."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We have our four carbon alkane, butane. What about if we had a four carbon ketone? This right here is a four carbon ketone. It wouldn't be butane. It's an O-N-E ending. It's an on so it would be butanone. We're going to drop the E here and add O-N-E. Butanone gives us the IUPAC name for this molecule."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It wouldn't be butane. It's an O-N-E ending. It's an on so it would be butanone. We're going to drop the E here and add O-N-E. Butanone gives us the IUPAC name for this molecule. There's an old way of naming ketones which is to think about the two alkyl groups. Let's go ahead and look at that. There's a methyl group on this side."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We're going to drop the E here and add O-N-E. Butanone gives us the IUPAC name for this molecule. There's an old way of naming ketones which is to think about the two alkyl groups. Let's go ahead and look at that. There's a methyl group on this side. Let's go ahead and write methyl right here. Then on this side there's an ethyl group, two carbons. Methyl, ethyl, and then ketone gives you the structure for this molecule too."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "There's a methyl group on this side. Let's go ahead and write methyl right here. Then on this side there's an ethyl group, two carbons. Methyl, ethyl, and then ketone gives you the structure for this molecule too. Methyl, ethyl, ketone is the old way of doing it. Normally you would put the E before the M just thinking about the alphabet. But for this one it doesn't really matter."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Methyl, ethyl, and then ketone gives you the structure for this molecule too. Methyl, ethyl, ketone is the old way of doing it. Normally you would put the E before the M just thinking about the alphabet. But for this one it doesn't really matter. Everyone knows what methyl ethyl ketone is. Also abbreviated M-E-K. Naming the alkyl groups is an old way of naming ketones which you will sometimes still see. Let's do another one here."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But for this one it doesn't really matter. Everyone knows what methyl ethyl ketone is. Also abbreviated M-E-K. Naming the alkyl groups is an old way of naming ketones which you will sometimes still see. Let's do another one here. Let's look at a longer ketone. We went to number to give the ketone, the carbonyl carbon, the lowest number possible. We start from the left."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one here. Let's look at a longer ketone. We went to number to give the ketone, the carbonyl carbon, the lowest number possible. We start from the left. One, two, three, four, five, and six. We have a six carbon ketone. It's going to be hexanone."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We start from the left. One, two, three, four, five, and six. We have a six carbon ketone. It's going to be hexanone. Let's go ahead and write that. This would be hexanone. We need to identify where the ketone is this time."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be hexanone. Let's go ahead and write that. This would be hexanone. We need to identify where the ketone is this time. It's on carbon three here. For longer molecules it's important to specify where it is. It would be three hexanone is one of the ways to name this molecule."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We need to identify where the ketone is this time. It's on carbon three here. For longer molecules it's important to specify where it is. It would be three hexanone is one of the ways to name this molecule. Again, this is the old way of doing it. The newer way would be to write, for this one, hexanethreon. If you wanted to use the alkyl way of naming this molecule, you could say this is an ethyl."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It would be three hexanone is one of the ways to name this molecule. Again, this is the old way of doing it. The newer way would be to write, for this one, hexanethreon. If you wanted to use the alkyl way of naming this molecule, you could say this is an ethyl. You're going to write ethyl here. Ethyl. On this side a three carbon alkyl group is propyl."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If you wanted to use the alkyl way of naming this molecule, you could say this is an ethyl. You're going to write ethyl here. Ethyl. On this side a three carbon alkyl group is propyl. Ethylpropyl ketone is just another way to name it. Several names for the same molecule. Ketones take priority over things like double bonds and alkyl groups and halogens once again."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "On this side a three carbon alkyl group is propyl. Ethylpropyl ketone is just another way to name it. Several names for the same molecule. Ketones take priority over things like double bonds and alkyl groups and halogens once again. If you look at this molecule, you want a number to give your ketone the lowest number possible and not your double bonds. We're going to start from the left. Again, one, two, three, four, five, and six."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Ketones take priority over things like double bonds and alkyl groups and halogens once again. If you look at this molecule, you want a number to give your ketone the lowest number possible and not your double bonds. We're going to start from the left. Again, one, two, three, four, five, and six. We have a six carbon ketone. That would be hexanone, but we have a double bond present. It's going to be hexenone."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Again, one, two, three, four, five, and six. We have a six carbon ketone. That would be hexanone, but we have a double bond present. It's going to be hexenone. We have hexenone as thinking about our base name here. Once again, two ways to think about naming it. I could do hexenone, so hexene, and then we have our ketone, a carbon two, so hexenetwoone."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be hexenone. We have hexenone as thinking about our base name here. Once again, two ways to think about naming it. I could do hexenone, so hexene, and then we have our ketone, a carbon two, so hexenetwoone. The two goes with the one to tell you where the ketone is. Then the double bond starts at carbon four right here, so four hexenetwoone. If you want to specify the stereochemistry of the double bond, once again, you have the hydrogens on opposite sides, so that would be trans or E. You could just write E here."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "I could do hexenone, so hexene, and then we have our ketone, a carbon two, so hexenetwoone. The two goes with the one to tell you where the ketone is. Then the double bond starts at carbon four right here, so four hexenetwoone. If you want to specify the stereochemistry of the double bond, once again, you have the hydrogens on opposite sides, so that would be trans or E. You could just write E here. E-4-hexenetwoone is one way to name this. Again, kind of the old way of doing it. If you wanted to name it another way, you could write E, and then you could write hex, and then four-ene."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If you want to specify the stereochemistry of the double bond, once again, you have the hydrogens on opposite sides, so that would be trans or E. You could just write E here. E-4-hexenetwoone is one way to name this. Again, kind of the old way of doing it. If you wanted to name it another way, you could write E, and then you could write hex, and then four-ene. Once again, the four is going with the ene part telling you where the double bond is. Then you have the twoone as well, so whichever way you prefer. All right, so let's look at two more examples here for nomenclature of ketones."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If you wanted to name it another way, you could write E, and then you could write hex, and then four-ene. Once again, the four is going with the ene part telling you where the double bond is. Then you have the twoone as well, so whichever way you prefer. All right, so let's look at two more examples here for nomenclature of ketones. What if you have a ketone in a cyclic molecule here? If I think about what I have, it would be cyclohexane, but I have a ketone in my ring, so it would be cyclohexanone. Let's go ahead and write that."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's look at two more examples here for nomenclature of ketones. What if you have a ketone in a cyclic molecule here? If I think about what I have, it would be cyclohexane, but I have a ketone in my ring, so it would be cyclohexanone. Let's go ahead and write that. Cyclohexanone, indicating we have a ketone present. We know that ketones take priority over alkyl groups, so we're going to give this carbon a number one on our ring. We're going to choose to go around clockwise, because that gives the lowest number to our alkyl group."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and write that. Cyclohexanone, indicating we have a ketone present. We know that ketones take priority over alkyl groups, so we're going to give this carbon a number one on our ring. We're going to choose to go around clockwise, because that gives the lowest number to our alkyl group. We go two and three. We have a methyl group of carbon three, so it would be three-methylcyclohexanone as the IUPAC name. Finally, what happens if you have two ketones in the same molecule here?"}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We're going to choose to go around clockwise, because that gives the lowest number to our alkyl group. We go two and three. We have a methyl group of carbon three, so it would be three-methylcyclohexanone as the IUPAC name. Finally, what happens if you have two ketones in the same molecule here? We want to give the lowest number possible, so we start numbering from the left side. One, two, three, four, five, and six. We have two ketones."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Finally, what happens if you have two ketones in the same molecule here? We want to give the lowest number possible, so we start numbering from the left side. One, two, three, four, five, and six. We have two ketones. That's going to be a dione. One way to name it would be to say, well, it would be hexane, and then we have two ketones, so dione, hexane-dione. The ketones are at two and four."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We have two ketones. That's going to be a dione. One way to name it would be to say, well, it would be hexane, and then we have two ketones, so dione, hexane-dione. The ketones are at two and four. The carbonyl carbons are at two and four, so two, four, hexane-dione is one way to do it. Of course, you could also do it the more modern way. You could just write hexane, and then after hexane."}, {"video_title": "Nomenclature of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The ketones are at two and four. The carbonyl carbons are at two and four, so two, four, hexane-dione is one way to do it. Of course, you could also do it the more modern way. You could just write hexane, and then after hexane. Hexane just tells you you have six carbons. You could write two, four, dione. That's it for nomenclature of aldehydes and ketones."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Oxidation involves an increase in the oxidation state, and reduction involves a decrease, or a reduction in the oxidation state. You might also remember that loss of electrons is oxidation, and gain of electrons is reduction. So Leo, the lion, goes grr is a good way to remember that. Let's look at this starting compound here in this reaction, and let's figure out whether the starting compound has been oxidized, reduced, or neither. If you look at these three carbons on the left and these three carbons on the right, there's no change to those carbons. So there should be no change in the oxidation states of those carbons. But when you look at this carbon right here, the one I just marked in yellow, on the left it's bonded to, let me go ahead and draw that out, it's double bonded to an oxygen."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at this starting compound here in this reaction, and let's figure out whether the starting compound has been oxidized, reduced, or neither. If you look at these three carbons on the left and these three carbons on the right, there's no change to those carbons. So there should be no change in the oxidation states of those carbons. But when you look at this carbon right here, the one I just marked in yellow, on the left it's bonded to, let me go ahead and draw that out, it's double bonded to an oxygen. And on the right it's bonded to a hydrogen, and on the left it's bonded to a carbon. But on the right here, that same carbon is now bonded to an oxygen on the right. So it's still double bonded to an oxygen, and has a bond to a carbon on the left."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "But when you look at this carbon right here, the one I just marked in yellow, on the left it's bonded to, let me go ahead and draw that out, it's double bonded to an oxygen. And on the right it's bonded to a hydrogen, and on the left it's bonded to a carbon. But on the right here, that same carbon is now bonded to an oxygen on the right. So it's still double bonded to an oxygen, and has a bond to a carbon on the left. But on the right now it has a bond to an OH. So that carbon has likely changed its oxidation state. And we can go ahead and find the oxidation state using what we learned in the last video."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it's still double bonded to an oxygen, and has a bond to a carbon on the left. But on the right now it has a bond to an OH. So that carbon has likely changed its oxidation state. And we can go ahead and find the oxidation state using what we learned in the last video. So remember, you need to put in your bonding electrons. So let's put in our bonding electrons here. Each bond consists of two electrons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "And we can go ahead and find the oxidation state using what we learned in the last video. So remember, you need to put in your bonding electrons. So let's put in our bonding electrons here. Each bond consists of two electrons. All right, so we know that when we're doing oxidation states, we need to think about electronegativity differences. And carbon is more electronegative than hydrogen, so we assign both of those electrons to carbon. Oxygen is more electronegative than carbon, so oxygen takes all four of those electrons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Each bond consists of two electrons. All right, so we know that when we're doing oxidation states, we need to think about electronegativity differences. And carbon is more electronegative than hydrogen, so we assign both of those electrons to carbon. Oxygen is more electronegative than carbon, so oxygen takes all four of those electrons. And when we get to carbon versus carbon, we can assume that those carbons have the same electronegativities. And therefore, for two electrons, we're gonna split those two electrons up. We're gonna give one electron to one carbon, and the other electron to the other carbon."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen is more electronegative than carbon, so oxygen takes all four of those electrons. And when we get to carbon versus carbon, we can assume that those carbons have the same electronegativities. And therefore, for two electrons, we're gonna split those two electrons up. We're gonna give one electron to one carbon, and the other electron to the other carbon. So this carbon is surrounded by three electrons, and we know carbon should have, we know carbon is supposed to have a number of four valence electrons around it. So carbon's supposed to have four. In our dot structure here, carbon only has three."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "We're gonna give one electron to one carbon, and the other electron to the other carbon. So this carbon is surrounded by three electrons, and we know carbon should have, we know carbon is supposed to have a number of four valence electrons around it. So carbon's supposed to have four. In our dot structure here, carbon only has three. Let me highlight them, one, two, and three. So four minus three gives us plus one, which is the oxidation state for this carbon. So this carbon right here has an oxidation state of plus one on the left."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "In our dot structure here, carbon only has three. Let me highlight them, one, two, and three. So four minus three gives us plus one, which is the oxidation state for this carbon. So this carbon right here has an oxidation state of plus one on the left. What about on the right? Let's go ahead and put in our bonding electrons. So we know that each bond consists of two electrons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here has an oxidation state of plus one on the left. What about on the right? Let's go ahead and put in our bonding electrons. So we know that each bond consists of two electrons. Let me put those in here. And again, we think about electronegativity. This time on the right, this carbon is bonded to an oxygen now, so carbon loses those two electrons to oxygen."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we know that each bond consists of two electrons. Let me put those in here. And again, we think about electronegativity. This time on the right, this carbon is bonded to an oxygen now, so carbon loses those two electrons to oxygen. Oxygen's more electronegative. This top oxygen takes those four electrons, and again, we split these two electrons. So this time, carbon is only surrounded by one."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "This time on the right, this carbon is bonded to an oxygen now, so carbon loses those two electrons to oxygen. Oxygen's more electronegative. This top oxygen takes those four electrons, and again, we split these two electrons. So this time, carbon is only surrounded by one. So to find carbon's oxidation state, we know that carbon's supposed to have four valence electrons, and from that, we subtract the number of valence electrons around carbon once we've accounted for electronegativity, and that's only one electron now. So four minus one gives us an oxidation state of plus three. So on the right, this same carbon now has an oxidation state of plus three."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this time, carbon is only surrounded by one. So to find carbon's oxidation state, we know that carbon's supposed to have four valence electrons, and from that, we subtract the number of valence electrons around carbon once we've accounted for electronegativity, and that's only one electron now. So four minus one gives us an oxidation state of plus three. So on the right, this same carbon now has an oxidation state of plus three. So what happened in this reaction? Carbon went from an oxidation state of plus one to an oxidation state of plus three. An increase in the oxidation state is oxidation, so our starting compound was oxidized here."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So on the right, this same carbon now has an oxidation state of plus three. So what happened in this reaction? Carbon went from an oxidation state of plus one to an oxidation state of plus three. An increase in the oxidation state is oxidation, so our starting compound was oxidized here. So our starting compound was oxidized. In order for this to be oxidized, we would need some sort of oxidizing agent. So I'll write that down here really quickly."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "An increase in the oxidation state is oxidation, so our starting compound was oxidized here. So our starting compound was oxidized. In order for this to be oxidized, we would need some sort of oxidizing agent. So I'll write that down here really quickly. We would need an oxidizing agent to accomplish this reaction, and it's the oxidizing agent that itself is being reduced, because remember, whenever something is oxidized, something else has to be reduced. Let's think about the other definitions for, the other definition, I should say, for oxidation. We know that oxidation involves the loss of electrons, so let's look at that here."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So I'll write that down here really quickly. We would need an oxidizing agent to accomplish this reaction, and it's the oxidizing agent that itself is being reduced, because remember, whenever something is oxidized, something else has to be reduced. Let's think about the other definitions for, the other definition, I should say, for oxidation. We know that oxidation involves the loss of electrons, so let's look at that here. On the left, carbon had three electrons around it, and when we added a bond to oxygen, carbon lost these two electrons. Over here, it had those two electrons. On the left, it had those two electrons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "We know that oxidation involves the loss of electrons, so let's look at that here. On the left, carbon had three electrons around it, and when we added a bond to oxygen, carbon lost these two electrons. Over here, it had those two electrons. On the left, it had those two electrons. On the right, it lost those two electrons. So now it's only surrounded by one, and we see an increase in the oxidation state. So it's like carbon lost electrons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "On the left, it had those two electrons. On the right, it lost those two electrons. So now it's only surrounded by one, and we see an increase in the oxidation state. So it's like carbon lost electrons. Also, there's a shortcut that you can use for this reaction. Let's think about what that shortcut would be. Let me use, I'll use red still."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it's like carbon lost electrons. Also, there's a shortcut that you can use for this reaction. Let's think about what that shortcut would be. Let me use, I'll use red still. On the left, this carbon with an oxidation state of plus one has two bonds to oxygen. Let me go ahead and highlight them here. So here's one bond, and here's the other bond."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let me use, I'll use red still. On the left, this carbon with an oxidation state of plus one has two bonds to oxygen. Let me go ahead and highlight them here. So here's one bond, and here's the other bond. On the right, that same carbon now has three bonds to oxygen. So here's one bond, here's two bonds, and then this bond over here. So we've increased in the number of bonds to oxygen, and that can tell us we have an oxidation really quickly without going through and doing all these oxidation states."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So here's one bond, and here's the other bond. On the right, that same carbon now has three bonds to oxygen. So here's one bond, here's two bonds, and then this bond over here. So we've increased in the number of bonds to oxygen, and that can tell us we have an oxidation really quickly without going through and doing all these oxidation states. Notice we lost a bond to hydrogen. So over here, we lost this bond to hydrogen, and we replaced it with a bond to oxygen. So an increase in the number of bonds to oxygen or a decrease in the number of bonds to hydrogen can tell you that your carbon is oxidized too."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we've increased in the number of bonds to oxygen, and that can tell us we have an oxidation really quickly without going through and doing all these oxidation states. Notice we lost a bond to hydrogen. So over here, we lost this bond to hydrogen, and we replaced it with a bond to oxygen. So an increase in the number of bonds to oxygen or a decrease in the number of bonds to hydrogen can tell you that your carbon is oxidized too. Let's do another one. So our goal is to figure out whether the starting compound has been oxidized, reduced, or neither. So again, these three carbons on the left are the same as these three carbons on the right, and there's no change in those carbons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So an increase in the number of bonds to oxygen or a decrease in the number of bonds to hydrogen can tell you that your carbon is oxidized too. Let's do another one. So our goal is to figure out whether the starting compound has been oxidized, reduced, or neither. So again, these three carbons on the left are the same as these three carbons on the right, and there's no change in those carbons. So therefore, we would expect no change in the oxidation state. Same with this carbon. So we're gonna focus in on this carbon."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So again, these three carbons on the left are the same as these three carbons on the right, and there's no change in those carbons. So therefore, we would expect no change in the oxidation state. Same with this carbon. So we're gonna focus in on this carbon. So on the left, that carbon has a double bond to oxygen, so we draw that in there, and then it's bonded to a carbon on the left and a carbon on the right. In our product, that same carbon is now bonded to only one oxygen. Let me go ahead and do that."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna focus in on this carbon. So on the left, that carbon has a double bond to oxygen, so we draw that in there, and then it's bonded to a carbon on the left and a carbon on the right. In our product, that same carbon is now bonded to only one oxygen. Let me go ahead and do that. Only one bond, I should say, to oxygen, and it's still bonded to a carbon on the left and still bonded to a carbon on the right, and it must have a bond to hydrogen in here. It's not drawn in, but it's assumed that you know that, so let's put in that bond to hydrogen, and let's find our oxidation states. So we start on the left, and we draw in our bonding electrons, and next, we think about electronegativity differences."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and do that. Only one bond, I should say, to oxygen, and it's still bonded to a carbon on the left and still bonded to a carbon on the right, and it must have a bond to hydrogen in here. It's not drawn in, but it's assumed that you know that, so let's put in that bond to hydrogen, and let's find our oxidation states. So we start on the left, and we draw in our bonding electrons, and next, we think about electronegativity differences. So we know that oxygen's more electronegative than carbon, so oxygen takes those four electrons. Over here, we assume that the electronegativities for these carbons are the same, and so we split up those two electrons. We give one electron to one carbon and one electron to the other carbon, and same over here."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we start on the left, and we draw in our bonding electrons, and next, we think about electronegativity differences. So we know that oxygen's more electronegative than carbon, so oxygen takes those four electrons. Over here, we assume that the electronegativities for these carbons are the same, and so we split up those two electrons. We give one electron to one carbon and one electron to the other carbon, and same over here. So the oxidation state for carbon would be four valence electrons minus, in this case, we have two around carbon, so here's two. Four minus two gives us an oxidation state of plus two. So on the left, this carbon has an oxidation state of plus two."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "We give one electron to one carbon and one electron to the other carbon, and same over here. So the oxidation state for carbon would be four valence electrons minus, in this case, we have two around carbon, so here's two. Four minus two gives us an oxidation state of plus two. So on the left, this carbon has an oxidation state of plus two. What about on the right? Let's put in our bonding electrons, and again, once we've done that, we think about electronegativity, and we can assign an oxidation state to that carbon. So we know that oxygen's more electronegative, so oxygen takes those two electrons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, this carbon has an oxidation state of plus two. What about on the right? Let's put in our bonding electrons, and again, once we've done that, we think about electronegativity, and we can assign an oxidation state to that carbon. So we know that oxygen's more electronegative, so oxygen takes those two electrons. Again, we assume that these carbons have the same electronegativity, so we split up those electrons, we split up these electrons, but now carbon is bonded to hydrogen, so carbon is a little bit more electronegative than hydrogen, so we just give both of those electrons to carbon. And now let's find our oxidation state. We know carbon is supposed to have four valence electrons around it, and in our drawing here, once we've accounted for electronegativity, carbon is surrounded by one, two, three, and four."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we know that oxygen's more electronegative, so oxygen takes those two electrons. Again, we assume that these carbons have the same electronegativity, so we split up those electrons, we split up these electrons, but now carbon is bonded to hydrogen, so carbon is a little bit more electronegative than hydrogen, so we just give both of those electrons to carbon. And now let's find our oxidation state. We know carbon is supposed to have four valence electrons around it, and in our drawing here, once we've accounted for electronegativity, carbon is surrounded by one, two, three, and four. So four minus four is equal to zero, and so this carbon now has an oxidation state of zero. So that carbon went from an oxidation state of plus two to an oxidation state of zero. That's a decrease in the oxidation state, so carbon was reduced."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "We know carbon is supposed to have four valence electrons around it, and in our drawing here, once we've accounted for electronegativity, carbon is surrounded by one, two, three, and four. So four minus four is equal to zero, and so this carbon now has an oxidation state of zero. So that carbon went from an oxidation state of plus two to an oxidation state of zero. That's a decrease in the oxidation state, so carbon was reduced. So in this reaction, our starting compound was reduced. So this was, our starting compound was reduced, which means we would need some sort of reducing agent to accomplish this reaction, and it's the reducing agent that would be oxidized, because if something is reduced, something else has to be oxidized. We could also think about our other definition for reduction."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "That's a decrease in the oxidation state, so carbon was reduced. So in this reaction, our starting compound was reduced. So this was, our starting compound was reduced, which means we would need some sort of reducing agent to accomplish this reaction, and it's the reducing agent that would be oxidized, because if something is reduced, something else has to be oxidized. We could also think about our other definition for reduction. Gain of electrons is reduction, and let's look at what happened here. So the electrons that were gained were these two electrons right here, because on the left we had this carbon with a double bond to this oxygen. On the right we have only now one bond to oxygen."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "We could also think about our other definition for reduction. Gain of electrons is reduction, and let's look at what happened here. So the electrons that were gained were these two electrons right here, because on the left we had this carbon with a double bond to this oxygen. On the right we have only now one bond to oxygen. So we gained these two electrons because we increased the number of bonds to hydrogen. So our other shortcut is to look at the number of bonds to oxygen, like we talked about in the last example. On the left, carbon has a double bond to oxygen, so two bonds to oxygen."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "On the right we have only now one bond to oxygen. So we gained these two electrons because we increased the number of bonds to hydrogen. So our other shortcut is to look at the number of bonds to oxygen, like we talked about in the last example. On the left, carbon has a double bond to oxygen, so two bonds to oxygen. On the right we have only one bond to oxygen, so we decreased in the number of bonds to oxygen, so that's a reduction, again, a fast way of figuring it out. Or we increased in number of bonds to hydrogen. Let's do one more of these problems."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "On the left, carbon has a double bond to oxygen, so two bonds to oxygen. On the right we have only one bond to oxygen, so we decreased in the number of bonds to oxygen, so that's a reduction, again, a fast way of figuring it out. Or we increased in number of bonds to hydrogen. Let's do one more of these problems. So our goal is to figure out if our starting compound on the left has been oxidized, reduced, or neither. And this time, these are the two carbons that we are going to analyze. Those are the two carbons that have changed in this reaction."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more of these problems. So our goal is to figure out if our starting compound on the left has been oxidized, reduced, or neither. And this time, these are the two carbons that we are going to analyze. Those are the two carbons that have changed in this reaction. So on the left, let's draw out what we have. We have a carbon with a double bond to another carbon, and both of those carbons are bonded to, directly bonded to two more carbons. On the right, those carbons now only have a single bond between them, and this carbon on the left is bonded to a hydrogen, this carbon on the right is directly bonded to an oxygen, and let's put in those other carbons, right?"}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Those are the two carbons that have changed in this reaction. So on the left, let's draw out what we have. We have a carbon with a double bond to another carbon, and both of those carbons are bonded to, directly bonded to two more carbons. On the right, those carbons now only have a single bond between them, and this carbon on the left is bonded to a hydrogen, this carbon on the right is directly bonded to an oxygen, and let's put in those other carbons, right? We have another carbon-carbon bond, let's put in those carbon-carbon bonds like that. Let's assign our oxidation states. So we draw in our bonding electrons, and we think about electronegativity differences, but we're going to assume, once again, that our carbons have the same electronegativity on the left, and so when we're assigning electrons, let's just pick one of those carbons."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "On the right, those carbons now only have a single bond between them, and this carbon on the left is bonded to a hydrogen, this carbon on the right is directly bonded to an oxygen, and let's put in those other carbons, right? We have another carbon-carbon bond, let's put in those carbon-carbon bonds like that. Let's assign our oxidation states. So we draw in our bonding electrons, and we think about electronegativity differences, but we're going to assume, once again, that our carbons have the same electronegativity on the left, and so when we're assigning electrons, let's just pick one of those carbons. So let's pick the carbon on the left, right? We divide up those two electrons in that bond, give one electron to one carbon, one electron to the other, we do the same thing here, and for the double bond with four electrons, we divide up those four electrons, we give two electrons to each carbon, and so carbon is supposed to have four valence electrons, and around it, this carbon has one, two, three, and four, so four minus four is an oxidation state of zero. So this carbon has an oxidation state of zero."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we draw in our bonding electrons, and we think about electronegativity differences, but we're going to assume, once again, that our carbons have the same electronegativity on the left, and so when we're assigning electrons, let's just pick one of those carbons. So let's pick the carbon on the left, right? We divide up those two electrons in that bond, give one electron to one carbon, one electron to the other, we do the same thing here, and for the double bond with four electrons, we divide up those four electrons, we give two electrons to each carbon, and so carbon is supposed to have four valence electrons, and around it, this carbon has one, two, three, and four, so four minus four is an oxidation state of zero. So this carbon has an oxidation state of zero. Same thing for the carbon on the right side of the double bond, right? So in the same situation, so it also has an oxidation state of zero. What about for our product, right?"}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon has an oxidation state of zero. Same thing for the carbon on the right side of the double bond, right? So in the same situation, so it also has an oxidation state of zero. What about for our product, right? Let's examine those two carbons, let's put in our bonding electrons, so we draw in our bonding electrons here, so we put those in, a lot more bonding electrons to draw, and we think about electronegativity differences, right? Things have changed. So now, now let's focus in on the carbon on the left."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "What about for our product, right? Let's examine those two carbons, let's put in our bonding electrons, so we draw in our bonding electrons here, so we put those in, a lot more bonding electrons to draw, and we think about electronegativity differences, right? Things have changed. So now, now let's focus in on the carbon on the left. So the carbon on the left is now bonded to a hydrogen, and carbon is a little bit more electronegative, so we give both of those electrons to carbon, and then we have carbon bonded to carbon, and carbon bonded to carbon, and carbon bonded to carbon. So what's the oxidation state of that carbon on the left now? Carbon is supposed to have four valence electrons, and how many do we have around it?"}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "So now, now let's focus in on the carbon on the left. So the carbon on the left is now bonded to a hydrogen, and carbon is a little bit more electronegative, so we give both of those electrons to carbon, and then we have carbon bonded to carbon, and carbon bonded to carbon, and carbon bonded to carbon. So what's the oxidation state of that carbon on the left now? Carbon is supposed to have four valence electrons, and how many do we have around it? Let's see, one, two, three, four, five. So four minus five gives us an oxidation state of minus one, so this carbon on the left now has an oxidation state of minus one. What about the carbon on the right?"}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Carbon is supposed to have four valence electrons, and how many do we have around it? Let's see, one, two, three, four, five. So four minus five gives us an oxidation state of minus one, so this carbon on the left now has an oxidation state of minus one. What about the carbon on the right? Well, we have a tie here, right? So we divide up those electrons, a tie here, but oxygen is more electronegative than carbon, so oxygen gets both of those electrons, so the carbon on the right should have an oxidation state of four minus three. Here are the three electrons around carbon once we've accounted for electronegativity."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "What about the carbon on the right? Well, we have a tie here, right? So we divide up those electrons, a tie here, but oxygen is more electronegative than carbon, so oxygen gets both of those electrons, so the carbon on the right should have an oxidation state of four minus three. Here are the three electrons around carbon once we've accounted for electronegativity. So that's an oxidation state of plus one, so this carbon has an oxidation state of plus one. So overall, what happened here? Well, let me use red."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Here are the three electrons around carbon once we've accounted for electronegativity. So that's an oxidation state of plus one, so this carbon has an oxidation state of plus one. So overall, what happened here? Well, let me use red. So the carbon on the left went from an oxidation state of zero to an oxidation state of minus one. That is a decrease in the oxidation state, so that carbon was reduced. What about the carbon on the right side of the double bond that we started off with?"}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "Well, let me use red. So the carbon on the left went from an oxidation state of zero to an oxidation state of minus one. That is a decrease in the oxidation state, so that carbon was reduced. What about the carbon on the right side of the double bond that we started off with? It started off with an oxidation state of zero, and it went to an oxidation state of plus one. That's an increase in the oxidation state, so that carbon was oxidized. What can we say about the starting compound overall?"}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "What about the carbon on the right side of the double bond that we started off with? It started off with an oxidation state of zero, and it went to an oxidation state of plus one. That's an increase in the oxidation state, so that carbon was oxidized. What can we say about the starting compound overall? Well, the starting compound overall, there's no net change in the oxidation states. So on the left for those two carbons, we had a total of zero. On the right, we had negative one and positive one for a total of zero."}, {"video_title": "Organic oxidation-reduction reactions Organic chemistry Khan Academy.mp3", "Sentence": "What can we say about the starting compound overall? Well, the starting compound overall, there's no net change in the oxidation states. So on the left for those two carbons, we had a total of zero. On the right, we had negative one and positive one for a total of zero. There's no net change, so this would be neither. The starting compound was overall, it wasn't oxidized or reduced. It's neither."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "Because on one level, it really defines our reality. It's maybe the most defining characteristic of our reality. Everything we see, how we perceive reality, is based on light bouncing off of objects, or bending around objects, or diffracting around objects, and then being sensed by our eyes, and then sending signals into our brain that create models of the world we see around us. So it really is almost the defining characteristic of our reality. But at the same time, when you really go down to experiment and observe with light, it starts to have a bunch of mysterious properties, and to a large degree, it is not fully understood yet. And probably the most amazing thing about light, well, actually there's tons of amazing things about light, but one of the mysterious things is when you really get down to it, and this is actually not just true of light, this is actually true of almost anything once you get onto a small enough quantum mechanical level, but light behaves as both a wave and a particle. And this is probably not that intuitive to you, because it's not that intuitive to me."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "So it really is almost the defining characteristic of our reality. But at the same time, when you really go down to experiment and observe with light, it starts to have a bunch of mysterious properties, and to a large degree, it is not fully understood yet. And probably the most amazing thing about light, well, actually there's tons of amazing things about light, but one of the mysterious things is when you really get down to it, and this is actually not just true of light, this is actually true of almost anything once you get onto a small enough quantum mechanical level, but light behaves as both a wave and a particle. And this is probably not that intuitive to you, because it's not that intuitive to me. In my life, I'm used to certain things behaving as waves, like sound waves, or the waves of an ocean, and I'm used to certain things behaving like particles, like basketballs, or my coffee cup. I'm not used to things behaving as both. And it really depends on what experiment you run and how you observe the light."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And this is probably not that intuitive to you, because it's not that intuitive to me. In my life, I'm used to certain things behaving as waves, like sound waves, or the waves of an ocean, and I'm used to certain things behaving like particles, like basketballs, or my coffee cup. I'm not used to things behaving as both. And it really depends on what experiment you run and how you observe the light. So when you observe it as a particle, and this comes out of Einstein's work with the photoelectric effect, and I won't go into the details here, maybe in a future video when we start thinking about quantum mechanics, you can view light as a train of particles moving at the speed of light, which I'll talk about in a second. We call these particles photons. If you view light in other ways, and you see it even when you see light being refracted by a prism here, it looks like it is a wave."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And it really depends on what experiment you run and how you observe the light. So when you observe it as a particle, and this comes out of Einstein's work with the photoelectric effect, and I won't go into the details here, maybe in a future video when we start thinking about quantum mechanics, you can view light as a train of particles moving at the speed of light, which I'll talk about in a second. We call these particles photons. If you view light in other ways, and you see it even when you see light being refracted by a prism here, it looks like it is a wave. And it has the properties of a wave. It has a frequency, and it has a wavelength. And like other waves, the velocity of that wave is the frequency times its wavelength."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "If you view light in other ways, and you see it even when you see light being refracted by a prism here, it looks like it is a wave. And it has the properties of a wave. It has a frequency, and it has a wavelength. And like other waves, the velocity of that wave is the frequency times its wavelength. Now, even if you ignore this particle aspect of light, if you just look at the wave aspect of light, it's still fascinating, because most waves require a medium to travel through. So for example, if I think about how sound travels through air, so let me draw a bunch of air particles. I'll draw a sound wave traveling through the air particles."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And like other waves, the velocity of that wave is the frequency times its wavelength. Now, even if you ignore this particle aspect of light, if you just look at the wave aspect of light, it's still fascinating, because most waves require a medium to travel through. So for example, if I think about how sound travels through air, so let me draw a bunch of air particles. I'll draw a sound wave traveling through the air particles. What happens in a sound wave is you compress some of the air particles, and those compress the ones next to them, and so you have points in the air that have higher pressure and points that have lower pressure. And you can plot that. So we have high pressure over here, high pressure, low pressure, high pressure, low pressure."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "I'll draw a sound wave traveling through the air particles. What happens in a sound wave is you compress some of the air particles, and those compress the ones next to them, and so you have points in the air that have higher pressure and points that have lower pressure. And you can plot that. So we have high pressure over here, high pressure, low pressure, high pressure, low pressure. And as these things bump into each other, and this wave essentially travels to the right, and if you were to plot that, you would see this waveform traveling to the right. But this is all predicated, or this is all based on, this energy traveling through a medium. And I'm used to visualizing waves in that way."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "So we have high pressure over here, high pressure, low pressure, high pressure, low pressure. And as these things bump into each other, and this wave essentially travels to the right, and if you were to plot that, you would see this waveform traveling to the right. But this is all predicated, or this is all based on, this energy traveling through a medium. And I'm used to visualizing waves in that way. But light needs no medium. Light will actually travel fastest through nothing, through a vacuum. And it will travel at an unimaginably fast speed, 3 times 10 to the 8th meters per second."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And I'm used to visualizing waves in that way. But light needs no medium. Light will actually travel fastest through nothing, through a vacuum. And it will travel at an unimaginably fast speed, 3 times 10 to the 8th meters per second. And just to give you a sense of this, this is 300 million meters per second. Or another way of thinking about it is it would take light less than a seventh of a second to travel around the Earth, or it would travel around the Earth more than seven times in one second. So unimaginably fast."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And it will travel at an unimaginably fast speed, 3 times 10 to the 8th meters per second. And just to give you a sense of this, this is 300 million meters per second. Or another way of thinking about it is it would take light less than a seventh of a second to travel around the Earth, or it would travel around the Earth more than seven times in one second. So unimaginably fast. And not only is this just a super fast rate, or a super fast speed, but once again, it tells us that light is something fundamental to our universe. Because it's not just an unimaginable fast speed, it is the fastest speed not just known to physics, but possible in physics. So once again, something very unintuitive to us in our everyday realm."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "So unimaginably fast. And not only is this just a super fast rate, or a super fast speed, but once again, it tells us that light is something fundamental to our universe. Because it's not just an unimaginable fast speed, it is the fastest speed not just known to physics, but possible in physics. So once again, something very unintuitive to us in our everyday realm. We always imagine that, okay, if something is going at some speed, maybe if there was an ant riding on top of that something, and it was moving in the same direction, it would be going even faster. But nothing can go faster than the speed of light. It's absolutely impossible based on our current understanding of physics."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "So once again, something very unintuitive to us in our everyday realm. We always imagine that, okay, if something is going at some speed, maybe if there was an ant riding on top of that something, and it was moving in the same direction, it would be going even faster. But nothing can go faster than the speed of light. It's absolutely impossible based on our current understanding of physics. So it's not just a fast speed, it is the fastest speed. It is the fastest speed possible. And this right here is an approximation."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "It's absolutely impossible based on our current understanding of physics. So it's not just a fast speed, it is the fastest speed. It is the fastest speed possible. And this right here is an approximation. It's actually 2.99 something something times 10 to the 8th meters per second. 3 times 10 to the 8th meters per second is a pretty good approximation. Now, within the visible light spectrum, and I'll talk about what's beyond the visible light spectrum in a second, you're probably familiar with the colors, maybe you imagine them as the colors of the rainbow."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And this right here is an approximation. It's actually 2.99 something something times 10 to the 8th meters per second. 3 times 10 to the 8th meters per second is a pretty good approximation. Now, within the visible light spectrum, and I'll talk about what's beyond the visible light spectrum in a second, you're probably familiar with the colors, maybe you imagine them as the colors of the rainbow. And rainbows really happen because the light from the sun, the white light, is being refracted by these little water particles. And you can see that in maybe a clearer way when you see light being refracted by a prism right over here. And the different wavelengths of light, so white light contains all of the visible wavelengths, but the different wavelengths get refracted differently by a prism."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "Now, within the visible light spectrum, and I'll talk about what's beyond the visible light spectrum in a second, you're probably familiar with the colors, maybe you imagine them as the colors of the rainbow. And rainbows really happen because the light from the sun, the white light, is being refracted by these little water particles. And you can see that in maybe a clearer way when you see light being refracted by a prism right over here. And the different wavelengths of light, so white light contains all of the visible wavelengths, but the different wavelengths get refracted differently by a prism. So in this case, the higher frequency wavelengths, the violet and the blue, gets refracted more, it gets bent, its direction gets bent more than the low frequency wavelengths, than the reds and the oranges right over here. And if you want to look at the wavelength of light, they're a visible light, it's between 400 nanometers and 700 nanometers, and the higher the frequency, the higher the energy of that light. And that actually goes into when you start talking about the quantum mechanics of it."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And the different wavelengths of light, so white light contains all of the visible wavelengths, but the different wavelengths get refracted differently by a prism. So in this case, the higher frequency wavelengths, the violet and the blue, gets refracted more, it gets bent, its direction gets bent more than the low frequency wavelengths, than the reds and the oranges right over here. And if you want to look at the wavelength of light, they're a visible light, it's between 400 nanometers and 700 nanometers, and the higher the frequency, the higher the energy of that light. And that actually goes into when you start talking about the quantum mechanics of it. That the higher frequency means that each of these photons have higher energy. They have a better ability to give kinetic energy to knock off electrons or whatever else they need to do. So higher frequency, let me write that down."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And that actually goes into when you start talking about the quantum mechanics of it. That the higher frequency means that each of these photons have higher energy. They have a better ability to give kinetic energy to knock off electrons or whatever else they need to do. So higher frequency, let me write that down. Higher frequency means higher energy. Now, I keep referring to this idea of the visible light. And you might say, what is beyond visible light?"}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "So higher frequency, let me write that down. Higher frequency means higher energy. Now, I keep referring to this idea of the visible light. And you might say, what is beyond visible light? And what you'll find is that light is just part of a much broader phenomenon, and it's just the part that we happen to observe. And if we want to broaden the discussion a little bit, light is just, or I should say visible light, is just really part of the electromagnetic spectrum. So light is really just electromagnetic radiation."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "And you might say, what is beyond visible light? And what you'll find is that light is just part of a much broader phenomenon, and it's just the part that we happen to observe. And if we want to broaden the discussion a little bit, light is just, or I should say visible light, is just really part of the electromagnetic spectrum. So light is really just electromagnetic radiation. And everything that I told you about light just now, it has a wave property, and it has particle properties. This is not just specific to visible light. This is true of all of electromagnetic radiation."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "So light is really just electromagnetic radiation. And everything that I told you about light just now, it has a wave property, and it has particle properties. This is not just specific to visible light. This is true of all of electromagnetic radiation. So at very low frequencies or very long wavelengths, we're talking about things like radio waves, the things that allow a radio to reach your car, the things that allow your cell phone to communicate with cell towers. Microwaves, the things that start vibrating water molecules in your food so that they heat up. Infrared, which is what our body releases, and that's why you can detect people through walls with infrared cameras."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "This is true of all of electromagnetic radiation. So at very low frequencies or very long wavelengths, we're talking about things like radio waves, the things that allow a radio to reach your car, the things that allow your cell phone to communicate with cell towers. Microwaves, the things that start vibrating water molecules in your food so that they heat up. Infrared, which is what our body releases, and that's why you can detect people through walls with infrared cameras. Visible light, ultraviolet light, the UV light coming from the sun that will give you sunburn. X-rays, the radiation that allows us to see through the soft material and just visualize the bones. Gamma rays, the super high energy that comes from quasars and other certain types of physical phenomena."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "Infrared, which is what our body releases, and that's why you can detect people through walls with infrared cameras. Visible light, ultraviolet light, the UV light coming from the sun that will give you sunburn. X-rays, the radiation that allows us to see through the soft material and just visualize the bones. Gamma rays, the super high energy that comes from quasars and other certain types of physical phenomena. These are all examples of the exact same thing. We just happen to perceive certain frequencies of this as visible light. You might say, hey Sal, how come we only perceive certain frequencies of this?"}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "Gamma rays, the super high energy that comes from quasars and other certain types of physical phenomena. These are all examples of the exact same thing. We just happen to perceive certain frequencies of this as visible light. You might say, hey Sal, how come we only perceive certain frequencies of this? How come we only see these frequencies? Literally, we can see those frequencies with our unaided eye. The reason, or at least my best guess of the reason of that, is that's the frequency where the sun dumps out a lot of electromagnetic radiation."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "You might say, hey Sal, how come we only perceive certain frequencies of this? How come we only see these frequencies? Literally, we can see those frequencies with our unaided eye. The reason, or at least my best guess of the reason of that, is that's the frequency where the sun dumps out a lot of electromagnetic radiation. It's inundating the earth, and if as a species you wanted to observe things based on reflected energy, a reflected electromagnetic energy, it is most useful to be able to perceive the things where there is the most electromagnetic radiation. It is possible that in other realities or other planets, there are species that perceive more in the ultraviolet range or in the infrared range, and even on earth there are some that perform better at either end of the range. But we see really well in the part of the spectrum where the sun just happens to dump a lot of radiation on us."}, {"video_title": "Introduction to light Electronic structure of atoms Chemistry Khan Academy.mp3", "Sentence": "The reason, or at least my best guess of the reason of that, is that's the frequency where the sun dumps out a lot of electromagnetic radiation. It's inundating the earth, and if as a species you wanted to observe things based on reflected energy, a reflected electromagnetic energy, it is most useful to be able to perceive the things where there is the most electromagnetic radiation. It is possible that in other realities or other planets, there are species that perceive more in the ultraviolet range or in the infrared range, and even on earth there are some that perform better at either end of the range. But we see really well in the part of the spectrum where the sun just happens to dump a lot of radiation on us. I'll leave you there. I think that's a pretty good overview of light. If any of this stuff seems kind of unintuitive or daunting or really on some level confusing, this wave-particle duality, this idea of a transfer of energy through nothing, and it seems unintuitive, don't worry."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Everything we've named so far has been an alkane. We've seen all single bonds. Let's see if we can expand our repertoire a little bit and do some alkenes. So let's look at this first carbon chain right here. And actually here I drew out all of the hydrogens. Just to remind you that everything we were doing before were just the lines. It really was representing something like this."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this first carbon chain right here. And actually here I drew out all of the hydrogens. Just to remind you that everything we were doing before were just the lines. It really was representing something like this. And when you start having the double bonds, and we'll explain it in more detail later on, it actually starts to matter a little bit more to draw the constituents, because there's actually different ways that you can arrange it, because these double bonds are, you can imagine, they're more rigid. You can't rotate around them as much. But don't think about that too much right now."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It really was representing something like this. And when you start having the double bonds, and we'll explain it in more detail later on, it actually starts to matter a little bit more to draw the constituents, because there's actually different ways that you can arrange it, because these double bonds are, you can imagine, they're more rigid. You can't rotate around them as much. But don't think about that too much right now. Let's just try to name these things. So like we always do, let's try to find the longest chain of carbons. And there's only one chain of carbons here."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But don't think about that too much right now. Let's just try to name these things. So like we always do, let's try to find the longest chain of carbons. And there's only one chain of carbons here. There's one, two, three, four, five, six, seven carbons in that chain. So we're going to be dealing with hept. That is seven carbons."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And there's only one chain of carbons here. There's one, two, three, four, five, six, seven carbons in that chain. So we're going to be dealing with hept. That is seven carbons. So we're going to have seven carbons. But it's not going to be a heptane. Heptane would mean that we have all single bonds."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That is seven carbons. So we're going to have seven carbons. But it's not going to be a heptane. Heptane would mean that we have all single bonds. Here we have a double bond. So this is going to be an alkene. So this tells us right here that we're dealing with an alkene, not an alkane."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Heptane would mean that we have all single bonds. Here we have a double bond. So this is going to be an alkene. So this tells us right here that we're dealing with an alkene, not an alkane. If you have a double bond, it's an alkene. Triple bond, alkyne. We'll talk about that in future videos."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this tells us right here that we're dealing with an alkene, not an alkane. If you have a double bond, it's an alkene. Triple bond, alkyne. We'll talk about that in future videos. So this is hept. And we'll put an ene here. We'll put an ene here."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'll talk about that in future videos. So this is hept. And we'll put an ene here. We'll put an ene here. But we haven't specified where the double bond is. And we haven't numbered our carbons. So when you see an alkene like this, you start numbering closest to the double bond."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'll put an ene here. But we haven't specified where the double bond is. And we haven't numbered our carbons. So when you see an alkene like this, you start numbering closest to the double bond. Just like as if it was a alkyl group, as if it was a side chain of carbons. So this side is closest to the double bond. So let's start numbering there."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when you see an alkene like this, you start numbering closest to the double bond. Just like as if it was a alkyl group, as if it was a side chain of carbons. So this side is closest to the double bond. So let's start numbering there. One, two, three, four, five, six, seven. The double bond is between two and three. And to specify its location, you start at the lowest of these numbers."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's start numbering there. One, two, three, four, five, six, seven. The double bond is between two and three. And to specify its location, you start at the lowest of these numbers. So this double bond is at two. So this is actually hept-2-ene. So this tells us that we have a seven-carbon chain that has a double bond starting."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And to specify its location, you start at the lowest of these numbers. So this double bond is at two. So this is actually hept-2-ene. So this tells us that we have a seven-carbon chain that has a double bond starting. The ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us, double bond between two carbons."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this tells us that we have a seven-carbon chain that has a double bond starting. The ene tells us a double bond. Let me write that down. So this double bond right there, that's what the ene tells us, double bond between two carbons. It's an alkene. And the double bond starts, if you start at this point, the double bond starts at number two carbon. And then it will go to the number three carbon."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this double bond right there, that's what the ene tells us, double bond between two carbons. It's an alkene. And the double bond starts, if you start at this point, the double bond starts at number two carbon. And then it will go to the number three carbon. Now, you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. So let's say I have something like, one, two, three, four, five, six, seven."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then it will go to the number three carbon. Now, you might be asking, well, what if I had more than one double bond here? So let me draw a quick example of that. So let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say I have something like, one, two, three, four, five, six, seven. So this is the same molecule again. One, two, three, four, five, six, seven. The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? Well, once again, we have seven carbons."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The way we drew it up here, it would look something like this. What if I had another double bond sitting right here? How would we specify this? Well, once again, we have seven carbons. One, two, three, four, five, six, seven. So we're still going to have a hept here. And it's still going to be an alkene."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, once again, we have seven carbons. One, two, three, four, five, six, seven. So we're still going to have a hept here. And it's still going to be an alkene. So we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And it's still going to be an alkene. So we put our ene here. But we start numbering it, once again, closest to the closest double bond. So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three. So it would be hept two. And we also have another double bond starting from four and going to five."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So one, two, three, four, five, six, seven. But now we have a double bond starting at two to three. So it would be hept two. And we also have another double bond starting from four and going to five. So hept two, comma, four, ene. That's what this molecule right there is. Now, sometimes, this is the, I guess, proper naming, but just so you're familiar with it if you ever see it."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we also have another double bond starting from four and going to five. So hept two, comma, four, ene. That's what this molecule right there is. Now, sometimes, this is the, I guess, proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene. They'll write that as 2-heptene. And probably because it's easier to say."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, sometimes, this is the, I guess, proper naming, but just so you're familiar with it if you ever see it. Sometimes someone would write hept-2-ene. They'll write that as 2-heptene. And probably because it's easier to say. 2-heptene. And from this, you would be able to draw this thing over here. So it's giving you the same amount of information."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And probably because it's easier to say. 2-heptene. And from this, you would be able to draw this thing over here. So it's giving you the same amount of information. And similarly, over here, they might say 2,4-heptene. But this is the specific, this is the correct way to write it. It lets you know that the two and the four apply to the ene, which you know applies to double bonds."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's giving you the same amount of information. And similarly, over here, they might say 2,4-heptene. But this is the specific, this is the correct way to write it. It lets you know that the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond and I also have some side chains, so let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons?"}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It lets you know that the two and the four apply to the ene, which you know applies to double bonds. Let's do a couple more. So over here, I have a double bond and I also have some side chains, so let's see if we can figure out how to deal with all of these things. So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. And we could go in either direction. It doesn't matter."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So first of all, what is our longest chain of carbons? So we have one, two, three, four, five, six. And we could go in either direction. It doesn't matter. Seven carbons or seven carbons. And let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't matter. Seven carbons or seven carbons. And let's start numbering closest to the double bond. The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence, well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The double bond actually will take precedence over any other groups that are attached to it. So let's take precedence, well, over any other groups in this case. There will be other groups that will take precedence in the future. But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case. We'd want to start numbering at this end. So it's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But the double bond takes precedence over this side chain, this methyl group. But it doesn't matter in this case. We'd want to start numbering at this end. So it's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. So we're dealing with a hept again. So we're dealing with a hept. And we have a double bond starting from the second carbon to the third carbon."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's one, two carbon, three carbon, four carbon, five carbon, six carbon, seven carbon. So we're dealing with a hept again. So we're dealing with a hept. And we have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept 2, 3 ene. Sorry, not 2, 3."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have a double bond starting from the second carbon to the third carbon. So this thing right here, this double bond from the second carbon to the third carbon. So it's hept 2, 3 ene. Sorry, not 2, 3. 2 ene. You don't write both endpoints. If there was a 3, then that would have been another double bond there."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Sorry, not 2, 3. 2 ene. You don't write both endpoints. If there was a 3, then that would have been another double bond there. It's hept dash 2 dash ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If there was a 3, then that would have been another double bond there. It's hept dash 2 dash ene. And then we have this methyl group here, which is also sitting on the second carbon. So this methyl group right there. So on the second carbon. So we would say 2-methyl hept 2 ene. So it's a hept 2 ene."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this methyl group right there. So on the second carbon. So we would say 2-methyl hept 2 ene. So it's a hept 2 ene. That's all of this part over here. Double bond starting on the 2 if we're numbering from the right. And then the methyl group is also attached to that second carbon."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's a hept 2 ene. That's all of this part over here. Double bond starting on the 2 if we're numbering from the right. And then the methyl group is also attached to that second carbon. Let's do one more of these. So we have a cycle here. And once again, the root is going to be the largest chain or the largest ring here."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then the methyl group is also attached to that second carbon. Let's do one more of these. So we have a cycle here. And once again, the root is going to be the largest chain or the largest ring here. And our main ring is the largest one. And we have 1, 2, 3, 4, 5, 6 carbons. So we are dealing with hex as our root for kind of the core of our structure."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And once again, the root is going to be the largest chain or the largest ring here. And our main ring is the largest one. And we have 1, 2, 3, 4, 5, 6 carbons. So we are dealing with hex as our root for kind of the core of our structure. And it's in a cycle. So it's going to be cyclohex. So let me write that."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we are dealing with hex as our root for kind of the core of our structure. And it's in a cycle. So it's going to be cyclohex. So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohexene."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me write that. So it's going to be cyclohex. But it has a double bond in it. So it's cyclohexene. Let me do this in a different color. So we have this double bond here. And that's why we know it is an ene."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's cyclohexene. Let me do this in a different color. So we have this double bond here. And that's why we know it is an ene. Now you're probably saying, hey, Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one endpoint of the double bond is your one carbon. So when you write just cyclohexene, you know."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that's why we know it is an ene. Now you're probably saying, hey, Sal, how come we didn't have to number where the ene is? So if you only have one double bond in a ring, it's assumed that one endpoint of the double bond is your one carbon. So when you write just cyclohexene, you know. So if cyclohexene would look just like this. You don't have to specify where it is. It's just one of these are going to be the double bond."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when you write just cyclohexene, you know. So if cyclohexene would look just like this. You don't have to specify where it is. It's just one of these are going to be the double bond. Now when you have other constituents on it, by definition, or I guess the proper naming mechanism, is one of the endpoints of the double bond will be the one carbon. And if any of those endpoints have something else on it, that will definitely be the one carbon. So these both are kind of the candidates for the one carbon."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's just one of these are going to be the double bond. Now when you have other constituents on it, by definition, or I guess the proper naming mechanism, is one of the endpoints of the double bond will be the one carbon. And if any of those endpoints have something else on it, that will definitely be the one carbon. So these both are kind of the candidates for the one carbon. But this point right here also has this methyl group. So we will start numbering there. One."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these both are kind of the candidates for the one carbon. But this point right here also has this methyl group. So we will start numbering there. One. And then you want to number in the direction of the other side of the double bond. One, two, three, four, five, six. So we have three methyl groups, one on one."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "One. And then you want to number in the direction of the other side of the double bond. One, two, three, four, five, six. So we have three methyl groups, one on one. So these are the, let me circle the methyl group. So that's a methyl group right there. That's a methyl group right there."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have three methyl groups, one on one. So these are the, let me circle the methyl group. So that's a methyl group right there. That's a methyl group right there. That's just one carbon. So we have three methyl groups. So this is going to be, it's at the one, the four, and the six, so it is one, comma, four, comma, six."}, {"video_title": "Naming alkenes examples Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's a methyl group right there. That's just one carbon. So we have three methyl groups. So this is going to be, it's at the one, the four, and the six, so it is one, comma, four, comma, six. We have three methyl groups, so it's trimethylcyclohexene. That's what that is. Hopefully you found that useful."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let me write this down. Amines tend to be smelly. And I can make the smelly smell a little bit. And in general, either in their gaseous state, they'll often smell like ammonia, so it's a very strong smell. Or in their liquid state, they usually smell like fish or even dead fish. They smell like fish. And in particular, if you look at trimethylamine, so this is H3C, you have your nitrogen, you have CH3, and then you have your CH3, and you could even draw the lone pair."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And in general, either in their gaseous state, they'll often smell like ammonia, so it's a very strong smell. Or in their liquid state, they usually smell like fish or even dead fish. They smell like fish. And in particular, if you look at trimethylamine, so this is H3C, you have your nitrogen, you have CH3, and then you have your CH3, and you could even draw the lone pair. This is trimethylamine. This is the fish smell. This right here is the fish smell."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And in particular, if you look at trimethylamine, so this is H3C, you have your nitrogen, you have CH3, and then you have your CH3, and you could even draw the lone pair. This is trimethylamine. This is the fish smell. This right here is the fish smell. Now, the whole reason why I brought this up is because in this video, we're going to introduce a new type of molecule, and that's an aldehyde. And I want to contrast their smells. Now, some of the smaller aldehydes still have a pretty strong smell, in particular, formaldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This right here is the fish smell. Now, the whole reason why I brought this up is because in this video, we're going to introduce a new type of molecule, and that's an aldehyde. And I want to contrast their smells. Now, some of the smaller aldehydes still have a pretty strong smell, in particular, formaldehyde. And we'll talk a little bit more about why this is an aldehyde. So this right here is formaldehyde, and that's the common name. And you've probably seen it used as kind of a preservative, maybe even in your biology class if there's like a dead frog in a solution."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now, some of the smaller aldehydes still have a pretty strong smell, in particular, formaldehyde. And we'll talk a little bit more about why this is an aldehyde. So this right here is formaldehyde, and that's the common name. And you've probably seen it used as kind of a preservative, maybe even in your biology class if there's like a dead frog in a solution. It's probably inside of formaldehyde. Let me write this down. Formaldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And you've probably seen it used as kind of a preservative, maybe even in your biology class if there's like a dead frog in a solution. It's probably inside of formaldehyde. Let me write this down. Formaldehyde. Once again, this is the common name. I'm going to teach you in a second how to systematically name these things. But when you have larger aldehydes, they actually can have a pretty sweet or even a rosy smell."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Formaldehyde. Once again, this is the common name. I'm going to teach you in a second how to systematically name these things. But when you have larger aldehydes, they actually can have a pretty sweet or even a rosy smell. So you have something like benzaldehyde, which you actually saw when we started the benzene derivatives. So this is benzaldehyde right over here. And then this molecule, let me write this down."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But when you have larger aldehydes, they actually can have a pretty sweet or even a rosy smell. So you have something like benzaldehyde, which you actually saw when we started the benzene derivatives. So this is benzaldehyde right over here. And then this molecule, let me write this down. This is benzaldehyde. And then this molecule here, just when I tell you what it's called, you'll probably guess what it smells like. So this has a benzene ring."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then this molecule, let me write this down. This is benzaldehyde. And then this molecule here, just when I tell you what it's called, you'll probably guess what it smells like. So this has a benzene ring. You have a benzene ring like this. And then you have one, two, three carbons. The last one is double bonded to an oxygen and then a hydrogen like there."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this has a benzene ring. You have a benzene ring like this. And then you have one, two, three carbons. The last one is double bonded to an oxygen and then a hydrogen like there. And they have a double bond right over here. And this is called cinnamaldehyde. And as you could guess, this smells like cinnamon."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The last one is double bonded to an oxygen and then a hydrogen like there. And they have a double bond right over here. And this is called cinnamaldehyde. And as you could guess, this smells like cinnamon. In fact, this is the molecule in cinnamon that gives it its smell, although you don't want to have large quantities of it or it might be poisonous. But in cinnamon, it's a very pleasant thing. So larger aldehydes tend to have kind of a nice, rosy, sweet, flowery smell."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And as you could guess, this smells like cinnamon. In fact, this is the molecule in cinnamon that gives it its smell, although you don't want to have large quantities of it or it might be poisonous. But in cinnamon, it's a very pleasant thing. So larger aldehydes tend to have kind of a nice, rosy, sweet, flowery smell. Smaller ones tend to be kind of pungent. If you open up that jar in your biology class and smell that frog, it will not be a pleasant smell. So now that I've talked about smell enough, I think, let's talk about what makes an aldehyde an aldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So larger aldehydes tend to have kind of a nice, rosy, sweet, flowery smell. Smaller ones tend to be kind of pungent. If you open up that jar in your biology class and smell that frog, it will not be a pleasant smell. So now that I've talked about smell enough, I think, let's talk about what makes an aldehyde an aldehyde. And you might even see a pattern here. In all of these aldehydes that I've drawn, we have a carbon double bonded to an oxygen. And actually, this part of it right here, and we'll see this over and over again, this is called a carbonyl group."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So now that I've talked about smell enough, I think, let's talk about what makes an aldehyde an aldehyde. And you might even see a pattern here. In all of these aldehydes that I've drawn, we have a carbon double bonded to an oxygen. And actually, this part of it right here, and we'll see this over and over again, this is called a carbonyl group. So that is a carbonyl group. This right here is a carbonyl group, and you even see it over here. But that by itself is not the distinctive feature of an aldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And actually, this part of it right here, and we'll see this over and over again, this is called a carbonyl group. So that is a carbonyl group. This right here is a carbonyl group, and you even see it over here. But that by itself is not the distinctive feature of an aldehyde. You'll see that in other types of molecules. Let me write it here, carbonyl. What's distinctive about an aldehyde is attached to the carbon in the carbonyl, you have a hydrogen."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But that by itself is not the distinctive feature of an aldehyde. You'll see that in other types of molecules. Let me write it here, carbonyl. What's distinctive about an aldehyde is attached to the carbon in the carbonyl, you have a hydrogen. Sorry, I forgot to draw this hydrogen here on benzaldehyde. You have a hydrogen here. So in general, an aldehyde is something that looks like this."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "What's distinctive about an aldehyde is attached to the carbon in the carbonyl, you have a hydrogen. Sorry, I forgot to draw this hydrogen here on benzaldehyde. You have a hydrogen here. So in general, an aldehyde is something that looks like this. You have a carbonyl group, you have a hydrogen, and then you just have some other type of carbon chain. This right here is the simplest possible aldehyde, and actually, this chain ends up being just another hydrogen. Now, these three that I've shown you right here, these are their common names, and these are the way that most people will talk about formaldehyde, benzaldehyde, or cinnamaldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So in general, an aldehyde is something that looks like this. You have a carbonyl group, you have a hydrogen, and then you just have some other type of carbon chain. This right here is the simplest possible aldehyde, and actually, this chain ends up being just another hydrogen. Now, these three that I've shown you right here, these are their common names, and these are the way that most people will talk about formaldehyde, benzaldehyde, or cinnamaldehyde. So they're just kind of good to know, and I'll show you one more. This is a pretty important one. Let me do it down here so we have some space."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now, these three that I've shown you right here, these are their common names, and these are the way that most people will talk about formaldehyde, benzaldehyde, or cinnamaldehyde. So they're just kind of good to know, and I'll show you one more. This is a pretty important one. Let me do it down here so we have some space. So another one looks like this, bonded to a CH3 over here, and then a hydrogen. This right here is called acetylaldehyde. And I just wanted to expose you to these common names because this is what people normally use for these molecules."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let me do it down here so we have some space. So another one looks like this, bonded to a CH3 over here, and then a hydrogen. This right here is called acetylaldehyde. And I just wanted to expose you to these common names because this is what people normally use for these molecules. Now, there is an IAPUC, I always forget the acronym. There is a systematic way to name them. It's actually pretty straightforward."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And I just wanted to expose you to these common names because this is what people normally use for these molecules. Now, there is an IAPUC, I always forget the acronym. There is a systematic way to name them. It's actually pretty straightforward. You just look at the longest carbon chain, and you always start numbering at the carbon that's in the carbonyl group, and it's always going to be at the end of the chain because that's always going to be bonded to a hydrogen. Actually, let me just make clear that this is also an aldehyde. You have your carbonyl group, and on one end you have a hydrogen, just like that."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's actually pretty straightforward. You just look at the longest carbon chain, and you always start numbering at the carbon that's in the carbonyl group, and it's always going to be at the end of the chain because that's always going to be bonded to a hydrogen. Actually, let me just make clear that this is also an aldehyde. You have your carbonyl group, and on one end you have a hydrogen, just like that. Now, the systematic way of it, you just look at the longest carbon chain. Over here, there's just one carbon. So here, you would call this, the systematic name, you would call this methanol, not methanol."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "You have your carbonyl group, and on one end you have a hydrogen, just like that. Now, the systematic way of it, you just look at the longest carbon chain. Over here, there's just one carbon. So here, you would call this, the systematic name, you would call this methanol, not methanol. Methanol means you would have an OH group, but since you have this double bond and you have a hydrogen, or you could say this double bond is at the end of the carbon chain, this is an aldehyde, so the systematic name is methanol, not methanol. Let me make that clear. This is methanol right there."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So here, you would call this, the systematic name, you would call this methanol, not methanol. Methanol means you would have an OH group, but since you have this double bond and you have a hydrogen, or you could say this double bond is at the end of the carbon chain, this is an aldehyde, so the systematic name is methanol, not methanol. Let me make that clear. This is methanol right there. This one, I won't do cinnamaldehyde or benzaldehyde because this is really the only way that people name it. Actually, frankly, formaldehyde, people never call it methanol. And acetylaldehyde, they'll never call it what I'm about to name it, but the systematic name is the longest carbon chain."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This is methanol right there. This one, I won't do cinnamaldehyde or benzaldehyde because this is really the only way that people name it. Actually, frankly, formaldehyde, people never call it methanol. And acetylaldehyde, they'll never call it what I'm about to name it, but the systematic name is the longest carbon chain. You have one, two carbons, so it is ethanol. I don't want to pronounce it incorrectly. Let's do a couple more of these just to make sure we have a decent understanding of the systematic names here."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And acetylaldehyde, they'll never call it what I'm about to name it, but the systematic name is the longest carbon chain. You have one, two carbons, so it is ethanol. I don't want to pronounce it incorrectly. Let's do a couple more of these just to make sure we have a decent understanding of the systematic names here. If we were to name, and in general, if you have a really long chain, the systematic names are what is used. This thing right here, what would you call it? It's clearly an aldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's do a couple more of these just to make sure we have a decent understanding of the systematic names here. If we were to name, and in general, if you have a really long chain, the systematic names are what is used. This thing right here, what would you call it? It's clearly an aldehyde. You have a carbonyl group, a hydrogen, and you have one, two, three carbons. Three carbons, the prefix is prop, so it's propanal. It's an aldehyde."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's clearly an aldehyde. You have a carbonyl group, a hydrogen, and you have one, two, three carbons. Three carbons, the prefix is prop, so it's propanal. It's an aldehyde. If we want to do something slightly more complicated, let's do something like this. What would this be called? Let's see."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It's an aldehyde. If we want to do something slightly more complicated, let's do something like this. What would this be called? Let's see. We have one, two, three, four, five. We have five carbons, so it will be pent. Then it's obviously an aldehyde, so it's pentanol."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's see. We have one, two, three, four, five. We have five carbons, so it will be pent. Then it's obviously an aldehyde, so it's pentanol. You always assume that you start numbering at the carbon in the carbonyl group. One, two, three, four, five. On the two carbon, you have a methyl group."}, {"video_title": "Aldehyde introduction Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Then it's obviously an aldehyde, so it's pentanol. You always assume that you start numbering at the carbon in the carbonyl group. One, two, three, four, five. On the two carbon, you have a methyl group. This is two methyl. You have a methyl group right here. This is going to be two methyl pentanol."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "So you actually should never see something called 2-propylheptane. And let me show you what I'm talking about. So when you see something like this, you might immediately say, and the way we drew it was actually correct. It just wouldn't be called 2-propylheptane. So you say heptane. So that is a 7-carbon alkane, no double bond. So 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "It just wouldn't be called 2-propylheptane. So you say heptane. So that is a 7-carbon alkane, no double bond. So 1, 2, 3, 4, 5, 6, 7. And then on the second carbon, so you have 1, 2, 3, 4, 5, 6, 7, we have a propyl group. Propyl, that is 3 carbons. So on the second carbon, we have a propyl group."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, 7. And then on the second carbon, so you have 1, 2, 3, 4, 5, 6, 7, we have a propyl group. Propyl, that is 3 carbons. So on the second carbon, we have a propyl group. That's 3 carbons. So that is 1, 2, 3. And so the way we drew it was just like this."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "So on the second carbon, we have a propyl group. That's 3 carbons. So that is 1, 2, 3. And so the way we drew it was just like this. And so if someone gave you 2-propylheptane, this would be what you would draw. But you wouldn't call this 2-propylheptane. Because remember, if you're given the molecule, you look for the longest chain."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "And so the way we drew it was just like this. And so if someone gave you 2-propylheptane, this would be what you would draw. But you wouldn't call this 2-propylheptane. Because remember, if you're given the molecule, you look for the longest chain. And the longest chain here is not the heptane chain. It is not 1, 2, 3. It is not this thing in this kind of mauve color."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "Because remember, if you're given the molecule, you look for the longest chain. And the longest chain here is not the heptane chain. It is not 1, 2, 3. It is not this thing in this kind of mauve color. It's this chain where you start over here. If you start over here, 1, 2, 3, 4, 5, 6, 7, 8, 9, you actually get a longer chain. So this would actually be the backbone of this molecule right over there."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "It is not this thing in this kind of mauve color. It's this chain where you start over here. If you start over here, 1, 2, 3, 4, 5, 6, 7, 8, 9, you actually get a longer chain. So this would actually be the backbone of this molecule right over there. That right over there would be the backbone. And so you would number it. And you would start numbering closest to the group that's attached."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "So this would actually be the backbone of this molecule right over there. That right over there would be the backbone. And so you would number it. And you would start numbering closest to the group that's attached. So 1, 2, 3, 4, 5, 6, 7, 8, 9. So you have nine carbons in your backbone. So we're dealing with nonane."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "And you would start numbering closest to the group that's attached. So 1, 2, 3, 4, 5, 6, 7, 8, 9. So you have nine carbons in your backbone. So we're dealing with nonane. And you have a methyl group, one carbon attached to the fourth carbon of our main backbone. So this is going to be 4-methyl. This is our methyl group right here."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "So we're dealing with nonane. And you have a methyl group, one carbon attached to the fourth carbon of our main backbone. So this is going to be 4-methyl. This is our methyl group right here. So it was brought up. I think the username is minoc2. And they correctly corrected me that there would never be such a thing as 2-propylheptane."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "This is our methyl group right here. So it was brought up. I think the username is minoc2. And they correctly corrected me that there would never be such a thing as 2-propylheptane. I just made that up. If someone were to kind of label this molecule, they would call it 4-methylnonane and ask you to draw it. But either way, both of these would point you to the right direction."}, {"video_title": "Correction - 2-propylheptane should never be the name! Organic chemistry Khan Academy.mp3", "Sentence": "And they correctly corrected me that there would never be such a thing as 2-propylheptane. I just made that up. If someone were to kind of label this molecule, they would call it 4-methylnonane and ask you to draw it. But either way, both of these would point you to the right direction. This would just be the incorrect name for it. Because you'd be looking at, if someone gave you this molecule and you named it this way, that would be incorrect. So I apologize for this."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "What I want to do in this video is really make some clarifications and go a little bit more in detail about the different layers of the Earth. So let me draw a cross-section of the Earth over here. And I'll try to do it, I won't be able to do it perfectly to scale, but I'll try to do a little bit better job at giving you a little bit of a sense of how thick these layers are. So let's say that this is the crust up here. And I'm going to make the continental crust a little bit thicker. So let's say that that is continental crust and this is continental crust. And then in between, let me put some oceanic crust, which is going to be thinner."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So let's say that this is the crust up here. And I'm going to make the continental crust a little bit thicker. So let's say that that is continental crust and this is continental crust. And then in between, let me put some oceanic crust, which is going to be thinner. So this right here is, actually let me do it in a different color. Let me do the oceanic crust in blue. But this isn't water, this is rock."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And then in between, let me put some oceanic crust, which is going to be thinner. So this right here is, actually let me do it in a different color. Let me do the oceanic crust in blue. But this isn't water, this is rock. I'll do it in purple, that's even better. I don't want it to be that thick. So let me draw the oceanic crust is thinner than the continental crust, which I'm trying to depict right over here."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "But this isn't water, this is rock. I'll do it in purple, that's even better. I don't want it to be that thick. So let me draw the oceanic crust is thinner than the continental crust, which I'm trying to depict right over here. So this right over here is oceanic crust. And up here is continental crust. And the thickness, or how deep you can go and still be in crust, it depends on where you are."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So let me draw the oceanic crust is thinner than the continental crust, which I'm trying to depict right over here. So this right over here is oceanic crust. And up here is continental crust. And the thickness, or how deep you can go and still be in crust, it depends on where you are. And we know that near hot spots, the oceanic crust can actually thin out a good bit. But roughly, when we talk about the crust, we're talking about something that's 30 to 60 kilometers deep. So if you are on a continent, which I'm assuming you are, and you dig for 20 kilometers, you will still be in the crust."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And the thickness, or how deep you can go and still be in crust, it depends on where you are. And we know that near hot spots, the oceanic crust can actually thin out a good bit. But roughly, when we talk about the crust, we're talking about something that's 30 to 60 kilometers deep. So if you are on a continent, which I'm assuming you are, and you dig for 20 kilometers, you will still be in the crust. 30 kilometers, probably still in the crust. If you dig for 70 kilometers or 100 kilometers, you will probably reach the mantle. And remember, what we're describing here when we talk about the crust, the mantle, and the core, we're talking about the chemical makeup."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So if you are on a continent, which I'm assuming you are, and you dig for 20 kilometers, you will still be in the crust. 30 kilometers, probably still in the crust. If you dig for 70 kilometers or 100 kilometers, you will probably reach the mantle. And remember, what we're describing here when we talk about the crust, the mantle, and the core, we're talking about the chemical makeup. Let me make this clear. We're talking about the chemical makeup. The crust is fundamentally different than the mantle based on the molecules that it is made up of, based on its composition."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And remember, what we're describing here when we talk about the crust, the mantle, and the core, we're talking about the chemical makeup. Let me make this clear. We're talking about the chemical makeup. The crust is fundamentally different than the mantle based on the molecules that it is made up of, based on its composition. So let's talk about the mantle now. So the mantle layer like this. And once again, this is not to scale, because the crust we're talking about 30 to 60 kilometers."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "The crust is fundamentally different than the mantle based on the molecules that it is made up of, based on its composition. So let's talk about the mantle now. So the mantle layer like this. And once again, this is not to scale, because the crust we're talking about 30 to 60 kilometers. The mantle we're talking about on the order of about 2,900 or 3,000 kilometers thick. So this right here is the entire mantle. So that's the mantle."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And once again, this is not to scale, because the crust we're talking about 30 to 60 kilometers. The mantle we're talking about on the order of about 2,900 or 3,000 kilometers thick. So this right here is the entire mantle. So that's the mantle. And this is 2,900 to 3,000 kilometers thick. So this isn't even 1 30th of that. So I would have to draw it even narrower than the way I've drawn it over here."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So that's the mantle. And this is 2,900 to 3,000 kilometers thick. So this isn't even 1 30th of that. So I would have to draw it even narrower than the way I've drawn it over here. And the mantle itself can be subdivided into the upper mantle and the lower mantle. So let me draw this division right over here. So this is the upper mantle."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So I would have to draw it even narrower than the way I've drawn it over here. And the mantle itself can be subdivided into the upper mantle and the lower mantle. So let me draw this division right over here. So this is the upper mantle. And there's different ways to define the boundary. But the upper mantle is roughly about 700 kilometers down. So these are huge distances."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So this is the upper mantle. And there's different ways to define the boundary. But the upper mantle is roughly about 700 kilometers down. So these are huge distances. I mean, this is going straight down. So this is the upper mantle. Let me write it on the actual mantle here."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So these are huge distances. I mean, this is going straight down. So this is the upper mantle. Let me write it on the actual mantle here. This is the upper mantle. And this over here is the lower mantle. And just to be clear on things."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "Let me write it on the actual mantle here. This is the upper mantle. And this over here is the lower mantle. And just to be clear on things. So the crust is solid. Now, when you go into the upper mantle, the upper part of the upper mantle, and we'll talk about that a little bit more, is cool enough to be solid. So there is a solid portion of the upper mantle."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And just to be clear on things. So the crust is solid. Now, when you go into the upper mantle, the upper part of the upper mantle, and we'll talk about that a little bit more, is cool enough to be solid. So there is a solid portion of the upper mantle. So all of this up here is solid because it's cool enough. It hasn't reached the melting point of those rocks just yet. And we learned in previous videos that the combination of the solid part of the upper mantle and the crust combined, we call that the lithosphere."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So there is a solid portion of the upper mantle. So all of this up here is solid because it's cool enough. It hasn't reached the melting point of those rocks just yet. And we learned in previous videos that the combination of the solid part of the upper mantle and the crust combined, we call that the lithosphere. And when we talk about the lithosphere, we're not talking about the mechanical makeup. We're not talking about what's solid and what's not solid. So this is the lithosphere."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And we learned in previous videos that the combination of the solid part of the upper mantle and the crust combined, we call that the lithosphere. And when we talk about the lithosphere, we're not talking about the mechanical makeup. We're not talking about what's solid and what's not solid. So this is the lithosphere. You go a little bit deeper, right below the lithosphere. Now the temperatures are high enough for, and I use the word liquid, but that's not exactly right. You can kind of think of it as kind of a deformable solid or a plastic solid or a magma."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So this is the lithosphere. You go a little bit deeper, right below the lithosphere. Now the temperatures are high enough for, and I use the word liquid, but that's not exactly right. You can kind of think of it as kind of a deformable solid or a plastic solid or a magma. And that's the asthenosphere. So this area right over here, the liquid part, actually I shouldn't use the word liquid, kind of the deformable, it deforms over long periods of time. But it is more fluid than what we normally associate with rock."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "You can kind of think of it as kind of a deformable solid or a plastic solid or a magma. And that's the asthenosphere. So this area right over here, the liquid part, actually I shouldn't use the word liquid, kind of the deformable, it deforms over long periods of time. But it is more fluid than what we normally associate with rock. Magma would be a good way to think about it. That's what we call the asthenosphere. It is fluid, just not as fluid as water."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "But it is more fluid than what we normally associate with rock. Magma would be a good way to think about it. That's what we call the asthenosphere. It is fluid, just not as fluid as water. It is more viscous than something like water. So this is the asthenosphere. Now the upper mantle is hot enough for the rock to melt and be fluid, and the pressure is low enough for it to still be able to kind of move past itself, to still be somewhat fluid."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "It is fluid, just not as fluid as water. It is more viscous than something like water. So this is the asthenosphere. Now the upper mantle is hot enough for the rock to melt and be fluid, and the pressure is low enough for it to still be able to kind of move past itself, to still be somewhat fluid. But then once you get even deeper into the lower mantle, you have higher pressure. And so it's still fluid, but it's less fluid. It's kind of thicker, I guess is the best way to think about it, in the lower mantle."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "Now the upper mantle is hot enough for the rock to melt and be fluid, and the pressure is low enough for it to still be able to kind of move past itself, to still be somewhat fluid. But then once you get even deeper into the lower mantle, you have higher pressure. And so it's still fluid, but it's less fluid. It's kind of thicker, I guess is the best way to think about it, in the lower mantle. It's thicker. So this whole area over here, you can kind of think of it as melted rock. It's fluid, but the upper part of the melted rock is more fluid."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "It's kind of thicker, I guess is the best way to think about it, in the lower mantle. It's thicker. So this whole area over here, you can kind of think of it as melted rock. It's fluid, but the upper part of the melted rock is more fluid. It's able to move easier because there's less pressure. And the pressure is just from all of the rock that's above it. Remember, gravity is pulling down on everything."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "It's fluid, but the upper part of the melted rock is more fluid. It's able to move easier because there's less pressure. And the pressure is just from all of the rock that's above it. Remember, gravity is pulling down on everything. Gravity, every molecule here wants to go downward because of gravity. So it's applying pressure downward. So the deeper you go, the more pressure you get."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "Remember, gravity is pulling down on everything. Gravity, every molecule here wants to go downward because of gravity. So it's applying pressure downward. So the deeper you go, the more pressure you get. Now, when we get even deeper than that, we get to the core, and the core is divided between the outer core and the inner core. So you have the outer core, and then, of course, you have the inner core. And just so we have a sense for distances, the width or the thickness of the outer core is approximately 2,300 kilometers."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So the deeper you go, the more pressure you get. Now, when we get even deeper than that, we get to the core, and the core is divided between the outer core and the inner core. So you have the outer core, and then, of course, you have the inner core. And just so we have a sense for distances, the width or the thickness of the outer core is approximately 2,300 kilometers. So these are huge distances when you think about thickness. You can go down another 2,300 kilometers, or once you go through the mantle, you go 2,300 kilometers to the outer core, and then you're in the inner core, and that essentially takes you to the rest. That's essentially the center of the Earth."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And just so we have a sense for distances, the width or the thickness of the outer core is approximately 2,300 kilometers. So these are huge distances when you think about thickness. You can go down another 2,300 kilometers, or once you go through the mantle, you go 2,300 kilometers to the outer core, and then you're in the inner core, and that essentially takes you to the rest. That's essentially the center of the Earth. And the inner core, and maybe I should draw the boundaries a little bit more to scale. Let me do it this way. It should actually look a little bit more like this because the outer core is thicker than the inner core."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "That's essentially the center of the Earth. And the inner core, and maybe I should draw the boundaries a little bit more to scale. Let me do it this way. It should actually look a little bit more like this because the outer core is thicker than the inner core. So the outer core is, as I said, let me rewrite it. The outer core is on the order, it's about 2,300 kilometers thick, and then you have your inner core. I shouldn't do it in blue."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "It should actually look a little bit more like this because the outer core is thicker than the inner core. So the outer core is, as I said, let me rewrite it. The outer core is on the order, it's about 2,300 kilometers thick, and then you have your inner core. I shouldn't do it in blue. I should do it in a hot color. So the inner core right over here just kind of takes us to the center of the Earth, and that's a little over 1,000 kilometers thick. So this is the inner core."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "I shouldn't do it in blue. I should do it in a hot color. So the inner core right over here just kind of takes us to the center of the Earth, and that's a little over 1,000 kilometers thick. So this is the inner core. The number I have is about 1,200 kilometers thick. And both the entire core, both the outer core and the inner core, is mainly nickel and iron. Think about when the Earth was forming."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So this is the inner core. The number I have is about 1,200 kilometers thick. And both the entire core, both the outer core and the inner core, is mainly nickel and iron. Think about when the Earth was forming. What happens is when this whole Earth was super hot and it was kind of in a fluid state, the heavier elements were allowed to sink down when everything was fluid. The things that are in between would kind of, or the things that were lighter would go up, and then the gases, things that would naturally be in the gaseous state, would kind of bubble up through that fluid, kind of the way actually carbon bubbles up in a soda. It would eventually bubble out of the fluid, and it would actually form the atmosphere."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "Think about when the Earth was forming. What happens is when this whole Earth was super hot and it was kind of in a fluid state, the heavier elements were allowed to sink down when everything was fluid. The things that are in between would kind of, or the things that were lighter would go up, and then the gases, things that would naturally be in the gaseous state, would kind of bubble up through that fluid, kind of the way actually carbon bubbles up in a soda. It would eventually bubble out of the fluid, and it would actually form the atmosphere. So that's why when you look at the composition of the Earth, you have the densest, the heaviest elements in the center, and then the lightest elements are forming the atmosphere. And the outer core and the inner core, they are made up predominantly of nickel and iron. And their makeup is actually very similar."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "It would eventually bubble out of the fluid, and it would actually form the atmosphere. So that's why when you look at the composition of the Earth, you have the densest, the heaviest elements in the center, and then the lightest elements are forming the atmosphere. And the outer core and the inner core, they are made up predominantly of nickel and iron. And their makeup is actually very similar. Chemically, they have a very similar composition. What's different about them is at the outer core, you have temperatures high enough that nickel and iron can melt, but the pressures are low enough that they can still be in a fluid state. So this is our liquid outer core, and this has a pretty low viscosity, especially even relative to the mantle."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "And their makeup is actually very similar. Chemically, they have a very similar composition. What's different about them is at the outer core, you have temperatures high enough that nickel and iron can melt, but the pressures are low enough that they can still be in a fluid state. So this is our liquid outer core, and this has a pretty low viscosity, especially even relative to the mantle. So that's why people kind of consider this in kind of a more traditional liquid state. But as you get deeper and deeper and deeper, the pressure becomes so huge as you get to the inner core. Remember, all of the weight of all of the rock above you, of these thousands of miles of rock above you, is all pushing down on the rock below it."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "So this is our liquid outer core, and this has a pretty low viscosity, especially even relative to the mantle. So that's why people kind of consider this in kind of a more traditional liquid state. But as you get deeper and deeper and deeper, the pressure becomes so huge as you get to the inner core. Remember, all of the weight of all of the rock above you, of these thousands of miles of rock above you, is all pushing down on the rock below it. So the inner core, even though the temperature is really, really, really hot, the pressure is so big that the molecules can't flow past each other. They can't be liquid. They're kind of jammed packed."}, {"video_title": "Structure of the Earth.mp3", "Sentence": "Remember, all of the weight of all of the rock above you, of these thousands of miles of rock above you, is all pushing down on the rock below it. So the inner core, even though the temperature is really, really, really hot, the pressure is so big that the molecules can't flow past each other. They can't be liquid. They're kind of jammed packed. And so the inner core, because of the high pressure, despite the high temperature, is solid. So the difference here is actually a mechanical one between the outer core and the inner core. They're made up of the same things, roughly the same chemical makeup."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Here's the general reaction for a hydroboration oxidation. We start with an alkene, and in the first step, this is our hydroboration step, we're adding borane, which is BH3, and tetrahydrofuran, which is THF. Our second step is the oxidation, where we add hydrogen peroxide and a source of hydroxide anions. And we can see what our product would be. We're gonna add an OH and an H across our double bond. So the double bond goes away, and the OH adds to one of the carbons, and an H adds to the other carbon. How do we know which carbon to add the OH and which carbon to add the H?"}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we can see what our product would be. We're gonna add an OH and an H across our double bond. So the double bond goes away, and the OH adds to one of the carbons, and an H adds to the other carbon. How do we know which carbon to add the OH and which carbon to add the H? We think about the regiochemistry for this reaction, and it turns out to be an anti-Markovnikov, which means the OH adds to the less substituted carbon. To understand this, we need to look at the mechanism for a hydroboration oxidation, which I put in the next video because it's way too long to fit into this video. Also in the next video, we'll go into the details about the stereochemistry for this reaction."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "How do we know which carbon to add the OH and which carbon to add the H? We think about the regiochemistry for this reaction, and it turns out to be an anti-Markovnikov, which means the OH adds to the less substituted carbon. To understand this, we need to look at the mechanism for a hydroboration oxidation, which I put in the next video because it's way too long to fit into this video. Also in the next video, we'll go into the details about the stereochemistry for this reaction. It turns out to be a syn addition, which means the H and the OH add to the same side. Let's look at an example of a hydroboration oxidation. We'll look at this alkene here."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Also in the next video, we'll go into the details about the stereochemistry for this reaction. It turns out to be a syn addition, which means the H and the OH add to the same side. Let's look at an example of a hydroboration oxidation. We'll look at this alkene here. We know we're gonna add an OH and an H across our double bond, and we know the OH adds to the less substituted carbon. I find my two carbons for my double bond, so this carbon right here and then this one. Which one of those carbons is the less substituted carbon?"}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'll look at this alkene here. We know we're gonna add an OH and an H across our double bond, and we know the OH adds to the less substituted carbon. I find my two carbons for my double bond, so this carbon right here and then this one. Which one of those carbons is the less substituted carbon? Obviously, the top carbon here is the less substituted carbon, so that's the one where we're going to add the OH. I can go ahead and draw my product. We're adding the OH to the top carbon right there."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Which one of those carbons is the less substituted carbon? Obviously, the top carbon here is the less substituted carbon, so that's the one where we're going to add the OH. I can go ahead and draw my product. We're adding the OH to the top carbon right there. That's this one in magenta. The hydrogen would add to this carbon down here in red. I'm just not gonna draw in the hydrogen, but that's where it would add."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're adding the OH to the top carbon right there. That's this one in magenta. The hydrogen would add to this carbon down here in red. I'm just not gonna draw in the hydrogen, but that's where it would add. There's our product. We don't really have to worry about stereochemistry for drawing our final product because this carbon in magenta, this is not a chiral center, and the carbon in red is also not a chiral center. Let's look at an example where we do have to worry about the stereochemistry when we're drawing our products."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm just not gonna draw in the hydrogen, but that's where it would add. There's our product. We don't really have to worry about stereochemistry for drawing our final product because this carbon in magenta, this is not a chiral center, and the carbon in red is also not a chiral center. Let's look at an example where we do have to worry about the stereochemistry when we're drawing our products. For this alkene right here, I look at my two carbons across my double bond, this carbon and this carbon. I know this is a hydroboration oxidation, so I'm going to add my OH to the less substituted carbon. This, of course, on the right would be the less substituted carbon."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an example where we do have to worry about the stereochemistry when we're drawing our products. For this alkene right here, I look at my two carbons across my double bond, this carbon and this carbon. I know this is a hydroboration oxidation, so I'm going to add my OH to the less substituted carbon. This, of course, on the right would be the less substituted carbon. Just thinking about regiochemistry, I'm gonna draw in my OH on that carbon. I'm gonna draw in my product, not thinking about stereochemistry yet. The OH added to the carbon in magenta, which was our less substituted carbon."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This, of course, on the right would be the less substituted carbon. Just thinking about regiochemistry, I'm gonna draw in my OH on that carbon. I'm gonna draw in my product, not thinking about stereochemistry yet. The OH added to the carbon in magenta, which was our less substituted carbon. When we look at our product, this is our carbon now. That carbon I just circled is a chiral center, so we do need to worry about stereochemistry. This carbon over here on the left is not a chiral center, so we don't have to worry about the carbon in red."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The OH added to the carbon in magenta, which was our less substituted carbon. When we look at our product, this is our carbon now. That carbon I just circled is a chiral center, so we do need to worry about stereochemistry. This carbon over here on the left is not a chiral center, so we don't have to worry about the carbon in red. We do have to worry about the carbon in magenta. Let's think about the stereochemistry with a syn addition, adding the H and the OH to the same side. We need to think about this alkene here and these carbons that I've marked in yellow."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This carbon over here on the left is not a chiral center, so we don't have to worry about the carbon in red. We do have to worry about the carbon in magenta. Let's think about the stereochemistry with a syn addition, adding the H and the OH to the same side. We need to think about this alkene here and these carbons that I've marked in yellow. Those are sp2 hybridized carbons, so the geometry around those carbons is planar. When we add our H and OH, we're turning those into sp3 hybridized carbons. Let's go ahead and sketch in one of our products here, so one of the possible products would be to add the H and the OH on the same side."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We need to think about this alkene here and these carbons that I've marked in yellow. Those are sp2 hybridized carbons, so the geometry around those carbons is planar. When we add our H and OH, we're turning those into sp3 hybridized carbons. Let's go ahead and sketch in one of our products here, so one of the possible products would be to add the H and the OH on the same side. I'm going to draw them as wedges. Here's our hydrogen as a wedge, and then here's the OH as a wedge. I already know which carbon to add which because we already figured that out."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and sketch in one of our products here, so one of the possible products would be to add the H and the OH on the same side. I'm going to draw them as wedges. Here's our hydrogen as a wedge, and then here's the OH as a wedge. I already know which carbon to add which because we already figured that out. We know the OH adds to this carbon, the one on the right, and we know the H added to the carbon over here in red. I'm showing a syn addition of my hydrogen, of my H, and of my OH here. Those are wedges."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I already know which carbon to add which because we already figured that out. We know the OH adds to this carbon, the one on the right, and we know the H added to the carbon over here in red. I'm showing a syn addition of my hydrogen, of my H, and of my OH here. Those are wedges. Let's think about what's attached to those carbons. Let me go back up to here. This is the carbon in magenta."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Those are wedges. Let's think about what's attached to those carbons. Let me go back up to here. This is the carbon in magenta. That's this carbon down here. What else is attached to that carbon? Well, there's a hydrogen."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is the carbon in magenta. That's this carbon down here. What else is attached to that carbon? Well, there's a hydrogen. Therefore, if I think about the stereochemistry, that hydrogen must be going away from me now for this product. We're going from an sp2 hybridized carbon in our alkene to an sp3 hybridized carbon. That hydrogen is going away from us."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, there's a hydrogen. Therefore, if I think about the stereochemistry, that hydrogen must be going away from me now for this product. We're going from an sp2 hybridized carbon in our alkene to an sp3 hybridized carbon. That hydrogen is going away from us. Let's think about this other carbon. Let me make this other carbon over here red, so this carbon right here. That's this carbon."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That hydrogen is going away from us. Let's think about this other carbon. Let me make this other carbon over here red, so this carbon right here. That's this carbon. What else is attached to this carbon? Well, there's a methyl group. If this hydrogen is coming out at me, this methyl group must be going away from me."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's this carbon. What else is attached to this carbon? Well, there's a methyl group. If this hydrogen is coming out at me, this methyl group must be going away from me. We have a methyl group going away from us in space. That's one of our possible products. I could draw this out the way we're used to seeing it where we would show the OH coming out at us in space."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If this hydrogen is coming out at me, this methyl group must be going away from me. We have a methyl group going away from us in space. That's one of our possible products. I could draw this out the way we're used to seeing it where we would show the OH coming out at us in space. The OH is coming out at us. That's one of our possible products. Our other products, I could show the H and the OH adding on the same side, but I can show them adding as dashes."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I could draw this out the way we're used to seeing it where we would show the OH coming out at us in space. The OH is coming out at us. That's one of our possible products. Our other products, I could show the H and the OH adding on the same side, but I can show them adding as dashes. I could draw in our carbons here. I can show this time, I can show the H adding as a dash and the OH adding as a dash. Now, that means that this hydrogen right here would have to be a wedge."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Our other products, I could show the H and the OH adding on the same side, but I can show them adding as dashes. I could draw in our carbons here. I can show this time, I can show the H adding as a dash and the OH adding as a dash. Now, that means that this hydrogen right here would have to be a wedge. That hydrogen would have to be a wedge, so I draw in the hydrogen here as a wedge. Then for the other carbon, the one in red, so this one, this methyl group would have to be coming out at us now. That methyl group is now coming out at us."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, that means that this hydrogen right here would have to be a wedge. That hydrogen would have to be a wedge, so I draw in the hydrogen here as a wedge. Then for the other carbon, the one in red, so this one, this methyl group would have to be coming out at us now. That methyl group is now coming out at us. I'm going to draw that as a wedge. Our methyl group CH3 would be here. This molecule, just another way to draw it, would be like this where your OH is going away from us in space because we already identified our chiral center."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That methyl group is now coming out at us. I'm going to draw that as a wedge. Our methyl group CH3 would be here. This molecule, just another way to draw it, would be like this where your OH is going away from us in space because we already identified our chiral center. This carbon up here, this carbon was our chiral center. The OH could be a wedge or the OH could be a dash. Of course, the relationship between our two products here, these are enantiomers of each other."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This molecule, just another way to draw it, would be like this where your OH is going away from us in space because we already identified our chiral center. This carbon up here, this carbon was our chiral center. The OH could be a wedge or the OH could be a dash. Of course, the relationship between our two products here, these are enantiomers of each other. Let's do one more example where we have to worry about the regiochemistry and the stereochemistry. Let's look at this one right here. Here's our alkene."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Of course, the relationship between our two products here, these are enantiomers of each other. Let's do one more example where we have to worry about the regiochemistry and the stereochemistry. Let's look at this one right here. Here's our alkene. We know this is a hydroboration oxidation. Here's our double bond. We have a carbon right here and we have a carbon right here."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Here's our alkene. We know this is a hydroboration oxidation. Here's our double bond. We have a carbon right here and we have a carbon right here. We know the OH adds to the less substituted carbon. This carbon is, of course, the less substituted carbon. Just thinking about regiochemistry, we can draw in, just thinking about regiochemistry right here, we know the OH adds to this carbon down here."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have a carbon right here and we have a carbon right here. We know the OH adds to the less substituted carbon. This carbon is, of course, the less substituted carbon. Just thinking about regiochemistry, we can draw in, just thinking about regiochemistry right here, we know the OH adds to this carbon down here. We have a methyl group up here. Let me go ahead and highlight our carbons once again. The carbon in magenta is this one and the carbon in red right here is this one."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Just thinking about regiochemistry, we can draw in, just thinking about regiochemistry right here, we know the OH adds to this carbon down here. We have a methyl group up here. Let me go ahead and highlight our carbons once again. The carbon in magenta is this one and the carbon in red right here is this one. The hydrogen would add to the carbon in red. We could go ahead and draw in CH3 here just to clarify. Notice that we have two chiral centers."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The carbon in magenta is this one and the carbon in red right here is this one. The hydrogen would add to the carbon in red. We could go ahead and draw in CH3 here just to clarify. Notice that we have two chiral centers. The carbon in red is a chiral center and the carbon in magenta is also a chiral center. Let's think about stereochemistry now. We're going to draw, let's sketch in one of our products here."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Notice that we have two chiral centers. The carbon in red is a chiral center and the carbon in magenta is also a chiral center. Let's think about stereochemistry now. We're going to draw, let's sketch in one of our products here. We know that this is a syn addition, which means that the hydrogen and the OH add, the H and the OH add on the same side. I already know the OH adds to this bottom carbon here. I'm going to show the OH adding as a wedge."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to draw, let's sketch in one of our products here. We know that this is a syn addition, which means that the hydrogen and the OH add, the H and the OH add on the same side. I already know the OH adds to this bottom carbon here. I'm going to show the OH adding as a wedge. I'm going to draw a wedge in here. That's my OH. My hydrogen adds to the other carbon, so this one right here."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to show the OH adding as a wedge. I'm going to draw a wedge in here. That's my OH. My hydrogen adds to the other carbon, so this one right here. Because it has to add on the same side as the OH, I draw it as a wedge. Let's think about this carbon in red. What else is attached to that carbon?"}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "My hydrogen adds to the other carbon, so this one right here. Because it has to add on the same side as the OH, I draw it as a wedge. Let's think about this carbon in red. What else is attached to that carbon? Of course, this methyl group. We're changing hybridization states. Now I can show that methyl group going away from us in space."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What else is attached to that carbon? Of course, this methyl group. We're changing hybridization states. Now I can show that methyl group going away from us in space. That's a CH3 right here. That's one of our possible products. Let's draw in the other one."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now I can show that methyl group going away from us in space. That's a CH3 right here. That's one of our possible products. Let's draw in the other one. The other possibility is the OH and the H add as dashes. I can show them adding from the opposite side. I can show the OH adding as a dash."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw in the other one. The other possibility is the OH and the H add as dashes. I can show them adding from the opposite side. I can show the OH adding as a dash. I can show the H adding as a dash. Now, when we think about this top carbon here and the stereochemistry, this carbon in red, this methyl group must be coming out at us now in space. For this, the methyl group is coming out at us in space right here."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I can show the OH adding as a dash. I can show the H adding as a dash. Now, when we think about this top carbon here and the stereochemistry, this carbon in red, this methyl group must be coming out at us now in space. For this, the methyl group is coming out at us in space right here. This would be a CH3. If you wanted to, you could add in the hydrogen for the carbon with the OH on it. I'm going to stop right there because we have our products."}, {"video_title": "Hydroboration-oxidation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "For this, the methyl group is coming out at us in space right here. This would be a CH3. If you wanted to, you could add in the hydrogen for the carbon with the OH on it. I'm going to stop right there because we have our products. What is the relationship between these molecules? They are enantiomers of each other. We have opposite absolute configurations at both of our chiral centers."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll take a look at the reaction of Grignard reagents with esters as well. So here is our alcohol. And we're going to think backwards. We're going to think using retrosynthesis and try to figure out how to make this alcohol using a Grignard reagent. And there are actually several different ways to do it. So if you understand the mechanism really well, you know that you add an alkyl group onto what used to be your carbonyl carbon. And your carbonyl carbon was the one connected to your oxygen."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "We're going to think using retrosynthesis and try to figure out how to make this alcohol using a Grignard reagent. And there are actually several different ways to do it. So if you understand the mechanism really well, you know that you add an alkyl group onto what used to be your carbonyl carbon. And your carbonyl carbon was the one connected to your oxygen. So if you're thinking backwards, you can think to yourself, this must be my carbonyl carbon. This must be the one. And I added an alkyl group onto that carbonyl carbon."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And your carbonyl carbon was the one connected to your oxygen. So if you're thinking backwards, you can think to yourself, this must be my carbonyl carbon. This must be the one. And I added an alkyl group onto that carbonyl carbon. So if you're thinking backwards, you can disconnect a bond. So I could say, oh, I'm going to make my alkyl group this group right here. So I'm going to say it's that methyl group."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And I added an alkyl group onto that carbonyl carbon. So if you're thinking backwards, you can disconnect a bond. So I could say, oh, I'm going to make my alkyl group this group right here. So I'm going to say it's that methyl group. So I'm going to disconnect that bond. And that will allow me to figure out my starting materials. So if that's the bond that isn't going to exist over here when I'm drawing my starting materials, I'm going to have my benzene ring right here."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to say it's that methyl group. So I'm going to disconnect that bond. And that will allow me to figure out my starting materials. So if that's the bond that isn't going to exist over here when I'm drawing my starting materials, I'm going to have my benzene ring right here. And the benzene ring is directly attached to what was my carbonyl carbon. So I'm going to go ahead and draw my carbonyl carbon in here like that. And I'm saying that this ethyl group over here on the right is still there."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So if that's the bond that isn't going to exist over here when I'm drawing my starting materials, I'm going to have my benzene ring right here. And the benzene ring is directly attached to what was my carbonyl carbon. So I'm going to go ahead and draw my carbonyl carbon in here like that. And I'm saying that this ethyl group over here on the right is still there. And what I added on in the mechanism was a methyl group. So the methyl group must have come from my Grignard reagent. So if I'm going to go ahead and write my Grignard reagent, it must be a methyl group."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And I'm saying that this ethyl group over here on the right is still there. And what I added on in the mechanism was a methyl group. So the methyl group must have come from my Grignard reagent. So if I'm going to go ahead and write my Grignard reagent, it must be a methyl group. And so I'll make methyl magnesium bromide, which is what we used in the previous video for our examples. So these two things on the right would be my starting materials to make the alcohol on the left. And if you're unsure of that, just think about working backwards."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm going to go ahead and write my Grignard reagent, it must be a methyl group. And so I'll make methyl magnesium bromide, which is what we used in the previous video for our examples. So these two things on the right would be my starting materials to make the alcohol on the left. And if you're unsure of that, just think about working backwards. So you could think to yourself, all right, well, once again, from the last video, we know that this methyl group is actually negatively charged as a carb anion. It's going to attack our carbonyl carbon like that. And then those electrons are going to kick off onto the oxygen."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And if you're unsure of that, just think about working backwards. So you could think to yourself, all right, well, once again, from the last video, we know that this methyl group is actually negatively charged as a carb anion. It's going to attack our carbonyl carbon like that. And then those electrons are going to kick off onto the oxygen. So when we add in our reagents, in the first step, when you add your Grignard reagent and your ketone up here, you would need to use ether as your solvent. And the second step, you need to protonate your alkoxides to form your alcohol over here on the left. So you need H3O+."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And then those electrons are going to kick off onto the oxygen. So when we add in our reagents, in the first step, when you add your Grignard reagent and your ketone up here, you would need to use ether as your solvent. And the second step, you need to protonate your alkoxides to form your alcohol over here on the left. So you need H3O+. So that's how to think backwards and come up with your starting materials used to synthesize the alcohol on the left. This isn't the only way to do it. Let's redraw this alcohol."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So you need H3O+. So that's how to think backwards and come up with your starting materials used to synthesize the alcohol on the left. This isn't the only way to do it. Let's redraw this alcohol. And let's look at another way that we can synthesize the same alcohol here. So let's go ahead and redraw the alcohol on the left. And think about how else could I make it."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "Let's redraw this alcohol. And let's look at another way that we can synthesize the same alcohol here. So let's go ahead and redraw the alcohol on the left. And think about how else could I make it. So here's my alcohol. And before, we first identified our carbonyl carbon. It's going to be the one attached to the oxygen right here."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And think about how else could I make it. So here's my alcohol. And before, we first identified our carbonyl carbon. It's going to be the one attached to the oxygen right here. And I have all these alkyl groups attached to it. So I'm just going to disconnect another alkyl group. This time, I'll disconnect the alkyl group on the right."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be the one attached to the oxygen right here. And I have all these alkyl groups attached to it. So I'm just going to disconnect another alkyl group. This time, I'll disconnect the alkyl group on the right. And so I'm thinking retrosynthesis. I'm thinking about working backwards. So if I'm going to go ahead and draw the carbonyl compound, that has my benzene ring untouched."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "This time, I'll disconnect the alkyl group on the right. And so I'm thinking retrosynthesis. I'm thinking about working backwards. So if I'm going to go ahead and draw the carbonyl compound, that has my benzene ring untouched. And the carbon in red is going to become my carbonyl carbon like that. And what's attached to that carbonyl? It's going to be a methyl group."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm going to go ahead and draw the carbonyl compound, that has my benzene ring untouched. And the carbon in red is going to become my carbonyl carbon like that. And what's attached to that carbonyl? It's going to be a methyl group. This time, the methyl group was left. So when I'm thinking about what should be my alkyl group here, it would be the ethyl group that I disconnected on the right. So my Grignard reagent is going to be the source of my ethyl group."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be a methyl group. This time, the methyl group was left. So when I'm thinking about what should be my alkyl group here, it would be the ethyl group that I disconnected on the right. So my Grignard reagent is going to be the source of my ethyl group. So I'm going to go ahead and draw an ethyl group attached to the magnesium. So we have ethyl, magnesium, a bromide like that. So this would give us the alcohol on the left."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So my Grignard reagent is going to be the source of my ethyl group. So I'm going to go ahead and draw an ethyl group attached to the magnesium. So we have ethyl, magnesium, a bromide like that. So this would give us the alcohol on the left. Again, same concept, same mechanism. This carbon right here would end up attacking my carbonyl, kick these electrons off. And once again, we need to specify that the first step of this synthesis is done in diethyl ether."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So this would give us the alcohol on the left. Again, same concept, same mechanism. This carbon right here would end up attacking my carbonyl, kick these electrons off. And once again, we need to specify that the first step of this synthesis is done in diethyl ether. And the second step, we protonate to form our alcohol like that. So let's look at yet another way to form this alcohol. So we're kind of ignoring stereochemistry here."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we need to specify that the first step of this synthesis is done in diethyl ether. And the second step, we protonate to form our alcohol like that. So let's look at yet another way to form this alcohol. So we're kind of ignoring stereochemistry here. So if some of you were thinking that carbon is a chirality center, we're just trying to think about how to make these alcohols here and not worry about what we're getting in terms of stereochemistry for our products. So if I start here with my benzene ring, and once again, we're going to make the exact same alcohol. And we're going to think backwards again."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So we're kind of ignoring stereochemistry here. So if some of you were thinking that carbon is a chirality center, we're just trying to think about how to make these alcohols here and not worry about what we're getting in terms of stereochemistry for our products. So if I start here with my benzene ring, and once again, we're going to make the exact same alcohol. And we're going to think backwards again. And once again, we know this is going to be my carbonyl carbon. This time, we're going to disconnect this bond. So that's the alkyl group that we're going to add on as our Grignard reagent."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to think backwards again. And once again, we know this is going to be my carbonyl carbon. This time, we're going to disconnect this bond. So that's the alkyl group that we're going to add on as our Grignard reagent. So there's a third way of doing it. So now this time, I've disconnected my benzene ring. And this is going to be my Grignard reagent this time."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So that's the alkyl group that we're going to add on as our Grignard reagent. So there's a third way of doing it. So now this time, I've disconnected my benzene ring. And this is going to be my Grignard reagent this time. So this carbon on my benzene ring is going to be ionically bonded to magnesium, so MgBr, like that. And so now it's called a phenyl group. So you would term this Grignard reagent phenyl magnesium bromide."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And this is going to be my Grignard reagent this time. So this carbon on my benzene ring is going to be ionically bonded to magnesium, so MgBr, like that. And so now it's called a phenyl group. So you would term this Grignard reagent phenyl magnesium bromide. And I can see the skeleton for the ketone that I'm going to use on the right there. So I have a carbonyl. And let's see, there's a methyl group on the left side and an ethyl group on the right side."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So you would term this Grignard reagent phenyl magnesium bromide. And I can see the skeleton for the ketone that I'm going to use on the right there. So I have a carbonyl. And let's see, there's a methyl group on the left side and an ethyl group on the right side. So once again, if you mix those two together in ether and then protonate it, you will end up with the alcohol on the left. So three different ways to synthesize the same alcohol. And it's just thinking about what's my alkyl group and how can I add it on."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And let's see, there's a methyl group on the left side and an ethyl group on the right side. So once again, if you mix those two together in ether and then protonate it, you will end up with the alcohol on the left. So three different ways to synthesize the same alcohol. And it's just thinking about what's my alkyl group and how can I add it on. All right, let's look at the general reaction for a Grignard reagent with esters. So there's my Grignard reagent. And I'm going to react it with an ester this time."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And it's just thinking about what's my alkyl group and how can I add it on. All right, let's look at the general reaction for a Grignard reagent with esters. So there's my Grignard reagent. And I'm going to react it with an ester this time. So I'll make this R prime. And I'll make this one R double prime to distinguish it from the R group on our Grignard reagent. And the second step, once again, we add H3O+."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to react it with an ester this time. So I'll make this R prime. And I'll make this one R double prime to distinguish it from the R group on our Grignard reagent. And the second step, once again, we add H3O+. In this situation, the Grignard reagent is going to add on twice to our carbonyl carbon. So the carbon attached to the OH for our product. This is another way to synthesize alcohols."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And the second step, once again, we add H3O+. In this situation, the Grignard reagent is going to add on twice to our carbonyl carbon. So the carbon attached to the OH for our product. This is another way to synthesize alcohols. It's going to have the R prime group still attached to it. And the R group from the Grignard reagent is going to add on twice. So if you look at this, you can see that the R prime group and this carbon came from our ester."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "This is another way to synthesize alcohols. It's going to have the R prime group still attached to it. And the R group from the Grignard reagent is going to add on twice. So if you look at this, you can see that the R prime group and this carbon came from our ester. So did this oxygen right here. And the R group, so this R group here from our Grignard reagent is going to add on twice in the mechanism. So let's take a look at what happens in our mechanism."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at this, you can see that the R prime group and this carbon came from our ester. So did this oxygen right here. And the R group, so this R group here from our Grignard reagent is going to add on twice in the mechanism. So let's take a look at what happens in our mechanism. We start with our ester. So R prime C double bond O, OR double prime. And we'll go ahead and put on our lone pairs of electrons."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a look at what happens in our mechanism. We start with our ester. So R prime C double bond O, OR double prime. And we'll go ahead and put on our lone pairs of electrons. We saw in the last video that that carbonyl is a polarized bond. The oxygen ends up being partially negative. And the carbon ends up being partially positive."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And we'll go ahead and put on our lone pairs of electrons. We saw in the last video that that carbonyl is a polarized bond. The oxygen ends up being partially negative. And the carbon ends up being partially positive. So the carbon is partially positive, making it an electrophile. The source of the electrons come from the Grignard reagent because we know that in that Grignard reagent, we're treating that R group as a carbanion. So this R group up here is actually negatively charged."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And the carbon ends up being partially positive. So the carbon is partially positive, making it an electrophile. The source of the electrons come from the Grignard reagent because we know that in that Grignard reagent, we're treating that R group as a carbanion. So this R group up here is actually negatively charged. So we saw that in the last video. So it is R with two electrons around it, giving it a negative 1 formal charge, making it a carbanion, which is a good nucleophile. So the nucleophile is going to attack our electrophile."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So this R group up here is actually negatively charged. So we saw that in the last video. So it is R with two electrons around it, giving it a negative 1 formal charge, making it a carbanion, which is a good nucleophile. So the nucleophile is going to attack our electrophile. Opposite charges attract. So nucleophilic attack will kick these electrons off onto your oxygen in the first step of the mechanism. So now we have R prime carbon with an oxygen up here, three lone pairs of electrons around it, giving it a negative 1 formal charge."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So the nucleophile is going to attack our electrophile. Opposite charges attract. So nucleophilic attack will kick these electrons off onto your oxygen in the first step of the mechanism. So now we have R prime carbon with an oxygen up here, three lone pairs of electrons around it, giving it a negative 1 formal charge. And over here on the right, we have oxygen with an R prime, and then our lone pairs of electrons like that, and then down our double prime, I should say. And then down here, we attached our R group. OK, so in the next step of the mechanism, lone pair of electrons are going to move in here to reform our pi bond, to reform our double bond."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So now we have R prime carbon with an oxygen up here, three lone pairs of electrons around it, giving it a negative 1 formal charge. And over here on the right, we have oxygen with an R prime, and then our lone pairs of electrons like that, and then down our double prime, I should say. And then down here, we attached our R group. OK, so in the next step of the mechanism, lone pair of electrons are going to move in here to reform our pi bond, to reform our double bond. And that's going to kick the electrons in this bond off onto the oxygen, since carbon cannot have five bonds. So let's go ahead and draw the product of that. So we're going to reform our carbonyl."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "OK, so in the next step of the mechanism, lone pair of electrons are going to move in here to reform our pi bond, to reform our double bond. And that's going to kick the electrons in this bond off onto the oxygen, since carbon cannot have five bonds. So let's go ahead and draw the product of that. So we're going to reform our carbonyl. So now we still have an R prime group attached to our carbonyl like that. We've just added an R group onto our carbonyl. And the leaving group will be this alkoxide over here."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to reform our carbonyl. So now we still have an R prime group attached to our carbonyl like that. We've just added an R group onto our carbonyl. And the leaving group will be this alkoxide over here. So R double prime with a negative 1 formal charge on the oxygen. So now we have a ketone. And if you have a molar excess of your Grignard reagents, the Grignard reagent is going to attack your ketone."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And the leaving group will be this alkoxide over here. So R double prime with a negative 1 formal charge on the oxygen. So now we have a ketone. And if you have a molar excess of your Grignard reagents, the Grignard reagent is going to attack your ketone. So once again, another molar equivalent of our carbanion from our Grignard reagent is going to attack our carbonyl carbon, just like before. Kick these electrons off onto the oxygen. So let's get some space here to show what happens."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And if you have a molar excess of your Grignard reagents, the Grignard reagent is going to attack your ketone. So once again, another molar equivalent of our carbanion from our Grignard reagent is going to attack our carbonyl carbon, just like before. Kick these electrons off onto the oxygen. So let's get some space here to show what happens. So our next intermediate is going to have our R prime attached to our carbon. This top oxygen here has three lone pairs of electrons and a negative 1 formal charge. We already added on one R group."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So let's get some space here to show what happens. So our next intermediate is going to have our R prime attached to our carbon. This top oxygen here has three lone pairs of electrons and a negative 1 formal charge. We already added on one R group. We're going to add on another R group from our Grignard reagent. And in the last step, it's just acid-base chemistry. We're going to add H3O+."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "We already added on one R group. We're going to add on another R group from our Grignard reagent. And in the last step, it's just acid-base chemistry. We're going to add H3O+. And this will allow us to protonate our alkoxide. So a lone pair of electrons on here. Pick up a proton like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "We're going to add H3O+. And this will allow us to protonate our alkoxide. So a lone pair of electrons on here. Pick up a proton like that. Kick those electrons off. And then we are done with our product, R prime. Carbon bonded to an oxygen bonded to a hydrogen, two lone pairs like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "Pick up a proton like that. Kick those electrons off. And then we are done with our product, R prime. Carbon bonded to an oxygen bonded to a hydrogen, two lone pairs like that. And then two R groups, which came from our Grignard reagent like that to form R alcohol. So let's look at a reaction. So let's react an ester with excess Grignard reagent and see what our product will be."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "Carbon bonded to an oxygen bonded to a hydrogen, two lone pairs like that. And then two R groups, which came from our Grignard reagent like that to form R alcohol. So let's look at a reaction. So let's react an ester with excess Grignard reagent and see what our product will be. So we'll do a nice simple ester over here on the left. So here is our ester. And in the first step, we add excess methyl magnesium bromide."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So let's react an ester with excess Grignard reagent and see what our product will be. So we'll do a nice simple ester over here on the left. So here is our ester. And in the first step, we add excess methyl magnesium bromide. And in the second step, we add our H3O+. So we need to think to ourself what's going to happen. Well, I know that in my mechanism, this negatively charged carbanion methyl group is going to add on to this carbon."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And in the first step, we add excess methyl magnesium bromide. And in the second step, we add our H3O+. So we need to think to ourself what's going to happen. Well, I know that in my mechanism, this negatively charged carbanion methyl group is going to add on to this carbon. This is going to leave. That's going to be our leaving group. And then the reaction happens again."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "Well, I know that in my mechanism, this negatively charged carbanion methyl group is going to add on to this carbon. This is going to leave. That's going to be our leaving group. And then the reaction happens again. And then the reaction happens again with adding on another methyl group. And then finally, protonated to form an alcohol. So when we draw the product, we know that this portion is going to come from our ester."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "And then the reaction happens again. And then the reaction happens again with adding on another methyl group. And then finally, protonated to form an alcohol. So when we draw the product, we know that this portion is going to come from our ester. We know we're going to form an alcohol. And we know that we're going to add our methyl group on twice. So all we have to do is put two methyl groups on there like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents II Organic chemistry Khan Academy.mp3", "Sentence": "So when we draw the product, we know that this portion is going to come from our ester. We know we're going to form an alcohol. And we know that we're going to add our methyl group on twice. So all we have to do is put two methyl groups on there like that. So that, of course, would form a tertiary alcohol. So this is tertiary right here. This carbon bonded to three other carbons."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So, one, two, three, four carbons. That's where the but comes in, and then it's a butanoil chloride. So it's an acyl halide, or it's an acyl chloride. Let me write this down just to give us some practice with namings. This is butanoil chloride. We saw a couple of videos ago that acyl halides, you just count the main number of carbons. One, two, three, four carbons."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Let me write this down just to give us some practice with namings. This is butanoil chloride. We saw a couple of videos ago that acyl halides, you just count the main number of carbons. One, two, three, four carbons. That's where the butan comes from. And then this is an acyl chloride. So the acyl chloride part, this part, gives us the oil, or I guess you could look even at the carbonyl."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "One, two, three, four carbons. That's where the butan comes from. And then this is an acyl chloride. So the acyl chloride part, this part, gives us the oil, or I guess you could look even at the carbonyl. We could even say this whole part over here. We know it's a carboxylic acid derivative. It's an acyl chloride, so that's why we have the oil here and the chloride."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So the acyl chloride part, this part, gives us the oil, or I guess you could look even at the carbonyl. We could even say this whole part over here. We know it's a carboxylic acid derivative. It's an acyl chloride, so that's why we have the oil here and the chloride. So let's think about what would happen if we have a molecule of butanoil chloride, or if we had a solution, even better, of butanoil chloride. And for every molecule of butanoil chloride, we had two molecules of dimethylamine. Let me draw that."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "It's an acyl chloride, so that's why we have the oil here and the chloride. So let's think about what would happen if we have a molecule of butanoil chloride, or if we had a solution, even better, of butanoil chloride. And for every molecule of butanoil chloride, we had two molecules of dimethylamine. Let me draw that. So dimethyl, so two methyl groups. One, two, and then we have one hydrogen. And then we're going to have two molecules of this for every one molecule of our butanoil chloride."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Let me draw that. So dimethyl, so two methyl groups. One, two, and then we have one hydrogen. And then we're going to have two molecules of this for every one molecule of our butanoil chloride. So let me copy and paste this. So edit, copy, and edit, paste. So let's think about what would happen in a solution with these reactants in this ratio right here."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And then we're going to have two molecules of this for every one molecule of our butanoil chloride. So let me copy and paste this. So edit, copy, and edit, paste. So let's think about what would happen in a solution with these reactants in this ratio right here. And just as a bit of review, this right here is dimethyl. We kind of have a tie for our longest chain attached to the nitrogen. Dimethylamine."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So let's think about what would happen in a solution with these reactants in this ratio right here. And just as a bit of review, this right here is dimethyl. We kind of have a tie for our longest chain attached to the nitrogen. Dimethylamine. You could pick one of these as the longest chain, and then it would actually be N-methylmethylamine. Either one. This is the one that you're more likely to see."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Dimethylamine. You could pick one of these as the longest chain, and then it would actually be N-methylmethylamine. Either one. This is the one that you're more likely to see. But this video isn't about naming. This video is to think about what's likely to happen. So what are we dealing with?"}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "This is the one that you're more likely to see. But this video isn't about naming. This video is to think about what's likely to happen. So what are we dealing with? We have an amine here, and then we have an acyl chloride. And if we look at our hierarchy of stability that we looked at in the last video, we saw that when an amine reacts, or a carboxylic derivative from an amine, which is an amide, is much more stable than an acyl chloride. This is the least stable."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So what are we dealing with? We have an amine here, and then we have an acyl chloride. And if we look at our hierarchy of stability that we looked at in the last video, we saw that when an amine reacts, or a carboxylic derivative from an amine, which is an amide, is much more stable than an acyl chloride. This is the least stable. So if we're starting with an acyl chloride, and we have the ingredients for an amide, this reaction will probably end up with amide. So let's see how that we can get there. Well, we know from previous videos when we studied amines that they are pretty good nucleophiles."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "This is the least stable. So if we're starting with an acyl chloride, and we have the ingredients for an amide, this reaction will probably end up with amide. So let's see how that we can get there. Well, we know from previous videos when we studied amines that they are pretty good nucleophiles. So they can have a nucleophile attack on this carbonyl carbon right over here. So you can imagine this amine right here, this nitrogen, gives an electron to this carbonyl carbon. This carbonyl carbon can then let go of this electron to this top oxygen over here."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Well, we know from previous videos when we studied amines that they are pretty good nucleophiles. So they can have a nucleophile attack on this carbonyl carbon right over here. So you can imagine this amine right here, this nitrogen, gives an electron to this carbonyl carbon. This carbonyl carbon can then let go of this electron to this top oxygen over here. It was already hogging it to begin with. And so the next step of our reaction would look like this. What was the butanoyl chloride will now look like 1, 2, 3, 4 carbons."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "This carbonyl carbon can then let go of this electron to this top oxygen over here. It was already hogging it to begin with. And so the next step of our reaction would look like this. What was the butanoyl chloride will now look like 1, 2, 3, 4 carbons. Now you only have a single bond. This oxygen up here now has a negative charge. Since it took an electron from this central carbon right over here, you are still bonded to this chlorine."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "What was the butanoyl chloride will now look like 1, 2, 3, 4 carbons. Now you only have a single bond. This oxygen up here now has a negative charge. Since it took an electron from this central carbon right over here, you are still bonded to this chlorine. And actually, let me do that in a different color so we can keep track of it. You are still bonded to this chlorine atom. And then the nitrogen has been bonded."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Since it took an electron from this central carbon right over here, you are still bonded to this chlorine. And actually, let me do that in a different color so we can keep track of it. You are still bonded to this chlorine atom. And then the nitrogen has been bonded. So this bond I'll draw in green because we're giving this green electron to form the bond. And then the rest of the dimethylamine is still over there. Let me do it in that same color."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And then the nitrogen has been bonded. So this bond I'll draw in green because we're giving this green electron to form the bond. And then the rest of the dimethylamine is still over there. Let me do it in that same color. The rest of the dimethylamine. So you have CH3, CH3, and then you just have your hydrogen. And then this nitrogen gave away an electron."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Let me do it in that same color. The rest of the dimethylamine. So you have CH3, CH3, and then you just have your hydrogen. And then this nitrogen gave away an electron. So it has a positive charge. Now the next step of this is chlorine. It's fairly electronegative."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And then this nitrogen gave away an electron. So it has a positive charge. Now the next step of this is chlorine. It's fairly electronegative. It's a not so bad leaving group. It will want to hog this electron that is on the central carbon. So you can imagine in the next step that this oxygen gives back an electron to reform the carbonyl group at the exact same time that this chlorine takes an electron and forms a chloride anion."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "It's fairly electronegative. It's a not so bad leaving group. It will want to hog this electron that is on the central carbon. So you can imagine in the next step that this oxygen gives back an electron to reform the carbonyl group at the exact same time that this chlorine takes an electron and forms a chloride anion. So the next step in this will look like... So the molecules will now look like... You have there 1, 2, 3, 4 carbons. That's one of the bonds of that oxygen."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So you can imagine in the next step that this oxygen gives back an electron to reform the carbonyl group at the exact same time that this chlorine takes an electron and forms a chloride anion. So the next step in this will look like... So the molecules will now look like... You have there 1, 2, 3, 4 carbons. That's one of the bonds of that oxygen. But now you've now formed another bond. That electron was given back to that central atom. And now you have the bond to the nitrogen."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "That's one of the bonds of that oxygen. But now you've now formed another bond. That electron was given back to that central atom. And now you have the bond to the nitrogen. This bond right here. And then the nitrogen is bound to one methyl group, two methyl groups, and a hydrogen. And it had given away an electron, so it has a positive charge."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And now you have the bond to the nitrogen. This bond right here. And then the nitrogen is bound to one methyl group, two methyl groups, and a hydrogen. And it had given away an electron, so it has a positive charge. And the chloride anion has been bumped off. The chloride anion, it took an electron from that central carbon. So now it has a negative charge."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And it had given away an electron, so it has a positive charge. And the chloride anion has been bumped off. The chloride anion, it took an electron from that central carbon. So now it has a negative charge. Now, we still have that other dimethylamine molecule. For every one of these, we have two of these. We've only used one of them."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So now it has a negative charge. Now, we still have that other dimethylamine molecule. For every one of these, we have two of these. We've only used one of them. So let me bring the other one into the mix. And this one could just be in the mix to make this other molecule neutral now. And this could go in either direction."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "We've only used one of them. So let me bring the other one into the mix. And this one could just be in the mix to make this other molecule neutral now. And this could go in either direction. This part right here could go in either direction. This guy's not so weak. Or it's not a bad base."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And this could go in either direction. This part right here could go in either direction. This guy's not so weak. Or it's not a bad base. So this guy could give an electron to this hydrogen proton. And then the hydrogen proton's electron, or that hydrogen atom's electron, can be taken back by this nitrogen to make it neutral. And I'll make this go in both directions."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "Or it's not a bad base. So this guy could give an electron to this hydrogen proton. And then the hydrogen proton's electron, or that hydrogen atom's electron, can be taken back by this nitrogen to make it neutral. And I'll make this go in both directions. So this could go in both directions. So then this molecule over here will look like one, two, three, four carbonyl group. And let me color code it the same way, just so we know what parts are which parts."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And I'll make this go in both directions. So this could go in both directions. So then this molecule over here will look like one, two, three, four carbonyl group. And let me color code it the same way, just so we know what parts are which parts. We have this green bond to this nitrogen, which has now lost a proton. It was able to take back an electron. So it is now neutral."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And let me color code it the same way, just so we know what parts are which parts. We have this green bond to this nitrogen, which has now lost a proton. It was able to take back an electron. So it is now neutral. It is bonded to one, two methyl groups. This dimethylamine took a hydrogen. So let me draw it over here."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So it is now neutral. It is bonded to one, two methyl groups. This dimethylamine took a hydrogen. So let me draw it over here. So it's nitrogen one, two. And then it already had one hydrogen. And now it's gaining another hydrogen."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So let me draw it over here. So it's nitrogen one, two. And then it already had one hydrogen. And now it's gaining another hydrogen. It's gaining this hydrogen right over here. So now it's bound to that hydrogen. It gave an electron to get that hydrogen nucleus, so it now has a positive charge."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And now it's gaining another hydrogen. It's gaining this hydrogen right over here. So now it's bound to that hydrogen. It gave an electron to get that hydrogen nucleus, so it now has a positive charge. And of course we can't forget about that chloride anion that's floating around. And these might be attracted to each other. One's positive, one's negative."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "It gave an electron to get that hydrogen nucleus, so it now has a positive charge. And of course we can't forget about that chloride anion that's floating around. And these might be attracted to each other. One's positive, one's negative. Form a salt. So what are we left with? And once again, this is a little bit of practice of naming."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "One's positive, one's negative. Form a salt. So what are we left with? And once again, this is a little bit of practice of naming. We started with an acyl chloride, and we ended with an amide, because we now have this nitrogen group attached. So what is this? This has one, two, three, four carbons."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "And once again, this is a little bit of practice of naming. We started with an acyl chloride, and we ended with an amide, because we now have this nitrogen group attached. So what is this? This has one, two, three, four carbons. So it's butan, but it's an amide. So this part tells us that we're dealing with an amide. That part, right."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "This has one, two, three, four carbons. So it's butan, but it's an amide. So this part tells us that we're dealing with an amide. That part, right. Butan, it's butanamide. But to specify what type of amide, we see that we have two methyl groups attached to the nitrogen there. So we would call this, let me put another color here, we have one, two methyl groups attached to the nitrogen."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "That part, right. Butan, it's butanamide. But to specify what type of amide, we see that we have two methyl groups attached to the nitrogen there. So we would call this, let me put another color here, we have one, two methyl groups attached to the nitrogen. So we would call this N,N dimethylbutanamide. This N,N lets us know the methyl groups are attached here, as opposed to on the butyl group. I guess we could view it this way."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So we would call this, let me put another color here, we have one, two methyl groups attached to the nitrogen. So we would call this N,N dimethylbutanamide. This N,N lets us know the methyl groups are attached here, as opposed to on the butyl group. I guess we could view it this way. So we end up with N,N dimethylbutanamide, which is an amide, one of the most stable carboxylic acid derivatives. And then this right here, this salt, what is this? This right here is dimethyl."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "I guess we could view it this way. So we end up with N,N dimethylbutanamide, which is an amide, one of the most stable carboxylic acid derivatives. And then this right here, this salt, what is this? This right here is dimethyl. And this is now no longer dimethylamine. We now are a positively charged cation. So this is dimethylammonium."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "This right here is dimethyl. And this is now no longer dimethylamine. We now are a positively charged cation. So this is dimethylammonium. We get that from the fact that we have four bonds, we have a positive charge. This is dimethylammonium, and then we have this negative anion. Dimethylammonium chloride."}, {"video_title": "Amide formation from acyl chloride Carboxylic acids and derivatives Khan Academy.mp3", "Sentence": "So this is dimethylammonium. We get that from the fact that we have four bonds, we have a positive charge. This is dimethylammonium, and then we have this negative anion. Dimethylammonium chloride. I'll just go to the next line. Dimethylammonium chloride. But I really just wanted to show you a mechanism here of how you could go from a less stable, or one of the least stable, carboxylic acid derivatives to one of the most stable, going from an acyl chloride, which butanoil chloride was, to an amide."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "We'll start with a name and see if we can actually draw the molecular structure of whatever this might be. So let's start. So when you first look at these, it's very daunting, but you always want to start really at the end, so you know what kind of the core of the structure is going to be. So if you look at the end of this, you have an ane, so there's not going to be any double or triple bonds here. It's all single bonds. It's a hexadecane. So let's think about that."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at the end of this, you have an ane, so there's not going to be any double or triple bonds here. It's all single bonds. It's a hexadecane. So let's think about that. So hexadec, that's 6 and 10. Hexadec was a prefix for 16. So this is 16 single bonded carbons."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about that. So hexadec, that's 6 and 10. Hexadec was a prefix for 16. So this is 16 single bonded carbons. And it's cyclohexadecane. So 16 single bonded carbons in a ring. So this part of it right here, so let me just do this."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So this is 16 single bonded carbons. And it's cyclohexadecane. So 16 single bonded carbons in a ring. So this part of it right here, so let me just do this. Let's draw that part first. It's not easy to draw even a 16 carbon ring. So let's start here."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So this part of it right here, so let me just do this. Let's draw that part first. It's not easy to draw even a 16 carbon ring. So let's start here. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and then 16. I think I got it. Let me count them again."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So let's start here. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and then 16. I think I got it. Let me count them again. Then I can connect them up in a cycle. So it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. All right."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Let me count them again. Then I can connect them up in a cycle. So it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. All right. I drew the cyclohexadecane part. Now, if we go back a little, let's see. We have a 2, 9 diisopropyl."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "All right. I drew the cyclohexadecane part. Now, if we go back a little, let's see. We have a 2, 9 diisopropyl. So what does this mean? This means that we have an isopropyl at the 2 and the 9 spot. Now, when you are drawing the structure from the name, you could just arbitrarily, on the sum of the weights on this ring, pick what your 1 through your 16 spots are."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "We have a 2, 9 diisopropyl. So what does this mean? This means that we have an isopropyl at the 2 and the 9 spot. Now, when you are drawing the structure from the name, you could just arbitrarily, on the sum of the weights on this ring, pick what your 1 through your 16 spots are. So I'll just arbitrarily pick them. Because I could have drawn this ring any which way. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Now, when you are drawing the structure from the name, you could just arbitrarily, on the sum of the weights on this ring, pick what your 1 through your 16 spots are. So I'll just arbitrarily pick them. Because I could have drawn this ring any which way. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. All right. So this next piece right here. Let me do this in magenta."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16. All right. So this next piece right here. Let me do this in magenta. 2, 9 diisopropyl. This is telling us that at spots number 2 and at spot number 9, I have isopropyls. The di is just saying I have 2 isopropyls at 2 and 9."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Let me do this in magenta. 2, 9 diisopropyl. This is telling us that at spots number 2 and at spot number 9, I have isopropyls. The di is just saying I have 2 isopropyls at 2 and 9. So you can kind of ignore the di. So I have an isopropyl. And you may or may not remember that an isopropyl looks like this."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "The di is just saying I have 2 isopropyls at 2 and 9. So you can kind of ignore the di. So I have an isopropyl. And you may or may not remember that an isopropyl looks like this. It's 3 carbons. So it's going to be 1, 2, 3. And the connection point to the main ring, in this case, is going to be in the middle carbon."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And you may or may not remember that an isopropyl looks like this. It's 3 carbons. So it's going to be 1, 2, 3. And the connection point to the main ring, in this case, is going to be in the middle carbon. So it kind of forms a Y. All of the isos, the isopropyl, isobutyl, they all look like Y's. So it's going to be linked right over here."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And the connection point to the main ring, in this case, is going to be in the middle carbon. So it kind of forms a Y. All of the isos, the isopropyl, isobutyl, they all look like Y's. So it's going to be linked right over here. It's going to be linked right over there. And that's also going to happen at the ninth carbon. So the ninth carbon, we're going to have another isopropyl."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to be linked right over here. It's going to be linked right over there. And that's also going to happen at the ninth carbon. So the ninth carbon, we're going to have another isopropyl. All right. We've taken care of the 2, 9 isopropyl. Then we have the 6-ethyl, which is just a 2 carbon."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So the ninth carbon, we're going to have another isopropyl. All right. We've taken care of the 2, 9 isopropyl. Then we have the 6-ethyl, which is just a 2 carbon. Remember, meth is 1, eth is 2, prop is 3. So this is, let me write this down. So this is going to be, so prop is equal to 3."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Then we have the 6-ethyl, which is just a 2 carbon. Remember, meth is 1, eth is 2, prop is 3. So this is, let me write this down. So this is going to be, so prop is equal to 3. Isoprop is equal to that type of shape right over there. In this case, eth is equal to 2. So it's a 6-ethyl group."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be, so prop is equal to 3. Isoprop is equal to that type of shape right over there. In this case, eth is equal to 2. So it's a 6-ethyl group. So at 6, we have an ethyl group. So 1, 2 carbons. And it's connected at the 6 carbon on the main ring."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So it's a 6-ethyl group. So at 6, we have an ethyl group. So 1, 2 carbons. And it's connected at the 6 carbon on the main ring. And then finally, we have a cyclopentyl. So if we look at, let me find a color I haven't used yet. Let's see, I've already used one this year."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And it's connected at the 6 carbon on the main ring. And then finally, we have a cyclopentyl. So if we look at, let me find a color I haven't used yet. Let's see, I've already used one this year. Cyclopentyl. So pent is 5, but it's 5 in a cycle. So this is a 5 carbon ring that's branching off of the main ring."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Let's see, I've already used one this year. Cyclopentyl. So pent is 5, but it's 5 in a cycle. So this is a 5 carbon ring that's branching off of the main ring. And it's at the first spot. So let me draw a 5 carbon ring. So pent is equal to 5."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So this is a 5 carbon ring that's branching off of the main ring. And it's at the first spot. So let me draw a 5 carbon ring. So pent is equal to 5. So it would look like this. 1, 2, 3, 4, 5. Looks just like a pentagon."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So pent is equal to 5. So it would look like this. 1, 2, 3, 4, 5. Looks just like a pentagon. That's a cyclopentyl group, and it's attached to the 1 carbon on my cyclohexadecane. So it is attached just like that. And we're done."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Looks just like a pentagon. That's a cyclopentyl group, and it's attached to the 1 carbon on my cyclohexadecane. So it is attached just like that. And we're done. We've drawn one cyclopentyl 6-ethyl-2-9-diisopropyl cyclohexadecane. Let's do another one. I think we're getting the hang of it."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And we're done. We've drawn one cyclopentyl 6-ethyl-2-9-diisopropyl cyclohexadecane. Let's do another one. I think we're getting the hang of it. So here, maybe we could do this one a little bit faster. Let's see, we have a tetramethyl dodecane. So the main root here is the dodecane."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "I think we're getting the hang of it. So here, maybe we could do this one a little bit faster. Let's see, we have a tetramethyl dodecane. So the main root here is the dodecane. Do for 2, dec for 10. This is a 12 carbon chain. It's not in a cycle."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So the main root here is the dodecane. Do for 2, dec for 10. This is a 12 carbon chain. It's not in a cycle. So let me just draw it out. So we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. And so we can just number them arbitrarily, just because I could have drawn this any which way."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "It's not in a cycle. So let me just draw it out. So we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. And so we can just number them arbitrarily, just because I could have drawn this any which way. So it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. That's the dodecane, all single bonds. And then we have a 3, 6, 9, 9 tetramethyl."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And so we can just number them arbitrarily, just because I could have drawn this any which way. So it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. That's the dodecane, all single bonds. And then we have a 3, 6, 9, 9 tetramethyl. And all this is telling us, remember, meth is 1 carbon. So all this is telling us is that the 3, the 6, the 9. So at the 3, the 6, and twice at the 9 spot, we have methyl groups, and we have 4 methyl groups."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a 3, 6, 9, 9 tetramethyl. And all this is telling us, remember, meth is 1 carbon. So all this is telling us is that the 3, the 6, the 9. So at the 3, the 6, and twice at the 9 spot, we have methyl groups, and we have 4 methyl groups. That's all the tetramethyl is saying. So it's a little bit redundant. We know we have 4 from here, 6, 3, 6, 9, 9."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So at the 3, the 6, and twice at the 9 spot, we have methyl groups, and we have 4 methyl groups. That's all the tetramethyl is saying. So it's a little bit redundant. We know we have 4 from here, 6, 3, 6, 9, 9. We have methyl groups at each of those places. So we have 1 methyl group at 3, and then that is bonded with the third carbon on the dodecane chain. We have 1 at 6, bonded to the 6 carbon on the dodecane chain."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "We know we have 4 from here, 6, 3, 6, 9, 9. We have methyl groups at each of those places. So we have 1 methyl group at 3, and then that is bonded with the third carbon on the dodecane chain. We have 1 at 6, bonded to the 6 carbon on the dodecane chain. And then we have 2 at 9. So we have 1, that's 1 at 9, and then we have another 1 at 9, bonded to the 9 carbon on the dodecane chain. And we're done."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "We have 1 at 6, bonded to the 6 carbon on the dodecane chain. And then we have 2 at 9. So we have 1, that's 1 at 9, and then we have another 1 at 9, bonded to the 9 carbon on the dodecane chain. And we're done. That's it. That's 3, 6, 9, 9 tetramethyldodecane. Let's do another one."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And we're done. That's it. That's 3, 6, 9, 9 tetramethyldodecane. Let's do another one. 1, 3-bis-1,1-dimethyl ethyl cyclopentane. So once again, just breathe slowly. It's very daunting right when you look at it, but just start with kind of the core."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one. 1, 3-bis-1,1-dimethyl ethyl cyclopentane. So once again, just breathe slowly. It's very daunting right when you look at it, but just start with kind of the core. Cyclopentane, that's just a simple 5-carbon ring. It's like a pentagon. 1, 2, 3, let me draw it a little bit better, 3, 4, 5."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "It's very daunting right when you look at it, but just start with kind of the core. Cyclopentane, that's just a simple 5-carbon ring. It's like a pentagon. 1, 2, 3, let me draw it a little bit better, 3, 4, 5. There you go. That is a 5-carbon ring, and we can number it however we want. So 1, 2, 3, 4, and 5."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, let me draw it a little bit better, 3, 4, 5. There you go. That is a 5-carbon ring, and we can number it however we want. So 1, 2, 3, 4, and 5. And this is telling us that the 1 and the 3 position, and the bis is kind of redundant. This is saying we have 2 of these things. Obviously we have 2."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, and 5. And this is telling us that the 1 and the 3 position, and the bis is kind of redundant. This is saying we have 2 of these things. Obviously we have 2. We have 1 at the 1, 1 at the 3. So you can kind of ignore the bis. That's just the convention, and we've seen that multiple times."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "Obviously we have 2. We have 1 at the 1, 1 at the 3. So you can kind of ignore the bis. That's just the convention, and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "That's just the convention, and we've seen that multiple times. But at each of those positions, we have a 1,1-dimethylethyl. So what's a dimethylethyl look like? So let's think about it a little bit. Let's think about it. Let me do it in orange. So this is going, they obviously named it using systematic naming."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about it a little bit. Let's think about it. Let me do it in orange. So this is going, they obviously named it using systematic naming. And what we have here, we have an ethyl is kind of the core of this side chain. So for an ethyl, ethyl is equal to 2 carbons. So this is 2 carbons right there."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So this is going, they obviously named it using systematic naming. And what we have here, we have an ethyl is kind of the core of this side chain. So for an ethyl, ethyl is equal to 2 carbons. So this is 2 carbons right there. So let me draw a 2-carbon 1, 2. That is 2 carbons right over there. I'm just drawing it at the 3 spot."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So this is 2 carbons right there. So let me draw a 2-carbon 1, 2. That is 2 carbons right over there. I'm just drawing it at the 3 spot. I'll draw it also at the 1 spot actually. So that is 2 carbons right there. That's the ethyl part."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "I'm just drawing it at the 3 spot. I'll draw it also at the 1 spot actually. So that is 2 carbons right there. That's the ethyl part. And then on 1, 1, so if we number them, we number where it's connected. So it's 1, 2. This is saying 1, 1-dimethyl."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "That's the ethyl part. And then on 1, 1, so if we number them, we number where it's connected. So it's 1, 2. This is saying 1, 1-dimethyl. So on this ethyl chain, you have 2 methyls. Remember, methyl is equal to 1. So you have 1 carbon."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "This is saying 1, 1-dimethyl. So on this ethyl chain, you have 2 methyls. Remember, methyl is equal to 1. So you have 1 carbon. That's what methyl is. But you have 2 of them. You have dimethyl."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "So you have 1 carbon. That's what methyl is. But you have 2 of them. You have dimethyl. You have it twice at the 1 spot. So you have 1 methyl here. And then you have another methyl there."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "You have dimethyl. You have it twice at the 1 spot. So you have 1 methyl here. And then you have another methyl there. Same thing over here. You have 1 methyl on the 1 spot. And then you have another 1 methyl on the 1 spot."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And then you have another methyl there. Same thing over here. You have 1 methyl on the 1 spot. And then you have another 1 methyl on the 1 spot. And then you are connected at positions 1 and positions 3. So you're connected there. And you are connected right over there."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And then you have another 1 methyl on the 1 spot. And then you are connected at positions 1 and positions 3. So you're connected there. And you are connected right over there. And you're done. That's it. That is our structure."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "And you are connected right over there. And you're done. That's it. That is our structure. Now, if you did this with common naming, instead of this group being a 1, 1-dimethyl ethyl, you might see that, OK, we're connected to a group that has 1, 2, 3, 4 carbons in it. The carbon that we're connected to branches off to 3 other carbons. It is a tert-butyl."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "That is our structure. Now, if you did this with common naming, instead of this group being a 1, 1-dimethyl ethyl, you might see that, OK, we're connected to a group that has 1, 2, 3, 4 carbons in it. The carbon that we're connected to branches off to 3 other carbons. It is a tert-butyl. So you could also call this a 1, 3. Let me just write it down. So another name for this would be 1, 3 tert."}, {"video_title": "Organic chemistry naming examples 4 Organic chemistry Khan Academy.mp3", "Sentence": "It is a tert-butyl. So you could also call this a 1, 3. Let me just write it down. So another name for this would be 1, 3 tert. Or sometimes people just write a T there. T-butyl cyclopentane. So that would be the common name."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So in the first step, we add a strong nucleophile to our epoxide. And in the second step, we add a proton source. And the nucleophile is going to end up opening the ring and adding an anti-fashion to the OH that is created here. And so in terms of regiochemistry, that nucleophile adds to the least substituted carbon. So it's going to add to this carbon right here. If we go back and look at our epoxide on the left, that's this carbon. And this carbon is attached to two other carbons, so it would be classified as a secondary carbon."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And so in terms of regiochemistry, that nucleophile adds to the least substituted carbon. So it's going to add to this carbon right here. If we go back and look at our epoxide on the left, that's this carbon. And this carbon is attached to two other carbons, so it would be classified as a secondary carbon. Whereas the carbon on the right is attached to three other carbons, and so that's a tertiary carbon. So the mechanism for this reaction is an SN2 type mechanism, which means that our strong nucleophile is going to attack the least sterically hindered carbon, which is the one on the left, since a hydrogen takes up less space than some bulky R group over here on the right. And so that's the regiochemistry here."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon is attached to two other carbons, so it would be classified as a secondary carbon. Whereas the carbon on the right is attached to three other carbons, and so that's a tertiary carbon. So the mechanism for this reaction is an SN2 type mechanism, which means that our strong nucleophile is going to attack the least sterically hindered carbon, which is the one on the left, since a hydrogen takes up less space than some bulky R group over here on the right. And so that's the regiochemistry here. The stereochemistry is an anti-opening of the ring for an anti-addition of the nucleophile. And this will affect possible chirality centers, as we will see in later examples. Let's look at some examples of strong nucleophiles."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And so that's the regiochemistry here. The stereochemistry is an anti-opening of the ring for an anti-addition of the nucleophile. And this will affect possible chirality centers, as we will see in later examples. Let's look at some examples of strong nucleophiles. So for your strong nucleophile, you could use an alkoxide anion, so something like this over here. You could use a Grignard reagent, which we know are sources of carbanions, so we could have a negatively charged carbon like this. And you didn't have to write your Grignard reagent like this."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at some examples of strong nucleophiles. So for your strong nucleophile, you could use an alkoxide anion, so something like this over here. You could use a Grignard reagent, which we know are sources of carbanions, so we could have a negatively charged carbon like this. And you didn't have to write your Grignard reagent like this. You could have written it like just showing it R-MgBr. Hydride itself is not a strong nucleophile, but if hydride's coming from something like lithium aluminum hydride, it functions like a strong nucleophile. So we could go ahead and show that hydride is really coming from lithium aluminum hydride."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And you didn't have to write your Grignard reagent like this. You could have written it like just showing it R-MgBr. Hydride itself is not a strong nucleophile, but if hydride's coming from something like lithium aluminum hydride, it functions like a strong nucleophile. So we could go ahead and show that hydride is really coming from lithium aluminum hydride. So if we go ahead and draw this in here, aluminum is going to have a negative 1 formal charge, and then lithium would have a plus 1 formal charge like that. So lithium aluminum hydride is a source of a strong nucleophile. It's just that the hydride anion on its own, we've already talked about, is not the best nucleophile."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So we could go ahead and show that hydride is really coming from lithium aluminum hydride. So if we go ahead and draw this in here, aluminum is going to have a negative 1 formal charge, and then lithium would have a plus 1 formal charge like that. So lithium aluminum hydride is a source of a strong nucleophile. It's just that the hydride anion on its own, we've already talked about, is not the best nucleophile. And then, of course, you could have something like hydrogen sulfide anion or the hydrosulfide anion right here, which also functions as a strong nucleophile. Let's take a look at the mechanism to form our alcohols here. So we're going to, once again, start with our epoxide."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "It's just that the hydride anion on its own, we've already talked about, is not the best nucleophile. And then, of course, you could have something like hydrogen sulfide anion or the hydrosulfide anion right here, which also functions as a strong nucleophile. Let's take a look at the mechanism to form our alcohols here. So we're going to, once again, start with our epoxide. And I'll use the same epoxide that we had before, meaning there's going to be an R prime here and R double prime going away from us, hydrogen coming out at us, and an R group going away from us like that. So when I think about electronegativity differences, I know that the oxygen is more electronegative than this carbon, for example. So the oxygen is going to pull the electrons in the bonds between the oxygen and the carbon closer to the oxygen, giving it a partial negative charge."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to, once again, start with our epoxide. And I'll use the same epoxide that we had before, meaning there's going to be an R prime here and R double prime going away from us, hydrogen coming out at us, and an R group going away from us like that. So when I think about electronegativity differences, I know that the oxygen is more electronegative than this carbon, for example. So the oxygen is going to pull the electrons in the bonds between the oxygen and the carbon closer to the oxygen, giving it a partial negative charge. So pulling some electron density away from this carbon will give this carbon down here a partially positive charge. And so that carbon is going to function as our electrophile. So when the nucleophile comes along, so the nucleophile will have a lone pair of electrons and a negative 1 formal charge."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So the oxygen is going to pull the electrons in the bonds between the oxygen and the carbon closer to the oxygen, giving it a partial negative charge. So pulling some electron density away from this carbon will give this carbon down here a partially positive charge. And so that carbon is going to function as our electrophile. So when the nucleophile comes along, so the nucleophile will have a lone pair of electrons and a negative 1 formal charge. It has two options. It could attack the carbon on the left, or it could attack the carbon on the right. And since it's a strong nucleophile, which proceeds via an SN2 type mechanism, it's going to attack the carbon on the left."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So when the nucleophile comes along, so the nucleophile will have a lone pair of electrons and a negative 1 formal charge. It has two options. It could attack the carbon on the left, or it could attack the carbon on the right. And since it's a strong nucleophile, which proceeds via an SN2 type mechanism, it's going to attack the carbon on the left. So once again, the carbon on the left is the least sterically hindered. That hydrogen is not very big compared to some alkyl group over here on the right. So the nucleophile is going to attack the electrophile like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And since it's a strong nucleophile, which proceeds via an SN2 type mechanism, it's going to attack the carbon on the left. So once again, the carbon on the left is the least sterically hindered. That hydrogen is not very big compared to some alkyl group over here on the right. So the nucleophile is going to attack the electrophile like that. And when that happens, it's going to kick these two electrons in here off onto your oxygen. So we can go ahead and draw the product of that nucleophilic attack. So now we broke our epoxide ring there."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So the nucleophile is going to attack the electrophile like that. And when that happens, it's going to kick these two electrons in here off onto your oxygen. So we can go ahead and draw the product of that nucleophilic attack. So now we broke our epoxide ring there. And we have an extra lone pair of electrons on that oxygen, giving that oxygen a negative 1 formal charge. These R groups on the carbon on the right haven't changed. So they are still there like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So now we broke our epoxide ring there. And we have an extra lone pair of electrons on that oxygen, giving that oxygen a negative 1 formal charge. These R groups on the carbon on the right haven't changed. So they are still there like that. We've now added our nucleophile like that. And that's going to push the hydrogen up relative to the plane. And then we have an R group over here like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So they are still there like that. We've now added our nucleophile like that. And that's going to push the hydrogen up relative to the plane. And then we have an R group over here like that. So the nucleophile has to attack from below, because the oxygen is on top. And the nucleophile is going to attack the least sterically hindered carbon. This ends up putting your nucleophile and your oxygen anti to each other."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And then we have an R group over here like that. So the nucleophile has to attack from below, because the oxygen is on top. And the nucleophile is going to attack the least sterically hindered carbon. This ends up putting your nucleophile and your oxygen anti to each other. And in the second step, we can add a proton source, so something like water. And we're going to get an acid-base reaction, where our alkoxide is going to take a proton from water, which kicks these electrons in here off onto the oxygen. And that will form our final product with an alcohol."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "This ends up putting your nucleophile and your oxygen anti to each other. And in the second step, we can add a proton source, so something like water. And we're going to get an acid-base reaction, where our alkoxide is going to take a proton from water, which kicks these electrons in here off onto the oxygen. And that will form our final product with an alcohol. So I'm going to have an alcohol right here. And once again, my R groups are still there. So I have R prime and R double prime."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And that will form our final product with an alcohol. So I'm going to have an alcohol right here. And once again, my R groups are still there. So I have R prime and R double prime. Our nucleophile has added. And our hydrogen is pushed up. And our R group is still back here."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So I have R prime and R double prime. Our nucleophile has added. And our hydrogen is pushed up. And our R group is still back here. So that is the mechanism for the SN2 type reaction. Let's do a problem which involves some stereochemistry. So we can see how the mechanism will determine the stereochemistry of the products here."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And our R group is still back here. So that is the mechanism for the SN2 type reaction. Let's do a problem which involves some stereochemistry. So we can see how the mechanism will determine the stereochemistry of the products here. So I'm going to start out. I'm going to put my epoxide coming out at me in space like that. And then at this top carbon here, I'm going to have a methyl group going away from me in space."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So we can see how the mechanism will determine the stereochemistry of the products here. So I'm going to start out. I'm going to put my epoxide coming out at me in space like that. And then at this top carbon here, I'm going to have a methyl group going away from me in space. So that's my reactant. And I'm going to use sodium ethoxide. So we'll go ahead and draw sodium ethoxide here."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And then at this top carbon here, I'm going to have a methyl group going away from me in space. So that's my reactant. And I'm going to use sodium ethoxide. So we'll go ahead and draw sodium ethoxide here. And we're going to use ethanol as the solvent. So we have ethanol here as our solvent like that. So if we look at this epoxide, it's a little bit hard to see the stereochemistry."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So we'll go ahead and draw sodium ethoxide here. And we're going to use ethanol as the solvent. So we have ethanol here as our solvent like that. So if we look at this epoxide, it's a little bit hard to see the stereochemistry. So let's go ahead and redraw this molecule. And we'll look at it from a different vantage point here. So if I look at that epoxide from a different vantage point, I'm looking at it from above and down a little bit."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So if we look at this epoxide, it's a little bit hard to see the stereochemistry. So let's go ahead and redraw this molecule. And we'll look at it from a different vantage point here. So if I look at that epoxide from a different vantage point, I'm looking at it from above and down a little bit. And that would put my epoxide going up like that. And then it was this carbon right here that's going to have the methyl group going down relative to the plane of the ring. And then this carbon over here on the left has a hydrogen."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at that epoxide from a different vantage point, I'm looking at it from above and down a little bit. And that would put my epoxide going up like that. And then it was this carbon right here that's going to have the methyl group going down relative to the plane of the ring. And then this carbon over here on the left has a hydrogen. So my first step would be to identify my nucleophile, which of course is the ethoxide anion. So I can go ahead and draw my ethoxide anion in here. And next I need to think about where that nucleophile will attack."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And then this carbon over here on the left has a hydrogen. So my first step would be to identify my nucleophile, which of course is the ethoxide anion. So I can go ahead and draw my ethoxide anion in here. And next I need to think about where that nucleophile will attack. So I know that this top oxygen here, being more electronegative than the carbons, is going to have a partial negative charge. And so I have two options. I could be the carbon on the left."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And next I need to think about where that nucleophile will attack. So I know that this top oxygen here, being more electronegative than the carbons, is going to have a partial negative charge. And so I have two options. I could be the carbon on the left. I could be the carbon on the right. And if I think about the fact that this is a strong nucleophile SN2 type mechanism, it's going to attack the less sterically hindered carbon, which of course is the carbon on the left, which is going to have a partial positive charge. And so we're going to get nucleophilic attack."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "I could be the carbon on the left. I could be the carbon on the right. And if I think about the fact that this is a strong nucleophile SN2 type mechanism, it's going to attack the less sterically hindered carbon, which of course is the carbon on the left, which is going to have a partial positive charge. And so we're going to get nucleophilic attack. My nucleophile is going to attack this carbon right here, which would kick these electrons in here off onto the oxygen. So I can go ahead and draw the product of that. So I have my ring like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to get nucleophilic attack. My nucleophile is going to attack this carbon right here, which would kick these electrons in here off onto the oxygen. So I can go ahead and draw the product of that. So I have my ring like that. And I opened up the epoxide ring. So now this bond is to an oxygen. It has three lone pairs of electrons like that, which give it a negative 1 formal charge."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So I have my ring like that. And I opened up the epoxide ring. So now this bond is to an oxygen. It has three lone pairs of electrons like that, which give it a negative 1 formal charge. There's still a methyl group going down at this carbon. And I've added my nucleophile to this carbon. So I can go ahead and draw my oxygen and then an ethyl like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "It has three lone pairs of electrons like that, which give it a negative 1 formal charge. There's still a methyl group going down at this carbon. And I've added my nucleophile to this carbon. So I can go ahead and draw my oxygen and then an ethyl like that. And then I pushed up the hydrogen. So when this reaction occurred, the hydrogen that was down relative to the plane of the ring got pushed up. So now there's a hydrogen up at that carbon."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and draw my oxygen and then an ethyl like that. And then I pushed up the hydrogen. So when this reaction occurred, the hydrogen that was down relative to the plane of the ring got pushed up. So now there's a hydrogen up at that carbon. And in my final step, I'm going to protonate the epoxide. So lone pair of electrons will pick up a proton from ethanol, which kicks these electrons off onto here. And I form my alcohol as my product."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So now there's a hydrogen up at that carbon. And in my final step, I'm going to protonate the epoxide. So lone pair of electrons will pick up a proton from ethanol, which kicks these electrons off onto here. And I form my alcohol as my product. So when I'm drawing my product over here, now I have an OH. And I still have a methyl group going down. And over here, I have an oxygen and then ethyl like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And I form my alcohol as my product. So when I'm drawing my product over here, now I have an OH. And I still have a methyl group going down. And over here, I have an oxygen and then ethyl like that. So if I wanted to draw my product as a two-dimensional dot structure, I could look this way. I could look down on it like that and just go ahead and draw what I see when I'm looking down. So I can draw my ring."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And over here, I have an oxygen and then ethyl like that. So if I wanted to draw my product as a two-dimensional dot structure, I could look this way. I could look down on it like that and just go ahead and draw what I see when I'm looking down. So I can draw my ring. So here is my ring. And if I look at this carbon right here, there's an OH coming out at me. So I can represent the OH coming out at me with a wedge right here."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So I can draw my ring. So here is my ring. And if I look at this carbon right here, there's an OH coming out at me. So I can represent the OH coming out at me with a wedge right here. And there's a methyl group going away from me at that carbon. So this methyl group is going away from me. So I can go ahead and put in a dash for that methyl group."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So I can represent the OH coming out at me with a wedge right here. And there's a methyl group going away from me at that carbon. So this methyl group is going away from me. So I can go ahead and put in a dash for that methyl group. And when I look at the other carbon here, I have an O and then ethyl going away from me. So I have this guy would be a dash going away from me like that. So that, of course, is this portion of the molecule."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and put in a dash for that methyl group. And when I look at the other carbon here, I have an O and then ethyl going away from me. So I have this guy would be a dash going away from me like that. So that, of course, is this portion of the molecule. So that's my product. When I look at what happened to the stereochemistry, let's first start with this top carbon right here. This top carbon had a methyl group going away and an oxygen coming out at you at that carbon."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So that, of course, is this portion of the molecule. So that's my product. When I look at what happened to the stereochemistry, let's first start with this top carbon right here. This top carbon had a methyl group going away and an oxygen coming out at you at that carbon. And if we look over here, that's the same thing that we have. We have a methyl group going away from us and then an oxygen coming out at us like that. So it's the same absolute configuration at that carbon."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "This top carbon had a methyl group going away and an oxygen coming out at you at that carbon. And if we look over here, that's the same thing that we have. We have a methyl group going away from us and then an oxygen coming out at us like that. So it's the same absolute configuration at that carbon. If we look at the other carbon, if we look at this carbon down here, on the left we had an oxygen coming out at us in space. And on the right at this carbon, now the oxygen is going away from us in space. So we actually see inversion of configuration."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So it's the same absolute configuration at that carbon. If we look at the other carbon, if we look at this carbon down here, on the left we had an oxygen coming out at us in space. And on the right at this carbon, now the oxygen is going away from us in space. So we actually see inversion of configuration. So right here we have inversion of configuration in terms of the stereochemistry. We went from a wedge to a dash. And that's, of course, due to the mechanism."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So we actually see inversion of configuration. So right here we have inversion of configuration in terms of the stereochemistry. We went from a wedge to a dash. And that's, of course, due to the mechanism. We had a nucleophilic attack at a chirality center. And so we observe inversion of configuration at that chirality center. Let's do one more example of a strong nucleophile attacking an epoxide."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And that's, of course, due to the mechanism. We had a nucleophilic attack at a chirality center. And so we observe inversion of configuration at that chirality center. Let's do one more example of a strong nucleophile attacking an epoxide. And so this time let's use a Grignard reagent. So I'm going to start with phenyl magnesium bromide. So we've already made phenyl magnesium bromide as our Grignard reagent."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more example of a strong nucleophile attacking an epoxide. And so this time let's use a Grignard reagent. So I'm going to start with phenyl magnesium bromide. So we've already made phenyl magnesium bromide as our Grignard reagent. And in the first step, we're going to add ethylene oxide, the simplest epoxide. And in our second step, we're going to add hydronium, so H3O plus. And that'll be our source of protons."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So we've already made phenyl magnesium bromide as our Grignard reagent. And in the first step, we're going to add ethylene oxide, the simplest epoxide. And in our second step, we're going to add hydronium, so H3O plus. And that'll be our source of protons. But that has to be in a second step. You can't add that at the same time because there would be an acid base reaction from your Grignard reagent with your hydronium ions. So if I'm looking at phenyl magnesium bromide, I could draw it like that."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And that'll be our source of protons. But that has to be in a second step. You can't add that at the same time because there would be an acid base reaction from your Grignard reagent with your hydronium ions. So if I'm looking at phenyl magnesium bromide, I could draw it like that. Or I could redraw it to identify my carb anion. So if I think about what's going to happen, the electrons in the bonds between this carbon and magnesium in here, carbon is more electronegative than magnesium. So you could think about those two electrons as being on that carbon like this, which would give that carbon a negative 1 formal charge."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm looking at phenyl magnesium bromide, I could draw it like that. Or I could redraw it to identify my carb anion. So if I think about what's going to happen, the electrons in the bonds between this carbon and magnesium in here, carbon is more electronegative than magnesium. So you could think about those two electrons as being on that carbon like this, which would give that carbon a negative 1 formal charge. And so now we have a carb anion. And then we'd have MgBr plus like that. And that's going to be the nucleophile."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about those two electrons as being on that carbon like this, which would give that carbon a negative 1 formal charge. And so now we have a carb anion. And then we'd have MgBr plus like that. And that's going to be the nucleophile. When we have our ethylene oxide that comes along, so same idea, oxygen's more electronegative, so partial negative on the oxygen and a partial positive on the carbons. They're symmetrical. So when I think about nucleophilic attack, these electrons in blue are going to be my nucleophile."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And that's going to be the nucleophile. When we have our ethylene oxide that comes along, so same idea, oxygen's more electronegative, so partial negative on the oxygen and a partial positive on the carbons. They're symmetrical. So when I think about nucleophilic attack, these electrons in blue are going to be my nucleophile. They're going to attack my electrophile, kicking these electrons off. So when I draw the product of the nucleophilic attack, I have my ring here. And now I'm going to end up with an alkoxide that looks like this."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So when I think about nucleophilic attack, these electrons in blue are going to be my nucleophile. They're going to attack my electrophile, kicking these electrons off. So when I draw the product of the nucleophilic attack, I have my ring here. And now I'm going to end up with an alkoxide that looks like this. So let me go ahead and highlight the electrons in blue. The electrons in blue formed a new bond, a carbon-carbon bond right here. So the electrons in blue form a new carbon-carbon bond, which is why organometallics are so useful."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And now I'm going to end up with an alkoxide that looks like this. So let me go ahead and highlight the electrons in blue. The electrons in blue formed a new bond, a carbon-carbon bond right here. So the electrons in blue form a new carbon-carbon bond, which is why organometallics are so useful. You can form these carbon-carbon bonds here. And then I can see that the two carbons over here on the right, this carbon and this carbon, those are the ones that came from my ethylene oxide like that. So in my last step, it's an acid-base reaction."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in blue form a new carbon-carbon bond, which is why organometallics are so useful. You can form these carbon-carbon bonds here. And then I can see that the two carbons over here on the right, this carbon and this carbon, those are the ones that came from my ethylene oxide like that. So in my last step, it's an acid-base reaction. And this alkoxide is going to pick up a proton from H3O plus in the second step. So a lone pair of electrons picks up a proton, kicks these electrons off. And I can go ahead and draw the product."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "So in my last step, it's an acid-base reaction. And this alkoxide is going to pick up a proton from H3O plus in the second step. So a lone pair of electrons picks up a proton, kicks these electrons off. And I can go ahead and draw the product. So once again, here's my ring. And I now protonated my alkoxide to form an alcohol. And the name of this alcohol, so if I think about this as being carbon 1 and this as being carbon 2, I have a phenyl group coming off of ethanol."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And I can go ahead and draw the product. So once again, here's my ring. And I now protonated my alkoxide to form an alcohol. And the name of this alcohol, so if I think about this as being carbon 1 and this as being carbon 2, I have a phenyl group coming off of ethanol. So 2-phenylethanol would be the name of the product that's formed. And 2-phenylethanol is famous for being a component of rose oil. So this molecule smells a lot like roses."}, {"video_title": "Ring-opening reactions of epoxides Strong nucleophiles Organic chemistry Khan Academy.mp3", "Sentence": "And the name of this alcohol, so if I think about this as being carbon 1 and this as being carbon 2, I have a phenyl group coming off of ethanol. So 2-phenylethanol would be the name of the product that's formed. And 2-phenylethanol is famous for being a component of rose oil. So this molecule smells a lot like roses. And it's used in the perfume industry. So this is one way to make 2-phenylethanol. There are actually several ways to do it."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This video is really just a continuation of the previous video, so make sure you've seen that one first. In that video, we saw that when you have an atom with a lone pair of electrons that's right next to the carbon on your benzene ring, that lone pair of electrons can participate in resonance and it can be conjugated into the pi system of your ring. That increases the electron density of your ring, which activates the ring towards electrophilic aromatic substitution, since the ring can be a better nucleophile and the positively charged sigma complex is better stabilized by the increased electron density. And so that's where the activated portion comes in for this molecule. So that lone pair of electrons can activate the ring towards electrophilic aromatic substitution. However, it's only a moderate activator and not a strong one, like we saw in the previous video. And the explanation for that observation comes from the fact that this lone pair of electrons can participate in resonance outside of the ring."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's where the activated portion comes in for this molecule. So that lone pair of electrons can activate the ring towards electrophilic aromatic substitution. However, it's only a moderate activator and not a strong one, like we saw in the previous video. And the explanation for that observation comes from the fact that this lone pair of electrons can participate in resonance outside of the ring. So if I go ahead and draw a resonance structure here, I have my benzene ring. I now have my nitrogen double bonded to this carbon. That carbon's bonded to an R group."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the explanation for that observation comes from the fact that this lone pair of electrons can participate in resonance outside of the ring. So if I go ahead and draw a resonance structure here, I have my benzene ring. I now have my nitrogen double bonded to this carbon. That carbon's bonded to an R group. The nitrogen's bonded to a hydrogen. And now my oxygen has three lone pairs of electrons around it, giving it a negative 1 formal charge. The nitrogen now has a plus 1 formal charge."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That carbon's bonded to an R group. The nitrogen's bonded to a hydrogen. And now my oxygen has three lone pairs of electrons around it, giving it a negative 1 formal charge. The nitrogen now has a plus 1 formal charge. So the electrons in magenta moved in here to form a pi bond. And because those electrons in magenta are tied up in resonance outside of the ring, they can't donate as much electron density to the ring. And so that's why it's only a moderate activator instead of a strong one."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The nitrogen now has a plus 1 formal charge. So the electrons in magenta moved in here to form a pi bond. And because those electrons in magenta are tied up in resonance outside of the ring, they can't donate as much electron density to the ring. And so that's why it's only a moderate activator instead of a strong one. Let's look at an example of a weak activator, so something like toluene here, so a methyl group on a benzene ring. All of the carbons on our benzene ring are sp2 hybridized. So all of those carbons have a free p orbital."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's why it's only a moderate activator instead of a strong one. Let's look at an example of a weak activator, so something like toluene here, so a methyl group on a benzene ring. All of the carbons on our benzene ring are sp2 hybridized. So all of those carbons have a free p orbital. So I'm going to go ahead and sketch in a free p orbital on all of my carbons. So I have my 6 pi electron system here in my benzene ring. And so those electrons are delocalized in my pi system like that."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So all of those carbons have a free p orbital. So I'm going to go ahead and sketch in a free p orbital on all of my carbons. So I have my 6 pi electron system here in my benzene ring. And so those electrons are delocalized in my pi system like that. I also have a methyl group. So let's go ahead and sketch in the methyl group here. So I have a carbon bonded to three hydrogens."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so those electrons are delocalized in my pi system like that. I also have a methyl group. So let's go ahead and sketch in the methyl group here. So I have a carbon bonded to three hydrogens. And it turns out one of those carbon hydrogen sigma bonds can actually interact with the pi system of our ring. So we could think about this sigma bond right here interacting with the pi system and therefore increasing or donating some electron density to that ring. And since that sigma bond can interact weakly and donate some electron density to that ring, we can think about the methyl group as being an activator, activating the ring towards electrophilic aromatic substitution."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I have a carbon bonded to three hydrogens. And it turns out one of those carbon hydrogen sigma bonds can actually interact with the pi system of our ring. So we could think about this sigma bond right here interacting with the pi system and therefore increasing or donating some electron density to that ring. And since that sigma bond can interact weakly and donate some electron density to that ring, we can think about the methyl group as being an activator, activating the ring towards electrophilic aromatic substitution. So you could call this interaction sigma conjugation. You could call it hyperconjugation, whatever you want to call it. But there is a little bit of electron donation to the pi system, therefore making it an activator."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And since that sigma bond can interact weakly and donate some electron density to that ring, we can think about the methyl group as being an activator, activating the ring towards electrophilic aromatic substitution. So you could call this interaction sigma conjugation. You could call it hyperconjugation, whatever you want to call it. But there is a little bit of electron donation to the pi system, therefore making it an activator. However, this sigma conjugation is nowhere near as strong as the full conjugation that we saw in the previous video with oxygen and nitrogen. And so since sigma conjugation is not as strong of an effect, that's why this is only a weak activator. Let's look at one more example of an ortho para director."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But there is a little bit of electron donation to the pi system, therefore making it an activator. However, this sigma conjugation is nowhere near as strong as the full conjugation that we saw in the previous video with oxygen and nitrogen. And so since sigma conjugation is not as strong of an effect, that's why this is only a weak activator. Let's look at one more example of an ortho para director. And this one is the exception. This is an ortho para director that is a deactivator here. So a halogen on a benzene ring will still direct substituents ortho and para to it because of the resonance structures that you can draw."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at one more example of an ortho para director. And this one is the exception. This is an ortho para director that is a deactivator here. So a halogen on a benzene ring will still direct substituents ortho and para to it because of the resonance structures that you can draw. However, it turns out to be a deactivator most of the time. The reactions are slower than that of benzene on its own. Let's first examine the inductive effect that the halogen has on our benzene ring."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So a halogen on a benzene ring will still direct substituents ortho and para to it because of the resonance structures that you can draw. However, it turns out to be a deactivator most of the time. The reactions are slower than that of benzene on its own. Let's first examine the inductive effect that the halogen has on our benzene ring. So let me go ahead and write induction here. If we think about the sigma bond between the carbon on the ring and the halogen, so this is our generic halogen here with an x, halogens are relatively electronegative, more electronegative than carbon. So they're going to withdraw some electron density from that ring."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's first examine the inductive effect that the halogen has on our benzene ring. So let me go ahead and write induction here. If we think about the sigma bond between the carbon on the ring and the halogen, so this is our generic halogen here with an x, halogens are relatively electronegative, more electronegative than carbon. So they're going to withdraw some electron density from that ring. Whenever you withdraw electron density from a benzene ring, you deactivate it towards electrophilic aromatic substitution. So the inductive effect would lead you to think that a halogen is a deactivator. However, the halogen does have lone pairs of electrons on it."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So they're going to withdraw some electron density from that ring. Whenever you withdraw electron density from a benzene ring, you deactivate it towards electrophilic aromatic substitution. So the inductive effect would lead you to think that a halogen is a deactivator. However, the halogen does have lone pairs of electrons on it. So you could also think about the resonance effect. So let me just go ahead and write the resonance effect here. And if I took one of these lone pairs of electrons and moved them in here to form a pi bond between that carbon and that halogen, that would push these electrons off onto this carbon."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "However, the halogen does have lone pairs of electrons on it. So you could also think about the resonance effect. So let me just go ahead and write the resonance effect here. And if I took one of these lone pairs of electrons and moved them in here to form a pi bond between that carbon and that halogen, that would push these electrons off onto this carbon. So we could go ahead and draw our resonance structure here. So we would have our halogen now double bonded to our carbon. The halogen has two lone pairs of electrons, a plus 1 formal charge."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if I took one of these lone pairs of electrons and moved them in here to form a pi bond between that carbon and that halogen, that would push these electrons off onto this carbon. So we could go ahead and draw our resonance structure here. So we would have our halogen now double bonded to our carbon. The halogen has two lone pairs of electrons, a plus 1 formal charge. And we now have a lone pair of electrons out on this carbon with a negative 1 formal charge like that. So let me just go ahead and highlight. One of these lone pairs of electrons can participate in resonance."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The halogen has two lone pairs of electrons, a plus 1 formal charge. And we now have a lone pair of electrons out on this carbon with a negative 1 formal charge like that. So let me just go ahead and highlight. One of these lone pairs of electrons can participate in resonance. And that would increase the electron density of your ring. And again, I'm not going to draw the other resonance structures. But increasing the electron density of your ring implies activation towards electrophilic aromatic substitution."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "One of these lone pairs of electrons can participate in resonance. And that would increase the electron density of your ring. And again, I'm not going to draw the other resonance structures. But increasing the electron density of your ring implies activation towards electrophilic aromatic substitution. So we have these two competing factors here. Induction says that the halogen is electron withdrawing and therefore a deactivator. Resonance suggests that the halogen might be electron donating and therefore an activator."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But increasing the electron density of your ring implies activation towards electrophilic aromatic substitution. So we have these two competing factors here. Induction says that the halogen is electron withdrawing and therefore a deactivator. Resonance suggests that the halogen might be electron donating and therefore an activator. Experimental results show that these halogens are actually deactivators here. So induction must be a more important effect than resonance. Let's see if we can think about why using different halogens."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Resonance suggests that the halogen might be electron donating and therefore an activator. Experimental results show that these halogens are actually deactivators here. So induction must be a more important effect than resonance. Let's see if we can think about why using different halogens. Let's start with fluorine here. So if your halogen is fluorine, fluorine is, of course, extremely electronegative. And so you could think about what the inductive effect, if the halogen is fluorine, it's withdrawing a lot of electron density from that benzene ring, deactivating the ring towards electrophilic aromatic substitution."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's see if we can think about why using different halogens. Let's start with fluorine here. So if your halogen is fluorine, fluorine is, of course, extremely electronegative. And so you could think about what the inductive effect, if the halogen is fluorine, it's withdrawing a lot of electron density from that benzene ring, deactivating the ring towards electrophilic aromatic substitution. And so the inductive effect is very strong when fluorine is on your benzene ring. When you think about resonance, so the fluorine forming a pi bond with a carbon on your ring. So let me show this carbon right here forming a pi bond with a halogen."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so you could think about what the inductive effect, if the halogen is fluorine, it's withdrawing a lot of electron density from that benzene ring, deactivating the ring towards electrophilic aromatic substitution. And so the inductive effect is very strong when fluorine is on your benzene ring. When you think about resonance, so the fluorine forming a pi bond with a carbon on your ring. So let me show this carbon right here forming a pi bond with a halogen. I'm saying the halogen is fluorine this time. So I can go ahead and show carbon bonded to fluorine. So we have p orbitals here, which are overlapping."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me show this carbon right here forming a pi bond with a halogen. I'm saying the halogen is fluorine this time. So I can go ahead and show carbon bonded to fluorine. So we have p orbitals here, which are overlapping. And so we can draw in the p orbitals on carbon. Carbon and fluorine are in the same period on the periodic table. Therefore, you can think about their p orbitals being pretty much the same size."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have p orbitals here, which are overlapping. And so we can draw in the p orbitals on carbon. Carbon and fluorine are in the same period on the periodic table. Therefore, you can think about their p orbitals being pretty much the same size. And that allows for good overlap. And so the fluorine can donate some electrons and increase the electron density of your ring. And so because there is some good overlap of your p orbitals there, fluorine turns out, fluorobenzene turns out to be the most reactive of your halogens because of that overlap."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, you can think about their p orbitals being pretty much the same size. And that allows for good overlap. And so the fluorine can donate some electrons and increase the electron density of your ring. And so because there is some good overlap of your p orbitals there, fluorine turns out, fluorobenzene turns out to be the most reactive of your halogens because of that overlap. But because fluorine is so electronegative, you could think about the inductive effect winning most of the time and still giving you a deactivator. However, there are some exceptions for fluorobenzene. But in general, we consider the halogens to be deactivators."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so because there is some good overlap of your p orbitals there, fluorine turns out, fluorobenzene turns out to be the most reactive of your halogens because of that overlap. But because fluorine is so electronegative, you could think about the inductive effect winning most of the time and still giving you a deactivator. However, there are some exceptions for fluorobenzene. But in general, we consider the halogens to be deactivators. Let's look at another halogen here. So for that example, induction wins because of the large electronegativity difference between fluorine and carbon. However, if the electronegativity difference decreases, such as when your halogen is something like chlorine, then you have to come up with a little bit different explanation because an atom like nitrogen has pretty much the same electronegativity as chlorine."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But in general, we consider the halogens to be deactivators. Let's look at another halogen here. So for that example, induction wins because of the large electronegativity difference between fluorine and carbon. However, if the electronegativity difference decreases, such as when your halogen is something like chlorine, then you have to come up with a little bit different explanation because an atom like nitrogen has pretty much the same electronegativity as chlorine. And so the inductive effect would be the same for nitrogen as it is for chlorine, approximately. And so there must be a slightly different explanation as to why chlorine is a deactivator. And if you think about the resonance effect, where you're chlorine forming a pi bond with your carbon."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "However, if the electronegativity difference decreases, such as when your halogen is something like chlorine, then you have to come up with a little bit different explanation because an atom like nitrogen has pretty much the same electronegativity as chlorine. And so the inductive effect would be the same for nitrogen as it is for chlorine, approximately. And so there must be a slightly different explanation as to why chlorine is a deactivator. And if you think about the resonance effect, where you're chlorine forming a pi bond with your carbon. So let's go ahead and show chlorine bonded to carbon here. Carbon being in the second period of the periodic table has a p orbital of a certain size. Chlorine, of course, is in a different period."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if you think about the resonance effect, where you're chlorine forming a pi bond with your carbon. So let's go ahead and show chlorine bonded to carbon here. Carbon being in the second period of the periodic table has a p orbital of a certain size. Chlorine, of course, is in a different period. It's in a third period. So it has a larger p orbital. And the size mismatch of those p orbitals means that you don't get as good of an overlap as you would with orbitals that are the same size."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Chlorine, of course, is in a different period. It's in a third period. So it has a larger p orbital. And the size mismatch of those p orbitals means that you don't get as good of an overlap as you would with orbitals that are the same size. And so therefore, the chlorine can't donate as much electron density via resonance. So resonance decreases in importance because of the poor overlap of those orbitals. And so that's a way of thinking about the fact that the resonance effect is decreased."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the size mismatch of those p orbitals means that you don't get as good of an overlap as you would with orbitals that are the same size. And so therefore, the chlorine can't donate as much electron density via resonance. So resonance decreases in importance because of the poor overlap of those orbitals. And so that's a way of thinking about the fact that the resonance effect is decreased. And so therefore, you have more of an inductive effect and the chlorine is a deactivator. I was talking about comparing chlorine to nitrogen. And the two, of course, have very similar electronegativities."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's a way of thinking about the fact that the resonance effect is decreased. And so therefore, you have more of an inductive effect and the chlorine is a deactivator. I was talking about comparing chlorine to nitrogen. And the two, of course, have very similar electronegativities. But the nitrogen, right, aniline in the previous video is observed to react much, much faster. And again, we can think about the size of those p orbitals, nitrogen being in the same period as carbon. And so you get better overlap of those orbitals."}, {"video_title": "Ortho-para directors III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the two, of course, have very similar electronegativities. But the nitrogen, right, aniline in the previous video is observed to react much, much faster. And again, we can think about the size of those p orbitals, nitrogen being in the same period as carbon. And so you get better overlap of those orbitals. The nitrogen is better able to donate electron density to that ring, making the nitrogen an activator here. So you can think about, again, large differences in electronegativity favoring induction. You can think about size mismatch of orbitals."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is a very nice picture here of a dinosaur enjoying a sunset at the beach. But unfortunately for the dinosaurs, about 65 million years ago, we believe that a huge meteorite struck the Earth and essentially wiped out the dinosaurs. And they probably wiped out a bunch of other species with it. Because you can imagine, the shockwave itself would just exterminate tons of species. Then you would have the tsunami of unimaginable size that would just envelop the continents for some period of time. And then you would have all of the soot that would go into the air and maybe make it impossible for most of the plant species to live because it would be blocking out all of the sunlight. And so in an environment like that, we could imagine that an animal like this would be well suited to survive."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because you can imagine, the shockwave itself would just exterminate tons of species. Then you would have the tsunami of unimaginable size that would just envelop the continents for some period of time. And then you would have all of the soot that would go into the air and maybe make it impossible for most of the plant species to live because it would be blocking out all of the sunlight. And so in an environment like that, we could imagine that an animal like this would be well suited to survive. It's sitting there underground. Maybe it can hibernate in some way so it doesn't need food for long periods of time. Maybe it has its own food stash under there someplace."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so in an environment like that, we could imagine that an animal like this would be well suited to survive. It's sitting there underground. Maybe it can hibernate in some way so it doesn't need food for long periods of time. Maybe it has its own food stash under there someplace. And so we believe that our ancient ancestors, after this mass extinction event, might have been something like this, kind of a mole-looking rodent animal that was protected from all of this craziness that was happening on the surface because they like to hang out underground and have all their food nearby them. And maybe they could hibernate in some way. So you can imagine once everything settled down, and now we're talking hundreds of years, thousands of years, even millions of years, some of this guy's descendants start to poke their head out of the ground."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe it has its own food stash under there someplace. And so we believe that our ancient ancestors, after this mass extinction event, might have been something like this, kind of a mole-looking rodent animal that was protected from all of this craziness that was happening on the surface because they like to hang out underground and have all their food nearby them. And maybe they could hibernate in some way. So you can imagine once everything settled down, and now we're talking hundreds of years, thousands of years, even millions of years, some of this guy's descendants start to poke their head out of the ground. They're like, you know what? There's food in trees and there's no one else in the trees. And trees are a good place to maybe get away from some of the other predators that have managed to survive this mass extinction event."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you can imagine once everything settled down, and now we're talking hundreds of years, thousands of years, even millions of years, some of this guy's descendants start to poke their head out of the ground. They're like, you know what? There's food in trees and there's no one else in the trees. And trees are a good place to maybe get away from some of the other predators that have managed to survive this mass extinction event. And some of its ancestors, or some of its descendants, I should say, that were good at climbing trees decide, hey, let's try this tree thing out. And so you started to have some selection for the descendants of this rodent that could climb trees well. They were able to find food where their ancestors couldn't."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And trees are a good place to maybe get away from some of the other predators that have managed to survive this mass extinction event. And some of its ancestors, or some of its descendants, I should say, that were good at climbing trees decide, hey, let's try this tree thing out. And so you started to have some selection for the descendants of this rodent that could climb trees well. They were able to find food where their ancestors couldn't. They could find protection in the trees where their ancestors couldn't. And so you could imagine that some subset of this guy's descendants evolved into something that might have looked like this guy. And all the pictures I'm showing you, these are of modern animals, except for, of course, the dinosaur."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They were able to find food where their ancestors couldn't. They could find protection in the trees where their ancestors couldn't. And so you could imagine that some subset of this guy's descendants evolved into something that might have looked like this guy. And all the pictures I'm showing you, these are of modern animals, except for, of course, the dinosaur. I'm sure this was kind of photoshopped in in some way. This is a modern bush baby. And I like this picture because it could have been what some of these primitive primates looked like."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And all the pictures I'm showing you, these are of modern animals, except for, of course, the dinosaur. I'm sure this was kind of photoshopped in in some way. This is a modern bush baby. And I like this picture because it could have been what some of these primitive primates looked like. Because a bush baby, it kind of climbs trees. It kind of looks like it's starting to get a hand here to start climbing the trees. But it also has rodent-like qualities."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I like this picture because it could have been what some of these primitive primates looked like. Because a bush baby, it kind of climbs trees. It kind of looks like it's starting to get a hand here to start climbing the trees. But it also has rodent-like qualities. But this is, of course, a modern version of it. So this bush baby's ancient, ancient, ancient ancestor might have been that primitive primate, or that species of primitive primate that was a descendant of rodents that started to say, hey, let's see if we can climb these trees and find some food. And some of its descendants might have had just the right adaptations, found their own little niche in the right ecosystems, and they would have evolved into monkeys."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it also has rodent-like qualities. But this is, of course, a modern version of it. So this bush baby's ancient, ancient, ancient ancestor might have been that primitive primate, or that species of primitive primate that was a descendant of rodents that started to say, hey, let's see if we can climb these trees and find some food. And some of its descendants might have had just the right adaptations, found their own little niche in the right ecosystems, and they would have evolved into monkeys. Once again, this is a modern monkey, but you could imagine some type of primitive monkey. And then some of those primitive monkeys' descendants, they turn into these modern monkeys eventually. But some of them, they grow larger in size, they spend more time outside of trees, they lose their tail, they don't need it as much for balance."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And some of its descendants might have had just the right adaptations, found their own little niche in the right ecosystems, and they would have evolved into monkeys. Once again, this is a modern monkey, but you could imagine some type of primitive monkey. And then some of those primitive monkeys' descendants, they turn into these modern monkeys eventually. But some of them, they grow larger in size, they spend more time outside of trees, they lose their tail, they don't need it as much for balance. Maybe it's actually a bad thing to have because someone else could grab it when you're in a fight or something like that. And they evolve into apes, and in particular, the great apes. So one of the great apes, the great apes involve gorillas and chimpanzees, and chimpanzees, and the ancestor, or really, the great apes also include humanity."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But some of them, they grow larger in size, they spend more time outside of trees, they lose their tail, they don't need it as much for balance. Maybe it's actually a bad thing to have because someone else could grab it when you're in a fight or something like that. And they evolve into apes, and in particular, the great apes. So one of the great apes, the great apes involve gorillas and chimpanzees, and chimpanzees, and the ancestor, or really, the great apes also include humanity. So let me just review back on this timeline just so that we don't get confused. Let me just review what we just talked about. So before this mass extinction event, 65 million years ago, you had all these types of species here."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So one of the great apes, the great apes involve gorillas and chimpanzees, and chimpanzees, and the ancestor, or really, the great apes also include humanity. So let me just review back on this timeline just so that we don't get confused. Let me just review what we just talked about. So before this mass extinction event, 65 million years ago, you had all these types of species here. Maybe this right up here, maybe this was, actually if I'm talking about species, maybe this was Tyrannosaurus rex, because the dinosaurs involve a whole bunch of, so this might have been T. rex. There's a bunch of species that we could list over here. But after that mass extinction event, that was an endpoint for a ton of species, except for maybe this primitive rodent mole-like thing that was, maybe a lot of them died in this event, but just enough of them survived because they were underground or just in the right place, or they were in a mountain someplace."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So before this mass extinction event, 65 million years ago, you had all these types of species here. Maybe this right up here, maybe this was, actually if I'm talking about species, maybe this was Tyrannosaurus rex, because the dinosaurs involve a whole bunch of, so this might have been T. rex. There's a bunch of species that we could list over here. But after that mass extinction event, that was an endpoint for a ton of species, except for maybe this primitive rodent mole-like thing that was, maybe a lot of them died in this event, but just enough of them survived because they were underground or just in the right place, or they were in a mountain someplace. Who knows where they were. And some of them were able to evolve into primitive primates. And some of those primitive primates, and this is, once again, these are pictures of primitive primates."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But after that mass extinction event, that was an endpoint for a ton of species, except for maybe this primitive rodent mole-like thing that was, maybe a lot of them died in this event, but just enough of them survived because they were underground or just in the right place, or they were in a mountain someplace. Who knows where they were. And some of them were able to evolve into primitive primates. And some of those primitive primates, and this is, once again, these are pictures of primitive primates. Some of those primitive, and when I say primitive, these are modern versions of them. So primitive doesn't necessarily mean worse, because obviously these guys were able to find, even in today's world, they have a niche for themselves. They're able to find food and reproduce in ways that don't get in the way of other people, and other way people don't get in the way of them."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And some of those primitive primates, and this is, once again, these are pictures of primitive primates. Some of those primitive, and when I say primitive, these are modern versions of them. So primitive doesn't necessarily mean worse, because obviously these guys were able to find, even in today's world, they have a niche for themselves. They're able to find food and reproduce in ways that don't get in the way of other people, and other way people don't get in the way of them. When I talk about primitive primate, I'm just talking about kind of an ancestral primate, maybe something that's not there today, although maybe some of its descendants look very much like it. But anyway, some of those primates evolve into primitive monkeys. Some of those primitive monkeys' descendants become modern monkeys."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're able to find food and reproduce in ways that don't get in the way of other people, and other way people don't get in the way of them. When I talk about primitive primate, I'm just talking about kind of an ancestral primate, maybe something that's not there today, although maybe some of its descendants look very much like it. But anyway, some of those primates evolve into primitive monkeys. Some of those primitive monkeys' descendants become modern monkeys. So this is, I'll call it M monkeys for modern monkeys. And some of them evolve into primitive apes. And apes, their distinctive characteristic is that they're like monkeys, but they don't have tails, and they're larger than most monkeys."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some of those primitive monkeys' descendants become modern monkeys. So this is, I'll call it M monkeys for modern monkeys. And some of them evolve into primitive apes. And apes, their distinctive characteristic is that they're like monkeys, but they don't have tails, and they're larger than most monkeys. And so these primitive apes, some of their descendants are modern gorillas. At some point, they break off. Some of these descendants are an ancestor of both modern chimpanzees and of human beings."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And apes, their distinctive characteristic is that they're like monkeys, but they don't have tails, and they're larger than most monkeys. And so these primitive apes, some of their descendants are modern gorillas. At some point, they break off. Some of these descendants are an ancestor of both modern chimpanzees and of human beings. And we think, just looking at the DNA evidence, we think that this departure right here, and the fossil evidence, was about 7 million years ago. That's our best guess for when we as human beings had a common ancestor with the chimpanzees. Now, you have that common ancestor."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some of these descendants are an ancestor of both modern chimpanzees and of human beings. And we think, just looking at the DNA evidence, we think that this departure right here, and the fossil evidence, was about 7 million years ago. That's our best guess for when we as human beings had a common ancestor with the chimpanzees. Now, you have that common ancestor. Some of that common ancestor's descendants became modern chimpanzees. And some of them, maybe they explored the right ecosystem where it was more advantageous to do so, started to walk on two legs. And the most famous fossil of this is the Australopithecine fossil of Lucy that was discovered 3.2 million years..."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, you have that common ancestor. Some of that common ancestor's descendants became modern chimpanzees. And some of them, maybe they explored the right ecosystem where it was more advantageous to do so, started to walk on two legs. And the most famous fossil of this is the Australopithecine fossil of Lucy that was discovered 3.2 million years... It was discovered more recently. It's 3.2 million years old. So the whole genus, and genus is kind of one level of categorization above species, the whole genus of Australopithecine, these were 4 to 2 million years ago."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the most famous fossil of this is the Australopithecine fossil of Lucy that was discovered 3.2 million years... It was discovered more recently. It's 3.2 million years old. So the whole genus, and genus is kind of one level of categorization above species, the whole genus of Australopithecine, these were 4 to 2 million years ago. And we never know. You could always find a fossil that's older than this, maybe newer than this. I read one account that says maybe 1 million years ago."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the whole genus, and genus is kind of one level of categorization above species, the whole genus of Australopithecine, these were 4 to 2 million years ago. And we never know. You could always find a fossil that's older than this, maybe newer than this. I read one account that says maybe 1 million years ago. But give or take, the Lucy fossil, which is the most well-established Australopithecine fossil, is about 3 million years old. And this is a reconstruction I have over here of Lucy. So this is probably what Lucy looked like."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I read one account that says maybe 1 million years ago. But give or take, the Lucy fossil, which is the most well-established Australopithecine fossil, is about 3 million years old. And this is a reconstruction I have over here of Lucy. So this is probably what Lucy looked like. And once again, there were many Lucys. It wasn't just there was one Lucy and we're all descended from Lucys. And it's actually not even clear that we are even descended directly from Australopithecine."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is probably what Lucy looked like. And once again, there were many Lucys. It wasn't just there was one Lucy and we're all descended from Lucys. And it's actually not even clear that we are even descended directly from Australopithecine. We might be a cousin species, or a cousin genus, I should say. Genus is the category right above species. So if you fast forward a little bit more, you go to about 2.3 to 1.4 million years ago."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's actually not even clear that we are even descended directly from Australopithecine. We might be a cousin species, or a cousin genus, I should say. Genus is the category right above species. So if you fast forward a little bit more, you go to about 2.3 to 1.4 million years ago. We see fossils that they're standing upright. The brain size is bigger. Because if you look at the Australopithecine fossils, they are standing upright."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you fast forward a little bit more, you go to about 2.3 to 1.4 million years ago. We see fossils that they're standing upright. The brain size is bigger. Because if you look at the Australopithecine fossils, they are standing upright. But their cranial capacity isn't that different than chimpanzees. You fast forward to 2.3 million to 1.4 million years ago. We start to see fossils where they're standing upright still."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because if you look at the Australopithecine fossils, they are standing upright. But their cranial capacity isn't that different than chimpanzees. You fast forward to 2.3 million to 1.4 million years ago. We start to see fossils where they're standing upright still. And the cranial capacity has grown. And you're starting to see primitive stone tools around the bone fossils. And so we believe that these are one of the first."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We start to see fossils where they're standing upright still. And the cranial capacity has grown. And you're starting to see primitive stone tools around the bone fossils. And so we believe that these are one of the first. And this is really just how we categorize it. But these are some of the first fossils that we categorize as belonging to the same genus as ours. And the genus is Homo."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we believe that these are one of the first. And this is really just how we categorize it. But these are some of the first fossils that we categorize as belonging to the same genus as ours. And the genus is Homo. And Homo just means man. So it's the group right above species of man. And we call them similar to man because it looks like they're starting to make primitive stone tools."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the genus is Homo. And Homo just means man. So it's the group right above species of man. And we call them similar to man because it looks like they're starting to make primitive stone tools. They stand upright like us. And they have larger cranial capacities than the Australopithecine fossils or modern chimpanzees. And once again, we don't know if Homo habilis, which literally means, so the Homo part means man."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we call them similar to man because it looks like they're starting to make primitive stone tools. They stand upright like us. And they have larger cranial capacities than the Australopithecine fossils or modern chimpanzees. And once again, we don't know if Homo habilis, which literally means, so the Homo part means man. Habilis means handy because he liked to, I guess, make tools or whatever else. We don't know if Homo habilis is a descendant of Lucy's species of Australopithecus or maybe a cousin species. Maybe they're both descendants from some common ancestor."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And once again, we don't know if Homo habilis, which literally means, so the Homo part means man. Habilis means handy because he liked to, I guess, make tools or whatever else. We don't know if Homo habilis is a descendant of Lucy's species of Australopithecus or maybe a cousin species. Maybe they're both descendants from some common ancestor. We're not quite sure. Then you fast forward a little bit more. We're talking now about 1.8."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe they're both descendants from some common ancestor. We're not quite sure. Then you fast forward a little bit more. We're talking now about 1.8. So now we're talking about 1.8 to 1.3 million years ago. And we start seeing fossils where the cranial capacity larger than Homo habilis getting closer in size to kind of what we, what our notion is of kind of a modern person's cranial capacity, at least relative to body size. And this is Homo erectus."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking now about 1.8. So now we're talking about 1.8 to 1.3 million years ago. And we start seeing fossils where the cranial capacity larger than Homo habilis getting closer in size to kind of what we, what our notion is of kind of a modern person's cranial capacity, at least relative to body size. And this is Homo erectus. And once again, we don't know if Homo erectus is a descendant of Homo habilis. Maybe they have a common ancestor. Who knows?"}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is Homo erectus. And once again, we don't know if Homo erectus is a descendant of Homo habilis. Maybe they have a common ancestor. Who knows? And it looks from the fossil evidence that there was, especially when you look at this range here, that there was some overlap where you had both Homo erectus and Homo habilis living on the same planet at the same time. Now you fast forward even more and we think about 600,000 to 300,000. Once again, you know, all of these are constantly being modified as we get better at finding new fossils or interpreting the fossils we have or we look at DNA evidence or whatever."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Who knows? And it looks from the fossil evidence that there was, especially when you look at this range here, that there was some overlap where you had both Homo erectus and Homo habilis living on the same planet at the same time. Now you fast forward even more and we think about 600,000 to 300,000. Once again, you know, all of these are constantly being modified as we get better at finding new fossils or interpreting the fossils we have or we look at DNA evidence or whatever. About 600,000 to 300,000 years ago, you have the Neanderthals appear. And Neanderthals are in the same genus as humans. So it's really Homo neanderthalensis."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once again, you know, all of these are constantly being modified as we get better at finding new fossils or interpreting the fossils we have or we look at DNA evidence or whatever. About 600,000 to 300,000 years ago, you have the Neanderthals appear. And Neanderthals are in the same genus as humans. So it's really Homo neanderthalensis. I always have trouble saying this. So this is still part of Homo. And a common misconception is that the Neanderthals are somehow a more primitive version of humans, that they're somehow cavemen and we're modern men."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's really Homo neanderthalensis. I always have trouble saying this. So this is still part of Homo. And a common misconception is that the Neanderthals are somehow a more primitive version of humans, that they're somehow cavemen and we're modern men. That's not the case. The belief is that Neanderthals are either a cousin species, we have a common ancestor, or that they're actually a subspecies of human beings. And there's some belief that they might have interbred with Homo sapiens and maybe some or a good number of us have Neanderthal genes."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And a common misconception is that the Neanderthals are somehow a more primitive version of humans, that they're somehow cavemen and we're modern men. That's not the case. The belief is that Neanderthals are either a cousin species, we have a common ancestor, or that they're actually a subspecies of human beings. And there's some belief that they might have interbred with Homo sapiens and maybe some or a good number of us have Neanderthal genes. It's nothing to be ashamed of. It's just something, you know, unfortunately, that Neanderthals just get a bad name because of, I guess, our popular culture, if anything. So this is a drawing of a Neanderthal brain."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there's some belief that they might have interbred with Homo sapiens and maybe some or a good number of us have Neanderthal genes. It's nothing to be ashamed of. It's just something, you know, unfortunately, that Neanderthals just get a bad name because of, I guess, our popular culture, if anything. So this is a drawing of a Neanderthal brain. They actually had a fairly large cranial capacity, although scientists say they kind of make one reason or another why we think that they might have been more primitive than Homo sapiens. But who knows? We don't know."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is a drawing of a Neanderthal brain. They actually had a fairly large cranial capacity, although scientists say they kind of make one reason or another why we think that they might have been more primitive than Homo sapiens. But who knows? We don't know. We're constantly learning things every day. But of course, the whole point of this is to talk about how humans showed up on this planet and the first really human fossils we find about 200,000 years ago. And this, remember, we're in the genus Homo, and now we finally found something that looks just like us anatomically at least."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We don't know. We're constantly learning things every day. But of course, the whole point of this is to talk about how humans showed up on this planet and the first really human fossils we find about 200,000 years ago. And this, remember, we're in the genus Homo, and now we finally found something that looks just like us anatomically at least. We can't study its behavior and all the rest. And now we get to Homo sapiens. The Homo part, once again, means man."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this, remember, we're in the genus Homo, and now we finally found something that looks just like us anatomically at least. We can't study its behavior and all the rest. And now we get to Homo sapiens. The Homo part, once again, means man. And the sapiens means thinking. So we can debate whether it's an appropriate title for our species, but it's thinking man. So once again, the Neanderthals, they were either a cousin species for a lot of this time, especially once Homo sapiens showed up."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The Homo part, once again, means man. And the sapiens means thinking. So we can debate whether it's an appropriate title for our species, but it's thinking man. So once again, the Neanderthals, they were either a cousin species for a lot of this time, especially once Homo sapiens showed up. And maybe Homo sapiens showed up before this. We just haven't found the fossils yet. They were maybe both inhabiting the same planet."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So once again, the Neanderthals, they were either a cousin species for a lot of this time, especially once Homo sapiens showed up. And maybe Homo sapiens showed up before this. We just haven't found the fossils yet. They were maybe both inhabiting the same planet. Maybe there was some interbreeding. But the Neanderthals disappeared about 30,000 years ago. 30,000 years ago, these guys disappeared."}, {"video_title": "Human evolution overview Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They were maybe both inhabiting the same planet. Maybe there was some interbreeding. But the Neanderthals disappeared about 30,000 years ago. 30,000 years ago, these guys disappeared. Maybe some of them kind of got mixed in with the Homo sapiens, started to interbreed with them, or they might have just been killed off because they were fighting over the same ecosystems. And I've made a little sample here of Homo sapiens just in case. Well, I'm assuming most of you watching this video are one, but just in case, here's my little sample."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So this molecule on the left undergoes an intramolecular Diels-Alder reaction to form the product on the right. And notice how we form two rings for our product. So this is a pretty cool reaction. The first thing we need to do is find our diene and our dienophile. Our diene is over here on the left. So let me highlight the pi electrons. These pi electrons in red are part of the diene, and so are these pi electrons in magenta."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "The first thing we need to do is find our diene and our dienophile. Our diene is over here on the left. So let me highlight the pi electrons. These pi electrons in red are part of the diene, and so are these pi electrons in magenta. Our dienophile is over here on the right. So these pi electrons, let me make them blue, these pi electrons are part of our dienophile. First, let's show how those two rings form."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "These pi electrons in red are part of the diene, and so are these pi electrons in magenta. Our dienophile is over here on the right. So these pi electrons, let me make them blue, these pi electrons are part of our dienophile. First, let's show how those two rings form. And we won't worry about stereochemistry right now. Let's just think about moving those six pi electrons to form our two rings. Our pi electrons in red move into here to form a bond between these two carbons."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "First, let's show how those two rings form. And we won't worry about stereochemistry right now. Let's just think about moving those six pi electrons to form our two rings. Our pi electrons in red move into here to form a bond between these two carbons. Our pi electrons in blue move into here to form a bond between these two carbons. Obviously, they'd have to be a lot closer together in space, in reality. And our pi electrons in magenta move into here."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "Our pi electrons in red move into here to form a bond between these two carbons. Our pi electrons in blue move into here to form a bond between these two carbons. Obviously, they'd have to be a lot closer together in space, in reality. And our pi electrons in magenta move into here. So let's draw our cyclohexene ring. So here's our cyclohexene ring. Our electrons in red formed this bond."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And our pi electrons in magenta move into here. So let's draw our cyclohexene ring. So here's our cyclohexene ring. Our electrons in red formed this bond. Our electrons in blue formed this bond. And the electrons in magenta formed this bond. All right, let's look at what's attached to those carbons."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "Our electrons in red formed this bond. Our electrons in blue formed this bond. And the electrons in magenta formed this bond. All right, let's look at what's attached to those carbons. So this carbon right here is this one in our product. And so there must be a CH2 and then an oxygen. So there's a CH2 and then an oxygen coming off of that carbon."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at what's attached to those carbons. So this carbon right here is this one in our product. And so there must be a CH2 and then an oxygen. So there's a CH2 and then an oxygen coming off of that carbon. Next, let's look at this carbon. So maybe I should change colors here. Let me make this blue."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So there's a CH2 and then an oxygen coming off of that carbon. Next, let's look at this carbon. So maybe I should change colors here. Let me make this blue. So this carbon right here is this one. And we know we have a carbonyl coming off of that carbon. And then we hit this oxygen."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "Let me make this blue. So this carbon right here is this one. And we know we have a carbonyl coming off of that carbon. And then we hit this oxygen. That's the same oxygen as before. So we have a carbonyl coming off of that carbon. And then that goes straight to the oxygen."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And then we hit this oxygen. That's the same oxygen as before. So we have a carbonyl coming off of that carbon. And then that goes straight to the oxygen. So already we see those two rings. The next carbon is this one. I'll make it green."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And then that goes straight to the oxygen. So already we see those two rings. The next carbon is this one. I'll make it green. And we can see we have a methyl group and a carboxylic acid coming off of that carbon. So again, I'll just draw these in without any stereochemistry. There's our methyl group."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "I'll make it green. And we can see we have a methyl group and a carboxylic acid coming off of that carbon. So again, I'll just draw these in without any stereochemistry. There's our methyl group. There is our carboxylic acid, CO2H. And then finally, our last carbon, and I'll just make this yellow here, this carbon has a methyl group coming off of it. So there's our methyl group."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "There's our methyl group. There is our carboxylic acid, CO2H. And then finally, our last carbon, and I'll just make this yellow here, this carbon has a methyl group coming off of it. So there's our methyl group. So now we can see the formation of these two rings. But the next thing we need to do is to account for the stereochemistry that we see in our product here. And notice we're told that this is the endo product."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So there's our methyl group. So now we can see the formation of these two rings. But the next thing we need to do is to account for the stereochemistry that we see in our product here. And notice we're told that this is the endo product. So the endo product needs to have an endo approach. And we've talked about this in earlier videos. If this is our dienophile, these carbonyls on the dienophile need to point towards the diene."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And notice we're told that this is the endo product. So the endo product needs to have an endo approach. And we've talked about this in earlier videos. If this is our dienophile, these carbonyls on the dienophile need to point towards the diene. So let's go to a video so we can see how to use the model set to better visualize the endo approach. So here's our molecule. And I'm going to rotate about this bond just so we can make our diene and our dienophile approach each other easier."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "If this is our dienophile, these carbonyls on the dienophile need to point towards the diene. So let's go to a video so we can see how to use the model set to better visualize the endo approach. So here's our molecule. And I'm going to rotate about this bond just so we can make our diene and our dienophile approach each other easier. So now we have our carbonyls pointing towards the diene. So this is the endo approach. And now if I hold it like this, you can see the diene and the dienophile."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to rotate about this bond just so we can make our diene and our dienophile approach each other easier. So now we have our carbonyls pointing towards the diene. So this is the endo approach. And now if I hold it like this, you can see the diene and the dienophile. On the left is a picture of what we saw in the video. And we know that a bond forms between this carbon and this one. So I'll draw in a dotted line here."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And now if I hold it like this, you can see the diene and the dienophile. On the left is a picture of what we saw in the video. And we know that a bond forms between this carbon and this one. So I'll draw in a dotted line here. On the right is a picture of our product. So that bond that forms is between this carbon and this one. And let me put it in on the drawing too."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw in a dotted line here. On the right is a picture of our product. So that bond that forms is between this carbon and this one. And let me put it in on the drawing too. So we're talking about this bond. We know that another bond forms between this carbon and this one. So I'll draw in a dotted line here."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And let me put it in on the drawing too. So we're talking about this bond. We know that another bond forms between this carbon and this one. So I'll draw in a dotted line here. And that's a bond between this carbon and this one. So I'll draw in that bond. And then on this drawing, we're talking about this bond in blue."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw in a dotted line here. And that's a bond between this carbon and this one. So I'll draw in that bond. And then on this drawing, we're talking about this bond in blue. Next, let's think about the stereochemistry of the diene. So here is our diene. And let's put in these two hydrogens."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And then on this drawing, we're talking about this bond in blue. Next, let's think about the stereochemistry of the diene. So here is our diene. And let's put in these two hydrogens. These two hydrogens are inside substituents. And we know that inside substituents go up. So if we find those two hydrogens on our picture, here they are."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "And let's put in these two hydrogens. These two hydrogens are inside substituents. And we know that inside substituents go up. So if we find those two hydrogens on our picture, here they are. And for our product, those two hydrogens are going up in space. So if we are staring down at the molecule from this direction, those two hydrogens are coming out at us. So in our drawing, here's one of the hydrogens coming out at us on a wedge."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So if we find those two hydrogens on our picture, here they are. And for our product, those two hydrogens are going up in space. So if we are staring down at the molecule from this direction, those two hydrogens are coming out at us. So in our drawing, here's one of the hydrogens coming out at us on a wedge. And at this carbon would be the other one. Let me go ahead and draw in this wedge. So here's our other hydrogen coming out at us in space."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So in our drawing, here's one of the hydrogens coming out at us on a wedge. And at this carbon would be the other one. Let me go ahead and draw in this wedge. So here's our other hydrogen coming out at us in space. The outside substituents go down. So let me use blue for this. So this methyl group is an outside substituent, and so is this CH2."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So here's our other hydrogen coming out at us in space. The outside substituents go down. So let me use blue for this. So this methyl group is an outside substituent, and so is this CH2. So on our picture, here is our methyl group, and here is that CH2. And for our product, those two are going away from us. So it's a little bit hard to see, but this methyl group is going away from us in space, and so is this CH2."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So this methyl group is an outside substituent, and so is this CH2. So on our picture, here is our methyl group, and here is that CH2. And for our product, those two are going away from us. So it's a little bit hard to see, but this methyl group is going away from us in space, and so is this CH2. So for our product, here is our methyl group going away from us, and here is our CH2 going away from us. Next, let's think about the stereochemistry of the dienophile. So I like to draw a line right here, and we analyze everything on both sides of that double bond."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So it's a little bit hard to see, but this methyl group is going away from us in space, and so is this CH2. So for our product, here is our methyl group going away from us, and here is our CH2 going away from us. Next, let's think about the stereochemistry of the dienophile. So I like to draw a line right here, and we analyze everything on both sides of that double bond. First, let's look at what's on the left side. So we have a carboxylic acid on the left side of that line, and we have this carbonyl. It's hard to see the carbonyl, but it's right back here."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So I like to draw a line right here, and we analyze everything on both sides of that double bond. First, let's look at what's on the left side. So we have a carboxylic acid on the left side of that line, and we have this carbonyl. It's hard to see the carbonyl, but it's right back here. And then this carboxylic acid, I've represented with this red atom here just to make it easier to work with in the model set. So those two are gonna end up on the same side. So these two are gonna end up on the same side, and they're to the left of this line that I drew."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "It's hard to see the carbonyl, but it's right back here. And then this carboxylic acid, I've represented with this red atom here just to make it easier to work with in the model set. So those two are gonna end up on the same side. So these two are gonna end up on the same side, and they're to the left of this line that I drew. So we know those two are going to end up down. So if we find them in our product, here is the carboxylic acid. It's actually going down away from us."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "So these two are gonna end up on the same side, and they're to the left of this line that I drew. So we know those two are going to end up down. So if we find them in our product, here is the carboxylic acid. It's actually going down away from us. And then this bond to that carbonyl is going away from us too. So let's find those on the drawing of our product. Well, here is the carbonyl, and you can see it's going away from us."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "It's actually going down away from us. And then this bond to that carbonyl is going away from us too. So let's find those on the drawing of our product. Well, here is the carbonyl, and you can see it's going away from us. It's on a dash for our drawing. And here is the carboxylic acid. It's also going away from us on a dash."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "Well, here is the carbonyl, and you can see it's going away from us. It's on a dash for our drawing. And here is the carboxylic acid. It's also going away from us on a dash. So those two end up on the same side of the ring, in this case, down in space. Let's look at what's on the right side of our double bond. We know there's a hydrogen here."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "It's also going away from us on a dash. So those two end up on the same side of the ring, in this case, down in space. Let's look at what's on the right side of our double bond. We know there's a hydrogen here. So let me change colors so we can see these things. So there's a hydrogen and we have a methyl group on the right side of our double bond. And here is the methyl group, which I've used as a yellow atom, and here is the hydrogen."}, {"video_title": "Diels-Alder intramolecular Organic chemistry Khan Academy.mp3", "Sentence": "We know there's a hydrogen here. So let me change colors so we can see these things. So there's a hydrogen and we have a methyl group on the right side of our double bond. And here is the methyl group, which I've used as a yellow atom, and here is the hydrogen. The stuff on the right ends up on the same side, and it goes up for our final product. So if we look at our product here, this hydrogen is actually going up in space, and it's really hard to see for this methyl group, but it's actually going up if you reorient this molecule. So these two are going up in space, and here is that hydrogen and here is that methyl group."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "Our goal in that video was to make more of our ester, so we shifted the equilibrium to the right to make more of our product. In this video, we're talking about the reverse reaction. We're gonna talk about ester hydrolysis. So if we increase the concentration of water, that would shift the equilibrium back to the left, and we would hydrolyze our ester and turn it into our alcohol and our carboxylic acid. So it's important to think about what bond we're going to break. You can see that we're going to break this bond in here, and this oxygen, and this R prime group, turn into our alcohol, and so we'll see that in our mechanism. So let's go ahead and look at the details of the mechanism where we're starting with an ester, and we're going to first think about what else is present, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So if we increase the concentration of water, that would shift the equilibrium back to the left, and we would hydrolyze our ester and turn it into our alcohol and our carboxylic acid. So it's important to think about what bond we're going to break. You can see that we're going to break this bond in here, and this oxygen, and this R prime group, turn into our alcohol, and so we'll see that in our mechanism. So let's go ahead and look at the details of the mechanism where we're starting with an ester, and we're going to first think about what else is present, right? Well, in solution, H2O and H plus would give us H3O plus. I'm going to go ahead and draw in the hydronium ion over here. So this is H3O plus."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and look at the details of the mechanism where we're starting with an ester, and we're going to first think about what else is present, right? Well, in solution, H2O and H plus would give us H3O plus. I'm going to go ahead and draw in the hydronium ion over here. So this is H3O plus. The first step of the mechanism is to protonate the carbonyl oxygen. So this lone pair of electrons picks up a proton from hydronium, and let's go ahead and show the protonated carbonyl. So now we have our oxygen, right, has been protonated, so it has a plus one formal charge."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So this is H3O plus. The first step of the mechanism is to protonate the carbonyl oxygen. So this lone pair of electrons picks up a proton from hydronium, and let's go ahead and show the protonated carbonyl. So now we have our oxygen, right, has been protonated, so it has a plus one formal charge. So let's show those electrons. So these electrons right here in magenta pick up this proton to form this bond right here, and this activates our carbonyl. So we've seen this in earlier videos, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our oxygen, right, has been protonated, so it has a plus one formal charge. So let's show those electrons. So these electrons right here in magenta pick up this proton to form this bond right here, and this activates our carbonyl. So we've seen this in earlier videos, right? So when you think about a resonance structure, that actually makes this carbon more positive, it's more electrophilic, and therefore that carbon is going to react with a nucleophile in our next step, and our nucleophile here is water. So water's gonna function as a nucleophile. The nucleophile attacks our electrophile, and that pushes these electrons off onto this oxygen."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So we've seen this in earlier videos, right? So when you think about a resonance structure, that actually makes this carbon more positive, it's more electrophilic, and therefore that carbon is going to react with a nucleophile in our next step, and our nucleophile here is water. So water's gonna function as a nucleophile. The nucleophile attacks our electrophile, and that pushes these electrons off onto this oxygen. So when we show the bond now between the oxygen and that carbon, let me go ahead and draw in the rest of this. This would be a plus one formal charge on this oxygen. And let me highlight those electrons here."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "The nucleophile attacks our electrophile, and that pushes these electrons off onto this oxygen. So when we show the bond now between the oxygen and that carbon, let me go ahead and draw in the rest of this. This would be a plus one formal charge on this oxygen. And let me highlight those electrons here. So these electrons in blue, right, are gonna form the bond between this carbon and this oxygen. We still have an oxygen over here on the left, and now that oxygen has two lone pairs of electrons. So let me go ahead and show those electrons here."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And let me highlight those electrons here. So these electrons in blue, right, are gonna form the bond between this carbon and this oxygen. We still have an oxygen over here on the left, and now that oxygen has two lone pairs of electrons. So let me go ahead and show those electrons here. So let me make them green. So these electrons right here in green, right, move off onto here, and are now a lone pair on that oxygen. We still have an oxygen bonded to this carbon and an R prime group."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and show those electrons here. So let me make them green. So these electrons right here in green, right, move off onto here, and are now a lone pair on that oxygen. We still have an oxygen bonded to this carbon and an R prime group. The next step we need to deprotonate, right? We need to get rid of that plus one formal charge. And so a molecule of water can come along, and this time function as a base."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "We still have an oxygen bonded to this carbon and an R prime group. The next step we need to deprotonate, right? We need to get rid of that plus one formal charge. And so a molecule of water can come along, and this time function as a base. So water's gonna function as a base. It's gonna take this proton, leaving these electrons behind on that oxygen. So let's get some space down here to show the deprotonation."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so a molecule of water can come along, and this time function as a base. So water's gonna function as a base. It's gonna take this proton, leaving these electrons behind on that oxygen. So let's get some space down here to show the deprotonation. So we would now have this carbon, right, with an OH on the left. And after we deprotonate, we're also gonna have an OH on the right. So let me go ahead and draw in that OH on the right."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So let's get some space down here to show the deprotonation. So we would now have this carbon, right, with an OH on the left. And after we deprotonate, we're also gonna have an OH on the right. So let me go ahead and draw in that OH on the right. So showing those electrons, right, let's make those red. So these electrons in here are gonna move off onto this oxygen, right? And then drawing in everything else, we have our R group on the left, and we have OR prime on the right."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw in that OH on the right. So showing those electrons, right, let's make those red. So these electrons in here are gonna move off onto this oxygen, right? And then drawing in everything else, we have our R group on the left, and we have OR prime on the right. And I'm gonna go ahead and put in lone pairs of electrons on this oxygen, because in the next step of the mechanism, we're going to protonate that oxygen. So hydronium is present, so H3O plus. So I'll go ahead and draw that in here, so H3O plus."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And then drawing in everything else, we have our R group on the left, and we have OR prime on the right. And I'm gonna go ahead and put in lone pairs of electrons on this oxygen, because in the next step of the mechanism, we're going to protonate that oxygen. So hydronium is present, so H3O plus. So I'll go ahead and draw that in here, so H3O plus. This oxygen is gonna pick up a proton, right, from hydronium, right, leaving these electrons behind. And so we're gonna protonate that oxygen. And the reason why we protonate that oxygen is it turns it into a better leaving group."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and draw that in here, so H3O plus. This oxygen is gonna pick up a proton, right, from hydronium, right, leaving these electrons behind. And so we're gonna protonate that oxygen. And the reason why we protonate that oxygen is it turns it into a better leaving group. So let me go ahead and draw in everything. So we have our OH groups at the top. We have our R group on the left."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why we protonate that oxygen is it turns it into a better leaving group. So let me go ahead and draw in everything. So we have our OH groups at the top. We have our R group on the left. We have our oxygen, right, which has been protonated now. So it's gonna have a plus one formal charge, right? So plus one formal charge on this oxygen."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "We have our R group on the left. We have our oxygen, right, which has been protonated now. So it's gonna have a plus one formal charge, right? So plus one formal charge on this oxygen. Let's show those electrons. So let's make those blue here. So I'm saying that this lone pair, right, it's gonna take this proton, right, forming this bond right here."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So plus one formal charge on this oxygen. Let's show those electrons. So let's make those blue here. So I'm saying that this lone pair, right, it's gonna take this proton, right, forming this bond right here. And if you look closely, right, you now have alcohol hiding as a leaving group, because in the next step of the mechanism, we're going to reform our carbonyl, and alcohol is going to leave. So if these electrons move into here to reform our carbonyl, right, that's too many bonds to carbon, and so these electrons are gonna come off onto the oxygen. And so here's the bond breaking step where the alcohol leaves."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So I'm saying that this lone pair, right, it's gonna take this proton, right, forming this bond right here. And if you look closely, right, you now have alcohol hiding as a leaving group, because in the next step of the mechanism, we're going to reform our carbonyl, and alcohol is going to leave. So if these electrons move into here to reform our carbonyl, right, that's too many bonds to carbon, and so these electrons are gonna come off onto the oxygen. And so here's the bond breaking step where the alcohol leaves. And so let's go ahead and draw our product, right? So we're going to form our carbonyl, and this oxygen is going to have a plus one formal charge now, and we have an R group, and over here would be an OH, because we're going to have, the alcohol is going to leave. So I'm gonna go ahead and draw in the alcohol here, so OR prime, right, with now two lone pairs of electrons."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so here's the bond breaking step where the alcohol leaves. And so let's go ahead and draw our product, right? So we're going to form our carbonyl, and this oxygen is going to have a plus one formal charge now, and we have an R group, and over here would be an OH, because we're going to have, the alcohol is going to leave. So I'm gonna go ahead and draw in the alcohol here, so OR prime, right, with now two lone pairs of electrons. So let's follow some electrons. So these electrons in blue are the same ones as blue up here, and let's make these electrons in here red. So these electrons in red are going to come off onto this oxygen, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna go ahead and draw in the alcohol here, so OR prime, right, with now two lone pairs of electrons. So let's follow some electrons. So these electrons in blue are the same ones as blue up here, and let's make these electrons in here red. So these electrons in red are going to come off onto this oxygen, right? So we lose our alcohol at this step, right? So loss of our alcohol gives us this over here. So we're almost done with our reaction, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in red are going to come off onto this oxygen, right? So we lose our alcohol at this step, right? So loss of our alcohol gives us this over here. So we're almost done with our reaction, right? So this is really close to a carboxylic acid. All we would have to do is deprotonate, and so we could think about another molecule of water coming along and acting as a base, right? So water acts as a base, takes this proton, leaves these electrons behind, and that of course gives us our carboxylic acid."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So we're almost done with our reaction, right? So this is really close to a carboxylic acid. All we would have to do is deprotonate, and so we could think about another molecule of water coming along and acting as a base, right? So water acts as a base, takes this proton, leaves these electrons behind, and that of course gives us our carboxylic acid. So if I just go ahead and draw in our R group, and then we have our OH, and let's follow those electrons. So let's make those red as well. So these electrons right here come off, and we get our carboxylic acid as our product."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So water acts as a base, takes this proton, leaves these electrons behind, and that of course gives us our carboxylic acid. So if I just go ahead and draw in our R group, and then we have our OH, and let's follow those electrons. So let's make those red as well. So these electrons right here come off, and we get our carboxylic acid as our product. So there's your mechanism for acid-catalyzed ester hydrolysis, right? Which produces a carboxylic acid, and it produces an alcohol. And when we look at some reactions in a second here, we're going to think about this part right here where we lose our alcohol as one of our products, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here come off, and we get our carboxylic acid as our product. So there's your mechanism for acid-catalyzed ester hydrolysis, right? Which produces a carboxylic acid, and it produces an alcohol. And when we look at some reactions in a second here, we're going to think about this part right here where we lose our alcohol as one of our products, right? To also give us our carboxylic acid. So let's look at a reaction. So over here on the left we have methyl salicylate, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And when we look at some reactions in a second here, we're going to think about this part right here where we lose our alcohol as one of our products, right? To also give us our carboxylic acid. So let's look at a reaction. So over here on the left we have methyl salicylate, right? Or oil of wintergreen. And we're looking for an ester, right? Because we have H2O and H+, giving us hydronium in solution."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the left we have methyl salicylate, right? Or oil of wintergreen. And we're looking for an ester, right? Because we have H2O and H+, giving us hydronium in solution. And so this is the portion of the molecule that's going to react, right? It's going to hydrolyze that ester. And we know from the mechanism, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "Because we have H2O and H+, giving us hydronium in solution. And so this is the portion of the molecule that's going to react, right? It's going to hydrolyze that ester. And we know from the mechanism, right? This is the bond. This is the bond that's going to break in our mechanism. And so we can go ahead and draw our products, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And we know from the mechanism, right? This is the bond. This is the bond that's going to break in our mechanism. And so we can go ahead and draw our products, right? So we're going to break that bond. And the whole left portion, right? So let me go ahead and draw that."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so we can go ahead and draw our products, right? So we're going to break that bond. And the whole left portion, right? So let me go ahead and draw that. This whole left portion here, right? Is going to form our carboxylic acid. So let's go ahead and draw in our carboxylic acid, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that. This whole left portion here, right? Is going to form our carboxylic acid. So let's go ahead and draw in our carboxylic acid, right? As one of our products. So we have our ring, right? We have our carboxylic acid."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in our carboxylic acid, right? As one of our products. So we have our ring, right? We have our carboxylic acid. So this OH on our carboxylic acid, right? So this OH came from H2O in the mechanism, right? And then we have this other OH, which produces, of course, salicylic acid."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "We have our carboxylic acid. So this OH on our carboxylic acid, right? So this OH came from H2O in the mechanism, right? And then we have this other OH, which produces, of course, salicylic acid. And then we're thinking about our alcohol product. Let me go ahead and use green for this, right? So this is going to leave, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And then we have this other OH, which produces, of course, salicylic acid. And then we're thinking about our alcohol product. Let me go ahead and use green for this, right? So this is going to leave, right? We can see the, after you protonate, right? You get loss of that as your alcohol. And so if we just add a proton onto this oxygen here, we can see that we would form methanol as our other product."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to leave, right? We can see the, after you protonate, right? You get loss of that as your alcohol. And so if we just add a proton onto this oxygen here, we can see that we would form methanol as our other product. So if I go ahead and draw in the methanol here, right? We have our two products. We have salicylic acid and we have methanol."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so if we just add a proton onto this oxygen here, we can see that we would form methanol as our other product. So if I go ahead and draw in the methanol here, right? We have our two products. We have salicylic acid and we have methanol. So acid-catalyzed ester hydrolysis. This reaction is at equilibrium technically. And so you could do things like push the equilibrium to the right."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "We have salicylic acid and we have methanol. So acid-catalyzed ester hydrolysis. This reaction is at equilibrium technically. And so you could do things like push the equilibrium to the right. And if you remember from the Fischer esterification video, this is what we used to make our wintergreen, right? We used methanol and we used salicylic acid to produce our wintergreen. And so everything depends on reaction conditions in terms of shifting the equilibrium."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so you could do things like push the equilibrium to the right. And if you remember from the Fischer esterification video, this is what we used to make our wintergreen, right? We used methanol and we used salicylic acid to produce our wintergreen. And so everything depends on reaction conditions in terms of shifting the equilibrium. Let's look at another practice problem here. So once again, we have our ester, right? And we have water."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so everything depends on reaction conditions in terms of shifting the equilibrium. Let's look at another practice problem here. So once again, we have our ester, right? And we have water. And I drew the water in a little bit of a weird way, and I'll show you why in a second. So acid-catalyzed ester hydrolysis. Once again, we think about what bond is gonna break."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And we have water. And I drew the water in a little bit of a weird way, and I'll show you why in a second. So acid-catalyzed ester hydrolysis. Once again, we think about what bond is gonna break. This bond is gonna break. And the left side is going to turn into our carboxylic acid, right? So we can go ahead and draw our carboxylic acid."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we think about what bond is gonna break. This bond is gonna break. And the left side is going to turn into our carboxylic acid, right? So we can go ahead and draw our carboxylic acid. So we would have this portion of the molecule. So our carboxylic acid like that. And once again, this OH, let me go ahead and highlight it, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and draw our carboxylic acid. So we would have this portion of the molecule. So our carboxylic acid like that. And once again, this OH, let me go ahead and highlight it, right? So this OH right here in our carboxylic acid came from our water molecule, right? So we could think about losing a proton off of water, and then we form our carboxylic acid on the right. Our other product, right?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And once again, this OH, let me go ahead and highlight it, right? So this OH right here in our carboxylic acid came from our water molecule, right? So we could think about losing a proton off of water, and then we form our carboxylic acid on the right. Our other product, right? Let's be consistent and stick with green here. So this is the portion that's going to turn into our alcohol. So we think about adding a proton onto that oxygen."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "Our other product, right? Let's be consistent and stick with green here. So this is the portion that's going to turn into our alcohol. So we think about adding a proton onto that oxygen. And once again, we have methanol as our other product. So acid-catalyzed ester hydrolysis. Now the reason I drew the water molecule in a weird way is because what would happen if instead of this hydrogen, what would happen if we had an R group, an alkyl group?"}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So we think about adding a proton onto that oxygen. And once again, we have methanol as our other product. So acid-catalyzed ester hydrolysis. Now the reason I drew the water molecule in a weird way is because what would happen if instead of this hydrogen, what would happen if we had an R group, an alkyl group? Well that would change this hydrogen into an alkyl group, and then we would form an ester as our product. So we'd be starting with one ester, and we'd be turning it into another ester. And that is called transesterification."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "Now the reason I drew the water molecule in a weird way is because what would happen if instead of this hydrogen, what would happen if we had an R group, an alkyl group? Well that would change this hydrogen into an alkyl group, and then we would form an ester as our product. So we'd be starting with one ester, and we'd be turning it into another ester. And that is called transesterification. So pretty much the same reaction that we've been talking about, except we wouldn't get a carboxylic acid as one of our products, right? We would get another ester. And so let's go ahead and do this again, starting with the same ester, but this time we're going to use butanol instead of water."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And that is called transesterification. So pretty much the same reaction that we've been talking about, except we wouldn't get a carboxylic acid as one of our products, right? We would get another ester. And so let's go ahead and do this again, starting with the same ester, but this time we're going to use butanol instead of water. So once again, we can think about losing, we're going to break that bond, and we're going to lose this proton. And then we can stick those portions of the molecule together. So if we stick this together with this, we can see the identity of the other ester that will form."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and do this again, starting with the same ester, but this time we're going to use butanol instead of water. So once again, we can think about losing, we're going to break that bond, and we're going to lose this proton. And then we can stick those portions of the molecule together. So if we stick this together with this, we can see the identity of the other ester that will form. So let's go ahead and draw in our product. So we would have an oxygen here, and then we'd have four carbons. One, two, three, four."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "So if we stick this together with this, we can see the identity of the other ester that will form. So let's go ahead and draw in our product. So we would have an oxygen here, and then we'd have four carbons. One, two, three, four. So these four carbons from butanol, which form our ester. Our other product, once again, we see this portion over here, that would give us methanol. So once again, we have methanol as our product."}, {"video_title": "Acid-catalyzed ester hydrolysis Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four. So these four carbons from butanol, which form our ester. Our other product, once again, we see this portion over here, that would give us methanol. So once again, we have methanol as our product. And even though everything is at equilibrium, methanol is a very low boiling point, so we could boil off the methanol and shift the equilibrium to the right to make more of our product. So we can see, once again, we're starting with an ester of methanol, and we're converting it into an ester of butanol, simply by changing the reaction conditions in a transesterification reaction. So this is a pretty useful reaction in industry."}, {"video_title": "Supernova clarification Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The supernova was first observed 1,000 years ago. Or the light from the first explosion was observed by astronomers, we believe, 1,000 years ago. But we have to be very clear here. Because the Crab Nebula, at its core, is roughly 6,500 light years away, even this light, even this image we see right here, is that nebula as it was 6,500 years ago. And so the supernova itself, if we think about when it actually occurred, it actually must have occurred about 7,500 years ago. So it must have occurred about 7,500 years ago. And that first light from that first explosion, from that first energetic event, reached us about 1,000 years ago."}, {"video_title": "Supernova clarification Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because the Crab Nebula, at its core, is roughly 6,500 light years away, even this light, even this image we see right here, is that nebula as it was 6,500 years ago. And so the supernova itself, if we think about when it actually occurred, it actually must have occurred about 7,500 years ago. So it must have occurred about 7,500 years ago. And that first light from that first explosion, from that first energetic event, reached us about 1,000 years ago. So it took 6,500 years to get to us and reached us 1,000 years ago. So first light 1,000 years ago. I just want to make that clear."}, {"video_title": "Supernova clarification Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that first light from that first explosion, from that first energetic event, reached us about 1,000 years ago. So it took 6,500 years to get to us and reached us 1,000 years ago. So first light 1,000 years ago. I just want to make that clear. It might have been obvious to some of you all. But always important to think about it. When I said 1,000 years ago, I really should have said it was first observed."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Over many videos now, we've been talking about how every interstellar object is moving away from Earth. And we've also been talking about how the further something is away from Earth, the faster it's moving. What I want to do in this video is to put a few numbers behind it, or even better conceptualize what we've been talking about. So one way to think about it is that if at an early stage in the universe, I were to pick some points. So that's one point, another point, another point, another point. Let me just pick 9 points so that I have a proper grid. So this is at an early stage in the universe."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So one way to think about it is that if at an early stage in the universe, I were to pick some points. So that's one point, another point, another point, another point. Let me just pick 9 points so that I have a proper grid. So this is at an early stage in the universe. If we fast forward a few billion years, and I'm clearly not drawing it to scale, all of these points have all moved away from each other. So this point is over here. Actually, let me draw another row, or actually another column, just to make it clear."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is at an early stage in the universe. If we fast forward a few billion years, and I'm clearly not drawing it to scale, all of these points have all moved away from each other. So this point is over here. Actually, let me draw another row, or actually another column, just to make it clear. So if we fast forward a few billion years, the universe has expanded, and so everything has moved away from everything. Let me color code it a little bit. Let me make this point magenta."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, let me draw another row, or actually another column, just to make it clear. So if we fast forward a few billion years, the universe has expanded, and so everything has moved away from everything. Let me color code it a little bit. Let me make this point magenta. So this point, the magenta point, is now here. This green point has now moved away from the magenta point. And now this blue point has now moved away from the magenta point in that direction."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me make this point magenta. So this point, the magenta point, is now here. This green point has now moved away from the magenta point. And now this blue point has now moved away from the magenta point in that direction. And we could keep going. This yellow point is maybe over here now. I think you get the general idea."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now this blue point has now moved away from the magenta point in that direction. And we could keep going. This yellow point is maybe over here now. I think you get the general idea. And I'll just draw the other yellow points. So they've all moved away from each other. So there's no center here."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I think you get the general idea. And I'll just draw the other yellow points. So they've all moved away from each other. So there's no center here. Everything is just expanding away from things next to it. And what you could see here is not only did this thing expand away from this, but this thing expanded away from this even further, because it had this expansion plus this expansion. Or another way to think about it is the apparent velocity with which something is expanding is going to be proportional to how far it is, because every point in between is also expanding away."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So there's no center here. Everything is just expanding away from things next to it. And what you could see here is not only did this thing expand away from this, but this thing expanded away from this even further, because it had this expansion plus this expansion. Or another way to think about it is the apparent velocity with which something is expanding is going to be proportional to how far it is, because every point in between is also expanding away. And just to review a little bit of the visualization of this, one way to think of this, if you think of the universe as an infinite flat sheet, you can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretching material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or another way to think about it is the apparent velocity with which something is expanding is going to be proportional to how far it is, because every point in between is also expanding away. And just to review a little bit of the visualization of this, one way to think of this, if you think of the universe as an infinite flat sheet, you can imagine that we're just taking a sheet of, I don't know, some type of sheet of stretching material and just stretching it out. We're just stretching it out. That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out so it has no boundaries, but we're still stretching it out. Another way to visualize it, and this is what we did earlier on, is you could imagine that the universe is the three-dimensional surface of a four-dimensional sphere or the three-dimensional surface of a hypersphere. So at an early stage in the universe, the sphere looked like this."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's if we kind of imagine a more infinite universe that just goes off in every direction. We're just stretching that infinite sheet out so it has no boundaries, but we're still stretching it out. Another way to visualize it, and this is what we did earlier on, is you could imagine that the universe is the three-dimensional surface of a four-dimensional sphere or the three-dimensional surface of a hypersphere. So at an early stage in the universe, the sphere looked like this. And these points here, that magenta point is right over here. The green point is right over there. Then we had the blue point up here."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So at an early stage in the universe, the sphere looked like this. And these points here, that magenta point is right over here. The green point is right over there. Then we had the blue point up here. And then let me just draw the rest of the yellow points. And the yellow points are here. They're all on the surface of this sphere."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then we had the blue point up here. And then let me just draw the rest of the yellow points. And the yellow points are here. They're all on the surface of this sphere. Obviously, I'm only dealing with two dimensions right now. It's nearly impossible, or maybe impossible, to imagine a three-dimensional surface of a four-dimensional sphere. But the analogy holds."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're all on the surface of this sphere. Obviously, I'm only dealing with two dimensions right now. It's nearly impossible, or maybe impossible, to imagine a three-dimensional surface of a four-dimensional sphere. But the analogy holds. If this is the surface of a balloon or the surface of a bubble, if the bubble were to expand over a few billion years, and once again, not drawn to scale, so now we have a bigger bubble here, this part of the surface is all going to expand. So once again, you have your magenta, you have your blue dot, you have your green dot right over here. And then let me just draw the rest in yellow."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the analogy holds. If this is the surface of a balloon or the surface of a bubble, if the bubble were to expand over a few billion years, and once again, not drawn to scale, so now we have a bigger bubble here, this part of the surface is all going to expand. So once again, you have your magenta, you have your blue dot, you have your green dot right over here. And then let me just draw the rest in yellow. So they will have all expanded away from each other on the surface of this sphere. And just to make it clear that this is a sphere, let me draw some contour lines. So this is a contour line."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then let me just draw the rest in yellow. So they will have all expanded away from each other on the surface of this sphere. And just to make it clear that this is a sphere, let me draw some contour lines. So this is a contour line. Just to make it clear that we are on the surface of an actual sphere. Now with that out of the way, let's think about how fast, or what is the apparent velocity with which things are moving away. And remember, we're going to have to say not only how far things are moving away, but we're going to say how far they're moving away from, if the observer is us, depending on how far they already are."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is a contour line. Just to make it clear that we are on the surface of an actual sphere. Now with that out of the way, let's think about how fast, or what is the apparent velocity with which things are moving away. And remember, we're going to have to say not only how far things are moving away, but we're going to say how far they're moving away from, if the observer is us, depending on how far they already are. So what we're going to do, what we could say is, let me write this down, all objects moving away from each other, and the velocity, and the apparent velocity, the apparent relative velocity is proportional to distance. And what I've just written down here, this is why I wrote it down, this is a rephrasing of essentially Hubble's Law. And he came up with this by just observing that when he looks, especially the further out he looks, the more red-shifted objects are."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And remember, we're going to have to say not only how far things are moving away, but we're going to say how far they're moving away from, if the observer is us, depending on how far they already are. So what we're going to do, what we could say is, let me write this down, all objects moving away from each other, and the velocity, and the apparent velocity, the apparent relative velocity is proportional to distance. And what I've just written down here, this is why I wrote it down, this is a rephrasing of essentially Hubble's Law. And he came up with this by just observing that when he looks, especially the further out he looks, the more red-shifted objects are. And not only were they moving faster and faster away from Earth, but they seemed to be moving faster and faster away from each other. So this is just a restating of Hubble's Law. Or another way to say it is, from any point, let's say from the Earth, the velocity that something appears to be moving is going to be some constant times the distance that it is away from the observer."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And he came up with this by just observing that when he looks, especially the further out he looks, the more red-shifted objects are. And not only were they moving faster and faster away from Earth, but they seemed to be moving faster and faster away from each other. So this is just a restating of Hubble's Law. Or another way to say it is, from any point, let's say from the Earth, the velocity that something appears to be moving is going to be some constant times the distance that it is away from the observer. In this case, we are the observer. And we put this little zero, so this h here is called Hubble's Constant. And it's a very non-constant constant, because this constant will change depending on where we are in the evolution of the universe."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or another way to say it is, from any point, let's say from the Earth, the velocity that something appears to be moving is going to be some constant times the distance that it is away from the observer. In this case, we are the observer. And we put this little zero, so this h here is called Hubble's Constant. And it's a very non-constant constant, because this constant will change depending on where we are in the evolution of the universe. So we put this little zero here, this little sub-zero right over here, to show that this is Hubble's Constant right now. And when we talk about distance, we're talking about the proper distance right now. And this has to be very important, because that proper distance is constantly changing as the universe expands."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's a very non-constant constant, because this constant will change depending on where we are in the evolution of the universe. So we put this little zero here, this little sub-zero right over here, to show that this is Hubble's Constant right now. And when we talk about distance, we're talking about the proper distance right now. And this has to be very important, because that proper distance is constantly changing as the universe expands. So the now will actually change slightly from the beginning of this video to the end of this video. But we can roughly say in our current period of time. And when we say proper distance, we're talking about if you actually had rulers, and if you were to just lay them down instantaneously."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this has to be very important, because that proper distance is constantly changing as the universe expands. So the now will actually change slightly from the beginning of this video to the end of this video. But we can roughly say in our current period of time. And when we say proper distance, we're talking about if you actually had rulers, and if you were to just lay them down instantaneously. Obviously we can't do something like that, but we can imagine doing something like that. So that's what we're talking about. So just to give a sense of, or do a little bit of math of how fast things are actually moving apart, the current Hubble Constant is 70.6 plus or minus 3.1."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when we say proper distance, we're talking about if you actually had rulers, and if you were to just lay them down instantaneously. Obviously we can't do something like that, but we can imagine doing something like that. So that's what we're talking about. So just to give a sense of, or do a little bit of math of how fast things are actually moving apart, the current Hubble Constant is 70.6 plus or minus 3.1. So we have observed some variation here. There is, I guess, some error to our actual measurements. So that's kilometers per second per megaparsec."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So just to give a sense of, or do a little bit of math of how fast things are actually moving apart, the current Hubble Constant is 70.6 plus or minus 3.1. So we have observed some variation here. There is, I guess, some error to our actual measurements. So that's kilometers per second per megaparsec. And remember, a parsec is roughly 3.2, 3.3 light years. So another way to think about it is, if this is where we are in the universe right now, and if this object right over here, if this distance right over here is 1 megaparsec, so 1 million parsecs, or 3.26 million light years from Earth, so this is roughly, just so we have a sense, this is 3.26 roughly, 3.26 million light years from Earth, then this object will appear to be moving away, although it's not moving in space. Just the space that it's in is stretching in such a way that it looks to be moving at 70, it looks like it's moving at 70, based on its redshift, 70.6 kilometers per second away from us."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's kilometers per second per megaparsec. And remember, a parsec is roughly 3.2, 3.3 light years. So another way to think about it is, if this is where we are in the universe right now, and if this object right over here, if this distance right over here is 1 megaparsec, so 1 million parsecs, or 3.26 million light years from Earth, so this is roughly, just so we have a sense, this is 3.26 roughly, 3.26 million light years from Earth, then this object will appear to be moving away, although it's not moving in space. Just the space that it's in is stretching in such a way that it looks to be moving at 70, it looks like it's moving at 70, based on its redshift, 70.6 kilometers per second away from us. So this is a huge velocity. 70.6 kilometers per second. This is a pretty fast velocity, but you have to remember, this is over 1 megaparsec."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just the space that it's in is stretching in such a way that it looks to be moving at 70, it looks like it's moving at 70, based on its redshift, 70.6 kilometers per second away from us. So this is a huge velocity. 70.6 kilometers per second. This is a pretty fast velocity, but you have to remember, this is over 1 megaparsec. The Andromeda Galaxy is not even a megaparsec away. It was about 2.5 million light years, so it's about, I don't know, 0.7 or 0.8 of a megaparsec. So if you look at a point in space a little bit further than the Andromeda Galaxy, it will look to be right now receding at about 70.6 kilometers per second."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is a pretty fast velocity, but you have to remember, this is over 1 megaparsec. The Andromeda Galaxy is not even a megaparsec away. It was about 2.5 million light years, so it's about, I don't know, 0.7 or 0.8 of a megaparsec. So if you look at a point in space a little bit further than the Andromeda Galaxy, it will look to be right now receding at about 70.6 kilometers per second. But what if you were to go twice that distance? If you were to look at something that's almost 7 light years away, 2 megaparsecs away. So if you were to look at this object over here, how fast would that be receding?"}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you look at a point in space a little bit further than the Andromeda Galaxy, it will look to be right now receding at about 70.6 kilometers per second. But what if you were to go twice that distance? If you were to look at something that's almost 7 light years away, 2 megaparsecs away. So if you were to look at this object over here, how fast would that be receding? Well, if you just look at it over here, it's 2 megaparsecs away, so it's going to be twice this. You're just going to multiply its distance, 2 megaparsecs times this. The megaparsecs cancel out."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you were to look at this object over here, how fast would that be receding? Well, if you just look at it over here, it's 2 megaparsecs away, so it's going to be twice this. You're just going to multiply its distance, 2 megaparsecs times this. The megaparsecs cancel out. So 70.6 times 2 is, so it's going to be moving, or it's going to look to be moving. It's not moving in space. Remember, space is just stretching."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The megaparsecs cancel out. So 70.6 times 2 is, so it's going to be moving, or it's going to look to be moving. It's not moving in space. Remember, space is just stretching. So its velocity, its apparent velocity, will be 70.6 times 2, so that's 141.2 kilometers per second. And this is, you know, one question. You say, well, how did Hubble know?"}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, space is just stretching. So its velocity, its apparent velocity, will be 70.6 times 2, so that's 141.2 kilometers per second. And this is, you know, one question. You say, well, how did Hubble know? You know, it's easy to, you could observe the redshift of objects moving away from us, but how did he know that they were moving away from each other? Well, if you were to look at the redshift of this object and say, wow, that's moving away at 70.6 kilometers per second, and then you were to look at the redshift of this and say, wow, that's moving away from us at 141.2 kilometers per second, then you also know that these two objects are moving away from each other at 70.6 kilometers per second. And we could keep doing this over different distances, but hopefully this gives you a little bit more, a little bit bigger sense of things."}, {"video_title": "Hubble's law Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You say, well, how did Hubble know? You know, it's easy to, you could observe the redshift of objects moving away from us, but how did he know that they were moving away from each other? Well, if you were to look at the redshift of this object and say, wow, that's moving away at 70.6 kilometers per second, and then you were to look at the redshift of this and say, wow, that's moving away from us at 141.2 kilometers per second, then you also know that these two objects are moving away from each other at 70.6 kilometers per second. And we could keep doing this over different distances, but hopefully this gives you a little bit more, a little bit bigger sense of things. And just remember, even though I said this is a huge distance, a megaparsec is further than it is to the Andromeda galaxy. The Andromeda galaxy is the nearest large galaxy to us. There are some smaller galaxies that are closer to us that are kind of satellite galaxies around the Milky Way, but the Andromeda is the nearest large galaxy to us."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "You can make mesylates and tosylates from alcohols. And you might want to do this because mesylates and tosylates are better leaving groups. So if we look at a general reaction to form a tosylate, you would start with an alcohol, and you'd add tosyl chloride and also pyridine, and you would form your tosylate over here on the right. When we look at the mechanism, we start with tosyl chloride. If you focus in on the sulfur here, the sulfur is bonded to two oxygens and a chlorine. And we know that oxygen and chlorine are more electronegative than sulfur, so they're going to withdraw some electron density from that sulfur. And so since sulfur's losing some electron density, the sulfur becomes partially positive, and we have an electrophilic center."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "When we look at the mechanism, we start with tosyl chloride. If you focus in on the sulfur here, the sulfur is bonded to two oxygens and a chlorine. And we know that oxygen and chlorine are more electronegative than sulfur, so they're going to withdraw some electron density from that sulfur. And so since sulfur's losing some electron density, the sulfur becomes partially positive, and we have an electrophilic center. So the sulfur wants electrons. It can get electrons from the alcohol. So a lone pair of electrons on the alcohol here can attack the sulfur."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And so since sulfur's losing some electron density, the sulfur becomes partially positive, and we have an electrophilic center. So the sulfur wants electrons. It can get electrons from the alcohol. So a lone pair of electrons on the alcohol here can attack the sulfur. So a nucleophile attacks the electrophile, and these electrons could kick off onto the chlorine here to form the chloride anion. And let's go ahead and draw what we would make. So we would have an R group."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on the alcohol here can attack the sulfur. So a nucleophile attacks the electrophile, and these electrons could kick off onto the chlorine here to form the chloride anion. And let's go ahead and draw what we would make. So we would have an R group. We would have an oxygen. The oxygen would now be bonded to sulfur. We would also have still a hydrogen attached to that oxygen, and we still have a lone pair of electrons on that oxygen."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So we would have an R group. We would have an oxygen. The oxygen would now be bonded to sulfur. We would also have still a hydrogen attached to that oxygen, and we still have a lone pair of electrons on that oxygen. So the oxygen gets a plus 1 formal charge. The sulfur is still double bonded to this oxygen and to another oxygen, and then we still have our ring attached to the sulfur. So I'll put in our pi electrons here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "We would also have still a hydrogen attached to that oxygen, and we still have a lone pair of electrons on that oxygen. So the oxygen gets a plus 1 formal charge. The sulfur is still double bonded to this oxygen and to another oxygen, and then we still have our ring attached to the sulfur. So I'll put in our pi electrons here. And oops, that was a bad one. Let me fix that. So we have our pi electrons right here, and then a methyl group like that."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So I'll put in our pi electrons here. And oops, that was a bad one. Let me fix that. So we have our pi electrons right here, and then a methyl group like that. So let's go ahead and follow those electrons. So the electrons in magenta right here on the oxygen formed a new bond to the sulfur. So here are those electrons in magenta."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So we have our pi electrons right here, and then a methyl group like that. So let's go ahead and follow those electrons. So the electrons in magenta right here on the oxygen formed a new bond to the sulfur. So here are those electrons in magenta. In the next step, we're going to take the proton off the oxygen here. And so the pyridine is going to function as a base. The lone pair of electrons on nitrogen is going to take this proton, leaving these electrons behind on the oxygen."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So here are those electrons in magenta. In the next step, we're going to take the proton off the oxygen here. And so the pyridine is going to function as a base. The lone pair of electrons on nitrogen is going to take this proton, leaving these electrons behind on the oxygen. So let's go ahead and draw what we would form. We would have our R group. We would have an oxygen, and our oxygen would have now two lone pairs of electrons."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "The lone pair of electrons on nitrogen is going to take this proton, leaving these electrons behind on the oxygen. So let's go ahead and draw what we would form. We would have our R group. We would have an oxygen, and our oxygen would have now two lone pairs of electrons. And we would have our sulfur double bonded to this oxygen, double bonded to this oxygen. And then we would have our ring like this. So let me just sketch that in really quickly."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "We would have an oxygen, and our oxygen would have now two lone pairs of electrons. And we would have our sulfur double bonded to this oxygen, double bonded to this oxygen. And then we would have our ring like this. So let me just sketch that in really quickly. So we would have electrons here, here, here, our methyl group. And let's follow some of those electrons. So the electrons in this bond now end up on the oxygen like that."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So let me just sketch that in really quickly. So we would have electrons here, here, here, our methyl group. And let's follow some of those electrons. So the electrons in this bond now end up on the oxygen like that. And we formed our toluene sulfonate ester, so also called a tosylate. So this is the exact same thing. So this compound and this compound are the same."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in this bond now end up on the oxygen like that. And we formed our toluene sulfonate ester, so also called a tosylate. So this is the exact same thing. So this compound and this compound are the same. The top way is just a way to abbreviate it. And so we formed our tosylate. So one reason to form a tosylate would be to have a better nucleophilic substitution reaction."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So this compound and this compound are the same. The top way is just a way to abbreviate it. And so we formed our tosylate. So one reason to form a tosylate would be to have a better nucleophilic substitution reaction. So let's look at first forming a tosylate from this alcohol over here on the left. And so we know that this carbon is a chiral center. So that is a chiral center."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So one reason to form a tosylate would be to have a better nucleophilic substitution reaction. So let's look at first forming a tosylate from this alcohol over here on the left. And so we know that this carbon is a chiral center. So that is a chiral center. But if we're forming a tosylate, the tosylate forms at this oxygen here. So let's go ahead and draw the product. We would now have, we'd still have a wedge here, because again, the reaction does not occur at the chirality center."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So that is a chiral center. But if we're forming a tosylate, the tosylate forms at this oxygen here. So let's go ahead and draw the product. We would now have, we'd still have a wedge here, because again, the reaction does not occur at the chirality center. The reaction occurs at the oxygen here. So the oxygen would now be bonded. So we'd form a tosylate group, which is a much better leaving group than this OH over here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "We would now have, we'd still have a wedge here, because again, the reaction does not occur at the chirality center. The reaction occurs at the oxygen here. So the oxygen would now be bonded. So we'd form a tosylate group, which is a much better leaving group than this OH over here. So the tosylate's an excellent leaving group for nucleophilic substitution reactions. So if we went ahead and did a nucleophilic substitution reaction, we could add something like sodium bromide, so NA plus and Br minus. So if this was an SN2 type mechanism, the bromide anion would attack this carbon right here, which is a little bit positive."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So we'd form a tosylate group, which is a much better leaving group than this OH over here. So the tosylate's an excellent leaving group for nucleophilic substitution reactions. So if we went ahead and did a nucleophilic substitution reaction, we could add something like sodium bromide, so NA plus and Br minus. So if this was an SN2 type mechanism, the bromide anion would attack this carbon right here, which is a little bit positive. So we have a partially positive carbon right here. And then we get nucleophilic attack from the bromide anion, so it attacks right here. In an SN2 type mechanism, you're going to get inversion of configuration."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So if this was an SN2 type mechanism, the bromide anion would attack this carbon right here, which is a little bit positive. So we have a partially positive carbon right here. And then we get nucleophilic attack from the bromide anion, so it attacks right here. In an SN2 type mechanism, you're going to get inversion of configuration. So you go ahead and draw your product like that. And so the formation of a tosylate just makes this process much easier. We could talk about formation of another good leaving group, and that's a mesylate, so very similar to a tosylate."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "In an SN2 type mechanism, you're going to get inversion of configuration. So you go ahead and draw your product like that. And so the formation of a tosylate just makes this process much easier. We could talk about formation of another good leaving group, and that's a mesylate, so very similar to a tosylate. So if we look at the general reaction, once again we start with an alcohol. This time we add mesyl chloride. And this time triethylamine is the base that we will use to form our mesylate over here on the right."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "We could talk about formation of another good leaving group, and that's a mesylate, so very similar to a tosylate. So if we look at the general reaction, once again we start with an alcohol. This time we add mesyl chloride. And this time triethylamine is the base that we will use to form our mesylate over here on the right. The mechanism is a little bit different from the formation of a tosylate. So let's go ahead and see what happens. First, the triethylamine is going to function as a base and take this proton right here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And this time triethylamine is the base that we will use to form our mesylate over here on the right. The mechanism is a little bit different from the formation of a tosylate. So let's go ahead and see what happens. First, the triethylamine is going to function as a base and take this proton right here. So these electrons are going to remain behind on this carbon. So triethylamine reacts with this mesyl chloride right here. So if we take a proton off, let's go ahead and draw what we would have."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "First, the triethylamine is going to function as a base and take this proton right here. So these electrons are going to remain behind on this carbon. So triethylamine reacts with this mesyl chloride right here. So if we take a proton off, let's go ahead and draw what we would have. We would have our sulfur double bonded to this oxygen, sulfur double bonded to another oxygen, a chlorine right here. And we would now have a carbon bonded to only two hydrogens and a lone pair of electrons on this carbon. So it's a carb anion."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So if we take a proton off, let's go ahead and draw what we would have. We would have our sulfur double bonded to this oxygen, sulfur double bonded to another oxygen, a chlorine right here. And we would now have a carbon bonded to only two hydrogens and a lone pair of electrons on this carbon. So it's a carb anion. So it's a negative 1 formal charge. So let's show those electrons. So these electrons in this bond were left behind on the carbon to form our carb anion."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So it's a carb anion. So it's a negative 1 formal charge. So let's show those electrons. So these electrons in this bond were left behind on the carbon to form our carb anion. So in the next step, these electrons in magenta are going to move in here to form a double bond between the sulfur and the carbon. And that would kick these electrons off onto chlorine to form the chloride anion. And let's go ahead and draw what we would make."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in this bond were left behind on the carbon to form our carb anion. So in the next step, these electrons in magenta are going to move in here to form a double bond between the sulfur and the carbon. And that would kick these electrons off onto chlorine to form the chloride anion. And let's go ahead and draw what we would make. So now we would have sulfur. Sulfur would be double bonded to an oxygen. Sulfur is double bonded to another oxygen."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and draw what we would make. So now we would have sulfur. Sulfur would be double bonded to an oxygen. Sulfur is double bonded to another oxygen. And now there's a double bond between sulfur and this carbon. This carbon is bonded to two hydrogens like that. Once again, we can think about sulfur as being electrophilic because this sulfur right here is bonded to these oxygens, which are more electronegative."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "Sulfur is double bonded to another oxygen. And now there's a double bond between sulfur and this carbon. This carbon is bonded to two hydrogens like that. Once again, we can think about sulfur as being electrophilic because this sulfur right here is bonded to these oxygens, which are more electronegative. They're going to withdraw electron density from that sulfur, leaving that sulfur partially positive. And so our electrophilic center, once again, we're going to have our alcohol function as a nucleophile. So a lone pair of electrons on our alcohol are going to go all the way to here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we can think about sulfur as being electrophilic because this sulfur right here is bonded to these oxygens, which are more electronegative. They're going to withdraw electron density from that sulfur, leaving that sulfur partially positive. And so our electrophilic center, once again, we're going to have our alcohol function as a nucleophile. So a lone pair of electrons on our alcohol are going to go all the way to here. They're going to attack our electrophile. And when they do that, they would push these electrons back off onto the carbon. So let's get a little more space down here so we can draw what happens."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on our alcohol are going to go all the way to here. They're going to attack our electrophile. And when they do that, they would push these electrons back off onto the carbon. So let's get a little more space down here so we can draw what happens. So this is a sulfene right here. So the alcohol attacks the sulfene, nucleophile, electrophile. And let's go ahead and draw what we would make."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So let's get a little more space down here so we can draw what happens. So this is a sulfene right here. So the alcohol attacks the sulfene, nucleophile, electrophile. And let's go ahead and draw what we would make. So now we would have an R group bonded to an oxygen. We would have a hydrogen. And this is the oxygen from the alcohol, which attacked the sulfur."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and draw what we would make. So now we would have an R group bonded to an oxygen. We would have a hydrogen. And this is the oxygen from the alcohol, which attacked the sulfur. So now there's a bond between the oxygen and the sulfur. There's still a lone pair of electrons on that oxygen, giving it a plus 1 formal charge. And the sulfur is double bonded to this oxygen, double bonded to this oxygen, and bonded to this carbon."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And this is the oxygen from the alcohol, which attacked the sulfur. So now there's a bond between the oxygen and the sulfur. There's still a lone pair of electrons on that oxygen, giving it a plus 1 formal charge. And the sulfur is double bonded to this oxygen, double bonded to this oxygen, and bonded to this carbon. This carbon has two hydrogens on it. It also has a lone pair of electrons. So that's a negative 1 formal charge."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And the sulfur is double bonded to this oxygen, double bonded to this oxygen, and bonded to this carbon. This carbon has two hydrogens on it. It also has a lone pair of electrons. So that's a negative 1 formal charge. So once again, let's identify those electrons. So these electrons right here moved off onto the carbon to form our carbanion. And we could identify some of the electrons on our alcohol, too."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So that's a negative 1 formal charge. So once again, let's identify those electrons. So these electrons right here moved off onto the carbon to form our carbanion. And we could identify some of the electrons on our alcohol, too. So let's say let's make it red here. These electrons right here on the oxygen, those are the ones that form the bond between the oxygen and the sulfur. So you could say that those electrons are these electrons in here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And we could identify some of the electrons on our alcohol, too. So let's say let's make it red here. These electrons right here on the oxygen, those are the ones that form the bond between the oxygen and the sulfur. So you could say that those electrons are these electrons in here. And so in the last step of our mechanism, the carbanion is going to function as a base. And this lone pair of electrons here is going to take this proton, leaving these electrons behind on our oxygen. So let's go ahead and draw what we would make."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So you could say that those electrons are these electrons in here. And so in the last step of our mechanism, the carbanion is going to function as a base. And this lone pair of electrons here is going to take this proton, leaving these electrons behind on our oxygen. So let's go ahead and draw what we would make. So we would have an R group over here. We would have an oxygen. And the oxygen would have two lone pairs of electrons."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw what we would make. So we would have an R group over here. We would have an oxygen. And the oxygen would have two lone pairs of electrons. The oxygen is bonded to a sulfur. The sulfur is double bonded to this oxygen, double bonded to this oxygen. And then we would have a CH3 group over here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And the oxygen would have two lone pairs of electrons. The oxygen is bonded to a sulfur. The sulfur is double bonded to this oxygen, double bonded to this oxygen. And then we would have a CH3 group over here. So a CH3 group. Because this carbon, let me go ahead and identify it in magenta. This carbon picked up this proton here."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have a CH3 group over here. So a CH3 group. Because this carbon, let me go ahead and identify it in magenta. This carbon picked up this proton here. So you could say it's that one if you wanted to. And we formed our mesylate. So this is the exact same thing."}, {"video_title": "Preparation of mesylates and tosylates Organic chemistry Khan Academy.mp3", "Sentence": "This carbon picked up this proton here. So you could say it's that one if you wanted to. And we formed our mesylate. So this is the exact same thing. We could just abbreviate it here. We could say it's RO and then an MS here. So that's how to form mesylates and tosylates, which are excellent leaving groups."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's really just the idea that the surface of the earth is made up of a bunch of these rigid plates. So it's broken up into a bunch of rigid plates. And these rigid plates move relative to each other. They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here, I have a picture I got off of Wikipedia of the actual plates."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They move relative to each other and take everything that's on them for a ride. And the things that are on them include the continents. So it literally is talking about the movement of these plates. And over here, I have a picture I got off of Wikipedia of the actual plates. And over here, you have the Pacific plate. Let me do that in a darker color. You have a Pacific plate."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And over here, I have a picture I got off of Wikipedia of the actual plates. And over here, you have the Pacific plate. Let me do that in a darker color. You have a Pacific plate. You have a Nazca plate. You have a South American plate. I could keep going on."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have a Pacific plate. You have a Nazca plate. You have a South American plate. I could keep going on. You have an Antarctic plate. It's actually, obviously, whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative to, say, North America or South America."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I could keep going on. You have an Antarctic plate. It's actually, obviously, whenever you do a projection onto two dimensions of a surface of a sphere, the stuff at the bottom and the top look much bigger than they actually are. Antarctica isn't this big relative to, say, North America or South America. It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic plate, North American plate. And you can see that they're actually moving relative to each other."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Antarctica isn't this big relative to, say, North America or South America. It's just that we've had to stretch it out to fill up the rectangle. But that's the Antarctic plate, North American plate. And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here, the Nazca plate and the Pacific plate are moving away from each other. New land is forming here."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you can see that they're actually moving relative to each other. And that's what these arrows are depicting. You see right over here, the Nazca plate and the Pacific plate are moving away from each other. New land is forming here. We'll talk more about that in other videos. You see right over here, in the middle of the Atlantic Ocean, the African plate and the South American plate meet each other. And they're moving away from each other, which means that new land, more plate material, I guess you could say, is somehow being created right here."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "New land is forming here. We'll talk more about that in other videos. You see right over here, in the middle of the Atlantic Ocean, the African plate and the South American plate meet each other. And they're moving away from each other, which means that new land, more plate material, I guess you could say, is somehow being created right here. And we'll talk about that in future videos in pushing these two plates apart. Now, before we go into the evidence for plate tonics, or even some of the more details about how plates are created, and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. Because sometimes people call them crustal plates, and that's not exactly right."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they're moving away from each other, which means that new land, more plate material, I guess you could say, is somehow being created right here. And we'll talk about that in future videos in pushing these two plates apart. Now, before we go into the evidence for plate tonics, or even some of the more details about how plates are created, and some theories as to why the plates might move, what I want to do is get a little bit of the terminology of plate tectonics out of the way. Because sometimes people call them crustal plates, and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth, and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. When I talk about chemical layers, I'm talking about what are the constituents of the different layers."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because sometimes people call them crustal plates, and that's not exactly right. And to show you the difference, what I want to do is show you two different ways of classifying the different layers of the Earth, and then think about how they might relate to each other. So what you traditionally see, and actually I've made a video that goes into a lot more detail of this, is a breakdown of the chemical layers of the Earth. When I talk about chemical layers, I'm talking about what are the constituents of the different layers. So when you talk of it in this term, the topmost layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When I talk about chemical layers, I'm talking about what are the constituents of the different layers. So when you talk of it in this term, the topmost layer, which is the thinnest layer, is the crust. Then below that is the mantle. Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the outer, the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that you have the mantle."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, let me show you the whole Earth, although I'm not going to draw it to scale. So if I were to draw the crust, the crust is the outer, the thinnest outer layer of the Earth. You can imagine the blue line itself is the crust. Then below that you have the mantle. So everything between the blue and the orange line is this over here is the mantle. Let me label the crust. The crust you can literally view as the actual blue pixels over here."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then below that you have the mantle. So everything between the blue and the orange line is this over here is the mantle. Let me label the crust. The crust you can literally view as the actual blue pixels over here. And then inside of the mantle you have the core. And when you do this very high level division, these are chemical divisions. This is saying that the crust is made up of different types of elements."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The crust you can literally view as the actual blue pixels over here. And then inside of the mantle you have the core. And when you do this very high level division, these are chemical divisions. This is saying that the crust is made up of different types of elements. Its makeup is different than the stuff that's in the mantle, which is made up of different things than what's inside of the core. It's not describing the mechanical properties of it. And when I talk about mechanical properties, I'm talking about whether something is."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is saying that the crust is made up of different types of elements. Its makeup is different than the stuff that's in the mantle, which is made up of different things than what's inside of the core. It's not describing the mechanical properties of it. And when I talk about mechanical properties, I'm talking about whether something is. So mechanical properties are whether something is solid and rigid. Or maybe it's so hot and melted, it's kind of a magma or kind of a plastic solid. So then this would be the most brittle stuff."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when I talk about mechanical properties, I'm talking about whether something is. So mechanical properties are whether something is solid and rigid. Or maybe it's so hot and melted, it's kind of a magma or kind of a plastic solid. So then this would be the most brittle stuff. If it gets warmed up, if rock starts to melt a little bit, then you have something like a magma, or you can view it as a deformable or a plastic solid. When we talk about plastic, I'm not talking about the stuff that the case of your cell phone is made up. I'm talking about it's deformable."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So then this would be the most brittle stuff. If it gets warmed up, if rock starts to melt a little bit, then you have something like a magma, or you can view it as a deformable or a plastic solid. When we talk about plastic, I'm not talking about the stuff that the case of your cell phone is made up. I'm talking about it's deformable. This rock is deformable because it's so hot and it's somewhat melted. It has somewhat, it kind of behaves like a fluid. It actually does behave like a fluid, but it's much more viscous."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'm talking about it's deformable. This rock is deformable because it's so hot and it's somewhat melted. It has somewhat, it kind of behaves like a fluid. It actually does behave like a fluid, but it's much more viscous. It's much thicker and slower moving than what we would normally associate with a fluid like water. So this is viscous. This is a viscous fluid."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It actually does behave like a fluid, but it's much more viscous. It's much thicker and slower moving than what we would normally associate with a fluid like water. So this is viscous. This is a viscous fluid. And then the most fluid would, of course, be the liquid state. This is what we mean when we talk about the mechanical properties. And when you look at this division over here, the crust is solid."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is a viscous fluid. And then the most fluid would, of course, be the liquid state. This is what we mean when we talk about the mechanical properties. And when you look at this division over here, the crust is solid. The mantle actually has some parts of it that are solid. So the uppermost part of the mantle is solid. Then below that, it has a kind of, the rest of the mantle is kind of in this magma, this deformable, somewhat fluid state."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when you look at this division over here, the crust is solid. The mantle actually has some parts of it that are solid. So the uppermost part of the mantle is solid. Then below that, it has a kind of, the rest of the mantle is kind of in this magma, this deformable, somewhat fluid state. And depending on what depth you go into the mantle, there are kind of different levels of fluidity. And then the core, the outer layer of the core, the outer core is liquid because the temperature is so high. The inner core is made up of the same things and the temperature is even higher."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then below that, it has a kind of, the rest of the mantle is kind of in this magma, this deformable, somewhat fluid state. And depending on what depth you go into the mantle, there are kind of different levels of fluidity. And then the core, the outer layer of the core, the outer core is liquid because the temperature is so high. The inner core is made up of the same things and the temperature is even higher. But since the pressure is so high, it's actually solid. So that's why the mantle, crust, and core differentiations don't tell you about mechanical, whether it's solid, whether it's magma, or whether it's really a liquid. It just really tells you what the makeup is."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The inner core is made up of the same things and the temperature is even higher. But since the pressure is so high, it's actually solid. So that's why the mantle, crust, and core differentiations don't tell you about mechanical, whether it's solid, whether it's magma, or whether it's really a liquid. It just really tells you what the makeup is. Now to think about the makeup, and this is important for plate tectonics. Because when we talk about these plates, we're not talking about just the crust. We're talking about the outer rigid layer."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It just really tells you what the makeup is. Now to think about the makeup, and this is important for plate tectonics. Because when we talk about these plates, we're not talking about just the crust. We're talking about the outer rigid layer. When we talk about that, let me just zoom in a little bit. Let me just zoom in. Let's say we zoomed in right over there."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking about the outer rigid layer. When we talk about that, let me just zoom in a little bit. Let me just zoom in. Let's say we zoomed in right over there. So now we have the crust zoomed in. This right here is the crust. And then everything below here, we're actually talking about the upper mantle."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say we zoomed in right over there. So now we have the crust zoomed in. This right here is the crust. And then everything below here, we're actually talking about the upper mantle. So we're talking about, we haven't gotten too deep in the mantle right here. So that's why we call it the upper mantle. Now right below the crust, the mantle is cool enough that it is also in real solid form."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then everything below here, we're actually talking about the upper mantle. So we're talking about, we haven't gotten too deep in the mantle right here. So that's why we call it the upper mantle. Now right below the crust, the mantle is cool enough that it is also in real solid form. So this right here is solid mantle. And when we talk about the plates, we're actually talking about the outer solid layer. So that includes both the crust and the solid part of the mantle."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now right below the crust, the mantle is cool enough that it is also in real solid form. So this right here is solid mantle. And when we talk about the plates, we're actually talking about the outer solid layer. So that includes both the crust and the solid part of the mantle. And we call that the lithosphere. When people talk about plate tectonics, they shouldn't say crustal plates. They should call these lithospheric plates."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that includes both the crust and the solid part of the mantle. And we call that the lithosphere. When people talk about plate tectonics, they shouldn't say crustal plates. They should call these lithospheric plates. Lithospheric. And then below the lithosphere, you have the least viscous part of the mantle. Because the temperature is high enough for the rock to melt, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They should call these lithospheric plates. Lithospheric. And then below the lithosphere, you have the least viscous part of the mantle. Because the temperature is high enough for the rock to melt, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other. Although still pretty viscous. It's still a magma. So this is still kind of in its magma state."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because the temperature is high enough for the rock to melt, but the pressure isn't so large as what will happen when you go into the lower part of the mantle that the fluid can actually kind of move past each other. Although still pretty viscous. It's still a magma. So this is still kind of in its magma state. And this fluid part of the mantle, we can't quite call it a liquid yet, but over large periods of time, it does have fluid properties. Essentially, the lithosphere is kind of riding on top of. We call this the asthenosphere."}, {"video_title": "Plate tectonics Difference between crust and lithosphere Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is still kind of in its magma state. And this fluid part of the mantle, we can't quite call it a liquid yet, but over large periods of time, it does have fluid properties. Essentially, the lithosphere is kind of riding on top of. We call this the asthenosphere. So when we talk about the lithosphere and asthenosphere, we're really talking about mechanical layers. The outer layer, the solid layer is the lithosphere. The more fluid layer right below that's the asthenosphere."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at some practice IR spectra. So here we have three molecules, a carboxylic acid, an alcohol, and an amine. And below there's an IR spectrum of one of these molecules. So let's figure out which molecule has this IR spectrum. So we could draw a line around 1500 and ignore the stuff to the right and focus in on the diagnostic region. And in here is your double bond region. And I don't see a signal at all in the double bond region."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's figure out which molecule has this IR spectrum. So we could draw a line around 1500 and ignore the stuff to the right and focus in on the diagnostic region. And in here is your double bond region. And I don't see a signal at all in the double bond region. I certainly don't see a very strong carbonyl stretch. And so the carboxylic acid is out. So I don't see any kind of a carbonyl stretch in here."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And I don't see a signal at all in the double bond region. I certainly don't see a very strong carbonyl stretch. And so the carboxylic acid is out. So I don't see any kind of a carbonyl stretch in here. We do see some signals over here to the left in the bond to hydrogen region. And so I could draw a line about 3000. And I know below that, we're talking about a carbon-hydrogen bond stretch where you have an sp3 hybridized carbon."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I don't see any kind of a carbonyl stretch in here. We do see some signals over here to the left in the bond to hydrogen region. And so I could draw a line about 3000. And I know below that, we're talking about a carbon-hydrogen bond stretch where you have an sp3 hybridized carbon. That doesn't help us out here at all. But this other signal does. So we have another signal."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And I know below that, we're talking about a carbon-hydrogen bond stretch where you have an sp3 hybridized carbon. That doesn't help us out here at all. But this other signal does. So we have another signal. So it's centered on a higher wave number. And it's extremely broad. So whenever you see that, you should think to yourself hydrogen bonding."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have another signal. So it's centered on a higher wave number. And it's extremely broad. So whenever you see that, you should think to yourself hydrogen bonding. And this is due to an OH bond stretch. So immediately we know that we must be talking about an alcohol here. A carboxylic acid has a similar OH bond stretch, so it has a broad signal due to that."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So whenever you see that, you should think to yourself hydrogen bonding. And this is due to an OH bond stretch. So immediately we know that we must be talking about an alcohol here. A carboxylic acid has a similar OH bond stretch, so it has a broad signal due to that. But there's no carbonyl, so it couldn't possibly be this molecule. It also couldn't possibly be the amine. Because even though we have nitrogen-hydrogen bond, a nitrogen-hydrogen bond stretch is going to be in a similar region, we would expect two signals for this."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "A carboxylic acid has a similar OH bond stretch, so it has a broad signal due to that. But there's no carbonyl, so it couldn't possibly be this molecule. It also couldn't possibly be the amine. Because even though we have nitrogen-hydrogen bond, a nitrogen-hydrogen bond stretch is going to be in a similar region, we would expect two signals for this. We would expect a symmetric stretch signal and an asymmetric stretching signal. And it wouldn't be as broad as what we're talking about here for the alcohol. So it's definitely not the amine."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Because even though we have nitrogen-hydrogen bond, a nitrogen-hydrogen bond stretch is going to be in a similar region, we would expect two signals for this. We would expect a symmetric stretch signal and an asymmetric stretching signal. And it wouldn't be as broad as what we're talking about here for the alcohol. So it's definitely not the amine. So this spectrum is the alcohol. Let's look at three more molecules in a different spectrum. So let's look at the spectrum here."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it's definitely not the amine. So this spectrum is the alcohol. Let's look at three more molecules in a different spectrum. So let's look at the spectrum here. We start with 1,500. So we draw a line here. We look in the double bond region."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at the spectrum here. We start with 1,500. So we draw a line here. We look in the double bond region. And so here's our double bond region. I do see a signal this time. And it doesn't look like it's a very strong signal either."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We look in the double bond region. And so here's our double bond region. I do see a signal this time. And it doesn't look like it's a very strong signal either. Let's see what the location of the signal is. So I drop down, and the signal shows up between 1,600 and 1,700. So we'll say approximately 1,650."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it doesn't look like it's a very strong signal either. Let's see what the location of the signal is. So I drop down, and the signal shows up between 1,600 and 1,700. So we'll say approximately 1,650. That's not very strong. Both of those things, location and the fact that it's not a very strong signal, clue me into the fact that this is probably a carbon-carbon double bond stretch. That's what this is talking about here."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we'll say approximately 1,650. That's not very strong. Both of those things, location and the fact that it's not a very strong signal, clue me into the fact that this is probably a carbon-carbon double bond stretch. That's what this is talking about here. So a carbonyl, we would expect that to be just past 1,700 and also much, much stronger. So we can rule out this molecule over here because I don't see any kind of a carbonyl stretch. So let's look in the triple bond region."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's what this is talking about here. So a carbonyl, we would expect that to be just past 1,700 and also much, much stronger. So we can rule out this molecule over here because I don't see any kind of a carbonyl stretch. So let's look in the triple bond region. So somewhere in here, I don't see any kind of a signal. So it couldn't possibly be this molecule. So we must be talking about cyclohexene here."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look in the triple bond region. So somewhere in here, I don't see any kind of a signal. So it couldn't possibly be this molecule. So we must be talking about cyclohexene here. And if we look over in the bond to hydrogen region and we draw a line, we can see that this signal, just higher than 3,000, this must be talking about our carbon-hydrogen bond stretch where the carbon is sp2 hybridized. So this is, of course, talking about our carbon-hydrogen stretch where we're talking about sp3 hybridized carbon. And so cyclohexene is the only thing that makes sense with this IR spectrum."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we must be talking about cyclohexene here. And if we look over in the bond to hydrogen region and we draw a line, we can see that this signal, just higher than 3,000, this must be talking about our carbon-hydrogen bond stretch where the carbon is sp2 hybridized. So this is, of course, talking about our carbon-hydrogen stretch where we're talking about sp3 hybridized carbon. And so cyclohexene is the only thing that makes sense with this IR spectrum. Let's do one more. So we have three molecules and an IR spectrum. So let's start analyzing."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so cyclohexene is the only thing that makes sense with this IR spectrum. Let's do one more. So we have three molecules and an IR spectrum. So let's start analyzing. Draw our line around 1,500 right here. Focus in to the left of that line. And this is our double bond region."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's start analyzing. Draw our line around 1,500 right here. Focus in to the left of that line. And this is our double bond region. So two signals, two clear signals in the double bond region. Let's look at this signal right here. So it's not as intense as the other one."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this is our double bond region. So two signals, two clear signals in the double bond region. Let's look at this signal right here. So it's not as intense as the other one. And it's pretty much between 1,600 and 1,700. So both those factors make me think carbon-carbon double bond stretch. So this is probably a carbon-carbon double bond stretch here."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it's not as intense as the other one. And it's pretty much between 1,600 and 1,700. So both those factors make me think carbon-carbon double bond stretch. So this is probably a carbon-carbon double bond stretch here. The signal next to it, if this is 1,600, this is 1,700. So this signal is just past 1,700. And it's very strong."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is probably a carbon-carbon double bond stretch here. The signal next to it, if this is 1,600, this is 1,700. So this signal is just past 1,700. And it's very strong. It's a very strong signal. So that makes me think carbonyl. So this makes me think carbonyl right here."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it's very strong. It's a very strong signal. So that makes me think carbonyl. So this makes me think carbonyl right here. So we can immediately rule out this one. So it couldn't possibly be that molecule. And that brings us to this, which is a conjugated ketone versus an unconjugated ketone."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this makes me think carbonyl right here. So we can immediately rule out this one. So it couldn't possibly be that molecule. And that brings us to this, which is a conjugated ketone versus an unconjugated ketone. So let's think about the unconjugated ketone for a minute here. So this carbonyl stretch we talked about in an earlier video, we'd expect to find that somewhere around 1,715, so past 1,700. This ketone over here, this conjugated ketone, we have resonance."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that brings us to this, which is a conjugated ketone versus an unconjugated ketone. So let's think about the unconjugated ketone for a minute here. So this carbonyl stretch we talked about in an earlier video, we'd expect to find that somewhere around 1,715, so past 1,700. This ketone over here, this conjugated ketone, we have resonance. And we know what resonance does to the carbonyl. So it decreases the strength of the carbonyl. Therefore, it decreases the force constant, k. That decreases the frequency of vibration."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This ketone over here, this conjugated ketone, we have resonance. And we know what resonance does to the carbonyl. So it decreases the strength of the carbonyl. Therefore, it decreases the force constant, k. That decreases the frequency of vibration. And we would expect this carbonyl signal to have a lower wave number than 1,715. Actually, it moves it under 1,700 to somewhere around 1,680 is where we would expect it to be. I don't know exactly where it is, but it's definitely less than 1,700."}, {"video_title": "IR spectra practice Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, it decreases the force constant, k. That decreases the frequency of vibration. And we would expect this carbonyl signal to have a lower wave number than 1,715. Actually, it moves it under 1,700 to somewhere around 1,680 is where we would expect it to be. I don't know exactly where it is, but it's definitely less than 1,700. And this is very clearly, let me go ahead and mark this here, this is very clearly the 1,700 line. And our signal is past that. So this must be talking about the unconjugated ketone over here on the right."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We know that an element is defined by the number of protons it has. For example, potassium, we look at the periodic table of elements, and I have a snapshot of not the entire table, but part of it here. Potassium has 19 protons, and we could write it like this. And this is a little bit redundant. We know that if it's potassium, that atom has 19 protons, and we know if an atom has 19 protons, it is going to be potassium. Now, we also know that not all members of, or not all of the atoms of a given element have the same number of neutrons. And when we talk about a given element, but we have different numbers of neutrons, we call them isotopes of that element."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is a little bit redundant. We know that if it's potassium, that atom has 19 protons, and we know if an atom has 19 protons, it is going to be potassium. Now, we also know that not all members of, or not all of the atoms of a given element have the same number of neutrons. And when we talk about a given element, but we have different numbers of neutrons, we call them isotopes of that element. So, for example, potassium can come in a form that has exactly 20 neutrons, and we call that potassium 39. And the 39, this mass number, it's a count of the 19 protons, 19 protons plus 20 neutrons. And this is actually the most common isotope of potassium."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when we talk about a given element, but we have different numbers of neutrons, we call them isotopes of that element. So, for example, potassium can come in a form that has exactly 20 neutrons, and we call that potassium 39. And the 39, this mass number, it's a count of the 19 protons, 19 protons plus 20 neutrons. And this is actually the most common isotope of potassium. It accounts for 93.3, I'm just rounding off. This is 93.3% of the potassium that you would find on Earth. Now, some of the other isotopes of potassium, you also have potassium, and once again, writing the K and the 19 are a little bit redundant."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is actually the most common isotope of potassium. It accounts for 93.3, I'm just rounding off. This is 93.3% of the potassium that you would find on Earth. Now, some of the other isotopes of potassium, you also have potassium, and once again, writing the K and the 19 are a little bit redundant. You also have potassium 41, so this would have 22 neutrons. 22 plus 19 is 41. This accounts for about 6.7% of the potassium on the planet."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, some of the other isotopes of potassium, you also have potassium, and once again, writing the K and the 19 are a little bit redundant. You also have potassium 41, so this would have 22 neutrons. 22 plus 19 is 41. This accounts for about 6.7% of the potassium on the planet. And then you have a very scarce isotope of potassium called potassium 40. Potassium 40 currently has 21 neutrons, and it's very, very, very, very scarce. It accounts for only 0.0117% of all the potassium."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This accounts for about 6.7% of the potassium on the planet. And then you have a very scarce isotope of potassium called potassium 40. Potassium 40 currently has 21 neutrons, and it's very, very, very, very scarce. It accounts for only 0.0117% of all the potassium. But this is also the isotope of potassium that's interesting to us from the point of view of dating old, old rock, and especially old volcanic rock. And as we'll see, when you can date old volcanic rock, it allows you to date other types of rock or other types of fossils that might be sandwiched in between old volcanic rock. And so what's really interesting about potassium 40 here is that it has a half-life of 1.25 billion years."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It accounts for only 0.0117% of all the potassium. But this is also the isotope of potassium that's interesting to us from the point of view of dating old, old rock, and especially old volcanic rock. And as we'll see, when you can date old volcanic rock, it allows you to date other types of rock or other types of fossils that might be sandwiched in between old volcanic rock. And so what's really interesting about potassium 40 here is that it has a half-life of 1.25 billion years. So the good thing about that, as opposed to something like carbon-14, it can be used to date really, really, really old things. And every 1.25 billion years, let me write it like this, that's its half-life, so 50% of any given sample will have decayed. And 11% will have decayed into argon-40."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so what's really interesting about potassium 40 here is that it has a half-life of 1.25 billion years. So the good thing about that, as opposed to something like carbon-14, it can be used to date really, really, really old things. And every 1.25 billion years, let me write it like this, that's its half-life, so 50% of any given sample will have decayed. And 11% will have decayed into argon-40. So argon is right over here. It has 18 protons. So when you think about it decaying into argon-40, what you see is that it lost a proton, but it has the same mass number."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And 11% will have decayed into argon-40. So argon is right over here. It has 18 protons. So when you think about it decaying into argon-40, what you see is that it lost a proton, but it has the same mass number. So one of the protons must have somehow turned into a neutron. It actually captures one of the inner electrons, and then emits other things, I won't go into all the quantum physics of it, but turns into argon-40, and 89% turn into calcium-40. And you see calcium on the periodic table right over here is 20 protons."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when you think about it decaying into argon-40, what you see is that it lost a proton, but it has the same mass number. So one of the protons must have somehow turned into a neutron. It actually captures one of the inner electrons, and then emits other things, I won't go into all the quantum physics of it, but turns into argon-40, and 89% turn into calcium-40. And you see calcium on the periodic table right over here is 20 protons. So this is a situation where one of the neutrons turns into a proton. This is a situation where one of the protons turns into a neutron. And what's really interesting to us is this part right over here."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you see calcium on the periodic table right over here is 20 protons. So this is a situation where one of the neutrons turns into a proton. This is a situation where one of the protons turns into a neutron. And what's really interesting to us is this part right over here. Because what's cool about argon, and we studied this a little bit in the chemistry playlist, it is a noble gas, it is unreactive. And so when it is embedded in something that's in a liquid state, it'll kind of just bubble out. It's not bonded to anything, and so it'll just bubble out and go out into the atmosphere."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's really interesting to us is this part right over here. Because what's cool about argon, and we studied this a little bit in the chemistry playlist, it is a noble gas, it is unreactive. And so when it is embedded in something that's in a liquid state, it'll kind of just bubble out. It's not bonded to anything, and so it'll just bubble out and go out into the atmosphere. So what's interesting about this whole situation is you can imagine what happens during a volcanic eruption. Let me draw a volcano here. So let's say that this is our volcano."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's not bonded to anything, and so it'll just bubble out and go out into the atmosphere. So what's interesting about this whole situation is you can imagine what happens during a volcanic eruption. Let me draw a volcano here. So let's say that this is our volcano. And it erupts at some time in the past. So it erupts, and you have all of this lava flowing. And while the lava, that lava will contain some amount of potassium-40, and actually it'll already contain some amount of argon-40."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say that this is our volcano. And it erupts at some time in the past. So it erupts, and you have all of this lava flowing. And while the lava, that lava will contain some amount of potassium-40, and actually it'll already contain some amount of argon-40. It'll already contain some amount of argon-40. But what's neat about argon-40 is that while it's lava, while it's in this liquid state, so let's imagine this lava right over here. It's a bunch of stuff right over here."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And while the lava, that lava will contain some amount of potassium-40, and actually it'll already contain some amount of argon-40. It'll already contain some amount of argon-40. But what's neat about argon-40 is that while it's lava, while it's in this liquid state, so let's imagine this lava right over here. It's a bunch of stuff right over here. But in that stuff, it is going to have, I'll do the potassium-40 in, let me do it in a color that I haven't used yet, I'll do the potassium-40 in magenta. It'll have some potassium-40 in it. I may be overdoing it, it's a very scarce isotope."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's a bunch of stuff right over here. But in that stuff, it is going to have, I'll do the potassium-40 in, let me do it in a color that I haven't used yet, I'll do the potassium-40 in magenta. It'll have some potassium-40 in it. I may be overdoing it, it's a very scarce isotope. But it'll have some potassium-40 in it, and it might already have some argon-40 in it. So it might already have some argon-40 in it. Just like that."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I may be overdoing it, it's a very scarce isotope. But it'll have some potassium-40 in it, and it might already have some argon-40 in it. So it might already have some argon-40 in it. Just like that. But argon-40 is a noble gas, it's not going to bond to anything, and while this lava is in a liquid state, it's going to be able to bubble out. It'll just float to the top, it has no bonds, and it'll just evaporate, I should say evaporate, it'll just bubble out, essentially, because it's not bonded to anything, and so it'll just seep out while we are in a liquid state. And what's really interesting about that is that when you have these volcanic eruptions, and because this argon-40 is seeping out, by the time this lava has hardened into volcanic rock, and I'll do that, let me do that volcanic rock in a different color, by the time it has hardened into volcanic rock, all of the argon-40 will be gone."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just like that. But argon-40 is a noble gas, it's not going to bond to anything, and while this lava is in a liquid state, it's going to be able to bubble out. It'll just float to the top, it has no bonds, and it'll just evaporate, I should say evaporate, it'll just bubble out, essentially, because it's not bonded to anything, and so it'll just seep out while we are in a liquid state. And what's really interesting about that is that when you have these volcanic eruptions, and because this argon-40 is seeping out, by the time this lava has hardened into volcanic rock, and I'll do that, let me do that volcanic rock in a different color, by the time it has hardened into volcanic rock, all of the argon-40 will be gone. It won't be there anymore. And so what's neat is, this volcanic event, the fact that this rock has become liquid, it kind of resets the amount of argon-40 there, so then you're only going to be left with potassium-40 here. You're going to be left with potassium-40."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's really interesting about that is that when you have these volcanic eruptions, and because this argon-40 is seeping out, by the time this lava has hardened into volcanic rock, and I'll do that, let me do that volcanic rock in a different color, by the time it has hardened into volcanic rock, all of the argon-40 will be gone. It won't be there anymore. And so what's neat is, this volcanic event, the fact that this rock has become liquid, it kind of resets the amount of argon-40 there, so then you're only going to be left with potassium-40 here. You're going to be left with potassium-40. And that's why the argon-40 is more interesting, because the calcium-40 won't necessarily have seeped out, and there might have already been calcium-40 here, so it won't necessarily seep out, but the argon-40 will seep out, so it kind of resets it. The volcanic event resets the amount of argon-40. So at future date, so right when the event happened, you shouldn't have any argon-40, right when that lava actually becomes solid."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You're going to be left with potassium-40. And that's why the argon-40 is more interesting, because the calcium-40 won't necessarily have seeped out, and there might have already been calcium-40 here, so it won't necessarily seep out, but the argon-40 will seep out, so it kind of resets it. The volcanic event resets the amount of argon-40. So at future date, so right when the event happened, you shouldn't have any argon-40, right when that lava actually becomes solid. And so if you fast-forward to some future date, and if you look at the sample, let me copy and paste it. So let me copy and let me paste it. So if you fast-forward to some future date, and you see that there is some argon-40 there, you see that there is some argon-40 in that sample, you know this is volcanic rock, you know that it was due to some previous volcanic event, you know that this argon-40 is from decayed potassium-40."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So at future date, so right when the event happened, you shouldn't have any argon-40, right when that lava actually becomes solid. And so if you fast-forward to some future date, and if you look at the sample, let me copy and paste it. So let me copy and let me paste it. So if you fast-forward to some future date, and you see that there is some argon-40 there, you see that there is some argon-40 in that sample, you know this is volcanic rock, you know that it was due to some previous volcanic event, you know that this argon-40 is from decayed potassium-40. And you know that it has decayed since that volcanic event, because if it was there before, it would have seeped out. So the only way that this would have been able to get trapped is if, while it was liquid, it would seep out, but once it's solid, it can get trapped inside the rock. And so you know that the only way this argon-40 can exist there is by decay from that potassium-40."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you fast-forward to some future date, and you see that there is some argon-40 there, you see that there is some argon-40 in that sample, you know this is volcanic rock, you know that it was due to some previous volcanic event, you know that this argon-40 is from decayed potassium-40. And you know that it has decayed since that volcanic event, because if it was there before, it would have seeped out. So the only way that this would have been able to get trapped is if, while it was liquid, it would seep out, but once it's solid, it can get trapped inside the rock. And so you know that the only way this argon-40 can exist there is by decay from that potassium-40. So you can look at the ratio. So you know for every one of these argon-40s, because it's only 11% of the decay products are argon-40, for every one of those, you must have on the order of about 9 calcium-40s that also decayed. And so for every one of these argon-40s, you know that there must have been 10 original potassium-40s."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you know that the only way this argon-40 can exist there is by decay from that potassium-40. So you can look at the ratio. So you know for every one of these argon-40s, because it's only 11% of the decay products are argon-40, for every one of those, you must have on the order of about 9 calcium-40s that also decayed. And so for every one of these argon-40s, you know that there must have been 10 original potassium-40s. And so what you can do is you can look at the ratio of the number of potassium-40s there are today to the number that there must have been, based on this evidence right over here, to actually date it. And in the next video, I'll actually go through the mathematical calculation to show you that you can actually date it. And the reason this is really useful is you can look at those ratios, and volcanic eruptions aren't happening every day, but if you start looking over millions and millions of years, on that timescale, they're actually happening reasonably frequent."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so for every one of these argon-40s, you know that there must have been 10 original potassium-40s. And so what you can do is you can look at the ratio of the number of potassium-40s there are today to the number that there must have been, based on this evidence right over here, to actually date it. And in the next video, I'll actually go through the mathematical calculation to show you that you can actually date it. And the reason this is really useful is you can look at those ratios, and volcanic eruptions aren't happening every day, but if you start looking over millions and millions of years, on that timescale, they're actually happening reasonably frequent. And so if you dig in the ground, and so let's dig in the ground. So let's say this is the ground right over here, and you dig enough, and you see a volcanic eruption. You see some volcanic rock right over there, and then you dig even more."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the reason this is really useful is you can look at those ratios, and volcanic eruptions aren't happening every day, but if you start looking over millions and millions of years, on that timescale, they're actually happening reasonably frequent. And so if you dig in the ground, and so let's dig in the ground. So let's say this is the ground right over here, and you dig enough, and you see a volcanic eruption. You see some volcanic rock right over there, and then you dig even more. There's another layer of volcanic rock right over there. So this is another layer of volcanic rock. And let's say that this one over here, so they're all going to have a certain amount of potassium-40 in it."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You see some volcanic rock right over there, and then you dig even more. There's another layer of volcanic rock right over there. So this is another layer of volcanic rock. And let's say that this one over here, so they're all going to have a certain amount of potassium-40 in it. This is going to have some amount of potassium-40 in it. And then let's say this one over here has more argon-40. This one has a little bit less."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say that this one over here, so they're all going to have a certain amount of potassium-40 in it. This is going to have some amount of potassium-40 in it. And then let's say this one over here has more argon-40. This one has a little bit less. And using the math that we're going to do in the next video, let's say you're able to say that this is, using the half-life and using the ratio of argon-40 that's left, versus what you, or using the ratio of the potassium-40 left to what you know was there before, you say that this must have solidified 100 million years ago, 100 million years before the present. And you know that this layer right over here solidified, let's say you know it solidified 150 million years before the present. And let's say you feel pretty good that this soil hasn't been dug up and mixed or anything like that."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This one has a little bit less. And using the math that we're going to do in the next video, let's say you're able to say that this is, using the half-life and using the ratio of argon-40 that's left, versus what you, or using the ratio of the potassium-40 left to what you know was there before, you say that this must have solidified 100 million years ago, 100 million years before the present. And you know that this layer right over here solidified, let's say you know it solidified 150 million years before the present. And let's say you feel pretty good that this soil hasn't been dug up and mixed or anything like that. It looks like it's been pretty untouched when you look at these soil samples right over here. And let's say you see some fossils in here. Then, even though carbon-14 dating is kind of useless, really when you get beyond 50,000 years, you see these fossils in between these two periods."}, {"video_title": "Potassium-argon (K-Ar) dating Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say you feel pretty good that this soil hasn't been dug up and mixed or anything like that. It looks like it's been pretty untouched when you look at these soil samples right over here. And let's say you see some fossils in here. Then, even though carbon-14 dating is kind of useless, really when you get beyond 50,000 years, you see these fossils in between these two periods. It's a pretty good indicator, if you can assume that this soil hasn't been dug around and mixed, that this fossil is between 100 million and 150 million years old. This event happened, then you have these fossils got deposited, these animals died, or they lived and they died, and then you had this other volcanic event. So it allows you, even though you're only directly dating the volcanic rock, it allows you, when you look at the layers, it allows you to relatively date things in between those layers."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So for this compound, we're gonna look down the C3-C4 bond and draw the most stable conformation. So let's start by numbering our carbons. This must be carbon one, two, three, four, five, and six. And if we look down the C3-C4 bond, that's this bond right here, we're gonna put our eye along this axis. Let me draw an eye in here. So we're going to stare down this way and draw what we see. Well, I'll show you in a video in a minute what we would actually see, but it's very important to be able to draw these Newman projections without the use of a model."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And if we look down the C3-C4 bond, that's this bond right here, we're gonna put our eye along this axis. Let me draw an eye in here. So we're going to stare down this way and draw what we see. Well, I'll show you in a video in a minute what we would actually see, but it's very important to be able to draw these Newman projections without the use of a model. So let's start by thinking about what is attached to this carbon three here. There's a methyl group coming out at us in space, which means there must also be a hydrogen going away from us in space. And what is attached to carbon four?"}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Well, I'll show you in a video in a minute what we would actually see, but it's very important to be able to draw these Newman projections without the use of a model. So let's start by thinking about what is attached to this carbon three here. There's a methyl group coming out at us in space, which means there must also be a hydrogen going away from us in space. And what is attached to carbon four? There must be a hydrogen coming out at us in space, so hydrogen here, and a hydrogen going away from us in space. So now that we've drawn that in, we can start to draw our Newman projection. And we start with a point or a dot to represent carbon three."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And what is attached to carbon four? There must be a hydrogen coming out at us in space, so hydrogen here, and a hydrogen going away from us in space. So now that we've drawn that in, we can start to draw our Newman projection. And we start with a point or a dot to represent carbon three. So that point there is carbon three. What is attached to carbon three? Well, there is a methyl group that would be going down and to the right."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And we start with a point or a dot to represent carbon three. So that point there is carbon three. What is attached to carbon three? Well, there is a methyl group that would be going down and to the right. So if your eye is here, you'll see a methyl group going down and to the right. So let's draw in a CH three down and to the right like that. And then we also have a hydrogen."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Well, there is a methyl group that would be going down and to the right. So if your eye is here, you'll see a methyl group going down and to the right. So let's draw in a CH three down and to the right like that. And then we also have a hydrogen. So this hydrogen will be going down and to the left. So let's draw in that hydrogen down and to the left. Next, we would have a CH two, CH three, or an ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then we also have a hydrogen. So this hydrogen will be going down and to the left. So let's draw in that hydrogen down and to the left. Next, we would have a CH two, CH three, or an ethyl group. And this would actually be going straight up if we're looking at it from this perspective. So this would be going straight up. So let's write CH two, CH three."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Next, we would have a CH two, CH three, or an ethyl group. And this would actually be going straight up if we're looking at it from this perspective. So this would be going straight up. So let's write CH two, CH three. Next, we need to think about carbon four. So we're thinking about this carbon right here. We wouldn't be able to see it because carbon three would be in the way."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let's write CH two, CH three. Next, we need to think about carbon four. So we're thinking about this carbon right here. We wouldn't be able to see it because carbon three would be in the way. But we know that carbon four is there. And we represent carbon four with a circle on our Newman projection. What is attached to carbon four?"}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "We wouldn't be able to see it because carbon three would be in the way. But we know that carbon four is there. And we represent carbon four with a circle on our Newman projection. What is attached to carbon four? Well, we know that we have a hydrogen going up and to the right. So let's draw that in. So there's a hydrogen going up and to the right attached to carbon four."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "What is attached to carbon four? Well, we know that we have a hydrogen going up and to the right. So let's draw that in. So there's a hydrogen going up and to the right attached to carbon four. There's also a hydrogen going up and to the left attached to carbon four. So let's draw that in on our Newman projection. And then finally, we would have a CH two, CH three."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So there's a hydrogen going up and to the right attached to carbon four. There's also a hydrogen going up and to the left attached to carbon four. So let's draw that in on our Newman projection. And then finally, we would have a CH two, CH three. And this would be going down. So a CH two, CH three going down. So let's draw that in, CH two, CH three."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, we would have a CH two, CH three. And this would be going down. So a CH two, CH three going down. So let's draw that in, CH two, CH three. Well, it gets annoying to draw in all these CH twos and CH threes. So let's redraw this Newman projection. Let's say that CH two, CH three, we know that's an ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw that in, CH two, CH three. Well, it gets annoying to draw in all these CH twos and CH threes. So let's redraw this Newman projection. Let's say that CH two, CH three, we know that's an ethyl group. So let's abbreviate that with ET. And a methyl group, let's just say that's ME. And then we have our hydrogen right here."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Let's say that CH two, CH three, we know that's an ethyl group. So let's abbreviate that with ET. And a methyl group, let's just say that's ME. And then we have our hydrogen right here. And then for our back carbon, we have a hydrogen here, we have a hydrogen here, and then we have an ethyl group going straight down. So it's just a little bit easier to see this way. And in the video, I'm gonna make an ethyl group red, so you're gonna see a red circle, a red sphere for an ethyl group in the video."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our hydrogen right here. And then for our back carbon, we have a hydrogen here, we have a hydrogen here, and then we have an ethyl group going straight down. So it's just a little bit easier to see this way. And in the video, I'm gonna make an ethyl group red, so you're gonna see a red circle, a red sphere for an ethyl group in the video. And I'll make the methyl group blue. So it's easier to show different conformations if you just represent it by a sphere. And you'll see what I mean."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And in the video, I'm gonna make an ethyl group red, so you're gonna see a red circle, a red sphere for an ethyl group in the video. And I'll make the methyl group blue. So it's easier to show different conformations if you just represent it by a sphere. And you'll see what I mean. And from the video, we're gonna figure out the most stable conformation. We know that has to be a staggered conformation from earlier videos. So we'll look at all the different staggered conformations and we'll pick which one is the most stable."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And you'll see what I mean. And from the video, we're gonna figure out the most stable conformation. We know that has to be a staggered conformation from earlier videos. So we'll look at all the different staggered conformations and we'll pick which one is the most stable. Here we have carbon one, and then carbon two, and then carbon three. Notice there's a methyl group coming out at us in space attached to carbon three. Then we have carbon four."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we'll look at all the different staggered conformations and we'll pick which one is the most stable. Here we have carbon one, and then carbon two, and then carbon three. Notice there's a methyl group coming out at us in space attached to carbon three. Then we have carbon four. And we're gonna stare down the carbon three, four bonds. So let's rotate the molecule here. Let's stare down the C three, C four bond."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Then we have carbon four. And we're gonna stare down the carbon three, four bonds. So let's rotate the molecule here. Let's stare down the C three, C four bond. And notice we have a staggered conformation. Up here we have an ethyl group. On the right here we have a methyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Let's stare down the C three, C four bond. And notice we have a staggered conformation. Up here we have an ethyl group. On the right here we have a methyl group. And then down here we have another ethyl group. So hopefully you see the staggered conformation. Remember that red is an ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "On the right here we have a methyl group. And then down here we have another ethyl group. So hopefully you see the staggered conformation. Remember that red is an ethyl group. So here's an ethyl group and here's an ethyl group. And blue represents the methyl group. It's just easier to work with the model set this way."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Remember that red is an ethyl group. So here's an ethyl group and here's an ethyl group. And blue represents the methyl group. It's just easier to work with the model set this way. So we're going for staggered conformation. So if I rotate the front carbon and keep the back carbon stationary, we get another staggered conformation. And if I rotate again, then we get another staggered conformation."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "It's just easier to work with the model set this way. So we're going for staggered conformation. So if I rotate the front carbon and keep the back carbon stationary, we get another staggered conformation. And if I rotate again, then we get another staggered conformation. Hopefully you can see the Newman projection that we drew matches the picture from the video. This is the same Newman projection. I just made the ethyl groups red and the methyl group blue."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And if I rotate again, then we get another staggered conformation. Hopefully you can see the Newman projection that we drew matches the picture from the video. This is the same Newman projection. I just made the ethyl groups red and the methyl group blue. In the video we moved the front carbon. We rotated the front carbon. We held the back carbon stationary."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "I just made the ethyl groups red and the methyl group blue. In the video we moved the front carbon. We rotated the front carbon. We held the back carbon stationary. So this ethyl group in red would move over to this position. The methyl group in blue would move over to this position. And finally this hydrogen right here, I'll make it green, would move over to this position."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "We held the back carbon stationary. So this ethyl group in red would move over to this position. The methyl group in blue would move over to this position. And finally this hydrogen right here, I'll make it green, would move over to this position. So that's this hydrogen in green. Let's go ahead and draw the next conformation, the next staggered conformation. If we held the back carbon stationary, we can go ahead and draw in the back carbon and what's attached to the back carbon, these hydrogens and this ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And finally this hydrogen right here, I'll make it green, would move over to this position. So that's this hydrogen in green. Let's go ahead and draw the next conformation, the next staggered conformation. If we held the back carbon stationary, we can go ahead and draw in the back carbon and what's attached to the back carbon, these hydrogens and this ethyl group. And next the ethyl group in red on the front carbon moved over to this position. The methyl group in blue moved over to this position and the hydrogen in green moved over to this position. So we can see that matches what we have for our picture down here."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "If we held the back carbon stationary, we can go ahead and draw in the back carbon and what's attached to the back carbon, these hydrogens and this ethyl group. And next the ethyl group in red on the front carbon moved over to this position. The methyl group in blue moved over to this position and the hydrogen in green moved over to this position. So we can see that matches what we have for our picture down here. We have our two ethyl groups are now gauche to each other and then we have our methyl group over here in blue. And our hydrogen in green, let me highlight that, the hydrogen in green is this hydrogen. We can rotate one more time to get our last staggered conformation."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we can see that matches what we have for our picture down here. We have our two ethyl groups are now gauche to each other and then we have our methyl group over here in blue. And our hydrogen in green, let me highlight that, the hydrogen in green is this hydrogen. We can rotate one more time to get our last staggered conformation. Our ethyl group in red can rotate over here. The methyl group in blue could rotate over here and that would mean the hydrogen in green has to rotate over to this position. So draw in the back carbon with the hydrogens and then the ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "We can rotate one more time to get our last staggered conformation. Our ethyl group in red can rotate over here. The methyl group in blue could rotate over here and that would mean the hydrogen in green has to rotate over to this position. So draw in the back carbon with the hydrogens and then the ethyl group. It doesn't matter if you rotate the front or the back carbon. I chose to rotate the front carbon here and that would move our ethyl group over to this position. So now our ethyl group is here."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So draw in the back carbon with the hydrogens and then the ethyl group. It doesn't matter if you rotate the front or the back carbon. I chose to rotate the front carbon here and that would move our ethyl group over to this position. So now our ethyl group is here. The methyl group in blue would move up to here and the hydrogen in green would move over to here. So hopefully we can see that. Let me highlight everything."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So now our ethyl group is here. The methyl group in blue would move up to here and the hydrogen in green would move over to here. So hopefully we can see that. Let me highlight everything. So here's an ethyl group. So down here we can see our ethyl groups, our methyl group in blue, and finally our hydrogen in green. So now we're finally able to choose the most stable conformation out of our three staggered ones here."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Let me highlight everything. So here's an ethyl group. So down here we can see our ethyl groups, our methyl group in blue, and finally our hydrogen in green. So now we're finally able to choose the most stable conformation out of our three staggered ones here. So we need to think about the gauche interactions that are present. And we'll start with this conformation on the right. Here we have an ethyl-ethyl gauche interaction."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So now we're finally able to choose the most stable conformation out of our three staggered ones here. So we need to think about the gauche interactions that are present. And we'll start with this conformation on the right. Here we have an ethyl-ethyl gauche interaction. And ethyl groups are pretty bulky so this gauche interaction would destabilize this conformation. Let's look at this conformation next. We have an ethyl-ethyl gauche interaction and we also have a methyl-ethyl gauche interaction."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Here we have an ethyl-ethyl gauche interaction. And ethyl groups are pretty bulky so this gauche interaction would destabilize this conformation. Let's look at this conformation next. We have an ethyl-ethyl gauche interaction and we also have a methyl-ethyl gauche interaction. So we have two gauche interactions. So that means that this conformation is even more unstable. And finally we have this one right here."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "We have an ethyl-ethyl gauche interaction and we also have a methyl-ethyl gauche interaction. So we have two gauche interactions. So that means that this conformation is even more unstable. And finally we have this one right here. We have only one gauche interaction and it's between an ethyl group and a methyl group. And since this methyl group is not as bulky as another ethyl group, that means this is the lowest energy conformation. This is the most stable."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And finally we have this one right here. We have only one gauche interaction and it's between an ethyl group and a methyl group. And since this methyl group is not as bulky as another ethyl group, that means this is the lowest energy conformation. This is the most stable. In part B, our goal is to draw the least stable conformation, or the one highest in energy. And that must be an eclipse conformation. So let's go to the video and let's start with the staggered and then go to an eclipse conformation."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "This is the most stable. In part B, our goal is to draw the least stable conformation, or the one highest in energy. And that must be an eclipse conformation. So let's go to the video and let's start with the staggered and then go to an eclipse conformation. And from that eclipse conformation we'll look at the others and we'll choose the one that's the highest energy. So here we have our staggered conformation. If I rotate the front carbon, we see one eclipsed conformation."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let's go to the video and let's start with the staggered and then go to an eclipse conformation. And from that eclipse conformation we'll look at the others and we'll choose the one that's the highest energy. So here we have our staggered conformation. If I rotate the front carbon, we see one eclipsed conformation. I can rotate again to get another eclipsed conformation. So here's another one. You can see this one has the two ethyl groups really close together."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "If I rotate the front carbon, we see one eclipsed conformation. I can rotate again to get another eclipsed conformation. So here's another one. You can see this one has the two ethyl groups really close together. I can rotate again to get our final eclipsed conformation. Here we have pictures of the three eclipse conformations from the video. And to save time, let's just analyze the pictures and then we'll draw the least stable conformation as a Newman projection."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "You can see this one has the two ethyl groups really close together. I can rotate again to get our final eclipsed conformation. Here we have pictures of the three eclipse conformations from the video. And to save time, let's just analyze the pictures and then we'll draw the least stable conformation as a Newman projection. Let's start with the conformation on the right. We can see we have an ethyl group eclipsing a hydrogen, a methyl group eclipsing a hydrogen, and a hydrogen eclipsing an ethyl group. So we don't have any of the alkyl groups eclipsing each other."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And to save time, let's just analyze the pictures and then we'll draw the least stable conformation as a Newman projection. Let's start with the conformation on the right. We can see we have an ethyl group eclipsing a hydrogen, a methyl group eclipsing a hydrogen, and a hydrogen eclipsing an ethyl group. So we don't have any of the alkyl groups eclipsing each other. So this one is definitely not the least stable. We're looking for bulky groups interfering with each other, so steric hindrance. Let's move over here to this conformation."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we don't have any of the alkyl groups eclipsing each other. So this one is definitely not the least stable. We're looking for bulky groups interfering with each other, so steric hindrance. Let's move over here to this conformation. We have two hydrogens eclipsing each other. We have an ethyl group eclipsing a hydrogen. And then we have a methyl group eclipsing an ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "Let's move over here to this conformation. We have two hydrogens eclipsing each other. We have an ethyl group eclipsing a hydrogen. And then we have a methyl group eclipsing an ethyl group. So there's a source of some strain. So that's going to increase in energy. So this conformation is definitely higher in energy than this conformation."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a methyl group eclipsing an ethyl group. So there's a source of some strain. So that's going to increase in energy. So this conformation is definitely higher in energy than this conformation. But let's compare this one to our center one. For the center one, we have two very bulky ethyl groups. So an ethyl group eclipsing another ethyl group."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this conformation is definitely higher in energy than this conformation. But let's compare this one to our center one. For the center one, we have two very bulky ethyl groups. So an ethyl group eclipsing another ethyl group. And that is very unstable. This increases the energy. That's a lot of steric strain."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So an ethyl group eclipsing another ethyl group. And that is very unstable. This increases the energy. That's a lot of steric strain. So this is the least stable conformation. These ethyl groups want to be as far away from each other as possible. And here we've put them very close together in space."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "That's a lot of steric strain. So this is the least stable conformation. These ethyl groups want to be as far away from each other as possible. And here we've put them very close together in space. So let's go ahead and draw the Newman projection for this conformation. And we start with this carbon. So that carbon is represented by a point."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And here we've put them very close together in space. So let's go ahead and draw the Newman projection for this conformation. And we start with this carbon. So that carbon is represented by a point. And attached to that carbon, we have a methyl group going up and to the left. We have a hydrogen going up and to the right. And we have an ethyl group going down."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon is represented by a point. And attached to that carbon, we have a methyl group going up and to the left. We have a hydrogen going up and to the right. And we have an ethyl group going down. For the back carbon, so here's the circle representing the back carbon. We have a hydrogen going up and to the left. So there's our hydrogen."}, {"video_title": "Newman projection practice 2 Organic chemistry Khan Academy.mp3", "Sentence": "And we have an ethyl group going down. For the back carbon, so here's the circle representing the back carbon. We have a hydrogen going up and to the left. So there's our hydrogen. We have another hydrogen back here. So I'll draw that one. And then finally we have an ethyl group going down."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have a two-carbon chain, and there's all single bonds, so we're dealing with an ethane. Actually, I'll write it all at once. And then we have, on the one carbon, we could call this the one carbon and call this the two carbon, we have a bromine and a fluorine. So we could call this 1-bromo, and we're putting the bromo instead of the fluoro because b comes before f alphabetically. 1-bromo, 1-fluoro, and then we're dealing with an ethane. We have a two-carbon chain, all single bonds, fluoroethane. That's the name of that molecule there."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "So we could call this 1-bromo, and we're putting the bromo instead of the fluoro because b comes before f alphabetically. 1-bromo, 1-fluoro, and then we're dealing with an ethane. We have a two-carbon chain, all single bonds, fluoroethane. That's the name of that molecule there. Just a review of some of the earlier organic, the nomenclature videos we had done. Now, we know immediately, based on the last few videos, that this is also a chiral carbon. And if we were to take its mirror image, we would get another enantiomer of this same molecule, or that they are enantiomers of each other."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "That's the name of that molecule there. Just a review of some of the earlier organic, the nomenclature videos we had done. Now, we know immediately, based on the last few videos, that this is also a chiral carbon. And if we were to take its mirror image, we would get another enantiomer of this same molecule, or that they are enantiomers of each other. So what does the mirror image of this 1-bromo, 1-fluoro ethane look like? Well, you'd have the carbon right here. You would still have, I want to get all the colors right, you would still have the bromine up above."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And if we were to take its mirror image, we would get another enantiomer of this same molecule, or that they are enantiomers of each other. So what does the mirror image of this 1-bromo, 1-fluoro ethane look like? Well, you'd have the carbon right here. You would still have, I want to get all the colors right, you would still have the bromine up above. You would have this methyl group that's attached to the carbon now pointing in the left direction, CH3. The fluorine would now still be behind the carbon. So the fluorine would still be behind the carbon."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "You would still have, I want to get all the colors right, you would still have the bromine up above. You would have this methyl group that's attached to the carbon now pointing in the left direction, CH3. The fluorine would now still be behind the carbon. So the fluorine would still be behind the carbon. And now the hydrogen would still pop out of the page, but it would now pop out and to the right. So it would now pop out and to the right, just like that. That is the hydrogen."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "So the fluorine would still be behind the carbon. And now the hydrogen would still pop out of the page, but it would now pop out and to the right. So it would now pop out and to the right, just like that. That is the hydrogen. Now, based on our naming so far, we would name this 1-bromo, 1-fluoro ethane, and we would also name this 1-bromo, 1-fluoro ethane, but these are fundamentally two different molecules, even though they have the same molecules in them, they have the same molecular formula, they have the same, I guess you could call it, constitution, in that this carbon is connected to a hydrogen, a fluorine, and a bromine, this carbon is connected to the same things, this carbon is connected to a carbon, 3 hydrogens, so is this one, these are stereoisomers. These are stereoisomers, and they're mirror images of each other, so they're enantiomers. And actually, they will, one, polarize light differently, and they can often have very different chemical properties in a chemical or a biological system."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "That is the hydrogen. Now, based on our naming so far, we would name this 1-bromo, 1-fluoro ethane, and we would also name this 1-bromo, 1-fluoro ethane, but these are fundamentally two different molecules, even though they have the same molecules in them, they have the same molecular formula, they have the same, I guess you could call it, constitution, in that this carbon is connected to a hydrogen, a fluorine, and a bromine, this carbon is connected to the same things, this carbon is connected to a carbon, 3 hydrogens, so is this one, these are stereoisomers. These are stereoisomers, and they're mirror images of each other, so they're enantiomers. And actually, they will, one, polarize light differently, and they can often have very different chemical properties in a chemical or a biological system. So it seems, I guess, not good that we have the same names for both of these. So we're going to focus on this video, is how do you differentiate between the two? Enantiomers."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And actually, they will, one, polarize light differently, and they can often have very different chemical properties in a chemical or a biological system. So it seems, I guess, not good that we have the same names for both of these. So we're going to focus on this video, is how do you differentiate between the two? Enantiomers. So how do we differentiate between the two? So the naming system we're going to use right here is called the Kahn-Ingold prelog system, but it's a different Kahn, it's not me, it's C-A-H-N, instead of K-H-A-N. Kahn-Ingold prelog system. And it's a way of differentiating between this enantiomer, which right now we would call 1-bromo-1-fluoroethane and this enantiomer, 1-bromo-1-fluoroethane."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Enantiomers. So how do we differentiate between the two? So the naming system we're going to use right here is called the Kahn-Ingold prelog system, but it's a different Kahn, it's not me, it's C-A-H-N, instead of K-H-A-N. Kahn-Ingold prelog system. And it's a way of differentiating between this enantiomer, which right now we would call 1-bromo-1-fluoroethane and this enantiomer, 1-bromo-1-fluoroethane. That's a pretty straightforward thing. Really, the hardest part is to just visualize rotating the molecules in the right way and figuring out in which direction, whether it's kind of a left-handed or right-handed molecule. We're going to take it step by step."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And it's a way of differentiating between this enantiomer, which right now we would call 1-bromo-1-fluoroethane and this enantiomer, 1-bromo-1-fluoroethane. That's a pretty straightforward thing. Really, the hardest part is to just visualize rotating the molecules in the right way and figuring out in which direction, whether it's kind of a left-handed or right-handed molecule. We're going to take it step by step. So the first thing you do in the Kahn-Ingold prelog system is to, one, identify your chiral molecule here. It's pretty obvious. It's this carbon right here."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "We're going to take it step by step. So the first thing you do in the Kahn-Ingold prelog system is to, one, identify your chiral molecule here. It's pretty obvious. It's this carbon right here. It is bonded to, well, we'll just focus on this left one, the one we started with first. It's bonded to three different groups. And then what you want to do is you want to rank the groups by atomic number."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "It's this carbon right here. It is bonded to, well, we'll just focus on this left one, the one we started with first. It's bonded to three different groups. And then what you want to do is you want to rank the groups by atomic number. So if you go up here, out of bromine, hydrogen, fluorine, and a carbon, this is what is bonded directly to this carbon, which has the highest atomic number? Well, you see bromine is over here. Let me do this in a darker color."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And then what you want to do is you want to rank the groups by atomic number. So if you go up here, out of bromine, hydrogen, fluorine, and a carbon, this is what is bonded directly to this carbon, which has the highest atomic number? Well, you see bromine is over here. Let me do this in a darker color. We have bromine at 35. We have fluorine at 9. We have carbon at 6."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Let me do this in a darker color. We have bromine at 35. We have fluorine at 9. We have carbon at 6. And then we have hydrogen at 1. So of all of these, bromine is the largest. We'll just call this number 1."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "We have carbon at 6. And then we have hydrogen at 1. So of all of these, bromine is the largest. We'll just call this number 1. Then after that, we have fluorine. That is number 2. Number 3 is the carbon."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "We'll just call this number 1. Then after that, we have fluorine. That is number 2. Number 3 is the carbon. And then hydrogen is the smallest. So that is number 4. So now that we've numbered them, the next step is to orient this molecule so that the smallest atomic number group is sitting into the page."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Number 3 is the carbon. And then hydrogen is the smallest. So that is number 4. So now that we've numbered them, the next step is to orient this molecule so that the smallest atomic number group is sitting into the page. It's sitting behind the molecule. So right now, this hydrogen is the smallest of all of them. Bromine is the largest."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "So now that we've numbered them, the next step is to orient this molecule so that the smallest atomic number group is sitting into the page. It's sitting behind the molecule. So right now, this hydrogen is the smallest of all of them. Bromine is the largest. Hydrogen is the smallest. So we want to orient it behind the molecule. The way it's drawn right now, it's oriented in front of the molecule."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Bromine is the largest. Hydrogen is the smallest. So we want to orient it behind the molecule. The way it's drawn right now, it's oriented in front of the molecule. So to orient it behind the molecule, and this really is the hardest part, is just to visualize it properly. Remember, this fluorine is behind. This is right in the plane of the paper."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "The way it's drawn right now, it's oriented in front of the molecule. So to orient it behind the molecule, and this really is the hardest part, is just to visualize it properly. Remember, this fluorine is behind. This is right in the plane of the paper. This is popping out of the paper. We would want to rotate. You can imagine we'd be rotating the molecule in this direction so that, let me redraw it, so that now we have the carbon here."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "This is right in the plane of the paper. This is popping out of the paper. We would want to rotate. You can imagine we'd be rotating the molecule in this direction so that, let me redraw it, so that now we have the carbon here. And now since we've rotated it like this, we've rotated it roughly about 1 third around the circle, so it's about 120 degrees, now this hydrogen is where the fluorine was. So that's where the hydrogen is. The fluorine is now where this methyl group is."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "You can imagine we'd be rotating the molecule in this direction so that, let me redraw it, so that now we have the carbon here. And now since we've rotated it like this, we've rotated it roughly about 1 third around the circle, so it's about 120 degrees, now this hydrogen is where the fluorine was. So that's where the hydrogen is. The fluorine is now where this methyl group is. These dotted lines show that we're behind now. This shows that we're in the plane. And the methyl group is now where the hydrogen is."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "The fluorine is now where this methyl group is. These dotted lines show that we're behind now. This shows that we're in the plane. And the methyl group is now where the hydrogen is. It's now popping out of the page. It's going to the left and out. So this methyl group is now popping out of the page, out and to the left."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And the methyl group is now where the hydrogen is. It's now popping out of the page. It's going to the left and out. So this methyl group is now popping out of the page, out and to the left. And that's where our methyl group is. So all we've done is we've just rotated this around about 120 degrees. That's the hardest."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "So this methyl group is now popping out of the page, out and to the left. And that's where our methyl group is. So all we've done is we've just rotated this around about 120 degrees. That's the hardest. We've just gotten this to go behind. And that's kind of the first step after we've identified the chiral carbon and ranked them by atomic number. And of course, the bromine is still going to be on top."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "That's the hardest. We've just gotten this to go behind. And that's kind of the first step after we've identified the chiral carbon and ranked them by atomic number. And of course, the bromine is still going to be on top. Now once you've put the smallest atomic number molecule in the back, then you want to look at the rankings of 1 through 3. I mean, we have four molecules here. So we look at the largest, which is bromine, number 1."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And of course, the bromine is still going to be on top. Now once you've put the smallest atomic number molecule in the back, then you want to look at the rankings of 1 through 3. I mean, we have four molecules here. So we look at the largest, which is bromine, number 1. Then number 2 is fluorine, number 2. And then number 3 is this methyl group. That's the carbon that's bonded to this carbon."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "So we look at the largest, which is bromine, number 1. Then number 2 is fluorine, number 2. And then number 3 is this methyl group. That's the carbon that's bonded to this carbon. So it's number 3 right there. And in the Kahn-Ingold pre-log system, we literally just think about what would it take to go from number 1 to number 2 to number 3. And in this case, we would go in this direction."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "That's the carbon that's bonded to this carbon. So it's number 3 right there. And in the Kahn-Ingold pre-log system, we literally just think about what would it take to go from number 1 to number 2 to number 3. And in this case, we would go in this direction. To go from number 1 to number 2 to number 3, we would go in the clockwise direction. We're just kind of ignoring the hydronite now. That's kind of sitting behind it."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And in this case, we would go in this direction. To go from number 1 to number 2 to number 3, we would go in the clockwise direction. We're just kind of ignoring the hydronite now. That's kind of sitting behind it. That was the first step, to orient it so it's sitting in the back, the smallest molecule. And then the three largest ones, you just say, what direction do we have to go to go from number 1 to number 2 to number 3? In this case, we have to go clockwise."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "That's kind of sitting behind it. That was the first step, to orient it so it's sitting in the back, the smallest molecule. And then the three largest ones, you just say, what direction do we have to go to go from number 1 to number 2 to number 3? In this case, we have to go clockwise. And if we go clockwise now, then we call this a right handed molecule, or we use the Latin word for right, which is rectus. And so we would call this, this molecule right here, is not just 1-bromo-1-fluoroethane, this is R. R for rectus. Or you can even think right, although we'll see left uses S, which is sinister."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "In this case, we have to go clockwise. And if we go clockwise now, then we call this a right handed molecule, or we use the Latin word for right, which is rectus. And so we would call this, this molecule right here, is not just 1-bromo-1-fluoroethane, this is R. R for rectus. Or you can even think right, although we'll see left uses S, which is sinister. So the Latin is really where the R comes from. But this is R-1-bromo-1-fluoroethane. That's this one right here."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Or you can even think right, although we'll see left uses S, which is sinister. So the Latin is really where the R comes from. But this is R-1-bromo-1-fluoroethane. That's this one right here. So you might guess, well this must be the opposite. This must be the counterclockwise version. We can do it really fast."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "That's this one right here. So you might guess, well this must be the opposite. This must be the counterclockwise version. We can do it really fast. So same idea. We know the largest one, bromine's number 1, that's the largest in terms of atomic number. Fluorine is number 2."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "We can do it really fast. So same idea. We know the largest one, bromine's number 1, that's the largest in terms of atomic number. Fluorine is number 2. Carbon is number 3. Hydrogen is number 4. What we want to do is put hydrogen in the back."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Fluorine is number 2. Carbon is number 3. Hydrogen is number 4. What we want to do is put hydrogen in the back. So what we're going to have to do is rotate it to the back, to where fluorine is right now. So if we were to redraw it, draw this molecule right here, you have your carbon still, you still have your bromine sitting on top. But we're going to put the hydrogen now to the back."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "What we want to do is put hydrogen in the back. So what we're going to have to do is rotate it to the back, to where fluorine is right now. So if we were to redraw it, draw this molecule right here, you have your carbon still, you still have your bromine sitting on top. But we're going to put the hydrogen now to the back. So the hydrogen is now where the fluorine used to be. The hydrogen's there. This chlorine is going to be moved to where the hydrogen now is, not the chlorine, sorry, this methyl group, this carbon with the 3 hydrogens, is going to be rotated to where the hydrogen used to be."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "But we're going to put the hydrogen now to the back. So the hydrogen is now where the fluorine used to be. The hydrogen's there. This chlorine is going to be moved to where the hydrogen now is, not the chlorine, sorry, this methyl group, this carbon with the 3 hydrogens, is going to be rotated to where the hydrogen used to be. It's now going to pop out of the page because we're rotating it in that direction. So this is our methyl group right there. And then this fluorine is going to be moved where the methyl group was."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "This chlorine is going to be moved to where the hydrogen now is, not the chlorine, sorry, this methyl group, this carbon with the 3 hydrogens, is going to be rotated to where the hydrogen used to be. It's now going to pop out of the page because we're rotating it in that direction. So this is our methyl group right there. And then this fluorine is going to be moved where the methyl group was. So this fluorine will go right here. And now, using the Kahn-Ingold prelog system, this is our number 1, this is our number 2, just by atomic number, this is number 3. You go from number 1 through number 2 to number 3."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "And then this fluorine is going to be moved where the methyl group was. So this fluorine will go right here. And now, using the Kahn-Ingold prelog system, this is our number 1, this is our number 2, just by atomic number, this is number 3. You go from number 1 through number 2 to number 3. You go in this direction. You're going counterclockwise. Or we are going to the left, or we use the Latin word for it, which is sinister."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "You go from number 1 through number 2 to number 3. You go in this direction. You're going counterclockwise. Or we are going to the left, or we use the Latin word for it, which is sinister. And the word sinister comes from the Latin word for left. So I guess right is good, and people thought either left handed people were bad, or if you're not going to the right it's bad. I don't know why sinister took on its sinister meaning now in common language."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "Or we are going to the left, or we use the Latin word for it, which is sinister. And the word sinister comes from the Latin word for left. So I guess right is good, and people thought either left handed people were bad, or if you're not going to the right it's bad. I don't know why sinister took on its sinister meaning now in common language. But it's now the sinister version of the molecule. So we would call this version, this enantiomer of 1-bromo-1 fluoroethane, we would call this S. S for sinister, or for left, or for counterclockwise. S-1-bromo-1-fluoroethane."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "I don't know why sinister took on its sinister meaning now in common language. But it's now the sinister version of the molecule. So we would call this version, this enantiomer of 1-bromo-1 fluoroethane, we would call this S. S for sinister, or for left, or for counterclockwise. S-1-bromo-1-fluoroethane. So now we can differentiate the names. We know that these are two different configurations. And that's what the S and the R tell us."}, {"video_title": "Cahn-Ingold-Prelog system for naming enantiomers Organic chemistry Khan Academy.mp3", "Sentence": "S-1-bromo-1-fluoroethane. So now we can differentiate the names. We know that these are two different configurations. And that's what the S and the R tell us. That if you had to go from this to this, you would literally have to detach and reattach different groups. You would actually have to break bonds. You would actually have to swap two of these groups in some way in order to get from this enantiomer to this enantiomer."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I threw out in the context of maybe the supermassive black holes at the galactic cores of each of those galaxies will start getting a little bit more material when that collision happens. And maybe quasars will happen. I don't know. But given the interest in that, what I wanted to do here is kind of an unconventional thing for the Khan Academy and actually show a video. And before I play the video, I have to give credit where credit is due. This is a supercomputer simulation made at the National Center for Supercomputing Applications in NASA. And it's by B. Robertson of Caltech and L. Hernquist of Harvard University."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But given the interest in that, what I wanted to do here is kind of an unconventional thing for the Khan Academy and actually show a video. And before I play the video, I have to give credit where credit is due. This is a supercomputer simulation made at the National Center for Supercomputing Applications in NASA. And it's by B. Robertson of Caltech and L. Hernquist of Harvard University. And what I want you to remember is this is super sped up in time. Just to give an idea, the amount of time it takes for a star about as far away as the sun to make one orbit around the galactic core is 250 million years. And you're going to see that this is happening multiple times over the course of this video."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's by B. Robertson of Caltech and L. Hernquist of Harvard University. And what I want you to remember is this is super sped up in time. Just to give an idea, the amount of time it takes for a star about as far away as the sun to make one orbit around the galactic core is 250 million years. And you're going to see that this is happening multiple times over the course of this video. So this video is actually spanning billions of years. But when you actually speed up time like that, you'll see that it really gives you a sense of the actual dynamics of these interactions. The other thing I want to talk about before I actually start the video is to make you realize when we talk about galaxies colliding, it doesn't mean that the stars are colliding."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you're going to see that this is happening multiple times over the course of this video. So this video is actually spanning billions of years. But when you actually speed up time like that, you'll see that it really gives you a sense of the actual dynamics of these interactions. The other thing I want to talk about before I actually start the video is to make you realize when we talk about galaxies colliding, it doesn't mean that the stars are colliding. In fact, there are going to be very few stars that actually collide. The probability of a star-star collision is very low. And that's because we learned, when we learned about interstellar scale, that there's mostly free space in between stars."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The other thing I want to talk about before I actually start the video is to make you realize when we talk about galaxies colliding, it doesn't mean that the stars are colliding. In fact, there are going to be very few stars that actually collide. The probability of a star-star collision is very low. And that's because we learned, when we learned about interstellar scale, that there's mostly free space in between stars. The closest star to us is 4.2 light years away. And that's roughly 30 million times the diameter of the sun. So you have a lot more free space than star space or even solar system space."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's because we learned, when we learned about interstellar scale, that there's mostly free space in between stars. The closest star to us is 4.2 light years away. And that's roughly 30 million times the diameter of the sun. So you have a lot more free space than star space or even solar system space. So let's start up this animation. It's pretty amazing. And what you're going to see here, so these are just the, obviously, so one rotation is actually 250 million years, give or take."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have a lot more free space than star space or even solar system space. So let's start up this animation. It's pretty amazing. And what you're going to see here, so these are just the, obviously, so one rotation is actually 250 million years, give or take. But now you see these stars right here are starting to get attracted to this core. And then they're actually attracted to that core. And then some of the stuff in that core was attracted to those stars."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what you're going to see here, so these are just the, obviously, so one rotation is actually 250 million years, give or take. But now you see these stars right here are starting to get attracted to this core. And then they're actually attracted to that core. And then some of the stuff in that core was attracted to those stars. And they get pulled away. That was the first pass of these two galaxies. Some stuff is just being thrown off into intergalactic space."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then some of the stuff in that core was attracted to those stars. And they get pulled away. That was the first pass of these two galaxies. Some stuff is just being thrown off into intergalactic space. And you might worry, maybe that'll happen to the Earth. And there's some probability that it would happen to the Earth, but it really wouldn't affect what happens within those stars' solar systems. This is happening so slow."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some stuff is just being thrown off into intergalactic space. And you might worry, maybe that'll happen to the Earth. And there's some probability that it would happen to the Earth, but it really wouldn't affect what happens within those stars' solar systems. This is happening so slow. You wouldn't feel like some type of acceleration or something. And then this is the second pass. So they passed one pass."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is happening so slow. You wouldn't feel like some type of acceleration or something. And then this is the second pass. So they passed one pass. And once again, we're doing this. This is occurring over hundreds of millions or billions of years. And on the second pass, they finally are able to merge."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they passed one pass. And once again, we're doing this. This is occurring over hundreds of millions or billions of years. And on the second pass, they finally are able to merge. And all of these interactions are just through the gravity over interstellar, almost you could call it, intergalactic distances. You can see they merge into what could be called as a milk-o-meta, or maybe the Andromedae way. I don't know, whatever you want to call it."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And on the second pass, they finally are able to merge. And all of these interactions are just through the gravity over interstellar, almost you could call it, intergalactic distances. You can see they merge into what could be called as a milk-o-meta, or maybe the Andromedae way. I don't know, whatever you want to call it. But even though they've merged, a lot of the stuff has still been thrown off into intergalactic space. But this is a pretty amazing animation to me. One, it's amazing to think about how this could happen over galactic space scales and time scales."}, {"video_title": "Galactic collisions Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't know, whatever you want to call it. But even though they've merged, a lot of the stuff has still been thrown off into intergalactic space. But this is a pretty amazing animation to me. One, it's amazing to think about how this could happen over galactic space scales and time scales. But it's also pretty neat how a supercomputer can do all of the computations to figure out what every particle, which is really a star, a cluster of stars, or group of stars, is actually doing to actually give us a sense of the actual dynamics here. But this is pretty neat. Look at that."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So first let's look at a nucleophile. The word nucleophile means nucleus loving. And since the nucleus is positively charged, you can think about a nucleophile as being negatively charged because opposite charges attract. So a nucleophile could have a full negative charge, which would be attracted to the positive charge of a nucleus, or it could have a partial negative charge, or you could just think about a nucleophile as having a region of high electron density. So let's look at some examples of nucleophiles. First let's start with the ethoxide anion here. And the ethoxide anion has an oxygen with a full negative charge."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So a nucleophile could have a full negative charge, which would be attracted to the positive charge of a nucleus, or it could have a partial negative charge, or you could just think about a nucleophile as having a region of high electron density. So let's look at some examples of nucleophiles. First let's start with the ethoxide anion here. And the ethoxide anion has an oxygen with a full negative charge. So obviously that is a nucleophile, and the oxygen is the nucleophilic center of ethoxide. Next let's look at ethanol. Well ethanol doesn't have a full negative charge, but we know that oxygen is more electronegative than hydrogen."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "And the ethoxide anion has an oxygen with a full negative charge. So obviously that is a nucleophile, and the oxygen is the nucleophilic center of ethoxide. Next let's look at ethanol. Well ethanol doesn't have a full negative charge, but we know that oxygen is more electronegative than hydrogen. So oxygen is going to pull the electrons in this bond closer to itself, giving it a partial negative charge. So this oxygen is the nucleophilic center of ethanol. Now the ethoxide anion is going to be a better nucleophile than ethanol because it has a full negative formal charge on the oxygen as opposed to only a partial negative."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Well ethanol doesn't have a full negative charge, but we know that oxygen is more electronegative than hydrogen. So oxygen is going to pull the electrons in this bond closer to itself, giving it a partial negative charge. So this oxygen is the nucleophilic center of ethanol. Now the ethoxide anion is going to be a better nucleophile than ethanol because it has a full negative formal charge on the oxygen as opposed to only a partial negative. Next let's look at methyl lithium. Let's think about the electronegativity difference between carbon and lithium. Carbon is more electronegative than lithium, so the two electrons in this bond are pulled closer to the carbon, giving the carbon a partial negative charge."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Now the ethoxide anion is going to be a better nucleophile than ethanol because it has a full negative formal charge on the oxygen as opposed to only a partial negative. Next let's look at methyl lithium. Let's think about the electronegativity difference between carbon and lithium. Carbon is more electronegative than lithium, so the two electrons in this bond are pulled closer to the carbon, giving the carbon a partial negative charge. And so the carbon is the nucleophilic center of methyl lithium. Since lithium is losing some electron density, we could draw a partial positive charge here on lithium. And here I've drawn it as a covalent bond, but really you could also show it as being an ionic bond."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Carbon is more electronegative than lithium, so the two electrons in this bond are pulled closer to the carbon, giving the carbon a partial negative charge. And so the carbon is the nucleophilic center of methyl lithium. Since lithium is losing some electron density, we could draw a partial positive charge here on lithium. And here I've drawn it as a covalent bond, but really you could also show it as being an ionic bond. So the difference in electronegativity is so great that I could show both of those electrons being on this carbon. So let me go ahead and put in the hydrogens here. Since carbon is more electronegative than lithium, I could take these two electrons in magenta and I could put them both on the carbon, which would give the carbon a negative one formal charge."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "And here I've drawn it as a covalent bond, but really you could also show it as being an ionic bond. So the difference in electronegativity is so great that I could show both of those electrons being on this carbon. So let me go ahead and put in the hydrogens here. Since carbon is more electronegative than lithium, I could take these two electrons in magenta and I could put them both on the carbon, which would give the carbon a negative one formal charge. So this carbon with a negative one formal charge would be the nucleophilic center. I took an electron away from lithium, giving it a plus one formal charge here. So here I've represented it as an ionic bond."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Since carbon is more electronegative than lithium, I could take these two electrons in magenta and I could put them both on the carbon, which would give the carbon a negative one formal charge. So this carbon with a negative one formal charge would be the nucleophilic center. I took an electron away from lithium, giving it a plus one formal charge here. So here I've represented it as an ionic bond. Here a little more covalent character. But this picture is useful because this is called a carbanion. Let me write this in here."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So here I've represented it as an ionic bond. Here a little more covalent character. But this picture is useful because this is called a carbanion. Let me write this in here. So a carbanion, which just means a negative charge on a carbon. And carbanions are excellent nucleophiles. Finally, let's look at cyclohexene."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Let me write this in here. So a carbanion, which just means a negative charge on a carbon. And carbanions are excellent nucleophiles. Finally, let's look at cyclohexene. And cyclohexene, we know, has a pi bond. So let me show the pi bond here. And pi bonds are regions of high electron density."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Finally, let's look at cyclohexene. And cyclohexene, we know, has a pi bond. So let me show the pi bond here. And pi bonds are regions of high electron density. So this pi bond can act like a nucleophile in an organic chemistry mechanism. Now let's look at electrophiles. So an electrophile is electron loving."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "And pi bonds are regions of high electron density. So this pi bond can act like a nucleophile in an organic chemistry mechanism. Now let's look at electrophiles. So an electrophile is electron loving. And since electrons are negatively charged, we're gonna think about an electrophile as having a region of low electron density. So it could have a full positive charge on it, because a positive charge will be attracted to electrons. Or you could be talking about a partial positive charge."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So an electrophile is electron loving. And since electrons are negatively charged, we're gonna think about an electrophile as having a region of low electron density. So it could have a full positive charge on it, because a positive charge will be attracted to electrons. Or you could be talking about a partial positive charge. So first let's look at this compound. We know that chlorine is more electronegative than carbon. So chlorine is going to withdraw some electron density."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Or you could be talking about a partial positive charge. So first let's look at this compound. We know that chlorine is more electronegative than carbon. So chlorine is going to withdraw some electron density. And if chlorine is withdrawing electron density away from this carbon, this carbon is partially positive. So this carbon is the electrophilic center of this compound. Next let's look at acetone."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So chlorine is going to withdraw some electron density. And if chlorine is withdrawing electron density away from this carbon, this carbon is partially positive. So this carbon is the electrophilic center of this compound. Next let's look at acetone. So oxygen is more electronegative than carbon. So oxygen is going to withdraw some electron density away from this carbon here. And this carbon would be partially positive."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "Next let's look at acetone. So oxygen is more electronegative than carbon. So oxygen is going to withdraw some electron density away from this carbon here. And this carbon would be partially positive. So this carbon is the electrophilic portion of this compound. Next let's look at a carbocation, where there's a full positive charge on this carbon. So this carbon has only three bonds to it, which gives it a full positive charge."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "And this carbon would be partially positive. So this carbon is the electrophilic portion of this compound. Next let's look at a carbocation, where there's a full positive charge on this carbon. So this carbon has only three bonds to it, which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract. So this carbon is the electrophilic portion of this ion."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So this carbon has only three bonds to it, which gives it a full positive charge. Obviously a full positive charge is going to love electrons. Opposite charges attract. So this carbon is the electrophilic portion of this ion. And finally let's look at this compound. We know that oxygen is more electronegative than carbon. So oxygen withdraws some electron density away from this carbon."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So this carbon is the electrophilic portion of this ion. And finally let's look at this compound. We know that oxygen is more electronegative than carbon. So oxygen withdraws some electron density away from this carbon. And we could even draw a resonance structure here. So let me take these pi electrons and move them out onto the oxygen. So let's draw a resonance structure."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So oxygen withdraws some electron density away from this carbon. And we could even draw a resonance structure here. So let me take these pi electrons and move them out onto the oxygen. So let's draw a resonance structure. So I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen, I would need three lone pairs of electrons on that top oxygen, giving it a negative one formal charge. I took a bond away from this carbon in magenta, which is this carbon, which gives it a plus one formal charge."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So let's draw a resonance structure. So I put in my double bond. Now if I'm showing those pi electrons moving off onto the oxygen, I would need three lone pairs of electrons on that top oxygen, giving it a negative one formal charge. I took a bond away from this carbon in magenta, which is this carbon, which gives it a plus one formal charge. So that's one of the possible resonance structures that you can draw. And of course we know the carbon in magenta is an electrophilic center. But I could draw another resonance structure here."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "I took a bond away from this carbon in magenta, which is this carbon, which gives it a plus one formal charge. So that's one of the possible resonance structures that you can draw. And of course we know the carbon in magenta is an electrophilic center. But I could draw another resonance structure here. So let me go ahead and do that. Put in my brackets over here. I could take these pi electrons, I showed on this one actually, these pi electrons and move them over to here."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "But I could draw another resonance structure here. So let me go ahead and do that. Put in my brackets over here. I could take these pi electrons, I showed on this one actually, these pi electrons and move them over to here. So let's draw the resulting resonance structure. So I have a double bond here now, an oxygen with a negative one formal charge. So let me put that in here, draw in the hydrogen, put in my bracket."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "I could take these pi electrons, I showed on this one actually, these pi electrons and move them over to here. So let's draw the resulting resonance structure. So I have a double bond here now, an oxygen with a negative one formal charge. So let me put that in here, draw in the hydrogen, put in my bracket. And I removed a bond, we took a bond away from, let me use blue for this, from this carbon. So this carbon now has a plus one formal charge. So the carbon in blue is this carbon over here."}, {"video_title": "Identifying nucleophilic and electrophilic centers.mp3", "Sentence": "So let me put that in here, draw in the hydrogen, put in my bracket. And I removed a bond, we took a bond away from, let me use blue for this, from this carbon. So this carbon now has a plus one formal charge. So the carbon in blue is this carbon over here. So let me draw in a plus one formal charge. So that is also electrophilic, right? A full positive charge is going to be attracted to a negative charge."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at two ways to prepare alkynes from alkyl halides. So here I have an alkyl halide. So this is a dihalide. And my two halogens are attached to one carbon. We call this a geminal dihalide. So this is going to be a geminal dihalide reacting with a very strong base, sodium amide. So this is going to give us an E2 elimination reaction."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And my two halogens are attached to one carbon. We call this a geminal dihalide. So this is going to be a geminal dihalide reacting with a very strong base, sodium amide. So this is going to give us an E2 elimination reaction. So we're going to get an E2 elimination reaction. And this E2 elimination reaction is actually going to occur twice. And we're going to end up with an alkyne as our final product."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to give us an E2 elimination reaction. So we're going to get an E2 elimination reaction. And this E2 elimination reaction is actually going to occur twice. And we're going to end up with an alkyne as our final product. So let's take a look at the mechanism of our double E2 elimination of a geminal dihalide. So let's start with our dihalide over here. And this time we're going to put in all of our lone pairs of electrons on our halogen like that."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to end up with an alkyne as our final product. So let's take a look at the mechanism of our double E2 elimination of a geminal dihalide. So let's start with our dihalide over here. And this time we're going to put in all of our lone pairs of electrons on our halogen like that. So let me go ahead and put all of those in there. And then I have two hydrogens on this carbon. OK, sodium amide is a source of amide anions, which we saw in our previous video."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this time we're going to put in all of our lone pairs of electrons on our halogen like that. So let me go ahead and put all of those in there. And then I have two hydrogens on this carbon. OK, sodium amide is a source of amide anions, which we saw in our previous video. It can function as a strong base. So a strong base means that a lone pair of electrons here on our nitrogen is going to take this proton. And these electrons in here are going to kick into form a double bond."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "OK, sodium amide is a source of amide anions, which we saw in our previous video. It can function as a strong base. So a strong base means that a lone pair of electrons here on our nitrogen is going to take this proton. And these electrons in here are going to kick into form a double bond. At the same time, these electrons kick off onto our halogen. So an E2 elimination mechanism. You can watch the previous videos on E2 elimination reactions for more details."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And these electrons in here are going to kick into form a double bond. At the same time, these electrons kick off onto our halogen. So an E2 elimination mechanism. You can watch the previous videos on E2 elimination reactions for more details. So we're going to form ammonia as one of our products. And our other product is going to be carbon double bonded to another carbon. And then we're going to still have our halogen down here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You can watch the previous videos on E2 elimination reactions for more details. So we're going to form ammonia as one of our products. And our other product is going to be carbon double bonded to another carbon. And then we're going to still have our halogen down here. And over here in the carbon on the right, we're still going to have a hydrogen like that. So we're not quite to our alkyne yet. So we've done one E2 elimination reaction."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we're going to still have our halogen down here. And over here in the carbon on the right, we're still going to have a hydrogen like that. So we're not quite to our alkyne yet. So we've done one E2 elimination reaction. And we're going to do one more. So we get another amide anion comes along. And it's negatively charged."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we've done one E2 elimination reaction. And we're going to do one more. So we get another amide anion comes along. And it's negatively charged. It's going to function as a base. It's going to take this proton this time. And these electrons are going to move in here to form our triple bond."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And it's negatively charged. It's going to function as a base. It's going to take this proton this time. And these electrons are going to move in here to form our triple bond. And these electrons are going to kick off onto our halogen like that. So that is going to finally form our alkyne here. So you always have to have your base in excess if you're trying to do this."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And these electrons are going to move in here to form our triple bond. And these electrons are going to kick off onto our halogen like that. So that is going to finally form our alkyne here. So you always have to have your base in excess if you're trying to do this. Let's look at a very similar reaction. A double E2 elimination. This time the halogens are not on the same carbons."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you always have to have your base in excess if you're trying to do this. Let's look at a very similar reaction. A double E2 elimination. This time the halogens are not on the same carbons. Let's go ahead and draw the general reaction for this. We have two carbons right here. And we have two halogens right here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This time the halogens are not on the same carbons. Let's go ahead and draw the general reaction for this. We have two carbons right here. And we have two halogens right here. And then hydrogen and then hydrogen. So this time we have our halogens, two halogens on adjacent carbons. So this is called vicinal dihalides."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have two halogens right here. And then hydrogen and then hydrogen. So this time we have our halogens, two halogens on adjacent carbons. So this is called vicinal dihalides. Let's go ahead and write that. So this is vicinal. And the one we did before was geminal."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is called vicinal dihalides. Let's go ahead and write that. So this is vicinal. And the one we did before was geminal. So a vicinal dihalide will react in a very similar way if you add a strong base like sodium amide and use ammonia for your solvent. So you're going to form an alkyne once again. So you're going to get an alkyne."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the one we did before was geminal. So a vicinal dihalide will react in a very similar way if you add a strong base like sodium amide and use ammonia for your solvent. So you're going to form an alkyne once again. So you're going to get an alkyne. It's going to be via a double E2 elimination reaction again. Let's look at the mechanism. So let's start with our vicinal dihalide down here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you're going to get an alkyne. It's going to be via a double E2 elimination reaction again. Let's look at the mechanism. So let's start with our vicinal dihalide down here. So let's go ahead and put our halogens in there, lone pairs of electrons on our halogens like that. And then we have hydrogen. And we have hydrogen right here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with our vicinal dihalide down here. So let's go ahead and put our halogens in there, lone pairs of electrons on our halogens like that. And then we have hydrogen. And we have hydrogen right here. So we have our amide anion. Once again, functions as a strong base. It's going to take a proton."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have hydrogen right here. So we have our amide anion. Once again, functions as a strong base. It's going to take a proton. So it's going to take this proton right here. These electrons are going to move in to form our double bond. At the same time, these electrons kick off onto our halogens."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to take a proton. So it's going to take this proton right here. These electrons are going to move in to form our double bond. At the same time, these electrons kick off onto our halogens. So that's our first E2 elimination reaction. So let's just go ahead and write E2 here to remind us this is yet another E2 reaction. And let's go ahead and draw the product of that."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "At the same time, these electrons kick off onto our halogens. So that's our first E2 elimination reaction. So let's just go ahead and write E2 here to remind us this is yet another E2 reaction. And let's go ahead and draw the product of that. So now we're going to have carbon double bonded to another carbon. And then we're going to have hydrogen right here. And then we're going to have our halogen up here like that."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and draw the product of that. So now we're going to have carbon double bonded to another carbon. And then we're going to have hydrogen right here. And then we're going to have our halogen up here like that. And then we're going to have one more reaction to form our alkyne. We're going to get another E2 elimination reaction. So sodium amide, another anion of sodium amide comes along."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we're going to have our halogen up here like that. And then we're going to have one more reaction to form our alkyne. We're going to get another E2 elimination reaction. So sodium amide, another anion of sodium amide comes along. So let's go ahead and put in those lone pairs of electrons like that. It's going to function as a base. Lone pair of electrons takes this proton."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So sodium amide, another anion of sodium amide comes along. So let's go ahead and put in those lone pairs of electrons like that. It's going to function as a base. Lone pair of electrons takes this proton. These electrons kick in here to form our triple bond. At the same time, our halogen leaves. And so we form our alkyne like that."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Lone pair of electrons takes this proton. These electrons kick in here to form our triple bond. At the same time, our halogen leaves. And so we form our alkyne like that. So you can produce alkynes from either viscinal or geminal dihalides via a double E2 elimination reaction. Let's see how we could use this in a synthesis reaction. So let's go ahead and try to make something, try to make an alkyne from an alkene."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we form our alkyne like that. So you can produce alkynes from either viscinal or geminal dihalides via a double E2 elimination reaction. Let's see how we could use this in a synthesis reaction. So let's go ahead and try to make something, try to make an alkyne from an alkene. So let's start with an alkene here. I'm going to put some benzene rings on this guy here. So here's a benzene ring like that."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and try to make something, try to make an alkyne from an alkene. So let's start with an alkene here. I'm going to put some benzene rings on this guy here. So here's a benzene ring like that. Put in my lone pairs. Sorry, put in my bonds like that. And then I'm going to put a double bond right here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here's a benzene ring like that. Put in my lone pairs. Sorry, put in my bonds like that. And then I'm going to put a double bond right here. And then I'm going to put another benzene ring attached like that. So this is 1,2-diphenylethylene. And I'm going to react this alkene with bromine."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then I'm going to put a double bond right here. And then I'm going to put another benzene ring attached like that. So this is 1,2-diphenylethylene. And I'm going to react this alkene with bromine. You could use a solvent like carbon tetrachloride or something like that. And we're reacting an alkene with a halogen. And we've seen this reaction before in the videos on reactions of alkenes."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to react this alkene with bromine. You could use a solvent like carbon tetrachloride or something like that. And we're reacting an alkene with a halogen. And we've seen this reaction before in the videos on reactions of alkenes. We're going to add two bromines across our double bond. So we're going to draw the product of this reaction. Our benzene rings aren't going to react as readily as our double bond will."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we've seen this reaction before in the videos on reactions of alkenes. We're going to add two bromines across our double bond. So we're going to draw the product of this reaction. Our benzene rings aren't going to react as readily as our double bond will. So let's go ahead and draw in our other benzene ring here like that. And we know that we're going to add a bromine to either side of our double bond. So let's go ahead and add a bromine to either side of our double bond here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Our benzene rings aren't going to react as readily as our double bond will. So let's go ahead and draw in our other benzene ring here like that. And we know that we're going to add a bromine to either side of our double bond. So let's go ahead and add a bromine to either side of our double bond here. And we also have a hydrogen bonded to each one of these carbons. That hydrogen was originally there as well over here on the left. And we form 1,2-dibromo-1,2-diphenylethane here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and add a bromine to either side of our double bond here. And we also have a hydrogen bonded to each one of these carbons. That hydrogen was originally there as well over here on the left. And we form 1,2-dibromo-1,2-diphenylethane here. And now we have a vicinal dihalide. So if we add a strong base to our vicinal dihalide, we can prepare an alkyne from that. So if we add an excess of sodium amide in ammonia, we know that we're going to get a double E2 elimination reaction."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we form 1,2-dibromo-1,2-diphenylethane here. And now we have a vicinal dihalide. So if we add a strong base to our vicinal dihalide, we can prepare an alkyne from that. So if we add an excess of sodium amide in ammonia, we know that we're going to get a double E2 elimination reaction. And those halogens are going to go away in our double E2 elimination reaction and form a triple bond. So we're going to form a triple bond. So those two carbons, the ones that form a triple bond, are these two carbons right here."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if we add an excess of sodium amide in ammonia, we know that we're going to get a double E2 elimination reaction. And those halogens are going to go away in our double E2 elimination reaction and form a triple bond. So we're going to form a triple bond. So those two carbons, the ones that form a triple bond, are these two carbons right here. So when you run through the mechanism, you're going to get an alkyne. And on either side of that alkyne, you're going to get phenyl groups. Let's go ahead and draw in our benzene rings like that."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So those two carbons, the ones that form a triple bond, are these two carbons right here. So when you run through the mechanism, you're going to get an alkyne. And on either side of that alkyne, you're going to get phenyl groups. Let's go ahead and draw in our benzene rings like that. So it doesn't really matter how we draw our electrons. We'll go ahead and do this. So this would be our product."}, {"video_title": "Preparation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw in our benzene rings like that. So it doesn't really matter how we draw our electrons. We'll go ahead and do this. So this would be our product. So let's go ahead and put in those electrons. So it would be diphenylacetylene. So you can synthesize alkynes from alkenes, or you could synthesize an alkyne from a dihalide."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I start with my alkyne over here, and I add to my alkyne one molar equivalent of a halogen, so X2. My solvent is going to be carbon tetrachloride. And I'm going to add those two halogen atoms across my triple bond in an anti-addition. So those two halogens end up on opposite sides from each other. So this is an anti-addition of my halogens like that. Now, I could add two molar equivalents of my halogen. And if that happens, each of these carbons is going to end up with two bonds to halogens like that for my product."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So those two halogens end up on opposite sides from each other. So this is an anti-addition of my halogens like that. Now, I could add two molar equivalents of my halogen. And if that happens, each of these carbons is going to end up with two bonds to halogens like that for my product. Now, the mechanism of the halogenation of alkynes is not completely understood. So because of that, we're just going to move on to one practice example instead of showing the mechanism. All right, so let's look at an alkyne."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And if that happens, each of these carbons is going to end up with two bonds to halogens like that for my product. Now, the mechanism of the halogenation of alkynes is not completely understood. So because of that, we're just going to move on to one practice example instead of showing the mechanism. All right, so let's look at an alkyne. So we'll go ahead and draw an alkyne over here. So here I have my carbon triple bonded to another carbon. I'm going to put a methyl group on one side here."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's look at an alkyne. So we'll go ahead and draw an alkyne over here. So here I have my carbon triple bonded to another carbon. I'm going to put a methyl group on one side here. And let's put one right at CH3 in here. And I'll put an ethyl group on this side, so CH2CH3. So to that alkyne, I'm going to add bromine."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to put a methyl group on one side here. And let's put one right at CH3 in here. And I'll put an ethyl group on this side, so CH2CH3. So to that alkyne, I'm going to add bromine. And I'm going to use carbon tetrachloride as my solvent. And I'm going to say one molar equivalent of my bromine is added. So when you do your stoichiometry, just one molar equivalent like that."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So to that alkyne, I'm going to add bromine. And I'm going to use carbon tetrachloride as my solvent. And I'm going to say one molar equivalent of my bromine is added. So when you do your stoichiometry, just one molar equivalent like that. So I'm going to add my two bromines on anti to each other. So let's go ahead and show my triple bond became a double bond. And my two bromines are going to add on anti to each other."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when you do your stoichiometry, just one molar equivalent like that. So I'm going to add my two bromines on anti to each other. So let's go ahead and show my triple bond became a double bond. And my two bromines are going to add on anti to each other. So they're going to add on opposite sides of my double bond like that. And I still have a methyl group over here on the connected to the carbon on the left. So I'll go ahead and put in my methyl group like that."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And my two bromines are going to add on anti to each other. So they're going to add on opposite sides of my double bond like that. And I still have a methyl group over here on the connected to the carbon on the left. So I'll go ahead and put in my methyl group like that. And the carbon on the right still has an ethyl group attached to it, so CH2CH3. So that would be the result of the halogenation of this alkyne. Let's take a look at one more reaction of alkynes."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and put in my methyl group like that. And the carbon on the right still has an ethyl group attached to it, so CH2CH3. So that would be the result of the halogenation of this alkyne. Let's take a look at one more reaction of alkynes. Let's look at ozonolysis. So the ozonolysis of alkynes. Let me go ahead and write ozonolysis here."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's take a look at one more reaction of alkynes. Let's look at ozonolysis. So the ozonolysis of alkynes. Let me go ahead and write ozonolysis here. And we've seen this reaction before, similar reaction, when we did this with alkenes. So this time we're going to do it with alkynes. Let's take a look at an alkyne."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and write ozonolysis here. And we've seen this reaction before, similar reaction, when we did this with alkenes. So this time we're going to do it with alkynes. Let's take a look at an alkyne. So there's an alkyne. I'm going to say it's an internal alkyne, meaning the triple bond is found in the interior of the molecule. It's not on the end of the molecule."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's take a look at an alkyne. So there's an alkyne. I'm going to say it's an internal alkyne, meaning the triple bond is found in the interior of the molecule. It's not on the end of the molecule. So let's look at ozonolysis of internal alkynes first. And when you're doing ozonolysis, you're adding ozone to the molecule in the first step. So we went in a very, very detailed mechanism for the ozonolysis of alkenes."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's not on the end of the molecule. So let's look at ozonolysis of internal alkynes first. And when you're doing ozonolysis, you're adding ozone to the molecule in the first step. So we went in a very, very detailed mechanism for the ozonolysis of alkenes. And you can go back and watch that video. For this video, we're not going to go through any kind of mechanism. We're just going to go for the products."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we went in a very, very detailed mechanism for the ozonolysis of alkenes. And you can go back and watch that video. For this video, we're not going to go through any kind of mechanism. We're just going to go for the products. So we add ozone in the first step. And in the second step, we're going to add water. And what this does is this cleaves your triple bond and gives you carboxylic acids as your product, so two of them."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're just going to go for the products. So we add ozone in the first step. And in the second step, we're going to add water. And what this does is this cleaves your triple bond and gives you carboxylic acids as your product, so two of them. So let's go ahead and draw those two carboxylic acids. And then we'll try to point out where everything comes from. So here's one of the carboxylic acids."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And what this does is this cleaves your triple bond and gives you carboxylic acids as your product, so two of them. So let's go ahead and draw those two carboxylic acids. And then we'll try to point out where everything comes from. So here's one of the carboxylic acids. And then here's going to be the carboxylic acid that's going to result on the right side. So I'll make this R prime over here. So let's go ahead and point out which carbons are what here."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here's one of the carboxylic acids. And then here's going to be the carboxylic acid that's going to result on the right side. So I'll make this R prime over here. So let's go ahead and point out which carbons are what here. So let's show that this carbon over here is bonded to an R group. So that's this carbon bonded to an R group. And then over here on the right, this carbon is the one bonded to an R prime."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and point out which carbons are what here. So let's show that this carbon over here is bonded to an R group. So that's this carbon bonded to an R group. And then over here on the right, this carbon is the one bonded to an R prime. This carbon is the one bonded to an R prime. So you cleave your triple bond. You break your triple bond."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then over here on the right, this carbon is the one bonded to an R prime. This carbon is the one bonded to an R prime. So you cleave your triple bond. You break your triple bond. And you're going to create two separate molecules from this. So we get carboxylic acids from this reaction. Let's look at ozonolysis of terminal alkynes now."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You break your triple bond. And you're going to create two separate molecules from this. So we get carboxylic acids from this reaction. Let's look at ozonolysis of terminal alkynes now. So instead of the triple bond being in the interior of the molecule, now the triple bond is on the end of the molecule. So that makes this a hydrogen right here. So let's add, once again, ozone in my first step."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at ozonolysis of terminal alkynes now. So instead of the triple bond being in the interior of the molecule, now the triple bond is on the end of the molecule. So that makes this a hydrogen right here. So let's add, once again, ozone in my first step. And water in my second step. And on the left side, it's going to give us the same product as before. So let's go ahead and identify that in blue."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's add, once again, ozone in my first step. And water in my second step. And on the left side, it's going to give us the same product as before. So let's go ahead and identify that in blue. This carbon in this R group, we saw before, that's going to give us a carboxylic acid like that. So let's go ahead and draw that again. So that portion of the molecule gives us the same product as before."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and identify that in blue. This carbon in this R group, we saw before, that's going to give us a carboxylic acid like that. So let's go ahead and draw that again. So that portion of the molecule gives us the same product as before. So we get a carboxylic acid. And once again, this is carbon in blue. And that R group in blue, those are the same ones on the left side of your reaction."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that portion of the molecule gives us the same product as before. So we get a carboxylic acid. And once again, this is carbon in blue. And that R group in blue, those are the same ones on the left side of your reaction. Now on the right side of my reaction, just go ahead and put that hydrogen in there. On the right side of the reaction, the terminal alkyne portion, you're actually going to get carbon dioxide. So now you only have only one carbon to think about."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that R group in blue, those are the same ones on the left side of your reaction. Now on the right side of my reaction, just go ahead and put that hydrogen in there. On the right side of the reaction, the terminal alkyne portion, you're actually going to get carbon dioxide. So now you only have only one carbon to think about. This is the only carbon you have right here. So you're going to get CO2 out of it. So let's go ahead and draw CO2 as your product like that."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now you only have only one carbon to think about. This is the only carbon you have right here. So you're going to get CO2 out of it. So let's go ahead and draw CO2 as your product like that. Remember, it's a linear molecule. And so this reaction was used decades ago for structure determination. If a terminal alkyne was present and you reacted it with ozone, then you would get some carbon dioxide."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw CO2 as your product like that. Remember, it's a linear molecule. And so this reaction was used decades ago for structure determination. If a terminal alkyne was present and you reacted it with ozone, then you would get some carbon dioxide. And you can also analyze the molecule by the carboxylic acids that you get. So for example, you could see how many carbons are in your R group. And then that would give you an idea about the structure and all those kinds of things."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If a terminal alkyne was present and you reacted it with ozone, then you would get some carbon dioxide. And you can also analyze the molecule by the carboxylic acids that you get. So for example, you could see how many carbons are in your R group. And then that would give you an idea about the structure and all those kinds of things. So this used to be a very important reaction in organic chemistry. Now, with all the spectroscopy stuff that organic chemistry has, this reaction isn't really used as much anymore. So let's look at one example of an ozonolysis reaction."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then that would give you an idea about the structure and all those kinds of things. So this used to be a very important reaction in organic chemistry. Now, with all the spectroscopy stuff that organic chemistry has, this reaction isn't really used as much anymore. So let's look at one example of an ozonolysis reaction. So let's look at this one right here like that. So we're going to take that terminal alkyne. We're going to add ozone to it in the first step."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at one example of an ozonolysis reaction. So let's look at this one right here like that. So we're going to take that terminal alkyne. We're going to add ozone to it in the first step. And then we're going to add water to it in the second step. And let's see how many carbons we're dealing with here. Sometimes that's what confuses students."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to add ozone to it in the first step. And then we're going to add water to it in the second step. And let's see how many carbons we're dealing with here. Sometimes that's what confuses students. So there's one carbon, two carbon, three carbons. And on this side is our fourth carbon. And that carbon on the far right is bonded to a hydrogen, making this a terminal alkyne."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes that's what confuses students. So there's one carbon, two carbon, three carbons. And on this side is our fourth carbon. And that carbon on the far right is bonded to a hydrogen, making this a terminal alkyne. So I have four carbons to worry about. And when this undergoes alkyne cleavage, it's going to cleave the molecule here. It's going to break that triple bond."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that carbon on the far right is bonded to a hydrogen, making this a terminal alkyne. So I have four carbons to worry about. And when this undergoes alkyne cleavage, it's going to cleave the molecule here. It's going to break that triple bond. And so you're going to get two products. You're going to get one product with three carbons over here on the left, and then one product with one carbon on the right. So the product with three carbons on the left is going to be a carboxylic acid."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to break that triple bond. And so you're going to get two products. You're going to get one product with three carbons over here on the left, and then one product with one carbon on the right. So the product with three carbons on the left is going to be a carboxylic acid. So all you have to do is draw a three-carbon carboxylic acid. So let's go ahead and do that. So here's our three-carbon carboxylic acid."}, {"video_title": "Halogenation and ozonolysis of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the product with three carbons on the left is going to be a carboxylic acid. So all you have to do is draw a three-carbon carboxylic acid. So let's go ahead and do that. So here's our three-carbon carboxylic acid. That is one of our products. And then terminal alkynes will give us CO2, which takes care of the other carbon on the right over here. So we're going to get CO2 as the other product of this reaction."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we're gonna focus on the stereochemistry of the diene. On the left is our diene, and on the right is our dienophile. So first, let's draw our product of this Diels-Alder reaction without worrying about stereochemistry. So we know that a Diels-Alder reaction involves a concerted movement of six pi electrons. So if these pi electrons move into here, we form a bond between these two carbons. If these pi electrons move into here, we form a bond between these two carbons, and then these pi electrons would move down. So let's draw the product, again, ignoring stereochemistry for the time being."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we know that a Diels-Alder reaction involves a concerted movement of six pi electrons. So if these pi electrons move into here, we form a bond between these two carbons. If these pi electrons move into here, we form a bond between these two carbons, and then these pi electrons would move down. So let's draw the product, again, ignoring stereochemistry for the time being. So let's put in these bonds. We'd have these methyls right here. And then I'm going to abbreviate these esters as CO2Me."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw the product, again, ignoring stereochemistry for the time being. So let's put in these bonds. We'd have these methyls right here. And then I'm going to abbreviate these esters as CO2Me. So CO2Me down here. So the reason that we chose this alkyne as our dienophile is because we don't have to worry about stereochemistry at these two carbons. We do need to worry about stereochemistry at these two carbons, and those came from our diene."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "And then I'm going to abbreviate these esters as CO2Me. So CO2Me down here. So the reason that we chose this alkyne as our dienophile is because we don't have to worry about stereochemistry at these two carbons. We do need to worry about stereochemistry at these two carbons, and those came from our diene. So let's look in more detail at the structure of the diene. So this is a trans-trans diene. So this double bond is trans, and so is this one."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "We do need to worry about stereochemistry at these two carbons, and those came from our diene. So let's look in more detail at the structure of the diene. So this is a trans-trans diene. So this double bond is trans, and so is this one. And let's focus in on these inside substituents first, so these two hydrogens. And let's look at them in the picture. So those two hydrogens are on the inside, and when we form a bond between these two carbons and between these two carbons, notice what happens to those hydrogens."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So this double bond is trans, and so is this one. And let's focus in on these inside substituents first, so these two hydrogens. And let's look at them in the picture. So those two hydrogens are on the inside, and when we form a bond between these two carbons and between these two carbons, notice what happens to those hydrogens. On the left here, those hydrogens are on sp2 hybridized carbons. So this carbon's sp2 hybridized, and so is this one. But on the right, notice how those two carbons are now sp3 hybridized."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So those two hydrogens are on the inside, and when we form a bond between these two carbons and between these two carbons, notice what happens to those hydrogens. On the left here, those hydrogens are on sp2 hybridized carbons. So this carbon's sp2 hybridized, and so is this one. But on the right, notice how those two carbons are now sp3 hybridized. So those two hydrogens, those inside substituents, go up in our product. So if we're looking down at our ring this way, those two hydrogens are coming out at us in space. So let's go ahead and draw that."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "But on the right, notice how those two carbons are now sp3 hybridized. So those two hydrogens, those inside substituents, go up in our product. So if we're looking down at our ring this way, those two hydrogens are coming out at us in space. So let's go ahead and draw that. So we have our ring, and we put in our double bonds here, and we have our esters. Let me go ahead and put that in, so CO2Me and CO2Me. And let's draw in those hydrogens coming out at us in space."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that. So we have our ring, and we put in our double bonds here, and we have our esters. Let me go ahead and put that in, so CO2Me and CO2Me. And let's draw in those hydrogens coming out at us in space. So those two carbons that I marked in red, those are these two carbons. So we should have a hydrogen coming out at us at both of those carbons. So here's a hydrogen coming out at us on a wedge, and then here is a hydrogen coming out at us on a wedge."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "And let's draw in those hydrogens coming out at us in space. So those two carbons that I marked in red, those are these two carbons. So we should have a hydrogen coming out at us at both of those carbons. So here's a hydrogen coming out at us on a wedge, and then here is a hydrogen coming out at us on a wedge. All right, next, let's look at the outside substituents. So let me use blue for this. So our outside substituents are these two."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So here's a hydrogen coming out at us on a wedge, and then here is a hydrogen coming out at us on a wedge. All right, next, let's look at the outside substituents. So let me use blue for this. So our outside substituents are these two. So in blue, these are the outside. So we have two methyl groups that are outside, and I made the methyl groups yellow in the model set here. So here are the two methyl groups."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So our outside substituents are these two. So in blue, these are the outside. So we have two methyl groups that are outside, and I made the methyl groups yellow in the model set here. So here are the two methyl groups. And when these two carbons, when these two carbons go from sp2 to sp3 hybridized, notice what happens to the methyl groups. If the inside substituents go up, the outside substituents go down. So these are our methyl groups, and they go down in space."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So here are the two methyl groups. And when these two carbons, when these two carbons go from sp2 to sp3 hybridized, notice what happens to the methyl groups. If the inside substituents go up, the outside substituents go down. So these are our methyl groups, and they go down in space. So looking down at our ring, both methyl groups will be going away from us in space. So let's go ahead and draw those in on a dash. So that methyl group is going away from us, and so is this one."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So these are our methyl groups, and they go down in space. So looking down at our ring, both methyl groups will be going away from us in space. So let's go ahead and draw those in on a dash. So that methyl group is going away from us, and so is this one. For this next problem, we have the same dienophile, but we have a different diene. This is cyclopentadiene, and you can see the double bonds are both cis. So when we think about our Diels-Alder reaction, we move our six pi electrons."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So that methyl group is going away from us, and so is this one. For this next problem, we have the same dienophile, but we have a different diene. This is cyclopentadiene, and you can see the double bonds are both cis. So when we think about our Diels-Alder reaction, we move our six pi electrons. So these pi electrons move into here form a bond between these two carbons. These pi electrons move into here, so we form a bond between these two carbons, and these pi electrons move over to here. So when we draw our products, let me go ahead and draw in this ring here, and let's highlight some of those electrons."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So when we think about our Diels-Alder reaction, we move our six pi electrons. So these pi electrons move into here form a bond between these two carbons. These pi electrons move into here, so we form a bond between these two carbons, and these pi electrons move over to here. So when we draw our products, let me go ahead and draw in this ring here, and let's highlight some of those electrons. So we're not done drawing the product yet, obviously. These electrons in red moved into here, and then we had electrons in blue. So these pi electrons in blue moved into here, and then finally our pi electrons in magenta moved over to here."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So when we draw our products, let me go ahead and draw in this ring here, and let's highlight some of those electrons. So we're not done drawing the product yet, obviously. These electrons in red moved into here, and then we had electrons in blue. So these pi electrons in blue moved into here, and then finally our pi electrons in magenta moved over to here. And then we still have a bridging CH2. So this bridging CH2, we need to draw that in on our product. So let me just sketch that in."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons in blue moved into here, and then finally our pi electrons in magenta moved over to here. And then we still have a bridging CH2. So this bridging CH2, we need to draw that in on our product. So let me just sketch that in. So this is actually a top view of our product, and then we still have our esters. So CO2Me coming off of this carbon, and CO2Me coming off of this carbon. So let me draw those in here."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let me just sketch that in. So this is actually a top view of our product, and then we still have our esters. So CO2Me coming off of this carbon, and CO2Me coming off of this carbon. So let me draw those in here. So this is one way to draw your product, but if you want to draw it from a more three-dimensional standpoint, let's look at the picture down here. So our bridging CH2 would be this carbon. So we have these bonds right here going to the bridging CH2, and those are these bonds."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw those in here. So this is one way to draw your product, but if you want to draw it from a more three-dimensional standpoint, let's look at the picture down here. So our bridging CH2 would be this carbon. So we have these bonds right here going to the bridging CH2, and those are these bonds. Those are inside substituents, and we know inside substituents go up. So when we form our bonds, so we formed our bond in red over here, so this one, that's the bond between this carbon and this carbon. So we form a bond between these two carbons, which is this bond."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we have these bonds right here going to the bridging CH2, and those are these bonds. Those are inside substituents, and we know inside substituents go up. So when we form our bonds, so we formed our bond in red over here, so this one, that's the bond between this carbon and this carbon. So we form a bond between these two carbons, which is this bond. And then we also formed our bond in blue over here, so that's the bond between this carbon and this one. So we form a bond between these two, and that's this bond right here. Those inside substituents, right, this bridging CH2 goes up in space."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we form a bond between these two carbons, which is this bond. And then we also formed our bond in blue over here, so that's the bond between this carbon and this one. So we form a bond between these two, and that's this bond right here. Those inside substituents, right, this bridging CH2 goes up in space. So this goes up, and we can sketch in our product from a different point of view. So I'm going to redraw what we see here in the picture. So this is a form of drawing our product that you see a lot."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "Those inside substituents, right, this bridging CH2 goes up in space. So this goes up, and we can sketch in our product from a different point of view. So I'm going to redraw what we see here in the picture. So this is a form of drawing our product that you see a lot. So let me sketch this in. So these take some practice. All right, and let's draw in our double bonds, and then we have our CO2Me and our CO2Me."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So this is a form of drawing our product that you see a lot. So let me sketch this in. So these take some practice. All right, and let's draw in our double bonds, and then we have our CO2Me and our CO2Me. So our bridging CH2 is right here, so that went up in space. And for our last problem, we have our same dienophile, but notice this time for our diene, we have cis at this double bond, but this double bond is trans. So let's first move our six pi electrons."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "All right, and let's draw in our double bonds, and then we have our CO2Me and our CO2Me. So our bridging CH2 is right here, so that went up in space. And for our last problem, we have our same dienophile, but notice this time for our diene, we have cis at this double bond, but this double bond is trans. So let's first move our six pi electrons. So these pi electrons move into here, so we form a bond between these two carbons. These pi electrons move into here, so we form a bond between those two carbons, and these pi electrons move down. So let's start to draw our product."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's first move our six pi electrons. So these pi electrons move into here, so we form a bond between these two carbons. These pi electrons move into here, so we form a bond between those two carbons, and these pi electrons move down. So let's start to draw our product. So we have our ring, and then we have two double bonds in our ring. Let's go ahead and put in our esters here. So we have CO2Me and then CO2Me."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's start to draw our product. So we have our ring, and then we have two double bonds in our ring. Let's go ahead and put in our esters here. So we have CO2Me and then CO2Me. Next, let's think about our substituents. So look at the inside ones first. So we have this hydrogen and this methyl group, and we go down here to the picture."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we have CO2Me and then CO2Me. Next, let's think about our substituents. So look at the inside ones first. So we have this hydrogen and this methyl group, and we go down here to the picture. Here is that hydrogen, and here is that methyl group. So when we form our bonds between these two carbons and these two carbons, our inside substituents go up. So here is that hydrogen, and here is that methyl group."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we have this hydrogen and this methyl group, and we go down here to the picture. Here is that hydrogen, and here is that methyl group. So when we form our bonds between these two carbons and these two carbons, our inside substituents go up. So here is that hydrogen, and here is that methyl group. So if we're staring down this way at our product, those two would be coming out at us in space. So let's go ahead and put in those groups. So the methyl group right here is on this carbon, and that's this one."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So here is that hydrogen, and here is that methyl group. So if we're staring down this way at our product, those two would be coming out at us in space. So let's go ahead and put in those groups. So the methyl group right here is on this carbon, and that's this one. So we have a methyl group coming out at us in space at this carbon, so let me draw that in. So we would have on a wedge a CH3, so let me fill in that wedge. And then for our other carbon, let me mark it right here, so this carbon, which is this one, we'd have a hydrogen coming out at us in space."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So the methyl group right here is on this carbon, and that's this one. So we have a methyl group coming out at us in space at this carbon, so let me draw that in. So we would have on a wedge a CH3, so let me fill in that wedge. And then for our other carbon, let me mark it right here, so this carbon, which is this one, we'd have a hydrogen coming out at us in space. So let me draw in that hydrogen coming out at us right here. And next, let's think about our outside substituents. So I'll make those in blue."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "And then for our other carbon, let me mark it right here, so this carbon, which is this one, we'd have a hydrogen coming out at us in space. So let me draw in that hydrogen coming out at us right here. And next, let's think about our outside substituents. So I'll make those in blue. So our outside substituents are this hydrogen and this methyl group. So let's find those on our picture. So here is that hydrogen and here is our methyl group."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So I'll make those in blue. So our outside substituents are this hydrogen and this methyl group. So let's find those on our picture. So here is that hydrogen and here is our methyl group. And we can see where they end up. So this hydrogen goes down in space, so we need to put that on a dash. So let's draw in our hydrogen on a dash at this carbon."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So here is that hydrogen and here is our methyl group. And we can see where they end up. So this hydrogen goes down in space, so we need to put that on a dash. So let's draw in our hydrogen on a dash at this carbon. So the outside substituents go down. And this methyl group right here, this one goes right here, so it's down relative to the hydrogen. So let's go ahead and put that in."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw in our hydrogen on a dash at this carbon. So the outside substituents go down. And this methyl group right here, this one goes right here, so it's down relative to the hydrogen. So let's go ahead and put that in. So let's put in our methyl group, our CH3, down at this carbon. So what would happen if you took your diene here and you looked at it from a different perspective? So let's go to the video so you see what I mean."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and put that in. So let's put in our methyl group, our CH3, down at this carbon. So what would happen if you took your diene here and you looked at it from a different perspective? So let's go to the video so you see what I mean. So here's our diene with the yellow representing the methyl groups. And if I just rotate it this way, now we can see we have this methyl group on as an inside substituent, and this methyl group would be an outside substituent. As we saw in the video, this is the same diene as before."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So let's go to the video so you see what I mean. So here's our diene with the yellow representing the methyl groups. And if I just rotate it this way, now we can see we have this methyl group on as an inside substituent, and this methyl group would be an outside substituent. As we saw in the video, this is the same diene as before. It's the same as this one. It's just viewed in a different perspective. So we move our six pi electrons around."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "As we saw in the video, this is the same diene as before. It's the same as this one. It's just viewed in a different perspective. So we move our six pi electrons around. So we form a bond between these two carbons. We move these pi electrons down. And let's draw in our ring."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we move our six pi electrons around. So we form a bond between these two carbons. We move these pi electrons down. And let's draw in our ring. So we have our ring. We have two double bonds in the ring. And then we have our esters coming off."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "And let's draw in our ring. So we have our ring. We have two double bonds in the ring. And then we have our esters coming off. So CO2Me and CO2Me. Next we think about our inside substituents. So we have a hydrogen and we have our methyl group."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our esters coming off. So CO2Me and CO2Me. Next we think about our inside substituents. So we have a hydrogen and we have our methyl group. The inside substituents go up. So the hydrogen and the methyl group are going to be on a wedge. So this hydrogen is on a wedge at this carbon, so we put that in."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we have a hydrogen and we have our methyl group. The inside substituents go up. So the hydrogen and the methyl group are going to be on a wedge. So this hydrogen is on a wedge at this carbon, so we put that in. And then this methyl group is on a wedge at this carbon. So fill in that wedge here and write CH3. Next we think about our outside substituents."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen is on a wedge at this carbon, so we put that in. And then this methyl group is on a wedge at this carbon. So fill in that wedge here and write CH3. Next we think about our outside substituents. So we have a methyl group and a hydrogen. Outside substituents go down. So we could draw in that methyl group is going away from us in space."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "Next we think about our outside substituents. So we have a methyl group and a hydrogen. Outside substituents go down. So we could draw in that methyl group is going away from us in space. So there's our CH3. And this hydrogen is also going away from us in space. So it's on a dash."}, {"video_title": "Diels-Alder stereochemistry of diene Organic chemistry Khan Academy.mp3", "Sentence": "So we could draw in that methyl group is going away from us in space. So there's our CH3. And this hydrogen is also going away from us in space. So it's on a dash. So now we have this as our product. Before, when we looked at our diene in this way, we got this as our product. And what is the relationship between this molecule and this molecule?"}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, we took a look at the mechanism for the oxidation of alcohols. In this video, we'll do specific examples for different types of alcohol. So we'll start with a primary alcohol. And we identified the carbon attached to the OH as my alpha carbon. And in order for this mechanism to work, we needed at least one hydrogen attached to our alpha carbon. So if we react our primary alcohol with sodium dichromate, sulfuric acid, and water, which we call the Jones reagent, in that mechanism we're going to oxidize our alpha carbon and lose one of those protons attached to the alpha carbon, which will give us an aldehyde functional group. So we increase the number of bonds of carbon to oxygen."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And we identified the carbon attached to the OH as my alpha carbon. And in order for this mechanism to work, we needed at least one hydrogen attached to our alpha carbon. So if we react our primary alcohol with sodium dichromate, sulfuric acid, and water, which we call the Jones reagent, in that mechanism we're going to oxidize our alpha carbon and lose one of those protons attached to the alpha carbon, which will give us an aldehyde functional group. So we increase the number of bonds of carbon to oxygen. We lost a bond of carbon to hydrogen. Now, the difficulty is trying to isolate this aldehyde. Usually, it's very difficult to isolate."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So we increase the number of bonds of carbon to oxygen. We lost a bond of carbon to hydrogen. Now, the difficulty is trying to isolate this aldehyde. Usually, it's very difficult to isolate. And oxidation will continue. And you get a second oxidation to produce a carboxylic acid as your final product. So if you react a primary alcohol with the Jones reagent, you're going to end up with a carboxylic acid."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "Usually, it's very difficult to isolate. And oxidation will continue. And you get a second oxidation to produce a carboxylic acid as your final product. So if you react a primary alcohol with the Jones reagent, you're going to end up with a carboxylic acid. Let's look at an example. And we'll use ethanol as our primary alcohol here. So if we react ethanol with the Jones reagent right here, the chromium in the sodium dichromate is chromium 6 plus, which has kind of an orangish color to it."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So if you react a primary alcohol with the Jones reagent, you're going to end up with a carboxylic acid. Let's look at an example. And we'll use ethanol as our primary alcohol here. So if we react ethanol with the Jones reagent right here, the chromium in the sodium dichromate is chromium 6 plus, which has kind of an orangish color to it. So when you're starting off with your reaction, it's going to look a little bit orangish due to that chromium present. And when we oxidize our primary alcohol, when we oxidize our ethanol, we're going to turn it into a carboxylic acid. We're not changing the number of carbons."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So if we react ethanol with the Jones reagent right here, the chromium in the sodium dichromate is chromium 6 plus, which has kind of an orangish color to it. So when you're starting off with your reaction, it's going to look a little bit orangish due to that chromium present. And when we oxidize our primary alcohol, when we oxidize our ethanol, we're going to turn it into a carboxylic acid. We're not changing the number of carbons. So there's still going to be two carbons like this. But we're now going to change it into a carboxylic acid. So acetic acid will be the product."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "We're not changing the number of carbons. So there's still going to be two carbons like this. But we're now going to change it into a carboxylic acid. So acetic acid will be the product. So we went from this carbon having one bond to oxygen. And we oxidized it. So this carbon now has three bonds to oxygen atoms."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So acetic acid will be the product. So we went from this carbon having one bond to oxygen. And we oxidized it. So this carbon now has three bonds to oxygen atoms. And in that process, if we oxidize that alpha carbon, we're going to reduce the chromium. So the chromium is going to go from an oxidation state of 6 plus. And eventually, it's going to reach an oxidation state of 3 plus, like we talked about in the last video, which has kind of a greenish color."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon now has three bonds to oxygen atoms. And in that process, if we oxidize that alpha carbon, we're going to reduce the chromium. So the chromium is going to go from an oxidation state of 6 plus. And eventually, it's going to reach an oxidation state of 3 plus, like we talked about in the last video, which has kind of a greenish color. So it's very easy to monitor this reaction by just looking for the color change. And this is a very, very fast reaction. So this was originally used for the breathalyzer tests to determine if ethanol is present."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And eventually, it's going to reach an oxidation state of 3 plus, like we talked about in the last video, which has kind of a greenish color. So it's very easy to monitor this reaction by just looking for the color change. And this is a very, very fast reaction. So this was originally used for the breathalyzer tests to determine if ethanol is present. So let's see. What would happen if you wanted to actually stop it at the aldehyde? You don't want the oxidation to continue to the carboxylic acid."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So this was originally used for the breathalyzer tests to determine if ethanol is present. So let's see. What would happen if you wanted to actually stop it at the aldehyde? You don't want the oxidation to continue to the carboxylic acid. Let's say you wanted to actually stop it at the aldehyde. Well, to do that, you would have to use a different reagent. So let's go ahead and look and see how we could stop the reaction after the first oxidation."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "You don't want the oxidation to continue to the carboxylic acid. Let's say you wanted to actually stop it at the aldehyde. Well, to do that, you would have to use a different reagent. So let's go ahead and look and see how we could stop the reaction after the first oxidation. So if we started with a primary alcohol. So I'll just redraw a primary alcohol really fast here, like this. And if we wanted to oxidize it only once, so that we end up with an aldehyde, the best reagent to use for this is something called pyridinium chlorochromate, or PCC."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and look and see how we could stop the reaction after the first oxidation. So if we started with a primary alcohol. So I'll just redraw a primary alcohol really fast here, like this. And if we wanted to oxidize it only once, so that we end up with an aldehyde, the best reagent to use for this is something called pyridinium chlorochromate, or PCC. So let's take a look at the structure of the PCC reagent really fast. So pyridinium, let's go ahead and show what that looks like. So it's derived from pyridine."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And if we wanted to oxidize it only once, so that we end up with an aldehyde, the best reagent to use for this is something called pyridinium chlorochromate, or PCC. So let's take a look at the structure of the PCC reagent really fast. So pyridinium, let's go ahead and show what that looks like. So it's derived from pyridine. So let's go ahead and sketch that in like that. So pyridine as a base is going to pick up a proton to form a positive charge here. And then we have CrO3 and then Cl, and then with a negative charge."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So it's derived from pyridine. So let's go ahead and sketch that in like that. So pyridine as a base is going to pick up a proton to form a positive charge here. And then we have CrO3 and then Cl, and then with a negative charge. So this would be the pyridinium part. So let's go ahead and write it. Pyridinium."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And then we have CrO3 and then Cl, and then with a negative charge. So this would be the pyridinium part. So let's go ahead and write it. Pyridinium. And then we have chlorochromate over here on the right. So I'll go ahead and write chlorochromate. And then that makes it easier to see where the PCC comes from."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "Pyridinium. And then we have chlorochromate over here on the right. So I'll go ahead and write chlorochromate. And then that makes it easier to see where the PCC comes from. So this is the PCC reagent, which is a mild. It's a much more mild agent than the Jones reagent. It will oxidize your primary alcohol and stop at your aldehyde."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And then that makes it easier to see where the PCC comes from. So this is the PCC reagent, which is a mild. It's a much more mild agent than the Jones reagent. It will oxidize your primary alcohol and stop at your aldehyde. So let's go ahead and react ethanol again. This time we'll use PCC instead of Jones. So if we started with ethanol and we added PCC, so here we go, we're going to end up with an aldehyde."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "It will oxidize your primary alcohol and stop at your aldehyde. So let's go ahead and react ethanol again. This time we'll use PCC instead of Jones. So if we started with ethanol and we added PCC, so here we go, we're going to end up with an aldehyde. And it's a two-carbon aldehyde. So we can say those two carbons are still there. And we are going to form a double bond."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So if we started with ethanol and we added PCC, so here we go, we're going to end up with an aldehyde. And it's a two-carbon aldehyde. So we can say those two carbons are still there. And we are going to form a double bond. And this time it's going to be an aldehyde. So this is ethanol or acetaldehyde, which will be the result of this oxidation reaction. So that takes care of primary alcohols."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And we are going to form a double bond. And this time it's going to be an aldehyde. So this is ethanol or acetaldehyde, which will be the result of this oxidation reaction. So that takes care of primary alcohols. Let's look at the oxidation of secondary alcohols now. So we'll start with a general reaction over here. So we'll have a secondary alcohol."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So that takes care of primary alcohols. Let's look at the oxidation of secondary alcohols now. So we'll start with a general reaction over here. So we'll have a secondary alcohol. So two different alkyl groups, or they could be the same, attached to our alpha carbon. Our alpha carbon is attached to an OH. And remember, for the mechanism to work, you must have a hydrogen attached to that alpha carbon."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So we'll have a secondary alcohol. So two different alkyl groups, or they could be the same, attached to our alpha carbon. Our alpha carbon is attached to an OH. And remember, for the mechanism to work, you must have a hydrogen attached to that alpha carbon. So this is my secondary alcohol like that. Now, for secondary alcohols, we can only get one product. We saw in the last video that when you oxidize a secondary alcohol, you are going to end up with a ketone."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And remember, for the mechanism to work, you must have a hydrogen attached to that alpha carbon. So this is my secondary alcohol like that. Now, for secondary alcohols, we can only get one product. We saw in the last video that when you oxidize a secondary alcohol, you are going to end up with a ketone. So for a secondary alcohol, you could use either Jones or you could use PCC. So either one of those two reagents will oxidize a secondary alcohol to a ketone. So let's take a look at an example."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "We saw in the last video that when you oxidize a secondary alcohol, you are going to end up with a ketone. So for a secondary alcohol, you could use either Jones or you could use PCC. So either one of those two reagents will oxidize a secondary alcohol to a ketone. So let's take a look at an example. So let's start with a secondary alcohol. So I'm just going to draw a benzene ring on here and then attach that benzene ring. There will be a secondary alcohol present."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a look at an example. So let's start with a secondary alcohol. So I'm just going to draw a benzene ring on here and then attach that benzene ring. There will be a secondary alcohol present. So there's my secondary alcohol. And if I were to add either Jones or PCC, I look at my secondary alcohol. I identify my alpha carbon."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "There will be a secondary alcohol present. So there's my secondary alcohol. And if I were to add either Jones or PCC, I look at my secondary alcohol. I identify my alpha carbon. It's the one attached to the OH. And I can see there is one hydrogen attached to that alpha carbon. This is a secondary alcohol."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "I identify my alpha carbon. It's the one attached to the OH. And I can see there is one hydrogen attached to that alpha carbon. This is a secondary alcohol. So when I draw the product, I'm going to convert that secondary alcohol into a ketone. So if I were to do that, I would just real quickly redraw my benzene ring here. And I would convert that alpha carbon into a ketone."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "This is a secondary alcohol. So when I draw the product, I'm going to convert that secondary alcohol into a ketone. So if I were to do that, I would just real quickly redraw my benzene ring here. And I would convert that alpha carbon into a ketone. So that would be my product. So let's look at a tertiary alcohol. So if I had a tertiary alcohol, so like tert-butanol here like that, and if I attempted to oxidize that tertiary alcohol with either Jones or PCC, we saw in the last video no reaction."}, {"video_title": "Oxidation of alcohols II Examples Organic chemistry Khan Academy.mp3", "Sentence": "And I would convert that alpha carbon into a ketone. So that would be my product. So let's look at a tertiary alcohol. So if I had a tertiary alcohol, so like tert-butanol here like that, and if I attempted to oxidize that tertiary alcohol with either Jones or PCC, we saw in the last video no reaction. Because if I find the alpha carbon, this carbon right here, there are no hydrogens attached to that alpha carbon. And again, we saw in the mechanism that that was necessary. So something like tert-butanol would not be able to be oxidized in this fashion."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw something crazy here. So let's see, let me draw a chain. Let me draw it like that. And so, like we've done in all of the examples, you want to find the longest chain. We could count from here, 1, 2, 3, 4, 5, 6, 7, 8. Or maybe it's 1, 2, 3, 4, 5, 6, 7. No, or maybe it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And so, like we've done in all of the examples, you want to find the longest chain. We could count from here, 1, 2, 3, 4, 5, 6, 7, 8. Or maybe it's 1, 2, 3, 4, 5, 6, 7. No, or maybe it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. That is our longest chain. Let me just make that in green. So our longest chain here is in green."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "No, or maybe it's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. That is our longest chain. Let me just make that in green. So our longest chain here is in green. So this backbone has 10 carbons in it. The prefix for 10 is dec. It is a alkane, since it has all single bonds."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So our longest chain here is in green. So this backbone has 10 carbons in it. The prefix for 10 is dec. It is a alkane, since it has all single bonds. So we can write decane for the backbone. And then it has a group right here. And this group consists of 1, 2 carbons attached to the backbone."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "It is a alkane, since it has all single bonds. So we can write decane for the backbone. And then it has a group right here. And this group consists of 1, 2 carbons attached to the backbone. The prefix for 2 carbons is ethyl. So this is an ethyl group. The yl is because it's a group attached to the main alkane chain."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And this group consists of 1, 2 carbons attached to the backbone. The prefix for 2 carbons is ethyl. So this is an ethyl group. The yl is because it's a group attached to the main alkane chain. So we call this ethyl decane. But we have to specify where the ethyl group is attached. And we want to give it as low of a number as possible."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "The yl is because it's a group attached to the main alkane chain. So we call this ethyl decane. But we have to specify where the ethyl group is attached. And we want to give it as low of a number as possible. So we start counting on the side closest to it. So it's 1, 2, 3, 4, 5. So this is 5-ethyl decane."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And we want to give it as low of a number as possible. So we start counting on the side closest to it. So it's 1, 2, 3, 4, 5. So this is 5-ethyl decane. Now let's complicate this a little bit more. So let me just copy and paste this. So I have pasted it there."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is 5-ethyl decane. Now let's complicate this a little bit more. So let me just copy and paste this. So I have pasted it there. And let me complicate this molecule a little bit more. Let me add another ethyl group to it. So let's say we have another ethyl group over there."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So I have pasted it there. And let me complicate this molecule a little bit more. Let me add another ethyl group to it. So let's say we have another ethyl group over there. Now what is this going to be? Well, the longest chain is still going to be that thing in green. So it's still going to be a decane."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's say we have another ethyl group over there. Now what is this going to be? Well, the longest chain is still going to be that thing in green. So it's still going to be a decane. But now we have 2 ethyl groups. One on the 5-carbon, 1, 2, 3, 4, 5. And then one on the 6-carbon."}, {"video_title": "Naming alkanes with ethyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So it's still going to be a decane. But now we have 2 ethyl groups. One on the 5-carbon, 1, 2, 3, 4, 5. And then one on the 6-carbon. So what we write here is, you might be tempted to write 5- ethyl, 6-ethyl decane, which really wouldn't be wrong, but it would just be maybe more letters than you want to write. Instead you write 5, 6-diethyl decane. The 5, 6 tells us the 2 carbons on the main backbone that the ethyl groups are attached to."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So if I started with an aldehyde or ketone, and I add something like potassium cyanide and a source of protons, I'm going to form a cyanohydrin over here. And so this is a cyano group. And notice we're forming a carbon-carbon bond, so that can be useful for synthesis reactions. So let's look in more detail at our carbon nucleophile, which is this from the cyanide anion here. So let's go ahead and draw the cyanide anion. So that would be carbon triple bonded to nitrogen. The lone pair of electrons on this carbon like that."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's look in more detail at our carbon nucleophile, which is this from the cyanide anion here. So let's go ahead and draw the cyanide anion. So that would be carbon triple bonded to nitrogen. The lone pair of electrons on this carbon like that. And so that gives this carbon a negative 1 formal charge, which makes it a nucleophile. It's attracted to positively charged things. And over here on our carbonyl, we know our oxygen is partially negative, and we know our carbonyl carbon is partially positive."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The lone pair of electrons on this carbon like that. And so that gives this carbon a negative 1 formal charge, which makes it a nucleophile. It's attracted to positively charged things. And over here on our carbonyl, we know our oxygen is partially negative, and we know our carbonyl carbon is partially positive. So the carbonyl carbon is our electrophile. And so our nucleophile is going to attack our electrophile. So these opposite charges attract here."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And over here on our carbonyl, we know our oxygen is partially negative, and we know our carbonyl carbon is partially positive. So the carbonyl carbon is our electrophile. And so our nucleophile is going to attack our electrophile. So these opposite charges attract here. And these electrons are going to attack right here at this carbon, pushing these electrons off onto the oxygen. So let's draw the intermediate. We would now form a carbon-carbon bond."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these opposite charges attract here. And these electrons are going to attack right here at this carbon, pushing these electrons off onto the oxygen. So let's draw the intermediate. We would now form a carbon-carbon bond. And then this carbon is triple bonded to a nitrogen like that. So following our electrons, these electrons right here are forming this carbon-carbon bond. And then on the left side, we would have our oxygen, this time with three lone pairs of electrons on it, which gives it a negative 1 formal charge."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We would now form a carbon-carbon bond. And then this carbon is triple bonded to a nitrogen like that. So following our electrons, these electrons right here are forming this carbon-carbon bond. And then on the left side, we would have our oxygen, this time with three lone pairs of electrons on it, which gives it a negative 1 formal charge. And if we started with an aldehyde, we would have an R and an H here. And so in the final step, all we have to do is protonate our oxygen. So a lone pair of electrons picks up this proton to form this OH."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then on the left side, we would have our oxygen, this time with three lone pairs of electrons on it, which gives it a negative 1 formal charge. And if we started with an aldehyde, we would have an R and an H here. And so in the final step, all we have to do is protonate our oxygen. So a lone pair of electrons picks up this proton to form this OH. And we formed our cyanohydrin as our product. Let's look at an example of cyanohydrin formation. If we started with something like acetone, so ketone, and we added potassium cyanide and hydrochloric acid, we would form our cyanohydrin."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons picks up this proton to form this OH. And we formed our cyanohydrin as our product. Let's look at an example of cyanohydrin formation. If we started with something like acetone, so ketone, and we added potassium cyanide and hydrochloric acid, we would form our cyanohydrin. So we would have our carbon triple bonded to our nitrogen over here. And then we would have our OH over here. And since we started with a ketone this time, we would have these two methyl groups right here like that."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we started with something like acetone, so ketone, and we added potassium cyanide and hydrochloric acid, we would form our cyanohydrin. So we would have our carbon triple bonded to our nitrogen over here. And then we would have our OH over here. And since we started with a ketone this time, we would have these two methyl groups right here like that. And so once you form a cyano group here, you can convert this into other functional groups. So that's another reason why it is useful for a synthesis. Let's look at another reaction where we have a carbon nucleophile."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And since we started with a ketone this time, we would have these two methyl groups right here like that. And so once you form a cyano group here, you can convert this into other functional groups. So that's another reason why it is useful for a synthesis. Let's look at another reaction where we have a carbon nucleophile. And this would be the organometallics. And specifically, we have a Grignard reagent here. So we have an alkyl group bonded to magnesium and then a halogen, so an organomagnesium compound."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at another reaction where we have a carbon nucleophile. And this would be the organometallics. And specifically, we have a Grignard reagent here. So we have an alkyl group bonded to magnesium and then a halogen, so an organomagnesium compound. We could have just as easily done something like R prime and then a bond to lithium, so an organolithium compound. So these are just organometallics. Now, the thing about organometallics is the electronegativity difference between the carbon that's bonded, in this case, to the magnesium, where the carbon bonded to lithium."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have an alkyl group bonded to magnesium and then a halogen, so an organomagnesium compound. We could have just as easily done something like R prime and then a bond to lithium, so an organolithium compound. So these are just organometallics. Now, the thing about organometallics is the electronegativity difference between the carbon that's bonded, in this case, to the magnesium, where the carbon bonded to lithium. Carbon is more electronegative than those atoms. And so if you think about the bond, let me go ahead and highlight it here. This bond between the carbon and our magnesium, carbon's more electronegative."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now, the thing about organometallics is the electronegativity difference between the carbon that's bonded, in this case, to the magnesium, where the carbon bonded to lithium. Carbon is more electronegative than those atoms. And so if you think about the bond, let me go ahead and highlight it here. This bond between the carbon and our magnesium, carbon's more electronegative. So those electrons are polarized towards the carbon. And so you could think about these bonds as being very polar covalent bonds. Or you could even think about those electrons in blue being on your carbons."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This bond between the carbon and our magnesium, carbon's more electronegative. So those electrons are polarized towards the carbon. And so you could think about these bonds as being very polar covalent bonds. Or you could even think about those electrons in blue being on your carbons. Let's go ahead and draw it that way. So another way to think about it would be it's so polarized that those electrons are on your carbon, which gives you a negative 1 formal charge. And you form a carb anion."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Or you could even think about those electrons in blue being on your carbons. Let's go ahead and draw it that way. So another way to think about it would be it's so polarized that those electrons are on your carbon, which gives you a negative 1 formal charge. And you form a carb anion. So either way of thinking about it is fine with me. Maybe some professors might care. But once you think about it as being a carb anion, it makes these reactions a little bit easier."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And you form a carb anion. So either way of thinking about it is fine with me. Maybe some professors might care. But once you think about it as being a carb anion, it makes these reactions a little bit easier. Because once again, we have a polarized carbonyl situation, partial negative, partial positive. And so your nucleophile can attack right here, pushing these electrons off onto your oxygen. And so you could have just as easily have thought about it this way, with a very polar covalent bond right here and attacking here."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But once you think about it as being a carb anion, it makes these reactions a little bit easier. Because once again, we have a polarized carbonyl situation, partial negative, partial positive. And so your nucleophile can attack right here, pushing these electrons off onto your oxygen. And so you could have just as easily have thought about it this way, with a very polar covalent bond right here and attacking here. Or you could think about it as being ionic, with these electrons on the carbon forming a carb anion. So you'll see either mechanism, depending on which textbook that you look in. And so once the nucleophile attacks, we would now have a carbon bonded to our R double prime group."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so you could have just as easily have thought about it this way, with a very polar covalent bond right here and attacking here. Or you could think about it as being ionic, with these electrons on the carbon forming a carb anion. So you'll see either mechanism, depending on which textbook that you look in. And so once the nucleophile attacks, we would now have a carbon bonded to our R double prime group. And then over here on the left, we would have our negatively charged oxygen like that. So an R group and then a hydrogen, if we started with an aldehyde. And so those electrons in blue, let's go ahead and change them, make them red here."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so once the nucleophile attacks, we would now have a carbon bonded to our R double prime group. And then over here on the left, we would have our negatively charged oxygen like that. So an R group and then a hydrogen, if we started with an aldehyde. And so those electrons in blue, let's go ahead and change them, make them red here. So these electrons are right here. So those are these electrons that formed this bond. So we formed a carbon-carbon bond now."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so those electrons in blue, let's go ahead and change them, make them red here. So these electrons are right here. So those are these electrons that formed this bond. So we formed a carbon-carbon bond now. And then in the next step of the mechanism, which needs to be kept separate, we're going to add a proton source or something like water. And so once again, we're going to protonate our alkoxide intermediate. So we pick up a proton here."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we formed a carbon-carbon bond now. And then in the next step of the mechanism, which needs to be kept separate, we're going to add a proton source or something like water. And so once again, we're going to protonate our alkoxide intermediate. So we pick up a proton here. And then that gives us our final product, which is an alcohol. And we can see we've added on our R double prime group and formed a carbon-carbon bond. So once again, a very useful reaction for synthesis."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we pick up a proton here. And then that gives us our final product, which is an alcohol. And we can see we've added on our R double prime group and formed a carbon-carbon bond. So once again, a very useful reaction for synthesis. Now, you have to do this reaction in two steps because if you try to do it in one step, if you tried to add on the proton at the same time you added your organometallic, these are good nucleophiles. They're also strong bases. And so if you think about a carbanion here reacting with water, this is going to pick up the proton from water to form an alkane."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So once again, a very useful reaction for synthesis. Now, you have to do this reaction in two steps because if you try to do it in one step, if you tried to add on the proton at the same time you added your organometallic, these are good nucleophiles. They're also strong bases. And so if you think about a carbanion here reacting with water, this is going to pick up the proton from water to form an alkane. So it's just going to go like that and form an alkane. And so you don't want that to happen, which is why you need to separate these steps. And this is an oversimplified mechanism of what's happening."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so if you think about a carbanion here reacting with water, this is going to pick up the proton from water to form an alkane. So it's just going to go like that and form an alkane. And so you don't want that to happen, which is why you need to separate these steps. And this is an oversimplified mechanism of what's happening. But it's just the best way to think about it for an undergraduate student. Let's look at some examples of an organometallic reaction here. So this time we're starting with an aldehyde."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And this is an oversimplified mechanism of what's happening. But it's just the best way to think about it for an undergraduate student. Let's look at some examples of an organometallic reaction here. So this time we're starting with an aldehyde. And we have methyl magnesium bromide here in our first step. In the second step, we need to add a proton source. And so once again, you could think about the bonds between the carbon and the magnesium as being very polar covalent."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this time we're starting with an aldehyde. And we have methyl magnesium bromide here in our first step. In the second step, we need to add a proton source. And so once again, you could think about the bonds between the carbon and the magnesium as being very polar covalent. Or you could think about that bond as being ionic. And those two electrons in magenta on the carbon to form your carbanion. So like this, so a negative 1 formal charge in that carbon."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, you could think about the bonds between the carbon and the magnesium as being very polar covalent. Or you could think about that bond as being ionic. And those two electrons in magenta on the carbon to form your carbanion. So like this, so a negative 1 formal charge in that carbon. And so that's just a simple way of thinking about it here for trying to figure out the product. So you would attack right here and push these electrons off onto your oxygen. So if we showed the intermediate, we would have our carbon right here is now bonded to a methyl group, still have a hydrogen."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So like this, so a negative 1 formal charge in that carbon. And so that's just a simple way of thinking about it here for trying to figure out the product. So you would attack right here and push these electrons off onto your oxygen. So if we showed the intermediate, we would have our carbon right here is now bonded to a methyl group, still have a hydrogen. And then we would have an alkoxide anion as our intermediate like that, so negative 1 formal charge. And so the bond that's formed is between this carbon and this carbon like that. And in the second step, we would protonate our alkoxide to form our alcohol."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So if we showed the intermediate, we would have our carbon right here is now bonded to a methyl group, still have a hydrogen. And then we would have an alkoxide anion as our intermediate like that, so negative 1 formal charge. And so the bond that's formed is between this carbon and this carbon like that. And in the second step, we would protonate our alkoxide to form our alcohol. So let me go ahead and draw out our alcohol here. So we would have an OH. And then you can see that we have increased the number of carbons."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And in the second step, we would protonate our alkoxide to form our alcohol. So let me go ahead and draw out our alcohol here. So we would have an OH. And then you can see that we have increased the number of carbons. If we look at how many carbons that we started with, we had 1, 2, 3, 4 carbons. So we started with butanol. And then when we are done, we ended with, let's see, 1, 2, 3, 4, 5 carbons."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then you can see that we have increased the number of carbons. If we look at how many carbons that we started with, we had 1, 2, 3, 4 carbons. So we started with butanol. And then when we are done, we ended with, let's see, 1, 2, 3, 4, 5 carbons. So we finished with 2 pentanol. And we added 1 carbon on. So this methyl group is the one that we added on in our organometallic reaction for our Grignard reagent."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then when we are done, we ended with, let's see, 1, 2, 3, 4, 5 carbons. So we finished with 2 pentanol. And we added 1 carbon on. So this methyl group is the one that we added on in our organometallic reaction for our Grignard reagent. So this is what we started with. So notice when you start with an aldehyde, let me just change colors again, when you start with an aldehyde, you're going to end up with a secondary alcohol. So this carbon that's bonded to your OH is bonded to two other carbons."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this methyl group is the one that we added on in our organometallic reaction for our Grignard reagent. So this is what we started with. So notice when you start with an aldehyde, let me just change colors again, when you start with an aldehyde, you're going to end up with a secondary alcohol. So this carbon that's bonded to your OH is bonded to two other carbons. So an aldehyde goes to a secondary alcohol. Let's look at one more example. So this time, we're going to start with a ketone."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon that's bonded to your OH is bonded to two other carbons. So an aldehyde goes to a secondary alcohol. Let's look at one more example. So this time, we're going to start with a ketone. So same reagent. But this time, we're starting with a ketone. And so once again, thinking about what happens."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this time, we're going to start with a ketone. So same reagent. But this time, we're starting with a ketone. And so once again, thinking about what happens. So this time, I could just show these electrons right here attacking this carbon, pushing these electrons off onto the oxygen. And then in the second step, we know that we would protonate the alkoxide anion. And so we can just go ahead and draw the final product."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, thinking about what happens. So this time, I could just show these electrons right here attacking this carbon, pushing these electrons off onto the oxygen. And then in the second step, we know that we would protonate the alkoxide anion. And so we can just go ahead and draw the final product. We'd have our ring. And we're adding on, again, a methyl group. So let me go ahead and put a CH3 right here."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we can just go ahead and draw the final product. We'd have our ring. And we're adding on, again, a methyl group. So let me go ahead and put a CH3 right here. And then, of course, we would form an OH like that. So let's again think about electrons. So these electrons right here formed this bond."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and put a CH3 right here. And then, of course, we would form an OH like that. So let's again think about electrons. So these electrons right here formed this bond. And then we just added on CH3 too. So that's one way to think about how to draw your final products. And so this time, if you think about our starting material, we start with a ketone."}, {"video_title": "Addition of carbon nucleophiles to aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here formed this bond. And then we just added on CH3 too. So that's one way to think about how to draw your final products. And so this time, if you think about our starting material, we start with a ketone. And then we finished with a tertiary alcohol. So if I look at this carbon, the one that's bonded to the OH right here, this carbon is bonded to one, two, three other carbons. So this is one way to make a tertiary alcohol."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So the steric number is equal to the number of sigma bonds. So here's a sigma bond, here's a sigma bond, and here's a sigma bond. So three sigma bonds plus the number of lone pairs of electrons. So there's one lone pair of electrons on that nitrogen. So I'll go ahead and highlight them there. So 3 plus 1 gives us a total of 4 for the steric number, which means four hybrid orbitals, which implies sp3 hybridization for that nitrogen. And from earlier videos, you know that sp3 hybridization means a trigonal pyramidal geometry for that nitrogen."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So there's one lone pair of electrons on that nitrogen. So I'll go ahead and highlight them there. So 3 plus 1 gives us a total of 4 for the steric number, which means four hybrid orbitals, which implies sp3 hybridization for that nitrogen. And from earlier videos, you know that sp3 hybridization means a trigonal pyramidal geometry for that nitrogen. And so that's one way of looking at this functional group and that lone pair of electrons being localized to that nitrogen. However, now that we know resonance structures, we know that that lone pair of electrons is not localized to that nitrogen. It's delocalized in resonance."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And from earlier videos, you know that sp3 hybridization means a trigonal pyramidal geometry for that nitrogen. And so that's one way of looking at this functional group and that lone pair of electrons being localized to that nitrogen. However, now that we know resonance structures, we know that that lone pair of electrons is not localized to that nitrogen. It's delocalized in resonance. So we could take that lone pair of electrons and move it in here to form a pi bond. So that would force us to push some pi electrons off onto this oxygen. So let's go ahead and draw the other resonance structure."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "It's delocalized in resonance. So we could take that lone pair of electrons and move it in here to form a pi bond. So that would force us to push some pi electrons off onto this oxygen. So let's go ahead and draw the other resonance structure. So this top oxygen would now have three lone pairs of electrons around it, a negative 1 formal charge. And there'd be a double bond between this carbon and this nitrogen. So let's go ahead and draw in everything."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the other resonance structure. So this top oxygen would now have three lone pairs of electrons around it, a negative 1 formal charge. And there'd be a double bond between this carbon and this nitrogen. So let's go ahead and draw in everything. This nitrogen now has a plus 1 formal charge. And we can go ahead and complete our resonance bracket here. So let's follow those electrons along."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in everything. This nitrogen now has a plus 1 formal charge. And we can go ahead and complete our resonance bracket here. So let's follow those electrons along. The electrons in magenta, the lone pair of electrons, moved in here to form our pi bond. And the pi electrons over here in blue came off onto the oxygen to give the oxygen a negative 1 formal charge. Let's now calculate a steric number for this nitrogen in our second resonance structure here."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons along. The electrons in magenta, the lone pair of electrons, moved in here to form our pi bond. And the pi electrons over here in blue came off onto the oxygen to give the oxygen a negative 1 formal charge. Let's now calculate a steric number for this nitrogen in our second resonance structure here. So a steric number is equal to number of sigma bonds. So here's a sigma bond. Here's a sigma bond."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Let's now calculate a steric number for this nitrogen in our second resonance structure here. So a steric number is equal to number of sigma bonds. So here's a sigma bond. Here's a sigma bond. And our double bond, we know one of them is a sigma bond and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen. Zero lone pairs of electrons, so 3 plus 0 gives us a steric number of 3."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Here's a sigma bond. And our double bond, we know one of them is a sigma bond and one of them is a pi bond. So we have a total of three sigma bonds around our nitrogen. Zero lone pairs of electrons, so 3 plus 0 gives us a steric number of 3. That implies three hybrid orbitals, which means sp2 hybridization, and a trigonal planar geometry around that nitrogen. So a planar geometry. And here we've shown our electrons being delocalized."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Zero lone pairs of electrons, so 3 plus 0 gives us a steric number of 3. That implies three hybrid orbitals, which means sp2 hybridization, and a trigonal planar geometry around that nitrogen. So a planar geometry. And here we've shown our electrons being delocalized. So the lone pair of electrons are delocalized due to resonance. And so experimental studies have shown that the amide functional group is planar. So these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And here we've shown our electrons being delocalized. So the lone pair of electrons are delocalized due to resonance. And so experimental studies have shown that the amide functional group is planar. So these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen. They are actually delocalized. And so that implies this nitrogen is sp2 hybridized and has a p orbital. And that allows that lone pair of electrons in magenta to be delocalized."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So these atoms actually are planar here, which means that the electrons in magenta are not localized to that nitrogen. They are actually delocalized. And so that implies this nitrogen is sp2 hybridized and has a p orbital. And that allows that lone pair of electrons in magenta to be delocalized. And so here's a situation where drawing resonance structure helps clue us into what's actually happening. That lone pair is participating in resonance, which makes this nitrogen sp2 hybridized. So it has a p orbital."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And that allows that lone pair of electrons in magenta to be delocalized. And so here's a situation where drawing resonance structure helps clue us into what's actually happening. That lone pair is participating in resonance, which makes this nitrogen sp2 hybridized. So it has a p orbital. Let's look at this example down here. And let's look first at this left side of the molecule. And so we can see our amide functional group."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So it has a p orbital. Let's look at this example down here. And let's look first at this left side of the molecule. And so we can see our amide functional group. And if I look at the lone pair of electrons on this nitrogen, we've just talked about the fact that this lone pair of electrons is actually delocalized. So this lone pair is participating in resonance. And so that, of course, affects the geometry and how you think about the hybridization of this nitrogen here."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And so we can see our amide functional group. And if I look at the lone pair of electrons on this nitrogen, we've just talked about the fact that this lone pair of electrons is actually delocalized. So this lone pair is participating in resonance. And so that, of course, affects the geometry and how you think about the hybridization of this nitrogen here. So the electrons in magenta are delocalized because they participate in resonance. And if I think about, let's make this a different color here. Let's make these electrons in here blue."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And so that, of course, affects the geometry and how you think about the hybridization of this nitrogen here. So the electrons in magenta are delocalized because they participate in resonance. And if I think about, let's make this a different color here. Let's make these electrons in here blue. So the electrons in blue on the amine, these electrons have nowhere to go. They can't participate in resonance. That lone pair of electrons in blue is localized to that nitrogen."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Let's make these electrons in here blue. So the electrons in blue on the amine, these electrons have nowhere to go. They can't participate in resonance. That lone pair of electrons in blue is localized to that nitrogen. And so this is why you can think about an amide being different from an amine in terms of functional group and in terms of how they react and how they behave. If we look at another example, so this molecule right here. And we assume the lone pair of electrons on that nitrogen is localized to that nitrogen."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "That lone pair of electrons in blue is localized to that nitrogen. And so this is why you can think about an amide being different from an amine in terms of functional group and in terms of how they react and how they behave. If we look at another example, so this molecule right here. And we assume the lone pair of electrons on that nitrogen is localized to that nitrogen. Let's go ahead and calculate the steric number. So the steric number would be equal to sigma bonds. So that's one, two, and three."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And we assume the lone pair of electrons on that nitrogen is localized to that nitrogen. Let's go ahead and calculate the steric number. So the steric number would be equal to sigma bonds. So that's one, two, and three. So three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen. So three plus one is four."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So that's one, two, and three. So three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen. So three plus one is four. So four hybrid orbitals, which implies SP3 hybridization on that nitrogen. But we know that that lone pair of electrons is not localized to that nitrogen. That lone pair of electrons is delocalized."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So three plus one is four. So four hybrid orbitals, which implies SP3 hybridization on that nitrogen. But we know that that lone pair of electrons is not localized to that nitrogen. That lone pair of electrons is delocalized. It participates in resonance because we have this pattern here. So this pattern of a lone pair of electrons in blue next to a pi bond, which I will make magenta right here. And so we could draw a resonance structure."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "That lone pair of electrons is delocalized. It participates in resonance because we have this pattern here. So this pattern of a lone pair of electrons in blue next to a pi bond, which I will make magenta right here. And so we could draw a resonance structure. So I could take the electrons in blue, move them into here. Too many bonds to this carbon, so I push electrons in magenta off onto this carbon. So we draw the resonance structure."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And so we could draw a resonance structure. So I could take the electrons in blue, move them into here. Too many bonds to this carbon, so I push electrons in magenta off onto this carbon. So we draw the resonance structure. So I have my ring here. Nitrogen's bonded to a hydrogen. The electrons in blue moved in to form a pi bond."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So we draw the resonance structure. So I have my ring here. Nitrogen's bonded to a hydrogen. The electrons in blue moved in to form a pi bond. And the electrons in magenta moved off onto this carbon right here to give that carbon a negative one formal charge. Let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. So the steric number would be equal to number of sigma bonds."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in blue moved in to form a pi bond. And the electrons in magenta moved off onto this carbon right here to give that carbon a negative one formal charge. Let's go ahead and calculate a steric number for the nitrogen here, which gets a plus one formal charge. So the steric number would be equal to number of sigma bonds. So there's one sigma bond. Here's another sigma bond. And then in our double bond, one of them is a sigma and one of them is a pi."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So the steric number would be equal to number of sigma bonds. So there's one sigma bond. Here's another sigma bond. And then in our double bond, one of them is a sigma and one of them is a pi. So I'm saying that's our sigma bond. So the steric number is equal to 3 plus the number of lone pairs of electrons, now zero. So that's a steric number of 3, which implies three hybrid orbitals, which says sp2 hybridization."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And then in our double bond, one of them is a sigma and one of them is a pi. So I'm saying that's our sigma bond. So the steric number is equal to 3 plus the number of lone pairs of electrons, now zero. So that's a steric number of 3, which implies three hybrid orbitals, which says sp2 hybridization. And since we know that that lone pair is delocalized, it's going to occupy a p orbital. And so therefore, this nitrogen is sp2 hybridized, because we know sp2 hybridization has three sp2 hybrid orbitals and one p orbital. So that lone pair in blue is actually delocalized."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So that's a steric number of 3, which implies three hybrid orbitals, which says sp2 hybridization. And since we know that that lone pair is delocalized, it's going to occupy a p orbital. And so therefore, this nitrogen is sp2 hybridized, because we know sp2 hybridization has three sp2 hybrid orbitals and one p orbital. So that lone pair in blue is actually delocalized. It's occupying a p orbital. And so let's go ahead and draw that down here. So let's say this is the nitrogen, and you're looking at it at a bit of an angle."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So that lone pair in blue is actually delocalized. It's occupying a p orbital. And so let's go ahead and draw that down here. So let's say this is the nitrogen, and you're looking at it at a bit of an angle. If that nitrogen is sp2 hybridized, that nitrogen has a p orbital. So we can go ahead and draw in a p orbital on that nitrogen. And so the electrons in blue, since those electrons are delocalized, those electrons are going to occupy that p orbital."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So let's say this is the nitrogen, and you're looking at it at a bit of an angle. If that nitrogen is sp2 hybridized, that nitrogen has a p orbital. So we can go ahead and draw in a p orbital on that nitrogen. And so the electrons in blue, since those electrons are delocalized, those electrons are going to occupy that p orbital. Those electrons can participate in resonance. So let's go ahead and draw a hydrogen here, just to finish that off. And I could have kept on going with resonance structures."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And so the electrons in blue, since those electrons are delocalized, those electrons are going to occupy that p orbital. Those electrons can participate in resonance. So let's go ahead and draw a hydrogen here, just to finish that off. And I could have kept on going with resonance structures. So I actually forgot to put in these electrons right here, because I could have kept on going. I could take these electrons in magenta, move them into here, and then push these electrons off onto that carbon. And I could keep on going."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And I could have kept on going with resonance structures. So I actually forgot to put in these electrons right here, because I could have kept on going. I could take these electrons in magenta, move them into here, and then push these electrons off onto that carbon. And I could keep on going. And so there are many more resonance structures that you could draw. I'm not going to do so for time purposes. Here I'm just trying to show how resonance affects the hybridization, or how you should think about the hybridization when you're drawing these imperfect dot structures."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And I could keep on going. And so there are many more resonance structures that you could draw. I'm not going to do so for time purposes. Here I'm just trying to show how resonance affects the hybridization, or how you should think about the hybridization when you're drawing these imperfect dot structures. Let's go ahead and finish our picture of the molecule, because we know that these four carbons here all have a double bond to them. And so those four carbons are therefore sp2 hybridized. And if those carbons are sp2 hybridized, each one of those carbons has a p orbital."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Here I'm just trying to show how resonance affects the hybridization, or how you should think about the hybridization when you're drawing these imperfect dot structures. Let's go ahead and finish our picture of the molecule, because we know that these four carbons here all have a double bond to them. And so those four carbons are therefore sp2 hybridized. And if those carbons are sp2 hybridized, each one of those carbons has a p orbital. So I could draw in a p orbital on all of my carbons here. And you could think about those electrons. Let me go ahead and highlight them."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And if those carbons are sp2 hybridized, each one of those carbons has a p orbital. So I could draw in a p orbital on all of my carbons here. And you could think about those electrons. Let me go ahead and highlight them. The electrons in red here. So these pi electrons are participating in resonance. And so we have a total of six pi electrons participating in resonance."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and highlight them. The electrons in red here. So these pi electrons are participating in resonance. And so we have a total of six pi electrons participating in resonance. Those four in red that I just highlighted here. And then these electrons here in blue, which occupy this p orbital. And so this is going to be extremely important when you talk about aromatic compounds later in the year."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And so we have a total of six pi electrons participating in resonance. Those four in red that I just highlighted here. And then these electrons here in blue, which occupy this p orbital. And so this is going to be extremely important when you talk about aromatic compounds later in the year. So for right now, just try to identify the skill of delocalized electrons versus localized. In this case, the electrons on the nitrogen are delocalized. They participate in resonance."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And so this is going to be extremely important when you talk about aromatic compounds later in the year. So for right now, just try to identify the skill of delocalized electrons versus localized. In this case, the electrons on the nitrogen are delocalized. They participate in resonance. Let's do one more example. So one final example here. And let's go ahead and start by calculating the steric number of this nitrogen."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "They participate in resonance. Let's do one more example. So one final example here. And let's go ahead and start by calculating the steric number of this nitrogen. So how I've drawn it here. So steric number is equal to number of sigma bonds. So here's a sigma bond."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and start by calculating the steric number of this nitrogen. So how I've drawn it here. So steric number is equal to number of sigma bonds. So here's a sigma bond. I have a double bond. One of them is a sigma bond. So I have two sigma bonds plus lone pairs of electrons."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So here's a sigma bond. I have a double bond. One of them is a sigma bond. So I have two sigma bonds plus lone pairs of electrons. There's one lone pair of electrons here on this nitrogen. So 2 plus 1 gives me a steric number of 3. That implies three hybrid orbitals, or sp2 hybridization."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So I have two sigma bonds plus lone pairs of electrons. There's one lone pair of electrons here on this nitrogen. So 2 plus 1 gives me a steric number of 3. That implies three hybrid orbitals, or sp2 hybridization. So if that nitrogen is sp2 hybridized, that nitrogen is going to have a p orbital. So let me go ahead and sketch in a p orbital here on this nitrogen. And let's think about the other carbons too."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "That implies three hybrid orbitals, or sp2 hybridization. So if that nitrogen is sp2 hybridized, that nitrogen is going to have a p orbital. So let me go ahead and sketch in a p orbital here on this nitrogen. And let's think about the other carbons too. So let's think about these carbons, I should say. So I have one, two, three, four, five carbons. All these carbons have a double bond to them."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And let's think about the other carbons too. So let's think about these carbons, I should say. So I have one, two, three, four, five carbons. All these carbons have a double bond to them. So all of those carbons are sp2 hybridized. So I can draw in a p orbital on all of those carbons. So all of those carbons have three sp2 hybrid orbitals and one unhybridized p orbital."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "All these carbons have a double bond to them. So all of those carbons are sp2 hybridized. So I can draw in a p orbital on all of those carbons. So all of those carbons have three sp2 hybrid orbitals and one unhybridized p orbital. So it's the same idea with this nitrogen here. If this nitrogen is sp2 hybridized, it has three sp2 hybrid orbitals and one unhybridized p orbital. I already drew in the unhybridized p orbital."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So all of those carbons have three sp2 hybrid orbitals and one unhybridized p orbital. So it's the same idea with this nitrogen here. If this nitrogen is sp2 hybridized, it has three sp2 hybrid orbitals and one unhybridized p orbital. I already drew in the unhybridized p orbital. Let's go ahead and put in those sp2 hybrid orbitals. And I'll use magenta. So one of those sp2 hybrid orbitals is forming this sigma bond over here."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "I already drew in the unhybridized p orbital. Let's go ahead and put in those sp2 hybrid orbitals. And I'll use magenta. So one of those sp2 hybrid orbitals is forming this sigma bond over here. Another one's forming this sigma bond back here. And then the third sp2 hybrid orbital must be here. And that's actually where that lone pair of electrons is."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So one of those sp2 hybrid orbitals is forming this sigma bond over here. Another one's forming this sigma bond back here. And then the third sp2 hybrid orbital must be here. And that's actually where that lone pair of electrons is. So this lone pair of electrons here in blue is localized to this sp2 hybrid orbital. So this is actually, let me go ahead and draw an arrow here. So the lone pair of electrons is actually localized to that sp2 hybrid orbital."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And that's actually where that lone pair of electrons is. So this lone pair of electrons here in blue is localized to this sp2 hybrid orbital. So this is actually, let me go ahead and draw an arrow here. So the lone pair of electrons is actually localized to that sp2 hybrid orbital. So that lone pair of electrons can't participate in resonance because that nitrogen already has a pi bond. So if I go over to here, these pi electrons are using that p orbital. And since those pi electrons in green are using this p orbital over here on the right, so this p orbital over here, that means that those electrons in blue can't use it."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So the lone pair of electrons is actually localized to that sp2 hybrid orbital. So that lone pair of electrons can't participate in resonance because that nitrogen already has a pi bond. So if I go over to here, these pi electrons are using that p orbital. And since those pi electrons in green are using this p orbital over here on the right, so this p orbital over here, that means that those electrons in blue can't use it. And so therefore, those electrons in blue are actually localized to that nitrogen, which might not be what you first thought. Because we talked about the pattern, of course, of the electrons in blue here, lone pair of electrons next to a pi bond. So you might think we could draw a resonance structure."}, {"video_title": "Resonance structures and hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And since those pi electrons in green are using this p orbital over here on the right, so this p orbital over here, that means that those electrons in blue can't use it. And so therefore, those electrons in blue are actually localized to that nitrogen, which might not be what you first thought. Because we talked about the pattern, of course, of the electrons in blue here, lone pair of electrons next to a pi bond. So you might think we could draw a resonance structure. However, the reason we can't is, once again, those electrons in green are actually participating in resonance. And so they're already using that p orbital. So this is kind of an exception to what we talked about before, where we said if there's a lone pair of electrons next to a pi bond, that lone pair is delocalized."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "One way to do it is to start by drawing two parallel lines that are offset from each other. So let me go ahead and show you what I mean. So here's one line, and then here is another line. They're parallel to each other, but they're offset a little bit. Next, we're gonna draw two horizontal dotted lines. So the top horizontal dotted line is going to be on level with that top point there. So I'm just gonna draw a dotted line here."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "They're parallel to each other, but they're offset a little bit. Next, we're gonna draw two horizontal dotted lines. So the top horizontal dotted line is going to be on level with that top point there. So I'm just gonna draw a dotted line here. I'm gonna draw another dotted line down here, which is gonna be level with this bottom point. So that's what we have so far. Next, we're gonna draw in another set of parallel lines."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just gonna draw a dotted line here. I'm gonna draw another dotted line down here, which is gonna be level with this bottom point. So that's what we have so far. Next, we're gonna draw in another set of parallel lines. And we're gonna start at this point right here, where this intersects with the dotted line. And we're gonna go from the top dotted line down to the bottom dotted line like that. And then we're gonna try to think about drawing a line parallel to that over here."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Next, we're gonna draw in another set of parallel lines. And we're gonna start at this point right here, where this intersects with the dotted line. And we're gonna go from the top dotted line down to the bottom dotted line like that. And then we're gonna try to think about drawing a line parallel to that over here. So we're starting at this dotted line, we're going up to this point. And then finally, we need to connect the dots. So we're gonna go from this point to here, and then from this point to here."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And then we're gonna try to think about drawing a line parallel to that over here. So we're starting at this dotted line, we're going up to this point. And then finally, we need to connect the dots. So we're gonna go from this point to here, and then from this point to here. And that's our carbon skeleton. So if you draw this properly, you should have three sets of parallel lines for your carbon skeleton. Let's go back up to this drawing here, and let me show you those parallel lines."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna go from this point to here, and then from this point to here. And that's our carbon skeleton. So if you draw this properly, you should have three sets of parallel lines for your carbon skeleton. Let's go back up to this drawing here, and let me show you those parallel lines. So we started with this set. So this bond and this bond. That's what we started with."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Let's go back up to this drawing here, and let me show you those parallel lines. So we started with this set. So this bond and this bond. That's what we started with. Next we drew in, let me change colors so you can see, we drew in this set of parallel lines. And then finally, we drew in the last set. So let me highlight those in blue."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "That's what we started with. Next we drew in, let me change colors so you can see, we drew in this set of parallel lines. And then finally, we drew in the last set. So let me highlight those in blue. So this line and this line are parallel. So that gives you your carbon skeleton. And next you need to think about putting in your groups."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So let me highlight those in blue. So this line and this line are parallel. So that gives you your carbon skeleton. And next you need to think about putting in your groups. In this case, we're talking about cyclohexane, so we're only dealing with hydrogens. So first, we're gonna put in the hydrogens that we call axial, or axial groups here. And axial groups, you can think about the Earth."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And next you need to think about putting in your groups. In this case, we're talking about cyclohexane, so we're only dealing with hydrogens. So first, we're gonna put in the hydrogens that we call axial, or axial groups here. And axial groups, you can think about the Earth. So here is the Earth. Think about this as being a globe, and this would be the axis going straight up and down. So hydrogens that go straight up and down are called axial hydrogens."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And axial groups, you can think about the Earth. So here is the Earth. Think about this as being a globe, and this would be the axis going straight up and down. So hydrogens that go straight up and down are called axial hydrogens. And we're gonna start at this carbon right here, which I'll call carbon one. So if this is carbon one, we're gonna draw in an axial hydrogen. So this is going straight up."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So hydrogens that go straight up and down are called axial hydrogens. And we're gonna start at this carbon right here, which I'll call carbon one. So if this is carbon one, we're gonna draw in an axial hydrogen. So this is going straight up. Next, we're gonna go to carbon two, so this one right here, and straight down would be an axial hydrogen. We're gonna go to carbon three, and it's straight up for an axial hydrogen. Carbon four is straight down for an axial hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So this is going straight up. Next, we're gonna go to carbon two, so this one right here, and straight down would be an axial hydrogen. We're gonna go to carbon three, and it's straight up for an axial hydrogen. Carbon four is straight down for an axial hydrogen. Carbon five back here would be up. And then carbon six would be down axial hydrogen. So you can see that we alternate."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Carbon four is straight down for an axial hydrogen. Carbon five back here would be up. And then carbon six would be down axial hydrogen. So you can see that we alternate. If we start at carbon one, we start up axial. So let's go ahead and put in the axial hydrogens on our cyclohexane, on our chair conformation. So here is carbon one."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So you can see that we alternate. If we start at carbon one, we start up axial. So let's go ahead and put in the axial hydrogens on our cyclohexane, on our chair conformation. So here is carbon one. We're gonna draw a line straight up and put in a hydrogen. We go to carbon two. We draw a line straight down and put in a hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So here is carbon one. We're gonna draw a line straight up and put in a hydrogen. We go to carbon two. We draw a line straight down and put in a hydrogen. Carbon three, we draw a line straight up and put in a hydrogen. Carbon four, we draw a line straight down and put in a hydrogen. Carbon five would be up, so we put in that hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "We draw a line straight down and put in a hydrogen. Carbon three, we draw a line straight up and put in a hydrogen. Carbon four, we draw a line straight down and put in a hydrogen. Carbon five would be up, so we put in that hydrogen. And finally, carbon six would be down for our hydrogen. All right, next, we need to think about the equatorial hydrogens. So that would be, starting at carbon one, that would be this hydrogen right here."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Carbon five would be up, so we put in that hydrogen. And finally, carbon six would be down for our hydrogen. All right, next, we need to think about the equatorial hydrogens. So that would be, starting at carbon one, that would be this hydrogen right here. Carbon two, it would be this hydrogen. And they're called equatorial because you could think about them as being along the equator of the ring. So if this is the Earth, we know that this is the equator."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So that would be, starting at carbon one, that would be this hydrogen right here. Carbon two, it would be this hydrogen. And they're called equatorial because you could think about them as being along the equator of the ring. So if this is the Earth, we know that this is the equator. So these are kind of out to the side. And they alternate, too. At carbon one, this equatorial hydrogen is down relative to the plane of the ring."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So if this is the Earth, we know that this is the equator. So these are kind of out to the side. And they alternate, too. At carbon one, this equatorial hydrogen is down relative to the plane of the ring. At carbon two, this equatorial hydrogen is up relative to the plane of the ring. So we start down and then up, and then at carbon three, this would be down. At carbon four, this would be up."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "At carbon one, this equatorial hydrogen is down relative to the plane of the ring. At carbon two, this equatorial hydrogen is up relative to the plane of the ring. So we start down and then up, and then at carbon three, this would be down. At carbon four, this would be up. At carbon five, this would be down. At carbon six, this would be up. So just imagine all of these hydrogens are along the equator of the ring."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "At carbon four, this would be up. At carbon five, this would be down. At carbon six, this would be up. So just imagine all of these hydrogens are along the equator of the ring. So let's put in those equatorial hydrogens into our chair conformation. We start at carbon one, and we start down. So down equatorial."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So just imagine all of these hydrogens are along the equator of the ring. So let's put in those equatorial hydrogens into our chair conformation. We start at carbon one, and we start down. So down equatorial. So we draw a line down here and put in our hydrogen. And then we go up. So at carbon two, it would be up."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So down equatorial. So we draw a line down here and put in our hydrogen. And then we go up. So at carbon two, it would be up. So we put that in. We alternate. So at carbon three, it would be down."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So at carbon two, it would be up. So we put that in. We alternate. So at carbon three, it would be down. So we draw in that hydrogen. At carbon four, it would be up. So we put in that hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So at carbon three, it would be down. So we draw in that hydrogen. At carbon four, it would be up. So we put in that hydrogen. At carbon five, it would be down. So we put that one in. And carbon six would be up."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we put in that hydrogen. At carbon five, it would be down. So we put that one in. And carbon six would be up. So we put that one in. These are the ones that students have the hardest time with. So you could think about the parallel lines that we drew earlier to help guide you."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And carbon six would be up. So we put that one in. These are the ones that students have the hardest time with. So you could think about the parallel lines that we drew earlier to help guide you. So let me go ahead and show you what I mean. So we started with the parallel lines in yellow over here. So if you look at those lines, you should have these, right?"}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about the parallel lines that we drew earlier to help guide you. So let me go ahead and show you what I mean. So we started with the parallel lines in yellow over here. So if you look at those lines, you should have these, right? These are also parallel. So this one is parallel and this one is parallel. Also, you could look at the ones in red next."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at those lines, you should have these, right? These are also parallel. So this one is parallel and this one is parallel. Also, you could look at the ones in red next. So let me highlight those again. So this set of parallel lines, these two right here. So that would be these hydrogens, right?"}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Also, you could look at the ones in red next. So let me highlight those again. So this set of parallel lines, these two right here. So that would be these hydrogens, right? Or these groups, I should say. Those should be parallel. And finally, let's look at the ones in blue."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So that would be these hydrogens, right? Or these groups, I should say. Those should be parallel. And finally, let's look at the ones in blue. So let me highlight those again really quickly. Let me use a darker blue so we can see it a little bit better. So this one and this one, right?"}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And finally, let's look at the ones in blue. So let me highlight those again really quickly. Let me use a darker blue so we can see it a little bit better. So this one and this one, right? Those were parallel lines. Well, we can also see that these, right? These would be parallel to those as well."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So this one and this one, right? Those were parallel lines. Well, we can also see that these, right? These would be parallel to those as well. There's another way of thinking about this which might be helpful. If you look at the drawing on the right, I'm highlighting some of the bonds in magenta. And when I'm done, this should look familiar."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "These would be parallel to those as well. There's another way of thinking about this which might be helpful. If you look at the drawing on the right, I'm highlighting some of the bonds in magenta. And when I'm done, this should look familiar. You've seen this before in the video on conformations of ethane. That was a sawhorse drawing for the conformation of ethane, the staggered conformation. Let me draw it over here so you can see it a little bit better."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And when I'm done, this should look familiar. You've seen this before in the video on conformations of ethane. That was a sawhorse drawing for the conformation of ethane, the staggered conformation. Let me draw it over here so you can see it a little bit better. So here we have the staggered conformation for ethane. And that's helpful sometimes when you're thinking about how to draw the different bonds in the chair conformation. So pretty much you can use any method you want."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw it over here so you can see it a little bit better. So here we have the staggered conformation for ethane. And that's helpful sometimes when you're thinking about how to draw the different bonds in the chair conformation. So pretty much you can use any method you want. I've seen lots of ways to draw chair conformations and lots of students come up with their own way of doing it. So whatever works for you, just be consistent and be careful. Take your time and you have to practice a lot."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So pretty much you can use any method you want. I've seen lots of ways to draw chair conformations and lots of students come up with their own way of doing it. So whatever works for you, just be consistent and be careful. Take your time and you have to practice a lot. Next, let's look at another chair conformation that you'll be drawing. We need some practice drawing the other chair conformation. We're gonna start the same way we did before with two parallel lines that are offset from each other."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Take your time and you have to practice a lot. Next, let's look at another chair conformation that you'll be drawing. We need some practice drawing the other chair conformation. We're gonna start the same way we did before with two parallel lines that are offset from each other. This time though, the lines will be going in a different direction than what we did before. So let's go ahead and put in one parallel line. So there's one and then another one like that."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "We're gonna start the same way we did before with two parallel lines that are offset from each other. This time though, the lines will be going in a different direction than what we did before. So let's go ahead and put in one parallel line. So there's one and then another one like that. So hopefully those are pretty much parallel. And then we're gonna draw in a dotted line, a horizontal dotted line that just touches this top point here on this top line that we drew. And these dotted lines, remember, are just guidelines."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So there's one and then another one like that. So hopefully those are pretty much parallel. And then we're gonna draw in a dotted line, a horizontal dotted line that just touches this top point here on this top line that we drew. And these dotted lines, remember, are just guidelines. They're not actually a part of our chair conformation. We draw another dotted line down here that intersects with this bottom point as our guideline. Next, we go from the top dotted line down to the bottom dotted line."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And these dotted lines, remember, are just guidelines. They're not actually a part of our chair conformation. We draw another dotted line down here that intersects with this bottom point as our guideline. Next, we go from the top dotted line down to the bottom dotted line. So from the top down to the bottom and then we go from the bottom back up to the top. And finally, we just connect the dots. So from this point, we draw a line to here and then we draw that line again, a parallel line at this point to this point."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Next, we go from the top dotted line down to the bottom dotted line. So from the top down to the bottom and then we go from the bottom back up to the top. And finally, we just connect the dots. So from this point, we draw a line to here and then we draw that line again, a parallel line at this point to this point. So make sure that everything is parallel. You have three sets of parallel lines. The first set of parallel lines we drew was this one up here on the drawing, so in yellow."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So from this point, we draw a line to here and then we draw that line again, a parallel line at this point to this point. So make sure that everything is parallel. You have three sets of parallel lines. The first set of parallel lines we drew was this one up here on the drawing, so in yellow. Next, in red, we went from this point down to this point and then we went from this point up to this point in red. Those should be parallel. And then finally, in blue, this line in blue should be parallel with this line in blue."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "The first set of parallel lines we drew was this one up here on the drawing, so in yellow. Next, in red, we went from this point down to this point and then we went from this point up to this point in red. Those should be parallel. And then finally, in blue, this line in blue should be parallel with this line in blue. So make sure you have three sets of parallel lines to draw your carbon skeleton. Next, this turns out to be carbon one right here and we'll see why in the next video. So we're gonna start to put in our groups."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, in blue, this line in blue should be parallel with this line in blue. So make sure you have three sets of parallel lines to draw your carbon skeleton. Next, this turns out to be carbon one right here and we'll see why in the next video. So we're gonna start to put in our groups. In this case, we're just gonna put in all hydrogens. And at carbon one, you can see that we're starting axial down this time. So we're gonna put a hydrogen axial down."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna start to put in our groups. In this case, we're just gonna put in all hydrogens. And at carbon one, you can see that we're starting axial down this time. So we're gonna put a hydrogen axial down. That's the opposite of what we did in the other chair conformation. At C1 in the other chair conformation, we started axial up. So we're gonna start axial down."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna put a hydrogen axial down. That's the opposite of what we did in the other chair conformation. At C1 in the other chair conformation, we started axial up. So we're gonna start axial down. So we put a hydrogen going down at this point. And we move over to carbon two. So this is carbon two right here."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna start axial down. So we put a hydrogen going down at this point. And we move over to carbon two. So this is carbon two right here. You can see now we have axial up for our hydrogen. So we go to carbon two and we have axial up. Carbon three would be this one."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So this is carbon two right here. You can see now we have axial up for our hydrogen. So we go to carbon two and we have axial up. Carbon three would be this one. You can see that it's axial down. So carbon three is axial down. And again, this just alternates."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Carbon three would be this one. You can see that it's axial down. So carbon three is axial down. And again, this just alternates. So carbon four should be axial up. So we draw in our hydrogen going up at carbon four. Carbon five, axial down."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And again, this just alternates. So carbon four should be axial up. So we draw in our hydrogen going up at carbon four. Carbon five, axial down. So we draw in our hydrogen here. And finally, carbon six is axial up. So let's put in that hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Carbon five, axial down. So we draw in our hydrogen here. And finally, carbon six is axial up. So let's put in that hydrogen. Next, we put in the equatorial hydrogens, which are a little bit harder to draw. So let me use a different color here so we can see. Actually, let me go ahead and keep it yellow because this equatorial hydrogen is parallel with this set of lines."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in that hydrogen. Next, we put in the equatorial hydrogens, which are a little bit harder to draw. So let me use a different color here so we can see. Actually, let me go ahead and keep it yellow because this equatorial hydrogen is parallel with this set of lines. And also this one over here, actually. So we're doing equatorial up for our first hydrogen. So this line is directed up relative to the plane."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me go ahead and keep it yellow because this equatorial hydrogen is parallel with this set of lines. And also this one over here, actually. So we're doing equatorial up for our first hydrogen. So this line is directed up relative to the plane. So that's how you have to think about it. So we start at carbon one and we go equatorial up. So we put in that line and put in the hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So this line is directed up relative to the plane. So that's how you have to think about it. So we start at carbon one and we go equatorial up. So we put in that line and put in the hydrogen. Next, at carbon two, we already had this hydrogen going up. So this must be equatorial down. So we have equatorial down."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we put in that line and put in the hydrogen. Next, at carbon two, we already had this hydrogen going up. So this must be equatorial down. So we have equatorial down. And this line should be parallel with the ones in red. So this line should be parallel with the ones in red that we've already seen. So let's put in that hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we have equatorial down. And this line should be parallel with the ones in red. So this line should be parallel with the ones in red that we've already seen. So let's put in that hydrogen. So this is equatorial down. At carbon three, we have equatorial up. Let me use a different color."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in that hydrogen. So this is equatorial down. At carbon three, we have equatorial up. Let me use a different color. We have equatorial up at carbon three. I used blue because, again, it's with these lines parallel in blue. So we have equatorial up at carbon three."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Let me use a different color. We have equatorial up at carbon three. I used blue because, again, it's with these lines parallel in blue. So we have equatorial up at carbon three. Go down to this drawing. Let's put it in. So here is our hydrogen going up."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So we have equatorial up at carbon three. Go down to this drawing. Let's put it in. So here is our hydrogen going up. And again, we're alternating. We started equatorial up, then equatorial down. Let me draw on that hydrogen."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So here is our hydrogen going up. And again, we're alternating. We started equatorial up, then equatorial down. Let me draw on that hydrogen. And then equatorial up. And then equatorial down is our next one. So equatorial down at carbon four."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw on that hydrogen. And then equatorial up. And then equatorial down is our next one. So equatorial down at carbon four. And then carbon five should be up. So here it is up. And this should be parallel with the ones in red."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So equatorial down at carbon four. And then carbon five should be up. So here it is up. And this should be parallel with the ones in red. This is one of those ones that's difficult for students. Again, it's parallel with the ones in red. So that should make it easier."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And this should be parallel with the ones in red. This is one of those ones that's difficult for students. Again, it's parallel with the ones in red. So that should make it easier. So let's draw that in. So we have our hydrogen going up at this point. And then finally, carbon six."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So that should make it easier. So let's draw that in. So we have our hydrogen going up at this point. And then finally, carbon six. You can see this is equatorial down. And I guess this should be blue. So let me be consistent here."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, carbon six. You can see this is equatorial down. And I guess this should be blue. So let me be consistent here. So this is parallel with everything else in blue. We have another hydrogen, which is equatorial down. So let me draw that one in."}, {"video_title": "Drawing chair conformations Organic chemistry Khan Academy.mp3", "Sentence": "So let me be consistent here. So this is parallel with everything else in blue. We have another hydrogen, which is equatorial down. So let me draw that one in. So actually, that wasn't very good. So let me do it again. This takes practice."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So on the left here is one possible Lewis dot structure that you can draw that has that molecular formula. There are three carbons, one, two, three. There's one oxygen right here, and if you count up the hydrogens, you will get eight. And this Lewis dot structure, let me go ahead and write that, this is a Lewis dot structure here, this one shows all of the bonds, right? So all of the bonds are drawn in. But it takes a lot of time to draw in all of the bonds, and so we could represent this molecule in different ways. We could condense this Lewis structure a little bit, so this is equal to the structure I'm about to draw."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And this Lewis dot structure, let me go ahead and write that, this is a Lewis dot structure here, this one shows all of the bonds, right? So all of the bonds are drawn in. But it takes a lot of time to draw in all of the bonds, and so we could represent this molecule in different ways. We could condense this Lewis structure a little bit, so this is equal to the structure I'm about to draw. And we'll focus in on the carbon that I just circled there in red. So that carbon is this one right here. That carbon is bonded to an OH, so we could say this is an OH right here, put in lone pairs of electrons on the oxygen."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We could condense this Lewis structure a little bit, so this is equal to the structure I'm about to draw. And we'll focus in on the carbon that I just circled there in red. So that carbon is this one right here. That carbon is bonded to an OH, so we could say this is an OH right here, put in lone pairs of electrons on the oxygen. And that carbon is also bonded to this hydrogen, so let me draw in that hydrogen. On the right side, that carbon in red is bonded to this carbon in magenta. And the carbon in magenta is bonded to three other hydrogens."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That carbon is bonded to an OH, so we could say this is an OH right here, put in lone pairs of electrons on the oxygen. And that carbon is also bonded to this hydrogen, so let me draw in that hydrogen. On the right side, that carbon in red is bonded to this carbon in magenta. And the carbon in magenta is bonded to three other hydrogens. So we could represent that as a CH3. So I could write CH3 here, and the carbon in red is this one, and the carbon in magenta is this one. On the left side, the carbon in red is bonded to another carbon in blue, and the carbon in blue is bonded to three hydrogens."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And the carbon in magenta is bonded to three other hydrogens. So we could represent that as a CH3. So I could write CH3 here, and the carbon in red is this one, and the carbon in magenta is this one. On the left side, the carbon in red is bonded to another carbon in blue, and the carbon in blue is bonded to three hydrogens. So there's another CH3 on the left side. So let me draw that in. So we have a CH3 on the left, and the carbon in blue is directly bonded to the carbon in red."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "On the left side, the carbon in red is bonded to another carbon in blue, and the carbon in blue is bonded to three hydrogens. So there's another CH3 on the left side. So let me draw that in. So we have a CH3 on the left, and the carbon in blue is directly bonded to the carbon in red. So this is called a partially condensed structure. So this is a partially condensed, partially condensed structure. We haven't shown all of the bonds here, but this structure has the same information as the Lewis structure on the left."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we have a CH3 on the left, and the carbon in blue is directly bonded to the carbon in red. So this is called a partially condensed structure. So this is a partially condensed, partially condensed structure. We haven't shown all of the bonds here, but this structure has the same information as the Lewis structure on the left. It's the same molecule, it's just a different way to represent that molecule. We could keep going. We could go for a fully condensed structure."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We haven't shown all of the bonds here, but this structure has the same information as the Lewis structure on the left. It's the same molecule, it's just a different way to represent that molecule. We could keep going. We could go for a fully condensed structure. So let's do that. Focus in on the carbon in red, so this one right here. So let me draw in that carbon over here."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We could go for a fully condensed structure. So let's do that. Focus in on the carbon in red, so this one right here. So let me draw in that carbon over here. So that's that carbon. That carbon is bonded to two CH3 groups. There's a CH3 group on the right, so there's a CH3 group on the right, and there's a CH3 group on the left."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw in that carbon over here. So that's that carbon. That carbon is bonded to two CH3 groups. There's a CH3 group on the right, so there's a CH3 group on the right, and there's a CH3 group on the left. So I could write CH3, and then I could write a two here, which indicates there are two CH3 groups bonded to, directly bonded to, the carbon in red. What else is bonded to the carbon in red? There's a hydrogen, so I'll put that in."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There's a CH3 group on the right, so there's a CH3 group on the right, and there's a CH3 group on the left. So I could write CH3, and then I could write a two here, which indicates there are two CH3 groups bonded to, directly bonded to, the carbon in red. What else is bonded to the carbon in red? There's a hydrogen, so I'll put that in. So the carbon is bonded to a hydrogen. The carbon is also bonded to an OH. So I will write in here an OH."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There's a hydrogen, so I'll put that in. So the carbon is bonded to a hydrogen. The carbon is also bonded to an OH. So I will write in here an OH. This is the fully condensed version. So this is completely condensed, and notice there are no bonds shown. There are no bonds drawn in here."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I will write in here an OH. This is the fully condensed version. So this is completely condensed, and notice there are no bonds shown. There are no bonds drawn in here. You have to infer the bonding from the condensed. All right, let's start with the condensed and go all the way to a Lewis structure. So we'll start with a condensed, and then we'll go to partially condensed structure, and then we'll go to a full Lewis structure, just to get some more practice here."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There are no bonds drawn in here. You have to infer the bonding from the condensed. All right, let's start with the condensed and go all the way to a Lewis structure. So we'll start with a condensed, and then we'll go to partially condensed structure, and then we'll go to a full Lewis structure, just to get some more practice here. So I'll draw in a condensed one. So we have CH3, three, and then COCH3. All right, let's turn that into a partially condensed structure."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start with a condensed, and then we'll go to partially condensed structure, and then we'll go to a full Lewis structure, just to get some more practice here. So I'll draw in a condensed one. So we have CH3, three, and then COCH3. All right, let's turn that into a partially condensed structure. So this carbon in red right here, we're gonna start with that carbon. So I'll start drawing in that carbon right here. What is bonded to that carbon?"}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's turn that into a partially condensed structure. So this carbon in red right here, we're gonna start with that carbon. So I'll start drawing in that carbon right here. What is bonded to that carbon? Well, we have CH3 groups, and we have three of them. So there are three CH3 groups directly bonded to that carbon. So let me draw them in."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "What is bonded to that carbon? Well, we have CH3 groups, and we have three of them. So there are three CH3 groups directly bonded to that carbon. So let me draw them in. So here's one CH3 group. Here is another CH3 group, and then finally here's the third CH3 group. So this carbon in red over here is this carbon."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw them in. So here's one CH3 group. Here is another CH3 group, and then finally here's the third CH3 group. So this carbon in red over here is this carbon. The carbon in red is also bonded to an oxygen. All right, so we need to draw in an oxygen next. So now we have our oxygen."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon in red over here is this carbon. The carbon in red is also bonded to an oxygen. All right, so we need to draw in an oxygen next. So now we have our oxygen. Notice the carbon in red now has an octet of electrons around it. The oxygen is bonded to another CH3 group. So the oxygen is bonded to another CH3."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our oxygen. Notice the carbon in red now has an octet of electrons around it. The oxygen is bonded to another CH3 group. So the oxygen is bonded to another CH3. Let's draw that in. So we have our CH3, and since we're doing a partially condensed, I won't draw in those bonds. We have CH3 like that."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the oxygen is bonded to another CH3. Let's draw that in. So we have our CH3, and since we're doing a partially condensed, I won't draw in those bonds. We have CH3 like that. I could put in my lone pairs of electrons on the oxygen to give the oxygen an octet of electrons. And now we have our partially condensed structure. If we wanted to expand it even more and draw the full Lewis structure, again we start with the carbon in red."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have CH3 like that. I could put in my lone pairs of electrons on the oxygen to give the oxygen an octet of electrons. And now we have our partially condensed structure. If we wanted to expand it even more and draw the full Lewis structure, again we start with the carbon in red. So here's the carbon in red, and that carbon is bonded to another carbon, and this carbon is bonded to three hydrogens. So I draw in those three hydrogens. So this CH3 group that I just drew is this one."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If we wanted to expand it even more and draw the full Lewis structure, again we start with the carbon in red. So here's the carbon in red, and that carbon is bonded to another carbon, and this carbon is bonded to three hydrogens. So I draw in those three hydrogens. So this CH3 group that I just drew is this one. All right, next we have a CH3 group on the left side. So I need to draw in a CH3 on the left. Hopefully I have enough room to do that."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this CH3 group that I just drew is this one. All right, next we have a CH3 group on the left side. So I need to draw in a CH3 on the left. Hopefully I have enough room to do that. I'll squeeze it in here. So we have our hydrogens, and that's our second CH3 group. So let me circle it in green here."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Hopefully I have enough room to do that. I'll squeeze it in here. So we have our hydrogens, and that's our second CH3 group. So let me circle it in green here. So here's a CH3. And then finally we have, let me make this blue down here. We have another CH3 group, so I'll draw that one in."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me circle it in green here. So here's a CH3. And then finally we have, let me make this blue down here. We have another CH3 group, so I'll draw that one in. So we have another CH3. We'll make room for all these hydrogens here, and that's the one in blue. So we're drawing out all of the bonds now in our full Lewis structure."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have another CH3 group, so I'll draw that one in. So we have another CH3. We'll make room for all these hydrogens here, and that's the one in blue. So we're drawing out all of the bonds now in our full Lewis structure. Next we have an oxygen. So we have an oxygen right in here with two lone pairs of electrons on the oxygen. The oxygen is bonded to another CH3."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we're drawing out all of the bonds now in our full Lewis structure. Next we have an oxygen. So we have an oxygen right in here with two lone pairs of electrons on the oxygen. The oxygen is bonded to another CH3. So let me pick a color here for that one. So we have another CH3 on the right, and let's draw it in. There's a carbon with three bonds, two hydrogens."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen is bonded to another CH3. So let me pick a color here for that one. So we have another CH3 on the right, and let's draw it in. There's a carbon with three bonds, two hydrogens. So that's our last CH3 group. So let me circle it. This one right here is this one."}, {"video_title": "Condensed structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There's a carbon with three bonds, two hydrogens. So that's our last CH3 group. So let me circle it. This one right here is this one. So that's an important skill, being able to go from a condensed to a partially condensed to a full Lewis dot structure, and also going the opposite direction, going from Lewis to partially condensed and finally to condensed. And usually you'll only see these used for small molecules. It's obviously easy to work with when you have small molecules."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "We've got a few more molecular structures to name. So let's look at this first one right here. So the first thing you always want to do is identify the longest chain. So if we start over here, we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 carbons. Looks pretty long. Now what if we start over here? This looks like it could also be a long chain."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So if we start over here, we have 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 carbons. Looks pretty long. Now what if we start over here? This looks like it could also be a long chain. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. So we have two different chains, depending on whether we want to go up here or whether we go over here, that have a length of 13. So you're probably asking, which chain do you choose?"}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "This looks like it could also be a long chain. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. So we have two different chains, depending on whether we want to go up here or whether we go over here, that have a length of 13. So you're probably asking, which chain do you choose? And you should always, if you have two chains of equal length, and they're the longest chains, you pick the one that will have more branches or more alkyl groups on it. So this group right here, if we pick this kind of from here to here as our chain, we only have one group on it, that group up there. If we pick this chain, starting over here and then going over here, we have two groups on it."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So you're probably asking, which chain do you choose? And you should always, if you have two chains of equal length, and they're the longest chains, you pick the one that will have more branches or more alkyl groups on it. So this group right here, if we pick this kind of from here to here as our chain, we only have one group on it, that group up there. If we pick this chain, starting over here and then going over here, we have two groups on it. We would have this group over here, and then we would have this group over here. So this is the better chain to use, because it has more groups on it. It has more groups, but the groups are smaller and simpler."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "If we pick this chain, starting over here and then going over here, we have two groups on it. We would have this group over here, and then we would have this group over here. So this is the better chain to use, because it has more groups on it. It has more groups, but the groups are smaller and simpler. So let's start counting. And the direction we want to count, we always want to start on the side of the chain where we're going to encounter something first. And everything is closer to this end of the chain, so we'll start counting here."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "It has more groups, but the groups are smaller and simpler. So let's start counting. And the direction we want to count, we always want to start on the side of the chain where we're going to encounter something first. And everything is closer to this end of the chain, so we'll start counting here. We'll start counting here. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. So we have 13 carbons on our main chain."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And everything is closer to this end of the chain, so we'll start counting here. We'll start counting here. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13. So we have 13 carbons on our main chain. Let me draw our main chain. So our main chain is this thing in orange. I'm drawing right here."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So we have 13 carbons on our main chain. Let me draw our main chain. So our main chain is this thing in orange. I'm drawing right here. That is our main chain. Just like that. 13, that's 3 and 10."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "I'm drawing right here. That is our main chain. Just like that. 13, that's 3 and 10. The prefix is tridec, and it's tridecane, because we have all single bonds here. So it is tridecane. And then we have two groups over here."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "13, that's 3 and 10. The prefix is tridec, and it's tridecane, because we have all single bonds here. So it is tridecane. And then we have two groups over here. This one in green, this only has one carbon branching off of the main chain. So its prefix will be meth, then it'll be a methyl group. So that is methyl, that is a methyl group right there."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have two groups over here. This one in green, this only has one carbon branching off of the main chain. So its prefix will be meth, then it'll be a methyl group. So that is methyl, that is a methyl group right there. And then this one down here, we have 1, 2, 3 carbons. The prefix is prop. So this is a propyl group."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So that is methyl, that is a methyl group right there. And then this one down here, we have 1, 2, 3 carbons. The prefix is prop. So this is a propyl group. And the methyl is sitting on the 3-carbon of our main chain, and the propyl group is sitting on the 4-carbon. 1, 2, 3, 4. Now, when we figure out what order to list them in, when we actually write out the name, m, we just do it in alphabetical order, m comes before p. So we write 3-methyl before 4-propyl."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So this is a propyl group. And the methyl is sitting on the 3-carbon of our main chain, and the propyl group is sitting on the 4-carbon. 1, 2, 3, 4. Now, when we figure out what order to list them in, when we actually write out the name, m, we just do it in alphabetical order, m comes before p. So we write 3-methyl before 4-propyl. So the entire compound here is, or the entire molecule is, 3-methyl-4-propyl-tridecane. And this is actually all going to be one word. If you don't use dashes to separate when you have numbers, but if you have a word followed by a word, it just becomes propyl-tridecane."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "Now, when we figure out what order to list them in, when we actually write out the name, m, we just do it in alphabetical order, m comes before p. So we write 3-methyl before 4-propyl. So the entire compound here is, or the entire molecule is, 3-methyl-4-propyl-tridecane. And this is actually all going to be one word. If you don't use dashes to separate when you have numbers, but if you have a word followed by a word, it just becomes propyl-tridecane. So 3-methyl-4-propyl-tridecane. And we're done. Let's do this one down here."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "If you don't use dashes to separate when you have numbers, but if you have a word followed by a word, it just becomes propyl-tridecane. So 3-methyl-4-propyl-tridecane. And we're done. Let's do this one down here. Now, this one seems a little bit more complex. So the first thing to see is what is the largest chain or the largest ring that we have in our structure. And the two candidates, we have this chain over here."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "Let's do this one down here. Now, this one seems a little bit more complex. So the first thing to see is what is the largest chain or the largest ring that we have in our structure. And the two candidates, we have this chain over here. This has 1, 2, 3, 4, 5, 6, 7 carbons. Let's see how many carbons our ring has. Our ring has 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And the two candidates, we have this chain over here. This has 1, 2, 3, 4, 5, 6, 7 carbons. Let's see how many carbons our ring has. Our ring has 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. So the ring is the largest, I guess you could say, core structure in this molecule. So that will be our core structure. And so we have 9 carbons."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "Our ring has 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. So the ring is the largest, I guess you could say, core structure in this molecule. So that will be our core structure. And so we have 9 carbons. Let me highlight it. So the ring is this right here. Has 9 carbons."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And so we have 9 carbons. Let me highlight it. So the ring is this right here. Has 9 carbons. The prefix for 9 is non. It's all single bonds, so it's nonane. And it is in a cycle."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "Has 9 carbons. The prefix for 9 is non. It's all single bonds, so it's nonane. And it is in a cycle. It's a ring. So it's cyclononane. So this part right here, that part right there, is cyclononane."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And it is in a cycle. It's a ring. So it's cyclononane. So this part right here, that part right there, is cyclononane. And then we have several things that branch off of the cyclononane. So let's look at them one at a time. And then we'll think about how we're going to number them on the ring."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So this part right here, that part right there, is cyclononane. And then we have several things that branch off of the cyclononane. So let's look at them one at a time. And then we'll think about how we're going to number them on the ring. So we looked already at this chain that has 7 carbons. 1, 2, 3, 4. Let me do that in other blue."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll think about how we're going to number them on the ring. So we looked already at this chain that has 7 carbons. 1, 2, 3, 4. Let me do that in other blue. I like the other blue better. 1, 2, 3, 4, 5, 6, 7. So that is a heptyl group."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "Let me do that in other blue. I like the other blue better. 1, 2, 3, 4, 5, 6, 7. So that is a heptyl group. Now this over here, let's see what we're dealing with. We have 1, 2, 3 carbons. So that is just a standard propyl group."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So that is a heptyl group. Now this over here, let's see what we're dealing with. We have 1, 2, 3 carbons. So that is just a standard propyl group. So this is just a standard propyl group. And then here we have 1, 2, 3, 4 carbons. So we could say this is a butyl group, but this isn't just any butyl group."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So that is just a standard propyl group. So this is just a standard propyl group. And then here we have 1, 2, 3, 4 carbons. So we could say this is a butyl group, but this isn't just any butyl group. If we use the common naming, the carbon we immediately touch on, or that we immediately get to when we go off of our main ring, that branches off into three other carbons. So this is tert, tert for 3. And the tert is usually written in italics."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So we could say this is a butyl group, but this isn't just any butyl group. If we use the common naming, the carbon we immediately touch on, or that we immediately get to when we go off of our main ring, that branches off into three other carbons. So this is tert, tert for 3. And the tert is usually written in italics. It's hard to differentiate that when you see it. I'll write it in cursive. This is a tert butyl group."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And the tert is usually written in italics. It's hard to differentiate that when you see it. I'll write it in cursive. This is a tert butyl group. Now the next question is, how do we specify where these different groups sit on this main ring? Now if you just had one group, you wouldn't have to specify. But when you have more than one, what you actually do is you figure out which one would be alphabetically first, and that would be number one."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "This is a tert butyl group. Now the next question is, how do we specify where these different groups sit on this main ring? Now if you just had one group, you wouldn't have to specify. But when you have more than one, what you actually do is you figure out which one would be alphabetically first, and that would be number one. Now we have an H in heptyl, a P in propyl. And in tert butyl, you might say, well, do I use the T or do I use the B? And this is just the convention."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "But when you have more than one, what you actually do is you figure out which one would be alphabetically first, and that would be number one. Now we have an H in heptyl, a P in propyl. And in tert butyl, you might say, well, do I use the T or do I use the B? And this is just the convention. You use the B. If you have sec butyl or tert butyl, you ignore the sec or the tert. If this was an isobutyl or an isopropyl, you actually would use the I."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And this is just the convention. You use the B. If you have sec butyl or tert butyl, you ignore the sec or the tert. If this was an isobutyl or an isopropyl, you actually would use the I. So it's a little bit, I guess the best way to think about it is there's a dash here, so you can kind of ignore it. But if this was isobutyl, it would all be one word. So you would consider the I."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "If this was an isobutyl or an isopropyl, you actually would use the I. So it's a little bit, I guess the best way to think about it is there's a dash here, so you can kind of ignore it. But if this was isobutyl, it would all be one word. So you would consider the I. So in this situation, we would consider the B. And B comes before a P or an H. So that is where we will start numbering one. And then to figure out which direction to keep numbering in, we just go in the direction where we're likely to encounter the first or where we will encounter the first side chain."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So you would consider the I. So in this situation, we would consider the B. And B comes before a P or an H. So that is where we will start numbering one. And then to figure out which direction to keep numbering in, we just go in the direction where we're likely to encounter the first or where we will encounter the first side chain. So we'll go in this direction because we get right to the propyl group. 1, 2, 3, 4, 5, 6, 7, 8, 9. So this compound, we're going to start with the alphabetically first side chain."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "And then to figure out which direction to keep numbering in, we just go in the direction where we're likely to encounter the first or where we will encounter the first side chain. So we'll go in this direction because we get right to the propyl group. 1, 2, 3, 4, 5, 6, 7, 8, 9. So this compound, we're going to start with the alphabetically first side chain. So it's one tert butyl. Let me write that in cursive. Let me get some more space here."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "So this compound, we're going to start with the alphabetically first side chain. So it's one tert butyl. Let me write that in cursive. Let me get some more space here. So it is one, I'll write this in cursive, tert butyl. Then the next one alphabetically is the heptyl group, that's H for heptyl. So then it is 5-heptyl."}, {"video_title": "Organic chemistry naming examples 3 Organic chemistry Khan Academy.mp3", "Sentence": "Let me get some more space here. So it is one, I'll write this in cursive, tert butyl. Then the next one alphabetically is the heptyl group, that's H for heptyl. So then it is 5-heptyl. And then we have the propyl. And then it is 2-propyl. 2-propyl."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The purpose of this video is to just begin to appreciate how vast and enormous the universe is. And frankly, our brains really can't grasp it. What we'll see in this video is that we can't even grasp things that are actually super small compared to the size of the universe. And we actually don't even know what the entire size of the universe is. But with that said, let's actually just try to appreciate how small we are. So this is me right over here. I am 5'9\", depending on whether I'm wearing shoes, maybe 5'10 with shoes."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we actually don't even know what the entire size of the universe is. But with that said, let's actually just try to appreciate how small we are. So this is me right over here. I am 5'9\", depending on whether I'm wearing shoes, maybe 5'10 with shoes. But for the sake of this video, let's just roughly approximate around 6 feet, or around roughly, I'm not going to go into the details of the math, around 2 meters. Now if I were to lie down 10 times in a row, you'd get about the length of an 18-wheeler. It's about 60 feet long, so this is times 10."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I am 5'9\", depending on whether I'm wearing shoes, maybe 5'10 with shoes. But for the sake of this video, let's just roughly approximate around 6 feet, or around roughly, I'm not going to go into the details of the math, around 2 meters. Now if I were to lie down 10 times in a row, you'd get about the length of an 18-wheeler. It's about 60 feet long, so this is times 10. Now if you were to put an 18-wheeler, if you were to make it tall as opposed to long, somehow stand it up, and you were to do that 10 times in a row, you'll get to the height of roughly a 60-story skyscraper. So once again, if you took me and you piled me up 100 times, you'll get about a 60-story skyscraper. Now if you took that skyscraper, and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's about 60 feet long, so this is times 10. Now if you were to put an 18-wheeler, if you were to make it tall as opposed to long, somehow stand it up, and you were to do that 10 times in a row, you'll get to the height of roughly a 60-story skyscraper. So once again, if you took me and you piled me up 100 times, you'll get about a 60-story skyscraper. Now if you took that skyscraper, and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge. And once again, I'm not giving you the exact numbers, it's not always going to be exactly 10, but we're now getting to about something that's a little on the order of a mile long. The Golden Gate Bridge is actually longer than a mile, but if you go within the twin spans, it's roughly about a mile. It's actually a little longer than that, but that gives you a sense of a mile."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now if you took that skyscraper, and if you were to lie it down 10 times in a row, you'd get something of the length of the Golden Gate Bridge. And once again, I'm not giving you the exact numbers, it's not always going to be exactly 10, but we're now getting to about something that's a little on the order of a mile long. The Golden Gate Bridge is actually longer than a mile, but if you go within the twin spans, it's roughly about a mile. It's actually a little longer than that, but that gives you a sense of a mile. Now, if you multiply that by 10, you get to the size of a large city. And this right here is a satellite photograph of San Francisco. This is the actual Golden Gate Bridge here, and when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles, just so you appreciate the scale."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's actually a little longer than that, but that gives you a sense of a mile. Now, if you multiply that by 10, you get to the size of a large city. And this right here is a satellite photograph of San Francisco. This is the actual Golden Gate Bridge here, and when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles, just so you appreciate the scale. And what's interesting here, and this picture's interesting, because this is the first time we can kind of relate to cities, but when you look at a city on this scale, it's starting to get larger than what we're used to processing on a daily basis. A bridge, we've been on a bridge, we know what a bridge looks like. We know that a bridge is huge, but it doesn't feel like something that we can't comprehend."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the actual Golden Gate Bridge here, and when I copy and pasted this picture, I tried to make it roughly 10 miles by 10 miles, just so you appreciate the scale. And what's interesting here, and this picture's interesting, because this is the first time we can kind of relate to cities, but when you look at a city on this scale, it's starting to get larger than what we're used to processing on a daily basis. A bridge, we've been on a bridge, we know what a bridge looks like. We know that a bridge is huge, but it doesn't feel like something that we can't comprehend. Already a city is something that we can't comprehend all at once. We can drive across a city, we can look at satellite imagery, but if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We know that a bridge is huge, but it doesn't feel like something that we can't comprehend. Already a city is something that we can't comprehend all at once. We can drive across a city, we can look at satellite imagery, but if I were to show a human on this, it would be unbelievably, unbelievably small. You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You wouldn't actually be able to see it. It would be less than a pixel on this image. A house is less than a pixel on this image. But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's keep multiplying by 10. If you multiply by 10 again, you get to something roughly the size of the San Francisco Bay Area. This whole square over here is roughly that square right over there. Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. Now you're including a big part of the western United States."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's multiply by 10 again. So this square is about 100 miles by 100 miles. So this one would be about 1,000 miles by 1,000 miles. Now you're including a big part of the western United States. You have California here, you have Nevada here, you have Arizona and New Mexico. So a big chunk of a big continent we're already including. Frankly, this is beyond the scale that we're used to operating."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now you're including a big part of the western United States. You have California here, you have Nevada here, you have Arizona and New Mexico. So a big chunk of a big continent we're already including. Frankly, this is beyond the scale that we're used to operating. We've seen maps, so maybe we're a little used to it, but if you ever had to walk across this type of distance, it would take you a while. To some degree, the fact that planes go so fast, almost unimaginably fast for us, that it's made it feel like things like continents aren't as big because you can fly across them in five or six hours. But these are already huge, huge, huge distances."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Frankly, this is beyond the scale that we're used to operating. We've seen maps, so maybe we're a little used to it, but if you ever had to walk across this type of distance, it would take you a while. To some degree, the fact that planes go so fast, almost unimaginably fast for us, that it's made it feel like things like continents aren't as big because you can fly across them in five or six hours. But these are already huge, huge, huge distances. But once again, you take this square that's about 1,000 miles by 1,000 miles and you multiply that by 10 and you get pretty close, a little bit over the diameter of the Earth. But once again, we're on the Earth. We kind of relate to the Earth."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But these are already huge, huge, huge distances. But once again, you take this square that's about 1,000 miles by 1,000 miles and you multiply that by 10 and you get pretty close, a little bit over the diameter of the Earth. But once again, we're on the Earth. We kind of relate to the Earth. If you look carefully at the horizon, you might see a little bit of a curvature, especially if you were to get into the plane. Even though this is, frankly, larger than my brain can really grasp, we can kind of relate to the Earth. Now, you multiply the diameter of Earth times 10 and you get to the diameter of Jupiter."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We kind of relate to the Earth. If you look carefully at the horizon, you might see a little bit of a curvature, especially if you were to get into the plane. Even though this is, frankly, larger than my brain can really grasp, we can kind of relate to the Earth. Now, you multiply the diameter of Earth times 10 and you get to the diameter of Jupiter. So if you were to sit Earth right next to Jupiter, obviously they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, you multiply the diameter of Earth times 10 and you get to the diameter of Jupiter. So if you were to sit Earth right next to Jupiter, obviously they're nowhere near that close. That would destroy both of the planets. Actually, it would definitely destroy Earth. It would probably just kind of be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that, right over there. So I would say that Jupiter is definitely, oh, you went on this kind of diagram that I'm drawing here, is definitely the first thing that I can't comprehend."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, it would definitely destroy Earth. It would probably just kind of be merged into Jupiter. So if you put Earth next to Jupiter, it would look something like that, right over there. So I would say that Jupiter is definitely, oh, you went on this kind of diagram that I'm drawing here, is definitely the first thing that I can't comprehend. The Earth itself is so vastly huge. Jupiter is 10 times bigger in diameter. It's much larger in terms of mass and volume and all the rest."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I would say that Jupiter is definitely, oh, you went on this kind of diagram that I'm drawing here, is definitely the first thing that I can't comprehend. The Earth itself is so vastly huge. Jupiter is 10 times bigger in diameter. It's much larger in terms of mass and volume and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the Sun."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's much larger in terms of mass and volume and all the rest. But just in terms of diameter, it is 10 times bigger. But let's keep going. 10 times Jupiter gets us to the Sun. This is times 10. So if this is the Sun and if I were to draw Jupiter, it would look something like, I'll do Jupiter in pink, Jupiter would be around that big. And then the Earth would be around that big, if you were to put them all next to each other."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "10 times Jupiter gets us to the Sun. This is times 10. So if this is the Sun and if I were to draw Jupiter, it would look something like, I'll do Jupiter in pink, Jupiter would be around that big. And then the Earth would be around that big, if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is kind of unimaginably huge, and the Sun is 100 times more unimaginably bigger."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then the Earth would be around that big, if you were to put them all next to each other. So the Sun, once again, is huge. Even though we see it almost every day, it is unimaginably huge. Even the Earth is kind of unimaginably huge, and the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth. You multiply that times 100, and that is the distance from the Earth to the Sun."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Even the Earth is kind of unimaginably huge, and the Sun is 100 times more unimaginably bigger. Now we're going to start getting really, really, really wacky. You multiply the diameter of the Sun, which is already 100 times the diameter of the Earth. You multiply that times 100, and that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel, and I didn't even draw the Earth as a pixel, because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is an unbelievable distance between the Earth and the Sun."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You multiply that times 100, and that is the distance from the Earth to the Sun. So I've drawn the Sun here as a little pixel, and I didn't even draw the Earth as a pixel, because a pixel would be way too large. It would have to be a hundredth of a pixel in order to draw the Earth properly. So this is an unbelievable distance between the Earth and the Sun. It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is an unbelievable distance between the Earth and the Sun. It's 100 times the distance of the diameter of the Sun itself. So it's massive, massive. But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star... So remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But once again, these things are relatively close compared to where we're about to go. Because if we want to get to the nearest star... So remember, the Sun is 100 times the diameter of the Earth. The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times... And once again, unimaginable."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The distance between the Sun and the Earth is 100 times that. Or you could say it's 10,000 times the diameter of the Earth. So these are unimaginable distances. But to get to the nearest star, which is 4.2 light years away, it's 200,000 times... And once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball, if the average star was about the size of a basketball, in our part of the galaxy, in a volume the size of the Earth... So if you had just like a big volume, the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But to get to the nearest star, which is 4.2 light years away, it's 200,000 times... And once again, unimaginable. It's 200,000 times the distance between the Earth and the Sun. And to give you a rough sense of how far apart these things are, if the Sun was roughly the size of a basketball, if the average star was about the size of a basketball, in our part of the galaxy, in a volume the size of the Earth... So if you had just like a big volume, the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though the galaxy looks like... Even though when you look at the galaxy, and this is just an artist's depiction of it, it looks like something that has kind of this spray of stars and it looks reasonably dense."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you had just like a big volume, the size of the Earth, if the stars were the sizes of basketballs, in our part of the galaxy, you would only have a handful of basketballs per that volume. So unbelievably sparse. Even though the galaxy looks like... Even though when you look at the galaxy, and this is just an artist's depiction of it, it looks like something that has kind of this spray of stars and it looks reasonably dense. There is actually a huge amount of space. The great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Even though when you look at the galaxy, and this is just an artist's depiction of it, it looks like something that has kind of this spray of stars and it looks reasonably dense. There is actually a huge amount of space. The great, great, great, great, great majority of the volume in the galaxy is just empty, empty space. There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy itself can be, you take this distance between the Sun or between our solar system and the nearest star. So that's 200,000 times the distance between the Earth and the Sun."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's no stars, no planets, no nothing. I mean, this is a huge jump that I'm talking about. And then if you really want to realize how large a galaxy itself can be, you take this distance between the Sun or between our solar system and the nearest star. So that's 200,000 times the distance between the Earth and the Sun. And you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within..."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's 200,000 times the distance between the Earth and the Sun. And you multiply that distance by 25,000. So if the Sun is right here, our nearest star will be in that same pixel. They'll actually be within... I mean, you'd have to get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance between the Sun and the nearest star."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They'll actually be within... I mean, you'd have to get a ton of stars within that one pixel, even though they're so far apart. And then this whole thing is 100,000 light years. It's 25,000 times the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances just for a galaxy. And now we're going to get our... Frankly, my brain is already well beyond anything that it can really process."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's 25,000 times the distance between the Sun and the nearest star. So we're talking about unimaginable, unfathomable distances just for a galaxy. And now we're going to get our... Frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But if we get a sense of the universe itself, the observable universe, and we have to be clear because we can only observe light that started leaving from its source 13.7 billion years ago because that's how old the universe is."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Frankly, my brain is already well beyond anything that it can really process. At this point, it almost just becomes abstract thinking. It just becomes playing with numbers and mathematics. But if we get a sense of the universe itself, the observable universe, and we have to be clear because we can only observe light that started leaving from its source 13.7 billion years ago because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if we get a sense of the universe itself, the observable universe, and we have to be clear because we can only observe light that started leaving from its source 13.7 billion years ago because that's how old the universe is. The observable universe is about 93 billion light years across. And the reason why it's larger than 13.7 billion is that the points in space that emitted light 13.7 billion years ago, those have been going away from us. So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get, and that we can observe just because we're getting electromagnetic radiation from those parts of the universe, you would have to multiply this number."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now they're on the order of 40 billion light years away. But this isn't about cosmology. This is just about scale and appreciating how huge the universe is. Just in the part of the universe that we can theoretically observe, you have to get, and that we can observe just because we're getting electromagnetic radiation from those parts of the universe, you would have to multiply this number. So let me make this clear. 100,000 light years, that's the diameter of the Milky Way. So 100,000 light years, that's the diameter of the Milky Way."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just in the part of the universe that we can theoretically observe, you have to get, and that we can observe just because we're getting electromagnetic radiation from those parts of the universe, you would have to multiply this number. So let me make this clear. 100,000 light years, that's the diameter of the Milky Way. So 100,000 light years, that's the diameter of the Milky Way. You would have to multiply it not by 1,000. 1,000 would get you to 100 million light years. This is 100,000 times 1,000 is 100 million."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So 100,000 light years, that's the diameter of the Milky Way. You would have to multiply it not by 1,000. 1,000 would get you to 100 million light years. This is 100,000 times 1,000 is 100 million. You have to multiply by 1,000 again to get to 100 billion light years. The universe, for all we know, might be much, much, much larger. It might even be infinite."}, {"video_title": "Scale of the large Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is 100,000 times 1,000 is 100 million. You have to multiply by 1,000 again to get to 100 billion light years. The universe, for all we know, might be much, much, much larger. It might even be infinite. Who knows? But to get from just the diameter of the Milky Way to the observable universe, you have to multiply by a million. Already, this is an unfathomable distance."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So in the top left here, we're starting with a primary alcohol. And the carbon that's attached to the OH group is your alpha carbon. To oxidize an alcohol, you must have alpha hydrogens. You must have hydrogens attached to that alpha carbon in order for the mechanism to work. So in that mechanism, you're actually going to lose one of those alpha hydrogens. And we'll take a look at the mechanism in a few minutes. So if I were to oxidize this primary alcohol, I'll add something to oxidize."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "You must have hydrogens attached to that alpha carbon in order for the mechanism to work. So in that mechanism, you're actually going to lose one of those alpha hydrogens. And we'll take a look at the mechanism in a few minutes. So if I were to oxidize this primary alcohol, I'll add something to oxidize. I'll oxidize my primary alcohol like that. One way to think about the oxidation of an alcohol is to think about the number of bonds of carbon to oxygen. On the left side here, we have one bond of our alpha carbon to this oxygen."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So if I were to oxidize this primary alcohol, I'll add something to oxidize. I'll oxidize my primary alcohol like that. One way to think about the oxidation of an alcohol is to think about the number of bonds of carbon to oxygen. On the left side here, we have one bond of our alpha carbon to this oxygen. In the mechanism, we're going to lose a bond of carbon to hydrogen, and we're going to gain another bond of carbon to oxygen. So you're increasing the number of bonds of carbon to oxygen. So that would, of course, give me two bonds of carbon to oxygen."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "On the left side here, we have one bond of our alpha carbon to this oxygen. In the mechanism, we're going to lose a bond of carbon to hydrogen, and we're going to gain another bond of carbon to oxygen. So you're increasing the number of bonds of carbon to oxygen. So that would, of course, give me two bonds of carbon to oxygen. If I oxidize my alcohol one time, and I'm going to lose one of those hydrogens. So one of those hydrogens is still left, and my alkyl group is still attached. So obviously, this would give me an aldehyde functional group."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So that would, of course, give me two bonds of carbon to oxygen. If I oxidize my alcohol one time, and I'm going to lose one of those hydrogens. So one of those hydrogens is still left, and my alkyl group is still attached. So obviously, this would give me an aldehyde functional group. So if you oxidize a primary alcohol one time, you will get an aldehyde. Let's take a look at the oxidation states of my alpha carbon and see what happened to it. So if I want to assign an oxidation state to my alpha carbon on the left, once again, I have to put in my electrons."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So obviously, this would give me an aldehyde functional group. So if you oxidize a primary alcohol one time, you will get an aldehyde. Let's take a look at the oxidation states of my alpha carbon and see what happened to it. So if I want to assign an oxidation state to my alpha carbon on the left, once again, I have to put in my electrons. Each bond consists of two electrons like that. And I need to think about electronegativity differences. Oxygen is more electronegative than carbon, so it's going to take those two electrons."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So if I want to assign an oxidation state to my alpha carbon on the left, once again, I have to put in my electrons. Each bond consists of two electrons like that. And I need to think about electronegativity differences. Oxygen is more electronegative than carbon, so it's going to take those two electrons. Carbon versus carbon is a tie, so each carbon will get one of those electrons. Carbon actually is slightly more electronegative than hydrogen, so carbon will win and take those electrons right there. Carbon normally has four valence electrons, and in this instance, it is being surrounded by five electrons."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen is more electronegative than carbon, so it's going to take those two electrons. Carbon versus carbon is a tie, so each carbon will get one of those electrons. Carbon actually is slightly more electronegative than hydrogen, so carbon will win and take those electrons right there. Carbon normally has four valence electrons, and in this instance, it is being surrounded by five electrons. So 4 minus 5 will give me an oxidation state of negative 1 for my alpha carbon. So let's look and see what happened to that alpha carbon after we oxidized it. So over here on the right, if I wanted to assign an oxidation state to what is now my carbonyl carbon, once again, I think about my electronegativity differences."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Carbon normally has four valence electrons, and in this instance, it is being surrounded by five electrons. So 4 minus 5 will give me an oxidation state of negative 1 for my alpha carbon. So let's look and see what happened to that alpha carbon after we oxidized it. So over here on the right, if I wanted to assign an oxidation state to what is now my carbonyl carbon, once again, I think about my electronegativity differences. And I know that oxygen is going to beat carbon. Carbon versus carbon is a tie. And carbon versus hydrogen, carbon will win."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the right, if I wanted to assign an oxidation state to what is now my carbonyl carbon, once again, I think about my electronegativity differences. And I know that oxygen is going to beat carbon. Carbon versus carbon is a tie. And carbon versus hydrogen, carbon will win. So the oxidation state of that carbon, normally four valence electrons, surrounded by three this time, so 4 minus 3 will give me plus 1. So I can see that my oxidation state went from negative 1 to plus 1. So an increase in the oxidation state is, of course, oxidation."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And carbon versus hydrogen, carbon will win. So the oxidation state of that carbon, normally four valence electrons, surrounded by three this time, so 4 minus 3 will give me plus 1. So I can see that my oxidation state went from negative 1 to plus 1. So an increase in the oxidation state is, of course, oxidation. So if you oxidize a primary alcohol one time, you will get an aldehyde. What about if you keep going? So if you form an aldehyde, and sometimes it's hard to stop the reaction mixture from continuing to oxidize."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So an increase in the oxidation state is, of course, oxidation. So if you oxidize a primary alcohol one time, you will get an aldehyde. What about if you keep going? So if you form an aldehyde, and sometimes it's hard to stop the reaction mixture from continuing to oxidize. So if you oxidize an aldehyde, you think about what functional group you would get. Well, again, a simple way of doing it would be to think, on the left side, I have two bonds of carbon to oxygen. Is there any kind of functional group where carbon is bonded three times to an oxygen?"}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So if you form an aldehyde, and sometimes it's hard to stop the reaction mixture from continuing to oxidize. So if you oxidize an aldehyde, you think about what functional group you would get. Well, again, a simple way of doing it would be to think, on the left side, I have two bonds of carbon to oxygen. Is there any kind of functional group where carbon is bonded three times to an oxygen? So that, of course, would be a carboxylic acid. So if I think about the structure of a carboxylic acid, I can see that carbon is actually bonded three times to an oxygen, if you will. Three bonds of carbon to oxygen."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Is there any kind of functional group where carbon is bonded three times to an oxygen? So that, of course, would be a carboxylic acid. So if I think about the structure of a carboxylic acid, I can see that carbon is actually bonded three times to an oxygen, if you will. Three bonds of carbon to oxygen. And over here, I have my alkyl group like that. So if you oxidize an aldehyde, you're going to get a carboxylic acid. Let's look again at the oxidation state of my carbonyl carbon."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Three bonds of carbon to oxygen. And over here, I have my alkyl group like that. So if you oxidize an aldehyde, you're going to get a carboxylic acid. Let's look again at the oxidation state of my carbonyl carbon. So once again, I put in my electrons here. And I think about electronegativity. So oxygen, of course, beats carbon."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Let's look again at the oxidation state of my carbonyl carbon. So once again, I put in my electrons here. And I think about electronegativity. So oxygen, of course, beats carbon. Tie between these two carbons, and oxygen beats carbon again. So in this case, normally four valence electrons. Now there's one."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen, of course, beats carbon. Tie between these two carbons, and oxygen beats carbon again. So in this case, normally four valence electrons. Now there's one. So 4 minus 1 gives us an oxidation state of plus 3. So once again, an increase in the oxidation state means oxidation. If you oxidize an aldehyde, you will get a carboxylic acid."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Now there's one. So 4 minus 1 gives us an oxidation state of plus 3. So once again, an increase in the oxidation state means oxidation. If you oxidize an aldehyde, you will get a carboxylic acid. Let's look at a secondary alcohol now. So we'll go down here to our secondary alcohol. And once again, identify the alpha carbon, the one attached to your OH group."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "If you oxidize an aldehyde, you will get a carboxylic acid. Let's look at a secondary alcohol now. So we'll go down here to our secondary alcohol. And once again, identify the alpha carbon, the one attached to your OH group. We need to have at least one hydrogen on that alpha carbon. And so we have one right here. So if we were to oxidize our secondary alcohol, so we're going to oxidize our secondary alcohol."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And once again, identify the alpha carbon, the one attached to your OH group. We need to have at least one hydrogen on that alpha carbon. And so we have one right here. So if we were to oxidize our secondary alcohol, so we're going to oxidize our secondary alcohol. Once again, a simple way of doing it is thinking my alpha carbon has one bond to oxygen. So I could increase that to two bonds. And that should be an oxidation reaction."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So if we were to oxidize our secondary alcohol, so we're going to oxidize our secondary alcohol. Once again, a simple way of doing it is thinking my alpha carbon has one bond to oxygen. So I could increase that to two bonds. And that should be an oxidation reaction. In the process, I'm going to lose a bond to my alpha hydrogen. So I'm now going to have two bonds of carbon to oxygen. And I'm going to lose the bond that that alpha carbon had with the hydrogen there."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And that should be an oxidation reaction. In the process, I'm going to lose a bond to my alpha hydrogen. So I'm now going to have two bonds of carbon to oxygen. And I'm going to lose the bond that that alpha carbon had with the hydrogen there. So that leaves my two alkyl groups like that. So now I have two alkyl groups. And of course, this would be a ketone functional group."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to lose the bond that that alpha carbon had with the hydrogen there. So that leaves my two alkyl groups like that. So now I have two alkyl groups. And of course, this would be a ketone functional group. So if you oxidize a secondary alcohol, you're going to end up with a ketone. I can assign oxidation states. So once again, let's show that this really is an oxidation reaction here."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And of course, this would be a ketone functional group. So if you oxidize a secondary alcohol, you're going to end up with a ketone. I can assign oxidation states. So once again, let's show that this really is an oxidation reaction here. And I go ahead and put in my electrons on my alpha carbon. And think about electronegativity differences. So once again, oxygen beats carbon."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So once again, let's show that this really is an oxidation reaction here. And I go ahead and put in my electrons on my alpha carbon. And think about electronegativity differences. So once again, oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus hydrogen, carbon wins. And then carbon versus carbon, of course, is a tie again."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So once again, oxygen beats carbon. Carbon versus carbon is a tie. Carbon versus hydrogen, carbon wins. And then carbon versus carbon, of course, is a tie again. So normally, four valence electrons. In this example, it's surrounded by four. So 4 minus 4 gives us an oxidation state of 0 for our secondary alcohol."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And then carbon versus carbon, of course, is a tie again. So normally, four valence electrons. In this example, it's surrounded by four. So 4 minus 4 gives us an oxidation state of 0 for our secondary alcohol. And when I oxidize it, I'm going to get this ketone over here on the right. So let's take a look at the oxidation state of the carbon that used to be our alpha carbon on the left, which is now our carbonyl carbon. So once again, we put in our electrons."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So 4 minus 4 gives us an oxidation state of 0 for our secondary alcohol. And when I oxidize it, I'm going to get this ketone over here on the right. So let's take a look at the oxidation state of the carbon that used to be our alpha carbon on the left, which is now our carbonyl carbon. So once again, we put in our electrons. And we think about electronegativity difference. So oxygen is going to beat carbon. So we go like that."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we put in our electrons. And we think about electronegativity difference. So oxygen is going to beat carbon. So we go like that. Carbon versus carbon is a tie. Carbon versus carbon is a tie. Once again, so normally 4 minus 2 this time around that carbon, giving us an oxidation state of plus 2."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So we go like that. Carbon versus carbon is a tie. Carbon versus carbon is a tie. Once again, so normally 4 minus 2 this time around that carbon, giving us an oxidation state of plus 2. So to go from a secondary alcohol to a ketone, we see there's an increase in the oxidation state. So this is definitely an oxidation reaction. Let's look now at a tertiary alcohol."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Once again, so normally 4 minus 2 this time around that carbon, giving us an oxidation state of plus 2. So to go from a secondary alcohol to a ketone, we see there's an increase in the oxidation state. So this is definitely an oxidation reaction. Let's look now at a tertiary alcohol. So here is my tertiary alcohol. And when I find my alpha carbon, I see that this time there are no hydrogens bonded to my alpha carbon. So according to the mechanism, which we'll see in a minute, there's no way we can oxidize this tertiary alcohol under normal conditions anyway."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Let's look now at a tertiary alcohol. So here is my tertiary alcohol. And when I find my alpha carbon, I see that this time there are no hydrogens bonded to my alpha carbon. So according to the mechanism, which we'll see in a minute, there's no way we can oxidize this tertiary alcohol under normal conditions anyway. So if we attempted to oxidize this, we would say there's no reaction here, since we are missing that alpha hydrogen. Let's take a look at the mechanism and see why we need to have that alpha hydrogen on our alpha carbon. So if I were to start my mechanism here with an alcohol, remember this must be either a primary or a secondary alcohol in order for this oxidation to work."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So according to the mechanism, which we'll see in a minute, there's no way we can oxidize this tertiary alcohol under normal conditions anyway. So if we attempted to oxidize this, we would say there's no reaction here, since we are missing that alpha hydrogen. Let's take a look at the mechanism and see why we need to have that alpha hydrogen on our alpha carbon. So if I were to start my mechanism here with an alcohol, remember this must be either a primary or a secondary alcohol in order for this oxidation to work. So I'm going to go ahead and show my alcohol there. So again, either primary, primary, or secondary, like that. And when we have our primary or secondary alcohol, it's going to be reacting with chromic acid."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So if I were to start my mechanism here with an alcohol, remember this must be either a primary or a secondary alcohol in order for this oxidation to work. So I'm going to go ahead and show my alcohol there. So again, either primary, primary, or secondary, like that. And when we have our primary or secondary alcohol, it's going to be reacting with chromic acid. So here is the dot structure for chromic acid, like that. So I'll just simplify it right here. I won't worry too much about my lone pairs of electrons."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And when we have our primary or secondary alcohol, it's going to be reacting with chromic acid. So here is the dot structure for chromic acid, like that. So I'll just simplify it right here. I won't worry too much about my lone pairs of electrons. And chromic acid can come from several different reagents. Probably the most common reagent would be sodium dichromate. So Na2Cr2O7, sulfuric acid, H2SO4, and water."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "I won't worry too much about my lone pairs of electrons. And chromic acid can come from several different reagents. Probably the most common reagent would be sodium dichromate. So Na2Cr2O7, sulfuric acid, H2SO4, and water. And all of this together is usually referred to as the Jones reagent. So a mixture of sodium dichromate, sulfuric acid, and water is called the Jones reagent. And that will mix together to give you chromic acid in solution."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So Na2Cr2O7, sulfuric acid, H2SO4, and water. And all of this together is usually referred to as the Jones reagent. So a mixture of sodium dichromate, sulfuric acid, and water is called the Jones reagent. And that will mix together to give you chromic acid in solution. So another way to do it would be you could start from chromium trioxide. So you could also use a different reagent, which consists of CrO3, chromium trioxide, and H3O plus, and acetone. And that will also generate chromic acid in solution."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And that will mix together to give you chromic acid in solution. So another way to do it would be you could start from chromium trioxide. So you could also use a different reagent, which consists of CrO3, chromium trioxide, and H3O plus, and acetone. And that will also generate chromic acid in solution. So whichever one you would like to use. The first step of the mechanism is similar to the formation of nitrate esters that we saw in the previous video. So this is going to be a reaction equilibrium, or it's reversible."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And that will also generate chromic acid in solution. So whichever one you would like to use. The first step of the mechanism is similar to the formation of nitrate esters that we saw in the previous video. So this is going to be a reaction equilibrium, or it's reversible. And if you remember, in the formation of nitrate esters, this is a similar mechanism for the formation of all inorganic esters here. And we're going to lose this hydrogen and this OH. And those are going to produce water."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be a reaction equilibrium, or it's reversible. And if you remember, in the formation of nitrate esters, this is a similar mechanism for the formation of all inorganic esters here. And we're going to lose this hydrogen and this OH. And those are going to produce water. And we can stick those two molecules together. And so we would get this as the initial product here. We're going to have the end result of putting that oxygen bonded to that chromium atom like this."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And those are going to produce water. And we can stick those two molecules together. And so we would get this as the initial product here. We're going to have the end result of putting that oxygen bonded to that chromium atom like this. So this is a chromate ester intermediate. So this is what we would make. So in the next step of the mechanism, we need something to function as a base."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "We're going to have the end result of putting that oxygen bonded to that chromium atom like this. So this is a chromate ester intermediate. So this is what we would make. So in the next step of the mechanism, we need something to function as a base. And water is going to do that for us. So water comes along like this. Two lone pairs of electrons."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So in the next step of the mechanism, we need something to function as a base. And water is going to do that for us. So water comes along like this. Two lone pairs of electrons. One of those lone pairs can function as a base. And it's going to take that alpha proton. Remember, this is our alpha hydrogen on that carbon."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Two lone pairs of electrons. One of those lone pairs can function as a base. And it's going to take that alpha proton. Remember, this is our alpha hydrogen on that carbon. And over here, we're going to take just the proton, just the nucleus of that hydrogen atom. And so this lone pair of electrons in here could take that proton. That's going to leave the electron that hydrogen brought to the dot structure behind."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Remember, this is our alpha hydrogen on that carbon. And over here, we're going to take just the proton, just the nucleus of that hydrogen atom. And so this lone pair of electrons in here could take that proton. That's going to leave the electron that hydrogen brought to the dot structure behind. And these two electrons are going to move into here to increase the number of bonds of carbon to oxygen. At the same time, that is going to kick these electrons and this bond off onto the chromium. So let's go ahead and draw the product of that reaction here."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "That's going to leave the electron that hydrogen brought to the dot structure behind. And these two electrons are going to move into here to increase the number of bonds of carbon to oxygen. At the same time, that is going to kick these electrons and this bond off onto the chromium. So let's go ahead and draw the product of that reaction here. So let's see if we can get some space. So right here. Well, we're going to lose that alpha hydrogen."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the product of that reaction here. So let's see if we can get some space. So right here. Well, we're going to lose that alpha hydrogen. So now our carbon still is bonded to two other things. We lost that alpha hydrogen. And now it's double bonded to that oxygen."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "Well, we're going to lose that alpha hydrogen. So now our carbon still is bonded to two other things. We lost that alpha hydrogen. And now it's double bonded to that oxygen. So that would be the mechanism. We went from one bond of carbon to oxygen on our primary or secondary alcohol. We've now increased it to two bonds of carbon to oxygen."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "And now it's double bonded to that oxygen. So that would be the mechanism. We went from one bond of carbon to oxygen on our primary or secondary alcohol. We've now increased it to two bonds of carbon to oxygen. So the other products here, we would make H3O plus, of course. So we'll go ahead and put H3O plus when water picks up that proton. We would form HCrO3 minus as our other product."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "We've now increased it to two bonds of carbon to oxygen. So the other products here, we would make H3O plus, of course. So we'll go ahead and put H3O plus when water picks up that proton. We would form HCrO3 minus as our other product. Now, if the alpha carbon is the one being oxidized, so if this carbon is oxidized to this carbon, it's the same carbon, but this carbon is being oxidized, something must be being reduced. So this is a redox reaction. If you oxidize something, something else is reduced."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "We would form HCrO3 minus as our other product. Now, if the alpha carbon is the one being oxidized, so if this carbon is oxidized to this carbon, it's the same carbon, but this carbon is being oxidized, something must be being reduced. So this is a redox reaction. If you oxidize something, something else is reduced. And that something else is chromium. So if you were to assign an oxidation state to chromium in the sodium dichromate over here, so in this guy over here, chromium has an oxidation state of 6 plus. So when we look at our products and we find chromium in our products here, if you were to assign an oxidation state to this chromium, you'd get 4 plus, so Cr4 plus."}, {"video_title": "Oxidation of alcohols I Mechanism and oxidation states Organic chemistry Khan Academy.mp3", "Sentence": "If you oxidize something, something else is reduced. And that something else is chromium. So if you were to assign an oxidation state to chromium in the sodium dichromate over here, so in this guy over here, chromium has an oxidation state of 6 plus. So when we look at our products and we find chromium in our products here, if you were to assign an oxidation state to this chromium, you'd get 4 plus, so Cr4 plus. And there's some other chemistry that goes on, which ends up converting the chromium from 4 plus into 3 plus. And so overall, you can see that you're starting out with 6 plus over here, and you're ending up with 3 plus over here. That's a decrease in the oxidation state, so chromium is being reduced."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Here's the general reaction for a hydrohalogenation. You have an alkene, and you react that with a hydrogen halide. And the hydrogen adds to one side of your double bond, and the halogen adds to the other side of your double bond. Let's look at the mechanism for this reaction. So here we have our alkene and our hydrogen halide. Think something like hydrochloric acid. So a strong acid donates protons in solution."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the mechanism for this reaction. So here we have our alkene and our hydrogen halide. Think something like hydrochloric acid. So a strong acid donates protons in solution. So the hydrogen halide is going to function as an acid. The alkene is going to function as a base. The electrons in this pi bond here are going to take this proton and leave these two electrons behind on your halogen."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So a strong acid donates protons in solution. So the hydrogen halide is going to function as an acid. The alkene is going to function as a base. The electrons in this pi bond here are going to take this proton and leave these two electrons behind on your halogen. So let's go ahead and draw the result of that acid-base reaction. So I'm going to say the proton adds to the carbon on the left. And the carbon on the right over here used to have four bonds to it."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in this pi bond here are going to take this proton and leave these two electrons behind on your halogen. So let's go ahead and draw the result of that acid-base reaction. So I'm going to say the proton adds to the carbon on the left. And the carbon on the right over here used to have four bonds to it. Now it has only three bonds to it, which means it's positively charged and is a carbocation. So it wants electrons. It wants to get an octet of electrons."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the carbon on the right over here used to have four bonds to it. Now it has only three bonds to it, which means it's positively charged and is a carbocation. So it wants electrons. It wants to get an octet of electrons. So it loves electrons. It's an electrophile. We have our halogen over here, which had three lone pairs of electrons."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It wants to get an octet of electrons. So it loves electrons. It's an electrophile. We have our halogen over here, which had three lone pairs of electrons. It got one more lone pair for a total of eight electrons around it, which gives it a negative 1 formal charge, which means that it likes nuclei. It is a nucleophile. So our nucleophile is going to attack our electrophile like that and form a new bond to give us our alkyl halide as our product."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have our halogen over here, which had three lone pairs of electrons. It got one more lone pair for a total of eight electrons around it, which gives it a negative 1 formal charge, which means that it likes nuclei. It is a nucleophile. So our nucleophile is going to attack our electrophile like that and form a new bond to give us our alkyl halide as our product. Let's follow some of these electrons through our mechanism. So the electrons in this pi bond here, these are the same electrons that form this bond right here with the proton. And let's follow these electrons in here."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So our nucleophile is going to attack our electrophile like that and form a new bond to give us our alkyl halide as our product. Let's follow some of these electrons through our mechanism. So the electrons in this pi bond here, these are the same electrons that form this bond right here with the proton. And let's follow these electrons in here. So the electrons in here are the ones that ended up on our halogen. And I happen to choose these very same electrons to form the bond between the carbon and the halogen like that. So make sure to be able to follow electrons in mechanisms."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's follow these electrons in here. So the electrons in here are the ones that ended up on our halogen. And I happen to choose these very same electrons to form the bond between the carbon and the halogen like that. So make sure to be able to follow electrons in mechanisms. Let's look at an actual problem and let's follow the mechanism through. So let's look at this reaction here. This is my alkene."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So make sure to be able to follow electrons in mechanisms. Let's look at an actual problem and let's follow the mechanism through. So let's look at this reaction here. This is my alkene. And I'm going to react this alkene with hydrochloric acid. First step of the mechanism, the pi electrons function as a base and take this proton here, kick these electrons off onto your chlorine. So now we have a problem."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is my alkene. And I'm going to react this alkene with hydrochloric acid. First step of the mechanism, the pi electrons function as a base and take this proton here, kick these electrons off onto your chlorine. So now we have a problem. Which side do I add my proton to? I have two options. I could add the proton to the top carbon."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now we have a problem. Which side do I add my proton to? I have two options. I could add the proton to the top carbon. So let's go ahead and do that. Let's add the proton to the top carbon there and see what we get. So if I add that proton there to the top carbon, here's my methyl group."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I could add the proton to the top carbon. So let's go ahead and do that. Let's add the proton to the top carbon there and see what we get. So if I add that proton there to the top carbon, here's my methyl group. Here is my proton that I added. Well, I just took a bond away from this carbon. So this is where my carbocation is going to be."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I add that proton there to the top carbon, here's my methyl group. Here is my proton that I added. Well, I just took a bond away from this carbon. So this is where my carbocation is going to be. What kind of carbocation is that? That carbon is connected to two other carbons. So that is, of course, a secondary carbocation."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is where my carbocation is going to be. What kind of carbocation is that? That carbon is connected to two other carbons. So that is, of course, a secondary carbocation. Let's see what would happen if I add the proton onto the other carbon. So what would I get if I add the proton onto that carbon right there? So what would happen if I add the proton onto that bottom carbon?"}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that is, of course, a secondary carbocation. Let's see what would happen if I add the proton onto the other carbon. So what would I get if I add the proton onto that carbon right there? So what would happen if I add the proton onto that bottom carbon? So my methyl group is still here. I add my proton onto that bottom carbon now. Now this is my carbocation."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what would happen if I add the proton onto that bottom carbon? So my methyl group is still here. I add my proton onto that bottom carbon now. Now this is my carbocation. That is my positively charged carbon right there. What kind of carbocation is that? Well, that positively charged carbon is bonded to one, two, three other carbons."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now this is my carbocation. That is my positively charged carbon right there. What kind of carbocation is that? Well, that positively charged carbon is bonded to one, two, three other carbons. So this is actually a tertiary carbocation. We know that a tertiary carbocation is more stable than a secondary carbocation. So the tertiary carbocation is going to form faster than the secondary carbocation."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, that positively charged carbon is bonded to one, two, three other carbons. So this is actually a tertiary carbocation. We know that a tertiary carbocation is more stable than a secondary carbocation. So the tertiary carbocation is going to form faster than the secondary carbocation. What happens in the second step of the mechanism? Well, our chlorine ends up being negatively charged. It is going to function as a nucleophile."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the tertiary carbocation is going to form faster than the secondary carbocation. What happens in the second step of the mechanism? Well, our chlorine ends up being negatively charged. It is going to function as a nucleophile. This lone pair of electrons is going to attack this carbon and form our product. So let's go ahead and draw our product here. Our product is going to have a methyl group and then a chlorine attached to that carbon."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It is going to function as a nucleophile. This lone pair of electrons is going to attack this carbon and form our product. So let's go ahead and draw our product here. Our product is going to have a methyl group and then a chlorine attached to that carbon. So the halogen adds to the more substituted carbon. And this is called Markovnikov's rule. So let's go ahead and write that."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Our product is going to have a methyl group and then a chlorine attached to that carbon. So the halogen adds to the more substituted carbon. And this is called Markovnikov's rule. So let's go ahead and write that. See if we can spell Markovnikov. So Markovnikov's rule, the halogen adds to the more substituted carbon. And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that. See if we can spell Markovnikov. So Markovnikov's rule, the halogen adds to the more substituted carbon. And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism. So let's do another mechanism here. Whenever you have a carbocation present, you could have rearrangement. So let's do one where there's some rearrangement."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the reason it does that is because the more substituted carbon is the one that was the more stable carbocation in the mechanism. So let's do another mechanism here. Whenever you have a carbocation present, you could have rearrangement. So let's do one where there's some rearrangement. So let's start out with this as our alkene. And we'll react that with hydrochloric acid once again. First step, pi electrons function as a base."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's do one where there's some rearrangement. So let's start out with this as our alkene. And we'll react that with hydrochloric acid once again. First step, pi electrons function as a base. These electrons kick off onto your chlorine. So which side do we add the proton to? We could add the proton to the left side of the double bond."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "First step, pi electrons function as a base. These electrons kick off onto your chlorine. So which side do we add the proton to? We could add the proton to the left side of the double bond. We could add the proton to the right side of the double bond. Well, we know that we want to form the most stable carbocation that we can. So it makes sense to add the proton to the right side of the double bond right here, because that's going to give us this as a carbocation."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We could add the proton to the left side of the double bond. We could add the proton to the right side of the double bond. Well, we know that we want to form the most stable carbocation that we can. So it makes sense to add the proton to the right side of the double bond right here, because that's going to give us this as a carbocation. What kind of carbocation is that? So let's identify this carbon as the one that has our positive charge. That carbon is bonded to two other carbons."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it makes sense to add the proton to the right side of the double bond right here, because that's going to give us this as a carbocation. What kind of carbocation is that? So let's identify this carbon as the one that has our positive charge. That carbon is bonded to two other carbons. So it is a secondary carbocation. If we had added on the proton to the left side of the double bond, we would have a primary carbocation here. So a secondary carbocation is more stable."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That carbon is bonded to two other carbons. So it is a secondary carbocation. If we had added on the proton to the left side of the double bond, we would have a primary carbocation here. So a secondary carbocation is more stable. Can we form a tertiary carbocation? Because we know tertiary carbocations are even more stable than secondary carbocations. And of course we can."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So a secondary carbocation is more stable. Can we form a tertiary carbocation? Because we know tertiary carbocations are even more stable than secondary carbocations. And of course we can. We're going to get a hydrogen attached to this carbon. And we saw in our earlier video on carbocations and rearrangements, we could get a hydride shift here. So the proton and these two electrons here are hydride anion."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And of course we can. We're going to get a hydrogen attached to this carbon. And we saw in our earlier video on carbocations and rearrangements, we could get a hydride shift here. So the proton and these two electrons here are hydride anion. And these two electrons are going to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well now, our hydride has shifted over here to that carbon."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the proton and these two electrons here are hydride anion. And these two electrons are going to move over here, shift over one carbon, and form a new covalent bond. So what would we get if we get a hydride shift in our mechanism? Well now, our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now this is where our positive charge is."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well now, our hydride has shifted over here to that carbon. This carbon no longer has a positive charge on it. We took a bond away from this carbon. So now this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now this is where our positive charge is. So we have a carbocation. How would we classify this carbocation? Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, one, two, three other carbons. So it's tertiary. It's more stable than our secondary carbocation. So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion negatively charged nucleophile. So nucleophilic attack on our carbocation. So right there."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in the final step of our mechanism, we had our chloride anion over here from the first step of our mechanism. So a chloride anion negatively charged nucleophile. So nucleophilic attack on our carbocation. So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up with our cyclohexane ring. Let's do another one."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So right there. And we're going to form a bond between that halogen and that carbon. So our final product is going to end up with our cyclohexane ring. Let's do another one. So we have our cyclohexane ring like that. And then we have our ethyl group attached to this carbon. And then our chlorine attached to that carbon like that."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one. So we have our cyclohexane ring like that. And then we have our ethyl group attached to this carbon. And then our chlorine attached to that carbon like that. So that's going to be our major product. Let's look at the stereochemistry of this reaction really fast. So let's look at what happens if we react this alkene with hydrochloric acid."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then our chlorine attached to that carbon like that. So that's going to be our major product. Let's look at the stereochemistry of this reaction really fast. So let's look at what happens if we react this alkene with hydrochloric acid. So what will we get? First step, pi electrons. Take the proton."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at what happens if we react this alkene with hydrochloric acid. So what will we get? First step, pi electrons. Take the proton. Kick the electrons off onto the chlorine. So once again, which side do we add our proton? So two possibilities."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Take the proton. Kick the electrons off onto the chlorine. So once again, which side do we add our proton? So two possibilities. I could add to the left side of the double bond or add to the right side of the double bond. It makes sense to add it to the right side of the double bond because that gives us a more stable carbocation. So if I add to the right side of the double bond, this carbon ends up being positively charged."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So two possibilities. I could add to the left side of the double bond or add to the right side of the double bond. It makes sense to add it to the right side of the double bond because that gives us a more stable carbocation. So if I add to the right side of the double bond, this carbon ends up being positively charged. What kind of a carbocation is that? That carbon is bonded to two other carbons. So it's a secondary carbocation here."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I add to the right side of the double bond, this carbon ends up being positively charged. What kind of a carbocation is that? That carbon is bonded to two other carbons. So it's a secondary carbocation here. So we have a secondary carbocation. There's no kind of rearrangement that we could get here to get a tertiary carbocation. So secondary carbocation is as stable as we're going to get."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's a secondary carbocation here. So we have a secondary carbocation. There's no kind of rearrangement that we could get here to get a tertiary carbocation. So secondary carbocation is as stable as we're going to get. Now, carbocations are carbons with three bonds to them, meaning that carbon is sp2 hybridized. So let's redraw this carbocation here. So I'm going to say that this carbon right here represents my carbocation."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So secondary carbocation is as stable as we're going to get. Now, carbocations are carbons with three bonds to them, meaning that carbon is sp2 hybridized. So let's redraw this carbocation here. So I'm going to say that this carbon right here represents my carbocation. What's bonded to it? Well, on the left side, there is an ethyl group, CH2CH3, right here. And I know that there's a methyl group bonded to it."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to say that this carbon right here represents my carbocation. What's bonded to it? Well, on the left side, there is an ethyl group, CH2CH3, right here. And I know that there's a methyl group bonded to it. I'm going to put the methyl group going back in space here. And then there's also a hydrogen attached to that carbon. We just didn't draw it in on our carbocation, so like this."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I know that there's a methyl group bonded to it. I'm going to put the methyl group going back in space here. And then there's also a hydrogen attached to that carbon. We just didn't draw it in on our carbocation, so like this. I know this carbon is sp2 hybridized, meaning there's a untouched, unhybridized p orbital on this carbon. So let me draw my p orbital in there like that. And let me go ahead and make sure that everyone realizes this is my carbocation."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We just didn't draw it in on our carbocation, so like this. I know this carbon is sp2 hybridized, meaning there's a untouched, unhybridized p orbital on this carbon. So let me draw my p orbital in there like that. And let me go ahead and make sure that everyone realizes this is my carbocation. So sp2 hybridized carbon means the atoms bonded to that sp2 hybridized carbon are in the same plane. So we can think about our carbocation being flat, because of the trigonal planar geometry like that. So when our chloride anion comes along, it sees a flat sp2 carbocation here."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let me go ahead and make sure that everyone realizes this is my carbocation. So sp2 hybridized carbon means the atoms bonded to that sp2 hybridized carbon are in the same plane. So we can think about our carbocation being flat, because of the trigonal planar geometry like that. So when our chloride anion comes along, it sees a flat sp2 carbocation here. So it's negatively charged. So when that chloride anion nucleophilic attacks our carbocation, it has two options. It could attack from the top of that plane."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when our chloride anion comes along, it sees a flat sp2 carbocation here. So it's negatively charged. So when that chloride anion nucleophilic attacks our carbocation, it has two options. It could attack from the top of that plane. So let's go ahead and draw the product that would result if it attacked from the top of that plane. So let's see what we'd have here. We would have our carbon form the new bond with our chlorine like that."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It could attack from the top of that plane. So let's go ahead and draw the product that would result if it attacked from the top of that plane. So let's see what we'd have here. We would have our carbon form the new bond with our chlorine like that. And then we have a CH2CH3 over here on the left side. And we have a hydrogen coming out at us. And we have a methyl group going away from us."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We would have our carbon form the new bond with our chlorine like that. And then we have a CH2CH3 over here on the left side. And we have a hydrogen coming out at us. And we have a methyl group going away from us. So that's one possibility. And what about if that chloride anion attacked from the bottom? So that's a possibility as well."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have a methyl group going away from us. So that's one possibility. And what about if that chloride anion attacked from the bottom? So that's a possibility as well. So since it's a plane, a flat sheet of paper, 50% chance it attacks from the top. 50% chance it attacks from below. So what would we get if it attacks from below?"}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's a possibility as well. So since it's a plane, a flat sheet of paper, 50% chance it attacks from the top. 50% chance it attacks from below. So what would we get if it attacks from below? Well, we would have a carbon forming a bond with that chlorine right here. And then we'd have an ethyl group over here on the left side, CH2CH3. We'd have a hydrogen coming out at us."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what would we get if it attacks from below? Well, we would have a carbon forming a bond with that chlorine right here. And then we'd have an ethyl group over here on the left side, CH2CH3. We'd have a hydrogen coming out at us. And we would have a methyl group going away from us like that. So if you look at these products here, this is a chirality center. This carbon right here has four different groups attached to it."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'd have a hydrogen coming out at us. And we would have a methyl group going away from us like that. So if you look at these products here, this is a chirality center. This carbon right here has four different groups attached to it. It's sp3 hybridized. Same with this group. So these two are actually mirror images of each other, non-superimposable mirror images, enantiomers of each other."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here has four different groups attached to it. It's sp3 hybridized. Same with this group. So these two are actually mirror images of each other, non-superimposable mirror images, enantiomers of each other. Let's see if we can draw them in a dot structure that's more familiar to you here. So if you need a molymod set, go ahead and take it out right now and put your eye right here and stare at your chirality center. So what enantiomer is this one over here?"}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these two are actually mirror images of each other, non-superimposable mirror images, enantiomers of each other. Let's see if we can draw them in a dot structure that's more familiar to you here. So if you need a molymod set, go ahead and take it out right now and put your eye right here and stare at your chirality center. So what enantiomer is this one over here? Well, here is my chirality center carbon. And if I were looking at it from this vantage point, there would be an ethyl group here on the left, CH2CH3. On the right, there would be a methyl group."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what enantiomer is this one over here? Well, here is my chirality center carbon. And if I were looking at it from this vantage point, there would be an ethyl group here on the left, CH2CH3. On the right, there would be a methyl group. And this hydrogen would be coming out at me in space, and the chlorine would be going away from me in space. So that is the enantiomer I'm looking at here. What about this one?"}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "On the right, there would be a methyl group. And this hydrogen would be coming out at me in space, and the chlorine would be going away from me in space. So that is the enantiomer I'm looking at here. What about this one? So same thing. I put my eye right here. I stare at my chirality center."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What about this one? So same thing. I put my eye right here. I stare at my chirality center. So here's my chirality center. I'd have an ethyl group on the left, methyl group on the right. This time, though, the chlorine would be coming out at me."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I stare at my chirality center. So here's my chirality center. I'd have an ethyl group on the left, methyl group on the right. This time, though, the chlorine would be coming out at me. So the chlorine would be coming out at me like that. And I'd get a racemic mixture of my products. So be careful when you have carbocations in your mechanism."}, {"video_title": "Hydrohalogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This time, though, the chlorine would be coming out at me. So the chlorine would be coming out at me like that. And I'd get a racemic mixture of my products. So be careful when you have carbocations in your mechanism. Think about the fact that they are trigonal planar. And think about, does this reaction form any chirality centers? If it forms chirality centers, you're going to get a racemic mixture for your products."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "But I thought it was about time that we actually devoted a video or two to ethers. And like all things that we've done in organic chemistry, a good way to familiarize ourselves with the molecules and how they look is to actually name them. So let's do a couple. And the first few you've seen already. So let's say we have this molecule right here. What I'm going to do is I'm going to teach you two ways to name it, the common name. And that's probably the more important one, especially with ethers, because as you can imagine, that is the more common name."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And the first few you've seen already. So let's say we have this molecule right here. What I'm going to do is I'm going to teach you two ways to name it, the common name. And that's probably the more important one, especially with ethers, because as you can imagine, that is the more common name. That is what people say. And then I'll also show you how to name it using the IUPAC name, so let me write this down. I-U-P-A-C name, which is the International Union of Pure and Applied Chemistry."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And that's probably the more important one, especially with ethers, because as you can imagine, that is the more common name. That is what people say. And then I'll also show you how to name it using the IUPAC name, so let me write this down. I-U-P-A-C name, which is the International Union of Pure and Applied Chemistry. And they come up with the official naming protocols for all of these organic molecules. And this is actually the convention that we used earlier when we did the alkanes and the alkenes. But in the case of ethers, the common name is more common."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "I-U-P-A-C name, which is the International Union of Pure and Applied Chemistry. And they come up with the official naming protocols for all of these organic molecules. And this is actually the convention that we used earlier when we did the alkanes and the alkenes. But in the case of ethers, the common name is more common. So the common name for this molecule right here, you look at the two carbon groups here. So let's see, you have this one right here. That is an ethyl group."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "But in the case of ethers, the common name is more common. So the common name for this molecule right here, you look at the two carbon groups here. So let's see, you have this one right here. That is an ethyl group. That's an ethyl group right there. You have one, two carbons. And then you look at the other carbon group right over there."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "That is an ethyl group. That's an ethyl group right there. You have one, two carbons. And then you look at the other carbon group right over there. That's also an ethyl group. You have one, two carbons. So you call this, let me just write this down."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And then you look at the other carbon group right over there. That's also an ethyl group. You have one, two carbons. So you call this, let me just write this down. That is also an ethyl group. So the common name for this is just diethyl ether. And the ether tells you, this part tells you, that you have an oxygen in between your two ethyl groups."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So you call this, let me just write this down. That is also an ethyl group. So the common name for this is just diethyl ether. And the ether tells you, this part tells you, that you have an oxygen in between your two ethyl groups. This is the common name. Now, the International Union of Pure and Applied Chemistry official name for it, you kind of do something similar to how we named other things before. You look for the longest chain."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And the ether tells you, this part tells you, that you have an oxygen in between your two ethyl groups. This is the common name. Now, the International Union of Pure and Applied Chemistry official name for it, you kind of do something similar to how we named other things before. You look for the longest chain. Let me redraw it. So maybe on the left-hand side, I'll do the common names. On the right-hand side, I'll do the IUPAC names."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "You look for the longest chain. Let me redraw it. So maybe on the left-hand side, I'll do the common names. On the right-hand side, I'll do the IUPAC names. So let me redraw the common name of the diethyl ether. Well, you look for the longest chain. In this case, there's two longest chains."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "On the right-hand side, I'll do the IUPAC names. So let me redraw the common name of the diethyl ether. Well, you look for the longest chain. In this case, there's two longest chains. There's this one that has one, two carbons. And then you have this one that has one, two carbons. You could pick either one."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "In this case, there's two longest chains. There's this one that has one, two carbons. And then you have this one that has one, two carbons. You could pick either one. I'll just pick this one as the main chain right over there. It has two carbons, no double bonds. It is an ethane."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "You could pick either one. I'll just pick this one as the main chain right over there. It has two carbons, no double bonds. It is an ethane. And then you say, OK, I have this alkoxy group. And we put the oxy at the end of it because it has this oxygen right here. But the alkyl part of it has two carbons, one, two."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "It is an ethane. And then you say, OK, I have this alkoxy group. And we put the oxy at the end of it because it has this oxygen right here. But the alkyl part of it has two carbons, one, two. So we call this right here. We call this ethoxy. So one other way to name this, we have this ethoxy group attached to the one carbon."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "But the alkyl part of it has two carbons, one, two. So we call this right here. We call this ethoxy. So one other way to name this, we have this ethoxy group attached to the one carbon. We're just going to start numbering on this side of the ethane just because that's where the group is. Attached to the one carbon. So we call this one ethoxy ethane."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So one other way to name this, we have this ethoxy group attached to the one carbon. We're just going to start numbering on this side of the ethane just because that's where the group is. Attached to the one carbon. So we call this one ethoxy ethane. So this is one ethoxy ethane. You'll almost never see it actually named this way, even though this is the official name. You're much more likely to see this as diethyl ether."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So we call this one ethoxy ethane. So this is one ethoxy ethane. You'll almost never see it actually named this way, even though this is the official name. You're much more likely to see this as diethyl ether. And at least in my brain, this resonates a lot more. You just say, what are the two groups? And you throw the ether at the end, you know that there's an oxygen in between."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "You're much more likely to see this as diethyl ether. And at least in my brain, this resonates a lot more. You just say, what are the two groups? And you throw the ether at the end, you know that there's an oxygen in between. Let's do a couple more of these. So let's say I have this molecule right here. The common way is you look at the two groups on either side of the oxygen."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And you throw the ether at the end, you know that there's an oxygen in between. Let's do a couple more of these. So let's say I have this molecule right here. The common way is you look at the two groups on either side of the oxygen. So this right here, let me do this in a different color. This group on the left right here, we have one, two, three carbons, it's a propyl group, but we're attached to that middle carbon, so this is an isopropyl group. And on the right-hand side right here, we just have one carbon."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "The common way is you look at the two groups on either side of the oxygen. So this right here, let me do this in a different color. This group on the left right here, we have one, two, three carbons, it's a propyl group, but we're attached to that middle carbon, so this is an isopropyl group. And on the right-hand side right here, we just have one carbon. So this is right here. I keep using that blue. This is a methyl group."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And on the right-hand side right here, we just have one carbon. So this is right here. I keep using that blue. This is a methyl group. So the common way of naming it, you just list both of these groups and then you write ether. And you list them in alphabetical order. I comes before M, so this is, the common name is isopropyl methyl ether."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "This is a methyl group. So the common way of naming it, you just list both of these groups and then you write ether. And you list them in alphabetical order. I comes before M, so this is, the common name is isopropyl methyl ether. Now, if we were to do the IUPAC naming, we look for the longest carbon chain. Let me redraw the molecule itself. So what's the longest carbon chain here?"}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "I comes before M, so this is, the common name is isopropyl methyl ether. Now, if we were to do the IUPAC naming, we look for the longest carbon chain. Let me redraw the molecule itself. So what's the longest carbon chain here? We have one, two, three carbons right there. We only have one carbon right there. So this thing right here is our longest carbon chain."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So what's the longest carbon chain here? We have one, two, three carbons right there. We only have one carbon right there. So this thing right here is our longest carbon chain. It has three carbons on it and it had no double bonds. So eth-meth-propyl, or it's actually propane. So this is our longest."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So this thing right here is our longest carbon chain. It has three carbons on it and it had no double bonds. So eth-meth-propyl, or it's actually propane. So this is our longest. So we write propane right there because we're using the IUPAC naming mechanism. And then we look at this methoxy group right here. And I call this a methoxy group because I have the O."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So this is our longest. So we write propane right there because we're using the IUPAC naming mechanism. And then we look at this methoxy group right here. And I call this a methoxy group because I have the O. That gives us the oxy and I just have a methyl group right here. So this is methoxy. You remember that meth is the prefix for just having only one carbon."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And I call this a methoxy group because I have the O. That gives us the oxy and I just have a methyl group right here. So this is methoxy. You remember that meth is the prefix for just having only one carbon. We add the oxy because that oxygen is there. And it's attached to the two carbon on the propane chain, no matter what direction you start naming from, or numbering from, one, two, three. So this is 2-methoxypropane."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "You remember that meth is the prefix for just having only one carbon. We add the oxy because that oxygen is there. And it's attached to the two carbon on the propane chain, no matter what direction you start naming from, or numbering from, one, two, three. So this is 2-methoxypropane. Let's do another one. Let's do one more. And I think you'll get the gist of at least the reasonably simple ethers to name."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So this is 2-methoxypropane. Let's do another one. Let's do one more. And I think you'll get the gist of at least the reasonably simple ethers to name. So let's put a ring over there. And then that's attached to an oxygen. And then we have another carbon chain right here."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And I think you'll get the gist of at least the reasonably simple ethers to name. So let's put a ring over there. And then that's attached to an oxygen. And then we have another carbon chain right here. And then we have another carbon chain right there. Let me just copy and paste that so that I don't have to redraw it. So let me copy and paste."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And then we have another carbon chain right here. And then we have another carbon chain right there. Let me just copy and paste that so that I don't have to redraw it. So let me copy and paste. So let's do the common name first. That always tends to be a little bit more fun. So on this side, we have one, two, three, four, five, six carbons in a cycle."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So let me copy and paste. So let's do the common name first. That always tends to be a little bit more fun. So on this side, we have one, two, three, four, five, six carbons in a cycle. This right here on the left-hand side is a cyclohexyl group. Cyclohexyl group. This on the right-hand side, we have one, two, three."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So on this side, we have one, two, three, four, five, six carbons in a cycle. This right here on the left-hand side is a cyclohexyl group. Cyclohexyl group. This on the right-hand side, we have one, two, three. This is just a straight-up propyl group. And so when you name the ether, you just put these two groups in alphabetical order. And you add an ether at the end."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "This on the right-hand side, we have one, two, three. This is just a straight-up propyl group. And so when you name the ether, you just put these two groups in alphabetical order. And you add an ether at the end. So it's cyclohexyl. C comes before P. So it's cyclohexylpropyl. Cyclohexylpropyl."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And you add an ether at the end. So it's cyclohexyl. C comes before P. So it's cyclohexylpropyl. Cyclohexylpropyl. Let me get that shade of yellow right. Cyclohexylpropyl ether. Now let's do the IUPAC way to name it."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "Cyclohexylpropyl. Let me get that shade of yellow right. Cyclohexylpropyl ether. Now let's do the IUPAC way to name it. So you look for the longest carbon chain here. In this case, it's going to be the cyclohexane right here. We have one, two, three, four, five, six carbons there."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "Now let's do the IUPAC way to name it. So you look for the longest carbon chain here. In this case, it's going to be the cyclohexane right here. We have one, two, three, four, five, six carbons there. We only have one, two, three there. So this is kind of our backbone. So we write down cyclohexane."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "We have one, two, three, four, five, six carbons there. We only have one, two, three there. So this is kind of our backbone. So we write down cyclohexane. No double bonds. So it's a hexane. So that's the cyclohexane right there."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So we write down cyclohexane. No double bonds. So it's a hexane. So that's the cyclohexane right there. If you just had these three carbons, it would be a propyl. But this is not just three carbons. It's three carbons and then an oxygen."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So that's the cyclohexane right there. If you just had these three carbons, it would be a propyl. But this is not just three carbons. It's three carbons and then an oxygen. So we would call it a propoxy. So this is propoxy group. Propoxy."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "It's three carbons and then an oxygen. So we would call it a propoxy. So this is propoxy group. Propoxy. And you don't have to number it because it can just be attached to any of these carbons. It would essentially be the same molecule. So you could just call this propoxycyclohexane."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "Propoxy. And you don't have to number it because it can just be attached to any of these carbons. It would essentially be the same molecule. So you could just call this propoxycyclohexane. Let me make it a little bit closer to the cyclohexane. But once again, the common name is what you're more likely to see. Now that we've named a few of them, let's think a little bit about their properties."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So you could just call this propoxycyclohexane. Let me make it a little bit closer to the cyclohexane. But once again, the common name is what you're more likely to see. Now that we've named a few of them, let's think a little bit about their properties. So what we've seen already is that we've used it several times, especially in our SN2 reactions and things like that, places where we didn't want protons floating around, we used actually diethyl ether. And in general, ethers do make for good solvents. They tend to be fairly unreactive."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "Now that we've named a few of them, let's think a little bit about their properties. So what we've seen already is that we've used it several times, especially in our SN2 reactions and things like that, places where we didn't want protons floating around, we used actually diethyl ether. And in general, ethers do make for good solvents. They tend to be fairly unreactive. So good solvents, especially when you're looking for an aprotic solvent. Remember, aprotic means you don't have hydrogens that can kind of lose their electron to maybe an electronegative atom like an oxygen, and then the proton just floats around and then can go and react with other things. This does not have any hydrogens directly bonded to an oxygen in any of these cases."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "They tend to be fairly unreactive. So good solvents, especially when you're looking for an aprotic solvent. Remember, aprotic means you don't have hydrogens that can kind of lose their electron to maybe an electronegative atom like an oxygen, and then the proton just floats around and then can go and react with other things. This does not have any hydrogens directly bonded to an oxygen in any of these cases. So it is an aprotic solvent. And because it doesn't have any hydrogens bonded to the oxygen, you also have no hydrogen bonding. And just as a bit of a review, you know that in water you have this situation."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "This does not have any hydrogens directly bonded to an oxygen in any of these cases. So it is an aprotic solvent. And because it doesn't have any hydrogens bonded to the oxygen, you also have no hydrogen bonding. And just as a bit of a review, you know that in water you have this situation. Let me draw some water molecules. In water you have this situation where the oxygen hogs the electrons. So it has a partial negative charge."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "And just as a bit of a review, you know that in water you have this situation. Let me draw some water molecules. In water you have this situation where the oxygen hogs the electrons. So it has a partial negative charge. Hydrogen gets its electrons hogged or taken away, or it spends less time with them. So it has a partial positive charge. So this oxygen will have a partial negative charge."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So it has a partial negative charge. Hydrogen gets its electrons hogged or taken away, or it spends less time with them. So it has a partial positive charge. So this oxygen will have a partial negative charge. And so the hydrogens with the partial positive charge are attracted to the oxygens with the partial negative charge, and you have this hydrogen bonding. And this hydrogen bonding makes water, it pulls the molecules together. So you need to put more energy into it for it to either melt or for it to actually boil and for the molecules to kind of get ripped away from each other."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen will have a partial negative charge. And so the hydrogens with the partial positive charge are attracted to the oxygens with the partial negative charge, and you have this hydrogen bonding. And this hydrogen bonding makes water, it pulls the molecules together. So you need to put more energy into it for it to either melt or for it to actually boil and for the molecules to kind of get ripped away from each other. And that's also true with alcohols. Alcohols only have one hydrogen to each oxygen, but they still have the hydrogen bonding going on. In the case of ethers, there is no hydrogen bonding."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "So you need to put more energy into it for it to either melt or for it to actually boil and for the molecules to kind of get ripped away from each other. And that's also true with alcohols. Alcohols only have one hydrogen to each oxygen, but they still have the hydrogen bonding going on. In the case of ethers, there is no hydrogen bonding. I'll represent each of the carbon chains with an R and an R. I'll write R prime right here to show that it could be a different carbon chain than this right here. And the R stands for radical, not to be confused with free radical, completely different things. This R just means really a carbon chain attached to this oxygen."}, {"video_title": "Ether naming and introduction Organic chemistry Khan Academy.mp3", "Sentence": "In the case of ethers, there is no hydrogen bonding. I'll represent each of the carbon chains with an R and an R. I'll write R prime right here to show that it could be a different carbon chain than this right here. And the R stands for radical, not to be confused with free radical, completely different things. This R just means really a carbon chain attached to this oxygen. But here, there's no hydrogen getting its electrons hogged by oxygens with partial positive and partial negative charges. So you're not going to have that type of hydrogen bonding. And because of that, ethers have much lower melting and boiling points."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we want to understand where we've come from, the stories that have led us to our present condition, if we want to understand our history, one of the prerequisites is to have a good sense of chronometry. And chronometry, very fancy word, but it really is just the science of the passage of time. Chrono, relating to time, metri, time measurement. And we take many, many things for granted these days. We assume that we know what happened the last 50 years, the last 100 years, and now we're starting to assume we know what happened 10,000 years ago, or what happened to our planet 100 million years ago, or a billion years ago. But these are all very, very, very new phenomena, this ability to kind of shine a light on the past. And even the traditional notions of history, the traditional stories of what led to what, the political nations that formed, the migrations of people, and when they happened, that traditional notion of history is even fairly new when you think about just the scope of how long we think humans have now been on this planet."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we take many, many things for granted these days. We assume that we know what happened the last 50 years, the last 100 years, and now we're starting to assume we know what happened 10,000 years ago, or what happened to our planet 100 million years ago, or a billion years ago. But these are all very, very, very new phenomena, this ability to kind of shine a light on the past. And even the traditional notions of history, the traditional stories of what led to what, the political nations that formed, the migrations of people, and when they happened, that traditional notion of history is even fairly new when you think about just the scope of how long we think humans have now been on this planet. And that first, that traditional notion of history, you can kind of use the first chronometric revolution. And that first chronometric revolution that gives us this kind of traditional notion of history really just comes out of humanity's ability to write. So the first, so writing, writing gives us our first chronometric revolution."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even the traditional notions of history, the traditional stories of what led to what, the political nations that formed, the migrations of people, and when they happened, that traditional notion of history is even fairly new when you think about just the scope of how long we think humans have now been on this planet. And that first, that traditional notion of history, you can kind of use the first chronometric revolution. And that first chronometric revolution that gives us this kind of traditional notion of history really just comes out of humanity's ability to write. So the first, so writing, writing gives us our first chronometric revolution. Because this was the first time, even though we think humans or human-like creatures have been around for hundreds of thousands of years at this point, they weren't able to keep their stories in a very exact way. They might have had an oral tradition. It might have gone from one generation to the other."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the first, so writing, writing gives us our first chronometric revolution. Because this was the first time, even though we think humans or human-like creatures have been around for hundreds of thousands of years at this point, they weren't able to keep their stories in a very exact way. They might have had an oral tradition. It might have gone from one generation to the other. But with those oral traditions, things would get lost. And the most important information would get lost is how long ago did these stories start up? And we weren't able, as a species, to really have a firm understanding of when things happened and how long ago things happened until writing became mainstream and until writing was done in a way that it became permanent."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It might have gone from one generation to the other. But with those oral traditions, things would get lost. And the most important information would get lost is how long ago did these stories start up? And we weren't able, as a species, to really have a firm understanding of when things happened and how long ago things happened until writing became mainstream and until writing was done in a way that it became permanent. And our best sense of when this happened the first time was by the Sumerians with cuneiform. And this happened right around the third millennia BC, so around 5,000 years before the present time. And this is what some of that earliest writing looked like."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we weren't able, as a species, to really have a firm understanding of when things happened and how long ago things happened until writing became mainstream and until writing was done in a way that it became permanent. And our best sense of when this happened the first time was by the Sumerians with cuneiform. And this happened right around the third millennia BC, so around 5,000 years before the present time. And this is what some of that earliest writing looked like. This is actually a letter from, I believe, a king. And you can see it's just highly symbolic carvings. This is what we more traditionally associate with cuneiform."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is what some of that earliest writing looked like. This is actually a letter from, I believe, a king. And you can see it's just highly symbolic carvings. This is what we more traditionally associate with cuneiform. And it was symbolic-based, as opposed to now most of our languages are based on phonetics, so you have fewer symbols that can represent more meanings. But this was a huge technological revolution, I could say, for humanity. Because now, with the advent of cuneiform, you now had permanent writing that someone could look at 1,000 years later, 2,000 years later, and if they can decipher the cuneiform, they can get a written testimony of what was happening at that time."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is what we more traditionally associate with cuneiform. And it was symbolic-based, as opposed to now most of our languages are based on phonetics, so you have fewer symbols that can represent more meanings. But this was a huge technological revolution, I could say, for humanity. Because now, with the advent of cuneiform, you now had permanent writing that someone could look at 1,000 years later, 2,000 years later, and if they can decipher the cuneiform, they can get a written testimony of what was happening at that time. And they didn't have to rely on an oral tradition or even guess when that oral story might have started. But writing, since it only happened about 5,000 years ago, so this is 5,000 years before the present, or you could say 3,000 years BC, give or take, that was a start. But this only gave us stories of about 5,000 years old."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because now, with the advent of cuneiform, you now had permanent writing that someone could look at 1,000 years later, 2,000 years later, and if they can decipher the cuneiform, they can get a written testimony of what was happening at that time. And they didn't have to rely on an oral tradition or even guess when that oral story might have started. But writing, since it only happened about 5,000 years ago, so this is 5,000 years before the present, or you could say 3,000 years BC, give or take, that was a start. But this only gave us stories of about 5,000 years old. And even then, it was a very spotty historical record. We didn't really get really deep history, depending on where you are in the world, until really the last few thousand years. But it was a start."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this only gave us stories of about 5,000 years old. And even then, it was a very spotty historical record. We didn't really get really deep history, depending on where you are in the world, until really the last few thousand years. But it was a start. This was the first chronometric revolution. But what you may or may not realize is that we are, frankly, I believe, at the very early stages of another chronometric revolution that has really just begun to accelerate in the last 50, 60, 70 years. And this second chronometric revolution, I should write revolution up here too, this was a revolution."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it was a start. This was the first chronometric revolution. But what you may or may not realize is that we are, frankly, I believe, at the very early stages of another chronometric revolution that has really just begun to accelerate in the last 50, 60, 70 years. And this second chronometric revolution, I should write revolution up here too, this was a revolution. It allowed us to keep time in a permanent way, to understand things, to not have to talk to the people to whom something happened. We can see their written testimony of it. But the second revolution really comes out of the advent of a lot of our understanding of modern science."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this second chronometric revolution, I should write revolution up here too, this was a revolution. It allowed us to keep time in a permanent way, to understand things, to not have to talk to the people to whom something happened. We can see their written testimony of it. But the second revolution really comes out of the advent of a lot of our understanding of modern science. So in the late 1800s, radioactivity gets discovered by Marie and Pierre Curie. So this is 1900 right here. So this is relatively recent."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the second revolution really comes out of the advent of a lot of our understanding of modern science. So in the late 1800s, radioactivity gets discovered by Marie and Pierre Curie. So this is 1900 right here. So this is relatively recent. Remember, we're talking about a species that has been around for several hundreds of thousands of years. And proto-humans have been around for millions of years. And now, only 5,000 years ago, at least as far as we know, was the first writing."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is relatively recent. Remember, we're talking about a species that has been around for several hundreds of thousands of years. And proto-humans have been around for millions of years. And now, only 5,000 years ago, at least as far as we know, was the first writing. And then only a little over 100 years ago was a discovery of radioactivity. So radioactivity. And then the ability to use radioactivity."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now, only 5,000 years ago, at least as far as we know, was the first writing. And then only a little over 100 years ago was a discovery of radioactivity. So radioactivity. And then the ability to use radioactivity. So radioactivity is interesting. It's this idea that, essentially, elements can change from one variation to another of an element over long periods of time, so through radioactivity. So they become kind of this natural clock."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then the ability to use radioactivity. So radioactivity is interesting. It's this idea that, essentially, elements can change from one variation to another of an element over long periods of time, so through radioactivity. So they become kind of this natural clock. No one had to go there and set up a timepiece for it. Luckily, there are these things that decay at a very predictable rate. So we discover radioactivity a little over 100 years ago."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they become kind of this natural clock. No one had to go there and set up a timepiece for it. Luckily, there are these things that decay at a very predictable rate. So we discover radioactivity a little over 100 years ago. And then over the course of the 20th century, we got better and better, more sophisticated at really understanding radioactivity to be able to use it to measure the times of things. And if you fast-forward to the second half of the 20th century, so now, say, we're at 1950, this is where the second chronometric revolution really took hold. This is where it really took hold, where we started to understand carbon-14 dating."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we discover radioactivity a little over 100 years ago. And then over the course of the 20th century, we got better and better, more sophisticated at really understanding radioactivity to be able to use it to measure the times of things. And if you fast-forward to the second half of the 20th century, so now, say, we're at 1950, this is where the second chronometric revolution really took hold. This is where it really took hold, where we started to understand carbon-14 dating. We started to understand some of the other techniques that we talk about, where we can start to date older and older things. And I want to be clear. The understanding of radioactivity was just the beginning of this second chronometric revolution."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is where it really took hold, where we started to understand carbon-14 dating. We started to understand some of the other techniques that we talk about, where we can start to date older and older things. And I want to be clear. The understanding of radioactivity was just the beginning of this second chronometric revolution. The second chronometric revolution, which, frankly, we are still a part of, isn't just radioactivity. It's also understanding the expansion of the universe, the constancy, kind of the speed limit of light that now lets us figure out, wow, that background radiation we're getting, that must have been traveling for 13.7 billion years ago. So we can now look at evidence from our environment."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The understanding of radioactivity was just the beginning of this second chronometric revolution. The second chronometric revolution, which, frankly, we are still a part of, isn't just radioactivity. It's also understanding the expansion of the universe, the constancy, kind of the speed limit of light that now lets us figure out, wow, that background radiation we're getting, that must have been traveling for 13.7 billion years ago. So we can now look at evidence from our environment. And our environment is not just the Earth itself. It's radiation bombarding us from space that gives us clues as to not just the age of us, of humanity, the age of species, the age of the planet, but the age of the universe itself. So it isn't just about radioactivity."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we can now look at evidence from our environment. And our environment is not just the Earth itself. It's radiation bombarding us from space that gives us clues as to not just the age of us, of humanity, the age of species, the age of the planet, but the age of the universe itself. So it isn't just about radioactivity. Radioactivity is a big part of our chronometric revolution. This is what allowed us for the first time, if we have layers on the Earth, people have known for a long time that if we assume that these layers haven't been jostled, that something at a lower layer down here is probably going to be older than at the upper layer because year after year you have deposits if it hasn't been messed up in some way. But no one knew."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it isn't just about radioactivity. Radioactivity is a big part of our chronometric revolution. This is what allowed us for the first time, if we have layers on the Earth, people have known for a long time that if we assume that these layers haven't been jostled, that something at a lower layer down here is probably going to be older than at the upper layer because year after year you have deposits if it hasn't been messed up in some way. But no one knew. They said, okay, well, this is relative dating. This is older, this is younger, but we had no way of knowing that, hey, is this 1,000 years old, or is this a million years old, or is this a billion years old? But now with radioactivity, now we could start to say, hey, we can date some of the rocks here that are 150 million years old, and some of the rocks here are about 100 million years old."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But no one knew. They said, okay, well, this is relative dating. This is older, this is younger, but we had no way of knowing that, hey, is this 1,000 years old, or is this a million years old, or is this a billion years old? But now with radioactivity, now we could start to say, hey, we can date some of the rocks here that are 150 million years old, and some of the rocks here are about 100 million years old. So maybe this fossil of a fish that we're finding or this primitive fish-like creature right over here, this would be between 100 and 150 million years old. And the only way we were able to do this was with being able to date things using radioactivity. But radioactivity is just a start."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But now with radioactivity, now we could start to say, hey, we can date some of the rocks here that are 150 million years old, and some of the rocks here are about 100 million years old. So maybe this fossil of a fish that we're finding or this primitive fish-like creature right over here, this would be between 100 and 150 million years old. And the only way we were able to do this was with being able to date things using radioactivity. But radioactivity is just a start. As I mentioned, we're getting better and better understanding of cosmology. We're getting better measurements of the universe itself. We're understanding physics at a deeper level."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But radioactivity is just a start. As I mentioned, we're getting better and better understanding of cosmology. We're getting better measurements of the universe itself. We're understanding physics at a deeper level. Now we can start to look at the genome and think about how the genome diverges from one species to another and how quickly it changes. So all of these things are just allowing us to get better and better refinements on the chronology. Obviously, this is a start, but you still don't know, plus or minus 50 million years, how old this is and how this relates to other things that you might find."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're understanding physics at a deeper level. Now we can start to look at the genome and think about how the genome diverges from one species to another and how quickly it changes. So all of these things are just allowing us to get better and better refinements on the chronology. Obviously, this is a start, but you still don't know, plus or minus 50 million years, how old this is and how this relates to other things that you might find. So I just wanted to point this out, that what we take for granted now, the age of the universe, the age of Earth at 4.5 million years old, humans being around for several hundreds of thousands of years, this understanding is a very, very, very new phenomenon. It's due to the second chronometric revolution that I think we are still a part of. And even the first chronometric revolution, this version of history, and I want to be clear, history was limited by this first chronometric revolution."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Obviously, this is a start, but you still don't know, plus or minus 50 million years, how old this is and how this relates to other things that you might find. So I just wanted to point this out, that what we take for granted now, the age of the universe, the age of Earth at 4.5 million years old, humans being around for several hundreds of thousands of years, this understanding is a very, very, very new phenomenon. It's due to the second chronometric revolution that I think we are still a part of. And even the first chronometric revolution, this version of history, and I want to be clear, history was limited by this first chronometric revolution. It was limited by whatever was documented. But now maybe we can expand our notion of history. And a lot of the videos that I've been working on have been for this big history project, which says, hey, before history was limited by the first chronometric revolution, to what was written, by what was testified by people and was made permanent in some way, now we have chronometry has taken us so that we can understand things into our deep past, before even the Earth has existed."}, {"video_title": "Chronometric revolution Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even the first chronometric revolution, this version of history, and I want to be clear, history was limited by this first chronometric revolution. It was limited by whatever was documented. But now maybe we can expand our notion of history. And a lot of the videos that I've been working on have been for this big history project, which says, hey, before history was limited by the first chronometric revolution, to what was written, by what was testified by people and was made permanent in some way, now we have chronometry has taken us so that we can understand things into our deep past, before even the Earth has existed. So why not redefine history in a big way for it to encompass everything, for it to be big history? Anyway, I'll leave you there. And actually, I want to also emphasize that the second chronometric revolution is a big deal."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's remind ourselves a little bit of what we already know about orbitals. And I've gone over this early on in the regular chemistry playlist. But let's say that this is the nucleus of our atom, super small, and around that we have our first orbital, the 1s orbital. And the 1s orbital, you can kind of just view it as a cloud around the nucleus. So you have your 1s orbital, and it can fit two electrons. So the first electron will go into the 1s orbital, and then the second electron will also go into the 1s orbital. So for example, hydrogen has only one electron, so it would go into 1s."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And the 1s orbital, you can kind of just view it as a cloud around the nucleus. So you have your 1s orbital, and it can fit two electrons. So the first electron will go into the 1s orbital, and then the second electron will also go into the 1s orbital. So for example, hydrogen has only one electron, so it would go into 1s. Helium has one more, so that will also go in the 1s orbital. After that is filled, then you move on to the 2s orbital. And the 2s orbital, you can view it as a shell around the 1s orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So for example, hydrogen has only one electron, so it would go into 1s. Helium has one more, so that will also go in the 1s orbital. After that is filled, then you move on to the 2s orbital. And the 2s orbital, you can view it as a shell around the 1s orbital. And all of these, you can't really view it in our conventional way of thinking. You can kind of view it as a probability cloud of where you might find the electrons. But for visualization purposes, just imagine it as kind of a shell cloud around the 1s orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And the 2s orbital, you can view it as a shell around the 1s orbital. And all of these, you can't really view it in our conventional way of thinking. You can kind of view it as a probability cloud of where you might find the electrons. But for visualization purposes, just imagine it as kind of a shell cloud around the 1s orbital. So imagine it as a fuzzy shell around the 1s orbital. So it's around the 1s orbital. And your next electron will go there, and then the fourth electron will also go there."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But for visualization purposes, just imagine it as kind of a shell cloud around the 1s orbital. So imagine it as a fuzzy shell around the 1s orbital. So it's around the 1s orbital. And your next electron will go there, and then the fourth electron will also go there. And I drew these arrows upward and downward because the first electron that goes in the 1s orbital has one spin, and then the next electron to go into the 1s orbital will have the opposite spin. And so they keep pairing up in that way. They have opposite spins."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And your next electron will go there, and then the fourth electron will also go there. And I drew these arrows upward and downward because the first electron that goes in the 1s orbital has one spin, and then the next electron to go into the 1s orbital will have the opposite spin. And so they keep pairing up in that way. They have opposite spins. Now, if we keep adding electrons, now we move to the 2p orbitals. And actually, you can view it as there are three 2p orbitals, and each of them can hold two electrons. So it can hold a total of six electrons in the 2p orbitals."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "They have opposite spins. Now, if we keep adding electrons, now we move to the 2p orbitals. And actually, you can view it as there are three 2p orbitals, and each of them can hold two electrons. So it can hold a total of six electrons in the 2p orbitals. And let me draw them for you just so you can visualize it. So if we were to label our axes here, so think in three dimensions, so imagine that that right there is the x axis, that is our x axis. Let me do this in different colors."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it can hold a total of six electrons in the 2p orbitals. And let me draw them for you just so you can visualize it. So if we were to label our axes here, so think in three dimensions, so imagine that that right there is the x axis, that is our x axis. Let me do this in different colors. Let's say that this right here is our y axis. That is our y axis. And then we have a z axis."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let me do this in different colors. Let's say that this right here is our y axis. That is our y axis. And then we have a z axis. I'll do that in blue. So let's say we have a z axis just like that. You actually have a p orbital that goes along each of those axes."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a z axis. I'll do that in blue. So let's say we have a z axis just like that. You actually have a p orbital that goes along each of those axes. So you could have your 2p sub x orbital. And so what that'll look like is a dumbbell shape that's going in the x direction. So let me draw my best attempt at drawing this."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "You actually have a p orbital that goes along each of those axes. So you could have your 2p sub x orbital. And so what that'll look like is a dumbbell shape that's going in the x direction. So let me draw my best attempt at drawing this. It's a dumbbell shape that goes in the x direction, in kind of both directions. And it's actually symmetric. I'm drawing this end bigger than that end, so it looks like it's coming out at you a little bit."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw my best attempt at drawing this. It's a dumbbell shape that goes in the x direction, in kind of both directions. And it's actually symmetric. I'm drawing this end bigger than that end, so it looks like it's coming out at you a little bit. But let me draw it a little bit better than that. Can do a better job. And maybe it comes out like that."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I'm drawing this end bigger than that end, so it looks like it's coming out at you a little bit. But let me draw it a little bit better than that. Can do a better job. And maybe it comes out like that. And remember, these are really just probability clouds, but it's helpful to kind of visualize them as maybe a little bit more things that we would see in our world. But I think probability clouds is the best way to think about it. So that is the 2px orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And maybe it comes out like that. And remember, these are really just probability clouds, but it's helpful to kind of visualize them as maybe a little bit more things that we would see in our world. But I think probability clouds is the best way to think about it. So that is the 2px orbital. And then I haven't talked about how they get filled yet. But then you also have your 2py orbital, which will go in this axis, but same idea. Kind of a dumbbell shape in the y direction, going in both along the y axis."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that is the 2px orbital. And then I haven't talked about how they get filled yet. But then you also have your 2py orbital, which will go in this axis, but same idea. Kind of a dumbbell shape in the y direction, going in both along the y axis. Going in that direction and in that direction. Then of course, let me do this. 2py."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Kind of a dumbbell shape in the y direction, going in both along the y axis. Going in that direction and in that direction. Then of course, let me do this. 2py. And then you also have your 2pz. And you also have your 2pz. And that goes in the z direction."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "2py. And then you also have your 2pz. And you also have your 2pz. And that goes in the z direction. Up like that, and then downwards like that. So when you keep adding electrons, so far we've added four electrons. If you add a fifth electron, you would expect it to go into the 2px orbital right there."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And that goes in the z direction. Up like that, and then downwards like that. So when you keep adding electrons, so far we've added four electrons. If you add a fifth electron, you would expect it to go into the 2px orbital right there. So even though this 2px orbital can fit two electrons, the first one goes there, the very next one won't go into that one, it actually wants to separate itself within the p orbital. So the very next electron that you add won't go into 2px, it'll go into 2py. And then the one after that won't go into 2py or 2px, it'll go into 2pz."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If you add a fifth electron, you would expect it to go into the 2px orbital right there. So even though this 2px orbital can fit two electrons, the first one goes there, the very next one won't go into that one, it actually wants to separate itself within the p orbital. So the very next electron that you add won't go into 2px, it'll go into 2py. And then the one after that won't go into 2py or 2px, it'll go into 2pz. They try to separate themselves. And then if you add another electron, if you add, let's see, we've added one, two, three, four, five, six, seven. If you add an eighth electron, that will then go into the 2px orbital, so the eighth electron would go there, but it would have the opposite spin."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then the one after that won't go into 2py or 2px, it'll go into 2pz. They try to separate themselves. And then if you add another electron, if you add, let's see, we've added one, two, three, four, five, six, seven. If you add an eighth electron, that will then go into the 2px orbital, so the eighth electron would go there, but it would have the opposite spin. So this is just a little bit of review with a little bit of visualization. Now given what we just reviewed, let's think about what's happening with carbon. Carbon has six electrons."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If you add an eighth electron, that will then go into the 2px orbital, so the eighth electron would go there, but it would have the opposite spin. So this is just a little bit of review with a little bit of visualization. Now given what we just reviewed, let's think about what's happening with carbon. Carbon has six electrons. It's electron configuration, it is 1s2, two electrons in the 1s orbital, then 2s2, then 2p2. It only has two left because it has a total of six electrons, two go here, then there, then two are left to fill the p orbitals. So if you go based on what we just drew and what we just talked about here, what you would expect for carbon, what you would expect for carbon, let me just draw it out the way I did this, so you have your 1s orbital, your 2s orbital, and then you have your 2px orbital, your 2py orbital, and then you have your 2pz orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Carbon has six electrons. It's electron configuration, it is 1s2, two electrons in the 1s orbital, then 2s2, then 2p2. It only has two left because it has a total of six electrons, two go here, then there, then two are left to fill the p orbitals. So if you go based on what we just drew and what we just talked about here, what you would expect for carbon, what you would expect for carbon, let me just draw it out the way I did this, so you have your 1s orbital, your 2s orbital, and then you have your 2px orbital, your 2py orbital, and then you have your 2pz orbital. If you just go straight from the electron configuration, you would expect carbon, so the 1s orbital fills first, so that's our first electron, our second electron, our third electron, and then we go to our 2s orbital, that fills next, third electron, then fourth electron, and then you would expect maybe your fifth electron to go in the 2px, we could have said 2py or 2z, it just depends on how you label the axes, but you would have your fifth electron go into one of the p orbitals, and then you would expect your sixth to go into another. So you would expect that to be kind of the configuration for carbon. And if we were to draw it, let me draw our axes, so that is our y-axis, and then this is our x-axis, that, let me draw it a little bit better than that, so that is the x-axis, and of course you have your z-axis, have to think in three dimensions a little bit, a little squiggly line there, so let me, then you have your z-axis, just like that."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if you go based on what we just drew and what we just talked about here, what you would expect for carbon, what you would expect for carbon, let me just draw it out the way I did this, so you have your 1s orbital, your 2s orbital, and then you have your 2px orbital, your 2py orbital, and then you have your 2pz orbital. If you just go straight from the electron configuration, you would expect carbon, so the 1s orbital fills first, so that's our first electron, our second electron, our third electron, and then we go to our 2s orbital, that fills next, third electron, then fourth electron, and then you would expect maybe your fifth electron to go in the 2px, we could have said 2py or 2z, it just depends on how you label the axes, but you would have your fifth electron go into one of the p orbitals, and then you would expect your sixth to go into another. So you would expect that to be kind of the configuration for carbon. And if we were to draw it, let me draw our axes, so that is our y-axis, and then this is our x-axis, that, let me draw it a little bit better than that, so that is the x-axis, and of course you have your z-axis, have to think in three dimensions a little bit, a little squiggly line there, so let me, then you have your z-axis, just like that. So first we fill the 1s orbital, so if our nucleus is sitting here, our 1s orbital gets filled with 2px, so 2 electrons, you can imagine that as a little cloud around the nucleus, then we fill the 2s orbital, and that would be a cloud around that, kind of a shell around that, and then we would put one electron in the 2px orbital, so one electron would start kind of jumping around or moving around, depending on how you want to think about it, in that orbital over there, 2px, and then you'd have the next electron jumping around or moving around in the 2py orbital. So it would be moving around like this. And if you went just off of this, you'd say, you know what, these guys, this guy over here and that guy over there is lonely, he's looking for a opposite spin partner, this would be the only places that bonds would form, you would expect some type of bonding to form with the x-orbitals or the y-orbitals."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And if we were to draw it, let me draw our axes, so that is our y-axis, and then this is our x-axis, that, let me draw it a little bit better than that, so that is the x-axis, and of course you have your z-axis, have to think in three dimensions a little bit, a little squiggly line there, so let me, then you have your z-axis, just like that. So first we fill the 1s orbital, so if our nucleus is sitting here, our 1s orbital gets filled with 2px, so 2 electrons, you can imagine that as a little cloud around the nucleus, then we fill the 2s orbital, and that would be a cloud around that, kind of a shell around that, and then we would put one electron in the 2px orbital, so one electron would start kind of jumping around or moving around, depending on how you want to think about it, in that orbital over there, 2px, and then you'd have the next electron jumping around or moving around in the 2py orbital. So it would be moving around like this. And if you went just off of this, you'd say, you know what, these guys, this guy over here and that guy over there is lonely, he's looking for a opposite spin partner, this would be the only places that bonds would form, you would expect some type of bonding to form with the x-orbitals or the y-orbitals. Now, that's what you would expect if you just straight up kind of stayed with this model of how things fill and how orbitals look. The reality of carbon, and I guess the simplest reality of carbon is if you look at a methane molecule, is very different than what you would expect here. First of all, what you would expect here is that carbon would probably maybe form two bonds, but we know carbon forms four bonds, that it wants to pretend like it has eight electrons, that frankly almost every atom wants to pretend like it has eight electrons."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And if you went just off of this, you'd say, you know what, these guys, this guy over here and that guy over there is lonely, he's looking for a opposite spin partner, this would be the only places that bonds would form, you would expect some type of bonding to form with the x-orbitals or the y-orbitals. Now, that's what you would expect if you just straight up kind of stayed with this model of how things fill and how orbitals look. The reality of carbon, and I guess the simplest reality of carbon is if you look at a methane molecule, is very different than what you would expect here. First of all, what you would expect here is that carbon would probably maybe form two bonds, but we know carbon forms four bonds, that it wants to pretend like it has eight electrons, that frankly almost every atom wants to pretend like it has eight electrons. So in order for that to happen, you have to think about a different reality. This isn't really what's happening when carbon bonds. So not what happens when carbon bonds."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "First of all, what you would expect here is that carbon would probably maybe form two bonds, but we know carbon forms four bonds, that it wants to pretend like it has eight electrons, that frankly almost every atom wants to pretend like it has eight electrons. So in order for that to happen, you have to think about a different reality. This isn't really what's happening when carbon bonds. So not what happens when carbon bonds. What's really happening when carbon bonds, and this will kind of go into the discussion of SP3 hybridization, but what you're going to see is it's not that complicated of a topic. It sounds very daunting, but it's actually pretty straightforward. What really happens when carbon bonds, because it wants to form four bonds with things, is its configuration, you can imagine, looks more like this."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So not what happens when carbon bonds. What's really happening when carbon bonds, and this will kind of go into the discussion of SP3 hybridization, but what you're going to see is it's not that complicated of a topic. It sounds very daunting, but it's actually pretty straightforward. What really happens when carbon bonds, because it wants to form four bonds with things, is its configuration, you can imagine, looks more like this. It looks more like this. So you have 1s, and we have two electrons there, and then you have your 2s, 2px, 2py, and 2pz. Now what you can imagine is it wants to form four bonds."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "What really happens when carbon bonds, because it wants to form four bonds with things, is its configuration, you can imagine, looks more like this. It looks more like this. So you have 1s, and we have two electrons there, and then you have your 2s, 2px, 2py, and 2pz. Now what you can imagine is it wants to form four bonds. It has four electrons that are willing to pair up with electrons from other molecules. In the case of methane, that other molecule is a hydrogen. So what you can imagine is that the electrons actually, maybe the hydrogen brings this electron right here into a higher energy state and puts it into 2z."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now what you can imagine is it wants to form four bonds. It has four electrons that are willing to pair up with electrons from other molecules. In the case of methane, that other molecule is a hydrogen. So what you can imagine is that the electrons actually, maybe the hydrogen brings this electron right here into a higher energy state and puts it into 2z. That's one way to visualize it. So this other guy here maybe ends up over there, and then these two guys are over there and over there. Now all of a sudden, it looks like you have four lonely guys and they are ready to bond."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So what you can imagine is that the electrons actually, maybe the hydrogen brings this electron right here into a higher energy state and puts it into 2z. That's one way to visualize it. So this other guy here maybe ends up over there, and then these two guys are over there and over there. Now all of a sudden, it looks like you have four lonely guys and they are ready to bond. And that's actually more accurate of how carbon bonds. It likes to bond with four other people. Now, it's a little bit arbitrary which electron ends up in each of these things."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now all of a sudden, it looks like you have four lonely guys and they are ready to bond. And that's actually more accurate of how carbon bonds. It likes to bond with four other people. Now, it's a little bit arbitrary which electron ends up in each of these things. And even if you had this type of bonding, you would expect things to bond along the x, y, and z-axis. The reality of carbon is that these four electrons in its second shell don't look like they're in just, the first one doesn't look like it's just in the s orbital and then the p's and y and z for the other three. They all look like they're a little bit in the s and a little bit in the p orbitals."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now, it's a little bit arbitrary which electron ends up in each of these things. And even if you had this type of bonding, you would expect things to bond along the x, y, and z-axis. The reality of carbon is that these four electrons in its second shell don't look like they're in just, the first one doesn't look like it's just in the s orbital and then the p's and y and z for the other three. They all look like they're a little bit in the s and a little bit in the p orbitals. So let me make that clear. So instead of this being a 2s, what it really looks like for carbon is that this looks like a 2sp3 orbital. This looks like a 2sp3 orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "They all look like they're a little bit in the s and a little bit in the p orbitals. So let me make that clear. So instead of this being a 2s, what it really looks like for carbon is that this looks like a 2sp3 orbital. This looks like a 2sp3 orbital. That looks like a 2sp3 orbital. They all look like they're kind of in the same orbital. And this special type of, it sounds very fancy, this sp3 hybridized orbital, what it actually looks like is something that's in between an s and a p orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This looks like a 2sp3 orbital. That looks like a 2sp3 orbital. They all look like they're kind of in the same orbital. And this special type of, it sounds very fancy, this sp3 hybridized orbital, what it actually looks like is something that's in between an s and a p orbital. It has a 25% s nature and a 75% p nature. You can imagine it as being a mixture of these four things. That's the behavior that carbon has."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And this special type of, it sounds very fancy, this sp3 hybridized orbital, what it actually looks like is something that's in between an s and a p orbital. It has a 25% s nature and a 75% p nature. You can imagine it as being a mixture of these four things. That's the behavior that carbon has. So when you mix them all, instead of having an s orbital, so if this is a nucleus and we do a cross-section, an s orbital looks like that, and a p orbital looks something like that in cross-section. This is an s and that is a p. When they get mixed up, the orbital looks like this. An sp3 orbital looks something like this."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That's the behavior that carbon has. So when you mix them all, instead of having an s orbital, so if this is a nucleus and we do a cross-section, an s orbital looks like that, and a p orbital looks something like that in cross-section. This is an s and that is a p. When they get mixed up, the orbital looks like this. An sp3 orbital looks something like this. This is a hybridized sp3 orbital. Hybrid just means a combination of two things. A hybrid car, it's a combination of gas and electric."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "An sp3 orbital looks something like this. This is a hybridized sp3 orbital. Hybrid just means a combination of two things. A hybrid car, it's a combination of gas and electric. Hybridized orbital, it's a combination of s and p. And a hybridized sp3 orbitals are the orbitals when carbon bonds with things like hydrogen, or really when it bonds with anything. And so if you look at a molecule of methane, and people talk about sp3 hybridized orbitals, all they're saying is that you have a carbon in the center. Maybe let me draw the, let's say that's the carbon nucleus right there."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "A hybrid car, it's a combination of gas and electric. Hybridized orbital, it's a combination of s and p. And a hybridized sp3 orbitals are the orbitals when carbon bonds with things like hydrogen, or really when it bonds with anything. And so if you look at a molecule of methane, and people talk about sp3 hybridized orbitals, all they're saying is that you have a carbon in the center. Maybe let me draw the, let's say that's the carbon nucleus right there. And instead of having one s and three p orbitals, it has four sp3 orbitals. So let me try my best at drawing the four sp3 orbitals. So let's say that this is the big lobe that's kind of pointing near us, and it has a small lobe in the back."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Maybe let me draw the, let's say that's the carbon nucleus right there. And instead of having one s and three p orbitals, it has four sp3 orbitals. So let me try my best at drawing the four sp3 orbitals. So let's say that this is the big lobe that's kind of pointing near us, and it has a small lobe in the back. And then you have another one that has a big lobe like that and a small lobe in the back. And then you have one that's going back behind the page. Let me draw that."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that this is the big lobe that's kind of pointing near us, and it has a small lobe in the back. And then you have another one that has a big lobe like that and a small lobe in the back. And then you have one that's going back behind the page. Let me draw that. You can kind of imagine a three-legged stool, and then its small lobe will come out like that. And then you have one where the big lobe is pointing straight up. And it has a small lobe going down."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw that. You can kind of imagine a three-legged stool, and then its small lobe will come out like that. And then you have one where the big lobe is pointing straight up. And it has a small lobe going down. You can imagine it as kind of a three-legged stool. One of them's behind like that, and it's pointing straight up. So a three-legged stool with something, I guess, it's kind of like a tripod, I guess, is the best way to think about it."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And it has a small lobe going down. You can imagine it as kind of a three-legged stool. One of them's behind like that, and it's pointing straight up. So a three-legged stool with something, I guess, it's kind of like a tripod, I guess, is the best way to think about it. And so that's the carbon nucleus in the center. And then you have the hydrogens. So that's our carbon right there."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So a three-legged stool with something, I guess, it's kind of like a tripod, I guess, is the best way to think about it. And so that's the carbon nucleus in the center. And then you have the hydrogens. So that's our carbon right there. And then you have your hydrogens. You have a hydrogen here. A hydrogen just has one electron in the 1s orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that's our carbon right there. And then you have your hydrogens. You have a hydrogen here. A hydrogen just has one electron in the 1s orbital. So the hydrogen just has 1s orbital. You have a hydrogen here, just has a 1s orbital. It has a hydrogen here, 1s orbital."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "A hydrogen just has one electron in the 1s orbital. So the hydrogen just has 1s orbital. You have a hydrogen here, just has a 1s orbital. It has a hydrogen here, 1s orbital. And so this is how the hydrogen orbital and the carbon orbitals get mixed. the hydrogen's 1s orbital bonds with each of the carbon's sp3 orbitals. And just so you get a little bit more notation."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It has a hydrogen here, 1s orbital. And so this is how the hydrogen orbital and the carbon orbitals get mixed. the hydrogen's 1s orbital bonds with each of the carbon's sp3 orbitals. And just so you get a little bit more notation. So when people talk about hybridized sp3 orbitals, all they're saying is, look, carbon doesn't bond. Once carbon, this right here is a molecule of methane, right, this is CH4 or methane. And it doesn't bond like you would expect if you just went with straight vanilla sp orbitals."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And just so you get a little bit more notation. So when people talk about hybridized sp3 orbitals, all they're saying is, look, carbon doesn't bond. Once carbon, this right here is a molecule of methane, right, this is CH4 or methane. And it doesn't bond like you would expect if you just went with straight vanilla sp orbitals. If you just went straight vanilla sp orbitals, the bonds would form, maybe the hydrogen might be there and there. And if it had four hydrogens, maybe there and there, depending on how you want to think about it. But the reality is it doesn't look like that."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And it doesn't bond like you would expect if you just went with straight vanilla sp orbitals. If you just went straight vanilla sp orbitals, the bonds would form, maybe the hydrogen might be there and there. And if it had four hydrogens, maybe there and there, depending on how you want to think about it. But the reality is it doesn't look like that. It looks more like a tripod. It has a tetrahedral shape. And the best way that that can be explained, if you, I guess, the shape of the structure, is if you have four equally, four of the same types of orbital shapes."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But the reality is it doesn't look like that. It looks more like a tripod. It has a tetrahedral shape. And the best way that that can be explained, if you, I guess, the shape of the structure, is if you have four equally, four of the same types of orbital shapes. And those four types of orbital shapes are hybrids between s's and p's. And now one other piece of notation to know, sometimes people think it's a very fancy term, but when you have a bond between two molecules where the orbitals are kind of pointing at each other. So you can imagine right here, this hydrogen orbital is pointing in that direction."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And the best way that that can be explained, if you, I guess, the shape of the structure, is if you have four equally, four of the same types of orbital shapes. And those four types of orbital shapes are hybrids between s's and p's. And now one other piece of notation to know, sometimes people think it's a very fancy term, but when you have a bond between two molecules where the orbitals are kind of pointing at each other. So you can imagine right here, this hydrogen orbital is pointing in that direction. This sp3 orbital is pointing in that direction. And they're overlapping right around here. This is called a sigma bond."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So you can imagine right here, this hydrogen orbital is pointing in that direction. This sp3 orbital is pointing in that direction. And they're overlapping right around here. This is called a sigma bond. Where kind of the overlap is along the same axis as if you connected the two molecules. Over here, you connect the two molecules, the overlap is on that same axis. This is the strongest form of covalent bonds."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This is called a sigma bond. Where kind of the overlap is along the same axis as if you connected the two molecules. Over here, you connect the two molecules, the overlap is on that same axis. This is the strongest form of covalent bonds. And this will be a good basis for discussion, maybe in the next video when we talk a little bit about pi bonds. But the big takeaway of this video is just understand what does it mean, what is a sp3 hybridized orbital? Nothing fancy, just a combination of s and p orbitals, has 25% s character, 75% p character, which makes sense, and it's what exists when carbon forms bonds, especially in the case of methane."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This is the strongest form of covalent bonds. And this will be a good basis for discussion, maybe in the next video when we talk a little bit about pi bonds. But the big takeaway of this video is just understand what does it mean, what is a sp3 hybridized orbital? Nothing fancy, just a combination of s and p orbitals, has 25% s character, 75% p character, which makes sense, and it's what exists when carbon forms bonds, especially in the case of methane. That's what describes its tetrahedral structure. That's why we have an angle between the various branches of 109.5 degrees, which some teachers might want you to know, so it's useful to know. If you take this angle right here, 109.5, that's the same thing as that angle."}, {"video_title": "sp3 hybridized orbitals and sigma bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Nothing fancy, just a combination of s and p orbitals, has 25% s character, 75% p character, which makes sense, and it's what exists when carbon forms bonds, especially in the case of methane. That's what describes its tetrahedral structure. That's why we have an angle between the various branches of 109.5 degrees, which some teachers might want you to know, so it's useful to know. If you take this angle right here, 109.5, that's the same thing as that angle. Or if you were to go behind it, that angle right there, 109.5 degrees. Explained by sp3 hybridization, the bonds themselves are sigma bonds. The overlap is along the axis connecting the hydrogen and the carbon."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It'll get a little bit math-y, usually involving a little bit of algebra or a little bit of exponential decay but to really show you how you can actually figure out the age of some volcanic rock using this technique, using a little bit of mathematics. So we know that anything that is experiencing radioactive decay, it's experiencing exponential decay, and we know that we can, there's a generalized way to describe that and we go into more depth and kind of prove it in other Khan Academy videos. But we know that the amount as a function of time, so if we say n is the amount of a radioactive sample we have at some time, we know that's equal to the initial amount we have, we'll call that n sub zero, times e to the negative kt, where this constant is particular to that thing's half-life. And to figure it out, and we're gonna do this for the example of potassium-40, we know that after, when time is 1.25 billion years, that the amount we have left is half of our initial amount. So let's write it that way. So let's say when we start with n naught, we start with n naught, whatever that might be, it might be one gram, kilogram, five grams, whatever it might be, whatever we start with, we take e to the negative k times 1.25 billion years, that's the half-life of potassium-40, so 1.25 billion years, we know after that long that half of the sample will be left. So we will have one half, one half n naught left."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And to figure it out, and we're gonna do this for the example of potassium-40, we know that after, when time is 1.25 billion years, that the amount we have left is half of our initial amount. So let's write it that way. So let's say when we start with n naught, we start with n naught, whatever that might be, it might be one gram, kilogram, five grams, whatever it might be, whatever we start with, we take e to the negative k times 1.25 billion years, that's the half-life of potassium-40, so 1.25 billion years, we know after that long that half of the sample will be left. So we will have one half, one half n naught left. Whatever we started with, we're going to have half left after 1.25 billion years. Divide both sides by n naught, divide both sides by n naught, and then to solve for k, we can take the natural log of both sides, so you get the natural log of one half, we don't have that n naught there anymore, is equal to the natural log of this thing. The natural log is just saying, to what power do I have to raise e to get e to the negative k times 1.25 billion?"}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we will have one half, one half n naught left. Whatever we started with, we're going to have half left after 1.25 billion years. Divide both sides by n naught, divide both sides by n naught, and then to solve for k, we can take the natural log of both sides, so you get the natural log of one half, we don't have that n naught there anymore, is equal to the natural log of this thing. The natural log is just saying, to what power do I have to raise e to get e to the negative k times 1.25 billion? So the natural log of this, the power that I have to raise e to to get to e to the negative k times 1.25 billion is just negative k times 1.25 billion. I could write it as negative 1.25, let me write it times 10 to the ninth, times 10 to the ninth k. That's the same thing as 1.25 billion, we have our negative sign and we have our k. And then to solve for k, we can divide both sides by negative 1.25 billion, and so we get k, and I'll just flip the sides here, k is equal to the natural log of one half times negative 1.25 times 10 to the ninth power. And what we can do is we can multiply the negative times the top, or you can view it as multiplying the numerator and the denominator by a negative so that the negative shows up at the top."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The natural log is just saying, to what power do I have to raise e to get e to the negative k times 1.25 billion? So the natural log of this, the power that I have to raise e to to get to e to the negative k times 1.25 billion is just negative k times 1.25 billion. I could write it as negative 1.25, let me write it times 10 to the ninth, times 10 to the ninth k. That's the same thing as 1.25 billion, we have our negative sign and we have our k. And then to solve for k, we can divide both sides by negative 1.25 billion, and so we get k, and I'll just flip the sides here, k is equal to the natural log of one half times negative 1.25 times 10 to the ninth power. And what we can do is we can multiply the negative times the top, or you can view it as multiplying the numerator and the denominator by a negative so that the negative shows up at the top. And so we could make this as over 1.25 times 10 to the ninth, this is 1.25 billion, and negative, let me write it over here in a different color. Negative, the negative natural log, well I could just write it this way. If I have a, a natural log of b, we know from our logarithm properties this is the same thing as the natural log of b to the a power, so the negative natural log of one half is the same thing as the natural log of one half to the negative one power, and so this is the same thing."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what we can do is we can multiply the negative times the top, or you can view it as multiplying the numerator and the denominator by a negative so that the negative shows up at the top. And so we could make this as over 1.25 times 10 to the ninth, this is 1.25 billion, and negative, let me write it over here in a different color. Negative, the negative natural log, well I could just write it this way. If I have a, a natural log of b, we know from our logarithm properties this is the same thing as the natural log of b to the a power, so the negative natural log of one half is the same thing as the natural log of one half to the negative one power, and so this is the same thing. Anything to the negative one power is just this multiplicative inverse, so this is just the natural log of two. So negative natural log of one half is just the natural log of two over here. So we were able to figure out our k, it's essentially the natural log of two over the half-life of this substance, so we could actually generalize this if we were talking about some other radioactive substance."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If I have a, a natural log of b, we know from our logarithm properties this is the same thing as the natural log of b to the a power, so the negative natural log of one half is the same thing as the natural log of one half to the negative one power, and so this is the same thing. Anything to the negative one power is just this multiplicative inverse, so this is just the natural log of two. So negative natural log of one half is just the natural log of two over here. So we were able to figure out our k, it's essentially the natural log of two over the half-life of this substance, so we could actually generalize this if we were talking about some other radioactive substance. And now let's think about a situation, now that we've figured out a k, let's think about a situation where we find in some sample, so let's say the potassium that we find, let's say it is one milligram, I'm just gonna make up these numbers, and let's say, and usually these aren't measured directly and you really care about the relative amounts, but let's say you're able to figure out the potassium is one milligram, and let's say that the argon, actually let me say the potassium 40 found, let's say the argon 40 found, let's say it is 0.01, 0.01 milligram. So how can we use this information, what we just figured out here, which is derived from the half-life, to figure out how old this sample right over here, how do we figure out how old this sample is right over there? Well, what we need to figure out, we know that n, the amount we were left with, is this thing right over here."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we were able to figure out our k, it's essentially the natural log of two over the half-life of this substance, so we could actually generalize this if we were talking about some other radioactive substance. And now let's think about a situation, now that we've figured out a k, let's think about a situation where we find in some sample, so let's say the potassium that we find, let's say it is one milligram, I'm just gonna make up these numbers, and let's say, and usually these aren't measured directly and you really care about the relative amounts, but let's say you're able to figure out the potassium is one milligram, and let's say that the argon, actually let me say the potassium 40 found, let's say the argon 40 found, let's say it is 0.01, 0.01 milligram. So how can we use this information, what we just figured out here, which is derived from the half-life, to figure out how old this sample right over here, how do we figure out how old this sample is right over there? Well, what we need to figure out, we know that n, the amount we were left with, is this thing right over here. So we know that we're left with one milligram, so we know that what we have left is one milligram, and that's going to be equal to some initial amount, it's going to be some initial amount, we'll use both of this information to figure that initial amount out, times e to the negative kT, and we know what k is and we'll figure it out later, so k is this thing right over here, so we need to figure out what our initial amount is, we know what k is and then we can solve for T, how old is this sample? And to figure out our initial amount, we just have to remember that for every argon-40 we see, that must have decayed from, when you have potassium-40, 11% decays, when it decays, 11% decays into argon-40, and the rest, 89%, decays into calcium-40, we saw that in the last video. So however much argon-40, that is 11% of the decay product, so if you want to think about the total number of potassium-40s that have decayed since this was kind of stuck in the lava and we learned that anything that was there before, any argon-40 that was there before would have been able to get out of the liquid lava before it froze or before it hardened, so to figure out how much potassium-40 this is derived from, we just derive it, we divide it by 11%, so maybe I could say k initial, the potassium-40 initial is going to be equal to the amount of potassium-40 we have today, one milligram, plus the amount of potassium-40 we needed to get this amount of argon-40."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, what we need to figure out, we know that n, the amount we were left with, is this thing right over here. So we know that we're left with one milligram, so we know that what we have left is one milligram, and that's going to be equal to some initial amount, it's going to be some initial amount, we'll use both of this information to figure that initial amount out, times e to the negative kT, and we know what k is and we'll figure it out later, so k is this thing right over here, so we need to figure out what our initial amount is, we know what k is and then we can solve for T, how old is this sample? And to figure out our initial amount, we just have to remember that for every argon-40 we see, that must have decayed from, when you have potassium-40, 11% decays, when it decays, 11% decays into argon-40, and the rest, 89%, decays into calcium-40, we saw that in the last video. So however much argon-40, that is 11% of the decay product, so if you want to think about the total number of potassium-40s that have decayed since this was kind of stuck in the lava and we learned that anything that was there before, any argon-40 that was there before would have been able to get out of the liquid lava before it froze or before it hardened, so to figure out how much potassium-40 this is derived from, we just derive it, we divide it by 11%, so maybe I could say k initial, the potassium-40 initial is going to be equal to the amount of potassium-40 we have today, one milligram, plus the amount of potassium-40 we needed to get this amount of argon-40. So we have this amount of argon-40,.01 milligrams, and that, the number of milligrams there, it's really just 11% of the original potassium-40 that it had to come from, the rest of it turned into calcium-40, so we divide it by 11%, or 0.11, and I'm doing, this isn't the exact number, but it'll get the general idea. And so our initial, which is really this thing right over here, I could call this N naught, I could call this N naught, this is going to be equal to, and I won't do any of the math, so we have one milligram we have left, is equal to one milligram, which is what we found, plus 0.01 milligram over 0.11, and then all of that times e to the negative kT. And what you see here is, when we want to solve for T, assuming we know k, and we do know k now, that it really, the absolute amount doesn't matter, what actually matters is the ratio, because if we're solving for T, you want to divide both sides of this equation by this quantity right over here, so you get this side, the left-hand side, divide both sides, you get one milligram over this quantity, I'll write it in blue, over this quantity is going to be one plus, I'm just going to assume, actually, that the units here are milligrams."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So however much argon-40, that is 11% of the decay product, so if you want to think about the total number of potassium-40s that have decayed since this was kind of stuck in the lava and we learned that anything that was there before, any argon-40 that was there before would have been able to get out of the liquid lava before it froze or before it hardened, so to figure out how much potassium-40 this is derived from, we just derive it, we divide it by 11%, so maybe I could say k initial, the potassium-40 initial is going to be equal to the amount of potassium-40 we have today, one milligram, plus the amount of potassium-40 we needed to get this amount of argon-40. So we have this amount of argon-40,.01 milligrams, and that, the number of milligrams there, it's really just 11% of the original potassium-40 that it had to come from, the rest of it turned into calcium-40, so we divide it by 11%, or 0.11, and I'm doing, this isn't the exact number, but it'll get the general idea. And so our initial, which is really this thing right over here, I could call this N naught, I could call this N naught, this is going to be equal to, and I won't do any of the math, so we have one milligram we have left, is equal to one milligram, which is what we found, plus 0.01 milligram over 0.11, and then all of that times e to the negative kT. And what you see here is, when we want to solve for T, assuming we know k, and we do know k now, that it really, the absolute amount doesn't matter, what actually matters is the ratio, because if we're solving for T, you want to divide both sides of this equation by this quantity right over here, so you get this side, the left-hand side, divide both sides, you get one milligram over this quantity, I'll write it in blue, over this quantity is going to be one plus, I'm just going to assume, actually, that the units here are milligrams. So you get one over this quantity, which is one plus 0.01 over the 11%, that is equal to e to the negative kT, and then you want to take, if you want to solve for T, you want to take the natural log of both sides, so then you get, so this is equal right over here, you want to take the natural log of both sides, so you get the natural log of one over one plus 0.01 over 0.11, or 11%, is equal to negative kT, and then to solve for T, you divide both sides by negative k, so I'll write it over here. You can see this is a little bit cumbersome mathematically, but we're getting to the answer. So we got the natural log of one over one plus 0.01 over 0.11 over negative k. Well, what is negative k?"}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what you see here is, when we want to solve for T, assuming we know k, and we do know k now, that it really, the absolute amount doesn't matter, what actually matters is the ratio, because if we're solving for T, you want to divide both sides of this equation by this quantity right over here, so you get this side, the left-hand side, divide both sides, you get one milligram over this quantity, I'll write it in blue, over this quantity is going to be one plus, I'm just going to assume, actually, that the units here are milligrams. So you get one over this quantity, which is one plus 0.01 over the 11%, that is equal to e to the negative kT, and then you want to take, if you want to solve for T, you want to take the natural log of both sides, so then you get, so this is equal right over here, you want to take the natural log of both sides, so you get the natural log of one over one plus 0.01 over 0.11, or 11%, is equal to negative kT, and then to solve for T, you divide both sides by negative k, so I'll write it over here. You can see this is a little bit cumbersome mathematically, but we're getting to the answer. So we got the natural log of one over one plus 0.01 over 0.11 over negative k. Well, what is negative k? We're just dividing both sides of this equation by negative k. Negative k is the negative of this over the negative natural log of two over 1.25 times 10 to the ninth. And now we can get our calculator out and just solve for what this time is, and it's going to be in years, because that's how we figured out this constant. So let's get out my handy TI-85."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we got the natural log of one over one plus 0.01 over 0.11 over negative k. Well, what is negative k? We're just dividing both sides of this equation by negative k. Negative k is the negative of this over the negative natural log of two over 1.25 times 10 to the ninth. And now we can get our calculator out and just solve for what this time is, and it's going to be in years, because that's how we figured out this constant. So let's get out my handy TI-85. And so first I'll do this part. So this is one divided by one plus 0.01 divided by 0.11. So that's this part right over here."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's get out my handy TI-85. And so first I'll do this part. So this is one divided by one plus 0.01 divided by 0.11. So that's this part right over here. That gives us that number. And then we want to take the natural log of that. So let's take the natural log of our, this is just our previous answer, so natural log of 0.9166667 gives us negative 0.087."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's this part right over here. That gives us that number. And then we want to take the natural log of that. So let's take the natural log of our, this is just our previous answer, so natural log of 0.9166667 gives us negative 0.087. So that's this numerator over here. And we're going to divide that. So this number is our numerator right over here."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's take the natural log of our, this is just our previous answer, so natural log of 0.9166667 gives us negative 0.087. So that's this numerator over here. And we're going to divide that. So this number is our numerator right over here. We're going to divide that by the negative, let me make, I'll use parentheses carefully, the negative natural log of two, the negative natural log of two, that's that there, divided by 1.25 times 10 to the ninth. So divided by, so it's negative natural log of two divided by 1.25. E nine means times 10 to the ninth."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this number is our numerator right over here. We're going to divide that by the negative, let me make, I'll use parentheses carefully, the negative natural log of two, the negative natural log of two, that's that there, divided by 1.25 times 10 to the ninth. So divided by, so it's negative natural log of two divided by 1.25. E nine means times 10 to the ninth. And I closed both parentheses. And now we need our drum roll. So this should give us our T in years."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "E nine means times 10 to the ninth. And I closed both parentheses. And now we need our drum roll. So this should give us our T in years. And we get, let's see how many, this is thousand, so it's 3,000. So we get 156 million, or 156.9 million years if we round. So this is approximately, or I could just say approximately 157 million years old sample."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this should give us our T in years. And we get, let's see how many, this is thousand, so it's 3,000. So we get 156 million, or 156.9 million years if we round. So this is approximately, or I could just say approximately 157 million years old sample. So the whole point of this, I know the math was a little bit involved, but it's something that you would actually see in kind of a pre-calculus class or an algebra two class when you're studying exponential growth and decay. But the whole point I wanted to do this is to show you that it's not some crazy voodoo here. And you know, Sal gave this very high level explanation."}, {"video_title": "K-Ar dating calculation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is approximately, or I could just say approximately 157 million years old sample. So the whole point of this, I know the math was a little bit involved, but it's something that you would actually see in kind of a pre-calculus class or an algebra two class when you're studying exponential growth and decay. But the whole point I wanted to do this is to show you that it's not some crazy voodoo here. And you know, Sal gave this very high level explanation. And then you say, oh, well, you know, there must be some super difficult mathematics after that. The mathematics really is something that you would see in high school. And if you saw a sample that had this ratio of argon 40 to potassium 40, you would actually be able to do that fairly, you know, that high school mathematics."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And it's just nabbing the proton itself. It's not grabbing the hydrogen and the electron. And then that electron goes to this carbon right over there. And then that allows that carbon to give away the electron that was forming a bond with the chlorine to go with the chlorine. This all happened at once. The chloride got eliminated. Now, one thing that might pop out in your brain is, why did I pick this hydrogen?"}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then that allows that carbon to give away the electron that was forming a bond with the chlorine to go with the chlorine. This all happened at once. The chloride got eliminated. Now, one thing that might pop out in your brain is, why did I pick this hydrogen? Why did I pick the hydrogen right over here? Why couldn't I have picked that hydrogen over there? And I'm going to introduce you a little bit of terminology, and then I'll introduce you to a rule."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now, one thing that might pop out in your brain is, why did I pick this hydrogen? Why did I pick the hydrogen right over here? Why couldn't I have picked that hydrogen over there? And I'm going to introduce you a little bit of terminology, and then I'll introduce you to a rule. And then I'll tell you a little bit about why people think this rule works. So in general, the carbon that has the functional group on it, that's the alpha carbon. So let me label it."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to introduce you a little bit of terminology, and then I'll introduce you to a rule. And then I'll tell you a little bit about why people think this rule works. So in general, the carbon that has the functional group on it, that's the alpha carbon. So let me label it. This carbon right here is the alpha carbon. And in order to have an E2 reaction, in this case, when we did the video on E2 reactions, actually the rule will hold as well for E1 reactions. But in order to have the E2 reaction, the hydrogen has to get swiped off of a beta carbon."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So let me label it. This carbon right here is the alpha carbon. And in order to have an E2 reaction, in this case, when we did the video on E2 reactions, actually the rule will hold as well for E1 reactions. But in order to have the E2 reaction, the hydrogen has to get swiped off of a beta carbon. And a beta carbon is just a carbon that's one away from the alpha carbon. So this is a beta carbon, and this is also a beta carbon. And so it's completely reasonable."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But in order to have the E2 reaction, the hydrogen has to get swiped off of a beta carbon. And a beta carbon is just a carbon that's one away from the alpha carbon. So this is a beta carbon, and this is also a beta carbon. And so it's completely reasonable. One would think that, well, it could swipe it from there, or it could swipe it from there. And let's think about this reaction. Let's just draw it out so you could visualize it a little bit better."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And so it's completely reasonable. One would think that, well, it could swipe it from there, or it could swipe it from there. And let's think about this reaction. Let's just draw it out so you could visualize it a little bit better. So here we swiped it from this beta carbon. Let me redraw the reaction where we're swiping it from the other beta carbon. So I want both reactions on the screen at the same time."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let's just draw it out so you could visualize it a little bit better. So here we swiped it from this beta carbon. Let me redraw the reaction where we're swiping it from the other beta carbon. So I want both reactions on the screen at the same time. So let me draw our methoxide. So we have our oxygen bonded to a CH3, to a methyl group. The oxygen has seven valence electrons."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So I want both reactions on the screen at the same time. So let me draw our methoxide. So we have our oxygen bonded to a CH3, to a methyl group. The oxygen has seven valence electrons. 1, 2, 3, 4, 5, 6, 7. I'll do the seventh one, the one that will bond with the hydrogen. I'll do it in green."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen has seven valence electrons. 1, 2, 3, 4, 5, 6, 7. I'll do the seventh one, the one that will bond with the hydrogen. I'll do it in green. Instead of attacking that hydrogen, or taking that proton, I should say, because it's not taking the electron with it, it does it to this one. So what it does is it takes this proton, or it bonds with that hydrogen proton. And then the hydrogen protons, or that hydrogen's electron, can then be taken by this molecule."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I'll do it in green. Instead of attacking that hydrogen, or taking that proton, I should say, because it's not taking the electron with it, it does it to this one. So what it does is it takes this proton, or it bonds with that hydrogen proton. And then the hydrogen protons, or that hydrogen's electron, can then be taken by this molecule. So then it goes to the alpha carbon to form a double bond between this beta carbon and the alpha carbon. And so now this alpha carbon got that electron. Doesn't need the electron that's bonded with the chloro group anymore."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then the hydrogen protons, or that hydrogen's electron, can then be taken by this molecule. So then it goes to the alpha carbon to form a double bond between this beta carbon and the alpha carbon. And so now this alpha carbon got that electron. Doesn't need the electron that's bonded with the chloro group anymore. And so that goes to the chlorine to form chloride. Chlorine was already way more electronegative. It was already hogging it."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Doesn't need the electron that's bonded with the chloro group anymore. And so that goes to the chlorine to form chloride. Chlorine was already way more electronegative. It was already hogging it. Now it gets to go there. And now when all is said and done, our products look like this. We still have the methanol, just like we had in the original reaction, because this grabbed this hydrogen."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It was already hogging it. Now it gets to go there. And now when all is said and done, our products look like this. We still have the methanol, just like we had in the original reaction, because this grabbed this hydrogen. Let me draw it. So you have your OCH3. And let me draw all of this."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We still have the methanol, just like we had in the original reaction, because this grabbed this hydrogen. Let me draw it. So you have your OCH3. And let me draw all of this. You have that pair right over there. You have this pair right over here. And then you have this purple electron."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And let me draw all of this. You have that pair right over there. You have this pair right over here. And then you have this purple electron. And now it's bonded with this green electron, which is now on the hydrogen. So it is now bonded with this green electron that has been given to the hydrogen. Oh, and not to make sure we don't forget it, with this oxygen over here, it's 7 valence electrons."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then you have this purple electron. And now it's bonded with this green electron, which is now on the hydrogen. So it is now bonded with this green electron that has been given to the hydrogen. Oh, and not to make sure we don't forget it, with this oxygen over here, it's 7 valence electrons. So it had a negative charge. Neutral oxygen would have 6. So this had a negative charge."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Oh, and not to make sure we don't forget it, with this oxygen over here, it's 7 valence electrons. So it had a negative charge. Neutral oxygen would have 6. So this had a negative charge. But now that it gave its electron to this hydrogen, it now has a neutral charge and is now methanol. Exactly what we saw when we first learned about E1 reactions. We also know that the chloro group, it took that electron."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this had a negative charge. But now that it gave its electron to this hydrogen, it now has a neutral charge and is now methanol. Exactly what we saw when we first learned about E1 reactions. We also know that the chloro group, it took that electron. It's now a chloride anion. So let me draw that. And that's exactly the same as what we saw in the first video on E1 reactions."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We also know that the chloro group, it took that electron. It's now a chloride anion. So let me draw that. And that's exactly the same as what we saw in the first video on E1 reactions. So it's now a chloride anion. It has grabbed this orange electron. So it now has a negative charge."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And that's exactly the same as what we saw in the first video on E1 reactions. So it's now a chloride anion. It has grabbed this orange electron. So it now has a negative charge. You can imagine that the negative charge has been transferred from the methoxide to the chloride anion. And now what's different this time is that the double bond is now between the 1 and the 2 carbon and not between the 2 and the 3 carbon. So now it's going to look like this."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it now has a negative charge. You can imagine that the negative charge has been transferred from the methoxide to the chloride anion. And now what's different this time is that the double bond is now between the 1 and the 2 carbon and not between the 2 and the 3 carbon. So now it's going to look like this. So now the result, this product, if the reaction went this way, would look like this. We have this carbon right here. It is bonded to 2 hydrogens."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So now it's going to look like this. So now the result, this product, if the reaction went this way, would look like this. We have this carbon right here. It is bonded to 2 hydrogens. Now it has a double bond with this carbon. And I'll do the second bond. I'll do the pi bond of the double bond in purple right over there."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It is bonded to 2 hydrogens. Now it has a double bond with this carbon. And I'll do the second bond. I'll do the pi bond of the double bond in purple right over there. I'll assume that that's the pi bond. That's the new double bond formed. And now this carbon, which was the alpha carbon, is right over here."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I'll do the pi bond of the double bond in purple right over there. I'll assume that that's the pi bond. That's the new double bond formed. And now this carbon, which was the alpha carbon, is right over here. This was the alpha carbon. It's bonded to 1 hydrogen. And then let me draw everything else."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And now this carbon, which was the alpha carbon, is right over here. This was the alpha carbon. It's bonded to 1 hydrogen. And then let me draw everything else. So then you have a carbon, carbon. This guy is bonded to 3 hydrogens. I could have just written a CH3 if I wanted."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then let me draw everything else. So then you have a carbon, carbon. This guy is bonded to 3 hydrogens. I could have just written a CH3 if I wanted. This guy is bonded to 2 hydrogens. And we're done. And so instead of, as we saw in the first video on E2 reactions, instead of forming but-2-ene, we now have still 1, 2, 3, 4 carbons, so still but."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I could have just written a CH3 if I wanted. This guy is bonded to 2 hydrogens. And we're done. And so instead of, as we saw in the first video on E2 reactions, instead of forming but-2-ene, we now have still 1, 2, 3, 4 carbons, so still but. But the double bond is on the 1 carbon. We'd start 1, 2, 3, 4. So we could call this but-1-ene or 1-but-ene either way."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And so instead of, as we saw in the first video on E2 reactions, instead of forming but-2-ene, we now have still 1, 2, 3, 4 carbons, so still but. But the double bond is on the 1 carbon. We'd start 1, 2, 3, 4. So we could call this but-1-ene or 1-but-ene either way. So let's call this but-1-ene. So the question is, which is more likely to happen? Do both happen?"}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So we could call this but-1-ene or 1-but-ene either way. So let's call this but-1-ene. So the question is, which is more likely to happen? Do both happen? Does 1 happen disproportionately? And the answer is that, yes, 1 happens disproportionately. This one, this is the dominant product."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Do both happen? Does 1 happen disproportionately? And the answer is that, yes, 1 happens disproportionately. This one, this is the dominant product. If you were to perform this reaction, this is what, and you were to analyze in your beaker wherever you're performing the reaction, what you see most of, the majority product, the great majority, is going to be the but-2-ene, not the but-1-ene. Maybe you see very, very little of this. And the question is why?"}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This one, this is the dominant product. If you were to perform this reaction, this is what, and you were to analyze in your beaker wherever you're performing the reaction, what you see most of, the majority product, the great majority, is going to be the but-2-ene, not the but-1-ene. Maybe you see very, very little of this. And the question is why? Or how would you even be able to determine that? Or how could you have predicted that? And to predict it, there's something called Zeitzel's rule."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And the question is why? Or how would you even be able to determine that? Or how could you have predicted that? And to predict it, there's something called Zeitzel's rule. And I'm sure I'm mispronouncing it, but let me write it down. So Zeitzel's rule. And it's kind of analogous to Markovnikov's rule, but for elimination reactions."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And to predict it, there's something called Zeitzel's rule. And I'm sure I'm mispronouncing it, but let me write it down. So Zeitzel's rule. And it's kind of analogous to Markovnikov's rule, but for elimination reactions. If you think about it, the addition reactions that we did many videos ago are the opposite of the elimination reactions. The addition reactions were adding the chlorogroup, and the elimination were taking it off. And so Zeitzel's rule is kind of analogous to Markovnikov's rule."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And it's kind of analogous to Markovnikov's rule, but for elimination reactions. If you think about it, the addition reactions that we did many videos ago are the opposite of the elimination reactions. The addition reactions were adding the chlorogroup, and the elimination were taking it off. And so Zeitzel's rule is kind of analogous to Markovnikov's rule. Now first I'll just tell you the rule, and then we can think a little bit about why it works. And the jury's not out on this. They think they know why it works, but they're not 100% sure."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And so Zeitzel's rule is kind of analogous to Markovnikov's rule. Now first I'll just tell you the rule, and then we can think a little bit about why it works. And the jury's not out on this. They think they know why it works, but they're not 100% sure. So Zeitzel's rule says the carbon that is going to lose the hydrogen is the one that has fewer hydrogens. So let me write it down over here. Carbon more likely to lose hydrogen is, I should say, the hydrogen proton, because it keeps the electrons still, is the one with fewer hydrogens."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "They think they know why it works, but they're not 100% sure. So Zeitzel's rule says the carbon that is going to lose the hydrogen is the one that has fewer hydrogens. So let me write it down over here. Carbon more likely to lose hydrogen is, I should say, the hydrogen proton, because it keeps the electrons still, is the one with fewer hydrogens. So if you were to look at this reaction right here, we have our alpha carbon. Either this beta carbon or this beta carbon could lose its hydrogen. This one has three hydrogens on it."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Carbon more likely to lose hydrogen is, I should say, the hydrogen proton, because it keeps the electrons still, is the one with fewer hydrogens. So if you were to look at this reaction right here, we have our alpha carbon. Either this beta carbon or this beta carbon could lose its hydrogen. This one has three hydrogens on it. This one only has two. So Zeitzel's rule tells us that this is the hydrogen, or actually the proton, that is more likely to be reacted with the base. You could almost view it as it is the more acidic proton."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This one has three hydrogens on it. This one only has two. So Zeitzel's rule tells us that this is the hydrogen, or actually the proton, that is more likely to be reacted with the base. You could almost view it as it is the more acidic proton. It is a lower-hanging fruit for this strong base to capture. Now, a more interesting question. That's a pretty easy rule to follow."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "You could almost view it as it is the more acidic proton. It is a lower-hanging fruit for this strong base to capture. Now, a more interesting question. That's a pretty easy rule to follow. And if they both had the same number, then you'd see equal products, depending on which side it gets. Now, the question is, why is this happening? And here, something called hyperconjugation comes into effect."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "That's a pretty easy rule to follow. And if they both had the same number, then you'd see equal products, depending on which side it gets. Now, the question is, why is this happening? And here, something called hyperconjugation comes into effect. I'm not going to go into details in it, into the quantum mechanics of it. Hyperconjugation. And hyperconjugation is the notion that the fact, so we said that the one with fewer hydrogens is the one that's less likely to lose, or the one with fewer hydrogens is the one more likely to lose the hydrogen proton."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And here, something called hyperconjugation comes into effect. I'm not going to go into details in it, into the quantum mechanics of it. Hyperconjugation. And hyperconjugation is the notion that the fact, so we said that the one with fewer hydrogens is the one that's less likely to lose, or the one with fewer hydrogens is the one more likely to lose the hydrogen proton. But the one with fewer hydrogens is also bonded to more carbons. This guy is bonded to one carbon. This outside of the alpha carbon, he's actually bonded to the alpha and this carbon."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And hyperconjugation is the notion that the fact, so we said that the one with fewer hydrogens is the one that's less likely to lose, or the one with fewer hydrogens is the one more likely to lose the hydrogen proton. But the one with fewer hydrogens is also bonded to more carbons. This guy is bonded to one carbon. This outside of the alpha carbon, he's actually bonded to the alpha and this carbon. This guy right here is only bonded to the alpha carbon. And hyperconjugation is the notion that not the beta carbon, but the carbons one over from that help stabilize the double bond that eventually forms. I almost think of it, you have more electrons over here, because carbons have more electrons to offer than hydrogens."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This outside of the alpha carbon, he's actually bonded to the alpha and this carbon. This guy right here is only bonded to the alpha carbon. And hyperconjugation is the notion that not the beta carbon, but the carbons one over from that help stabilize the double bond that eventually forms. I almost think of it, you have more electrons over here, because carbons have more electrons to offer than hydrogens. So you kind of have, at the end of the day, this guy is more likely able to donate electrons to form from the right-hand side to make a double bond than from the left-hand side. I won't go to the detail of hyperconjugation, but it's all based on the notion that the more stable double bond will be formed if we have other carbons near the double bond. Another way to think about it is to look at the products."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I almost think of it, you have more electrons over here, because carbons have more electrons to offer than hydrogens. So you kind of have, at the end of the day, this guy is more likely able to donate electrons to form from the right-hand side to make a double bond than from the left-hand side. I won't go to the detail of hyperconjugation, but it's all based on the notion that the more stable double bond will be formed if we have other carbons near the double bond. Another way to think about it is to look at the products. So we saw, or Zaitsev's rule tells us, that but-2-ene is a more likely product than but-1-ene. And if you look at but-1-ene, we could rewrite it like this. We could draw the double bond like this."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Another way to think about it is to look at the products. So we saw, or Zaitsev's rule tells us, that but-2-ene is a more likely product than but-1-ene. And if you look at but-1-ene, we could rewrite it like this. We could draw the double bond like this. This carbon is what was the alpha carbon. We could draw a carbon right here. And then it is bonded to a hydrogen."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We could draw the double bond like this. This carbon is what was the alpha carbon. We could draw a carbon right here. And then it is bonded to a hydrogen. It's bonded to this hydrogen. And then it's bonded to just a chain of carbons. And we'll just write R that."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then it is bonded to a hydrogen. It's bonded to this hydrogen. And then it's bonded to just a chain of carbons. And we'll just write R that. And then this guy is just bonded to two hydrogens. So this is but-1-ene. I just put an R here, but this is how it could be represented."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And we'll just write R that. And then this guy is just bonded to two hydrogens. So this is but-1-ene. I just put an R here, but this is how it could be represented. Now the but-2-ene, if we wanted to draw it like this, would look like this. If we wanted to call, we could call this right here. We could call this R prime."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I just put an R here, but this is how it could be represented. Now the but-2-ene, if we wanted to draw it like this, would look like this. If we wanted to call, we could call this right here. We could call this R prime. It's just a chain of carbons. And then we could call this. And then we could call this R prime prime."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We could call this R prime. It's just a chain of carbons. And then we could call this. And then we could call this R prime prime. It's another, it's not even a chain. It's just one carbon. But if we call it that, then the but-2-ene, let me draw it down here where I have more real estate."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then we could call this R prime prime. It's another, it's not even a chain. It's just one carbon. But if we call it that, then the but-2-ene, let me draw it down here where I have more real estate. The but-2-ene would look like this. You have your carbon-carbon double bond. Now the left-hand carbon is bonded to a hydrogen, that hydrogen right there, and to R prime."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But if we call it that, then the but-2-ene, let me draw it down here where I have more real estate. The but-2-ene would look like this. You have your carbon-carbon double bond. Now the left-hand carbon is bonded to a hydrogen, that hydrogen right there, and to R prime. And the right-hand carbon is bonded to a hydrogen and R prime prime. So it's bonded to a hydrogen and R prime prime. All I did here, let me see if I can fit it all on the same screen, is I just redrew this."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now the left-hand carbon is bonded to a hydrogen, that hydrogen right there, and to R prime. And the right-hand carbon is bonded to a hydrogen and R prime prime. So it's bonded to a hydrogen and R prime prime. All I did here, let me see if I can fit it all on the same screen, is I just redrew this. And I just abstracted away the chain as it goes away from the double bond. And I did that so that we can look at it this way. We can just have the double bond kind of as our focus of attention and think about what's going on around it."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "All I did here, let me see if I can fit it all on the same screen, is I just redrew this. And I just abstracted away the chain as it goes away from the double bond. And I did that so that we can look at it this way. We can just have the double bond kind of as our focus of attention and think about what's going on around it. Over here for the but-1-ene, we already said this is the lesser product. So this is the greater product. This is the greater or the dominant product."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We can just have the double bond kind of as our focus of attention and think about what's going on around it. Over here for the but-1-ene, we already said this is the lesser product. So this is the greater product. This is the greater or the dominant product. In the lesser product, if we go off of the double bond, we only have one alkyl group right there, that R right here. Over here we have two. And we say that this is more substituted."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This is the greater or the dominant product. In the lesser product, if we go off of the double bond, we only have one alkyl group right there, that R right here. Over here we have two. And we say that this is more substituted. And when we say substituted, you're imagining that you're substituting hydrogens with carbon chains, with alkyl groups. So this one right here is more substituted. And hyperconjugation, so the idea would have it, is that these carbon chains that are near the double bond help stabilize it."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And we say that this is more substituted. And when we say substituted, you're imagining that you're substituting hydrogens with carbon chains, with alkyl groups. So this one right here is more substituted. And hyperconjugation, so the idea would have it, is that these carbon chains that are near the double bond help stabilize it. They're able to kind of, some of their sigma electrons and sigma orbitals are there to somehow help stabilize the pi orbitals. Now this is getting into quantum mechanics and all of that. The world isn't 100% clear whether that's definitely the mechanism, although people have run the experiment."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And hyperconjugation, so the idea would have it, is that these carbon chains that are near the double bond help stabilize it. They're able to kind of, some of their sigma electrons and sigma orbitals are there to somehow help stabilize the pi orbitals. Now this is getting into quantum mechanics and all of that. The world isn't 100% clear whether that's definitely the mechanism, although people have run the experiment. And they know that the more substituted product is what you're going to see more of, as opposed to the less substituted. And then that all comes from Zaitsev's rule. So hopefully you at least get Zaitsev's rule."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The world isn't 100% clear whether that's definitely the mechanism, although people have run the experiment. And they know that the more substituted product is what you're going to see more of, as opposed to the less substituted. And then that all comes from Zaitsev's rule. So hopefully you at least get Zaitsev's rule. The hyperconjugation, that's kind of a deeper concept. And the jury's not even out on it. That's a belief of why Zaitsev's rule works."}, {"video_title": "Zaitsev's rule Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So hopefully you at least get Zaitsev's rule. The hyperconjugation, that's kind of a deeper concept. And the jury's not even out on it. That's a belief of why Zaitsev's rule works. But the rule itself is pretty straightforward. If you're trying to pick between two beta carbons, the one that's going to lose the hydrogen is the one that already has fewer hydrogens. Or the one that's bonded to more carbons."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When I say huge cloud, huge both in distance and in mass. If you were to combine all of the hydrogen atoms, it would just be this really, really massive thing. So you have this huge cloud. Well, we know that gravity would make the atoms actually attracted to each other. We normally don't think about the gravity of atoms, but it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, we know that gravity would make the atoms actually attracted to each other. We normally don't think about the gravity of atoms, but it would slowly affect these atoms. And they would slowly draw close to each other. It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would slowly condense. They'd slowly move towards the center of mass of all of the atoms. They'd slowly move in. So if we fast forward, this cloud's going to get denser and denser. The hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. So it's going to get denser and denser and denser. Remember, it was a huge mass of hydrogen atoms."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we fast forward, this cloud's going to get denser and denser. The hydrogen atoms are going to start bumping into each other and rubbing up against each other and interacting with each other. So it's going to get denser and denser and denser. Remember, it was a huge mass of hydrogen atoms. So the temperature is going up. And they'll keep condensing. They'll just keep condensing and condensing until something really interesting happens."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, it was a huge mass of hydrogen atoms. So the temperature is going up. And they'll keep condensing. They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They'll just keep condensing and condensing until something really interesting happens. So let's imagine that they've gotten really dense here in the center. And there's a bunch of hydrogen atoms all over. It's really dense. I could never draw the actual number of atoms here. This is really giving you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's really dense. I could never draw the actual number of atoms here. This is really giving you an idea. There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. To kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's a huge amount of inward pressure from gravity, from everything that wants to get to that center of mass of our entire cloud. The temperature here is approaching 10 million Kelvin. And at that point, something neat happens. To kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. And even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "To kind of realize the neat thing that's happening, let's remember what a hydrogen atom looks like. And even more, I'm just going to focus on the hydrogen nucleus. So the hydrogen nucleus is a proton. If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. This distance is also not to scale."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you want to think about a hydrogen atom, it also has an electron orbiting around or floating around. And let's draw another hydrogen atom over here. And obviously this distance isn't to scale. This distance is also not to scale. The nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This distance is also not to scale. The nucleus of atoms are actually much, much, much, much smaller than the actual radius of an atom. And so is the electron. But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from what we learned about the four forces that if they did get close enough to each other, that if somehow under huge temperatures and huge pressures, you were able to get these two protons close enough to each other, then all of a sudden the strong force will overtake. It's much stronger than the Coulomb force, and that these two hydrogens will actually, these nucleuses would actually fuse, or is it nuclei?"}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But anyway, this just gives you an idea. So we know from the Coulomb forces, from electromagnetic forces, that these two positively charged nucleuses will not want to get anywhere near each other. But we do know from what we learned about the four forces that if they did get close enough to each other, that if somehow under huge temperatures and huge pressures, you were able to get these two protons close enough to each other, then all of a sudden the strong force will overtake. It's much stronger than the Coulomb force, and that these two hydrogens will actually, these nucleuses would actually fuse, or is it nuclei? Well anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring these protons close enough to each other for fusion to occur."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's much stronger than the Coulomb force, and that these two hydrogens will actually, these nucleuses would actually fuse, or is it nuclei? Well anyway, they would actually fuse together. And so that is what actually happens once this gets hot and dense enough. You now have enough pressure and enough temperature to overcome the Coulomb force and bring these protons close enough to each other for fusion to occur. For fusion. Fusion. Ignition."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You now have enough pressure and enough temperature to overcome the Coulomb force and bring these protons close enough to each other for fusion to occur. For fusion. Fusion. Ignition. And the reason why is, and I want to be very careful, it's not ignition, it's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion, it's ignition."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Ignition. And the reason why is, and I want to be very careful, it's not ignition, it's not combustion in the traditional sense. It's not like you're burning a carbon molecule in the presence of oxygen. It's not combustion, it's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure. Obviously this would not happen with just the Coulomb forces."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's not combustion, it's ignition. And the reason why it's called ignition is because when two of these protons, or two of the nucleuses fuse, the resulting nucleus has a slightly smaller mass. And so in the first stage of this, you actually have two protons under enough pressure. Obviously this would not happen with just the Coulomb forces. With enough pressure, they're close enough, and then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original, by a little bit."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Obviously this would not happen with just the Coulomb forces. With enough pressure, they're close enough, and then the strong interaction actually keeps them together. One of these guys degrades into a neutron. And the resulting mass of the combined protons is lower than the mass of each of the original, by a little bit. But that little bit of mass results in a lot of energy. Plus energy. And this energy is why we call it ignition."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the resulting mass of the combined protons is lower than the mass of each of the original, by a little bit. But that little bit of mass results in a lot of energy. Plus energy. And this energy is why we call it ignition. So what this energy does is it provides a little bit of outward pressure so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs, and then that energy provides outward pressure to balance what is now a star. So now we actually have the ignition at the center, and we still have all of the other molecules trying to get in, providing the pressure for this fusion ignition."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this energy is why we call it ignition. So what this energy does is it provides a little bit of outward pressure so that this thing doesn't keep collapsing. So once you get pressure enough, the fusion occurs, and then that energy provides outward pressure to balance what is now a star. So now we actually have the ignition at the center, and we still have all of the other molecules trying to get in, providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction, and I'm just kind of doing the most basic type of fusion that happens in stars, the hydrogen gets fused into deuterium, which is another way of calling heavy hydrogen. This is still hydrogen because it has one proton and one neutron now."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now we actually have the ignition at the center, and we still have all of the other molecules trying to get in, providing the pressure for this fusion ignition. Now, what is the hydrogen being fused into? Well, in the first step of the reaction, and I'm just kind of doing the most basic type of fusion that happens in stars, the hydrogen gets fused into deuterium, which is another way of calling heavy hydrogen. This is still hydrogen because it has one proton and one neutron now. It does not have helium yet. It does not have two protons. But then the deuterium keeps fusing, and then we eventually end up with helium."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is still hydrogen because it has one proton and one neutron now. It does not have helium yet. It does not have two protons. But then the deuterium keeps fusing, and then we eventually end up with helium. And we can even see that on the periodic table. I lost my periodic table. Well, I'll show you in the next video."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But then the deuterium keeps fusing, and then we eventually end up with helium. And we can even see that on the periodic table. I lost my periodic table. Well, I'll show you in the next video. But we know hydrogen in its atomic state has an atomic number of one, and it also has a mass of one. It only has one nucleon in its nucleus. But it's being fused."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, I'll show you in the next video. But we know hydrogen in its atomic state has an atomic number of one, and it also has a mass of one. It only has one nucleon in its nucleus. But it's being fused. It goes to hydrogen two, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused, and I'm not going into the detail of the reaction, into helium. And by definition, helium has two protons and two neutrons."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's being fused. It goes to hydrogen two, which is deuterium, which is one neutron, one proton in its nucleus, two nucleons. And then that eventually gets fused, and I'm not going into the detail of the reaction, into helium. And by definition, helium has two protons and two neutrons. We're talking about helium-4 in particular, that isotope of helium. It has an atomic mass of four. And this process releases a ton of energy because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And by definition, helium has two protons and two neutrons. We're talking about helium-4 in particular, that isotope of helium. It has an atomic mass of four. And this process releases a ton of energy because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy from the fusion, but it needs super high pressure, super high temperatures to happen, keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen, it is fusing hydrogen in its core where the pressure and the temperature is the most to form helium, it is now in its main sequence. This is now a main sequence star."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this process releases a ton of energy because the atomic mass of the helium that gets produced is slightly lower than four times the atomic mass of each of the constituent hydrogens. So all of this energy from the fusion, but it needs super high pressure, super high temperatures to happen, keeps the star from collapsing. And once a star is in this stage, once it is using hydrogen, it is fusing hydrogen in its core where the pressure and the temperature is the most to form helium, it is now in its main sequence. This is now a main sequence star. And that's actually where the sun is right now. Now, there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is now a main sequence star. And that's actually where the sun is right now. Now, there's questions of, well, what if there just wasn't enough mass to get to this level over here? And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite make the threshold of stars that only fuse to this level, so they are generating some of their heat. Or there are even smaller objects that just get to the point there's a huge temperature and pressure, but fusion is not actually occurring inside of the core. And something like Jupiter would be an example."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there actually are things that never get to quite that threshold to fuse all the way into helium. There are a few things that don't quite make the threshold of stars that only fuse to this level, so they are generating some of their heat. Or there are even smaller objects that just get to the point there's a huge temperature and pressure, but fusion is not actually occurring inside of the core. And something like Jupiter would be an example. And you can go several masses above Jupiter where you get something like that. So you have to reach a certain threshold where the mass, where the pressure, and the temperature due to the heavy mass gets so large that you start this fusion. But the smaller you are above that threshold, the slower the fusion will occur."}, {"video_title": "Birth of stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And something like Jupiter would be an example. And you can go several masses above Jupiter where you get something like that. So you have to reach a certain threshold where the mass, where the pressure, and the temperature due to the heavy mass gets so large that you start this fusion. But the smaller you are above that threshold, the slower the fusion will occur. But if you're supermassive, the fusion will occur really, really fast. So that's a general idea of just how stars get formed and why they don't collapse on themselves and why they are these kind of little balls of fusion reactions existing in the universe. In the next few videos, we'll talk about what happens once that hydrogen fuel in the core starts to run out."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Here you can see a carbon-oxygen double bond in this functional group, and here's another one in this functional group, and here's another one down here. For the first functional group, this is called an aldehyde. An aldehyde has an R group on one side of the carbonyl, and a hydrogen on the other side of the carbonyl. As an example, let's look at this compound. Here is our carbonyl, our carbon-oxygen double bond, and then on this side we have a hydrogen directly bonded to this carbonyl carbon. On the left side we have this R group. The carbonyl carbon gets a number one."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "As an example, let's look at this compound. Here is our carbonyl, our carbon-oxygen double bond, and then on this side we have a hydrogen directly bonded to this carbonyl carbon. On the left side we have this R group. The carbonyl carbon gets a number one. It gets the lowest number possible. Then this carbon would get a number two. This carbon would get a number three, and this carbon would get a number four."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The carbonyl carbon gets a number one. It gets the lowest number possible. Then this carbon would get a number two. This carbon would get a number three, and this carbon would get a number four. We have a four-carbon aldehyde. How do we name a four-carbon aldehyde? A four-carbon alkane we know is butane."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "This carbon would get a number three, and this carbon would get a number four. We have a four-carbon aldehyde. How do we name a four-carbon aldehyde? A four-carbon alkane we know is butane. One, two, three, and four, we know that this compound is called butane. An aldehyde has an AL ending. If you see an AL ending, that indicates the presence of an aldehyde."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "A four-carbon alkane we know is butane. One, two, three, and four, we know that this compound is called butane. An aldehyde has an AL ending. If you see an AL ending, that indicates the presence of an aldehyde. We're going to lose the E from butane and add on AL. The name of this molecule becomes butanal. Let me write that out here."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "If you see an AL ending, that indicates the presence of an aldehyde. We're going to lose the E from butane and add on AL. The name of this molecule becomes butanal. Let me write that out here. This is butanal. Next, let's look at a ketone. A ketone has an R group on one side of the carbonyl and an R group on the other side of the carbonyl."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Let me write that out here. This is butanal. Next, let's look at a ketone. A ketone has an R group on one side of the carbonyl and an R group on the other side of the carbonyl. For our ketone example, let's look at this compound. We're going to number to give our carbonyl the lowest number possible. We're going to start numbering from the left side of our compound."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "A ketone has an R group on one side of the carbonyl and an R group on the other side of the carbonyl. For our ketone example, let's look at this compound. We're going to number to give our carbonyl the lowest number possible. We're going to start numbering from the left side of our compound. This would get carbon number one. The carbonyl carbon would be carbon number two. This is carbon number three."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We're going to start numbering from the left side of our compound. This would get carbon number one. The carbonyl carbon would be carbon number two. This is carbon number three. This is carbon number four. This is carbon number five. We have a five-carbon ketone."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "This is carbon number three. This is carbon number four. This is carbon number five. We have a five-carbon ketone. A five-carbon alkane, let me go ahead and draw out a five-carbon alkane here. This would be called pentane. That would be pentane."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have a five-carbon ketone. A five-carbon alkane, let me go ahead and draw out a five-carbon alkane here. This would be called pentane. That would be pentane. A ketone has an ONE ending. ONE indicates the presence of a ketone. We're going to lose the E from pentane and add ONE."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "That would be pentane. A ketone has an ONE ending. ONE indicates the presence of a ketone. We're going to lose the E from pentane and add ONE. This would be pentanone. This is pentanone. We also need to indicate the position of the carbonyls."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We're going to lose the E from pentane and add ONE. This would be pentanone. This is pentanone. We also need to indicate the position of the carbonyls. The carbonyl is at carbon two. We need to write two pentanone. That's because we could draw another molecule that's pentanone, but this could have the carbonyl at carbon three."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We also need to indicate the position of the carbonyls. The carbonyl is at carbon two. We need to write two pentanone. That's because we could draw another molecule that's pentanone, but this could have the carbonyl at carbon three. Let me go ahead and write that in here, one, two, three. This compound would be three pentanone. Three pentanone and two pentanone are different molecules."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "That's because we could draw another molecule that's pentanone, but this could have the carbonyl at carbon three. Let me go ahead and write that in here, one, two, three. This compound would be three pentanone. Three pentanone and two pentanone are different molecules. That's why we need to indicate the position of the carbonyl. For the aldehyde up here, we didn't need to indicate that the aldehyde is at carbon one. That's the only place that it could possibly be."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Three pentanone and two pentanone are different molecules. That's why we need to indicate the position of the carbonyl. For the aldehyde up here, we didn't need to indicate that the aldehyde is at carbon one. That's the only place that it could possibly be. Notice that an aldehyde has this hydrogen directly bonded to the carbonyl carbon, whereas a ketone has, if we look at the carbonyl group right here, we have a carbon on one side, a CH3 on one side in this case, and we have a carbon on this side directly bonded to that carbonyl carbon. That's the difference between an aldehyde and a ketone. A lot of students have a hard time telling the difference between an aldehyde and a ketone."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "That's the only place that it could possibly be. Notice that an aldehyde has this hydrogen directly bonded to the carbonyl carbon, whereas a ketone has, if we look at the carbonyl group right here, we have a carbon on one side, a CH3 on one side in this case, and we have a carbon on this side directly bonded to that carbonyl carbon. That's the difference between an aldehyde and a ketone. A lot of students have a hard time telling the difference between an aldehyde and a ketone. Look for this hydrogen directly bonded to the carbonyl carbon, and that's an aldehyde. Next, we have a carboxylic acid. A carboxylic acid has an R group on one side of the carbonyl and an OH on the other side."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "A lot of students have a hard time telling the difference between an aldehyde and a ketone. Look for this hydrogen directly bonded to the carbonyl carbon, and that's an aldehyde. Next, we have a carboxylic acid. A carboxylic acid has an R group on one side of the carbonyl and an OH on the other side. Some students think that this is an alcohol at first, but this is not an alcohol. This OH is right next to this carbonyl, and that changes the properties. Carboxylic acids have different properties from alcohols."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "A carboxylic acid has an R group on one side of the carbonyl and an OH on the other side. Some students think that this is an alcohol at first, but this is not an alcohol. This OH is right next to this carbonyl, and that changes the properties. Carboxylic acids have different properties from alcohols. For an example of a carboxylic acid, over here on the right, we know that our carbonyl carbon gets number one, so that's carbon number one. This is carbon number two, carbon number three, and carbon number four. A four-carbon carboxylic acid, so a four-carbon alkane up here was butane, so we're going to have oic acid as our ending."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Carboxylic acids have different properties from alcohols. For an example of a carboxylic acid, over here on the right, we know that our carbonyl carbon gets number one, so that's carbon number one. This is carbon number two, carbon number three, and carbon number four. A four-carbon carboxylic acid, so a four-carbon alkane up here was butane, so we're going to have oic acid as our ending. It would be butanoic acid for the name for this compound. It was just one of the names. One of the names for this molecule would be butanoic acid."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "A four-carbon carboxylic acid, so a four-carbon alkane up here was butane, so we're going to have oic acid as our ending. It would be butanoic acid for the name for this compound. It was just one of the names. One of the names for this molecule would be butanoic acid. Probably the most famous carboxylic acid is a two-carbon carboxylic acid, so I can just draw it like this really quickly. This is called acetic acid. Vinegar is just a dilute solution of acetic acid in water."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "One of the names for this molecule would be butanoic acid. Probably the most famous carboxylic acid is a two-carbon carboxylic acid, so I can just draw it like this really quickly. This is called acetic acid. Vinegar is just a dilute solution of acetic acid in water. Our next functional group is called an acyl halide, or we could also call it an acid halide. You're pretty much talking about acid chlorides in organic chemistry, and so here is our carbon-oxygen double bond, and then on this side we have an R group, and the right side we have a chlorine. Acyl halides come from carboxylic acid."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Vinegar is just a dilute solution of acetic acid in water. Our next functional group is called an acyl halide, or we could also call it an acid halide. You're pretty much talking about acid chlorides in organic chemistry, and so here is our carbon-oxygen double bond, and then on this side we have an R group, and the right side we have a chlorine. Acyl halides come from carboxylic acid. This is an example of a carboxylic acid derivative. An example of an acyl halide on the right here, we have our carbonyl, and then we have our halogen. The reason I like to call it an acyl halide is this right here on the left, this is an acyl group, and on the right we have our halogen."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Acyl halides come from carboxylic acid. This is an example of a carboxylic acid derivative. An example of an acyl halide on the right here, we have our carbonyl, and then we have our halogen. The reason I like to call it an acyl halide is this right here on the left, this is an acyl group, and on the right we have our halogen. This would be acetyl chloride is one of the names for this, but you could also call them an acid halide. These are carboxylic acid derivatives. Some professors prefer one term, and some professors prefer the other term."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The reason I like to call it an acyl halide is this right here on the left, this is an acyl group, and on the right we have our halogen. This would be acetyl chloride is one of the names for this, but you could also call them an acid halide. These are carboxylic acid derivatives. Some professors prefer one term, and some professors prefer the other term. Our next functional group is an acid anhydride. Here is our carbonyl. Here is the R group, and instead of a chlorine like we had up here, we have all of this over here on the right."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Some professors prefer one term, and some professors prefer the other term. Our next functional group is an acid anhydride. Here is our carbonyl. Here is the R group, and instead of a chlorine like we had up here, we have all of this over here on the right. This is an acid anhydride, or just an anhydride. This is also a carboxylic acid derivative. You can form an acid anhydride from a carboxylic acid."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Here is the R group, and instead of a chlorine like we had up here, we have all of this over here on the right. This is an acid anhydride, or just an anhydride. This is also a carboxylic acid derivative. You can form an acid anhydride from a carboxylic acid. For the example over here on the right, we have acetic anhydride. Next functional group is called an ester. An ester is another carboxylic acid derivative."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "You can form an acid anhydride from a carboxylic acid. For the example over here on the right, we have acetic anhydride. Next functional group is called an ester. An ester is another carboxylic acid derivative. A carboxylic acid had a hydrogen in this position, but now we have an R group. We have our carbonyl, we have an OR, and then we have an R on the left. That's an ester."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "An ester is another carboxylic acid derivative. A carboxylic acid had a hydrogen in this position, but now we have an R group. We have our carbonyl, we have an OR, and then we have an R on the left. That's an ester. Here's an example of an ester over here. We have an ethyl group instead of a hydrogen. For acetic acid, let me go ahead and draw acetic acid again."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "That's an ester. Here's an example of an ester over here. We have an ethyl group instead of a hydrogen. For acetic acid, let me go ahead and draw acetic acid again. Acetic acid had a hydrogen right here, and now we have an ethyl group. Instantly we know we're talking about an ester. We're not talking about a carboxylic acid."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "For acetic acid, let me go ahead and draw acetic acid again. Acetic acid had a hydrogen right here, and now we have an ethyl group. Instantly we know we're talking about an ester. We're not talking about a carboxylic acid. If you wanted to name this ester, you have this ethyl group, so that would be ethyl, and then this part would be acetate. Ethyl acetate would be the name. For the names of all of these, you're going to get to them later in the course, so it's probably not that important to name these compounds, but it's good to start hearing these things now."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We're not talking about a carboxylic acid. If you wanted to name this ester, you have this ethyl group, so that would be ethyl, and then this part would be acetate. Ethyl acetate would be the name. For the names of all of these, you're going to get to them later in the course, so it's probably not that important to name these compounds, but it's good to start hearing these things now. Finally, our last functional group is called an amide. At least that's what I call it, but there are several different ways to pronounce this functional group. The dictionary says, at least my dictionary says, amide or amide are the two preferred ways to pronounce this."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "For the names of all of these, you're going to get to them later in the course, so it's probably not that important to name these compounds, but it's good to start hearing these things now. Finally, our last functional group is called an amide. At least that's what I call it, but there are several different ways to pronounce this functional group. The dictionary says, at least my dictionary says, amide or amide are the two preferred ways to pronounce this. You'll probably hear amid as one of the most common ways to pronounce this. A lot of professors say amid, and some professors get pretty upset if you don't pronounce this the way that they want. I've heard many more ways to pronounce this functional group too, like amide and all kinds of things."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The dictionary says, at least my dictionary says, amide or amide are the two preferred ways to pronounce this. You'll probably hear amid as one of the most common ways to pronounce this. A lot of professors say amid, and some professors get pretty upset if you don't pronounce this the way that they want. I've heard many more ways to pronounce this functional group too, like amide and all kinds of things. Use whatever pronunciation your professor wants. That's always the way to do things. Now we have a carbonyl."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "I've heard many more ways to pronounce this functional group too, like amide and all kinds of things. Use whatever pronunciation your professor wants. That's always the way to do things. Now we have a carbonyl. Notice we have a nitrogen right next to the carbonyl carbon. A nitrogen directly bonded to the carbonyl carbon, lone pair of electrons on the nitrogen. That's an amide functional group or an amid functional group."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now we have a carbonyl. Notice we have a nitrogen right next to the carbonyl carbon. A nitrogen directly bonded to the carbonyl carbon, lone pair of electrons on the nitrogen. That's an amide functional group or an amid functional group. These Rs could be hydrogens. You could be talking about hydrogens, or you could be talking about alkyl groups. There's lots of different examples of amides or amids."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "That's an amide functional group or an amid functional group. These Rs could be hydrogens. You could be talking about hydrogens, or you could be talking about alkyl groups. There's lots of different examples of amides or amids. On the right is an example. Our carbonyl carbon gets a number one, so this would be one. This would be number two."}, {"video_title": "More functional groups Organic chemistry Khan Academy.mp3", "Sentence": "There's lots of different examples of amides or amids. On the right is an example. Our carbonyl carbon gets a number one, so this would be one. This would be number two. This would be number three. This would be a number four. A four carbon amide would be butanamide."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "You often see two different ways to name alkyl halides. And so we'll start with the common way first. So think about alkyl halides. First you want to think about an alkyl group, and this alkyl group is an ethyl group. There are two carbons on it, so we write in here ethyl. And then since it's alkyl halides, you want to think about the halogen you have and end in i. So this is chlorine, so it's going to end in i, so chloride."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "First you want to think about an alkyl group, and this alkyl group is an ethyl group. There are two carbons on it, so we write in here ethyl. And then since it's alkyl halides, you want to think about the halogen you have and end in i. So this is chlorine, so it's going to end in i, so chloride. So ethyl chloride would be the name for this compound. Now let's name the same molecule using IUPAC nomenclature. In this case, it's going to be named as a haloalkane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So this is chlorine, so it's going to end in i, so chloride. So ethyl chloride would be the name for this compound. Now let's name the same molecule using IUPAC nomenclature. In this case, it's going to be named as a haloalkane. So for a two-carbon alkane, that would be ethane. So I write in here ethane. And of course, our halogen is chlorine, so this would be chloro."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "In this case, it's going to be named as a haloalkane. So for a two-carbon alkane, that would be ethane. So I write in here ethane. And of course, our halogen is chlorine, so this would be chloro. So chloroethane is the name of this molecule. If I had fluorine instead of chlorine, it would be fluoroethane. So let me write in here fluoro."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And of course, our halogen is chlorine, so this would be chloro. So chloroethane is the name of this molecule. If I had fluorine instead of chlorine, it would be fluoroethane. So let me write in here fluoro. Notice the spelling on that. If I had a bromine instead of the chlorine, it would be bromoethane. And finally, if I had an iodine instead of the chlorine, it would be iodoethane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So let me write in here fluoro. Notice the spelling on that. If I had a bromine instead of the chlorine, it would be bromoethane. And finally, if I had an iodine instead of the chlorine, it would be iodoethane. So let me write in here iodo. Let's name this compound using our common system. So again, think about the alkyl group that is present."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And finally, if I had an iodine instead of the chlorine, it would be iodoethane. So let me write in here iodo. Let's name this compound using our common system. So again, think about the alkyl group that is present. So we saw in earlier videos, this alkyl group is isopropyl. So I write in here isopropyl. And again, we have chlorine attached to that."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So again, think about the alkyl group that is present. So we saw in earlier videos, this alkyl group is isopropyl. So I write in here isopropyl. And again, we have chlorine attached to that. So it would be isopropyl chloride using the common system. If I'm naming this using the IUPAC system, I look for my longest carbon chain. So that would be 1, 2, and 3."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And again, we have chlorine attached to that. So it would be isopropyl chloride using the common system. If I'm naming this using the IUPAC system, I look for my longest carbon chain. So that would be 1, 2, and 3. I know that is propane, so I write in here propane. And we have a chlorine attached to carbon 2. So that would be 2-chloropropane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So that would be 1, 2, and 3. I know that is propane, so I write in here propane. And we have a chlorine attached to carbon 2. So that would be 2-chloropropane. Let's look at how to classify alkyl halides. We find the carbon that's directly bonded to our halogen. And we see how many alkyl groups are attached to that carbon."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So that would be 2-chloropropane. Let's look at how to classify alkyl halides. We find the carbon that's directly bonded to our halogen. And we see how many alkyl groups are attached to that carbon. There's only one alkyl group, this methyl group here, attached to this carbon. So that's called primary. So ethyl chloride is an example of a primary alkyl halide."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And we see how many alkyl groups are attached to that carbon. There's only one alkyl group, this methyl group here, attached to this carbon. So that's called primary. So ethyl chloride is an example of a primary alkyl halide. If we look at isopropyl chloride down here, this is the carbon that's bonded to our halogen. And that carbon is bonded to two alkyl groups. So that's said to be a secondary alkyl halide."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So ethyl chloride is an example of a primary alkyl halide. If we look at isopropyl chloride down here, this is the carbon that's bonded to our halogen. And that carbon is bonded to two alkyl groups. So that's said to be a secondary alkyl halide. And let me draw in an example of another one here really fast. So for this compound, the carbon that is bonded to our halogen is bonded to three alkyl groups, so three methyl groups here. So that's called a tertiary alkyl halide."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So that's said to be a secondary alkyl halide. And let me draw in an example of another one here really fast. So for this compound, the carbon that is bonded to our halogen is bonded to three alkyl groups, so three methyl groups here. So that's called a tertiary alkyl halide. And the name of this compound is tert-butyl chloride. So that's the common name for it. And that's the one that you see used most of the time."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So that's called a tertiary alkyl halide. And the name of this compound is tert-butyl chloride. So that's the common name for it. And that's the one that you see used most of the time. For larger molecules, it's usually easier to use the IUPAC system. So let's name this compound using the IUPAC system. It's just like naming alkanes."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And that's the one that you see used most of the time. For larger molecules, it's usually easier to use the IUPAC system. So let's name this compound using the IUPAC system. It's just like naming alkanes. First, you want to find your longest carbon chain and name it. So that would be 1, 2, 3, 4, 5, 6, and 7. A 7-carbon alkane we know is heptane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "It's just like naming alkanes. First, you want to find your longest carbon chain and name it. So that would be 1, 2, 3, 4, 5, 6, and 7. A 7-carbon alkane we know is heptane. So I'm going to write in here heptane. Next, let's think about how to number this carbon chain. Do I want to number it from the left, or do I want to number it from the right?"}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "A 7-carbon alkane we know is heptane. So I'm going to write in here heptane. Next, let's think about how to number this carbon chain. Do I want to number it from the left, or do I want to number it from the right? So let's try numbering it from the left first. So 1, 2, 3, 4, 5, 6, and 7. If I'm numbering it from the left, I would have a bromine at 2, a methyl at 4, and another methyl at 5."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Do I want to number it from the left, or do I want to number it from the right? So let's try numbering it from the left first. So 1, 2, 3, 4, 5, 6, and 7. If I'm numbering it from the left, I would have a bromine at 2, a methyl at 4, and another methyl at 5. So 2, 4, and 5. Let's try numbering this compound from the right. So numbering our carbon chain from the right would mean this is carbon 1, 2, 3, 4, 5, 6, and 7."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "If I'm numbering it from the left, I would have a bromine at 2, a methyl at 4, and another methyl at 5. So 2, 4, and 5. Let's try numbering this compound from the right. So numbering our carbon chain from the right would mean this is carbon 1, 2, 3, 4, 5, 6, and 7. This is giving me a methyl group at 3, a methyl group at 4, and a bromine at 6. So 3, 4, and 6. Our goal is to give the lowest number possible to our first substituent."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So numbering our carbon chain from the right would mean this is carbon 1, 2, 3, 4, 5, 6, and 7. This is giving me a methyl group at 3, a methyl group at 4, and a bromine at 6. So 3, 4, and 6. Our goal is to give the lowest number possible to our first substituent. So on the left, that would be a 2 for the lowest number. And on the right, that would be a 3. Obviously, 2 is lower than 3."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Our goal is to give the lowest number possible to our first substituent. So on the left, that would be a 2 for the lowest number. And on the right, that would be a 3. Obviously, 2 is lower than 3. So we're going to stick with the numbering system on the left. So we've already identified our substituents. We have bromine, and we have two methyl groups."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Obviously, 2 is lower than 3. So we're going to stick with the numbering system on the left. So we've already identified our substituents. We have bromine, and we have two methyl groups. That would be dimethyl. And you also want to think about alphabetical order. So the bromine's going to go first."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "We have bromine, and we have two methyl groups. That would be dimethyl. And you also want to think about alphabetical order. So the bromine's going to go first. So just for spacing purposes, I'm going to put in dimethyl here. So dimethyl. I'm going to write that first."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So the bromine's going to go first. So just for spacing purposes, I'm going to put in dimethyl here. So dimethyl. I'm going to write that first. And that's at carbon 4 and 5. So the 4 and 5 just identify where those two methyl groups are. And then we have a bromine at carbon 2."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "I'm going to write that first. And that's at carbon 4 and 5. So the 4 and 5 just identify where those two methyl groups are. And then we have a bromine at carbon 2. So 2-bromo. So our full IUPAC name is 2-bromo-4,5-dimethyl-heptane. Let's name this compound."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And then we have a bromine at carbon 2. So 2-bromo. So our full IUPAC name is 2-bromo-4,5-dimethyl-heptane. Let's name this compound. So we approach it the same way. 1, 2, 3, 4, 5, 6, and 7. So again, heptane would be our parent name here."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Let's name this compound. So we approach it the same way. 1, 2, 3, 4, 5, 6, and 7. So again, heptane would be our parent name here. Now, numbering from the left or from the right. 1, 2, 3, 4, 5, 6, and 7. This gives us a methyl group at 2, another methyl group at 5, and a bromine at 5."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So again, heptane would be our parent name here. Now, numbering from the left or from the right. 1, 2, 3, 4, 5, 6, and 7. This gives us a methyl group at 2, another methyl group at 5, and a bromine at 5. So 2, 5, 5. For the one on the right, again, same compound. I'm just going to number this one starting from the right to see if this gives me a lower number or not."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "This gives us a methyl group at 2, another methyl group at 5, and a bromine at 5. So 2, 5, 5. For the one on the right, again, same compound. I'm just going to number this one starting from the right to see if this gives me a lower number or not. So this would give me a bromine at 3, a methyl at 3, and another methyl group at 6. So 3, 3, 6. Obviously, the one on the left would win again, because 2 is lower than 3."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "I'm just going to number this one starting from the right to see if this gives me a lower number or not. So this would give me a bromine at 3, a methyl at 3, and another methyl group at 6. So 3, 3, 6. Obviously, the one on the left would win again, because 2 is lower than 3. Our goal is to give the lowest number possible to our first substituent. So if we're numbering it from the left, let's see. We have a methyl group at 2 and a methyl group at 5."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Obviously, the one on the left would win again, because 2 is lower than 3. Our goal is to give the lowest number possible to our first substituent. So if we're numbering it from the left, let's see. We have a methyl group at 2 and a methyl group at 5. So it's like the previous situation. We have two methyl groups, so dimethyl. So let me write that in here."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "We have a methyl group at 2 and a methyl group at 5. So it's like the previous situation. We have two methyl groups, so dimethyl. So let me write that in here. So dimethyl. This time, we have a methyl group at 2 and 5. So I'm going to write in here 2 and 5."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So let me write that in here. So dimethyl. This time, we have a methyl group at 2 and 5. So I'm going to write in here 2 and 5. And then a bromine at 5. So it would be 5-bromo. So our full name would be 5-bromo-2,5-dimethylheptane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So I'm going to write in here 2 and 5. And then a bromine at 5. So it would be 5-bromo. So our full name would be 5-bromo-2,5-dimethylheptane. Now let's name this compound. So it's the same approach. Let's count up our carbons here in our longest carbon chain."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So our full name would be 5-bromo-2,5-dimethylheptane. Now let's name this compound. So it's the same approach. Let's count up our carbons here in our longest carbon chain. 1, 2, 3, 4, and 5. So a 5-carbon alkane is pentane. So I write in here pentane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Let's count up our carbons here in our longest carbon chain. 1, 2, 3, 4, and 5. So a 5-carbon alkane is pentane. So I write in here pentane. Next, I think about how do I number my carbon chain to give the lowest number possible to my first substituent? If I number it from the left, I give the bromine here a 1. So that's the best way to number it."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So I write in here pentane. Next, I think about how do I number my carbon chain to give the lowest number possible to my first substituent? If I number it from the left, I give the bromine here a 1. So that's the best way to number it. So that would be 1, 2, 3, 4, and 5. So I have a methyl group at 4, a chlorine at 3, and a bromine at 1. And we want to put these in alphabetical order."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So that's the best way to number it. So that would be 1, 2, 3, 4, and 5. So I have a methyl group at 4, a chlorine at 3, and a bromine at 1. And we want to put these in alphabetical order. So it's going to be the bromine first, then the chlorine, and then the methyl. So I'll go ahead and put the methyl in here. So this is at carbon 4, so 4-methyl."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "And we want to put these in alphabetical order. So it's going to be the bromine first, then the chlorine, and then the methyl. So I'll go ahead and put the methyl in here. So this is at carbon 4, so 4-methyl. Next, I have my chlorine at 3, so that would be 3-chloro. And finally, my bromine at 1, so 1-bromo. So the full name is 1-bromo, 3-chloro, 4-methylpentane."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So this is at carbon 4, so 4-methyl. Next, I have my chlorine at 3, so that would be 3-chloro. And finally, my bromine at 1, so 1-bromo. So the full name is 1-bromo, 3-chloro, 4-methylpentane. Now let's look at this compound. So we have 1, 2, 3, 4, 5, and 6. So 6 carbons in our chain."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So the full name is 1-bromo, 3-chloro, 4-methylpentane. Now let's look at this compound. So we have 1, 2, 3, 4, 5, and 6. So 6 carbons in our chain. A 6-carbon alkane is hexane. So I write in here hexane. Now I have two substituents."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So 6 carbons in our chain. A 6-carbon alkane is hexane. So I write in here hexane. Now I have two substituents. I have a bromine, and I have a methyl group. So let's number from the left and see what happens. 1, 2, 3, 4, 5, and 6."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "Now I have two substituents. I have a bromine, and I have a methyl group. So let's number from the left and see what happens. 1, 2, 3, 4, 5, and 6. That gives me a bromine at 2 and a methyl at 5. Now let's number this from the right. So this would be 1, 2, 3, 4, 5, 6."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "1, 2, 3, 4, 5, and 6. That gives me a bromine at 2 and a methyl at 5. Now let's number this from the right. So this would be 1, 2, 3, 4, 5, 6. This gives me a methyl at 2 and a bromine at 5. So just looking at numbers, we can't decide who wins here. So we have 2 versus 2, which is a tie, 5 versus 5, so that's a tie."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So this would be 1, 2, 3, 4, 5, 6. This gives me a methyl at 2 and a bromine at 5. So just looking at numbers, we can't decide who wins here. So we have 2 versus 2, which is a tie, 5 versus 5, so that's a tie. So the way to break the tie is to think about the alphabet for your substituents. So we have bromine versus methyl, so B versus M. Obviously, B comes before M in the alphabet. So the bromine's going to win."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So we have 2 versus 2, which is a tie, 5 versus 5, so that's a tie. So the way to break the tie is to think about the alphabet for your substituents. So we have bromine versus methyl, so B versus M. Obviously, B comes before M in the alphabet. So the bromine's going to win. We're going to give the bromine the lower number. And that, of course, is the example on the left where the bromine is coming off of carbon 2. So we're going to choose the system on the left or the way of numbering the carbon chain from the left, which means we have a methyl group at 5."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So the bromine's going to win. We're going to give the bromine the lower number. And that, of course, is the example on the left where the bromine is coming off of carbon 2. So we're going to choose the system on the left or the way of numbering the carbon chain from the left, which means we have a methyl group at 5. So 5-methylhexane and then bromine at 2, so 2-bromo. So 2-bromo, 5-methylhexane would be the name. So what do we do if we have some stereochemistry in our compound?"}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So we're going to choose the system on the left or the way of numbering the carbon chain from the left, which means we have a methyl group at 5. So 5-methylhexane and then bromine at 2, so 2-bromo. So 2-bromo, 5-methylhexane would be the name. So what do we do if we have some stereochemistry in our compound? Well, first, let's ignore the stereochemistry. And let's just name this how we've been naming the other molecules. We find our longest carbon chain, so it'll be 1, 2, 3, and 4."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So what do we do if we have some stereochemistry in our compound? Well, first, let's ignore the stereochemistry. And let's just name this how we've been naming the other molecules. We find our longest carbon chain, so it'll be 1, 2, 3, and 4. So that's butane. So I write in here butane. We want to number our chain to give the lowest number possible to our substituents."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "We find our longest carbon chain, so it'll be 1, 2, 3, and 4. So that's butane. So I write in here butane. We want to number our chain to give the lowest number possible to our substituents. That's 1, 2, 3, and 4. So our substituent is the bromine coming off of carbon 2. So 2-bromobutane would be the name."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "We want to number our chain to give the lowest number possible to our substituents. That's 1, 2, 3, and 4. So our substituent is the bromine coming off of carbon 2. So 2-bromobutane would be the name. But now we have to worry about our stereochemistry. We know that we have one chiral center. So here is our chiral center."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So 2-bromobutane would be the name. But now we have to worry about our stereochemistry. We know that we have one chiral center. So here is our chiral center. We know there's a hydrogen going away from us in space if there's a bromine coming out at us. And we need to assign priority to those four groups that are bonded to our chiral center. So I showed you how to do this in an earlier video, so I won't go into too much detail."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So here is our chiral center. We know there's a hydrogen going away from us in space if there's a bromine coming out at us. And we need to assign priority to those four groups that are bonded to our chiral center. So I showed you how to do this in an earlier video, so I won't go into too much detail. But let's assign priority really quickly. We know that bromine, right, with the highest atomic number is going to get the highest priority. So this group gets a number 1."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So I showed you how to do this in an earlier video, so I won't go into too much detail. But let's assign priority really quickly. We know that bromine, right, with the highest atomic number is going to get the highest priority. So this group gets a number 1. The ethyl group would get the second highest priority. The methyl group would get the third highest. And finally, this hydrogen going away from us is the lowest priority group."}, {"video_title": "Nomenclature and classification of alkyl halides.mp3", "Sentence": "So this group gets a number 1. The ethyl group would get the second highest priority. The methyl group would get the third highest. And finally, this hydrogen going away from us is the lowest priority group. So this gets a number 4. So we have 1, 2, and 3 going around clockwise. And we know that is R. So to complete the name, in parentheses here, I put R. And R-2-bromobutane is the IUPAC name."}, {"video_title": "Alkene stability.mp3", "Sentence": "So a disubstituted alkene is more stable than a monosubstituted. A trisubstituted is more stable than a disubstituted, and a tetrasubstituted is the most stable of them all. So on the left, we have a monosubstituted alkene. We have one alkyl group bonded to this carbon of our double bond. On the right is a disubstituted alkene. Now we have two alkyl groups. And the disubstituted alkene is more stable than the monosubstituted alkene."}, {"video_title": "Alkene stability.mp3", "Sentence": "We have one alkyl group bonded to this carbon of our double bond. On the right is a disubstituted alkene. Now we have two alkyl groups. And the disubstituted alkene is more stable than the monosubstituted alkene. To explain why, we need to go back to the idea of stability of carbocations that we saw in an earlier video. On the right, we have a secondary carbocation. So this positively charged carbon is directly bonded to two alkyl groups."}, {"video_title": "Alkene stability.mp3", "Sentence": "And the disubstituted alkene is more stable than the monosubstituted alkene. To explain why, we need to go back to the idea of stability of carbocations that we saw in an earlier video. On the right, we have a secondary carbocation. So this positively charged carbon is directly bonded to two alkyl groups. So this is a secondary carbocation. And the positively charged carbon is sp2 hybridized. So this carbon is sp2 hybridized."}, {"video_title": "Alkene stability.mp3", "Sentence": "So this positively charged carbon is directly bonded to two alkyl groups. So this is a secondary carbocation. And the positively charged carbon is sp2 hybridized. So this carbon is sp2 hybridized. And so the geometry around it is planar. So let's go back to the picture on the left, and we can see the geometry around that carbon is planar. And this positively charged sp2 hybridized carbon, let me just go ahead and mark this down here as being sp2 hybridized, should have an unhybridized p orbital."}, {"video_title": "Alkene stability.mp3", "Sentence": "So this carbon is sp2 hybridized. And so the geometry around it is planar. So let's go back to the picture on the left, and we can see the geometry around that carbon is planar. And this positively charged sp2 hybridized carbon, let me just go ahead and mark this down here as being sp2 hybridized, should have an unhybridized p orbital. And on the picture here, that's what the paddles are supposed to represent. This is our unhybridized p orbital. And to stabilize this positive charge on this carbon, we have two methyl groups."}, {"video_title": "Alkene stability.mp3", "Sentence": "And this positively charged sp2 hybridized carbon, let me just go ahead and mark this down here as being sp2 hybridized, should have an unhybridized p orbital. And on the picture here, that's what the paddles are supposed to represent. This is our unhybridized p orbital. And to stabilize this positive charge on this carbon, we have two methyl groups. So this methyl group and this methyl group are both electron donating through an effect that's called hyperconjugation. So here's one of the methyl groups on the left side. And notice the orientation of this methyl group."}, {"video_title": "Alkene stability.mp3", "Sentence": "And to stabilize this positive charge on this carbon, we have two methyl groups. So this methyl group and this methyl group are both electron donating through an effect that's called hyperconjugation. So here's one of the methyl groups on the left side. And notice the orientation of this methyl group. This carbon hydrogen bond is able to donate electron density into the p orbital on this sp2 hybridized carbon. And that stabilizes the carbocation. Same thing for this methyl group over here."}, {"video_title": "Alkene stability.mp3", "Sentence": "And notice the orientation of this methyl group. This carbon hydrogen bond is able to donate electron density into the p orbital on this sp2 hybridized carbon. And that stabilizes the carbocation. Same thing for this methyl group over here. It can donate some electron density into the p orbital on this sp2 hybridized carbon, stabilizing the positive charge. And that's an effect called hyperconjugation. This hydrogen here can't do anything because of the geometry."}, {"video_title": "Alkene stability.mp3", "Sentence": "Same thing for this methyl group over here. It can donate some electron density into the p orbital on this sp2 hybridized carbon, stabilizing the positive charge. And that's an effect called hyperconjugation. This hydrogen here can't do anything because of the geometry. So this bond doesn't have the right geometry to help stabilize the carbocation. So alkyl groups help to stabilize the positive charge on a carbocation. That's a similar idea with our alkene."}, {"video_title": "Alkene stability.mp3", "Sentence": "This hydrogen here can't do anything because of the geometry. So this bond doesn't have the right geometry to help stabilize the carbocation. So alkyl groups help to stabilize the positive charge on a carbocation. That's a similar idea with our alkene. So here we have our alkene. And this carbon is sp2 hybridized. And so we have these alkyl groups, which we know are electron donating."}, {"video_title": "Alkene stability.mp3", "Sentence": "That's a similar idea with our alkene. So here we have our alkene. And this carbon is sp2 hybridized. And so we have these alkyl groups, which we know are electron donating. And we know that they can donate some electron density to this sp2 hybridized carbon. Sp2 hybridized carbons are more electronegative than sp3 hybridized carbons. So donating electron density can help stabilize this sp2 hybridized carbon, which stabilizes the overall alkene."}, {"video_title": "Alkene stability.mp3", "Sentence": "And so we have these alkyl groups, which we know are electron donating. And we know that they can donate some electron density to this sp2 hybridized carbon. Sp2 hybridized carbons are more electronegative than sp3 hybridized carbons. So donating electron density can help stabilize this sp2 hybridized carbon, which stabilizes the overall alkene. And so that's why we see more substituted alkenes being more stable than less substituted alkenes. Now let's do some examples. Let's break these three alkenes in order of stability."}, {"video_title": "Alkene stability.mp3", "Sentence": "So donating electron density can help stabilize this sp2 hybridized carbon, which stabilizes the overall alkene. And so that's why we see more substituted alkenes being more stable than less substituted alkenes. Now let's do some examples. Let's break these three alkenes in order of stability. So let's start by classifying them according to their degrees of substitution. So we'll start with the first alkene right here. And we look for the two carbons across our double bond."}, {"video_title": "Alkene stability.mp3", "Sentence": "Let's break these three alkenes in order of stability. So let's start by classifying them according to their degrees of substitution. So we'll start with the first alkene right here. And we look for the two carbons across our double bond. And sometimes it helps to draw in hydrogens. So once we've done that, it's clear that we have two alkyl groups bonded to this carbon. So this must be a disubstituted alkene."}, {"video_title": "Alkene stability.mp3", "Sentence": "And we look for the two carbons across our double bond. And sometimes it helps to draw in hydrogens. So once we've done that, it's clear that we have two alkyl groups bonded to this carbon. So this must be a disubstituted alkene. So let me write that down here. Let's move on to the middle one. So here are the two carbons across our double bond."}, {"video_title": "Alkene stability.mp3", "Sentence": "So this must be a disubstituted alkene. So let me write that down here. Let's move on to the middle one. So here are the two carbons across our double bond. And again, I think it's often helpful to put in your hydrogen. So when you do that, it's clear you have only one alkyl group this time. And so this is a monosubstituted alkene."}, {"video_title": "Alkene stability.mp3", "Sentence": "So here are the two carbons across our double bond. And again, I think it's often helpful to put in your hydrogen. So when you do that, it's clear you have only one alkyl group this time. And so this is a monosubstituted alkene. And then let's look at the one on the right. So here are the two carbons across our double bond. And the carbon on the left would have only one hydrogen here."}, {"video_title": "Alkene stability.mp3", "Sentence": "And so this is a monosubstituted alkene. And then let's look at the one on the right. So here are the two carbons across our double bond. And the carbon on the left would have only one hydrogen here. So that's one, two, three alkyl groups. So this is a trisubstituted alkene. We know that in general, the more substituted alkenes are more stable than the less substituted alkenes."}, {"video_title": "Alkene stability.mp3", "Sentence": "And the carbon on the left would have only one hydrogen here. So that's one, two, three alkyl groups. So this is a trisubstituted alkene. We know that in general, the more substituted alkenes are more stable than the less substituted alkenes. So out of these three, the most substituted would be the trisubstituted. So this is the most stable of these three. So next would be the disubstituted alkene."}, {"video_title": "Alkene stability.mp3", "Sentence": "We know that in general, the more substituted alkenes are more stable than the less substituted alkenes. So out of these three, the most substituted would be the trisubstituted. So this is the most stable of these three. So next would be the disubstituted alkene. And finally, the least stable one would be the monosubstituted alkene. So this one would be the least stable. And the trisubstituted would be the most stable."}, {"video_title": "Alkene stability.mp3", "Sentence": "So next would be the disubstituted alkene. And finally, the least stable one would be the monosubstituted alkene. So this one would be the least stable. And the trisubstituted would be the most stable. Next, let's look at two isomers of each other. So we've talked about cis-2-butene and trans-2-butene. They are both disubstituted alkenes."}, {"video_title": "Alkene stability.mp3", "Sentence": "And the trisubstituted would be the most stable. Next, let's look at two isomers of each other. So we've talked about cis-2-butene and trans-2-butene. They are both disubstituted alkenes. So it turns out that trans-2-butene is the more stable of the two. So this one is more stable. And we can think about that in terms of steric hindrance."}, {"video_title": "Alkene stability.mp3", "Sentence": "They are both disubstituted alkenes. So it turns out that trans-2-butene is the more stable of the two. So this one is more stable. And we can think about that in terms of steric hindrance. If we look at cis-2-butene, we have these methyl groups, relatively bulky, and they would sterically interfere with each other if they're on the same side of the double bond. So this steric hindrance destabilizes the cis-2-butene molecules. We have increased steric hindrance, decreasing the stability of cis-2-butene."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "On the left, we have the dot structure for methane. And we've seen in an earlier video that this carbon is sp3 hybridized, which means that the atoms around that central carbon atom are arranged in a tetrahedral geometry. It's very difficult to see tetrahedral geometry on a two-dimensional Lewis dot structure. So it's much easier to see it over here on the right with the three-dimensional representation of the methane molecule. So if I'm trying to see the four sides of the tetrahedron, I can find my first side by connecting these hydrogen atoms like that. So there's the first side of my tetrahedron. And if I'm going to find the second side, I could connect these hydrogen atoms like that."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it's much easier to see it over here on the right with the three-dimensional representation of the methane molecule. So if I'm trying to see the four sides of the tetrahedron, I can find my first side by connecting these hydrogen atoms like that. So there's the first side of my tetrahedron. And if I'm going to find the second side, I could connect these hydrogen atoms like that. And there's my second side. And to find my last two sides, if I connect this hydrogen atom to this one down here, I can now see the four sides of my tetrahedron. We're also concerned with the bond angle."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And if I'm going to find the second side, I could connect these hydrogen atoms like that. And there's my second side. And to find my last two sides, if I connect this hydrogen atom to this one down here, I can now see the four sides of my tetrahedron. We're also concerned with the bond angle. So what is the bond angle? What is the angle between that top hydrogen, the central carbon, and this hydrogen over here on the left? It turns out that bond angle is 109.5 degrees."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We're also concerned with the bond angle. So what is the bond angle? What is the angle between that top hydrogen, the central carbon, and this hydrogen over here on the left? It turns out that bond angle is 109.5 degrees. And it's the same all the way around. So you could say that this angle is 109.5 degrees, or this angle back here. It's all the same."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It turns out that bond angle is 109.5 degrees. And it's the same all the way around. So you could say that this angle is 109.5 degrees, or this angle back here. It's all the same. And so an sp3 bond angle is 109.5. And the proof for this was shown to me by two of my students. So Anthony Grebe and Andrew Foster came up with a very nice proof to show that the bond angle of an sp3 hybridized carbon is 109.5 degrees."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It's all the same. And so an sp3 bond angle is 109.5. And the proof for this was shown to me by two of my students. So Anthony Grebe and Andrew Foster came up with a very nice proof to show that the bond angle of an sp3 hybridized carbon is 109.5 degrees. And what they did was they said, let's go ahead and take that tetrahedron. And let's go ahead and put on x, y, z axes. And let's put carbon at the center here."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So Anthony Grebe and Andrew Foster came up with a very nice proof to show that the bond angle of an sp3 hybridized carbon is 109.5 degrees. And what they did was they said, let's go ahead and take that tetrahedron. And let's go ahead and put on x, y, z axes. And let's put carbon at the center here. And we can choose any four points to represent the four hydrogen atoms of our tetrahedron if we satisfy two conditions. Each point that we choose for our hydrogens is equidistant from the other three points. And also, each point that we choose for our hydrogens is equidistant from the central carbon atom itself."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And let's put carbon at the center here. And we can choose any four points to represent the four hydrogen atoms of our tetrahedron if we satisfy two conditions. Each point that we choose for our hydrogens is equidistant from the other three points. And also, each point that we choose for our hydrogens is equidistant from the central carbon atom itself. And if you fulfill those two criteria, you guarantee that the points that you choose form a tetrahedron. And so here we have the tetrahedron on our axes. And let's go ahead and look at the first point, so this point right here."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And also, each point that we choose for our hydrogens is equidistant from the central carbon atom itself. And if you fulfill those two criteria, you guarantee that the points that you choose form a tetrahedron. And so here we have the tetrahedron on our axes. And let's go ahead and look at the first point, so this point right here. And they chose this point to be at square root of 2, 1, and 0, meaning positive square root of 2 on the x-axis, positive 1 on the y-axis, and 0 on the z-axis. And then this point over here on the left, they were very clever and said this point is going to be in the same plane. So this point on the left is in the same plane as the point we just talked about, the xy-plane."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and look at the first point, so this point right here. And they chose this point to be at square root of 2, 1, and 0, meaning positive square root of 2 on the x-axis, positive 1 on the y-axis, and 0 on the z-axis. And then this point over here on the left, they were very clever and said this point is going to be in the same plane. So this point on the left is in the same plane as the point we just talked about, the xy-plane. And therefore, the coordinates for that point would be negative square root of 2, 1, and 0. We go to the hydrogen down here. So this point of our tetrahedron is located at 0, negative 1, and square root of 2."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this point on the left is in the same plane as the point we just talked about, the xy-plane. And therefore, the coordinates for that point would be negative square root of 2, 1, and 0. We go to the hydrogen down here. So this point of our tetrahedron is located at 0, negative 1, and square root of 2. And then finally, this point going away from us right here would be at 0, negative 1, and negative square root of 2. So once again, you could choose any points that you want as long as you meet that criteria. And orienting the molecule in this way allows us to find this bond angle."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this point of our tetrahedron is located at 0, negative 1, and square root of 2. And then finally, this point going away from us right here would be at 0, negative 1, and negative square root of 2. So once again, you could choose any points that you want as long as you meet that criteria. And orienting the molecule in this way allows us to find this bond angle. So this is the bond angle that we are going for. And we don't know that bond angle yet. But we can figure out this angle right here."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And orienting the molecule in this way allows us to find this bond angle. So this is the bond angle that we are going for. And we don't know that bond angle yet. But we can figure out this angle right here. So I'm going to call this theta for this triangle that's formed. And I know that this x distance down here is positive square root of 2. And then we go up 1 on the y-axis, and then 0 on the z-axis."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But we can figure out this angle right here. So I'm going to call this theta for this triangle that's formed. And I know that this x distance down here is positive square root of 2. And then we go up 1 on the y-axis, and then 0 on the z-axis. So I can find out what theta is because I know that tan of theta is equal to opposite over adjacent. So for this triangle I have here, the opposite side would be 1, and the adjacent side would be square root of 2. So to find theta, all I have to do is take inverse tan."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then we go up 1 on the y-axis, and then 0 on the z-axis. So I can find out what theta is because I know that tan of theta is equal to opposite over adjacent. So for this triangle I have here, the opposite side would be 1, and the adjacent side would be square root of 2. So to find theta, all I have to do is take inverse tan. So I take inverse tan of 1 over square root of 2 on my calculator. And I get 35.26 degrees. So I know that theta, this angle right in here, is 35.26 degrees."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So to find theta, all I have to do is take inverse tan. So I take inverse tan of 1 over square root of 2 on my calculator. And I get 35.26 degrees. So I know that theta, this angle right in here, is 35.26 degrees. And therefore, this angle is also 35.26 degrees. So this is also going to be theta in here. And if I want to find my bond angle in here, I know that those three angles have to add up to equal 180 degrees since they're all in the same plane here."}, {"video_title": "Tetrahedral bond angle proof Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I know that theta, this angle right in here, is 35.26 degrees. And therefore, this angle is also 35.26 degrees. So this is also going to be theta in here. And if I want to find my bond angle in here, I know that those three angles have to add up to equal 180 degrees since they're all in the same plane here. So to find my bond angle, all I have to do is take 180 degrees. And from that, we're going to subtract 2 times 35.26 degrees. And we, of course, come out with a bond angle of 109.5 degrees."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm going to start with gravity. And it might surprise some of you that gravity is actually the weakest of the four fundamental forces. And that's surprising because you say, wow, that's what keeps us glued. Not glued, but it keeps us from jumping off the planet. It's what keeps the moon in orbit around the Earth. The Earth in orbit around the sun. The sun in orbit around the center of the Milky Way galaxy."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Not glued, but it keeps us from jumping off the planet. It's what keeps the moon in orbit around the Earth. The Earth in orbit around the sun. The sun in orbit around the center of the Milky Way galaxy. So it's a little bit surprising that it's actually the weakest of the forces. And that starts to make sense when you actually think about things on maybe more of a human scale, or a molecular scale, or even an atomic scale. Even on a human scale, your computer monitor and you have some type of gravitational attraction."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The sun in orbit around the center of the Milky Way galaxy. So it's a little bit surprising that it's actually the weakest of the forces. And that starts to make sense when you actually think about things on maybe more of a human scale, or a molecular scale, or even an atomic scale. Even on a human scale, your computer monitor and you have some type of gravitational attraction. But you don't notice it. Or your cell phone and your wallet. There's gravitational attraction, but you don't see them being drawn to each other the way you might see two magnets drawn to each other or repelled from each other."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Even on a human scale, your computer monitor and you have some type of gravitational attraction. But you don't notice it. Or your cell phone and your wallet. There's gravitational attraction, but you don't see them being drawn to each other the way you might see two magnets drawn to each other or repelled from each other. And if you go to even a smaller scale, you'll see that it matters even less. We never even talk about gravity in chemistry. Although the gravity is there."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's gravitational attraction, but you don't see them being drawn to each other the way you might see two magnets drawn to each other or repelled from each other. And if you go to even a smaller scale, you'll see that it matters even less. We never even talk about gravity in chemistry. Although the gravity is there. But at those scales, the other forces really, really, really start to dominate. So gravities are weakest. So if we move up a little bit from that, we get, and this is maybe the hardest force for us to visualize, or at least the least intuitive force for me, is actually the weak force."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Although the gravity is there. But at those scales, the other forces really, really, really start to dominate. So gravities are weakest. So if we move up a little bit from that, we get, and this is maybe the hardest force for us to visualize, or at least the least intuitive force for me, is actually the weak force. The weak, sometimes called the weak interaction. And it's what's responsible for radioactive decay. In particular, beta minus and beta plus decay."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we move up a little bit from that, we get, and this is maybe the hardest force for us to visualize, or at least the least intuitive force for me, is actually the weak force. The weak, sometimes called the weak interaction. And it's what's responsible for radioactive decay. In particular, beta minus and beta plus decay. And just to give you an example of the actual weak interaction, if I had some cesium, 137, 137 means it has 137 nucleons. A nucleon is either a proton or a neutron. You add up the protons and neutrons of cesium, you get 137."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In particular, beta minus and beta plus decay. And just to give you an example of the actual weak interaction, if I had some cesium, 137, 137 means it has 137 nucleons. A nucleon is either a proton or a neutron. You add up the protons and neutrons of cesium, you get 137. And it is cesium because it has exactly 55 protons. Now, the weak interaction is what's responsible for one of the neutrons, essentially one of its quarks flipping and turning into a proton. And I'm not going to go into detail of what a quark is and all of that, and the math can get pretty hairy."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You add up the protons and neutrons of cesium, you get 137. And it is cesium because it has exactly 55 protons. Now, the weak interaction is what's responsible for one of the neutrons, essentially one of its quarks flipping and turning into a proton. And I'm not going to go into detail of what a quark is and all of that, and the math can get pretty hairy. But I just want to give you an example of what the weak interaction does. So if one of these neutrons turns into a proton, then we're going to have one extra proton, but we're going to have the same number of nucleons. Instead of an extra neutron here, you now have an extra proton here."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm not going to go into detail of what a quark is and all of that, and the math can get pretty hairy. But I just want to give you an example of what the weak interaction does. So if one of these neutrons turns into a proton, then we're going to have one extra proton, but we're going to have the same number of nucleons. Instead of an extra neutron here, you now have an extra proton here. And so now this is a different atom. It is now barium. And in that flipping, it will actually emit an electron and an anti-electron neutrino."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Instead of an extra neutron here, you now have an extra proton here. And so now this is a different atom. It is now barium. And in that flipping, it will actually emit an electron and an anti-electron neutrino. And I'm not going to go into the details of what an anti-electron neutrino is. These are fundamental particles. But this is just what the weak interaction is."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And in that flipping, it will actually emit an electron and an anti-electron neutrino. And I'm not going to go into the details of what an anti-electron neutrino is. These are fundamental particles. But this is just what the weak interaction is. It's not something that's completely obvious to us. It's not kind of this traditional thing pulling or pushing away from each other like we associate with the other forces. Now the next strongest force, and just to give a sense of how weak gravity is even relative to the weak interaction, the weak interaction is 10 to the 25th times the strength of gravity."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is just what the weak interaction is. It's not something that's completely obvious to us. It's not kind of this traditional thing pulling or pushing away from each other like we associate with the other forces. Now the next strongest force, and just to give a sense of how weak gravity is even relative to the weak interaction, the weak interaction is 10 to the 25th times the strength of gravity. And you might be thinking, if this is so strong, how come this doesn't operate on planets or us relative to the Earth, or why doesn't this apply to intergalactic distances the way gravity does? And the reason is the weak interaction really applies to very small distances. So it can be much stronger than gravity, but only over very, very, and it really only applies on the subatomic scale."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now the next strongest force, and just to give a sense of how weak gravity is even relative to the weak interaction, the weak interaction is 10 to the 25th times the strength of gravity. And you might be thinking, if this is so strong, how come this doesn't operate on planets or us relative to the Earth, or why doesn't this apply to intergalactic distances the way gravity does? And the reason is the weak interaction really applies to very small distances. So it can be much stronger than gravity, but only over very, very, and it really only applies on the subatomic scale. You go anything beyond that, it kind of disappears as an actual force, as an actual interaction. Now the next force up the hierarchy, which is one that we are more familiar with, it is something, it's what actually dominates most of the chemistry that we deal with, and electromagnetism that we deal with, and that's the electromagnetic force. Let me write it in magenta."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it can be much stronger than gravity, but only over very, very, and it really only applies on the subatomic scale. You go anything beyond that, it kind of disappears as an actual force, as an actual interaction. Now the next force up the hierarchy, which is one that we are more familiar with, it is something, it's what actually dominates most of the chemistry that we deal with, and electromagnetism that we deal with, and that's the electromagnetic force. Let me write it in magenta. Electromagnetic force. And just to give a sense, this is 10 to the 36th times the strength of gravity, so it kind of puts the weak force in its place. It's 10 to the 12th times stronger than the weak force."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me write it in magenta. Electromagnetic force. And just to give a sense, this is 10 to the 36th times the strength of gravity, so it kind of puts the weak force in its place. It's 10 to the 12th times stronger than the weak force. These are huge numbers that we're talking about, either this relative to that, or even this relative to gravity. And so you might be saying, well, the electromagnetic force, that's unbelievably strong. Why doesn't that apply over these kind of macro scales like gravity?"}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's 10 to the 12th times stronger than the weak force. These are huge numbers that we're talking about, either this relative to that, or even this relative to gravity. And so you might be saying, well, the electromagnetic force, that's unbelievably strong. Why doesn't that apply over these kind of macro scales like gravity? Let me write there, macro scales. Why doesn't it apply to macro scales? And actually, there's nothing about the electromagnetic force why it can't, or it actually does, apply over large distances."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Why doesn't that apply over these kind of macro scales like gravity? Let me write there, macro scales. Why doesn't it apply to macro scales? And actually, there's nothing about the electromagnetic force why it can't, or it actually does, apply over large distances. The reality, though, is you don't have these huge concentrations of either Coulomb charges or magnetism, the way you do mass. So the mass, since you have such huge concentrations, it can operate over huge, huge distances, even though it's way, way, way weaker than the electromagnetic force. The electromagnetic force, what happens is because it's both attractive and repulsive, it tends to kind of sort itself out so you don't have these huge, huge, huge concentrations of charge."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And actually, there's nothing about the electromagnetic force why it can't, or it actually does, apply over large distances. The reality, though, is you don't have these huge concentrations of either Coulomb charges or magnetism, the way you do mass. So the mass, since you have such huge concentrations, it can operate over huge, huge distances, even though it's way, way, way weaker than the electromagnetic force. The electromagnetic force, what happens is because it's both attractive and repulsive, it tends to kind of sort itself out so you don't have these huge, huge, huge concentrations of charge. Now, the other thing you might be wondering about is why is it called the electromagnetic force? In our everyday life, there's things like the Coulomb force or the electrostatic force, which we're familiar with. Positive charges or like charges want to repel."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The electromagnetic force, what happens is because it's both attractive and repulsive, it tends to kind of sort itself out so you don't have these huge, huge, huge concentrations of charge. Now, the other thing you might be wondering about is why is it called the electromagnetic force? In our everyday life, there's things like the Coulomb force or the electrostatic force, which we're familiar with. Positive charges or like charges want to repel. If both of these were negative, the same thing would be happening. And different charges like to attract. We've seen this multiple times."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Positive charges or like charges want to repel. If both of these were negative, the same thing would be happening. And different charges like to attract. We've seen this multiple times. This is the Coulomb force or the electrostatic force. And then on the other side of the word, I guess, you have the magnetic part. And magnets, you've played with magnets on your fridge."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We've seen this multiple times. This is the Coulomb force or the electrostatic force. And then on the other side of the word, I guess, you have the magnetic part. And magnets, you've played with magnets on your fridge. If they're the same side of the magnet, they're going to repel each other. If they're the opposite sides, opposite poles, they're going to attract each other. So why is it called one force?"}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And magnets, you've played with magnets on your fridge. If they're the same side of the magnet, they're going to repel each other. If they're the opposite sides, opposite poles, they're going to attract each other. So why is it called one force? And it's called one force. And once again, I'm not going to go into detail here. It's called one force because it turns out that the Coulomb force, the electrostatic force, and the magnetic force are actually the same thing viewed in different frames of references."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So why is it called one force? And it's called one force. And once again, I'm not going to go into detail here. It's called one force because it turns out that the Coulomb force, the electrostatic force, and the magnetic force are actually the same thing viewed in different frames of references. So I won't go into a lot of detail, but just keep that in the back of your mind, that they are connected. In a future video, I'll go more into the intuition of how they are connected. It's more apparent when the charges are moving at relativistic frames."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's called one force because it turns out that the Coulomb force, the electrostatic force, and the magnetic force are actually the same thing viewed in different frames of references. So I won't go into a lot of detail, but just keep that in the back of your mind, that they are connected. In a future video, I'll go more into the intuition of how they are connected. It's more apparent when the charges are moving at relativistic frames. Well, I won't go into a lot of detail there. But just keep in mind that they really are the same force just viewed from different frames of reference. Now, the strongest of the force is probably the best named of them all."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's more apparent when the charges are moving at relativistic frames. Well, I won't go into a lot of detail there. But just keep in mind that they really are the same force just viewed from different frames of reference. Now, the strongest of the force is probably the best named of them all. And that's the strong force. And although you probably haven't seen this yet in chemistry classes, it actually applies very strongly in chemistry. Because from the get-go, when you first learn about atoms, let me draw a helium atom."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, the strongest of the force is probably the best named of them all. And that's the strong force. And although you probably haven't seen this yet in chemistry classes, it actually applies very strongly in chemistry. Because from the get-go, when you first learn about atoms, let me draw a helium atom. A helium atom has two protons in its nucleus, and it has two neutrons, and then it also has two electrons circulating around. So it has an electron. I could draw the electrons as much smaller."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because from the get-go, when you first learn about atoms, let me draw a helium atom. A helium atom has two protons in its nucleus, and it has two neutrons, and then it also has two electrons circulating around. So it has an electron. I could draw the electrons as much smaller. Well, I won't try to do anything in relative size, but it has two electrons floating around. And one question that may or may not have jumped into your mind when you first saw this model of an atom is like, well, I see why the electrons are attracted to the nucleus. It has a negative Coulomb charge."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I could draw the electrons as much smaller. Well, I won't try to do anything in relative size, but it has two electrons floating around. And one question that may or may not have jumped into your mind when you first saw this model of an atom is like, well, I see why the electrons are attracted to the nucleus. It has a negative Coulomb charge. The nucleus has a net positive Coulomb charge. But what's not so obvious, and what tends not to sometimes be explained in chemistry class, is these two positive charges are sitting right next to each other. If the electromagnetic force was the only force in play, if the Coulomb force was the only thing happening, these guys would just run away from each other."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It has a negative Coulomb charge. The nucleus has a net positive Coulomb charge. But what's not so obvious, and what tends not to sometimes be explained in chemistry class, is these two positive charges are sitting right next to each other. If the electromagnetic force was the only force in play, if the Coulomb force was the only thing happening, these guys would just run away from each other. They would repel each other. And so the only reason why they're able to stick to each other is that there's an even stronger force than the electromagnetic force operating at these very, very, very small distances. So if you get two of these protons close enough together, and the strong force only applies over very, very, very small distances, subatomic, or I should even say subnucleic distances, then the strong interaction comes into play."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If the electromagnetic force was the only force in play, if the Coulomb force was the only thing happening, these guys would just run away from each other. They would repel each other. And so the only reason why they're able to stick to each other is that there's an even stronger force than the electromagnetic force operating at these very, very, very small distances. So if you get two of these protons close enough together, and the strong force only applies over very, very, very small distances, subatomic, or I should even say subnucleic distances, then the strong interaction comes into play. So then you have the strong interaction actually keeping these charges together. And once again, just to keep it in mind relative to gravity, it is 10 to the 38th times the strength of gravity. Or it's about 100 times stronger than the electromagnetic force."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you get two of these protons close enough together, and the strong force only applies over very, very, very small distances, subatomic, or I should even say subnucleic distances, then the strong interaction comes into play. So then you have the strong interaction actually keeping these charges together. And once again, just to keep it in mind relative to gravity, it is 10 to the 38th times the strength of gravity. Or it's about 100 times stronger than the electromagnetic force. So once again, the reason why you don't see the strong force, which is the strongest of all the forces, or the weak interaction applying over huge scales, is that their strength dies off super, super fast, even when you start going to large radius nucleuses of atoms, the strength starts to die off, especially for the strong force. The reason why you don't see the electromagnetic force operating over large distances, even though in theory it can, like gravity, is that you don't see the type of charge concentrations the way you see mass concentrations in the universe. Because the charge concentrations tend to sort them out."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or it's about 100 times stronger than the electromagnetic force. So once again, the reason why you don't see the strong force, which is the strongest of all the forces, or the weak interaction applying over huge scales, is that their strength dies off super, super fast, even when you start going to large radius nucleuses of atoms, the strength starts to die off, especially for the strong force. The reason why you don't see the electromagnetic force operating over large distances, even though in theory it can, like gravity, is that you don't see the type of charge concentrations the way you see mass concentrations in the universe. Because the charge concentrations tend to sort them out. They start to equalize. If I have some positive, a huge positive charge there and a huge negative charge there, they will attract each other and then become essentially a big lump of neutral charge. And once they're a big lump of neutral charge, they won't interact with anything else."}, {"video_title": "Four fundamental forces Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because the charge concentrations tend to sort them out. They start to equalize. If I have some positive, a huge positive charge there and a huge negative charge there, they will attract each other and then become essentially a big lump of neutral charge. And once they're a big lump of neutral charge, they won't interact with anything else. In gravity, if you have one mass and another mass and they attract each other, then you have another mass that's even better at attracting each other, at other masses. And so it'll keep attracting things to it. So it kind of snowballs the process."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "If HX donates this proton, we're left with the conjugate base, which is X minus. And we saw from the previous video, the more stable the conjugate base, the more likely HX is to donate a proton. So the more stable the conjugate base, the stronger the acid. If we take a look at these four binary acids here, we have hydrofluoric acid, with a pKa value of approximately positive three, hydrochloric acid, with an approximate pKa of negative seven, hydrobromic acid at negative nine, and hydroiodic acid at negative 10. We know the lower the pKa value, the stronger the acid. So as we go down this way, we are decreasing in pKa values, and therefore we are increasing in acid strength. So we're increasing in acidity."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "If we take a look at these four binary acids here, we have hydrofluoric acid, with a pKa value of approximately positive three, hydrochloric acid, with an approximate pKa of negative seven, hydrobromic acid at negative nine, and hydroiodic acid at negative 10. We know the lower the pKa value, the stronger the acid. So as we go down this way, we are decreasing in pKa values, and therefore we are increasing in acid strength. So we're increasing in acidity. Therefore, hydroiodic acid is the strongest acid out of these four, because hydroiodic acid has the lowest value for the pKa. If hydroiodic acid is our strongest acid, the conjugate base must be the most stable. So the conjugate base to hydroiodic acid would be the iodide anion, I minus."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So we're increasing in acidity. Therefore, hydroiodic acid is the strongest acid out of these four, because hydroiodic acid has the lowest value for the pKa. If hydroiodic acid is our strongest acid, the conjugate base must be the most stable. So the conjugate base to hydroiodic acid would be the iodide anion, I minus. So here we have all the different conjugate bases. We have the fluoride anion, which is the conjugate base to HF. We have the chloride anion, which is the conjugate base to HCl."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So the conjugate base to hydroiodic acid would be the iodide anion, I minus. So here we have all the different conjugate bases. We have the fluoride anion, which is the conjugate base to HF. We have the chloride anion, which is the conjugate base to HCl. We have the bromide anion, which is the conjugate base to HBr, and of course, again, the iodide anion, the conjugate base to HI. The iodide anion must be the most stable, because HI is our strongest acid. So we can explain the stability of this conjugate base in terms of the size of the ion."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "We have the chloride anion, which is the conjugate base to HCl. We have the bromide anion, which is the conjugate base to HBr, and of course, again, the iodide anion, the conjugate base to HI. The iodide anion must be the most stable, because HI is our strongest acid. So we can explain the stability of this conjugate base in terms of the size of the ion. Remember, as you go down a group on the periodic table, you would increase in the size of the anion. So let me go ahead and write anion instead of ion here. So we increase in the size, in the size of the anion."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So we can explain the stability of this conjugate base in terms of the size of the ion. Remember, as you go down a group on the periodic table, you would increase in the size of the anion. So let me go ahead and write anion instead of ion here. So we increase in the size, in the size of the anion. So why does that help to stabilize the conjugate base? Well, we need to think about this negative charge here. So we have a negative charge in the iodide anion, and we have this charge spread out over a large volume of space, and that makes the anion more stable."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So we increase in the size, in the size of the anion. So why does that help to stabilize the conjugate base? Well, we need to think about this negative charge here. So we have a negative charge in the iodide anion, and we have this charge spread out over a large volume of space, and that makes the anion more stable. So remember, electrons repel each other, but if you can spread out the negative charge over a large amount of space, then you can better stabilize that negative charge. So this is more stable than, for example, the fluoride anion. The fluoride anion has a negative charge that's concentrated in a small volume of space, and so that destabilizes this anion compared to the iodide anion, and the iodide anion becomes the most stable, and therefore, HI is the most likely to donate a proton, and therefore, HI is our strongest acid out of these four."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So we have a negative charge in the iodide anion, and we have this charge spread out over a large volume of space, and that makes the anion more stable. So remember, electrons repel each other, but if you can spread out the negative charge over a large amount of space, then you can better stabilize that negative charge. So this is more stable than, for example, the fluoride anion. The fluoride anion has a negative charge that's concentrated in a small volume of space, and so that destabilizes this anion compared to the iodide anion, and the iodide anion becomes the most stable, and therefore, HI is the most likely to donate a proton, and therefore, HI is our strongest acid out of these four. Notice this is different from the previous video where we talked about electronegativity. There, we were comparing elements in the same period on the periodic table. So we were moving horizontally across our periodic table this way, and in that video, the fluoride anion was the most stable one because fluorines are most electronegative element, and therefore, best able to stabilize a negative charge."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "The fluoride anion has a negative charge that's concentrated in a small volume of space, and so that destabilizes this anion compared to the iodide anion, and the iodide anion becomes the most stable, and therefore, HI is the most likely to donate a proton, and therefore, HI is our strongest acid out of these four. Notice this is different from the previous video where we talked about electronegativity. There, we were comparing elements in the same period on the periodic table. So we were moving horizontally across our periodic table this way, and in that video, the fluoride anion was the most stable one because fluorines are most electronegative element, and therefore, best able to stabilize a negative charge. But as you go down a group on the periodic table, your electronegativity decreases. So that can't be the dominant trend because if your electronegativity decreases as you go down, just thinking about electronegativity, that would predict HF to be the strongest acid, and that's not what we observe. So as you go down a group on the periodic table, it's the size of the anion that determines the stability of the conjugate base."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So we were moving horizontally across our periodic table this way, and in that video, the fluoride anion was the most stable one because fluorines are most electronegative element, and therefore, best able to stabilize a negative charge. But as you go down a group on the periodic table, your electronegativity decreases. So that can't be the dominant trend because if your electronegativity decreases as you go down, just thinking about electronegativity, that would predict HF to be the strongest acid, and that's not what we observe. So as you go down a group on the periodic table, it's the size of the anion that determines the stability of the conjugate base. So the larger the anion, the better it is to stabilize a negative charge, and therefore, the more stable the conjugate base. The more stable the conjugate base, the more likely HX is to donate a proton, and therefore, the stronger the acid. Another very important factor to think about is the strength of the bond."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So as you go down a group on the periodic table, it's the size of the anion that determines the stability of the conjugate base. So the larger the anion, the better it is to stabilize a negative charge, and therefore, the more stable the conjugate base. The more stable the conjugate base, the more likely HX is to donate a proton, and therefore, the stronger the acid. Another very important factor to think about is the strength of the bond. We've already said that hydroiodic acid is our strongest acid with the lowest pKa value. So this bond right here must be the easiest to break. If it's easy to break this bond, that makes it easy to donate this proton."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "Another very important factor to think about is the strength of the bond. We've already said that hydroiodic acid is our strongest acid with the lowest pKa value. So this bond right here must be the easiest to break. If it's easy to break this bond, that makes it easy to donate this proton. So we can get an idea of the bond strengths for our binary acids by looking at bond dissociation energies. So you could also call these bond energies or bond enthalpies. So remember, bond dissociation energy measures the amount of energy that's needed to break a bond in the gaseous state."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "If it's easy to break this bond, that makes it easy to donate this proton. So we can get an idea of the bond strengths for our binary acids by looking at bond dissociation energies. So you could also call these bond energies or bond enthalpies. So remember, bond dissociation energy measures the amount of energy that's needed to break a bond in the gaseous state. So if we look at our hydrogen halides and we think about our bonds, notice what happens to the bond energy. It's the hardest to break this bond, the bond between hydrogen and fluorine. This takes the most energy to break this bond."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "So remember, bond dissociation energy measures the amount of energy that's needed to break a bond in the gaseous state. So if we look at our hydrogen halides and we think about our bonds, notice what happens to the bond energy. It's the hardest to break this bond, the bond between hydrogen and fluorine. This takes the most energy to break this bond. And as you go down, we see we decrease in bond dissociation energy. So it only takes 299 kilojoules per mole to break this bond between H and I. And I should say these are approximate bond energies and you'll see several different values in different textbooks."}, {"video_title": "Acid strength, anion size, and bond energy Chemistry Khan Academy.mp3", "Sentence": "This takes the most energy to break this bond. And as you go down, we see we decrease in bond dissociation energy. So it only takes 299 kilojoules per mole to break this bond between H and I. And I should say these are approximate bond energies and you'll see several different values in different textbooks. So if the HI bond is the easiest to break, that means when you're thinking about the acids, this bond is the easiest to break, therefore it's the most likely to donate a proton. And therefore it has the lowest value for the pKa. And hydroiodic acid is the strongest out of these four binary acids."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the computer calculates the area under the signal, so for example, for this signal, the area under the signal is calculated by the computer, and gives us this number. The computer gives us 57.9. For this signal, the computer gives us 23.1, and finally for this signal, we get an integration value of 35.4. Let's go back up here to the dot structure of benzyl acetate, and let's see how many protons that we need to account for in our proton NMR spectrum. This carbon right here has three protons. Let me go ahead and draw those protons in. This carbon has two protons on it, and that's five so far."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's go back up here to the dot structure of benzyl acetate, and let's see how many protons that we need to account for in our proton NMR spectrum. This carbon right here has three protons. Let me go ahead and draw those protons in. This carbon has two protons on it, and that's five so far. And then on our ring, we have five more protons. So going around the ring here, we have five more, for a total of 10. So we need to account for 10 protons in our spectrum."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This carbon has two protons on it, and that's five so far. And then on our ring, we have five more protons. So going around the ring here, we have five more, for a total of 10. So we need to account for 10 protons in our spectrum. All right, so going back to the integration values, you find the smallest integration value. So out of those three numbers, 23.1 is the smallest integration value. And we're going to divide all three integration values by the smallest one."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we need to account for 10 protons in our spectrum. All right, so going back to the integration values, you find the smallest integration value. So out of those three numbers, 23.1 is the smallest integration value. And we're going to divide all three integration values by the smallest one. And we'll start with 57.9. So 57.9 divided by 23.1. So let's get out the calculator here."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to divide all three integration values by the smallest one. And we'll start with 57.9. So 57.9 divided by 23.1. So let's get out the calculator here. 57.9, divide that by 23.1, and we get 2.5. So I'll write 2.5 right here. 23.1 divided by 23.1 is obviously equal to one."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's get out the calculator here. 57.9, divide that by 23.1, and we get 2.5. So I'll write 2.5 right here. 23.1 divided by 23.1 is obviously equal to one. And then finally, 35.4, we need to divide that by the smallest integration value. So 35.4 divided by 23.1 gives us about 1.5. So we have 1.5 here."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "23.1 divided by 23.1 is obviously equal to one. And then finally, 35.4, we need to divide that by the smallest integration value. So 35.4 divided by 23.1 gives us about 1.5. So we have 1.5 here. And this gives us a ratio of the protons that are giving these three signals. So the ratio would be 2.5 to one to 1.5. But you can't have 2.5 protons, right?"}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have 1.5 here. And this gives us a ratio of the protons that are giving these three signals. So the ratio would be 2.5 to one to 1.5. But you can't have 2.5 protons, right? You can't have half a proton here. And so those aren't the exact number of protons, right? We need to account for 10 protons in our molecule."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But you can't have 2.5 protons, right? You can't have half a proton here. And so those aren't the exact number of protons, right? We need to account for 10 protons in our molecule. And so if you think about it, if you multiply these numbers by two, right, then that gives us what we want. Because if you multiply 2.5 by two, that gives us five. If you multiply one by two, that gives us two."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We need to account for 10 protons in our molecule. And so if you think about it, if you multiply these numbers by two, right, then that gives us what we want. Because if you multiply 2.5 by two, that gives us five. If you multiply one by two, that gives us two. If you multiply 1.5 by two, that gives us three. And obviously, five plus two plus three gives us 10. And 10 protons is how many protons that we need to account for for our molecule."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you multiply one by two, that gives us two. If you multiply 1.5 by two, that gives us three. And obviously, five plus two plus three gives us 10. And 10 protons is how many protons that we need to account for for our molecule. And so therefore, this signal right here corresponds to five protons. This signal corresponds to two protons. And this signal corresponds to three protons."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And 10 protons is how many protons that we need to account for for our molecule. And so therefore, this signal right here corresponds to five protons. This signal corresponds to two protons. And this signal corresponds to three protons. So if we go back up here to our dot structure, and I look at these protons, right, so we have three equivalent protons. The chemical shift for these protons, we're next to a carbonyl, so we would expect the chemical shift to be just past two. And that's, of course, what we see right here."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this signal corresponds to three protons. So if we go back up here to our dot structure, and I look at these protons, right, so we have three equivalent protons. The chemical shift for these protons, we're next to a carbonyl, so we would expect the chemical shift to be just past two. And that's, of course, what we see right here. So the shift is just past two. This signal represents three protons, and it's these three protons right here. All right, next, let's look at these two protons."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that's, of course, what we see right here. So the shift is just past two. This signal represents three protons, and it's these three protons right here. All right, next, let's look at these two protons. So these two protons are next to an oxygen, right? So the oxygen deshields that. Those two protons are also next to this benzene ring over here."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, next, let's look at these two protons. So these two protons are next to an oxygen, right? So the oxygen deshields that. Those two protons are also next to this benzene ring over here. So we would expect a higher chemical shift, right? And we have two protons, and of course, it's this signal which corresponds to two protons. Finally, we have five nearly equivalent protons on our ring, so they might not be exactly the same."}, {"video_title": "Integration Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Those two protons are also next to this benzene ring over here. So we would expect a higher chemical shift, right? And we have two protons, and of course, it's this signal which corresponds to two protons. Finally, we have five nearly equivalent protons on our ring, so they might not be exactly the same. But for this signal here, right, we have five protons giving us this signal, and it's a little bit more complex than the other ones, but notice where it is, right? We're in the aromatic region in terms of a chemical shift. And so this signal must represent these five aromatic protons on our ring."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "In this video, we're going to look at primary substrates and figure out if the reaction is a substitution or an elimination. So for this primary alkyl halide, we know an SN1 reaction is out because that would require a stable carbocation and we can't make one from this primary alkyl halide. An SN2 reaction is possible because of the decreased steric hindrance of our primary alkyl halide. An E1 reaction is out, again, for the same reason as SN1. We can't form a stable carbocation. An E2 mechanism is possible. So now, the next step is to look at our reagent and figure out what the reagent is going to do."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "An E1 reaction is out, again, for the same reason as SN1. We can't form a stable carbocation. An E2 mechanism is possible. So now, the next step is to look at our reagent and figure out what the reagent is going to do. So for this reaction, we have a sulfur nucleophile, which we know is going to act only as a nucleophile and not as a base. And since it's going to act as a nucleophile, that means that E2 is out, right? We need a strong base for an E2 reaction."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So now, the next step is to look at our reagent and figure out what the reagent is going to do. So for this reaction, we have a sulfur nucleophile, which we know is going to act only as a nucleophile and not as a base. And since it's going to act as a nucleophile, that means that E2 is out, right? We need a strong base for an E2 reaction. So this must be an SN2 mechanism, which we know is a concerted mechanism. Our nucleophile attacks our electrophile at the same time that we get loss of a leaving group. So we're going to form a bond between the sulfur and this carbon here in red."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "We need a strong base for an E2 reaction. So this must be an SN2 mechanism, which we know is a concerted mechanism. Our nucleophile attacks our electrophile at the same time that we get loss of a leaving group. So we're going to form a bond between the sulfur and this carbon here in red. At the same time, these electrons come off to form the iodide leaving group. So let's draw in our final product. So we have four carbons and then we have a bond to our sulfur, which is bonded to a hydrogen."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So we're going to form a bond between the sulfur and this carbon here in red. At the same time, these electrons come off to form the iodide leaving group. So let's draw in our final product. So we have four carbons and then we have a bond to our sulfur, which is bonded to a hydrogen. So the carbon in red is this one. And so this is our product and we don't have to worry about any stereochemistry here since we don't have any chiral centers. So again, we have a primary alkyl halide and we've seen that SN1 and E1 are out."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So we have four carbons and then we have a bond to our sulfur, which is bonded to a hydrogen. So the carbon in red is this one. And so this is our product and we don't have to worry about any stereochemistry here since we don't have any chiral centers. So again, we have a primary alkyl halide and we've seen that SN1 and E1 are out. So we only have to decide between SN2 and E2. So we look at our reagent and we know that DBN is a strong base. So therefore, this is going to be an E2 reaction and not an SN2 reaction."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So again, we have a primary alkyl halide and we've seen that SN1 and E1 are out. So we only have to decide between SN2 and E2. So we look at our reagent and we know that DBN is a strong base. So therefore, this is going to be an E2 reaction and not an SN2 reaction. So DBN has to function as a strong base. And we know that the halogen is directly connected to this carbon, so that must be our alpha carbon. And in the E2 mechanism, we're going to take a proton from a beta carbon."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So therefore, this is going to be an E2 reaction and not an SN2 reaction. So DBN has to function as a strong base. And we know that the halogen is directly connected to this carbon, so that must be our alpha carbon. And in the E2 mechanism, we're going to take a proton from a beta carbon. So this is a beta carbon right here and there's one hydrogen on this beta carbon. So DBN, we know, is a neutral base. So I'll draw in here just a generic base with a lone pair of electrons."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "And in the E2 mechanism, we're going to take a proton from a beta carbon. So this is a beta carbon right here and there's one hydrogen on this beta carbon. So DBN, we know, is a neutral base. So I'll draw in here just a generic base with a lone pair of electrons. This base is going to take this proton and these electrons are going to move into here. At the same time, these electrons come off to form the iodide anion. So let's draw the final product."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So I'll draw in here just a generic base with a lone pair of electrons. This base is going to take this proton and these electrons are going to move into here. At the same time, these electrons come off to form the iodide anion. So let's draw the final product. We have this long carbon chain here, so a double bond in this position. And then we're also going to form a double bond in this position. So let's look at those electrons."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So let's draw the final product. We have this long carbon chain here, so a double bond in this position. And then we're also going to form a double bond in this position. So let's look at those electrons. So these electrons here in magenta would move in here to form our double bond. So this primary alkyl halide, again, we've seen SN1 and E1 are out, so we're deciding between SN2 and E2. So next we look at our reagent."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So let's look at those electrons. So these electrons here in magenta would move in here to form our double bond. So this primary alkyl halide, again, we've seen SN1 and E1 are out, so we're deciding between SN2 and E2. So next we look at our reagent. We have potassium hydroxide. And we know that the hydroxide ion is a strong nucleophile and a strong base. So that's the category that we saw in an earlier video."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So next we look at our reagent. We have potassium hydroxide. And we know that the hydroxide ion is a strong nucleophile and a strong base. So that's the category that we saw in an earlier video. And for a primary alkyl halide that is unhindered, the SN2 reaction is going to win out. So a small nucleophile can attack this carbon at the same time these electrons come off onto the bromine to form our bromide anion. So let's see, our carbon in red is this one."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So that's the category that we saw in an earlier video. And for a primary alkyl halide that is unhindered, the SN2 reaction is going to win out. So a small nucleophile can attack this carbon at the same time these electrons come off onto the bromine to form our bromide anion. So let's see, our carbon in red is this one. So let's draw out our carbon chain. So we have one, two, three, four, five carbons. And this carbon is the one in red."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So let's see, our carbon in red is this one. So let's draw out our carbon chain. So we have one, two, three, four, five carbons. And this carbon is the one in red. And attached to the carbon in red is going to be our OH. So we're going to form an alcohol from this SN2 reaction. For this alkyl halide, again, we're deciding between SN2 and E2."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "And this carbon is the one in red. And attached to the carbon in red is going to be our OH. So we're going to form an alcohol from this SN2 reaction. For this alkyl halide, again, we're deciding between SN2 and E2. Now we have potassium tert-butoxide as our reagent. And this looks similar to the previous problem where we had potassium hydroxide, right? We have negative charge on this oxygen."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "For this alkyl halide, again, we're deciding between SN2 and E2. Now we have potassium tert-butoxide as our reagent. And this looks similar to the previous problem where we had potassium hydroxide, right? We have negative charge on this oxygen. So you would think this could act as a strong nucleophile or a strong base. But potassium tert-butoxide is sterically hindered because of this large group over here. And because it's sterically hindered, it can't get close enough to act as a nucleophile."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "We have negative charge on this oxygen. So you would think this could act as a strong nucleophile or a strong base. But potassium tert-butoxide is sterically hindered because of this large group over here. And because it's sterically hindered, it can't get close enough to act as a nucleophile. So an SN2 reaction is out, which means this must be an E2 reaction. So let me redraw our alkyl halide here just so we can see things a little bit better. I'm gonna put the bromine going down in this direction."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "And because it's sterically hindered, it can't get close enough to act as a nucleophile. So an SN2 reaction is out, which means this must be an E2 reaction. So let me redraw our alkyl halide here just so we can see things a little bit better. I'm gonna put the bromine going down in this direction. And we know that the carbon directly connected to the bromine here is our alpha carbon, so I'll mark that. And the carbon bonded to the alpha carbon is our beta carbon. So that's our beta carbon."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "I'm gonna put the bromine going down in this direction. And we know that the carbon directly connected to the bromine here is our alpha carbon, so I'll mark that. And the carbon bonded to the alpha carbon is our beta carbon. So that's our beta carbon. And we have two beta hydrogens. I'll just draw one in here. And we know in our E2 mechanism, our strong base is going to take that beta proton."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So that's our beta carbon. And we have two beta hydrogens. I'll just draw one in here. And we know in our E2 mechanism, our strong base is going to take that beta proton. So our base takes this proton. At the same time, these electrons move into here. And these electrons come off to form our bromide ion."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "And we know in our E2 mechanism, our strong base is going to take that beta proton. So our base takes this proton. At the same time, these electrons move into here. And these electrons come off to form our bromide ion. So our final product, let's draw in our carbons here. So we should have five carbons and then a double bond between these two carbons. So our electrons in magenta moved in here to form our double bond."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "And these electrons come off to form our bromide ion. So our final product, let's draw in our carbons here. So we should have five carbons and then a double bond between these two carbons. So our electrons in magenta moved in here to form our double bond. For our last example, let's look at ethyl bromide reacting with water. We know it's not SN1, we know it's not E1, so we're deciding between SN2 and E2. We know that water is a weak nucleophile, and it's also a weak base."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So our electrons in magenta moved in here to form our double bond. For our last example, let's look at ethyl bromide reacting with water. We know it's not SN1, we know it's not E1, so we're deciding between SN2 and E2. We know that water is a weak nucleophile, and it's also a weak base. And since water is a weak base, an E2 reaction is out. So this must be an SN2 reaction. So our nucleophile attacks our alkyl halide, and a bond forms between the oxygen and this carbon here, which I'll mark in red."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "We know that water is a weak nucleophile, and it's also a weak base. And since water is a weak base, an E2 reaction is out. So this must be an SN2 reaction. So our nucleophile attacks our alkyl halide, and a bond forms between the oxygen and this carbon here, which I'll mark in red. So let's draw the result of that. This happens at the same time, these electrons come off onto the bromine to form the bromide anion. So let's draw what we have in here."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So our nucleophile attacks our alkyl halide, and a bond forms between the oxygen and this carbon here, which I'll mark in red. So let's draw the result of that. This happens at the same time, these electrons come off onto the bromine to form the bromide anion. So let's draw what we have in here. So this oxygen forms a bond with the carbon in red, and a lone pair of electrons on the oxygen, which I'll make magenta, would form this bond. The oxygen is still bonded to two hydrogens, so I'll draw in these two hydrogens here, and we still have a lone pair of electrons on that oxygen, so that gives the oxygen a plus one formal charge. To get to a neutral product, we need to deprotonate this."}, {"video_title": "Elimination vs substitution primary substrate.mp3", "Sentence": "So let's draw what we have in here. So this oxygen forms a bond with the carbon in red, and a lone pair of electrons on the oxygen, which I'll make magenta, would form this bond. The oxygen is still bonded to two hydrogens, so I'll draw in these two hydrogens here, and we still have a lone pair of electrons on that oxygen, so that gives the oxygen a plus one formal charge. To get to a neutral product, we need to deprotonate this. And so another molecule of water comes along and acts as a base to take, let's say, this proton here, leaving these electrons behind on the oxygen to give us our final product. So I'll just draw an OH in here, so we'd form ethanol. Now this reaction would need a lot of heat, it would need a lot of time, so it's not exactly the most practical reaction because an SN2 reaction needs a strong nucleophile, and water is not that great of a nucleophile."}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That we had 1 BC and then we had that theoretical birth of Jesus, and most historians don't think that he was born right exactly on January 1, 1 AD. But there is no year zero. Right after that you go from December 31, 1 BC to January 1, 1 AD. There's no year zero. And despite the fact that I emphasized that in the last video, I didn't take that into consideration when I calculated how many years there were between Plato's birth and Columbus discovering the New World. And the reason why I didn't take that into consideration is that the year 1492, whether you want to call it AD 1492, Anno Domini 1492, whether you want to call it that or whether you want to call it 1492 in the Common Era, it's not 1492 years since the theoretical birth of Jesus, which we know is not the actual birth. He was probably born before that."}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's no year zero. And despite the fact that I emphasized that in the last video, I didn't take that into consideration when I calculated how many years there were between Plato's birth and Columbus discovering the New World. And the reason why I didn't take that into consideration is that the year 1492, whether you want to call it AD 1492, Anno Domini 1492, whether you want to call it that or whether you want to call it 1492 in the Common Era, it's not 1492 years since the theoretical birth of Jesus, which we know is not the actual birth. He was probably born before that. It is 1491 years since the birth of Jesus. And to think about it this way, let's just assume, I'll keep emphasizing, it's a theoretical date that we're talking about, this theoretical event, this kind of birth of Jesus that our calendars revolve around. If we talk about January 1, let's think about it this way, so January 1, 1 in the Common Era, how long is that since the birth of Jesus?"}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "He was probably born before that. It is 1491 years since the birth of Jesus. And to think about it this way, let's just assume, I'll keep emphasizing, it's a theoretical date that we're talking about, this theoretical event, this kind of birth of Jesus that our calendars revolve around. If we talk about January 1, let's think about it this way, so January 1, 1 in the Common Era, how long is that since the birth of Jesus? It's not one year. You wouldn't just look at this and say it's been one year, because this is theoretically the day that he was born, so this is zero years or almost zero years since that theoretical birth of Jesus. Another way to think about it is, how long after January 1, the year 1 before the Common Era, and I could have called this AD and I could have called this BC, what's the time difference between these two dates?"}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we talk about January 1, let's think about it this way, so January 1, 1 in the Common Era, how long is that since the birth of Jesus? It's not one year. You wouldn't just look at this and say it's been one year, because this is theoretically the day that he was born, so this is zero years or almost zero years since that theoretical birth of Jesus. Another way to think about it is, how long after January 1, the year 1 before the Common Era, and I could have called this AD and I could have called this BC, what's the time difference between these two dates? The way I calculated it before, I said this is one year after that theoretical birth, that's wrong, this is during that theoretical birth. But the way I did it in the last video, I said that's one year after, one year before, you add them together and you would get 2. But that's wrong, because there is no year zero, so January 1, 1 AD or 1 in the Common Era is right over here, and then January 1, 1 BCE is exactly one year before that, so there's only one year difference."}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Another way to think about it is, how long after January 1, the year 1 before the Common Era, and I could have called this AD and I could have called this BC, what's the time difference between these two dates? The way I calculated it before, I said this is one year after that theoretical birth, that's wrong, this is during that theoretical birth. But the way I did it in the last video, I said that's one year after, one year before, you add them together and you would get 2. But that's wrong, because there is no year zero, so January 1, 1 AD or 1 in the Common Era is right over here, and then January 1, 1 BCE is exactly one year before that, so there's only one year difference. The reason why the math is strange is because there is no year zero. If there was a year zero, then my calculation in the last video was correct. So really, the way that you would calculate the time between Plato's birth at 428 BC and Columbus sailing across the Atlantic in 1492, you would say, okay, this is 428 years before that theoretical birth of Christ."}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But that's wrong, because there is no year zero, so January 1, 1 AD or 1 in the Common Era is right over here, and then January 1, 1 BCE is exactly one year before that, so there's only one year difference. The reason why the math is strange is because there is no year zero. If there was a year zero, then my calculation in the last video was correct. So really, the way that you would calculate the time between Plato's birth at 428 BC and Columbus sailing across the Atlantic in 1492, you would say, okay, this is 428 years before that theoretical birth of Christ. But this isn't 1492 years after that theoretical birth, this is 1492 minus 1. So what you would do is you would add these two numbers, this is 428 before, this is 1492 minus 1 years after, so you would add them and then subtract 1. So the correct answer, this is the correction part, it isn't 1920 years between Plato's birth and Columbus, we want to subtract 1 from that."}, {"video_title": "Correction calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So really, the way that you would calculate the time between Plato's birth at 428 BC and Columbus sailing across the Atlantic in 1492, you would say, okay, this is 428 years before that theoretical birth of Christ. But this isn't 1492 years after that theoretical birth, this is 1492 minus 1. So what you would do is you would add these two numbers, this is 428 before, this is 1492 minus 1 years after, so you would add them and then subtract 1. So the correct answer, this is the correction part, it isn't 1920 years between Plato's birth and Columbus, we want to subtract 1 from that. It is 1920 minus 1 years, so that is 1919 years. The same way that the difference between 1 AD and 1 BCE, you would say, you could almost view it as positive and negative numbers, you would say, oh, this is positive 1 minus negative 1, that would give me 2, but there is no zero, so you subtract another 1. So this is exactly 1 year difference."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video, we learned that 380,000 years after the Big Bang, which is still roughly 13.7 billion years ago, every point, I shouldn't say every point, every atom in space that was kind of at this roughly 3,000 Kelvin temperature was emitting this electromagnetic radiation. Since every point in space was, there were points in space, or there was points in the universe, that that radiation is only just now reaching us. It has been traveling for 13.7 billion years. So when we look at radiation that's been traveling for that long, we can look in any direction and we'll see this uniform radiation. And that radiation has been redshifted into the microwave range from the higher frequencies that it was actually emitted at. Now, a question that might pop in your brain is, well, what happens if we wait a billion years? Because if we wait a billion years, if we have a billion 380,000 years after the beginning of the universe, this stuff won't just be atoms anymore."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when we look at radiation that's been traveling for that long, we can look in any direction and we'll see this uniform radiation. And that radiation has been redshifted into the microwave range from the higher frequencies that it was actually emitted at. Now, a question that might pop in your brain is, well, what happens if we wait a billion years? Because if we wait a billion years, if we have a billion 380,000 years after the beginning of the universe, this stuff won't just be atoms anymore. It will have started to condense into actual stars. We would not, the universe at every point in space will no longer be this uniform. We'll actually start having condensation into stars."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because if we wait a billion years, if we have a billion 380,000 years after the beginning of the universe, this stuff won't just be atoms anymore. It will have started to condense into actual stars. We would not, the universe at every point in space will no longer be this uniform. We'll actually start having condensation into stars. So if we go, the universe, if we move forward a little bit, the universe will expand. Maybe I'll just draw half of it since it's expanded. It's obviously expanded much more."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We'll actually start having condensation into stars. So if we go, the universe, if we move forward a little bit, the universe will expand. Maybe I'll just draw half of it since it's expanded. It's obviously expanded much more. But now all of a sudden, we actually have stars. These are no longer just uniform atoms spread through the universe. We actually have condensation into stars."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's obviously expanded much more. But now all of a sudden, we actually have stars. These are no longer just uniform atoms spread through the universe. We actually have condensation into stars. And so if you look at what is being emitted from the points in space from which we're only now getting this cosmic background radiation, if we wait a billion years, the light that we see from those points in space will not look like this uniform radiation. It'll start to look a little bit more like the more mature parts of the universe. It'll start to, we'll essentially be looking at the universe a billion years after the Big Bang when stars have formed, other structures have formed."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We actually have condensation into stars. And so if you look at what is being emitted from the points in space from which we're only now getting this cosmic background radiation, if we wait a billion years, the light that we see from those points in space will not look like this uniform radiation. It'll start to look a little bit more like the more mature parts of the universe. It'll start to, we'll essentially be looking at the universe a billion years after the Big Bang when stars have formed, other structures have formed. So the question is, in a billion years, will this cosmic microwave background radiation disappear? And I'm using the billion just because that's just to arbitrarily use a number. But will it eventually disappear?"}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It'll start to, we'll essentially be looking at the universe a billion years after the Big Bang when stars have formed, other structures have formed. So the question is, in a billion years, will this cosmic microwave background radiation disappear? And I'm using the billion just because that's just to arbitrarily use a number. But will it eventually disappear? And there's kind of, the answer to that is yes and no. So to think about it, it is true that this point in space will mature. It will mature in a billion years."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But will it eventually disappear? And there's kind of, the answer to that is yes and no. So to think about it, it is true that this point in space will mature. It will mature in a billion years. It will no longer be this uniform haze of hot hydrogen atoms. But what you have to think about is, there were further points in the universe. There were further points in the universe."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It will mature in a billion years. It will no longer be this uniform haze of hot hydrogen atoms. But what you have to think about is, there were further points in the universe. There were further points in the universe. At that same time, there were further points that were also emitting this radiation. And the original photons from those original points still haven't gotten to us. So from those further out points, right now the observable universe is, we can only see electromagnetic radiation that's been traveling for 13.7 billion years."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There were further points in the universe. At that same time, there were further points that were also emitting this radiation. And the original photons from those original points still haven't gotten to us. So from those further out points, right now the observable universe is, we can only see electromagnetic radiation that's been traveling for 13.7 billion years. In another billion years, the universe will be a billion years older. And then there will be radiation that has been traveling for 14.7 billion years. And so we will start to observe that."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So from those further out points, right now the observable universe is, we can only see electromagnetic radiation that's been traveling for 13.7 billion years. In another billion years, the universe will be a billion years older. And then there will be radiation that has been traveling for 14.7 billion years. And so we will start to observe that. And we'll start to observe that radiation from the same time period in the universe. It will just be from further out. Now, what I want to make clear is that since those points were even further out, where that radiation was emitted, the stuff that we'll see in a billion years, it will be even more redshifted."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we will start to observe that. And we'll start to observe that radiation from the same time period in the universe. It will just be from further out. Now, what I want to make clear is that since those points were even further out, where that radiation was emitted, the stuff that we'll see in a billion years, it will be even more redshifted. So at that point, the cosmic background radiation we see will have longer wavelengths than the radio spectrum. It will be redder. And I should say redder, because we're already more."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, what I want to make clear is that since those points were even further out, where that radiation was emitted, the stuff that we'll see in a billion years, it will be even more redshifted. So at that point, the cosmic background radiation we see will have longer wavelengths than the radio spectrum. It will be redder. And I should say redder, because we're already more. Would redder have two D's? I've never written redder. Well, it would be more red than the microwave radiation."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I should say redder, because we're already more. Would redder have two D's? I've never written redder. Well, it would be more red than the microwave radiation. And of course, that's a funny thing, because microwave radiation is already more red than actual visible red light. It has a longer wavelength. Now, this will keep happening."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, it would be more red than the microwave radiation. And of course, that's a funny thing, because microwave radiation is already more red than actual visible red light. It has a longer wavelength. Now, this will keep happening. And I don't know what, you know, it'll keep happening. We'll keep getting radiation as we go further and further into the future. We'll keep getting radiation from further out points in space."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, this will keep happening. And I don't know what, you know, it'll keep happening. We'll keep getting radiation as we go further and further into the future. We'll keep getting radiation from further out points in space. And it'll get more and more redshifted. The actual wavelengths of that electromagnetic light will be bigger and bigger and bigger, until we really aren't able to even see it as electromagnetic light, because it'll be redshifted to infinity. It'll have an infinite wavelength."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We'll keep getting radiation from further out points in space. And it'll get more and more redshifted. The actual wavelengths of that electromagnetic light will be bigger and bigger and bigger, until we really aren't able to even see it as electromagnetic light, because it'll be redshifted to infinity. It'll have an infinite wavelength. And to make that point clear, I want to show you that there'll even be points, at some point, there'll be kind of a threshold where we can't even get radiation from further out. And let me show you. Let me draw a diagram of that."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It'll have an infinite wavelength. And to make that point clear, I want to show you that there'll even be points, at some point, there'll be kind of a threshold where we can't even get radiation from further out. And let me show you. Let me draw a diagram of that. So let's say that this is the universe 13.7 billion years ago, right when that radiation, what we now see as cosmic microwave background radiation, right when it started to be emitted. And let's say that this is the point in the universe where we are now. So this is us."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me draw a diagram of that. So let's say that this is the universe 13.7 billion years ago, right when that radiation, what we now see as cosmic microwave background radiation, right when it started to be emitted. And let's say that this is the point in the universe where we are now. So this is us. Let's say that this is the point in the universe that where we now observe the background radiation. Or this is one of the points. We obviously could form a circle around us."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is us. Let's say that this is the point in the universe that where we now observe the background radiation. Or this is one of the points. We obviously could form a circle around us. It could be any of these points over here. We're only where the photons, the electromagnetic radiation that were emitted from this point 380,000 years after the beginning of the universe, is only just now reaching us. So this is the point in the universe from which we are observing the cosmic background radiation."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We obviously could form a circle around us. It could be any of these points over here. We're only where the photons, the electromagnetic radiation that were emitted from this point 380,000 years after the beginning of the universe, is only just now reaching us. So this is the point in the universe from which we are observing the cosmic background radiation. And let me be very clear. That point in the universe has now matured into things that look into stars and galaxies and planets. And if they were to look at our point in space, they are also going to see cosmic background radiation from us."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the point in the universe from which we are observing the cosmic background radiation. And let me be very clear. That point in the universe has now matured into things that look into stars and galaxies and planets. And if they were to look at our point in space, they are also going to see cosmic background radiation from us. It's not like this is some type of permanently old place. It's just the light we're getting from them right now is old light, light that that point in space emitted way before it was able to mature into actual structures. So this is the point in space from which we are receiving cosmic background radiation right now."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if they were to look at our point in space, they are also going to see cosmic background radiation from us. It's not like this is some type of permanently old place. It's just the light we're getting from them right now is old light, light that that point in space emitted way before it was able to mature into actual structures. So this is the point in space from which we are receiving cosmic background radiation right now. I don't want to write all that. It'll take me forever. Now, let's take another point in space that's whatever this distance is."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the point in space from which we are receiving cosmic background radiation right now. I don't want to write all that. It'll take me forever. Now, let's take another point in space that's whatever this distance is. Well, it's actually estimated to be about now. It's estimated to be about 46 billion light years. At that time, when things were just beginning to be emitted, this was only about 36 million light years."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, let's take another point in space that's whatever this distance is. Well, it's actually estimated to be about now. It's estimated to be about 46 billion light years. At that time, when things were just beginning to be emitted, this was only about 36 million light years. And this is a very rough estimate. I shouldn't even write it down, because that's really based on how fast we assume the universe is expanding and all of that type of thing. But it was just a lot smaller than 46 billion light years."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "At that time, when things were just beginning to be emitted, this was only about 36 million light years. And this is a very rough estimate. I shouldn't even write it down, because that's really based on how fast we assume the universe is expanding and all of that type of thing. But it was just a lot smaller than 46 billion light years. Now, let's go that same distance again from this point in space. So let me make it clear. This is 380,000 years ago."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it was just a lot smaller than 46 billion light years. Now, let's go that same distance again from this point in space. So let me make it clear. This is 380,000 years ago. Now, let's fast forward. Let's fast forward. Sorry, not 380,000 years ago."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is 380,000 years ago. Now, let's fast forward. Let's fast forward. Sorry, not 380,000 years ago. 380,000 years after the Big Bang, which is approximately, it's still 13.7 billion years ago. So that's then. Now, let's look at now."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Sorry, not 380,000 years ago. 380,000 years after the Big Bang, which is approximately, it's still 13.7 billion years ago. So that's then. Now, let's look at now. Now, I'll just draw it a little bit bigger. It's actually going to be much, much bigger now. Now, if we do it a little bit bigger, so when I draw it like this, this is where we are now."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, let's look at now. Now, I'll just draw it a little bit bigger. It's actually going to be much, much bigger now. Now, if we do it a little bit bigger, so when I draw it like this, this is where we are now. This point in space is this point in space from which we are only now receiving that cosmic background radiation is over here. And then this other point in space is going to be over here. And we saw in the video on the actual size of the observable universe, not just what it appears to be based on how long the light's been traveling."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, if we do it a little bit bigger, so when I draw it like this, this is where we are now. This point in space is this point in space from which we are only now receiving that cosmic background radiation is over here. And then this other point in space is going to be over here. And we saw in the video on the actual size of the observable universe, not just what it appears to be based on how long the light's been traveling. This is now on the order of 46 or 47 billion light years. And so this distance is also going to be 46 billion light years. Now, every point in space back then was emitting this radiation."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we saw in the video on the actual size of the observable universe, not just what it appears to be based on how long the light's been traveling. This is now on the order of 46 or 47 billion light years. And so this distance is also going to be 46 billion light years. Now, every point in space back then was emitting this radiation. We have this uniform radiation. It was just hydrogen atoms everywhere, these hot hydrogen atoms. Now, maybe I should just do it in the color of the radiation."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, every point in space back then was emitting this radiation. We have this uniform radiation. It was just hydrogen atoms everywhere, these hot hydrogen atoms. Now, maybe I should just do it in the color of the radiation. So this guy's receiving. I'm just showing it's coming from this guy. We're only now, 13.7 billion years in the future, receiving photons from this guy."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, maybe I should just do it in the color of the radiation. So this guy's receiving. I'm just showing it's coming from this guy. We're only now, 13.7 billion years in the future, receiving photons from this guy. Only now are we receiving it. And frankly, this green guy only now is going to be receiving photons. When he looks at the points in space, or the things that he thinks are the points in space out there, he will see that uniform radiation."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're only now, 13.7 billion years in the future, receiving photons from this guy. Only now are we receiving it. And frankly, this green guy only now is going to be receiving photons. When he looks at the points in space, or the things that he thinks are the points in space out there, he will see that uniform radiation. And likewise, this guy over here will only now be receiving photons from the point in space for where we are now. He'll see the universe where we are now as it was 380,000 years after the Big Bang. And same thing from that point in space."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When he looks at the points in space, or the things that he thinks are the points in space out there, he will see that uniform radiation. And likewise, this guy over here will only now be receiving photons from the point in space for where we are now. He'll see the universe where we are now as it was 380,000 years after the Big Bang. And same thing from that point in space. The photons will only just now reach. Now, let's think about it. It took this guy's photons."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And same thing from that point in space. The photons will only just now reach. Now, let's think about it. It took this guy's photons. Let me make it clear. It took him 13.7 billion years to reach this point over here, which is now 46 billion light years away from us. And the universe continues to expand."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It took this guy's photons. Let me make it clear. It took him 13.7 billion years to reach this point over here, which is now 46 billion light years away from us. And the universe continues to expand. Depending on if the universe expands fast enough, there's no way that that photon that got to this guy will eventually get to this. The universe is expanding faster that the light can never even catch up to us, and this light will never, ever, ever get to us. And so there is some threshold, some distance from which we will never get light during this time period, or actually from which we will never, ever get any electromagnetic radiation."}, {"video_title": "Cosmic background radiation 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the universe continues to expand. Depending on if the universe expands fast enough, there's no way that that photon that got to this guy will eventually get to this. The universe is expanding faster that the light can never even catch up to us, and this light will never, ever, ever get to us. And so there is some threshold, some distance from which we will never get light during this time period, or actually from which we will never, ever get any electromagnetic radiation. So the simple answer is the cosmic background radiation from this point, yes, it will start to mature. It won't be as uniform if we go fast forward 400 million years or a billion years, but we will get uniform radiation from further out, but it'll be even more redshifted. And the further forward we get into the future, the background radiation we get will be from further and further out, and it will be more and more and more redshifted until some point where it's going to be so redshifted that we won't even observe it as electromagnetic radiation."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And what I want to do in this video is show a mechanism for actually creating one. In particular, we're going to create a ketone. So let's say we've got ourselves some benzene. And the first step of this reaction, the benzene is just going to sit and watch. We've got some benzene and we've got some acetyl chloride. So it looks almost like an aldehyde or a ketone. But instead of having a carbon chain here or a hydrogen, we're going to have a chlorine atom right over there."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the first step of this reaction, the benzene is just going to sit and watch. We've got some benzene and we've got some acetyl chloride. So it looks almost like an aldehyde or a ketone. But instead of having a carbon chain here or a hydrogen, we're going to have a chlorine atom right over there. So this is acetyl chloride, sometimes called acyl chloride. And we're going to have an aluminum chloride catalyst. And a catalyst means that it participates in the reaction, but it enters the reaction and it exits the reaction."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But instead of having a carbon chain here or a hydrogen, we're going to have a chlorine atom right over there. So this is acetyl chloride, sometimes called acyl chloride. And we're going to have an aluminum chloride catalyst. And a catalyst means that it participates in the reaction, but it enters the reaction and it exits the reaction. It is the same molecule. So it just catalyzes it. It doesn't disappear."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And a catalyst means that it participates in the reaction, but it enters the reaction and it exits the reaction. It is the same molecule. So it just catalyzes it. It doesn't disappear. It just changes halfway, but then goes back to what it was before. So we have some aluminum chloride. And it's bonded to 1, 2, 3 chlorines."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't disappear. It just changes halfway, but then goes back to what it was before. So we have some aluminum chloride. And it's bonded to 1, 2, 3 chlorines. Now the first step of this reaction is to turn this acetyl chloride into a good electrophile. Turn it into something that's really good at nabbing electrons. So good that it can break the aromaticity of the benzene ring and essentially add itself to the benzene ring."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it's bonded to 1, 2, 3 chlorines. Now the first step of this reaction is to turn this acetyl chloride into a good electrophile. Turn it into something that's really good at nabbing electrons. So good that it can break the aromaticity of the benzene ring and essentially add itself to the benzene ring. This is actually the same mechanism we saw with electrophilic aromatic substitution. I always have trouble remembering the names, but I always imagine it's electrophilic substitution. Either way, it's a very similar mechanism."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So good that it can break the aromaticity of the benzene ring and essentially add itself to the benzene ring. This is actually the same mechanism we saw with electrophilic aromatic substitution. I always have trouble remembering the names, but I always imagine it's electrophilic substitution. Either way, it's a very similar mechanism. And actually, what we're going to show in this video is called Friedel-Crafts acylation. Because this right here is called an acyl group. And we're essentially going to acylate the benzene ring."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Either way, it's a very similar mechanism. And actually, what we're going to show in this video is called Friedel-Crafts acylation. Because this right here is called an acyl group. And we're essentially going to acylate the benzene ring. We're going to add this group right here to the benzene ring. So enough of what's going to happen. Let's actually see it happen."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're essentially going to acylate the benzene ring. We're going to add this group right here to the benzene ring. So enough of what's going to happen. Let's actually see it happen. So the first thing to realize, this aluminum chloride, the aluminum in it is electron deficient. At a first cut, if you just look at the periodic table, you have these chlorines over here, pretty electronegative. Aluminum is in the same row, but chlorine's way more to the right, so it's more electronegative."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's actually see it happen. So the first thing to realize, this aluminum chloride, the aluminum in it is electron deficient. At a first cut, if you just look at the periodic table, you have these chlorines over here, pretty electronegative. Aluminum is in the same row, but chlorine's way more to the right, so it's more electronegative. So the chlorines are going to hog the electrons in this molecule. The chlorines are going to hog the electrons, so the aluminum's going to have a partial positive charge. Chlorines will have slightly partial negative charge."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Aluminum is in the same row, but chlorine's way more to the right, so it's more electronegative. So the chlorines are going to hog the electrons in this molecule. The chlorines are going to hog the electrons, so the aluminum's going to have a partial positive charge. Chlorines will have slightly partial negative charge. On top of that, you see aluminum is a group 3 element. 1, 2, 3. So it has 3 valence electrons."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Chlorines will have slightly partial negative charge. On top of that, you see aluminum is a group 3 element. 1, 2, 3. So it has 3 valence electrons. You see that right here. 1, 2, 3. Nowhere close to the magic number of 8."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it has 3 valence electrons. You see that right here. 1, 2, 3. Nowhere close to the magic number of 8. Even when it covalently bonds with these chlorines, it can only pretend like it has 6 electrons. It can kind of pretend like it has these chlorines electrons over here, but that only gets it to 6. So it would really like to have more electrons to get closer to that magic 8 number."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Nowhere close to the magic number of 8. Even when it covalently bonds with these chlorines, it can only pretend like it has 6 electrons. It can kind of pretend like it has these chlorines electrons over here, but that only gets it to 6. So it would really like to have more electrons to get closer to that magic 8 number. So what you can imagine is a situation where this chlorine on the acetyl chloride, it's already hogging this green electron from this carbon. It was already doing that. It's more electronegative."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it would really like to have more electrons to get closer to that magic 8 number. So what you can imagine is a situation where this chlorine on the acetyl chloride, it's already hogging this green electron from this carbon. It was already doing that. It's more electronegative. So this thing over here will actually be given to the aluminum. And so it will then have a bond with the chloride. So if that happened, what does our reaction look like?"}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's more electronegative. So this thing over here will actually be given to the aluminum. And so it will then have a bond with the chloride. So if that happened, what does our reaction look like? So if that happened, what does everything look like? Our aluminum chloride, or what was formerly aluminum chloride, now just gained an electron. And with it, it is now bonded to another chlorine."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if that happened, what does our reaction look like? So if that happened, what does everything look like? Our aluminum chloride, or what was formerly aluminum chloride, now just gained an electron. And with it, it is now bonded to another chlorine. And since it gained an electron, let me make it very clear. This is an L. My penmanship is deteriorating. That's an L. And since this aluminum gained an electron, it now has a negative charge."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And with it, it is now bonded to another chlorine. And since it gained an electron, let me make it very clear. This is an L. My penmanship is deteriorating. That's an L. And since this aluminum gained an electron, it now has a negative charge. And normally, a negatively charged thing isn't that stable. But these guys are electronegative. So they might hog a lot of that negative charge."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That's an L. And since this aluminum gained an electron, it now has a negative charge. And normally, a negatively charged thing isn't that stable. But these guys are electronegative. So they might hog a lot of that negative charge. And on top of it, aluminum can now pretend like it has 8 electrons. It has 1, 2, 3, 4, 5, 6, 7, 8. When you covalently bond to someone, you can kind of pretend like you have their electrons as well."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So they might hog a lot of that negative charge. And on top of it, aluminum can now pretend like it has 8 electrons. It has 1, 2, 3, 4, 5, 6, 7, 8. When you covalently bond to someone, you can kind of pretend like you have their electrons as well. So now you have this anion that was formed from the aluminum chloride. And now the acetyl chloride will look like this. Let me scroll down a little bit."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "When you covalently bond to someone, you can kind of pretend like you have their electrons as well. So now you have this anion that was formed from the aluminum chloride. And now the acetyl chloride will look like this. Let me scroll down a little bit. Let me make it clear. We're in the next step of the reaction. What was formerly the acetyl chloride has now lost the chloride."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me scroll down a little bit. Let me make it clear. We're in the next step of the reaction. What was formerly the acetyl chloride has now lost the chloride. So it's now really just an acyl group. So we have the carbonyl bonded to a CH3, a methyl, just like that. This guy lost his electrons."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "What was formerly the acetyl chloride has now lost the chloride. So it's now really just an acyl group. So we have the carbonyl bonded to a CH3, a methyl, just like that. This guy lost his electrons. So now he has a positive charge. And this is actually not that stable. And you're going to see it's actually highly reactive."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This guy lost his electrons. So now he has a positive charge. And this is actually not that stable. And you're going to see it's actually highly reactive. It's a very good electrophile. It wants to steal other people's electrons. But it can exist for a short amount of time, especially because it is resonance stabilized."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you're going to see it's actually highly reactive. It's a very good electrophile. It wants to steal other people's electrons. But it can exist for a short amount of time, especially because it is resonance stabilized. You say, how is it resonance stabilized? Well, this oxygen over here has 2 electron pairs that I didn't draw before. And let me draw the second one a different color."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But it can exist for a short amount of time, especially because it is resonance stabilized. You say, how is it resonance stabilized? Well, this oxygen over here has 2 electron pairs that I didn't draw before. And let me draw the second one a different color. It has 2 electron pairs like that. So you can imagine a situation. This carbon is already, he has a positive charge and this oxygen is more electronegative."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let me draw the second one a different color. It has 2 electron pairs like that. So you can imagine a situation. This carbon is already, he has a positive charge and this oxygen is more electronegative. It's already hogging his electrons. Maybe he wants to give back a little bit. Say, hey, this is positive."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is already, he has a positive charge and this oxygen is more electronegative. It's already hogging his electrons. Maybe he wants to give back a little bit. Say, hey, this is positive. All the electrons are hanging out here. They'd be attracted to the positive. And you can imagine one of these electrons being given back to the carbon."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Say, hey, this is positive. All the electrons are hanging out here. They'd be attracted to the positive. And you can imagine one of these electrons being given back to the carbon. And if that happened, then we have another resonance form that looks like this. So this was our original molecule. Or that's our original or what it looked like."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you can imagine one of these electrons being given back to the carbon. And if that happened, then we have another resonance form that looks like this. So this was our original molecule. Or that's our original or what it looked like. We still have this double bond right over there. But now this or that pair of electrons, now this pair of electrons now forms another bond. This electron over here is now on the oxygen end."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Or that's our original or what it looked like. We still have this double bond right over there. But now this or that pair of electrons, now this pair of electrons now forms another bond. This electron over here is now on the oxygen end. This electron over here is now on the carbon end. And now they have a triple bond. And what happened?"}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This electron over here is now on the oxygen end. This electron over here is now on the carbon end. And now they have a triple bond. And what happened? This positive carbon gained an electron, so it's now neutral. And the neutral oxygen lost an electron. So it is now positive."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And what happened? This positive carbon gained an electron, so it's now neutral. And the neutral oxygen lost an electron. So it is now positive. And you can imagine this is not a very stable. You wouldn't see this just floating around by itself. But it does stabilize this entire configuration."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it is now positive. And you can imagine this is not a very stable. You wouldn't see this just floating around by itself. But it does stabilize this entire configuration. It stabilizes this molecule. So you can show that these are alternates, resonantly stabilized structures right there. But as I said, these aren't super stable."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But it does stabilize this entire configuration. It stabilizes this molecule. So you can show that these are alternates, resonantly stabilized structures right there. But as I said, these aren't super stable. This guy really, really, really wants to react. And now this is where benzene comes into the mix. And actually, let me draw a little dividing line here, just so we know that this was a separate stage of our reaction."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But as I said, these aren't super stable. This guy really, really, really wants to react. And now this is where benzene comes into the mix. And actually, let me draw a little dividing line here, just so we know that this was a separate stage of our reaction. So that was the first stage. Then we go over here. And now benzene comes into the mix."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And actually, let me draw a little dividing line here, just so we know that this was a separate stage of our reaction. So that was the first stage. Then we go over here. And now benzene comes into the mix. The benzene was floating around. So we have our benzene floating around just like that. And then I'm going to draw one of the hydrogens on one of the benzenes."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now benzene comes into the mix. The benzene was floating around. So we have our benzene floating around just like that. And then I'm going to draw one of the hydrogens on one of the benzenes. All of these carbons have hydrogens on them. I just won't draw them all. This makes things complicated."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then I'm going to draw one of the hydrogens on one of the benzenes. All of these carbons have hydrogens on them. I just won't draw them all. This makes things complicated. But this guy, we said, is a really good electrophile. And you have to be a really good electrophile to attract electrons from a benzene ring to break its aromaticity. But if it bumps into this guy in just the right way, just the right angle, you could imagine that this electron on this carbon right here gets swiped by the acyl group."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This makes things complicated. But this guy, we said, is a really good electrophile. And you have to be a really good electrophile to attract electrons from a benzene ring to break its aromaticity. But if it bumps into this guy in just the right way, just the right angle, you could imagine that this electron on this carbon right here gets swiped by the acyl group. So then what do we have? So now I will go back in this direction. So you have what was a benzene ring."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But if it bumps into this guy in just the right way, just the right angle, you could imagine that this electron on this carbon right here gets swiped by the acyl group. So then what do we have? So now I will go back in this direction. So you have what was a benzene ring. So what was a benzene ring? It still has, we could draw the double bonds here and here. And we, of course, have this hydrogen."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you have what was a benzene ring. So what was a benzene ring? It still has, we could draw the double bonds here and here. And we, of course, have this hydrogen. But now this bond, which was a double bond there, is now bonded to the acyl group. So it has that blue electron that the acyl group nabbed. And let me draw the acyl group."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we, of course, have this hydrogen. But now this bond, which was a double bond there, is now bonded to the acyl group. So it has that blue electron that the acyl group nabbed. And let me draw the acyl group. And I'll flip it over so that we have the methyl on the right-hand side. So it's a carbonyl bonded to a CH3. It was positive."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let me draw the acyl group. And I'll flip it over so that we have the methyl on the right-hand side. So it's a carbonyl bonded to a CH3. It was positive. It now gained an electron. It is now neutral. This carbon over here lost an electron."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It was positive. It now gained an electron. It is now neutral. This carbon over here lost an electron. So it is now positive. Now, we mentioned the aluminum chloride is a catalyst. So it won't just sit around there as the anion."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This carbon over here lost an electron. So it is now positive. Now, we mentioned the aluminum chloride is a catalyst. So it won't just sit around there as the anion. It has to go back to being aluminum chloride. So let's bring the aluminum chloride back into the scene. So we have our aluminum chloride."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it won't just sit around there as the anion. It has to go back to being aluminum chloride. So let's bring the aluminum chloride back into the scene. So we have our aluminum chloride. Let me copy and paste it. Copy and paste. So we have our aluminum chloride here."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our aluminum chloride. Let me copy and paste it. Copy and paste. So we have our aluminum chloride here. And so you can imagine that the benzene ring wants to go back to being aromatic. So this electron right here on the hydrogen might really want to go back to this carbon right over here, this carbocation. And at the same time, if this anion now passes the hydrogen at just the right angle at the right time while this guy is attracted to this carbon, this chlorine can give this green electron to the hydrogen nucleus, which is really just a proton."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our aluminum chloride here. And so you can imagine that the benzene ring wants to go back to being aromatic. So this electron right here on the hydrogen might really want to go back to this carbon right over here, this carbocation. And at the same time, if this anion now passes the hydrogen at just the right angle at the right time while this guy is attracted to this carbon, this chlorine can give this green electron to the hydrogen nucleus, which is really just a proton. And then the hydrogen's electron can be taken up by what was this carbocation. And then what did we have? Then we have a situation where our benzene ring is reformed."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And at the same time, if this anion now passes the hydrogen at just the right angle at the right time while this guy is attracted to this carbon, this chlorine can give this green electron to the hydrogen nucleus, which is really just a proton. And then the hydrogen's electron can be taken up by what was this carbocation. And then what did we have? Then we have a situation where our benzene ring is reformed. We have the aromaticity again. We have that double bond, that double bond. And now we have this double bond again, although now it's using the electron from the hydrogen."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Then we have a situation where our benzene ring is reformed. We have the aromaticity again. We have that double bond, that double bond. And now we have this double bond again, although now it's using the electron from the hydrogen. And now we've substituted this hydrogen, which is essentially this acyl group right over here. So we have a carbon double bonded to an oxygen, bonded to a methyl group. And now the aluminum, or this anion, lost its electron."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now we have this double bond again, although now it's using the electron from the hydrogen. And now we've substituted this hydrogen, which is essentially this acyl group right over here. So we have a carbon double bonded to an oxygen, bonded to a methyl group. And now the aluminum, or this anion, lost its electron. So it goes back to just being straight up aluminum chloride. And we're done. We've just acylized this benzene ring."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now the aluminum, or this anion, lost its electron. So it goes back to just being straight up aluminum chloride. And we're done. We've just acylized this benzene ring. And that's why this mechanism is called Friedel-Crafts acylation. And Friedel is actually a former president of MIT. And I did some reading on this."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We've just acylized this benzene ring. And that's why this mechanism is called Friedel-Crafts acylation. And Friedel is actually a former president of MIT. And I did some reading on this. And apparently, he did not have a PhD. But because he discovered Friedel-Crafts acylation, this Friedel-Crafts actually alkylation as well, they said, hey, this guy's a smart dude. Let's make him the president of MIT."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I did some reading on this. And apparently, he did not have a PhD. But because he discovered Friedel-Crafts acylation, this Friedel-Crafts actually alkylation as well, they said, hey, this guy's a smart dude. Let's make him the president of MIT. But I just wanted to show you that this is a reaction for creating a ketone. So this ketone that we've created is acetophenone, which we've seen before, which we learned is a common name for this molecule that we learned in the first ketone video. And I'll write it in purple."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's make him the president of MIT. But I just wanted to show you that this is a reaction for creating a ketone. So this ketone that we've created is acetophenone, which we've seen before, which we learned is a common name for this molecule that we learned in the first ketone video. And I'll write it in purple. Acetophenone. Acetophenone. And we're done."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so the difference is, instead of adding water, this time we're adding an alcohol. And so this reaction is at equilibrium. And then over here on the right, our product this time is a hemiacetal. So this is a hemiacetal right here. So in terms of the mechanism for the formation of a hemiacetal, it's completely analogous to the formation of a hydrate. We have our carbonyl situation over here on the left for our aldehyde or ketone, with the oxygen being more electronegative and withdrawing some electron density away from our carbonyl carbon. So we have a partially negative oxygen and a partially positive carbonyl."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is a hemiacetal right here. So in terms of the mechanism for the formation of a hemiacetal, it's completely analogous to the formation of a hydrate. We have our carbonyl situation over here on the left for our aldehyde or ketone, with the oxygen being more electronegative and withdrawing some electron density away from our carbonyl carbon. So we have a partially negative oxygen and a partially positive carbonyl. So this carbonyl carbon right here is partially positive. So it is electrophilic. And so alcohol can act as a nucleophile."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have a partially negative oxygen and a partially positive carbonyl. So this carbonyl carbon right here is partially positive. So it is electrophilic. And so alcohol can act as a nucleophile. So our alcohol molecule, I'm going to go ahead and draw it out here, is going to function as our nucleophile in this reaction. And so a lone pair of electrons on our oxygen is going to attack our carbonyl carbon right here and push these pi electrons in here off onto our oxygen. So our oxygen already had two lone pairs of electrons like that."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so alcohol can act as a nucleophile. So our alcohol molecule, I'm going to go ahead and draw it out here, is going to function as our nucleophile in this reaction. And so a lone pair of electrons on our oxygen is going to attack our carbonyl carbon right here and push these pi electrons in here off onto our oxygen. So our oxygen already had two lone pairs of electrons like that. So let's go ahead and show the result of that nucleophilic attack. So the oxygen right now is bonded to our carbon. So here's our oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So our oxygen already had two lone pairs of electrons like that. So let's go ahead and show the result of that nucleophilic attack. So the oxygen right now is bonded to our carbon. So here's our oxygen. It's formed a bond to our carbon. And there was a hydrogen attached to our oxygen and also an R double prime group like that. And there was still one lone pair of electrons on our oxygen, giving our oxygen a plus one formal charge."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So here's our oxygen. It's formed a bond to our carbon. And there was a hydrogen attached to our oxygen and also an R double prime group like that. And there was still one lone pair of electrons on our oxygen, giving our oxygen a plus one formal charge. This carbon was bonded to another oxygen, which used to have two lone pairs of electrons. It just picked up another lone pair. So it now has a negative one formal charge like that."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And there was still one lone pair of electrons on our oxygen, giving our oxygen a plus one formal charge. This carbon was bonded to another oxygen, which used to have two lone pairs of electrons. It just picked up another lone pair. So it now has a negative one formal charge like that. And then we still had an R group and then a hydrogen. So let's follow some electrons. So these electrons right here in magenta on our alcohol formed a bond between the oxygen and the carbon."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it now has a negative one formal charge like that. And then we still had an R group and then a hydrogen. So let's follow some electrons. So these electrons right here in magenta on our alcohol formed a bond between the oxygen and the carbon. So these electrons right in there. And then our pi electrons in our carbonyl here moved out onto our oxygen. So it doesn't really matter which lone pair we say it is."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here in magenta on our alcohol formed a bond between the oxygen and the carbon. So these electrons right in there. And then our pi electrons in our carbonyl here moved out onto our oxygen. So it doesn't really matter which lone pair we say it is. Let's just say it's that one right there. And so this is our intermediate. Next step would be to deprotonate this intermediate here."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it doesn't really matter which lone pair we say it is. Let's just say it's that one right there. And so this is our intermediate. Next step would be to deprotonate this intermediate here. So we could have another molecule of alcohol. So let's go ahead and draw that in here. So another molecule of an alcohol, which in this case can function as a base."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Next step would be to deprotonate this intermediate here. So we could have another molecule of alcohol. So let's go ahead and draw that in here. So another molecule of an alcohol, which in this case can function as a base. So we could think about a lone pair of electrons on this alcohol taking this proton, leaving these electrons behind on this oxygen. So let's go ahead and draw the result of that acid-base reaction. So we deprotonate."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So another molecule of an alcohol, which in this case can function as a base. So we could think about a lone pair of electrons on this alcohol taking this proton, leaving these electrons behind on this oxygen. So let's go ahead and draw the result of that acid-base reaction. So we deprotonate. And so now we have our oxygen bonded to our R double prime group. And then we still have this oxygen over here on the right with a negative one formal charge. We still have an R and a hydrogen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we deprotonate. And so now we have our oxygen bonded to our R double prime group. And then we still have this oxygen over here on the right with a negative one formal charge. We still have an R and a hydrogen. So on this oxygen, there is now another lone pair of electrons, which came from the deprotonation step. So let's go ahead and show that. These electrons in here between the oxygen and the hydrogen are now on the oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We still have an R and a hydrogen. So on this oxygen, there is now another lone pair of electrons, which came from the deprotonation step. So let's go ahead and show that. These electrons in here between the oxygen and the hydrogen are now on the oxygen. So that takes away the plus one formal charge from the oxygen. And we have this. So we're obviously extremely close to our product."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "These electrons in here between the oxygen and the hydrogen are now on the oxygen. So that takes away the plus one formal charge from the oxygen. And we have this. So we're obviously extremely close to our product. We only need one more acid-base reaction. And alcohols are amphoteric like water. So they can function as acids and donate protons as well."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we're obviously extremely close to our product. We only need one more acid-base reaction. And alcohols are amphoteric like water. So they can function as acids and donate protons as well. So let's go ahead and show that. So we have another molecule of alcohol come along. And a lone pair of electrons takes this proton, leaves these electrons behind."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So they can function as acids and donate protons as well. So let's go ahead and show that. So we have another molecule of alcohol come along. And a lone pair of electrons takes this proton, leaves these electrons behind. And then we protonate and we can form our hemiacetal. So once again, we could have carried through an R prime group like that. If we had started with a ketone, and we would get this right here instead of the hydrogen for our hemiacetal product."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And a lone pair of electrons takes this proton, leaves these electrons behind. And then we protonate and we can form our hemiacetal. So once again, we could have carried through an R prime group like that. If we had started with a ketone, and we would get this right here instead of the hydrogen for our hemiacetal product. So both are hemiacetals. And so this is the general mechanism to form a hemiacetal. Now, this is not acid or base catalyzed, so I will cover that in the next video."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we had started with a ketone, and we would get this right here instead of the hydrogen for our hemiacetal product. So both are hemiacetals. And so this is the general mechanism to form a hemiacetal. Now, this is not acid or base catalyzed, so I will cover that in the next video. And so the formation of hemiacetals, usually the equilibrium is actually, favors the formation of your aldehyde or ketone. So it's usually back here to the left. However, for the formation of five or six-membered rings in an intramolecular hemiacetal formation, the equilibrium is actually to the right."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now, this is not acid or base catalyzed, so I will cover that in the next video. And so the formation of hemiacetals, usually the equilibrium is actually, favors the formation of your aldehyde or ketone. So it's usually back here to the left. However, for the formation of five or six-membered rings in an intramolecular hemiacetal formation, the equilibrium is actually to the right. And so this is a very important reaction. So let's take a look at it right here. So here we have over here an aldehyde and an alcohol in the same molecule."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "However, for the formation of five or six-membered rings in an intramolecular hemiacetal formation, the equilibrium is actually to the right. And so this is a very important reaction. So let's take a look at it right here. So here we have over here an aldehyde and an alcohol in the same molecule. So this is going to be an intramolecular hemiacetal reaction. So we're gonna push the equilibrium to the right. So it's actually to the right for these."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So here we have over here an aldehyde and an alcohol in the same molecule. So this is going to be an intramolecular hemiacetal reaction. So we're gonna push the equilibrium to the right. So it's actually to the right for these. We form a cyclic hemiacetal over here on the right. Let's go ahead and number our carbon so we can try to figure out what happened here. This carbon right here would be number one, two, three, four, and five, like this."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it's actually to the right for these. We form a cyclic hemiacetal over here on the right. Let's go ahead and number our carbon so we can try to figure out what happened here. This carbon right here would be number one, two, three, four, and five, like this. And we know what happens in the mechanism. We know that the alcohol functions as a nucleophile and attacks the carbonyl carbon. So this oxygen right here must be able to swing around and attack this carbonyl carbon and push these electrons off onto this oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here would be number one, two, three, four, and five, like this. And we know what happens in the mechanism. We know that the alcohol functions as a nucleophile and attacks the carbonyl carbon. So this oxygen right here must be able to swing around and attack this carbonyl carbon and push these electrons off onto this oxygen. And so let's go ahead and follow our carbons over here on the right. So this carbon turns out to be number one. This carbon's number two, three, four, and five."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen right here must be able to swing around and attack this carbonyl carbon and push these electrons off onto this oxygen. And so let's go ahead and follow our carbons over here on the right. So this carbon turns out to be number one. This carbon's number two, three, four, and five. And so this oxygen right here on our ring was this oxygen. And then let's go ahead and label the other oxygen as well. So this oxygen becomes this oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This carbon's number two, three, four, and five. And so this oxygen right here on our ring was this oxygen. And then let's go ahead and label the other oxygen as well. So this oxygen becomes this oxygen. And so formation of cyclic hemiacetals is extremely important in biochemistry and carbohydrate chemistry. And so let's go ahead and go through this in a little bit more detail here. And so here I have the exact same molecule, except I've shown it in a different conformation."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen becomes this oxygen. And so formation of cyclic hemiacetals is extremely important in biochemistry and carbohydrate chemistry. And so let's go ahead and go through this in a little bit more detail here. And so here I have the exact same molecule, except I've shown it in a different conformation. I've shown these sigma bonds here rotate a little bit differently. So all these sigma bonds in here have free rotation. And so they're gonna rotate to allow our nucleophile to attack our electrophile a little bit easier."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so here I have the exact same molecule, except I've shown it in a different conformation. I've shown these sigma bonds here rotate a little bit differently. So all these sigma bonds in here have free rotation. And so they're gonna rotate to allow our nucleophile to attack our electrophile a little bit easier. So I can think about lone pairs of electrons on our oxygen. And now it's a little bit easier to see the nucleophilic attack. Gonna attack right here at this carbon and push these electrons in here off onto this oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so they're gonna rotate to allow our nucleophile to attack our electrophile a little bit easier. So I can think about lone pairs of electrons on our oxygen. And now it's a little bit easier to see the nucleophilic attack. Gonna attack right here at this carbon and push these electrons in here off onto this oxygen. So let's go ahead and show the result of that. So now I have my oxygen. So it is bonded to this carbon now and I've formed my ring."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Gonna attack right here at this carbon and push these electrons in here off onto this oxygen. So let's go ahead and show the result of that. So now I have my oxygen. So it is bonded to this carbon now and I've formed my ring. So let's follow those electrons. These electrons right here in magenta now form this bond right here. And I'm gonna say that this oxygen right here in red is actually going to go up in the plane."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it is bonded to this carbon now and I've formed my ring. So let's follow those electrons. These electrons right here in magenta now form this bond right here. And I'm gonna say that this oxygen right here in red is actually going to go up in the plane. So we'll come back to that in a minute. So this oxygen right here is going to go up. It had two lone pairs of electrons around it."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And I'm gonna say that this oxygen right here in red is actually going to go up in the plane. So we'll come back to that in a minute. So this oxygen right here is going to go up. It had two lone pairs of electrons around it. It picked up another one. So that's where our negative one formal charge is. The oxygen in our ring still has a hydrogen bonded to it and a lone pair of electrons."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "It had two lone pairs of electrons around it. It picked up another one. So that's where our negative one formal charge is. The oxygen in our ring still has a hydrogen bonded to it and a lone pair of electrons. So it gets a plus one formal charge like that. And then I'm saying this hydrogen here is down. So this is our intermediate."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen in our ring still has a hydrogen bonded to it and a lone pair of electrons. So it gets a plus one formal charge like that. And then I'm saying this hydrogen here is down. So this is our intermediate. And without showing all of the acid-base steps here, let's just think about what happens. We know that we next deprotonate. So a base comes along, takes this proton, and then these electrons in here move off onto our oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is our intermediate. And without showing all of the acid-base steps here, let's just think about what happens. We know that we next deprotonate. So a base comes along, takes this proton, and then these electrons in here move off onto our oxygen. And then we know after that, we protonate our negative charge right here. So let's go ahead and draw one of the possible products. So we have our oxygen as part of our ring here."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So a base comes along, takes this proton, and then these electrons in here move off onto our oxygen. And then we know after that, we protonate our negative charge right here. So let's go ahead and draw one of the possible products. So we have our oxygen as part of our ring here. Drawn in a chair conformation. We now have an OH right here. And then, so we have an OH equatorial."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have our oxygen as part of our ring here. Drawn in a chair conformation. We now have an OH right here. And then, so we have an OH equatorial. And then we have a hydrogen axial here as one of our possible products. And then two lone pairs of electrons on this oxygen now. So hopefully you can see how this is one of the possible products."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then, so we have an OH equatorial. And then we have a hydrogen axial here as one of our possible products. And then two lone pairs of electrons on this oxygen now. So hopefully you can see how this is one of the possible products. And here we drew it with like a flat plane. And here we've drawn it in more of a chair conformation. So the OH equatorial for this one."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So hopefully you can see how this is one of the possible products. And here we drew it with like a flat plane. And here we've drawn it in more of a chair conformation. So the OH equatorial for this one. But let's go back to our original situation over here on the left where the nucleophile attacked the carbonyl carbon. We know that the geometry at our carbonyl carbon is trigonal planar. So it's possible the nucleophile could attack from the opposite side."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So the OH equatorial for this one. But let's go back to our original situation over here on the left where the nucleophile attacked the carbonyl carbon. We know that the geometry at our carbonyl carbon is trigonal planar. So it's possible the nucleophile could attack from the opposite side. And if that happened, then the oxygen would go down relative to the plane. So let's go ahead and show that now. So we have another possibility here."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it's possible the nucleophile could attack from the opposite side. And if that happened, then the oxygen would go down relative to the plane. So let's go ahead and show that now. So we have another possibility here. Alright, so let's go ahead and draw it out. So we have our oxygen is now going to be part of our ring. So we've formed our ring."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have another possibility here. Alright, so let's go ahead and draw it out. So we have our oxygen is now going to be part of our ring. So we've formed our ring. This time, instead of putting the oxygen going equatorial, we're gonna put the oxygen going down. We're gonna put it going axial. And so it has three lone pairs of electrons around it."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we've formed our ring. This time, instead of putting the oxygen going equatorial, we're gonna put the oxygen going down. We're gonna put it going axial. And so it has three lone pairs of electrons around it. So negative one formal charge. And the hydrogen goes equatorial like that. And then we still have a lone pair of electrons on this oxygen."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so it has three lone pairs of electrons around it. So negative one formal charge. And the hydrogen goes equatorial like that. And then we still have a lone pair of electrons on this oxygen. And then hydrogen and a plus one formal charge. And so this is another possibility. Alright, so once again, thinking about the mechanism, we know a base deprotonates."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we still have a lone pair of electrons on this oxygen. And then hydrogen and a plus one formal charge. And so this is another possibility. Alright, so once again, thinking about the mechanism, we know a base deprotonates. These electrons kick off onto here. And then we know that next we protonate our negative one charge like that. And that forms our hemiacetal."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Alright, so once again, thinking about the mechanism, we know a base deprotonates. These electrons kick off onto here. And then we know that next we protonate our negative one charge like that. And that forms our hemiacetal. So let's go ahead and draw it over here. Alright, so here's another product. Alright, so let's go ahead and this time we have our hydrogen equatorial."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And that forms our hemiacetal. So let's go ahead and draw it over here. Alright, so here's another product. Alright, so let's go ahead and this time we have our hydrogen equatorial. And our OH is now axial. And so these are our two possibilities. When I think about this carbon right here, and I think about the stereochemistry of this carbon, this is a chiral center."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Alright, so let's go ahead and this time we have our hydrogen equatorial. And our OH is now axial. And so these are our two possibilities. When I think about this carbon right here, and I think about the stereochemistry of this carbon, this is a chiral center. It's chiral. So we have these two as our possible products. Now this is extremely important when you get into carbohydrate chemistry."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "When I think about this carbon right here, and I think about the stereochemistry of this carbon, this is a chiral center. It's chiral. So we have these two as our possible products. Now this is extremely important when you get into carbohydrate chemistry. So these acetals, alright, so differ at carbon one. So this would be the carbon one position. And when you do carbohydrate chemistry, this is called the anomeric carbon."}, {"video_title": "Formation of hemiacetals and hemiketals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now this is extremely important when you get into carbohydrate chemistry. So these acetals, alright, so differ at carbon one. So this would be the carbon one position. And when you do carbohydrate chemistry, this is called the anomeric carbon. And if the OH is up or down, those are different anomers. And this is extremely important, once again, for something like glucose. And so we'll definitely talk about that in a later video."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a few. We'll start with the free radical bromination of alkyl benzenes. And so here is my alkyl benzene. So a benzene ring, I have an alkyl group attached to that. So this is a carbon. And on that carbon, this is the benzylic position. So this would be a benzylic hydrogen attached to that carbon, which is necessary for this reaction."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So a benzene ring, I have an alkyl group attached to that. So this is a carbon. And on that carbon, this is the benzylic position. So this would be a benzylic hydrogen attached to that carbon, which is necessary for this reaction. And then there are two other things attached to that carbon here. So if we add some NBS and some heat, and this is a free radical mechanism, so we'd usually add something to initiate this, like peroxide. You can see that we get a bromine added on to the benzylic position in the place of that hydrogen there."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this would be a benzylic hydrogen attached to that carbon, which is necessary for this reaction. And then there are two other things attached to that carbon here. So if we add some NBS and some heat, and this is a free radical mechanism, so we'd usually add something to initiate this, like peroxide. You can see that we get a bromine added on to the benzylic position in the place of that hydrogen there. So let's look at a quick reaction. So if I took propyl benzene, so I'm going to go ahead and draw out propyl benzene, so three carbons coming off like that. And to it, I add some NBS and some heat."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You can see that we get a bromine added on to the benzylic position in the place of that hydrogen there. So let's look at a quick reaction. So if I took propyl benzene, so I'm going to go ahead and draw out propyl benzene, so three carbons coming off like that. And to it, I add some NBS and some heat. And you could do this in a solvent like carbon tetrachloride and some peroxide here. Add in some peroxide to start your radical mechanism. And we have three carbons here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And to it, I add some NBS and some heat. And you could do this in a solvent like carbon tetrachloride and some peroxide here. Add in some peroxide to start your radical mechanism. And we have three carbons here. So there's one, two, three carbons for propyl benzene on our alkyl group. But the only one that is at the benzylic position is, of course, this one right here. So that's where we're going to end up adding our bromine."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have three carbons here. So there's one, two, three carbons for propyl benzene on our alkyl group. But the only one that is at the benzylic position is, of course, this one right here. So that's where we're going to end up adding our bromine. So when I draw the product, I go ahead and I still have all three of those carbons, but the bromine adds exclusively to the benzylic position right here like that. And if I look at the hydrogens that I have attached to this carbon, so we started off with two hydrogens attached to that carbon, and we're left with only one of those hydrogens. So let me just go ahead and highlight the hydrogen in magenta here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's where we're going to end up adding our bromine. So when I draw the product, I go ahead and I still have all three of those carbons, but the bromine adds exclusively to the benzylic position right here like that. And if I look at the hydrogens that I have attached to this carbon, so we started off with two hydrogens attached to that carbon, and we're left with only one of those hydrogens. So let me just go ahead and highlight the hydrogen in magenta here. So if we go back up here to this general reaction, so this benzylic hydrogen, I could say that it's either one of these two. I'm going to go ahead and say that it's this one right here. And so the other hydrogen is attached to that carbon, and it is still left for our final product."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me just go ahead and highlight the hydrogen in magenta here. So if we go back up here to this general reaction, so this benzylic hydrogen, I could say that it's either one of these two. I'm going to go ahead and say that it's this one right here. And so the other hydrogen is attached to that carbon, and it is still left for our final product. So the bromine took the place of one of those benzylic hydrogens there. Now, the reason why the reaction occurs only at the benzylic position is because in this mechanism, you form a radical. So let me just go ahead and draw what the radical would look like for this reaction here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the other hydrogen is attached to that carbon, and it is still left for our final product. So the bromine took the place of one of those benzylic hydrogens there. Now, the reason why the reaction occurs only at the benzylic position is because in this mechanism, you form a radical. So let me just go ahead and draw what the radical would look like for this reaction here. So a radical for the reaction of propyl benzene. So we have our benzene ring. And we already lost the hydrogen in magenta."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me just go ahead and draw what the radical would look like for this reaction here. So a radical for the reaction of propyl benzene. So we have our benzene ring. And we already lost the hydrogen in magenta. So now we're left with just the hydrogen in blue here like that. And we're going to have an unpaired electron right here on this carbon. And because this radical, this electron, is right next to this benzene ring, we can draw some resonance structures."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we already lost the hydrogen in magenta. So now we're left with just the hydrogen in blue here like that. And we're going to have an unpaired electron right here on this carbon. And because this radical, this electron, is right next to this benzene ring, we can draw some resonance structures. So I could take this one electron and show it moving into here with my fishhook arrow. And then from this pi bond, I could take one of those electrons and move it into here. And the other one could come off onto this carbon."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And because this radical, this electron, is right next to this benzene ring, we can draw some resonance structures. So I could take this one electron and show it moving into here with my fishhook arrow. And then from this pi bond, I could take one of those electrons and move it into here. And the other one could come off onto this carbon. So for a resonance structure here, I can draw my ring. And I can show these pi bonds over here. Now I can show a double bond into here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the other one could come off onto this carbon. So for a resonance structure here, I can draw my ring. And I can show these pi bonds over here. Now I can show a double bond into here. There's still a hydrogen attached at this point. And then I have my other carbons like that. And then there is an electron out on this carbon."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now I can show a double bond into here. There's still a hydrogen attached at this point. And then I have my other carbons like that. And then there is an electron out on this carbon. So the benzylic radical is resonance stabilized. And if you formed a radical at the other two carbons on the ring, so not on the ring, on the alkyl group right here, these are not resonance stabilized because they're not right next to the ring. And so that's just a little bit of insight as to why the reaction occurs only at the benzylic position."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then there is an electron out on this carbon. So the benzylic radical is resonance stabilized. And if you formed a radical at the other two carbons on the ring, so not on the ring, on the alkyl group right here, these are not resonance stabilized because they're not right next to the ring. And so that's just a little bit of insight as to why the reaction occurs only at the benzylic position. It's because of resonance stabilization of the benzylic radical. So let's look at another type of reaction. And this, of course, would be a substitution reaction of a benzylic halide."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's just a little bit of insight as to why the reaction occurs only at the benzylic position. It's because of resonance stabilization of the benzylic radical. So let's look at another type of reaction. And this, of course, would be a substitution reaction of a benzylic halide. So I put a bromine here, but you can imagine a different halogen. And if you add a nucleophile and a solvent, depending on the classification of your alkyl halide portion of the molecule and what solvent you use, the mechanism could be either SN1 or SN2. But of course, the end result would be for your nucleophile to substitute in for your halogen at the benzylic position."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this, of course, would be a substitution reaction of a benzylic halide. So I put a bromine here, but you can imagine a different halogen. And if you add a nucleophile and a solvent, depending on the classification of your alkyl halide portion of the molecule and what solvent you use, the mechanism could be either SN1 or SN2. But of course, the end result would be for your nucleophile to substitute in for your halogen at the benzylic position. So if we were to look at a reaction here, so let me just go ahead and show one. So let's say we started with benzyl bromide here. So this is our benzyl bromide molecule."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But of course, the end result would be for your nucleophile to substitute in for your halogen at the benzylic position. So if we were to look at a reaction here, so let me just go ahead and show one. So let's say we started with benzyl bromide here. So this is our benzyl bromide molecule. And to benzyl bromide, we added some sodium hydroxide. So I have Na plus and OH minus. I'm going to go ahead and draw my lone pairs on my hydroxide anion, which is going to function as a nucleophile."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is our benzyl bromide molecule. And to benzyl bromide, we added some sodium hydroxide. So I have Na plus and OH minus. I'm going to go ahead and draw my lone pairs on my hydroxide anion, which is going to function as a nucleophile. So this is our nucleophile. And if I look at the structure of this alkyl halide over here, so this carbon that's bonded to my bromine is bonded to one other carbon. So that's primary."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and draw my lone pairs on my hydroxide anion, which is going to function as a nucleophile. So this is our nucleophile. And if I look at the structure of this alkyl halide over here, so this carbon that's bonded to my bromine is bonded to one other carbon. So that's primary. And if we're talking about a primary alkyl halide, decreased steric hindrance. And so your mechanism would be SN2. So if you think about an SN2 type mechanism for this, which is a concerted mechanism, your nucleophile is going to attack this carbon at the same time that these electrons kick off onto your bromine to form the bromide anion as a relatively stable leaving group."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's primary. And if we're talking about a primary alkyl halide, decreased steric hindrance. And so your mechanism would be SN2. So if you think about an SN2 type mechanism for this, which is a concerted mechanism, your nucleophile is going to attack this carbon at the same time that these electrons kick off onto your bromine to form the bromide anion as a relatively stable leaving group. And so this is a concerted mechanism. Everything happens at the same time in an SN2 type mechanism. And we end up with an OH replacing RBr."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if you think about an SN2 type mechanism for this, which is a concerted mechanism, your nucleophile is going to attack this carbon at the same time that these electrons kick off onto your bromine to form the bromide anion as a relatively stable leaving group. And so this is a concerted mechanism. Everything happens at the same time in an SN2 type mechanism. And we end up with an OH replacing RBr. So we form benzyl alcohol for this reaction. So that's SN2. If you thought about an SN1 type reaction, let's go ahead and think about that."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we end up with an OH replacing RBr. So we form benzyl alcohol for this reaction. So that's SN2. If you thought about an SN1 type reaction, let's go ahead and think about that. If I started with a tertiary alkyl halide, so something like this. So if I were to analyze that, put my lone pairs of electrons on this bromine, the carbon that is attached to my bromine is attached to one, two, and three other carbons. So therefore, it's tertiary."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If you thought about an SN1 type reaction, let's go ahead and think about that. If I started with a tertiary alkyl halide, so something like this. So if I were to analyze that, put my lone pairs of electrons on this bromine, the carbon that is attached to my bromine is attached to one, two, and three other carbons. So therefore, it's tertiary. So you could think about an SN1 type mechanism. And we know in an SN1 type mechanism, let me go ahead and just write that here. So in an SN1 type mechanism, your first step is dissociation."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So therefore, it's tertiary. So you could think about an SN1 type mechanism. And we know in an SN1 type mechanism, let me go ahead and just write that here. So in an SN1 type mechanism, your first step is dissociation. So you're going to get these electrons in here kicking off onto your halogen, onto your bromine to form your bromide anion. And you would therefore be taking a bond away from this carbon here at the benzylic position. So let me go ahead and draw the resulting carbocation."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So in an SN1 type mechanism, your first step is dissociation. So you're going to get these electrons in here kicking off onto your halogen, onto your bromine to form your bromide anion. And you would therefore be taking a bond away from this carbon here at the benzylic position. So let me go ahead and draw the resulting carbocation. So I have my ring. I have these two methyl groups coming off of that carbon. And that is where I have a plus 1 formal charge."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw the resulting carbocation. So I have my ring. I have these two methyl groups coming off of that carbon. And that is where I have a plus 1 formal charge. So I have a benzylic carbocation here, which is resonance stabilized. So I could think about these electrons moving into here. And I could go ahead and draw a resonance structure."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that is where I have a plus 1 formal charge. So I have a benzylic carbocation here, which is resonance stabilized. So I could think about these electrons moving into here. And I could go ahead and draw a resonance structure. So for this resonance structure, these pi electrons are over here. Now there's a double bond between those two carbons. And let me show those electrons here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I could go ahead and draw a resonance structure. So for this resonance structure, these pi electrons are over here. Now there's a double bond between those two carbons. And let me show those electrons here. So the electrons in blue have moved out to here to form a pi bond, took a bond away from this carbon. So that carbon gets our plus 1 formal charge. And you could, of course, continue."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let me show those electrons here. So the electrons in blue have moved out to here to form a pi bond, took a bond away from this carbon. So that carbon gets our plus 1 formal charge. And you could, of course, continue. But the point is that the presence of our benzene ring allows for resonance stabilization of our benzylic carbocation, which means that the substitution is going to occur at the benzylic position because of that resonance stabilization here. So let's look at one more reaction. And this is oxidation of alkyl benzenes."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you could, of course, continue. But the point is that the presence of our benzene ring allows for resonance stabilization of our benzylic carbocation, which means that the substitution is going to occur at the benzylic position because of that resonance stabilization here. So let's look at one more reaction. And this is oxidation of alkyl benzenes. And so here I have my alkyl benzene. So once again, carbon with a benzylic hydrogen. And then two other things attached to my carbon here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this is oxidation of alkyl benzenes. And so here I have my alkyl benzene. So once again, carbon with a benzylic hydrogen. And then two other things attached to my carbon here. If I added some sodium dichromate, a source of protons, so something like sulfuric acid, and I heat up my reaction, I can oxidize my alkyl side chain to a carboxylic acid functional group over here. So this would, of course, be the benzoic acid molecule. You could also do this oxidation with something like permanganate and some heat as well."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then two other things attached to my carbon here. If I added some sodium dichromate, a source of protons, so something like sulfuric acid, and I heat up my reaction, I can oxidize my alkyl side chain to a carboxylic acid functional group over here. So this would, of course, be the benzoic acid molecule. You could also do this oxidation with something like permanganate and some heat as well. So that's another possibility. Let's go ahead and do an example. Let's start with, let's do butyl benzene this time."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You could also do this oxidation with something like permanganate and some heat as well. So that's another possibility. Let's go ahead and do an example. Let's start with, let's do butyl benzene this time. So if I have butyl benzene, so four carbons for my alkyl group here. And instead of rewriting all that stuff, I'm just going to put some ditto marks here. So we add some sodium dichromate, a source of protons, sulfuric acid, and some heat."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with, let's do butyl benzene this time. So if I have butyl benzene, so four carbons for my alkyl group here. And instead of rewriting all that stuff, I'm just going to put some ditto marks here. So we add some sodium dichromate, a source of protons, sulfuric acid, and some heat. We're going to oxidize our side chain. And it doesn't really matter the length of the alkyl group you have. In this case, we have four carbons."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we add some sodium dichromate, a source of protons, sulfuric acid, and some heat. We're going to oxidize our side chain. And it doesn't really matter the length of the alkyl group you have. In this case, we have four carbons. We're still going to get a carboxylic acid. The reaction is still going to occur at the benzylic position. And we get a carboxylic acid here."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In this case, we have four carbons. We're still going to get a carboxylic acid. The reaction is still going to occur at the benzylic position. And we get a carboxylic acid here. Again, a complicated mechanism, so I won't go into it. But again, the length of carbons that you have doesn't matter. So you could add on a bunch more, and you would still get this as your product."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we get a carboxylic acid here. Again, a complicated mechanism, so I won't go into it. But again, the length of carbons that you have doesn't matter. So you could add on a bunch more, and you would still get this as your product. What about if you try to do this reaction with something like tert-butyl benzene? So instead of butyl benzene, now we have tert-butyl benzene like that. And we went ahead and added sodium dichromate and sulfuric acid and heat again."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could add on a bunch more, and you would still get this as your product. What about if you try to do this reaction with something like tert-butyl benzene? So instead of butyl benzene, now we have tert-butyl benzene like that. And we went ahead and added sodium dichromate and sulfuric acid and heat again. We heated it up. We wouldn't get any kind of reaction, so no reaction this time. And that's because we don't have any benzylic hydrogens present."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we went ahead and added sodium dichromate and sulfuric acid and heat again. We heated it up. We wouldn't get any kind of reaction, so no reaction this time. And that's because we don't have any benzylic hydrogens present. So we go back here to our generic reaction. There's a benzylic hydrogen. If I go down to here, here's a methyl group, here's a methyl group, here's a methyl group, and then this is a carbon."}, {"video_title": "Reactions at the benzylic position Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that's because we don't have any benzylic hydrogens present. So we go back here to our generic reaction. There's a benzylic hydrogen. If I go down to here, here's a methyl group, here's a methyl group, here's a methyl group, and then this is a carbon. So we don't have any benzylic hydrogens. There are no hydrogens attached to this carbon here, which is necessary for the reaction to occur. And so we see no reaction in that case."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "However, you can get them to hydrolyze if you use harsh reaction conditions. So if you use strong acid or strong base and you heat things up for several hours, you can hydrolyze amides. And so if we take a look at this amide right here, we can break this bond, right? So using acid and heat and form a carboxylic acid. Let's look at the mechanism for acid-catalyzed hydrolysis of amides. And so the first step is protonation of the carbonyl oxygen. So a lone pair of electrons on this oxygen picks up a proton from hydronium, leaving these electrons behind."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So using acid and heat and form a carboxylic acid. Let's look at the mechanism for acid-catalyzed hydrolysis of amides. And so the first step is protonation of the carbonyl oxygen. So a lone pair of electrons on this oxygen picks up a proton from hydronium, leaving these electrons behind. And so we've seen in previous videos, once again, that protonating your carbonyl oxygen makes your carbonyl carbon more electrophilic. And so let me just draw in the NH2 right here. We can follow those electrons along."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on this oxygen picks up a proton from hydronium, leaving these electrons behind. And so we've seen in previous videos, once again, that protonating your carbonyl oxygen makes your carbonyl carbon more electrophilic. And so let me just draw in the NH2 right here. We can follow those electrons along. So the lone pair of electrons right here on the oxygen is going to pick up a proton from hydronium, forming this bond right here, and giving the oxygen a plus one formal charge. If you think about a resonance structure for this, it's going to withdraw some electron density from our carbonyl carbon here, making that carbon more electrophilic. And so therefore, it's going to react with a nucleophile."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "We can follow those electrons along. So the lone pair of electrons right here on the oxygen is going to pick up a proton from hydronium, forming this bond right here, and giving the oxygen a plus one formal charge. If you think about a resonance structure for this, it's going to withdraw some electron density from our carbonyl carbon here, making that carbon more electrophilic. And so therefore, it's going to react with a nucleophile. And the nucleophile that's present would be water. So I'm going to go ahead and draw water in here. If you deprotonate hydronium, you form H2O."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, it's going to react with a nucleophile. And the nucleophile that's present would be water. So I'm going to go ahead and draw water in here. If you deprotonate hydronium, you form H2O. And so water's going to function as a nucleophile, attack our electrophile, which is our carbon, which would push these electrons off onto the oxygen. So if we go ahead and draw that, we would now have on the left side, we would have an OH, so with two lone pairs of electrons. So let's follow those electrons in here."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "If you deprotonate hydronium, you form H2O. And so water's going to function as a nucleophile, attack our electrophile, which is our carbon, which would push these electrons off onto the oxygen. So if we go ahead and draw that, we would now have on the left side, we would have an OH, so with two lone pairs of electrons. So let's follow those electrons in here. So these electrons in green move off onto the oxygen. And we're going to form a bond between the carbon and this oxygen here. This oxygen has also two hydrogens on it, lone pair of electrons, plus one formal charge."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons in here. So these electrons in green move off onto the oxygen. And we're going to form a bond between the carbon and this oxygen here. This oxygen has also two hydrogens on it, lone pair of electrons, plus one formal charge. So let's say the electrons in blue, so these electrons right here in blue, form a bond between the carbon and that oxygen. That gives the oxygen a plus one formal charge. And then we also have our NH2 down here like that."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen has also two hydrogens on it, lone pair of electrons, plus one formal charge. So let's say the electrons in blue, so these electrons right here in blue, form a bond between the carbon and that oxygen. That gives the oxygen a plus one formal charge. And then we also have our NH2 down here like that. And the next step, we have a plus one formal charge, so we're going to deprotonate to get rid of that plus one charge. And water's going to come along again, this time function as a base. So we just saw water act as a nucleophile, and now water's going to act as a base."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And then we also have our NH2 down here like that. And the next step, we have a plus one formal charge, so we're going to deprotonate to get rid of that plus one charge. And water's going to come along again, this time function as a base. So we just saw water act as a nucleophile, and now water's going to act as a base. So it's going to take a proton, leaving those electrons behind on the oxygen, and that gets rid of our formal charge. So on the left side, we would have an OH, and we would also have an OH now on the right side. So let me go ahead and show those electrons."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we just saw water act as a nucleophile, and now water's going to act as a base. So it's going to take a proton, leaving those electrons behind on the oxygen, and that gets rid of our formal charge. So on the left side, we would have an OH, and we would also have an OH now on the right side. So let me go ahead and show those electrons. So if we're going to deprotonate, take that proton, leave these electrons behind on the oxygen, so those are those electrons right there. I could draw these electrons in blue to remind us of which ones they are. And then we still have an NH2."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and show those electrons. So if we're going to deprotonate, take that proton, leave these electrons behind on the oxygen, so those are those electrons right there. I could draw these electrons in blue to remind us of which ones they are. And then we still have an NH2. This time when I draw the NH2, I'm going to draw it out. So I'm going to show a lone pair of electrons on the nitrogen, I'm going to draw in those hydrogens. Because in the next step, this amine is going to function as a base."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And then we still have an NH2. This time when I draw the NH2, I'm going to draw it out. So I'm going to show a lone pair of electrons on the nitrogen, I'm going to draw in those hydrogens. Because in the next step, this amine is going to function as a base. And we have hydronium present, which we know is of course an acid. So H3O+, I'm drawing everything here. So the amine is going to act as a base, and hydronium is going to act as an acid."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "Because in the next step, this amine is going to function as a base. And we have hydronium present, which we know is of course an acid. So H3O+, I'm drawing everything here. So the amine is going to act as a base, and hydronium is going to act as an acid. The amine is going to take a proton from H3O+, leaving these electrons behind on the oxygen. So let's get some more room here. So we have another acid-base reaction."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So the amine is going to act as a base, and hydronium is going to act as an acid. The amine is going to take a proton from H3O+, leaving these electrons behind on the oxygen. So let's get some more room here. So we have another acid-base reaction. So we're going to protonate the amine, because that's going to give us a better leaving group. So let me go ahead and draw in everything else. So once again, we have our OHs on both sides."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we have another acid-base reaction. So we're going to protonate the amine, because that's going to give us a better leaving group. So let me go ahead and draw in everything else. So once again, we have our OHs on both sides. So I'm going to draw those in. We still have our R group. And then over here, now we have our nitrogen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we have our OHs on both sides. So I'm going to draw those in. We still have our R group. And then over here, now we have our nitrogen. Now has four bonds. That gives nitrogen a plus one formal charge. So let's follow those electrons."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And then over here, now we have our nitrogen. Now has four bonds. That gives nitrogen a plus one formal charge. So let's follow those electrons. So let's make them magenta. So these electrons right here on the nitrogen are going to take this proton. So we could say it's this one right here, giving the nitrogen a plus one formal charge."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons. So let's make them magenta. So these electrons right here on the nitrogen are going to take this proton. So we could say it's this one right here, giving the nitrogen a plus one formal charge. And this gives us a good leaving group. Because if you look over here, you can see ammonia is hiding as our leaving group. And in our next step, we're going to reform our carbonyl."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we could say it's this one right here, giving the nitrogen a plus one formal charge. And this gives us a good leaving group. Because if you look over here, you can see ammonia is hiding as our leaving group. And in our next step, we're going to reform our carbonyl. And so these electrons right here are going to move in to reform our carbonyl. Once again, too many bonds to this carbon. And so these electrons have to come off onto the nitrogen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And in our next step, we're going to reform our carbonyl. And so these electrons right here are going to move in to reform our carbonyl. Once again, too many bonds to this carbon. And so these electrons have to come off onto the nitrogen. So let's go ahead and draw what we would form. So we would have an R group. We would have a carbon now double bonded to this oxygen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so these electrons have to come off onto the nitrogen. So let's go ahead and draw what we would form. So we would have an R group. We would have a carbon now double bonded to this oxygen. A plus one formal charge on this oxygen. So let's show those electrons. So the electrons in here, in red, move in to reform our carbonyl."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "We would have a carbon now double bonded to this oxygen. A plus one formal charge on this oxygen. So let's show those electrons. So the electrons in here, in red, move in to reform our carbonyl. And we still have an OH bonded to our carbon. But we pushed the electrons in here in blue off onto the nitrogen. So we form ammonia."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in here, in red, move in to reform our carbonyl. And we still have an OH bonded to our carbon. But we pushed the electrons in here in blue off onto the nitrogen. So we form ammonia. So let me go ahead and draw in ammonia over here. So we have ammonia now. So NH3 with a lone pair of electrons on that nitrogen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we form ammonia. So let me go ahead and draw in ammonia over here. So we have ammonia now. So NH3 with a lone pair of electrons on that nitrogen. And so I'll say that the electrons in blue were the ones that came off right here onto the nitrogen. And so we're almost to our final product of a carboxylic acid. All we have to do is deprotonate."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So NH3 with a lone pair of electrons on that nitrogen. And so I'll say that the electrons in blue were the ones that came off right here onto the nitrogen. And so we're almost to our final product of a carboxylic acid. All we have to do is deprotonate. And so ammonia is gonna function as a base. It's going to take this proton, leaving these electrons behind on the oxygen. And of course that's gonna give us our carboxylic acid."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "All we have to do is deprotonate. And so ammonia is gonna function as a base. It's going to take this proton, leaving these electrons behind on the oxygen. And of course that's gonna give us our carboxylic acid. So let me go ahead and draw in the carboxylic acid here. And we could say, let's make these electrons, let's say they're green here. So these electrons right here in green come off onto the oxygen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And of course that's gonna give us our carboxylic acid. So let me go ahead and draw in the carboxylic acid here. And we could say, let's make these electrons, let's say they're green here. So these electrons right here in green come off onto the oxygen. And we now have our carboxylic acid. Let me draw in the OH here. And then if I take ammonia and I add a proton to it, then I would form NH4 plus, ammonium."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here in green come off onto the oxygen. And we now have our carboxylic acid. Let me draw in the OH here. And then if I take ammonia and I add a proton to it, then I would form NH4 plus, ammonium. Let me see if I can squeeze that in here. So we would have the ammonium ion, so NH4 plus. And so the electrons in blue, so these electrons in blue take this proton, so we could form, let's say, that bond."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And then if I take ammonia and I add a proton to it, then I would form NH4 plus, ammonium. Let me see if I can squeeze that in here. So we would have the ammonium ion, so NH4 plus. And so the electrons in blue, so these electrons in blue take this proton, so we could form, let's say, that bond. And we have our products. So we formed a carboxylic acid and we formed ammonium. So this last step here favors the formation of the products."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so the electrons in blue, so these electrons in blue take this proton, so we could form, let's say, that bond. And we have our products. So we formed a carboxylic acid and we formed ammonium. So this last step here favors the formation of the products. And this is what helps to drive this reaction to completion, to make our carboxylic acid. Let's look at another way to hydrolyze an amide. This time we're gonna use base."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So this last step here favors the formation of the products. And this is what helps to drive this reaction to completion, to make our carboxylic acid. Let's look at another way to hydrolyze an amide. This time we're gonna use base. So we just did it using acid. And so this time we're gonna look at the mechanism using base. And so here once again we have our amides."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "This time we're gonna use base. So we just did it using acid. And so this time we're gonna look at the mechanism using base. And so here once again we have our amides. We're gonna add sodium hydroxide this time. The hydroxide anion is a better nucleophile than water. So for this mechanism, right away our hydroxide anion is gonna function as a nucleophile."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so here once again we have our amides. We're gonna add sodium hydroxide this time. The hydroxide anion is a better nucleophile than water. So for this mechanism, right away our hydroxide anion is gonna function as a nucleophile. It's going to attack our carbonyl carbon right here, pushing these electrons off onto the oxygen. So let's go ahead and show what we would make. So we would now have on the left side, we would have an oxygen with three lone pairs of electrons, so negative one formal charge."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So for this mechanism, right away our hydroxide anion is gonna function as a nucleophile. It's going to attack our carbonyl carbon right here, pushing these electrons off onto the oxygen. So let's go ahead and show what we would make. So we would now have on the left side, we would have an oxygen with three lone pairs of electrons, so negative one formal charge. So these electrons in here come off onto the oxygen. And let's see, what else do we have bonded to this carbon? We have our R group, we have our NH2, and we have our OH."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we would now have on the left side, we would have an oxygen with three lone pairs of electrons, so negative one formal charge. So these electrons in here come off onto the oxygen. And let's see, what else do we have bonded to this carbon? We have our R group, we have our NH2, and we have our OH. So I'm gonna go ahead and draw in the OH here. So let's highlight a pair of electrons on our sodium hydroxide. So let's say these electrons right here on hydroxide are the ones that form this bond between the carbon and the oxygen like that."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "We have our R group, we have our NH2, and we have our OH. So I'm gonna go ahead and draw in the OH here. So let's highlight a pair of electrons on our sodium hydroxide. So let's say these electrons right here on hydroxide are the ones that form this bond between the carbon and the oxygen like that. Okay, so now we can reform our carbonyl. So we're gonna reform our carbonyl here. And I should put in a lone pair of electrons on this nitrogen here, because when we reform our carbonyl, so let's say these electrons move in here to form our carbonyl, then these electrons would come off onto the nitrogen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's say these electrons right here on hydroxide are the ones that form this bond between the carbon and the oxygen like that. Okay, so now we can reform our carbonyl. So we're gonna reform our carbonyl here. And I should put in a lone pair of electrons on this nitrogen here, because when we reform our carbonyl, so let's say these electrons move in here to form our carbonyl, then these electrons would come off onto the nitrogen. And let's go ahead and draw what we would have now. So we'd have an R group, we would have our carbonyl has been reformed, right? We would have our oxygen and our hydrogen like that."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And I should put in a lone pair of electrons on this nitrogen here, because when we reform our carbonyl, so let's say these electrons move in here to form our carbonyl, then these electrons would come off onto the nitrogen. And let's go ahead and draw what we would have now. So we'd have an R group, we would have our carbonyl has been reformed, right? We would have our oxygen and our hydrogen like that. So we formed our carboxylic acid. Let me go ahead and put these electrons here in blue so we can see which ones they are like that. And we lost, let's think about what we lost here."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "We would have our oxygen and our hydrogen like that. So we formed our carboxylic acid. Let me go ahead and put these electrons here in blue so we can see which ones they are like that. And we lost, let's think about what we lost here. We lost NH2. Let me go ahead and change colors. Let me make this yellow again."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And we lost, let's think about what we lost here. We lost NH2. Let me go ahead and change colors. Let me make this yellow again. So we have NH2, but let's think about our electrons. We already had a lone pair of electrons on the nitrogen, right? And we got another lone pair."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "Let me make this yellow again. So we have NH2, but let's think about our electrons. We already had a lone pair of electrons on the nitrogen, right? And we got another lone pair. So let me go ahead and show where that came from. So these electrons right here came off onto the nitrogen too. So let's make them these right here."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And we got another lone pair. So let me go ahead and show where that came from. So these electrons right here came off onto the nitrogen too. So let's make them these right here. That gives the nitrogen a negative one formal charge. And so this is the amide anion, which is a pretty strong base. And so we have a strong base and we have an acid."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's make them these right here. That gives the nitrogen a negative one formal charge. And so this is the amide anion, which is a pretty strong base. And so we have a strong base and we have an acid. And so we're going to have an acid-base reaction. The base is going to take a proton from the acid. So let's say it's these electrons in red."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so we have a strong base and we have an acid. And so we're going to have an acid-base reaction. The base is going to take a proton from the acid. So let's say it's these electrons in red. Take a proton from the acid, leave these electrons behind. And let's go ahead and draw the product. We would have the conjugate base to a carboxylic acid, which is a carboxylate anion."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's say it's these electrons in red. Take a proton from the acid, leave these electrons behind. And let's go ahead and draw the product. We would have the conjugate base to a carboxylic acid, which is a carboxylate anion. So I'm going to draw the lone pairs on the oxygen here, which gives this oxygen a negative one formal charge. Let's use green here. So these electrons in here come off onto our oxygen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "We would have the conjugate base to a carboxylic acid, which is a carboxylate anion. So I'm going to draw the lone pairs on the oxygen here, which gives this oxygen a negative one formal charge. Let's use green here. So these electrons in here come off onto our oxygen. So we form our carboxylate anion. Let's think about the other product. If we have NH2 minus and we're adding a proton to that, then we would form NH3."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here come off onto our oxygen. So we form our carboxylate anion. Let's think about the other product. If we have NH2 minus and we're adding a proton to that, then we would form NH3. So we form ammonia. So let me go ahead and draw in ammonia here. Put my lone pair of electrons on the nitrogen."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "If we have NH2 minus and we're adding a proton to that, then we would form NH3. So we form ammonia. So let me go ahead and draw in ammonia here. Put my lone pair of electrons on the nitrogen. And so these electrons here in red pick up this proton. So let's say it forms this bond right here. And so we make ammonia."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "Put my lone pair of electrons on the nitrogen. And so these electrons here in red pick up this proton. So let's say it forms this bond right here. And so we make ammonia. Once again, this acid-base equilibrium favors the formation of our products. And so this helps to drive the reaction forward. So we drive the reaction forward."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so we make ammonia. Once again, this acid-base equilibrium favors the formation of our products. And so this helps to drive the reaction forward. So we drive the reaction forward. And finally, let's think about how we would get from a carboxylate anion to our carboxylic acid product. And so in the first step, we added our sodium hydroxide. That gives us this."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we drive the reaction forward. And finally, let's think about how we would get from a carboxylate anion to our carboxylic acid product. And so in the first step, we added our sodium hydroxide. That gives us this. And so if we wanted to go from our carboxylate anion to a carboxylic acid, we need a source of protons. So the carboxylate anion picks up a proton and then forms our carboxylic acid. So that's two ways to hydrolyze an amide."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "That gives us this. And so if we wanted to go from our carboxylate anion to a carboxylic acid, we need a source of protons. So the carboxylate anion picks up a proton and then forms our carboxylic acid. So that's two ways to hydrolyze an amide. And let's take a look at a reaction here. Let's say our goal was to go from benzamide to benzoic acid. So we've just learned two ways to do that."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So that's two ways to hydrolyze an amide. And let's take a look at a reaction here. Let's say our goal was to go from benzamide to benzoic acid. So we've just learned two ways to do that. One way to do that would be to add water and an acid. So we get H3O plus. And if we heat things up, we know that we could hydrolyze our amide that way and give us our benzoic acid."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we've just learned two ways to do that. One way to do that would be to add water and an acid. So we get H3O plus. And if we heat things up, we know that we could hydrolyze our amide that way and give us our benzoic acid. So that's one possibility. So that's acid-catalyzed amide hydrolysis. Or we could use base."}, {"video_title": "Acid and base-catalyzed hydrolysis of amides Organic chemistry Khan Academy.mp3", "Sentence": "And if we heat things up, we know that we could hydrolyze our amide that way and give us our benzoic acid. So that's one possibility. So that's acid-catalyzed amide hydrolysis. Or we could use base. Or if we wanted to form a carboxylic acid, first thing we could do is sodium hydroxide and heat things up. So Na plus, OH minus. Once again, heat things up."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Here's the hydroboration oxidation of alkyne reaction. So we start with our alkyne. And once again, usually it's a terminal alkyne. So here's your hydrogen here. And on the other side of our triple bond, let's say there's some R group attached to this carbon. So to our alkyne, we're going to add, in the first step, borane, BH3, in THF as our solvent. And in the second step, we're going to add hydrogen peroxide and hydroxide anion OH minus."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here's your hydrogen here. And on the other side of our triple bond, let's say there's some R group attached to this carbon. So to our alkyne, we're going to add, in the first step, borane, BH3, in THF as our solvent. And in the second step, we're going to add hydrogen peroxide and hydroxide anion OH minus. And we're going to get addition of water across our double bond. So it's very similar to the last reaction. The difference here is the regiochemistry."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And in the second step, we're going to add hydrogen peroxide and hydroxide anion OH minus. And we're going to get addition of water across our double bond. So it's very similar to the last reaction. The difference here is the regiochemistry. This regiochemistry is actually anti-Markovnikov addition. So this is anti-Mark's rule. So anti-Markovnikov, meaning the OH is going to add to the least substituted carbon, which, of course, must be the carbon over here on the left."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The difference here is the regiochemistry. This regiochemistry is actually anti-Markovnikov addition. So this is anti-Mark's rule. So anti-Markovnikov, meaning the OH is going to add to the least substituted carbon, which, of course, must be the carbon over here on the left. So we end up adding H2O water across our triple bond. So now there's a double bond present. And then there's an OH in our molecule."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So anti-Markovnikov, meaning the OH is going to add to the least substituted carbon, which, of course, must be the carbon over here on the left. So we end up adding H2O water across our triple bond. So now there's a double bond present. And then there's an OH in our molecule. So just like in the last video, we form an enol here like that. And the enol, of course, is not the most stable form. We're going to get some rearrangement."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then there's an OH in our molecule. So just like in the last video, we form an enol here like that. And the enol, of course, is not the most stable form. We're going to get some rearrangement. And we're going to end up with an aldehyde here. So we can see we get an aldehyde functional group from a terminal alkyne undergoing this reaction. Now, sometimes if you use borane in this first step over here, sometimes borane is so reactive that it's actually going to add two molecules across your triple bond since your triple bond consists of two pi bonds."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to get some rearrangement. And we're going to end up with an aldehyde here. So we can see we get an aldehyde functional group from a terminal alkyne undergoing this reaction. Now, sometimes if you use borane in this first step over here, sometimes borane is so reactive that it's actually going to add two molecules across your triple bond since your triple bond consists of two pi bonds. So sometimes you'll see a different molecule used. Sometimes you'll see a dialkyl borane. So put two alkyl groups."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, sometimes if you use borane in this first step over here, sometimes borane is so reactive that it's actually going to add two molecules across your triple bond since your triple bond consists of two pi bonds. So sometimes you'll see a different molecule used. Sometimes you'll see a dialkyl borane. So put two alkyl groups. I'll put R2BH. And if you make them really bulky, this steric hindrance prevents the second addition of the borane because in the mechanism you want only one addition of the borane. And the borane is going to add on to the same carbon that the OH does."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So put two alkyl groups. I'll put R2BH. And if you make them really bulky, this steric hindrance prevents the second addition of the borane because in the mechanism you want only one addition of the borane. And the borane is going to add on to the same carbon that the OH does. So check out the video on hydroboration oxidation of alkenes for much more detail of that first mechanism that we discussed. So the net result of this reaction is to form an aldehyde from your terminal alkyne. So let's look at a reaction here."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the borane is going to add on to the same carbon that the OH does. So check out the video on hydroboration oxidation of alkenes for much more detail of that first mechanism that we discussed. So the net result of this reaction is to form an aldehyde from your terminal alkyne. So let's look at a reaction here. So let's make our triple bond right here like that. And we'll make it a terminal alkyne. And we'll make this side a benzene ring."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a reaction here. So let's make our triple bond right here like that. And we'll make it a terminal alkyne. And we'll make this side a benzene ring. So we'll make a rather large R group here. So let me put in some electrons on my benzene ring. So let's just go like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we'll make this side a benzene ring. So we'll make a rather large R group here. So let me put in some electrons on my benzene ring. So let's just go like that. So if I react this alkyne with either a dialkyl borane, or I'll just put borane here. You can use a dialkyl borane if you want to, whatever your professor wants to use, in THF. And second step, hydrogen peroxide and OH minus like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's just go like that. So if I react this alkyne with either a dialkyl borane, or I'll just put borane here. You can use a dialkyl borane if you want to, whatever your professor wants to use, in THF. And second step, hydrogen peroxide and OH minus like that. So the first thing you have to think about is, OK, what's going to happen? I'm going to be adding water across my triple bond. And I need to think about what side gets the hydrogen and what side gets the OH."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And second step, hydrogen peroxide and OH minus like that. So the first thing you have to think about is, OK, what's going to happen? I'm going to be adding water across my triple bond. And I need to think about what side gets the hydrogen and what side gets the OH. So once again, I have two choices. I could add the OH onto the carbon on the right, or I could add the OH to the carbon on the left. And the regiochemistry here is anti-Markovnikov."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I need to think about what side gets the hydrogen and what side gets the OH. So once again, I have two choices. I could add the OH onto the carbon on the right, or I could add the OH to the carbon on the left. And the regiochemistry here is anti-Markovnikov. So I'm going to add that OH to the least substituted carbon, which of course is the carbon on the right side. So I'm going to end up adding the OH to the carbon on the right side. So let's go ahead and draw our double bond in here."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the regiochemistry here is anti-Markovnikov. So I'm going to add that OH to the least substituted carbon, which of course is the carbon on the right side. So I'm going to end up adding the OH to the carbon on the right side. So let's go ahead and draw our double bond in here. First, let's go ahead and draw our benzene ring. So I'll go ahead and make it like this. And that benzene ring is connected to a carbon."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw our double bond in here. First, let's go ahead and draw our benzene ring. So I'll go ahead and make it like this. And that benzene ring is connected to a carbon. This carbon right here, there's now a double bond present, like that. And I added the OH to the least substituted carbon. I added the OH to the carbon on the right, like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that benzene ring is connected to a carbon. This carbon right here, there's now a double bond present, like that. And I added the OH to the least substituted carbon. I added the OH to the carbon on the right, like that. And I added a hydrogen to the carbon on the left. So that's adding my H2O across my triple bond. And then there was still a hydrogen on this carbon on the right, like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I added the OH to the carbon on the right, like that. And I added a hydrogen to the carbon on the left. So that's adding my H2O across my triple bond. And then there was still a hydrogen on this carbon on the right, like that. So that's my enol intermediate, like that. So let's take a look at what's going to happen to that enol intermediate. It's going to rearrange."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then there was still a hydrogen on this carbon on the right, like that. So that's my enol intermediate, like that. So let's take a look at what's going to happen to that enol intermediate. It's going to rearrange. And in the last video, we saw an acid-catalyzed tautomerization. In this video, base is present. So we have our hydroxide anions in solution."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to rearrange. And in the last video, we saw an acid-catalyzed tautomerization. In this video, base is present. So we have our hydroxide anions in solution. So let's go ahead and redraw our enol and show the enol reacting with base. So let's go ahead and draw the enol here again. So I'll draw that enol on the right side."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have our hydroxide anions in solution. So let's go ahead and redraw our enol and show the enol reacting with base. So let's go ahead and draw the enol here again. So I'll draw that enol on the right side. So there's my benzene ring. And I have my double bond in here, like that. And I know I have an OH, like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw that enol on the right side. So there's my benzene ring. And I have my double bond in here, like that. And I know I have an OH, like that. And put in my lone pairs of electrons. And I'm going to react this with hydroxide anions. So hydroxide anions are floating around."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I know I have an OH, like that. And put in my lone pairs of electrons. And I'm going to react this with hydroxide anions. So hydroxide anions are floating around. Hydroxide's negatively charged. So hydroxide will function as a base. So let me get in my lone pairs of electrons here on my oxygen."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So hydroxide anions are floating around. Hydroxide's negatively charged. So hydroxide will function as a base. So let me get in my lone pairs of electrons here on my oxygen. And I'm going to pick up that proton and kick these electrons off onto this oxygen. So let's go ahead and draw the results of that acid-base reaction. So it's an equilibrium."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me get in my lone pairs of electrons here on my oxygen. And I'm going to pick up that proton and kick these electrons off onto this oxygen. So let's go ahead and draw the results of that acid-base reaction. So it's an equilibrium. So let me go ahead and draw my equilibrium arrows like that. So what's going to happen? I have my benzene ring still."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's an equilibrium. So let me go ahead and draw my equilibrium arrows like that. So what's going to happen? I have my benzene ring still. And I have my double bond. And let's see. Now I have my oxygen with three lone pairs of electrons around it like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I have my benzene ring still. And I have my double bond. And let's see. Now I have my oxygen with three lone pairs of electrons around it like that. So negatively charged. So now I can draw a resonance structure for this molecule. So let's go ahead and put in our resonance brackets here, like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now I have my oxygen with three lone pairs of electrons around it like that. So negatively charged. So now I can draw a resonance structure for this molecule. So let's go ahead and put in our resonance brackets here, like that. So what can I do to spread that formal charge out throughout the molecule? Well, I could take a lone pair of electrons right here. And I could move it in to here to form a double bond."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and put in our resonance brackets here, like that. So what can I do to spread that formal charge out throughout the molecule? Well, I could take a lone pair of electrons right here. And I could move it in to here to form a double bond. And that would give too many bonds to this carbon. So the pi bond in here is going to break. And the two electrons here are going to move out onto this carbon like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I could move it in to here to form a double bond. And that would give too many bonds to this carbon. So the pi bond in here is going to break. And the two electrons here are going to move out onto this carbon like that. So let's go ahead and draw the results of the movement of those electrons. I still have my benzene ring like that. And there's my carbon skeleton here."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the two electrons here are going to move out onto this carbon like that. So let's go ahead and draw the results of the movement of those electrons. I still have my benzene ring like that. And there's my carbon skeleton here. Now I have a carbon double bonded to an oxygen. There used to be three lone pairs of electrons around this top oxygen. One moved in to form the double bonds."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And there's my carbon skeleton here. Now I have a carbon double bonded to an oxygen. There used to be three lone pairs of electrons around this top oxygen. One moved in to form the double bonds. Now there's only two lone pairs like that. And the electrons that used to be in this pi bond have moved onto this carbon right here. So this carbon right here is still bonded to another hydrogen."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "One moved in to form the double bonds. Now there's only two lone pairs like that. And the electrons that used to be in this pi bond have moved onto this carbon right here. So this carbon right here is still bonded to another hydrogen. So the two electrons moving out onto it are actually going to form a carbanion here. So this is actually a negatively charged carbanion like that. And water is now present."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here is still bonded to another hydrogen. So the two electrons moving out onto it are actually going to form a carbanion here. So this is actually a negatively charged carbanion like that. And water is now present. OH minus fixable proton to become H2O. So now water is floating around like that. So now water is actually going to function as an acid."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And water is now present. OH minus fixable proton to become H2O. So now water is floating around like that. So now water is actually going to function as an acid. It's going to donate a proton. And this lone pair of electrons are going to take that proton. So they take that proton and kick these electrons in here off onto the oxygen atom like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now water is actually going to function as an acid. It's going to donate a proton. And this lone pair of electrons are going to take that proton. So they take that proton and kick these electrons in here off onto the oxygen atom like that. This is an acid-base reaction. So this is going to be at equilibrium. We need to go ahead and put in our other resonance bracket like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So they take that proton and kick these electrons in here off onto the oxygen atom like that. This is an acid-base reaction. So this is going to be at equilibrium. We need to go ahead and put in our other resonance bracket like that. So what are we going to get? We still have our benzene ring untouched over here. And let's see, now we have our carbonyl like that."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We need to go ahead and put in our other resonance bracket like that. So what are we going to get? We still have our benzene ring untouched over here. And let's see, now we have our carbonyl like that. And we just added on a proton to this carbon right here like that. So now we are done. And you can see this is an aldehyde."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's see, now we have our carbonyl like that. And we just added on a proton to this carbon right here like that. So now we are done. And you can see this is an aldehyde. I didn't draw in the hydrogen over here. So you can leave that out if you want to. But if you want to draw in your hydrogen, there's a hydrogen there."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And you can see this is an aldehyde. I didn't draw in the hydrogen over here. So you can leave that out if you want to. But if you want to draw in your hydrogen, there's a hydrogen there. There's also a hydrogen on this carbon right here. I just didn't draw it in to show some clarity in the mechanism. So the end result is we form an aldehyde from a terminal alkyne."}, {"video_title": "Hydroboration-oxidation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But if you want to draw in your hydrogen, there's a hydrogen there. There's also a hydrogen on this carbon right here. I just didn't draw it in to show some clarity in the mechanism. So the end result is we form an aldehyde from a terminal alkyne. And we showed this tautomerization using base. So let me just go ahead and write one more thing here. This is an example of a base-catalyzed tautomerization where we go from the enol form to, in this case, an aldehyde as your product, which is more stable because of the carbon-oxygen double bond."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "We'll start with regioselectivity. So the reaction I've shown here is a regioselective reaction. This alcohol gets dehydrated to form two products, the alkene on the left and the alkene on the right. These two alkenes are regioisomers. Let me write this down here. So they are regioisomers. They're isomers of each other, but they differ in terms of the region or the location of the double bond."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "These two alkenes are regioisomers. Let me write this down here. So they are regioisomers. They're isomers of each other, but they differ in terms of the region or the location of the double bond. The isomer on the left has a double bond here, and the isomer on the right has the double bond here. This is a regioselective reaction. One regioisomer is favored over the other, and in this case, the tri-substituted alkene, the one on the left here, is the major product, whereas the di-substituted alkene, the one on the right, is the minor product."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "They're isomers of each other, but they differ in terms of the region or the location of the double bond. The isomer on the left has a double bond here, and the isomer on the right has the double bond here. This is a regioselective reaction. One regioisomer is favored over the other, and in this case, the tri-substituted alkene, the one on the left here, is the major product, whereas the di-substituted alkene, the one on the right, is the minor product. So that's regioselectivity. Let's compare that to stereoselectivity. So for this next example, this alcohol, it's another dehydration reaction, reacts with sulfuric acid to give us two alkenes."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "One regioisomer is favored over the other, and in this case, the tri-substituted alkene, the one on the left here, is the major product, whereas the di-substituted alkene, the one on the right, is the minor product. So that's regioselectivity. Let's compare that to stereoselectivity. So for this next example, this alcohol, it's another dehydration reaction, reacts with sulfuric acid to give us two alkenes. The mechanism for this reaction, first we protonated the OH to form water as a good leaving group, so watered left, and that gave us a carbocation with a plus one formal charge on this carbon, this benzylic carbon. So we have a benzylic carbocation in our mechanism. Let me just go ahead and draw that in here."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "So for this next example, this alcohol, it's another dehydration reaction, reacts with sulfuric acid to give us two alkenes. The mechanism for this reaction, first we protonated the OH to form water as a good leaving group, so watered left, and that gave us a carbocation with a plus one formal charge on this carbon, this benzylic carbon. So we have a benzylic carbocation in our mechanism. Let me just go ahead and draw that in here. So here's our benzylic carbocation, so plus one formal charge on this carbon in magenta. And from this carbocation, we had a choice of which proton we wanted to take. There are two hydrogens on this carbon, and depending on which proton we took, we got one of these different products."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "Let me just go ahead and draw that in here. So here's our benzylic carbocation, so plus one formal charge on this carbon in magenta. And from this carbocation, we had a choice of which proton we wanted to take. There are two hydrogens on this carbon, and depending on which proton we took, we got one of these different products. So this would be the trans product, so let me write that down here, this is the trans product, and this would be the cis product. These are stereoisomers, so let me write that down. These two are stereoisomers, that's one term that you could use to describe them."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "There are two hydrogens on this carbon, and depending on which proton we took, we got one of these different products. So this would be the trans product, so let me write that down here, this is the trans product, and this would be the cis product. These are stereoisomers, so let me write that down. These two are stereoisomers, that's one term that you could use to describe them. And this reaction is stereoselective. One stereoisomer is favored over the other. In this case, the trans product, right, is the most stable product, so this is the major product."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "These two are stereoisomers, that's one term that you could use to describe them. And this reaction is stereoselective. One stereoisomer is favored over the other. In this case, the trans product, right, is the most stable product, so this is the major product. This is favored over the cis, so this would be the minor product. Now let's look at stereospecificity. In a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry of the product, and the E2 reaction can be a good example of a stereospecific reaction."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "In this case, the trans product, right, is the most stable product, so this is the major product. This is favored over the cis, so this would be the minor product. Now let's look at stereospecificity. In a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry of the product, and the E2 reaction can be a good example of a stereospecific reaction. On the left, we have our substrate, and we have these two phenyl groups here. We have a bromine, but notice the stereochemistry at this carbon. We have a methyl group coming out at us in space, and a hydrogen going away from us."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "In a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry of the product, and the E2 reaction can be a good example of a stereospecific reaction. On the left, we have our substrate, and we have these two phenyl groups here. We have a bromine, but notice the stereochemistry at this carbon. We have a methyl group coming out at us in space, and a hydrogen going away from us. When our strong base takes our beta proton, we end up with the E alkene, so there's stereochemistry in our product. We would have the two phenyl groups on opposite sides of the double bond. Look at this reaction now."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "We have a methyl group coming out at us in space, and a hydrogen going away from us. When our strong base takes our beta proton, we end up with the E alkene, so there's stereochemistry in our product. We would have the two phenyl groups on opposite sides of the double bond. Look at this reaction now. We have the phenyl groups, we have our bromine, those are all the same. The difference is the stereochemistry at this carbon. Now we have a hydrogen coming out at us, and a methyl group going away from us."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "Look at this reaction now. We have the phenyl groups, we have our bromine, those are all the same. The difference is the stereochemistry at this carbon. Now we have a hydrogen coming out at us, and a methyl group going away from us. Our strong base takes our proton, our beta proton in the mechanism, but this time we get the Z alkene. So the stereochemistry of the substrate determined the stereochemistry of the product. There's no choice because of the mechanism."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "Now we have a hydrogen coming out at us, and a methyl group going away from us. Our strong base takes our proton, our beta proton in the mechanism, but this time we get the Z alkene. So the stereochemistry of the substrate determined the stereochemistry of the product. There's no choice because of the mechanism. You could also think about that going backwards. If you look at this product here, the Z alkene, because you know the product is a Z alkene, you know the stereochemistry at this carbon. It must have this particular stereochemistry."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "There's no choice because of the mechanism. You could also think about that going backwards. If you look at this product here, the Z alkene, because you know the product is a Z alkene, you know the stereochemistry at this carbon. It must have this particular stereochemistry. Same thing for the other reaction. We form only the E alkene, and because we form only the E alkene, we know the stereochemistry at this carbon. So the stereochemical information is kept in a stereospecific reaction."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "It must have this particular stereochemistry. Same thing for the other reaction. We form only the E alkene, and because we form only the E alkene, we know the stereochemistry at this carbon. So the stereochemical information is kept in a stereospecific reaction. Finally, let's directly compare stereoselectivity with stereospecificity. We just said in a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry of the product. So the stereochemical information is preserved because of the mechanism."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "So the stereochemical information is kept in a stereospecific reaction. Finally, let's directly compare stereoselectivity with stereospecificity. We just said in a stereospecific reaction, the stereochemistry of the substrate determines the stereochemistry of the product. So the stereochemical information is preserved because of the mechanism. That's not the case in this stereoselective reaction. If we look at the stereochemistry of this OH here, at this carbon, the stereochemistry is not preserved in our products. The stereochemical information is lost when we formed our carbocation."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "So the stereochemical information is preserved because of the mechanism. That's not the case in this stereoselective reaction. If we look at the stereochemistry of this OH here, at this carbon, the stereochemistry is not preserved in our products. The stereochemical information is lost when we formed our carbocation. For example, we could have started out with, let me go ahead and draw this in here. We could have started out with the OH on a dash, and we would have ended up with the same products. So the stereochemical information in the substrate is not preserved."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "The stereochemical information is lost when we formed our carbocation. For example, we could have started out with, let me go ahead and draw this in here. We could have started out with the OH on a dash, and we would have ended up with the same products. So the stereochemical information in the substrate is not preserved. Let's think about that concept going backwards again. So for this E alkene, because this is an E alkene with the phenyl groups on opposite sides, we know the stereochemistry at this carbon in this stereospecific reaction. But if we look at the products here, if we look at this trans product and this cis product, that does not tell us the stereochemistry of our substrate."}, {"video_title": "Regioselectivity, stereoselectivity, and stereospecificity.mp3", "Sentence": "So the stereochemical information in the substrate is not preserved. Let's think about that concept going backwards again. So for this E alkene, because this is an E alkene with the phenyl groups on opposite sides, we know the stereochemistry at this carbon in this stereospecific reaction. But if we look at the products here, if we look at this trans product and this cis product, that does not tell us the stereochemistry of our substrate. We do not know what the stereochemistry is at this carbon. Is the OH on a dash or is the OH on a wedge? So this is not a stereospecific reaction."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "In the video on sp2 hybridization, we saw that carbon was bonded to three atoms. And in this video, we're going to look at the type of hybridization that's present when carbon is bonded to two atoms. So if I look at this carbon right here and the ethine or the acetylene molecule, this carbon is bonded to a hydrogen and it's also bonded to another carbon. So we have carbon bonded to only two atoms. And the shape of the acetylene molecule has been determined to be linear. So we have a linear geometry. We also have a bond angle here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have carbon bonded to only two atoms. And the shape of the acetylene molecule has been determined to be linear. So we have a linear geometry. We also have a bond angle here. This bond angle is 180 degrees. And so we must have a different hybridization for this carbon. We have a different geometry, a different bond angle, and a different number of atoms that this carbon is bonded to."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "We also have a bond angle here. This bond angle is 180 degrees. And so we must have a different hybridization for this carbon. We have a different geometry, a different bond angle, and a different number of atoms that this carbon is bonded to. And so to find our new type of hybridization, we look at our electron configuration already in the excited state. So we have carbons, four valence electrons represented here in the excited state. So one, two, three, and four."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "We have a different geometry, a different bond angle, and a different number of atoms that this carbon is bonded to. And so to find our new type of hybridization, we look at our electron configuration already in the excited state. So we have carbons, four valence electrons represented here in the excited state. So one, two, three, and four. And we're looking for two hybrid orbitals since carbon is bonded to two atoms. So we're going to take an s orbital, so we're going to promote an s orbital in terms of energy, and we're going to demote a p orbital, only one p orbital this time. So we have an s orbital with one electron, a p orbital with one electron."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So one, two, three, and four. And we're looking for two hybrid orbitals since carbon is bonded to two atoms. So we're going to take an s orbital, so we're going to promote an s orbital in terms of energy, and we're going to demote a p orbital, only one p orbital this time. So we have an s orbital with one electron, a p orbital with one electron. That's going to leave behind two p orbitals. Each one of those p orbitals has one electron in it, so we have carbons, four valence electrons. But this is no longer an s orbital because we're going to hybridize it with the p orbital to make an sp hybrid orbital."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have an s orbital with one electron, a p orbital with one electron. That's going to leave behind two p orbitals. Each one of those p orbitals has one electron in it, so we have carbons, four valence electrons. But this is no longer an s orbital because we're going to hybridize it with the p orbital to make an sp hybrid orbital. This is no longer a p orbital because we're going to hybridize it to form our sp hybrid orbital. And so this is called sp hybridization. So this is sp hybridization because our new hybrid orbitals came from one s orbital and one p orbital like that."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "But this is no longer an s orbital because we're going to hybridize it with the p orbital to make an sp hybrid orbital. This is no longer a p orbital because we're going to hybridize it to form our sp hybrid orbital. And so this is called sp hybridization. So this is sp hybridization because our new hybrid orbitals came from one s orbital and one p orbital like that. So this carbon right here is sp hybridized since it's bonded to two atoms. And this carbon right here is also sp hybridized. Let's think about the shape of our new sp hybrid orbital."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So this is sp hybridization because our new hybrid orbitals came from one s orbital and one p orbital like that. So this carbon right here is sp hybridized since it's bonded to two atoms. And this carbon right here is also sp hybridized. Let's think about the shape of our new sp hybrid orbital. So let's get a little bit of room down here. So once again, we know an s orbital is shaped like a sphere, right? So we took one s orbital and we took one p orbital, which is shaped like a dumbbell, and we hybridized these two orbitals together to give us two new hybrid orbitals, two sp hybrid orbitals."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Let's think about the shape of our new sp hybrid orbital. So let's get a little bit of room down here. So once again, we know an s orbital is shaped like a sphere, right? So we took one s orbital and we took one p orbital, which is shaped like a dumbbell, and we hybridized these two orbitals together to give us two new hybrid orbitals, two sp hybrid orbitals. I'm gonna go ahead and draw on an sp hybrid orbital here. And once again, we're going to ignore the small lobe, right? So we're going to ignore the small lobe and when we draw our picture, only think about this bigger frontal lobe here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we took one s orbital and we took one p orbital, which is shaped like a dumbbell, and we hybridized these two orbitals together to give us two new hybrid orbitals, two sp hybrid orbitals. I'm gonna go ahead and draw on an sp hybrid orbital here. And once again, we're going to ignore the small lobe, right? So we're going to ignore the small lobe and when we draw our picture, only think about this bigger frontal lobe here. And when I think about the percentage of s character, right, so we use one s orbital and one p orbital. So that means it's 50% s character and 50% p character. And this is more s character than in the previous videos, right, so in the video on sp3 hybridization, we were talking about 25% s character."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we're going to ignore the small lobe and when we draw our picture, only think about this bigger frontal lobe here. And when I think about the percentage of s character, right, so we use one s orbital and one p orbital. So that means it's 50% s character and 50% p character. And this is more s character than in the previous videos, right, so in the video on sp3 hybridization, we were talking about 25% s character. The video on sp2 hybridization, we talked about 33% s character. And then for these hybrid orbitals, we have even more s character, up to 50%. And since the electron density for an s orbital, right, there's increased electron density closer to the nucleus for an s orbital than for a p orbital, that means that this lobe here, right, has increased electron density closer to the nucleus, which is one way to think about why these bonds get shorter as you increase in s character."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And this is more s character than in the previous videos, right, so in the video on sp3 hybridization, we were talking about 25% s character. The video on sp2 hybridization, we talked about 33% s character. And then for these hybrid orbitals, we have even more s character, up to 50%. And since the electron density for an s orbital, right, there's increased electron density closer to the nucleus for an s orbital than for a p orbital, that means that this lobe here, right, has increased electron density closer to the nucleus, which is one way to think about why these bonds get shorter as you increase in s character. So in general, as you increase in s character, you're going to get shorter bonds because you have smaller hybrid orbitals here. All right, so let's go back up here to this picture of acetylene. Let's see if we can draw it."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And since the electron density for an s orbital, right, there's increased electron density closer to the nucleus for an s orbital than for a p orbital, that means that this lobe here, right, has increased electron density closer to the nucleus, which is one way to think about why these bonds get shorter as you increase in s character. So in general, as you increase in s character, you're going to get shorter bonds because you have smaller hybrid orbitals here. All right, so let's go back up here to this picture of acetylene. Let's see if we can draw it. Now we now know that both of these carbons in acetylene are sp hybridized. So let's go back down here and let me go ahead and redraw the dot structure really fast. All right, so we have acetylene here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Let's see if we can draw it. Now we now know that both of these carbons in acetylene are sp hybridized. So let's go back down here and let me go ahead and redraw the dot structure really fast. All right, so we have acetylene here. So we have carbon triple bonded to another carbon. We know each of those carbons is sp hybridized. All right, so if each of those carbons is sp hybridized, each carbon has two sp hybrid orbitals."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "All right, so we have acetylene here. So we have carbon triple bonded to another carbon. We know each of those carbons is sp hybridized. All right, so if each of those carbons is sp hybridized, each carbon has two sp hybrid orbitals. Let me go ahead and draw in one sp hybrid orbital. Again, I'm ignoring the smaller back lobe. And here's our other sp hybrid orbital on this carbon."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "All right, so if each of those carbons is sp hybridized, each carbon has two sp hybrid orbitals. Let me go ahead and draw in one sp hybrid orbital. Again, I'm ignoring the smaller back lobe. And here's our other sp hybrid orbital on this carbon. So let me go back and look at our diagram again. All right, so each carbon, let me use red for this. All right, so each sp hybridized carbon has an sp orbital with one valence electron in it."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And here's our other sp hybrid orbital on this carbon. So let me go back and look at our diagram again. All right, so each carbon, let me use red for this. All right, so each sp hybridized carbon has an sp orbital with one valence electron in it. I'm gonna put that in here. And then there's another one. All right, so there's another sp hybrid orbital with one valence electron in it."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "All right, so each sp hybridized carbon has an sp orbital with one valence electron in it. I'm gonna put that in here. And then there's another one. All right, so there's another sp hybrid orbital with one valence electron in it. So I'm gonna go ahead and put the other electron over here in this hybrid orbital. All right, and also notice, all right, if you're dealing with an sp hybridized carbon, you also have two p orbitals, two unhybridized p orbitals, each p orbital with a valence electron. So let me go back down here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "All right, so there's another sp hybrid orbital with one valence electron in it. So I'm gonna go ahead and put the other electron over here in this hybrid orbital. All right, and also notice, all right, if you're dealing with an sp hybridized carbon, you also have two p orbitals, two unhybridized p orbitals, each p orbital with a valence electron. So let me go back down here. And I'm gonna draw in, here's one p orbital. All right, there's one p orbital with one valence electron. And then here's another p orbital right here with another one of those valence electrons."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let me go back down here. And I'm gonna draw in, here's one p orbital. All right, there's one p orbital with one valence electron. And then here's another p orbital right here with another one of those valence electrons. All right, so now we have our picture of an sp hybridized carbon. So let's say that was this carbon over here on the left. And so now let's go ahead and draw in this carbon on the right."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then here's another p orbital right here with another one of those valence electrons. All right, so now we have our picture of an sp hybridized carbon. So let's say that was this carbon over here on the left. And so now let's go ahead and draw in this carbon on the right. This carbon on the right is also sp hybridized. Therefore, this carbon on the right has an sp hybrid orbital, right, with one valence electron in here. And then another sp hybrid orbital with one valence electron here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so now let's go ahead and draw in this carbon on the right. This carbon on the right is also sp hybridized. Therefore, this carbon on the right has an sp hybrid orbital, right, with one valence electron in here. And then another sp hybrid orbital with one valence electron here. This carbon, since it's sp hybridized, right, so once again, go back up here to this diagram. This carbon is also going to have a p orbital with a valence electron and another p orbital with another electron in it. So go back down to here and we draw in those p orbitals."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then another sp hybrid orbital with one valence electron here. This carbon, since it's sp hybridized, right, so once again, go back up here to this diagram. This carbon is also going to have a p orbital with a valence electron and another p orbital with another electron in it. So go back down to here and we draw in those p orbitals. So here's one p orbital, right, here's one p orbital with one valence electron. And then here's another p orbital with an electron in here like that. All right, finally, we have to add in hydrogen, right."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So go back down to here and we draw in those p orbitals. So here's one p orbital, right, here's one p orbital with one valence electron. And then here's another p orbital with an electron in here like that. All right, finally, we have to add in hydrogen, right. So we have a hydrogen on either side here. So we know that hydrogen has one valence electron in an unhybridized s orbital. So over here, we have a hydrogen with one valence electron in an unhybridized s orbital."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "All right, finally, we have to add in hydrogen, right. So we have a hydrogen on either side here. So we know that hydrogen has one valence electron in an unhybridized s orbital. So over here, we have a hydrogen with one valence electron in an unhybridized s orbital. And now we can finally analyze the bonding that's present. All right, so we know that if we have head-on overlap of orbitals like right in here, that's a sigma bond. So there's one sigma bond."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So over here, we have a hydrogen with one valence electron in an unhybridized s orbital. And now we can finally analyze the bonding that's present. All right, so we know that if we have head-on overlap of orbitals like right in here, that's a sigma bond. So there's one sigma bond. Here's a head-on overlap of orbitals between our two carbons, so that's a sigma bond. Then finally, we have a head-on overlap of orbitals here, so that's another sigma bond. So we have a total of three sigma bonds in the acetylene molecule."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So there's one sigma bond. Here's a head-on overlap of orbitals between our two carbons, so that's a sigma bond. Then finally, we have a head-on overlap of orbitals here, so that's another sigma bond. So we have a total of three sigma bonds in the acetylene molecule. In the video on sp2 hybridization, we saw how to make a pi bond, right. So we had this side-by-side overlap of orbitals. So here we have one pi bond, right."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have a total of three sigma bonds in the acetylene molecule. In the video on sp2 hybridization, we saw how to make a pi bond, right. So we had this side-by-side overlap of orbitals. So here we have one pi bond, right. Here we have interaction above and below, so that's one pi bond. And then we have another pi bond here. So we have side-by-side overlap of orbitals here as well."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So here we have one pi bond, right. Here we have interaction above and below, so that's one pi bond. And then we have another pi bond here. So we have side-by-side overlap of orbitals here as well. And so we have two pi bonds present. So we have two pi bonds present in the acetylene molecule. So let's look at our dot structure again."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have side-by-side overlap of orbitals here as well. And so we have two pi bonds present. So we have two pi bonds present in the acetylene molecule. So let's look at our dot structure again. So we saw the bond between this carbon and this hydrogen was a sigma bond. And we saw there was one sigma bond between our two carbons. So I'm just going to pick the one in the middle here, and say that's a sigma bond."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let's look at our dot structure again. So we saw the bond between this carbon and this hydrogen was a sigma bond. And we saw there was one sigma bond between our two carbons. So I'm just going to pick the one in the middle here, and say that's a sigma bond. And then this bond over here, we said, was a sigma bond. So there's our three sigma bonds. And then we had a triple bond present."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I'm just going to pick the one in the middle here, and say that's a sigma bond. And then this bond over here, we said, was a sigma bond. So there's our three sigma bonds. And then we had a triple bond present. And so there were two pi bonds also present. So two of these are pi bonds here. So a total of two pi bonds and three sigma bonds for the acetylene molecule here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then we had a triple bond present. And so there were two pi bonds also present. So two of these are pi bonds here. So a total of two pi bonds and three sigma bonds for the acetylene molecule here. Remember, pi bonds prevent free rotation. So we can't rotate about the sigma bond between the two carbons because of the pi bonds. So there's no free rotation for our triple bond."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So a total of two pi bonds and three sigma bonds for the acetylene molecule here. Remember, pi bonds prevent free rotation. So we can't rotate about the sigma bond between the two carbons because of the pi bonds. So there's no free rotation for our triple bond. So we have a linear shape. So let me go ahead and draw that line in here. So you see this linear geometry for this molecule like that."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So there's no free rotation for our triple bond. So we have a linear shape. So let me go ahead and draw that line in here. So you see this linear geometry for this molecule like that. Also in terms of bond length, so the distance between these two carbons. The distance between this carbon and this carbon. Let me go ahead and circle them."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So you see this linear geometry for this molecule like that. Also in terms of bond length, so the distance between these two carbons. The distance between this carbon and this carbon. Let me go ahead and circle them. The distance between these two carbons turns out to be approximately 1.20 angstroms. So an even shorter bond length than in our previous video. So once again, that's due to the increased s character."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and circle them. The distance between these two carbons turns out to be approximately 1.20 angstroms. So an even shorter bond length than in our previous video. So once again, that's due to the increased s character. So increased s character gives you these smaller orbitals. And so that's one way to think about the shorter bond distance in a triple bond compared to a double bond or a single bond. So that's a lot that we've covered here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So once again, that's due to the increased s character. So increased s character gives you these smaller orbitals. And so that's one way to think about the shorter bond distance in a triple bond compared to a double bond or a single bond. So that's a lot that we've covered here. Let's go ahead and draw the dot structure one more time and analyze it using steric numbers. So we have our triple bonds. And if we're doing steric number to find out the hybridization state, we know to do steric number, you take the number of sigma bonds."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So that's a lot that we've covered here. Let's go ahead and draw the dot structure one more time and analyze it using steric numbers. So we have our triple bonds. And if we're doing steric number to find out the hybridization state, we know to do steric number, you take the number of sigma bonds. So let's say our goal was to figure out the steric number for this carbon. The number of sigma bonds, I know this is a sigma bond. I know in a triple bond, I have one sigma bond and two pi bonds."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And if we're doing steric number to find out the hybridization state, we know to do steric number, you take the number of sigma bonds. So let's say our goal was to figure out the steric number for this carbon. The number of sigma bonds, I know this is a sigma bond. I know in a triple bond, I have one sigma bond and two pi bonds. So there are two sigma bonds here and zero lone pairs of electrons. 2 plus 0 gives me 2. So I need two hybrid orbitals, which you make from one s orbital and one p orbital."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "I know in a triple bond, I have one sigma bond and two pi bonds. So there are two sigma bonds here and zero lone pairs of electrons. 2 plus 0 gives me 2. So I need two hybrid orbitals, which you make from one s orbital and one p orbital. So if you get a steric number of 2, you think sp hybridization. So this carbon is sp hybridized and so is this carbon as well. And so that's how to think about it using steric numbers."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I need two hybrid orbitals, which you make from one s orbital and one p orbital. So if you get a steric number of 2, you think sp hybridization. So this carbon is sp hybridized and so is this carbon as well. And so that's how to think about it using steric numbers. So once again, a linear geometry with a bond angles of 180 degrees. Let's do one more example using steric number to analyze the molecule. Let's do carbon dioxide."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so that's how to think about it using steric numbers. So once again, a linear geometry with a bond angles of 180 degrees. Let's do one more example using steric number to analyze the molecule. Let's do carbon dioxide. So if we wanted to figure out the hybridization of the carbon there, let's go ahead and do that. So using steric number. So the hybridization of this carbon."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Let's do carbon dioxide. So if we wanted to figure out the hybridization of the carbon there, let's go ahead and do that. So using steric number. So the hybridization of this carbon. So the steric number is equal to the number of sigma bonds. So if I focus in on the double bond between one of these oxygens and this carbon, I know that one of these bonds is a sigma bond from our previous video. So I have one sigma bond here."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So the hybridization of this carbon. So the steric number is equal to the number of sigma bonds. So if I focus in on the double bond between one of these oxygens and this carbon, I know that one of these bonds is a sigma bond from our previous video. So I have one sigma bond here. And then for this other double bond on the right, I know that one of them is a sigma bond. So I have two sigma bonds here and 0 lone pairs of electrons around the carbon. So 2 plus 0 gives me a steric number of 2."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I have one sigma bond here. And then for this other double bond on the right, I know that one of them is a sigma bond. So I have two sigma bonds here and 0 lone pairs of electrons around the carbon. So 2 plus 0 gives me a steric number of 2. So I need two hybrid orbitals for that carbon. And of course, that must mean this carbon is sp hybridized. So this carbon here is sp hybridized as well."}, {"video_title": "sp hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So 2 plus 0 gives me a steric number of 2. So I need two hybrid orbitals for that carbon. And of course, that must mean this carbon is sp hybridized. So this carbon here is sp hybridized as well. And therefore, we know that this is a linear molecule with a bond angle of 180 degrees. And so once again, steric number is just a nice way of analyzing the hybridization and also the geometry of the molecules. And so in the next video, we'll look at a couple of examples of organic molecules in different hybridization states."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And he or she says, if I understood this correctly, precession changes the time of the seasons over long periods of time, and obliquity changes the strengths of the season over long periods of time. And so this is a good comment. So the first comment isn't exactly true, and that's what I'm going to focus on in this video. He says, or she says, precession changes the time of the seasons over long periods of time. This is kind of true, but I don't think in the sense that Vic Soma is referring to it. The second part is roughly true. Obliquity changes the strength of the season over long periods of time."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "He says, or she says, precession changes the time of the seasons over long periods of time. This is kind of true, but I don't think in the sense that Vic Soma is referring to it. The second part is roughly true. Obliquity changes the strength of the season over long periods of time. If you are more tilted towards the sun in the extreme, then yes, you will have a bigger disparity than when you are less tilted, when you are tilted away from the sun. Or I guess you'd say, if you are more tilted to or away from the sun, the disparity between summer and winter will be greater than if you are less tilted. So the second statement is true, although you always have to be careful with things as complicated as the climate, because it can really depend from parts of the globe to parts on the globe, depending on what are all of the other factors that play into it."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Obliquity changes the strength of the season over long periods of time. If you are more tilted towards the sun in the extreme, then yes, you will have a bigger disparity than when you are less tilted, when you are tilted away from the sun. Or I guess you'd say, if you are more tilted to or away from the sun, the disparity between summer and winter will be greater than if you are less tilted. So the second statement is true, although you always have to be careful with things as complicated as the climate, because it can really depend from parts of the globe to parts on the globe, depending on what are all of the other factors that play into it. So you would want to run some type of simulation or something like that. But the second part is roughly true. But I want to focus on the first part, because I think it'll really give us a better understanding of what precession is."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the second statement is true, although you always have to be careful with things as complicated as the climate, because it can really depend from parts of the globe to parts on the globe, depending on what are all of the other factors that play into it. So you would want to run some type of simulation or something like that. But the second part is roughly true. But I want to focus on the first part, because I think it'll really give us a better understanding of what precession is. And so this last statement Vic Soma makes is actually not true. He says, I'm assuming that he means the she, so in several thousand years, if we still use the same calendar system, summer and winter will happen in different months, and they will be more mild or more harsh. And what we're going to see is that the second statement is not true, because our calendar is actually based on when we are most tilted away from or towards the sun."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I want to focus on the first part, because I think it'll really give us a better understanding of what precession is. And so this last statement Vic Soma makes is actually not true. He says, I'm assuming that he means the she, so in several thousand years, if we still use the same calendar system, summer and winter will happen in different months, and they will be more mild or more harsh. And what we're going to see is that the second statement is not true, because our calendar is actually based on when we are most tilted away from or towards the sun. So our calendar is actually, to some degree, you could say, takes precession into account. And what actually does change according to our calendar is when we are closest or furthest away from the sun, the perihelion or the aphelion. And we're going to think about that in this video."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what we're going to see is that the second statement is not true, because our calendar is actually based on when we are most tilted away from or towards the sun. So our calendar is actually, to some degree, you could say, takes precession into account. And what actually does change according to our calendar is when we are closest or furthest away from the sun, the perihelion or the aphelion. And we're going to think about that in this video. So let me draw the sun here. Let me draw the sun right over here. And let me draw Earth's orbit around the sun."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we're going to think about that in this video. So let me draw the sun here. Let me draw the sun right over here. And let me draw Earth's orbit around the sun. And I'm going to draw it with some eccentricity. And just so you know what I'm talking about when I say eccentricity, a circle has no eccentricity. An ellipse, this ellipse right here, has more eccentricity than the circle, which has no eccentricity."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me draw Earth's orbit around the sun. And I'm going to draw it with some eccentricity. And just so you know what I'm talking about when I say eccentricity, a circle has no eccentricity. An ellipse, this ellipse right here, has more eccentricity than the circle, which has no eccentricity. And an even more eccentric ellipse would look like this. So you can really think about eccentricity as a measure of how far you are away from being perfectly circular. So Earth's orbit around the sun is pretty close to circular, but it has some eccentricity."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "An ellipse, this ellipse right here, has more eccentricity than the circle, which has no eccentricity. And an even more eccentric ellipse would look like this. So you can really think about eccentricity as a measure of how far you are away from being perfectly circular. So Earth's orbit around the sun is pretty close to circular, but it has some eccentricity. It is slightly elliptical. And I'm going to exaggerate it a bit in this drawing right over here. So let's say that this is the closest point that Earth gets to the sun."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So Earth's orbit around the sun is pretty close to circular, but it has some eccentricity. It is slightly elliptical. And I'm going to exaggerate it a bit in this drawing right over here. So let's say that this is the closest point that Earth gets to the sun. So that is the perihelion. And let's say that this is the furthest point. And obviously, it's not this big of a difference."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say that this is the closest point that Earth gets to the sun. So that is the perihelion. And let's say that this is the furthest point. And obviously, it's not this big of a difference. It's actually only a 3% difference right now. But we'll also learn that that's changing. But it never becomes this dramatic."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And obviously, it's not this big of a difference. It's actually only a 3% difference right now. But we'll also learn that that's changing. But it never becomes this dramatic. But this will help us visualize it. So Earth's orbit might look something like this. Let me make it do a better job than that."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it never becomes this dramatic. But this will help us visualize it. So Earth's orbit might look something like this. Let me make it do a better job than that. Earth's orbit might look something like this. Obviously, I'm exaggerating the eccentricity here. So let's say that's Earth's orbit."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me make it do a better job than that. Earth's orbit might look something like this. Obviously, I'm exaggerating the eccentricity here. So let's say that's Earth's orbit. This is the point in orbit where we're closest. So that's perihelion. And this is when we are furthest."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say that's Earth's orbit. This is the point in orbit where we're closest. So that's perihelion. And this is when we are furthest. This is aphelion. And we saw in the first video when we discussed this that right now, this perihelion is occurring in January. And this will change over time, as we'll see in this video."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is when we are furthest. This is aphelion. And we saw in the first video when we discussed this that right now, this perihelion is occurring in January. And this will change over time, as we'll see in this video. So January right now, and aphelion right now occurs in July. Now, the time where we are most tilted towards the sun is not at the perihelion right now. It actually occurs a few weeks before the perihelion."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this will change over time, as we'll see in this video. So January right now, and aphelion right now occurs in July. Now, the time where we are most tilted towards the sun is not at the perihelion right now. It actually occurs a few weeks before the perihelion. So when we are most tilted to the sun, this is our winter solstice. And this is when we are, actually, in the case of the northern hemisphere, this is when we are most tilted away from the sun, I should say. So if we were to draw our tilt here, if I were to come out of the North Pole, it would look like that."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It actually occurs a few weeks before the perihelion. So when we are most tilted to the sun, this is our winter solstice. And this is when we are, actually, in the case of the northern hemisphere, this is when we are most tilted away from the sun, I should say. So if we were to draw our tilt here, if I were to come out of the North Pole, it would look like that. And this right over here, depending on the year and where you are and your time zone and everything, this is usually December 21 or 22. I'll just go with December 22 for now. And when the northern hemisphere is most tilted towards the sun occurs on June 20 or 21."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we were to draw our tilt here, if I were to come out of the North Pole, it would look like that. And this right over here, depending on the year and where you are and your time zone and everything, this is usually December 21 or 22. I'll just go with December 22 for now. And when the northern hemisphere is most tilted towards the sun occurs on June 20 or 21. So roughly six months later. Or really, exactly six months later. It's just that all the months have different numbers of days and you have leap years sometimes, with February sometimes having 28 or 29 days."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when the northern hemisphere is most tilted towards the sun occurs on June 20 or 21. So roughly six months later. Or really, exactly six months later. It's just that all the months have different numbers of days and you have leap years sometimes, with February sometimes having 28 or 29 days. But if you go half a year away, then you are at, in the case of the northern hemisphere, the summer solstice. And this is when we are most tilted towards the sun. And once again, it occurs right now, a few weeks before the aphelion, before we are furthest away from the sun."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's just that all the months have different numbers of days and you have leap years sometimes, with February sometimes having 28 or 29 days. But if you go half a year away, then you are at, in the case of the northern hemisphere, the summer solstice. And this is when we are most tilted towards the sun. And once again, it occurs right now, a few weeks before the aphelion, before we are furthest away from the sun. Now, I want to zoom in right over here on December 22. So right over here, let me zoom in. So let's say that this is the Earth."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And once again, it occurs right now, a few weeks before the aphelion, before we are furthest away from the sun. Now, I want to zoom in right over here on December 22. So right over here, let me zoom in. So let's say that this is the Earth. I'm zooming in on this circle right over here. So let me box it to show this is the one I'm focused on right over here. And let me draw the axis of rotation."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say that this is the Earth. I'm zooming in on this circle right over here. So let me box it to show this is the one I'm focused on right over here. And let me draw the axis of rotation. And we know that that angle versus the vertical, that you could call the tilt or the obliquity. And we know this is 23.4 degrees relative to the vertical, relative to perpendicular, compared to the plane of our orbit, I guess. So if our orbital axis was straight up and down, it would look something like this."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me draw the axis of rotation. And we know that that angle versus the vertical, that you could call the tilt or the obliquity. And we know this is 23.4 degrees relative to the vertical, relative to perpendicular, compared to the plane of our orbit, I guess. So if our orbital axis was straight up and down, it would look something like this. It's not. It tilts. And this angle right over here is 23.4 degrees."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if our orbital axis was straight up and down, it would look something like this. It's not. It tilts. And this angle right over here is 23.4 degrees. And when I say straight up and down, I'm saying relative to the plane of Earth's orbit around the sun. So this right here is the obliquity. And as Viksoma mentioned, this does change."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this angle right over here is 23.4 degrees. And when I say straight up and down, I'm saying relative to the plane of Earth's orbit around the sun. So this right here is the obliquity. And as Viksoma mentioned, this does change. It kind of goes between 22 degrees and 24 and 1 1\u20442 degrees over very long periods of time. But it does, I think it's 41,000 years if I remember that correctly. So it will affect, on some level, the severity."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as Viksoma mentioned, this does change. It kind of goes between 22 degrees and 24 and 1 1\u20442 degrees over very long periods of time. But it does, I think it's 41,000 years if I remember that correctly. So it will affect, on some level, the severity. So this tilt is kind of going between that, where actually it's reducing right now. And it'll get to a minimum in a few thousand years. So it'll get to some minimum, and then eventually it'll get back to some maximum tilt."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it will affect, on some level, the severity. So this tilt is kind of going between that, where actually it's reducing right now. And it'll get to a minimum in a few thousand years. So it'll get to some minimum, and then eventually it'll get back to some maximum tilt. So it kind of goes back and forth between those two over the course of several tens of thousands of years. But anyway, this is as it is zoomed in right now. And as we mentioned, precession, you can kind of view it as if this arrow right here actually existed, it would trace out a circle."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it'll get to some minimum, and then eventually it'll get back to some maximum tilt. So it kind of goes back and forth between those two over the course of several tens of thousands of years. But anyway, this is as it is zoomed in right now. And as we mentioned, precession, you can kind of view it as if this arrow right here actually existed, it would trace out a circle. And it's tracing out the circle over a huge period of time, over 26,000 years. Let me make everything clear here. Right now I'm just going to assume Earth is rotating."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as we mentioned, precession, you can kind of view it as if this arrow right here actually existed, it would trace out a circle. And it's tracing out the circle over a huge period of time, over 26,000 years. Let me make everything clear here. Right now I'm just going to assume Earth is rotating. The orbital direction is in that direction. I'm going to assume, let me make that a little bit more curved, Earth, this is the rotation of the Earth. It is in this direction right over here."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Right now I'm just going to assume Earth is rotating. The orbital direction is in that direction. I'm going to assume, let me make that a little bit more curved, Earth, this is the rotation of the Earth. It is in this direction right over here. And what we learned about precession, and actually to be particular, it's axial precession. There's multiple types of precession we'll talk about. If someone says just precession, they're usually referring to axial precession."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is in this direction right over here. And what we learned about precession, and actually to be particular, it's axial precession. There's multiple types of precession we'll talk about. If someone says just precession, they're usually referring to axial precession. It's this idea that over 26,000 years, the tip of this arrow, or you can even imagine the poles themselves, will kind of trace out a circle. If you look at the same point in our orbit at any period in time, the circle it's tracing out is going to be going in this direction right over here. So if we wait 1,800 years, and I want to make sure I get this right because it's important to see what happens to our calendar."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If someone says just precession, they're usually referring to axial precession. It's this idea that over 26,000 years, the tip of this arrow, or you can even imagine the poles themselves, will kind of trace out a circle. If you look at the same point in our orbit at any period in time, the circle it's tracing out is going to be going in this direction right over here. So if we wait 1,800 years, and I want to make sure I get this right because it's important to see what happens to our calendar. If we wait 1,800 years, we might, this arrow, instead of, it'll still have a tilt of 23.4 degrees, but instead of pointing in this direction, it might be pointing in this direction. Or in fact, it is likely, it will be pointing in this direction. I'm obviously not drawing it that exact."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we wait 1,800 years, and I want to make sure I get this right because it's important to see what happens to our calendar. If we wait 1,800 years, we might, this arrow, instead of, it'll still have a tilt of 23.4 degrees, but instead of pointing in this direction, it might be pointing in this direction. Or in fact, it is likely, it will be pointing in this direction. I'm obviously not drawing it that exact. And then the bottom of the arrow will come out over here. So if you think about that, if you wait 1,800 years, and once again, the tilt hasn't changed, or it's changed a little bit, but what the precession has done, tracing out the circle, has changed the direction of this arrow, changed the direction of our axis of rotation. And if you wait 1,800 years, when will the northern hemisphere be pointed most away from the sun?"}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'm obviously not drawing it that exact. And then the bottom of the arrow will come out over here. So if you think about that, if you wait 1,800 years, and once again, the tilt hasn't changed, or it's changed a little bit, but what the precession has done, tracing out the circle, has changed the direction of this arrow, changed the direction of our axis of rotation. And if you wait 1,800 years, when will the northern hemisphere be pointed most away from the sun? Well, now it won't be pointed most away from the sun at this point in space relative to the sun anymore, because now its axis of rotation looks something like this. So now, if we wait, or I should say, in 1,800 years, it'll be most pointed away, or the northern hemisphere will be most pointed away from the sun about a month earlier. So about a month earlier."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if you wait 1,800 years, when will the northern hemisphere be pointed most away from the sun? Well, now it won't be pointed most away from the sun at this point in space relative to the sun anymore, because now its axis of rotation looks something like this. So now, if we wait, or I should say, in 1,800 years, it'll be most pointed away, or the northern hemisphere will be most pointed away from the sun about a month earlier. So about a month earlier. It'll be most pointed away from the sun about a month earlier. So this is when it will be most pointed away from the sun. But if we, to today's time, we would say, no, that's still not the most pointed away."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So about a month earlier. It'll be most pointed away from the sun about a month earlier. So this is when it will be most pointed away from the sun. But if we, to today's time, we would say, no, that's still not the most pointed away. But since we have this precession, since the direction of the tilt, I guess we could say, or the direction of our axis, our rotational axis is changing, we are now at a different point in orbit where we're most pointed away from the sun. So this is 1,800 years later, approximately. So now, based on this, and I think this is what Vick's almost might have been hinting at, you say, look, OK, it's earlier in our orbit."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if we, to today's time, we would say, no, that's still not the most pointed away. But since we have this precession, since the direction of the tilt, I guess we could say, or the direction of our axis, our rotational axis is changing, we are now at a different point in orbit where we're most pointed away from the sun. So this is 1,800 years later, approximately. So now, based on this, and I think this is what Vick's almost might have been hinting at, you say, look, OK, it's earlier in our orbit. Wouldn't this now be like November? And the answer is no. It will still be December 22nd."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now, based on this, and I think this is what Vick's almost might have been hinting at, you say, look, OK, it's earlier in our orbit. Wouldn't this now be like November? And the answer is no. It will still be December 22nd. This will still be December 21st or 22nd, depending on the year, still be December 22nd, still be the same date. And that's because our calendar is based on when we are most tilted away or when we are most tilted towards the sun. So by definition, this is when we're most tilted away."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It will still be December 22nd. This will still be December 21st or 22nd, depending on the year, still be December 22nd, still be the same date. And that's because our calendar is based on when we are most tilted away or when we are most tilted towards the sun. So by definition, this is when we're most tilted away. So this will be the winter solstice. So what happens is, every year, so the way I drew it right over here, and actually, this perihelion actually changes over time as well. There's a precession of the perihelion as well, but I'm not going to go into that right now."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So by definition, this is when we're most tilted away. So this will be the winter solstice. So what happens is, every year, so the way I drew it right over here, and actually, this perihelion actually changes over time as well. There's a precession of the perihelion as well, but I'm not going to go into that right now. So if you fast forward 1,800 years, all that's going to happen is that what we consider by our calendar to be December 22nd, in an absolute point in our orbit, will be earlier in our orbit. But we're still going to call it December 22nd. And so the perihelion is going to be further away from that December 22nd."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's a precession of the perihelion as well, but I'm not going to go into that right now. So if you fast forward 1,800 years, all that's going to happen is that what we consider by our calendar to be December 22nd, in an absolute point in our orbit, will be earlier in our orbit. But we're still going to call it December 22nd. And so the perihelion is going to be further away from that December 22nd. It's actually going to be a month further away. So the perihelion 1,800 years from now won't be in January. It will be in February."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the perihelion is going to be further away from that December 22nd. It's actually going to be a month further away. So the perihelion 1,800 years from now won't be in January. It will be in February. So the real takeaway here is that our calendar is based not on the exact point in space relative to the sun. Our calendar is based on the maximum tilt towards or away from the sun. And that, as we see, that is slightly changing in terms of where it occurs in the absolute point in space."}, {"video_title": "Precession causing perihelion to happen later Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It will be in February. So the real takeaway here is that our calendar is based not on the exact point in space relative to the sun. Our calendar is based on the maximum tilt towards or away from the sun. And that, as we see, that is slightly changing in terms of where it occurs in the absolute point in space. I think it's changing by roughly 20 minutes a year. So every year, the perihelion is getting 20 minutes later. If we wanted to use the perihelion, if we wanted to use the exact point in space as our calendar, our year would actually be about that, I don't know what it is, roughly 20 or 25 minutes longer."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "The Diels-Alder reaction is a very important reaction because it's used a lot in synthesis to make complicated molecules. On the left we have our diene, so we have two double bonds in that molecule. On the right is our dienophile, so let's take a look at that word. We know that phile means love, so the dienophile loves the diene. And the dienophile usually has at least one electron withdrawing group, which withdraws electron density from this double bond. So the dienophile is relatively electron poor. The diene, on the other hand, is relatively electron rich, so we have four pi electrons."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "We know that phile means love, so the dienophile loves the diene. And the dienophile usually has at least one electron withdrawing group, which withdraws electron density from this double bond. So the dienophile is relatively electron poor. The diene, on the other hand, is relatively electron rich, so we have four pi electrons. So you can think about the electrons flowing from the diene to the dienophile. And this is what's called a pericyclic reaction. This is a one-step reaction that proceeds through a cyclic transition state."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "The diene, on the other hand, is relatively electron rich, so we have four pi electrons. So you can think about the electrons flowing from the diene to the dienophile. And this is what's called a pericyclic reaction. This is a one-step reaction that proceeds through a cyclic transition state. And if we think about electron density flowing from the diene to the dienophile, we could start with these pi electrons moving into here, so we form a bond between these two carbons. Next, these pi electrons would move into here to form a bond between these two carbons. And then finally, these pi electrons would move over to here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "This is a one-step reaction that proceeds through a cyclic transition state. And if we think about electron density flowing from the diene to the dienophile, we could start with these pi electrons moving into here, so we form a bond between these two carbons. Next, these pi electrons would move into here to form a bond between these two carbons. And then finally, these pi electrons would move over to here. So think about all of those six pi electrons moving at the same time. And that would give us our product on the right over here, which is a cyclohexene ring. So if we follow our pi electrons, we'll start with these pi electrons in red."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, these pi electrons would move over to here. So think about all of those six pi electrons moving at the same time. And that would give us our product on the right over here, which is a cyclohexene ring. So if we follow our pi electrons, we'll start with these pi electrons in red. So these pi electrons formed this bond. Next, let's follow these pi electrons on the dienophile. So they start on the dienophile, and they end up forming this bond between those two carbons."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "So if we follow our pi electrons, we'll start with these pi electrons in red. So these pi electrons formed this bond. Next, let's follow these pi electrons on the dienophile. So they start on the dienophile, and they end up forming this bond between those two carbons. And then finally, the electrons in magenta right here on the diene move down to here to form the double bond and to give us our cyclohexene ring. But all this happens at once. Now, I drew my electrons going around in a counterclockwise fashion."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "So they start on the dienophile, and they end up forming this bond between those two carbons. And then finally, the electrons in magenta right here on the diene move down to here to form the double bond and to give us our cyclohexene ring. But all this happens at once. Now, I drew my electrons going around in a counterclockwise fashion. I showed my electrons going around in this direction. But it doesn't matter. You could have drawn your electrons going around in a clockwise fashion."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Now, I drew my electrons going around in a counterclockwise fashion. I showed my electrons going around in this direction. But it doesn't matter. You could have drawn your electrons going around in a clockwise fashion. What matters is thinking about moving your six pi electrons to give you your product on the right. Let's get some practice with some simple Diels-Alder reactions. We won't worry about stereochemistry in this video."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "You could have drawn your electrons going around in a clockwise fashion. What matters is thinking about moving your six pi electrons to give you your product on the right. Let's get some practice with some simple Diels-Alder reactions. We won't worry about stereochemistry in this video. First, you need to recognize the diene and the dienophile. On the left is our diene. On the right is our dienophile."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "We won't worry about stereochemistry in this video. First, you need to recognize the diene and the dienophile. On the left is our diene. On the right is our dienophile. If we think about electron density flowing from the diene to the dienophile, I can move these electrons into here, so we form a bond between these two carbons. And these electrons move into here to form a bond between these two carbons. And then these electrons down to give us our cyclohexene ring."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "On the right is our dienophile. If we think about electron density flowing from the diene to the dienophile, I can move these electrons into here, so we form a bond between these two carbons. And these electrons move into here to form a bond between these two carbons. And then these electrons down to give us our cyclohexene ring. We could draw our product right away. We know we get a cyclohexene ring here. And then we would have our aldehyde coming off of that carbon."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons down to give us our cyclohexene ring. We could draw our product right away. We know we get a cyclohexene ring here. And then we would have our aldehyde coming off of that carbon. Following our electrons, I'll be consistent with the colors that we used before. These pi electrons are red, and those electrons move over here to form this bond. Next, these pi electrons in blue moved into here to form this bond."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have our aldehyde coming off of that carbon. Following our electrons, I'll be consistent with the colors that we used before. These pi electrons are red, and those electrons move over here to form this bond. Next, these pi electrons in blue moved into here to form this bond. And then finally, the pi electrons in magenta moved into here to form this bond. All six pi electrons move at the same time in this one-step reaction. Let's do the next problem."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Next, these pi electrons in blue moved into here to form this bond. And then finally, the pi electrons in magenta moved into here to form this bond. All six pi electrons move at the same time in this one-step reaction. Let's do the next problem. Down here on the left, this is our diene. On the right is our dienophile. This on the left, this is a diene, but notice that it has an interesting conformation."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Let's do the next problem. Down here on the left, this is our diene. On the right is our dienophile. This on the left, this is a diene, but notice that it has an interesting conformation. Up here, we had our diene in what's called the S-cis conformation. S refers to this single or sigma bond here. But here we have the S-trans conformation."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "This on the left, this is a diene, but notice that it has an interesting conformation. Up here, we had our diene in what's called the S-cis conformation. S refers to this single or sigma bond here. But here we have the S-trans conformation. We have our double bonds trans about this single bond. We have to rotate about this single bond here to go from the S-trans conformation to the S-cis conformation. You have to do that."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "But here we have the S-trans conformation. We have our double bonds trans about this single bond. We have to rotate about this single bond here to go from the S-trans conformation to the S-cis conformation. You have to do that. If you got this problem on a test, you could just do that in your head. Now you have your diene in the S-cis conformation. It needs to be in this conformation in order to undergo a Diels-Alder reaction."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "You have to do that. If you got this problem on a test, you could just do that in your head. Now you have your diene in the S-cis conformation. It needs to be in this conformation in order to undergo a Diels-Alder reaction. Now we're ready for our reaction. We think about our six pi electrons. We're going to move these electrons into here to form a bond between these two carbons."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "It needs to be in this conformation in order to undergo a Diels-Alder reaction. Now we're ready for our reaction. We think about our six pi electrons. We're going to move these electrons into here to form a bond between these two carbons. Notice this time we're dealing with a triple bond. Up here, we only had a double bond, but alkynes can act as dienophiles too. We're going to take these pi electrons and move them into here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "We're going to move these electrons into here to form a bond between these two carbons. Notice this time we're dealing with a triple bond. Up here, we only had a double bond, but alkynes can act as dienophiles too. We're going to take these pi electrons and move them into here. There's a bond that forms between these two carbons. Then finally move these electrons into here. Let's draw our product."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "We're going to take these pi electrons and move them into here. There's a bond that forms between these two carbons. Then finally move these electrons into here. Let's draw our product. For this time, we have two double bonds in the ring like that. Then we would have this group coming off of this carbon, which is this one right here. Let's draw that in."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw our product. For this time, we have two double bonds in the ring like that. Then we would have this group coming off of this carbon, which is this one right here. Let's draw that in. We have our ester, so we put that in. Then the same thing down here. Let's follow our electrons again."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw that in. We have our ester, so we put that in. Then the same thing down here. Let's follow our electrons again. We'll start with the electrons in red. These electrons moved into here to form this bond. Next, let's look at these electrons right here on our alkyne."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Let's follow our electrons again. We'll start with the electrons in red. These electrons moved into here to form this bond. Next, let's look at these electrons right here on our alkyne. These pi electrons move into here to form this bond. Then finally, our electrons in magenta move into here. This is our product."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at these electrons right here on our alkyne. These pi electrons move into here to form this bond. Then finally, our electrons in magenta move into here. This is our product. Let's do another one. Let's go down here and let's look at this Diels-Alder reaction. On the left, we have our diene."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "This is our product. Let's do another one. Let's go down here and let's look at this Diels-Alder reaction. On the left, we have our diene. On the right, we have our dienophile. We can start moving our electrons around because we already have an S-cis conformation. If we move these electrons into here, then we form a bond right here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "On the left, we have our diene. On the right, we have our dienophile. We can start moving our electrons around because we already have an S-cis conformation. If we move these electrons into here, then we form a bond right here. Then we move these pi electrons into here to form a bond here. Then move these electrons. We can draw our product."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "If we move these electrons into here, then we form a bond right here. Then we move these pi electrons into here to form a bond here. Then move these electrons. We can draw our product. We would have this ring on the left. Then on the right, we would have this. We'll draw in this portion, draw in our carbonyls here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "We can draw our product. We would have this ring on the left. Then on the right, we would have this. We'll draw in this portion, draw in our carbonyls here. This is our product. Following our electrons as usual, electrons in red moved into here. Our pi electrons in blue moved into here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "We'll draw in this portion, draw in our carbonyls here. This is our product. Following our electrons as usual, electrons in red moved into here. Our pi electrons in blue moved into here. Our electrons in magenta moved over to here. Six pi electrons moving at the same time. What if you were given the product and asked to come up with the necessary diene and dienophile?"}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Our pi electrons in blue moved into here. Our electrons in magenta moved over to here. Six pi electrons moving at the same time. What if you were given the product and asked to come up with the necessary diene and dienophile? Let's do a problem like that. Thinking backwards, let's say you were given this on the right and asked what combination of diene and dienophile do you need? Think about moving those electrons in reverse."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "What if you were given the product and asked to come up with the necessary diene and dienophile? Let's do a problem like that. Thinking backwards, let's say you were given this on the right and asked what combination of diene and dienophile do you need? Think about moving those electrons in reverse. Let's look at our product here. Let's start with these electrons. I'll make them the same colors we've been using."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Think about moving those electrons in reverse. Let's look at our product here. Let's start with these electrons. I'll make them the same colors we've been using. Move them in the reverse order this time. These electrons would move over to here. Then these blue electrons in this bond would move over to here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "I'll make them the same colors we've been using. Move them in the reverse order this time. These electrons would move over to here. Then these blue electrons in this bond would move over to here. Then finally, these electrons in red would move over to here. Let's go ahead and draw our diene and our dienophile. On the left, our diene, we would have our double bonds looking like that."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "Then these blue electrons in this bond would move over to here. Then finally, these electrons in red would move over to here. Let's go ahead and draw our diene and our dienophile. On the left, our diene, we would have our double bonds looking like that. Then on the right for our dienophile, let me go ahead and draw our ring here and put in the carbonyls. Let's follow our electrons along here. The electrons in magenta moved over to here."}, {"video_title": "Diels-Alder reaction Organic chemistry Khan Academy.mp3", "Sentence": "On the left, our diene, we would have our double bonds looking like that. Then on the right for our dienophile, let me go ahead and draw our ring here and put in the carbonyls. Let's follow our electrons along here. The electrons in magenta moved over to here. The electrons in blue were over to here. Our electrons in red moved over to here. To check yourself on a problem like this, you can just take the diene and the dienophile that you drew and double check and make sure they give you the product on the right."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "If we have this molecule right over here, I won't go through the trouble of naming it. It would take up too much time in this video. But it's dissolved in methanol. And we talk about what type of reactions. We're going to pick between SN2, SN1, E2, and E1 reactions. Now, maybe the first place to start, or the place I like to start, is to just look at the solvent itself. And when we're trying to decide what type of reaction will occur, the important thing to think about is, is the solvent protic or aprotic?"}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And we talk about what type of reactions. We're going to pick between SN2, SN1, E2, and E1 reactions. Now, maybe the first place to start, or the place I like to start, is to just look at the solvent itself. And when we're trying to decide what type of reaction will occur, the important thing to think about is, is the solvent protic or aprotic? And if you look at this solvent right here, this is methanol, it is protic. And in case you don't remember what protic means, it means that there are protons flying around in the solvent, that they can kind of go loose and jump around from one molecule to another. And the reason why I know that methanol is protic is because you have hydrogen bonded to a very electronegative atom in oxygen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And when we're trying to decide what type of reaction will occur, the important thing to think about is, is the solvent protic or aprotic? And if you look at this solvent right here, this is methanol, it is protic. And in case you don't remember what protic means, it means that there are protons flying around in the solvent, that they can kind of go loose and jump around from one molecule to another. And the reason why I know that methanol is protic is because you have hydrogen bonded to a very electronegative atom in oxygen. So every now and then, in one of the methanol molecules, the oxygen can steal hydrogen's electron. And then the hydrogen itself, without the electron, the hydrogen proton, will be flying around because it doesn't have a neutron. So this is a protic solvent."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And the reason why I know that methanol is protic is because you have hydrogen bonded to a very electronegative atom in oxygen. So every now and then, in one of the methanol molecules, the oxygen can steal hydrogen's electron. And then the hydrogen itself, without the electron, the hydrogen proton, will be flying around because it doesn't have a neutron. So this is a protic solvent. Now, you might say, well, does anything with a hydrogen, would that be protic? And the answer is no. If you have a bunch of hydrogens bonded to just carbons, that is not protic."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So this is a protic solvent. Now, you might say, well, does anything with a hydrogen, would that be protic? And the answer is no. If you have a bunch of hydrogens bonded to just carbons, that is not protic. Carbon is not so electronegative that it could steal a hydrogen's electron and have the hydrogens floating around. So a big giveaway is hydrogen bonded to a very electronegative atom like oxygen. So this is protic."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "If you have a bunch of hydrogens bonded to just carbons, that is not protic. Carbon is not so electronegative that it could steal a hydrogen's electron and have the hydrogens floating around. So a big giveaway is hydrogen bonded to a very electronegative atom like oxygen. So this is protic. And when we think about protic, out of all the reactions we studied, that favors, so this makes us, well, even a better way to think about it, is it disfavors. So it tells us that it's unlikely to have an SN2 or an E2 reaction. And the logic there is an SN2 reaction needs a strong nucleophile."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So this is protic. And when we think about protic, out of all the reactions we studied, that favors, so this makes us, well, even a better way to think about it, is it disfavors. So it tells us that it's unlikely to have an SN2 or an E2 reaction. And the logic there is an SN2 reaction needs a strong nucleophile. An E2 reaction needs a strong base. Now, if you have protons flying around, the nucleophile or the base is likely to react with the proton. So it would not be likely to react with the substrate itself."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And the logic there is an SN2 reaction needs a strong nucleophile. An E2 reaction needs a strong base. Now, if you have protons flying around, the nucleophile or the base is likely to react with the proton. So it would not be likely to react with the substrate itself. So protic solution, you're unlikely to have an SN2 or E2. What you are likely to have is an SN1 or an E1 reaction. Both of these need the leaving group to leave on its own."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So it would not be likely to react with the substrate itself. So protic solution, you're unlikely to have an SN2 or E2. What you are likely to have is an SN1 or an E1 reaction. Both of these need the leaving group to leave on its own. And actually having protons around might help to stabilize the leaving group to some degree. So it makes SN2E2 unlikely, SN1E1 a little more likely. So so far, these are good candidates."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Both of these need the leaving group to leave on its own. And actually having protons around might help to stabilize the leaving group to some degree. So it makes SN2E2 unlikely, SN1E1 a little more likely. So so far, these are good candidates. Now, the next thing to think about is to just look at the leaving group itself or see if there is even a leaving group. And over here, everything we see on this molecule is either a carbon or hydrogen, except for this iodine right here, and we know that iodide is a good leaving group. So this so far, well, a good leaving group does not make it any less likely that you'd have SN2 or E2."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So so far, these are good candidates. Now, the next thing to think about is to just look at the leaving group itself or see if there is even a leaving group. And over here, everything we see on this molecule is either a carbon or hydrogen, except for this iodine right here, and we know that iodide is a good leaving group. So this so far, well, a good leaving group does not make it any less likely that you'd have SN2 or E2. Both of those can do well with a good leaving group. But it's a necessary requirement for an SN1 or an E1 reaction. Remember, an SN1 and E1, in both of them, the first step is that the leaving group leaves on its own."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So this so far, well, a good leaving group does not make it any less likely that you'd have SN2 or E2. Both of those can do well with a good leaving group. But it's a necessary requirement for an SN1 or an E1 reaction. Remember, an SN1 and E1, in both of them, the first step is that the leaving group leaves on its own. That is the rate-determining step. So that's a requirement for SN1 or E1. So it still looks likely."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Remember, an SN1 and E1, in both of them, the first step is that the leaving group leaves on its own. That is the rate-determining step. So that's a requirement for SN1 or E1. So it still looks likely. We haven't seen anything that would make us think that we wouldn't have an SN1 or E1. This is a good leaving group. Let me write it here."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So it still looks likely. We haven't seen anything that would make us think that we wouldn't have an SN1 or E1. This is a good leaving group. Let me write it here. Good leaving group. Now, the last thing we can think about right here is the carbon that we might be leaving from. So far, everything is pointing the SN1 and E1 direction."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Let me write it here. Good leaving group. Now, the last thing we can think about right here is the carbon that we might be leaving from. So far, everything is pointing the SN1 and E1 direction. The final thing is when this leaving group leaves, it's going to form a carbocation from the carbon that it's bonded to right now. And in order for that carbocation to be reasonably stable, at minimum, it should be a secondary carbocation bonded to at least two carbons, but ideally, it would be bonded to three carbons, be a tertiary carbon. Now, the carbon that the leaving group is bonded to is a tertiary carbon."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So far, everything is pointing the SN1 and E1 direction. The final thing is when this leaving group leaves, it's going to form a carbocation from the carbon that it's bonded to right now. And in order for that carbocation to be reasonably stable, at minimum, it should be a secondary carbocation bonded to at least two carbons, but ideally, it would be bonded to three carbons, be a tertiary carbon. Now, the carbon that the leaving group is bonded to is a tertiary carbon. It's bonded to one, two, three carbons. So it is a tertiary carbon. So it can actually be a stable carbocation."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Now, the carbon that the leaving group is bonded to is a tertiary carbon. It's bonded to one, two, three carbons. So it is a tertiary carbon. So it can actually be a stable carbocation. It's a tertiary carbon, which once again, favors SN1 and E1. So all of the clues here tell us that SN1 and E1 are going to happen, and actually, they'll both happen. So let's think about the mechanism."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So it can actually be a stable carbocation. It's a tertiary carbon, which once again, favors SN1 and E1. So all of the clues here tell us that SN1 and E1 are going to happen, and actually, they'll both happen. So let's think about the mechanism. So the very first step, the leaving group leaves in either one of these reactions. So if we look at the iodine, it already has seven valence electrons, one, two, three, four, five, six, seven. And in the first step, the rate determining step of the SN1 or E1 reaction, the iodine's going to nab an electron off the carbon."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So let's think about the mechanism. So the very first step, the leaving group leaves in either one of these reactions. So if we look at the iodine, it already has seven valence electrons, one, two, three, four, five, six, seven. And in the first step, the rate determining step of the SN1 or E1 reaction, the iodine's going to nab an electron off the carbon. I'll do that electron in green right over there. That's going to get nabbed onto the iodine to make iodide, and then that carbon is going to lose an electron and become a carbocation. So after that very first step, we have something that looks like this."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And in the first step, the rate determining step of the SN1 or E1 reaction, the iodine's going to nab an electron off the carbon. I'll do that electron in green right over there. That's going to get nabbed onto the iodine to make iodide, and then that carbon is going to lose an electron and become a carbocation. So after that very first step, we have something that looks like this. So that's our molecule. So you can just so clear what we did, this arrow I'll do all the way over here. So it's clear that this is our next step."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So after that very first step, we have something that looks like this. So that's our molecule. So you can just so clear what we did, this arrow I'll do all the way over here. So it's clear that this is our next step. And what's happened here? Our tertiary carbon has lost its electron. So now this carbon right here has a positive charge."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So it's clear that this is our next step. And what's happened here? Our tertiary carbon has lost its electron. So now this carbon right here has a positive charge. It's a tertiary carbocation. And now the iodine has become iodide. It has left the molecule."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So now this carbon right here has a positive charge. It's a tertiary carbocation. And now the iodine has become iodide. It has left the molecule. So it had its original seven valence electrons, one, two, three, four, five, six, seven. It nabbed one more electron from the carbon. And now it is, I wanted to do that in green, nabbed one more electron from the carbon."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "It has left the molecule. So it had its original seven valence electrons, one, two, three, four, five, six, seven. It nabbed one more electron from the carbon. And now it is, I wanted to do that in green, nabbed one more electron from the carbon. Now it is the iodide anion. So this step right here is common to both SN1 and E1 reaction. The leaving group has to leave."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And now it is, I wanted to do that in green, nabbed one more electron from the carbon. Now it is the iodide anion. So this step right here is common to both SN1 and E1 reaction. The leaving group has to leave. Now, after this, they start to diverge. In an SN1, the leaving group essentially gets substituted with a weak nucleophile. In E1, a weak base strips off one of the beta hydrogens and forms an alkene."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "The leaving group has to leave. Now, after this, they start to diverge. In an SN1, the leaving group essentially gets substituted with a weak nucleophile. In E1, a weak base strips off one of the beta hydrogens and forms an alkene. So let's do them separately. So over here I'm going to do the SN1. And on the right-hand side, I will do the E1 reaction."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "In E1, a weak base strips off one of the beta hydrogens and forms an alkene. So let's do them separately. So over here I'm going to do the SN1. And on the right-hand side, I will do the E1 reaction. So let me start over here. So the SN1 is starting over here at this step. I'll just redo this step over here."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And on the right-hand side, I will do the E1 reaction. So let me start over here. So the SN1 is starting over here at this step. I'll just redo this step over here. So this has a positive charge. The iodide has left. I don't have to draw all of its valence electrons anymore."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "I'll just redo this step over here. So this has a positive charge. The iodide has left. I don't have to draw all of its valence electrons anymore. And what's going to happen next? We're going to get substituted with the weak base. And the weak base here is actually the methanol."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "I don't have to draw all of its valence electrons anymore. And what's going to happen next? We're going to get substituted with the weak base. And the weak base here is actually the methanol. So let me draw some methanol here. It's got two unbonded pairs of electrons. And one of them, it's a weak base."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And the weak base here is actually the methanol. So let me draw some methanol here. It's got two unbonded pairs of electrons. And one of them, it's a weak base. It was willing to give an electron, but it has a partial negative charge over here because oxygen is electronegative. But it doesn't have a full negative charge, so it's not a strong nucleophile. But it can donate an electron to this carbocation."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And one of them, it's a weak base. It was willing to give an electron, but it has a partial negative charge over here because oxygen is electronegative. But it doesn't have a full negative charge, so it's not a strong nucleophile. But it can donate an electron to this carbocation. And that's what is going to happen. It will donate an electron to this carbocation. And then after that happens, it will look like this."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "But it can donate an electron to this carbocation. And that's what is going to happen. It will donate an electron to this carbocation. And then after that happens, it will look like this. So that's our original molecule. Now this magenta electron has been donated to the carbocation. The other end of it is this blue electron right here on the oxygen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And then after that happens, it will look like this. So that's our original molecule. Now this magenta electron has been donated to the carbocation. The other end of it is this blue electron right here on the oxygen. It is now bonded. That is our oxygen. Here's that other pair of electrons on that oxygen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "The other end of it is this blue electron right here on the oxygen. It is now bonded. That is our oxygen. Here's that other pair of electrons on that oxygen. And it is bonded to a hydrogen and a methyl group. And then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there. Oh, I want to be very clear here."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Here's that other pair of electrons on that oxygen. And it is bonded to a hydrogen and a methyl group. And then the last step of this is another weak base might be able to come and nab off the hydrogen proton right there. Oh, I want to be very clear here. The oxygen was neutral. The methanol here is neutral. It is giving away an electron to the carbocation."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Oh, I want to be very clear here. The oxygen was neutral. The methanol here is neutral. It is giving away an electron to the carbocation. The carbocation had a positive charge because it had lost it originally. Now it gets an electron back. It becomes neutral."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "It is giving away an electron to the carbocation. The carbocation had a positive charge because it had lost it originally. Now it gets an electron back. It becomes neutral. The methanol, on the other hand, was neutral, gives away an electron. So now it becomes positive. So now you might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "It becomes neutral. The methanol, on the other hand, was neutral, gives away an electron. So now it becomes positive. So now you might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion. So this one right here, it would nab it or it would bond with it. It would give the electron to the hydrogen proton, really. The hydrogen's electron gets nabbed by the oxygen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So now you might have another methanol molecule sitting out here someplace that might also nab the proton off of this positive ion. So this one right here, it would nab it or it would bond with it. It would give the electron to the hydrogen proton, really. The hydrogen's electron gets nabbed by the oxygen. And so then that becomes neutral. So then the final step, it'll all look like this. We have that over here."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "The hydrogen's electron gets nabbed by the oxygen. And so then that becomes neutral. So then the final step, it'll all look like this. We have that over here. The methanol that it originally bonded has lost its hydrogen. So it looks like this. We just have the oxygen and the CH3 there."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "We have that over here. The methanol that it originally bonded has lost its hydrogen. So it looks like this. We just have the oxygen and the CH3 there. It is now neutral because it gained an electron when that hydrogen proton was nabbed. So if you wanted to draw it, it has actually those two extra electrons just like that. And then if you want to draw this last methanol, it's now a positive cation."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "We just have the oxygen and the CH3 there. It is now neutral because it gained an electron when that hydrogen proton was nabbed. So if you wanted to draw it, it has actually those two extra electrons just like that. And then if you want to draw this last methanol, it's now a positive cation. So it looks like this. So it's OHCH3H. And then it has unbonded pair right there."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And then if you want to draw this last methanol, it's now a positive cation. So it looks like this. So it's OHCH3H. And then it has unbonded pair right there. And now this has a positive charge. So that was the SN1 reaction. Now the other reaction that's going to occur is the E1 reaction."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And then it has unbonded pair right there. And now this has a positive charge. So that was the SN1 reaction. Now the other reaction that's going to occur is the E1 reaction. So E1 reaction. Once again, our first step looks like that. So that is our first step."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Now the other reaction that's going to occur is the E1 reaction. So E1 reaction. Once again, our first step looks like that. So that is our first step. Let me get everything straight. So the leaving group had left. So in each situation, the leaving group had left."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So that is our first step. Let me get everything straight. So the leaving group had left. So in each situation, the leaving group had left. Our iodide is up here. And in an E1 reaction, you don't get substituted. What happens is one of the beta carbons gets a hydrogen swiped off of it by a weak base."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So in each situation, the leaving group had left. Our iodide is up here. And in an E1 reaction, you don't get substituted. What happens is one of the beta carbons gets a hydrogen swiped off of it by a weak base. Now let's think about what a beta carbon is. The alpha carbon is the carbocation carbon. That's right over there."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "What happens is one of the beta carbons gets a hydrogen swiped off of it by a weak base. Now let's think about what a beta carbon is. The alpha carbon is the carbocation carbon. That's right over there. That's the alpha carbon. Beta carbons are one carbon away. So this is a beta carbon."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "That's right over there. That's the alpha carbon. Beta carbons are one carbon away. So this is a beta carbon. This is a beta carbon. And then that is a beta carbon. Now this carbon over here is not bonded to any hydrogen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So this is a beta carbon. This is a beta carbon. And then that is a beta carbon. Now this carbon over here is not bonded to any hydrogen. It's only bonded to other carbons. So that one cannot lose any hydrogens. And then we have to pick between this carbon that's bonded to three hydrogens and this carbon that's actually bonded to two hydrogens."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Now this carbon over here is not bonded to any hydrogen. It's only bonded to other carbons. So that one cannot lose any hydrogens. And then we have to pick between this carbon that's bonded to three hydrogens and this carbon that's actually bonded to two hydrogens. I didn't draw it before, but it's implicitly there. It's bonded to two hydrogens. Now Zaitsev's rule tells us that the dominating product is going to be produced when the carbon that has less hydrogens loses a hydrogen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "And then we have to pick between this carbon that's bonded to three hydrogens and this carbon that's actually bonded to two hydrogens. I didn't draw it before, but it's implicitly there. It's bonded to two hydrogens. Now Zaitsev's rule tells us that the dominating product is going to be produced when the carbon that has less hydrogens loses a hydrogen. So out of this, this carbon has two hydrogens. This one has three. So the one that's going to lose a hydrogen to the base, or more likely to lose a hydrogen to the base, is the one that has two hydrogens, not three."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Now Zaitsev's rule tells us that the dominating product is going to be produced when the carbon that has less hydrogens loses a hydrogen. So out of this, this carbon has two hydrogens. This one has three. So the one that's going to lose a hydrogen to the base, or more likely to lose a hydrogen to the base, is the one that has two hydrogens, not three. The fewer hydrogens. So our base in this case is once again the methanol, acted as a nucleophile in the SN1, acted as a weak nucleophile in SN1, now it'll act as a weak base. So we have methanol right over here."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So the one that's going to lose a hydrogen to the base, or more likely to lose a hydrogen to the base, is the one that has two hydrogens, not three. The fewer hydrogens. So our base in this case is once again the methanol, acted as a nucleophile in the SN1, acted as a weak nucleophile in SN1, now it'll act as a weak base. So we have methanol right over here. That's a hydrogen. CH3 has some electrons right over here. Has a partial negative charge."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "So we have methanol right over here. That's a hydrogen. CH3 has some electrons right over here. Has a partial negative charge. It will give one of the electrons to the hydrogen, just to the hydrogen proton. The hydrogen is not going to take its electron with it. That electron is then going to be given to the carbocation to make it neutral, and it will form a double bond."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Has a partial negative charge. It will give one of the electrons to the hydrogen, just to the hydrogen proton. The hydrogen is not going to take its electron with it. That electron is then going to be given to the carbocation to make it neutral, and it will form a double bond. So after that happens, we are left with something that looks like this. This was our original molecule. Now this carbon right here, in yellow, it has lost its hydrogen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "That electron is then going to be given to the carbocation to make it neutral, and it will form a double bond. So after that happens, we are left with something that looks like this. This was our original molecule. Now this carbon right here, in yellow, it has lost its hydrogen. This hydrogen back here is still there. I could draw it if I like. I don't need to."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "Now this carbon right here, in yellow, it has lost its hydrogen. This hydrogen back here is still there. I could draw it if I like. I don't need to. It just makes the thing messy. But it has now formed a double bond with the primary carbon. That electron that it had bonded with the hydrogen was now given to the carbocation."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "I don't need to. It just makes the thing messy. But it has now formed a double bond with the primary carbon. That electron that it had bonded with the hydrogen was now given to the carbocation. It has a double bond with that. And then you have your methanol has now turned into a positive ion. CH3, and now it has this bond with the hydrogen."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "That electron that it had bonded with the hydrogen was now given to the carbocation. It has a double bond with that. And then you have your methanol has now turned into a positive ion. CH3, and now it has this bond with the hydrogen. I'll even make that electron that it gave away in magenta. And then it has that extra lone pair of electrons. So in this circumstance, we looked at all of the clues."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 3.mp3", "Sentence": "CH3, and now it has this bond with the hydrogen. I'll even make that electron that it gave away in magenta. And then it has that extra lone pair of electrons. So in this circumstance, we looked at all of the clues. All of the clues were against SN2 and E2. They favored SN1, E1. So if you were to actually try to see what happens in this reaction, you would get products of both SN1 and E1 reactions, these products and these products over here."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Or I guess in the simplest form, this could just be a hydrogen over here. And then they definitely will have at least one hydrogen bonded to the carbon in the carbonyl group. So this was an aldehyde. Now very closely related to an aldehyde is a type of molecule called a ketone. Now let's draw a couple of ketones just to make things clear. And then we'll think about what the difference is between a ketone and an aldehyde. So this right here, CH3, carbon right over here, this right here is a ketone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now very closely related to an aldehyde is a type of molecule called a ketone. Now let's draw a couple of ketones just to make things clear. And then we'll think about what the difference is between a ketone and an aldehyde. So this right here, CH3, carbon right over here, this right here is a ketone. And you say, hey, Sal, it looks very similar to an aldehyde. I have a carbonyl group in both. So this right here is a carbonyl group."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this right here, CH3, carbon right over here, this right here is a ketone. And you say, hey, Sal, it looks very similar to an aldehyde. I have a carbonyl group in both. So this right here is a carbonyl group. This right here is a carbonyl group. Let me write this, we have a carbonyl group in both. We have a carbon chain over here."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this right here is a carbonyl group. This right here is a carbonyl group. Let me write this, we have a carbonyl group in both. We have a carbon chain over here. This could be a general carbon chain. Here we have a methyl group. So how is this different?"}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We have a carbon chain over here. This could be a general carbon chain. Here we have a methyl group. So how is this different? And I think you'll see. It's this part that I haven't highlighted yet. In an aldehyde, there's two ways to think about an aldehyde."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So how is this different? And I think you'll see. It's this part that I haven't highlighted yet. In an aldehyde, there's two ways to think about an aldehyde. You could either say that, look, the carbonyl group is at the end of a carbon chain. So the next thing over is going to be a hydrogen. Or you could say that in an aldehyde, you have at least one hydrogen bonded to the carbonyl carbon."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "In an aldehyde, there's two ways to think about an aldehyde. You could either say that, look, the carbonyl group is at the end of a carbon chain. So the next thing over is going to be a hydrogen. Or you could say that in an aldehyde, you have at least one hydrogen bonded to the carbonyl carbon. And the way I remember that, so you have a hydrogen there. And in the ketone, you don't. The carbonyl group is embedded in a carbon chain."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Or you could say that in an aldehyde, you have at least one hydrogen bonded to the carbonyl carbon. And the way I remember that, so you have a hydrogen there. And in the ketone, you don't. The carbonyl group is embedded in a carbon chain. It is bonded to a carbon on either side. So over here, you have a carbon. And the way that I remember the difference, and this is really just a little bit of a mnemonic just to memorize it, is aldehyde has an H in it."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The carbonyl group is embedded in a carbon chain. It is bonded to a carbon on either side. So over here, you have a carbon. And the way that I remember the difference, and this is really just a little bit of a mnemonic just to memorize it, is aldehyde has an H in it. There's an H right over there. And an aldehyde has a hydrogen bonded to this carbon. So now that we at least have a reasonable understanding of what a ketone is, let's name a few just to familiarize ourselves."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And the way that I remember the difference, and this is really just a little bit of a mnemonic just to memorize it, is aldehyde has an H in it. There's an H right over there. And an aldehyde has a hydrogen bonded to this carbon. So now that we at least have a reasonable understanding of what a ketone is, let's name a few just to familiarize ourselves. So this right here is the simplest possible ketone. And it's called acetone. Or that's its common name."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So now that we at least have a reasonable understanding of what a ketone is, let's name a few just to familiarize ourselves. So this right here is the simplest possible ketone. And it's called acetone. Or that's its common name. It is called acetone. And actually, the word ketone comes from the German word for acetone, which was, I think, instead of a C there, they had a K, so it was like aketone. And they said, oh, this is a ketone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Or that's its common name. It is called acetone. And actually, the word ketone comes from the German word for acetone, which was, I think, instead of a C there, they had a K, so it was like aketone. And they said, oh, this is a ketone. So this is a common name. Another kind of traditional or common way of naming a ketone is to name each of these groups. And it's similar to the way that we named ethers, but instead of writing the word ether at the end, we write the word ketone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And they said, oh, this is a ketone. So this is a common name. Another kind of traditional or common way of naming a ketone is to name each of these groups. And it's similar to the way that we named ethers, but instead of writing the word ether at the end, we write the word ketone. So here we have a methyl group and we have another methyl group. So this would be, so we have two methyl groups right over here. So this would be dimethylketone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And it's similar to the way that we named ethers, but instead of writing the word ether at the end, we write the word ketone. So here we have a methyl group and we have another methyl group. So this would be, so we have two methyl groups right over here. So this would be dimethylketone. This right here is dimethylketone. And then if you wanted the systematic way of naming it, you just look at the longest carbon chain, which is one, two, three carbons. So it's propa."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this would be dimethylketone. This right here is dimethylketone. And then if you wanted the systematic way of naming it, you just look at the longest carbon chain, which is one, two, three carbons. So it's propa. And instead of calling it propane, we get rid of that E over there, and we'd call it propanone. We could call it propanone. That tells us that this right here is a ketone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it's propa. And instead of calling it propane, we get rid of that E over there, and we'd call it propanone. We could call it propanone. That tells us that this right here is a ketone. And you have to know where this double bond, and actually for propanone, you don't have to specify it. Because if you know it's a ketone, you know that it has to have a carbon on either side of the carbonyl group. So you actually don't even have to specify where the carbonyl group is."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "That tells us that this right here is a ketone. And you have to know where this double bond, and actually for propanone, you don't have to specify it. Because if you know it's a ketone, you know that it has to have a carbon on either side of the carbonyl group. So you actually don't even have to specify where the carbonyl group is. But if you wanted to, you could say, OK, that's going to be on the two carbon. No matter what direction you start counting from, it's going to be on the two carbon. But the two is kind of optional for propanone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So you actually don't even have to specify where the carbonyl group is. But if you wanted to, you could say, OK, that's going to be on the two carbon. No matter what direction you start counting from, it's going to be on the two carbon. But the two is kind of optional for propanone. The two is optional for propanone. Let's do a couple of other ones. So let's say we had a molecule that looked like this."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But the two is kind of optional for propanone. The two is optional for propanone. Let's do a couple of other ones. So let's say we had a molecule that looked like this. Let's say we have a molecule that looks like this. So the traditional way of naming it, you'd say, OK, on this end of the ketone, I have one, two, three carbons. So on that end, I have three carbons."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's say we had a molecule that looked like this. Let's say we have a molecule that looks like this. So the traditional way of naming it, you'd say, OK, on this end of the ketone, I have one, two, three carbons. So on that end, I have three carbons. That is a propyl group. And on this other side of the ketone, right over here, I have only one carbon. That is a methyl group."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So on that end, I have three carbons. That is a propyl group. And on this other side of the ketone, right over here, I have only one carbon. That is a methyl group. So then you would just name them. And you name them in order of increasing chain size or carbon, or I guess molecule size or group size. So this one, you'd write methyl first."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "That is a methyl group. So then you would just name them. And you name them in order of increasing chain size or carbon, or I guess molecule size or group size. So this one, you'd write methyl first. Methyl, because it's only one carbon. So this is methylpropyl ketone. This is kind of the traditional or the common way, so often kind of the most used way, of naming this molecule."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this one, you'd write methyl first. Methyl, because it's only one carbon. So this is methylpropyl ketone. This is kind of the traditional or the common way, so often kind of the most used way, of naming this molecule. But the systematic way of naming it, you look at the longest carbon chain. You say, OK, I have one, two, three, four, five carbons. So it's going to be pent."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This is kind of the traditional or the common way, so often kind of the most used way, of naming this molecule. But the systematic way of naming it, you look at the longest carbon chain. You say, OK, I have one, two, three, four, five carbons. So it's going to be pent. And then you want to start numbering it so that the ketone carbon, or the carbonyl carbon, I should say, has the lowest possible number. So you want to start numbering on the right side. One, two, three, four, and five."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to be pent. And then you want to start numbering it so that the ketone carbon, or the carbonyl carbon, I should say, has the lowest possible number. So you want to start numbering on the right side. One, two, three, four, and five. So this right here, so we said the prefix would be pent. So it's penton. And instead of saying it's pentane, you say it's pentanone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, and five. So this right here, so we said the prefix would be pent. So it's penton. And instead of saying it's pentane, you say it's pentanone. And to specify where the carbonyl group is, you say it's two. This is two pentanone. And you might also see it written like this."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And instead of saying it's pentane, you say it's pentanone. And to specify where the carbonyl group is, you say it's two. This is two pentanone. And you might also see it written like this. You might also see it written as pentanetuone. Either one of these right here would be acceptable. Let's do a slightly more complicated example."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And you might also see it written like this. You might also see it written as pentanetuone. Either one of these right here would be acceptable. Let's do a slightly more complicated example. Let's say we had something that looked like this. So we have something, a molecule that looked like this. Let me stick some chlorines over here."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's do a slightly more complicated example. Let's say we had something that looked like this. So we have something, a molecule that looked like this. Let me stick some chlorines over here. So what would this be? Well, our longest chain, once again, is this cyclohexane. One, two, three, four, five, six carbons."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let me stick some chlorines over here. So what would this be? Well, our longest chain, once again, is this cyclohexane. One, two, three, four, five, six carbons. And I'll just name this systematically right here. And the more complicated things get, the more systematic people will want to name it. So if we have six carbons right here, and they're in a chain, so this is cyclohexane."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six carbons. And I'll just name this systematically right here. And the more complicated things get, the more systematic people will want to name it. So if we have six carbons right here, and they're in a chain, so this is cyclohexane. You would put the e there if this carbonyl group wasn't there, but since it is, we would call this cyclohexanone. So this right here tells us to name it cyclohexanone. And then in a ring like this, this would implicitly be the number one carbon."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So if we have six carbons right here, and they're in a chain, so this is cyclohexane. You would put the e there if this carbonyl group wasn't there, but since it is, we would call this cyclohexanone. So this right here tells us to name it cyclohexanone. And then in a ring like this, this would implicitly be the number one carbon. So if this is the number one carbon, and we want to number in the direction so that the next groups have the lowest possible number, so we want to make this the two carbon. So this is 2,2-dichlorocyclohexanone. So this is 2,2-dichlorocyclohexanone."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then in a ring like this, this would implicitly be the number one carbon. So if this is the number one carbon, and we want to number in the direction so that the next groups have the lowest possible number, so we want to make this the two carbon. So this is 2,2-dichlorocyclohexanone. So this is 2,2-dichlorocyclohexanone. Now, there's two more, and I'll just show these to you because these tend to be referred to by their common name, so I just want to show them to you real fast. One is this molecule right here, where we have a methyl group on this side. CH3, and over here we have a benzene ring."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is 2,2-dichlorocyclohexanone. Now, there's two more, and I'll just show these to you because these tend to be referred to by their common name, so I just want to show them to you real fast. One is this molecule right here, where we have a methyl group on this side. CH3, and over here we have a benzene ring. Now, that first super simple ketone that we saw, we call this acetone. And so the common name here is actually derived from acetone, instead of calling it acetone because it doesn't have just a methyl group here, this is called acetone. And instead of acetone, it's acetophenone because we have this phenol group, that benzene ring right there."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "CH3, and over here we have a benzene ring. Now, that first super simple ketone that we saw, we call this acetone. And so the common name here is actually derived from acetone, instead of calling it acetone because it doesn't have just a methyl group here, this is called acetone. And instead of acetone, it's acetophenone because we have this phenol group, that benzene ring right there. Acetophenone, which is a pretty common molecule you'll see, and you'll see it referred to this way. Now, the other one that you might see every now and then, and I just want to expose it to you, is a molecule that looks like this. It has two benzene rings on it."}, {"video_title": "Ketone naming Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And instead of acetone, it's acetophenone because we have this phenol group, that benzene ring right there. Acetophenone, which is a pretty common molecule you'll see, and you'll see it referred to this way. Now, the other one that you might see every now and then, and I just want to expose it to you, is a molecule that looks like this. It has two benzene rings on it. So it has two benzene rings on it. Looks like that. And this is benzophenone."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We've already seen how to name alkanes. For example, this alkane is five carbons, one, two, three, four, five. We know it's called pentane. So what do we do if we're trying to name alkenes? So this compound has a double bond present, but there's also five carbons, one, two, three, four, and five. So we start with pent, indicating five carbons present, but since there's a double bond, we need to change our ending from ane to ene. So let me write in here ene, and we'd call this molecule pentene."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what do we do if we're trying to name alkenes? So this compound has a double bond present, but there's also five carbons, one, two, three, four, and five. So we start with pent, indicating five carbons present, but since there's a double bond, we need to change our ending from ane to ene. So let me write in here ene, and we'd call this molecule pentene. We want to number it to give the lowest number possible to our double bond. So we start from the left. This would be carbon one, this would be carbon two, three, four, and five."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me write in here ene, and we'd call this molecule pentene. We want to number it to give the lowest number possible to our double bond. So we start from the left. This would be carbon one, this would be carbon two, three, four, and five. So our double bond starts at carbon one, and we could write one pentene. If you leave the one off, it's implied that the double bond starts at carbon one. What if we tried to number this starting from the right?"}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This would be carbon one, this would be carbon two, three, four, and five. So our double bond starts at carbon one, and we could write one pentene. If you leave the one off, it's implied that the double bond starts at carbon one. What if we tried to number this starting from the right? So if we said that this was carbon one, that would make this carbon two, carbon three, carbon four, and carbon five. So you're saying the double bond would start at carbon four, and this would be four pentene. That's incorrect, because our goal is to give the double bond the lowest number possible."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What if we tried to number this starting from the right? So if we said that this was carbon one, that would make this carbon two, carbon three, carbon four, and carbon five. So you're saying the double bond would start at carbon four, and this would be four pentene. That's incorrect, because our goal is to give the double bond the lowest number possible. So we call it one pentene, or just pentene. Let's look at this next example here. So our goal is to find the longest carbon chain that includes a double bond, and we want to give the double bond the lowest number possible."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's incorrect, because our goal is to give the double bond the lowest number possible. So we call it one pentene, or just pentene. Let's look at this next example here. So our goal is to find the longest carbon chain that includes a double bond, and we want to give the double bond the lowest number possible. So the longest carbon chain that includes the double bond, we would have to make this carbon one. And this would be carbon two, this would be carbon three, this is four, and this is five. So a five-carbon alkene would be pentene."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So our goal is to find the longest carbon chain that includes a double bond, and we want to give the double bond the lowest number possible. So the longest carbon chain that includes the double bond, we would have to make this carbon one. And this would be carbon two, this would be carbon three, this is four, and this is five. So a five-carbon alkene would be pentene. You could put the one in there, or you could leave it off. Next, you think about the substituents. In this case, we have only one group coming off our carbon chain."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So a five-carbon alkene would be pentene. You could put the one in there, or you could leave it off. Next, you think about the substituents. In this case, we have only one group coming off our carbon chain. It's a two-carbon alkyl group, and we call that ethyl. So with an ethyl group coming off carbon two, we would call this two-ethyl pentene for the name. But that's not the longest carbon chain, so someone might look at this molecule and say, wait a second, shouldn't this be carbon one, and then this two, three, four, five, and six, because that's six carbons, and six carbons is, of course, more than five."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "In this case, we have only one group coming off our carbon chain. It's a two-carbon alkyl group, and we call that ethyl. So with an ethyl group coming off carbon two, we would call this two-ethyl pentene for the name. But that's not the longest carbon chain, so someone might look at this molecule and say, wait a second, shouldn't this be carbon one, and then this two, three, four, five, and six, because that's six carbons, and six carbons is, of course, more than five. The problem with this is the double bond, the alkene, is not a part of our longest carbon chain. And that's one of our rules. We need to make the longest carbon chain that includes the double bond, and that's why we name this as a pentene derivative, and not numbering it according to the way I did on the right here."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But that's not the longest carbon chain, so someone might look at this molecule and say, wait a second, shouldn't this be carbon one, and then this two, three, four, five, and six, because that's six carbons, and six carbons is, of course, more than five. The problem with this is the double bond, the alkene, is not a part of our longest carbon chain. And that's one of our rules. We need to make the longest carbon chain that includes the double bond, and that's why we name this as a pentene derivative, and not numbering it according to the way I did on the right here. So this is incorrect. So we find the longest carbon chain that includes your double bond, and give the lowest number possible to your double bond. That's true even if you have some alkyl substituents."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We need to make the longest carbon chain that includes the double bond, and that's why we name this as a pentene derivative, and not numbering it according to the way I did on the right here. So this is incorrect. So we find the longest carbon chain that includes your double bond, and give the lowest number possible to your double bond. That's true even if you have some alkyl substituents. So let's look at this next example, and let's name this compound. So first, let's try numbering from the left to the right. This would be carbon one, this would be carbon two, this is carbon three, carbon four, carbon five, and carbon six."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's true even if you have some alkyl substituents. So let's look at this next example, and let's name this compound. So first, let's try numbering from the left to the right. This would be carbon one, this would be carbon two, this is carbon three, carbon four, carbon five, and carbon six. A six-carbon alkene would be called hexene. So let me write that in here. So hexene, and our double bond starts at carbon two."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This would be carbon one, this would be carbon two, this is carbon three, carbon four, carbon five, and carbon six. A six-carbon alkene would be called hexene. So let me write that in here. So hexene, and our double bond starts at carbon two. So here we can see our double bond starting at carbon two. So I'm gonna put a two in front of this, so two-hexene. Next, let's think about the substituents coming off of that longest carbon chain including the double bond."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So hexene, and our double bond starts at carbon two. So here we can see our double bond starting at carbon two. So I'm gonna put a two in front of this, so two-hexene. Next, let's think about the substituents coming off of that longest carbon chain including the double bond. We have a methyl group coming off carbon two, another methyl group on carbon four, and another methyl group on carbon five. So three methyl groups would be called trimethyl. So let me write that in here."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's think about the substituents coming off of that longest carbon chain including the double bond. We have a methyl group coming off carbon two, another methyl group on carbon four, and another methyl group on carbon five. So three methyl groups would be called trimethyl. So let me write that in here. So we have trimethyl, and those methyl groups are on two, four, and five. So two, four, five, trimethyl-two-hexene would be the name for this compound. There's another way to name this molecule."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me write that in here. So we have trimethyl, and those methyl groups are on two, four, and five. So two, four, five, trimethyl-two-hexene would be the name for this compound. There's another way to name this molecule. Instead of putting this two in front of the hexene, you can put the two between the hex and the ene. So let me write out that way. So we'd still have methyl groups at two, four, five, so two, four, five, trimethyl."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "There's another way to name this molecule. Instead of putting this two in front of the hexene, you can put the two between the hex and the ene. So let me write out that way. So we'd still have methyl groups at two, four, five, so two, four, five, trimethyl. Trimethyl, let's write hex, and then we'll put the two in here, hex-two-ene. So again, the two right here indicates the double bond starts at carbon two. This first way of naming it, two, four, five, trimethyl-two-hexene, is more of the old school way of naming this alkene."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we'd still have methyl groups at two, four, five, so two, four, five, trimethyl. Trimethyl, let's write hex, and then we'll put the two in here, hex-two-ene. So again, the two right here indicates the double bond starts at carbon two. This first way of naming it, two, four, five, trimethyl-two-hexene, is more of the old school way of naming this alkene. Two, four, five, trimethyl-hex-two-ene is more of the newer way, but currently most people would accept both ways to name this alkene. What if you tried to number this from the right to the left? If you said that this was carbon one, and this was carbon two, and this was carbon three, carbon four, carbon five, and carbon six?"}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This first way of naming it, two, four, five, trimethyl-two-hexene, is more of the old school way of naming this alkene. Two, four, five, trimethyl-hex-two-ene is more of the newer way, but currently most people would accept both ways to name this alkene. What if you tried to number this from the right to the left? If you said that this was carbon one, and this was carbon two, and this was carbon three, carbon four, carbon five, and carbon six? You might think this would be correct because this would give you methyl groups at two, three, and five. And thinking about the first point of difference rule, we have two here and a two here, but we have a three here versus a four, and we know that three is, of course, a lower number. So you might think that the alkyl groups would lead you to number it starting from the right."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If you said that this was carbon one, and this was carbon two, and this was carbon three, carbon four, carbon five, and carbon six? You might think this would be correct because this would give you methyl groups at two, three, and five. And thinking about the first point of difference rule, we have two here and a two here, but we have a three here versus a four, and we know that three is, of course, a lower number. So you might think that the alkyl groups would lead you to number it starting from the right. The problem with this way is this would give you an alkene that starts at carbon four, a double bond that starts at carbon four. And our goal is to give the lowest number possible to the double bond. And we want to give the double bond the lowest number possible, and that's over any alkyl groups, which is why we number it the way we did."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you might think that the alkyl groups would lead you to number it starting from the right. The problem with this way is this would give you an alkene that starts at carbon four, a double bond that starts at carbon four. And our goal is to give the lowest number possible to the double bond. And we want to give the double bond the lowest number possible, and that's over any alkyl groups, which is why we number it the way we did. And 2,4,5-trimethyl-2-hexene is the correct IUPAC name for this compound. Next, let's look at cyclic examples. So down here on the left, we already know this is cyclohexane with an A-N-E ending."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we want to give the double bond the lowest number possible, and that's over any alkyl groups, which is why we number it the way we did. And 2,4,5-trimethyl-2-hexene is the correct IUPAC name for this compound. Next, let's look at cyclic examples. So down here on the left, we already know this is cyclohexane with an A-N-E ending. If we put a double bond in, we have cyclohexene. So let me write that in here. This is called cyclohexene, so E-N-E."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So down here on the left, we already know this is cyclohexane with an A-N-E ending. If we put a double bond in, we have cyclohexene. So let me write that in here. This is called cyclohexene, so E-N-E. So either one of these carbons could be carbon one. So this carbon or this carbon, either one of them could be. Let's just say this one is."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is called cyclohexene, so E-N-E. So either one of these carbons could be carbon one. So this carbon or this carbon, either one of them could be. Let's just say this one is. If that's carbon one, then this must be carbon two, and so on around the ring, three, four, five, and six. What if you have substituents? Let's look at this example down here on the left first."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's just say this one is. If that's carbon one, then this must be carbon two, and so on around the ring, three, four, five, and six. What if you have substituents? Let's look at this example down here on the left first. So we want to give our substituent the lowest number possible. So here's our double bond, so we can make this one carbon one, or the one below it carbon one. To give this methyl group the lowest number possible, we make this top carbon here carbon one."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at this example down here on the left first. So we want to give our substituent the lowest number possible. So here's our double bond, so we can make this one carbon one, or the one below it carbon one. To give this methyl group the lowest number possible, we make this top carbon here carbon one. So that's carbon one, and this is carbon two, and so on around the ring. So this is a cyclohexene ring. So let me write this down here."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "To give this methyl group the lowest number possible, we make this top carbon here carbon one. So that's carbon one, and this is carbon two, and so on around the ring. So this is a cyclohexene ring. So let me write this down here. So cyclohexene, and we have this methyl group coming off carbon one. So we would write one methyl cyclohexene. What about this one right here?"}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me write this down here. So cyclohexene, and we have this methyl group coming off carbon one. So we would write one methyl cyclohexene. What about this one right here? So again, we have a cyclohexene ring. So let me start by writing that in. So cyclohexene is our parent name here, and we have a choice."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What about this one right here? So again, we have a cyclohexene ring. So let me start by writing that in. So cyclohexene is our parent name here, and we have a choice. Here's our double bond. We could make this one carbon one, or we could make this one carbon one. Well, if we make the bottom one carbon one, that gives this substituent a lower number."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So cyclohexene is our parent name here, and we have a choice. Here's our double bond. We could make this one carbon one, or we could make this one carbon one. Well, if we make the bottom one carbon one, that gives this substituent a lower number. So this would be carbon one. We have to follow the double bond around the ring. So if that's carbon one, this is carbon two, and that makes this carbon three."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, if we make the bottom one carbon one, that gives this substituent a lower number. So this would be carbon one. We have to follow the double bond around the ring. So if that's carbon one, this is carbon two, and that makes this carbon three. So let's think about those substituents. At carbon three, we have two methyl groups. So that's dimethyl."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if that's carbon one, this is carbon two, and that makes this carbon three. So let's think about those substituents. At carbon three, we have two methyl groups. So that's dimethyl. So let's write that in here. So dimethylcyclohexene, and those two methyl groups are both on carbon three. So three dimethyl."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's dimethyl. So let's write that in here. So dimethylcyclohexene, and those two methyl groups are both on carbon three. So three dimethyl. And then we have this ethyl group here coming off of carbon one. So this is an ethyl group. So let's write that in."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So three dimethyl. And then we have this ethyl group here coming off of carbon one. So this is an ethyl group. So let's write that in. So we have one ethyl. So one ethyl, three, three, dimethylcyclohexene. And don't forget about the alphabet rule."}, {"video_title": "Alkene nomenclature Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's write that in. So we have one ethyl. So one ethyl, three, three, dimethylcyclohexene. And don't forget about the alphabet rule. So ethyl comes before methyl, right? E comes before M, which is why we put the ethyl first in the name. So one ethyl, three, three, dimethylcyclohexene."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And they use the letter R, and I've used it before. R stands for radical. And I don't want you to confuse this R with free radical. It means completely different things. R in this form really just means a functional group or a chain of carbons here. It doesn't mean a free radical. This just means it could be just something attached to this OH right there."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It means completely different things. R in this form really just means a functional group or a chain of carbons here. It doesn't mean a free radical. This just means it could be just something attached to this OH right there. Now, another point of clarification. Do not think that anything that fits this pattern is drinkable. Do not associate it with the traditional alcohol that you may or may not have been exposed to."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This just means it could be just something attached to this OH right there. Now, another point of clarification. Do not think that anything that fits this pattern is drinkable. Do not associate it with the traditional alcohol that you may or may not have been exposed to. Traditional drinking alcohol is actually ethanol. Let me write out the molecular formula. CH3, CH2, and then OH."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Do not associate it with the traditional alcohol that you may or may not have been exposed to. Traditional drinking alcohol is actually ethanol. Let me write out the molecular formula. CH3, CH2, and then OH. This is what is inside of wine and beer and hard liquor or whatever you might want. You do not want to drink, and maybe you might not actually want to drink this either, but you definitely do not want to drink something like methanol. It might kill you."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "CH3, CH2, and then OH. This is what is inside of wine and beer and hard liquor or whatever you might want. You do not want to drink, and maybe you might not actually want to drink this either, but you definitely do not want to drink something like methanol. It might kill you. So you do not want to do something like this. You do not want to ingest that. It might kill or blind you."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It might kill you. So you do not want to do something like this. You do not want to ingest that. It might kill or blind you. This might do it in a more indirect way. So I want to get that out of the way. And just so that we get kind of a little bit more comfortable with alcohols."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It might kill or blind you. This might do it in a more indirect way. So I want to get that out of the way. And just so that we get kind of a little bit more comfortable with alcohols. And we've seen them involved in other reactions. We've seen hydroxides act as nucleophiles and SN2 substitution reactions create alcohols. But what I want to do is just learn to get comfortable and really make sure we know how to name these things."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And just so that we get kind of a little bit more comfortable with alcohols. And we've seen them involved in other reactions. We've seen hydroxides act as nucleophiles and SN2 substitution reactions create alcohols. But what I want to do is just learn to get comfortable and really make sure we know how to name these things. So let's just name these molecules that I drew right before I pressed record right over here. So over here, like everything else, we always want to define the longest carbon chain. We have one, two, three, four, five carbons."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But what I want to do is just learn to get comfortable and really make sure we know how to name these things. So let's just name these molecules that I drew right before I pressed record right over here. So over here, like everything else, we always want to define the longest carbon chain. We have one, two, three, four, five carbons. So it's going to be pent. And there's no double bonds. So it's a pentane."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We have one, two, three, four, five carbons. So it's going to be pent. And there's no double bonds. So it's a pentane. So I'll just write pentane right there. And we're not going to just write a pentane. Because actually, the fact that it's an alcohol, that takes precedence over the fact that it is an alkane."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it's a pentane. So I'll just write pentane right there. And we're not going to just write a pentane. Because actually, the fact that it's an alcohol, that takes precedence over the fact that it is an alkane. So actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that it's an alcohol."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Because actually, the fact that it's an alcohol, that takes precedence over the fact that it is an alkane. So actually, the suffix of the word will involve the alcohol part. So it is pentanol. That tells us that it's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So one, two, three, four, five. So sometimes it'll be called 2-pentanol."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That tells us that it's an alcohol. And to know where the OH is grouped, we'll start numbering closest to the OH. So one, two, three, four, five. So sometimes it'll be called 2-pentanol. And this is pretty clear, because we only have one group here. Only one OH. So we know that that is what the 2 applies to."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So sometimes it'll be called 2-pentanol. And this is pretty clear, because we only have one group here. Only one OH. So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan 2-ol. And this way is more useful, especially if you have multiple functional groups. So you know exactly where they sit."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we know that that is what the 2 applies to. But a lot of times, if people want to be a little bit more particular, they might write pentan 2-ol. And this way is more useful, especially if you have multiple functional groups. So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try to name this beast right over here."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So you know exactly where they sit. This one is harder to say. 2-pentanol is pretty straightforward. Now let's try to name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now let's try to name this beast right over here. So we have a couple of things going on. This is an alkyne. We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We have a triple bond. It's an alkyne. We have two bromo groups here. And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And it's also an alcohol. And alcohol takes precedence on all of them. So we want to start numbering closest to the alcohol. So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. So this is an oct. We want to call it an octine. But because we have an alcohol there, we want to call this an octine."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we want to start numbering from this end of the carbon chain. And we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. So this is an oct. We want to call it an octine. But because we have an alcohol there, we want to call this an octine. Let me make it very clear. So oct tells us that we have 8 carbons. Now we have to specify where that triple bond is."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But because we have an alcohol there, we want to call this an octine. Let me make it very clear. So oct tells us that we have 8 carbons. Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that triple bond. So it is oct-5-5-ine that tells us that's where the triple bond is."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now we have to specify where that triple bond is. The triple bond is on the 5 carbon. You always specify the lower number of the carbons on that triple bond. So it is oct-5-5-ine that tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-4-4-ol. And now we have these two bromo groups here on the 7 carbon."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it is oct-5-5-ine that tells us that's where the triple bond is. And then we have the OH on the 4 carbon. So 4-4-4-ol. And now we have these two bromo groups here on the 7 carbon. So it's 7-7-dibromo oct-5-ine-4-ol. And this would all be one word. Let me make sure that you realize that this should be connected."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And now we have these two bromo groups here on the 7 carbon. So it's 7-7-dibromo oct-5-ine-4-ol. And this would all be one word. Let me make sure that you realize that this should be connected. I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let me make sure that you realize that this should be connected. I just ran out of space. So that's probably about as messy of a thing you'll have to name, but just showing you that these things can be named. Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex, and there's all single bonds. So it's a hexane."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now let's think about this one over here in green. So we have 1, 2, 3, 4, 5, 6 carbons. So it's going to be a hex, and there's all single bonds. So it's a hexane. It's a cyclohexane. But then, of course, the hydroxide, or the hydroxy group, I should call it, takes dominance. So it's a hexanol."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it's a hexane. It's a cyclohexane. But then, of course, the hydroxide, or the hydroxy group, I should call it, takes dominance. So it's a hexanol. So this is a cyclohexanol. And once again, that comes from the OH right there. And you don't have to number it."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it's a hexanol. So this is a cyclohexanol. And once again, that comes from the OH right there. And you don't have to number it. Because no matter what carbon it's on, it's on the same one. If you had more than one of these OH groups, then we would have to worry about numbering them. Now let's just do this one right over here."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And you don't have to number it. Because no matter what carbon it's on, it's on the same one. If you had more than one of these OH groups, then we would have to worry about numbering them. Now let's just do this one right over here. So once again, what is our carbon chain? We have 1, 2, 3 carbons. And we have the hydroxy group attached to the 1 and the 3 carbon."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now let's just do this one right over here. So once again, what is our carbon chain? We have 1, 2, 3 carbons. And we have the hydroxy group attached to the 1 and the 3 carbon. So this is going to be, prope is our prefix. It is an alkane. So we would call this."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we have the hydroxy group attached to the 1 and the 3 carbon. So this is going to be, prope is our prefix. It is an alkane. So we would call this. And there's a couple of ways to do this. We could call this 1, 3 propanediol. Actually, I don't have to put a dash there."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we would call this. And there's a couple of ways to do this. We could call this 1, 3 propanediol. Actually, I don't have to put a dash there. And over here, we would add the E, because we have the D right there. So it's propanediol. If it wasn't diol, it would be propanol."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Actually, I don't have to put a dash there. And over here, we would add the E, because we have the D right there. So it's propanediol. If it wasn't diol, it would be propanol. You wouldn't have the E, D, and the I there. So this was specified. We're at the 1 and the 3 carbons."}, {"video_title": "Alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If it wasn't diol, it would be propanol. You wouldn't have the E, D, and the I there. So this was specified. We're at the 1 and the 3 carbons. We have the hydroxy group. Or this could also be written as propan-1, 3-diol. And once again, the di is telling us that we have two of the hydroxy groups attached to this thing."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And to make an imine, we started with an aldehyde or ketone, added an amine, used an acid catalyst, and we formed our imine. And if this Y here, this Y is equal to a hydrogen or an alkyl group, we called it an imine. If that Y is equal to an OH, we would call it an oxime. So let me go ahead and write that. This would be an oxime as your product. And so the word oxime is kind of like a combination of oxygen and imine. And then another derivative would be if the Y group was equal to NH2, or if the Y group were equal to NH and then have another alkyl group in there, so R double prime, we would call it a hydrazone."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write that. This would be an oxime as your product. And so the word oxime is kind of like a combination of oxygen and imine. And then another derivative would be if the Y group was equal to NH2, or if the Y group were equal to NH and then have another alkyl group in there, so R double prime, we would call it a hydrazone. So this would be a hydrazone derivative. And so again, the mechanism would be the same as what we talked about before. And if we wanted to, if we formed any of these, an imine, an oxime, or a hydrazone, and we wanted to go from those products back to our original starting aldehyde or ketone, we could just increase the concentration of water."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then another derivative would be if the Y group was equal to NH2, or if the Y group were equal to NH and then have another alkyl group in there, so R double prime, we would call it a hydrazone. So this would be a hydrazone derivative. And so again, the mechanism would be the same as what we talked about before. And if we wanted to, if we formed any of these, an imine, an oxime, or a hydrazone, and we wanted to go from those products back to our original starting aldehyde or ketone, we could just increase the concentration of water. So an excess of water, again, an acid catalyzed reaction, could push your equilibrium back this way and give you back your amine and your aldehyde or ketone. So let's take a look at an example of that. So here we have our imine."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And if we wanted to, if we formed any of these, an imine, an oxime, or a hydrazone, and we wanted to go from those products back to our original starting aldehyde or ketone, we could just increase the concentration of water. So an excess of water, again, an acid catalyzed reaction, could push your equilibrium back this way and give you back your amine and your aldehyde or ketone. So let's take a look at an example of that. So here we have our imine. And to it, we're adding aqueous acid, so an excess of water here. And so we're going to get back our original aldehyde or ketone. If we think about it, if we look at our product over here, we have a nitrogen double bond to a carbon, then we have an R group and an R group."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So here we have our imine. And to it, we're adding aqueous acid, so an excess of water here. And so we're going to get back our original aldehyde or ketone. If we think about it, if we look at our product over here, we have a nitrogen double bond to a carbon, then we have an R group and an R group. So nitrogen double bond to a carbon, R group, and R group. So it's R and R prime. So going backwards, we'd be starting with a ketone here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we think about it, if we look at our product over here, we have a nitrogen double bond to a carbon, then we have an R group and an R group. So nitrogen double bond to a carbon, R group, and R group. So it's R and R prime. So going backwards, we'd be starting with a ketone here. And so we would get cyclohexanone. So let's go ahead and draw cyclohexanone in as one of our products. So for our hydrolysis of an imine, so we would get cyclohexanone."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So going backwards, we'd be starting with a ketone here. And so we would get cyclohexanone. So let's go ahead and draw cyclohexanone in as one of our products. So for our hydrolysis of an imine, so we would get cyclohexanone. And then our other product, we could see right in here, this portion, we would get back our imine. And so we would get back our nitrogen. And then think about the mechanism."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So for our hydrolysis of an imine, so we would get cyclohexanone. And then our other product, we could see right in here, this portion, we would get back our imine. And so we would get back our nitrogen. And then think about the mechanism. We had lost two protons in that mechanism. So we get back our primary imine like that. And so this reaction works, of course, for oximes or hydrazones as well."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then think about the mechanism. We had lost two protons in that mechanism. So we get back our primary imine like that. And so this reaction works, of course, for oximes or hydrazones as well. So let's look into the formation of oximes and hydrazones here. So let's look at another reaction. Once again, we're starting with cyclohexanone."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so this reaction works, of course, for oximes or hydrazones as well. So let's look into the formation of oximes and hydrazones here. So let's look at another reaction. Once again, we're starting with cyclohexanone. But this time, we're dealing with hydroxyl amine. So this guy right here, once again, with an acid catalyst. And so here, our OH is our Y group."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we're starting with cyclohexanone. But this time, we're dealing with hydroxyl amine. So this guy right here, once again, with an acid catalyst. And so here, our OH is our Y group. So let's go back up here and identify that again. Let me use a different color. So our OH is our Y."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so here, our OH is our Y group. So let's go back up here and identify that again. Let me use a different color. So our OH is our Y. So right here. So it's this portion. And so we're going to put an OH on the nitrogen that's double bonded to our carbon."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So our OH is our Y. So right here. So it's this portion. And so we're going to put an OH on the nitrogen that's double bonded to our carbon. So let's go ahead and draw the products for our oxime here. So we're going to start with our ring. So we have our ring here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to put an OH on the nitrogen that's double bonded to our carbon. So let's go ahead and draw the products for our oxime here. So we're going to start with our ring. So we have our ring here. And then we're going to have our carbon double bonded to a nitrogen. And this time, instead of having a hydrogen or an alkyl group, we're going to have an OH. So we have an OH right here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have our ring here. And then we're going to have our carbon double bonded to a nitrogen. And this time, instead of having a hydrogen or an alkyl group, we're going to have an OH. So we have an OH right here. And let's go ahead and put in some lone pairs of electrons. And so this would be formation of an oxime. So this is our oxime product."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have an OH right here. And let's go ahead and put in some lone pairs of electrons. And so this would be formation of an oxime. So this is our oxime product. And oximes are more stable than imines. And the reason for that has to do with the fact that we have this oxygen here with a lone pair of electrons. So we could think about moving in this lone pair of electrons into here and pushing these electrons off onto this carbon."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is our oxime product. And oximes are more stable than imines. And the reason for that has to do with the fact that we have this oxygen here with a lone pair of electrons. So we could think about moving in this lone pair of electrons into here and pushing these electrons off onto this carbon. So if we were to draw a resonance structure, we could go ahead and show, once again, here's our ring. And then here we have our carbon bonded to our nitrogen. And then now there would be a double bond between this nitrogen and this oxygen."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we could think about moving in this lone pair of electrons into here and pushing these electrons off onto this carbon. So if we were to draw a resonance structure, we could go ahead and show, once again, here's our ring. And then here we have our carbon bonded to our nitrogen. And then now there would be a double bond between this nitrogen and this oxygen. And this oxygen would still have a lone pair of electrons on it, giving it a plus 1 formal charge. This nitrogen would have a lone pair of electrons on it. And then we move some electrons out onto this carbon."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then now there would be a double bond between this nitrogen and this oxygen. And this oxygen would still have a lone pair of electrons on it, giving it a plus 1 formal charge. This nitrogen would have a lone pair of electrons on it. And then we move some electrons out onto this carbon. So this carbon right here gets a negative 1 formal charge. So let's show some of those electrons. So these electrons right here on the oxygen moved into here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we move some electrons out onto this carbon. So this carbon right here gets a negative 1 formal charge. So let's show some of those electrons. So these electrons right here on the oxygen moved into here. And then we could show these electrons, kicking off onto this carbon to form a carbanion here for our resonance structure. And so we can delocalize some of those electrons. And you could think about this resonance structure over here with this little bit more negative charge, helping to stabilize this carbon, which we know is partially positive."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here on the oxygen moved into here. And then we could show these electrons, kicking off onto this carbon to form a carbanion here for our resonance structure. And so we can delocalize some of those electrons. And you could think about this resonance structure over here with this little bit more negative charge, helping to stabilize this carbon, which we know is partially positive. So it's partially positive over here on the left. And so that electron density from that resonance structure kind of helps to stabilize it a little bit. And so this is one way to look at why an oxime is more stable than an imine."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And you could think about this resonance structure over here with this little bit more negative charge, helping to stabilize this carbon, which we know is partially positive. So it's partially positive over here on the left. And so that electron density from that resonance structure kind of helps to stabilize it a little bit. And so this is one way to look at why an oxime is more stable than an imine. Let's do another reaction. So let's look at this ketone. So so far, we talked about using a symmetrical ketone with R groups that are the same on both sides."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so this is one way to look at why an oxime is more stable than an imine. Let's do another reaction. So let's look at this ketone. So so far, we talked about using a symmetrical ketone with R groups that are the same on both sides. This time, we're dealing with an unsymmetrical ketone. And so when we add our hydroxyl amine, we're going to get an oxime product. But we get two possible products here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So so far, we talked about using a symmetrical ketone with R groups that are the same on both sides. This time, we're dealing with an unsymmetrical ketone. And so when we add our hydroxyl amine, we're going to get an oxime product. But we get two possible products here. So let me go ahead and draw them out. So because we're dealing with an unsymmetrical ketone to start with, we could show lone pair of electrons and nitrogen on one side and the OH on the other side. So that's one of our possible products."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But we get two possible products here. So let me go ahead and draw them out. So because we're dealing with an unsymmetrical ketone to start with, we could show lone pair of electrons and nitrogen on one side and the OH on the other side. So that's one of our possible products. And then we could show a stereoisomer to that. We could show the lone pair on the left side this time and the OH on the right side. And so these are stereoisomers."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So that's one of our possible products. And then we could show a stereoisomer to that. We could show the lone pair on the left side this time and the OH on the right side. And so these are stereoisomers. So if we examine them a little bit more in detail, the OH is opposite side from this. If you're thinking about the double bond, the double bond here, you could think about the OH being on the same side as this. So stereoisomers are possible if you're not starting with a symmetrical ketone here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so these are stereoisomers. So if we examine them a little bit more in detail, the OH is opposite side from this. If you're thinking about the double bond, the double bond here, you could think about the OH being on the same side as this. So stereoisomers are possible if you're not starting with a symmetrical ketone here. And the same thing happens with imines. But oximes are more stable than imines. And so it's easier to isolate the oxime isomers once they are formed."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So stereoisomers are possible if you're not starting with a symmetrical ketone here. And the same thing happens with imines. But oximes are more stable than imines. And so it's easier to isolate the oxime isomers once they are formed. And so that's something to look out for on reactions. All right, let's do another reaction. And this one's a little bit different than before."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so it's easier to isolate the oxime isomers once they are formed. And so that's something to look out for on reactions. All right, let's do another reaction. And this one's a little bit different than before. So instead of having an OH here, we have an NH2 as our Y component here. So let's go back up and look at our generic reaction again. So an NH2 is now our Y right here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And this one's a little bit different than before. So instead of having an OH here, we have an NH2 as our Y component here. So let's go back up and look at our generic reaction again. So an NH2 is now our Y right here. And so when the Y is equal to an NH2, we're dealing with a hydrazone as our product. So once again, same mechanism. But let's think about what the product would look like here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So an NH2 is now our Y right here. And so when the Y is equal to an NH2, we're dealing with a hydrazone as our product. So once again, same mechanism. But let's think about what the product would look like here. So a reaction of hydrazine, so this guy right here is called hydrazine, with cyclohexanone is going to give us our ring. So a double bond to this nitrogen here. And then this time, our Y is going to be NH2."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But let's think about what the product would look like here. So a reaction of hydrazine, so this guy right here is called hydrazine, with cyclohexanone is going to give us our ring. So a double bond to this nitrogen here. And then this time, our Y is going to be NH2. So we go ahead and put NH2 coming off of here like that. And so this would be a hydrazone. So I'll put in my lone pairs of electrons right here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then this time, our Y is going to be NH2. So we go ahead and put NH2 coming off of here like that. And so this would be a hydrazone. So I'll put in my lone pairs of electrons right here. And so once again, hydrazones are also more stable than amines. And once again, this has to do with the fact that this nitrogen here has a lone pair of electrons. And so we could do the same thing that we did before."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'll put in my lone pairs of electrons right here. And so once again, hydrazones are also more stable than amines. And once again, this has to do with the fact that this nitrogen here has a lone pair of electrons. And so we could do the same thing that we did before. So this is a hydrazone product. So hydrazone like that. Let's do another example of a formation of a hydrazone."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we could do the same thing that we did before. So this is a hydrazone product. So hydrazone like that. Let's do another example of a formation of a hydrazone. So let's look at this guy. Little bit more complicated looking. But you can see that we have here our NH2."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another example of a formation of a hydrazone. So let's look at this guy. Little bit more complicated looking. But you can see that we have here our NH2. And then we have all of this. We have all of this up here as well. And so we can think about this part reacting with our carbonyl, so acid catalyzed reaction again."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But you can see that we have here our NH2. And then we have all of this. We have all of this up here as well. And so we can think about this part reacting with our carbonyl, so acid catalyzed reaction again. And so it looks a little bit intimidating. Let's get a little more space here. But really, we're just going to form a hydrazone."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we can think about this part reacting with our carbonyl, so acid catalyzed reaction again. And so it looks a little bit intimidating. Let's get a little more space here. But really, we're just going to form a hydrazone. So you could think about all this as being an R double prime group if you want to. So let's get a little bit more space. And let's draw the product."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But really, we're just going to form a hydrazone. So you could think about all this as being an R double prime group if you want to. So let's get a little bit more space. And let's draw the product. So we're going to have our ring like that. And then we're going to have it double bonded to our nitrogen. So you think about this nitrogen right here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And let's draw the product. So we're going to have our ring like that. And then we're going to have it double bonded to our nitrogen. So you think about this nitrogen right here. And then we can just go ahead and draw the rest of the molecule here. So since it's symmetrical, it doesn't really matter which side I put it. So I'll just put it on the left here."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So you think about this nitrogen right here. And then we can just go ahead and draw the rest of the molecule here. So since it's symmetrical, it doesn't really matter which side I put it. So I'll just put it on the left here. So nitrogen bonded to a hydrogen. And then we have our benzene ring here. So go ahead and put that in like that."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'll just put it on the left here. So nitrogen bonded to a hydrogen. And then we have our benzene ring here. So go ahead and put that in like that. And then we have these nitro groups coming off of our benzene ring. And so this is a famous historical reaction. So this compound that we are reacting our cyclohexanone with is 2,4-DNP, or 2,4-dinitrophenylhydrazine."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So go ahead and put that in like that. And then we have these nitro groups coming off of our benzene ring. And so this is a famous historical reaction. So this compound that we are reacting our cyclohexanone with is 2,4-DNP, or 2,4-dinitrophenylhydrazine. So this would be 2-nitro, 4-nitro, dinitrophenylhydrazine. So you can see this derivative here of hydrazine. And the reason why this is useful historically is this is a diagnostic test for an aldehyde or a ketone because it reacts with aldehydes or ketones usually to give an orangish, orange, or red precipitate."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this compound that we are reacting our cyclohexanone with is 2,4-DNP, or 2,4-dinitrophenylhydrazine. So this would be 2-nitro, 4-nitro, dinitrophenylhydrazine. So you can see this derivative here of hydrazine. And the reason why this is useful historically is this is a diagnostic test for an aldehyde or a ketone because it reacts with aldehydes or ketones usually to give an orangish, orange, or red precipitate. And that solid usually has a good melting point. So you can characterize aldehydes and ketones. And so before the advent of things like NMR, this helped with structure determination."}, {"video_title": "Formation of oximes and hydrazones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why this is useful historically is this is a diagnostic test for an aldehyde or a ketone because it reacts with aldehydes or ketones usually to give an orangish, orange, or red precipitate. And that solid usually has a good melting point. So you can characterize aldehydes and ketones. And so before the advent of things like NMR, this helped with structure determination. And there are all kinds of tables listing these hydrazone derivatives. And so this is more of a historical reaction than anything else. But it is kind of interesting to look at it here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We know oxygen is more electronegative, so this oxygen here gets a partial negative, and this hydrogen here gets a partial positive. Same thing for this alcohol over here. That sets up the opportunity for hydrogen bonding. So this partially positive hydrogen can be attracted to this partially negative oxygen here. And this attractive force, if this hydrogen is being attracted, that's going to weaken this oxygen-hydrogen bond. So hydrogen bonding weakens the oxygen-hydrogen bond, and we know if we're weakening the strength of a bond, that's like decreasing the force constant or decreasing the spring constant. And we saw in an earlier video, if you decrease k, you're going to decrease the frequency or decrease the wave number."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this partially positive hydrogen can be attracted to this partially negative oxygen here. And this attractive force, if this hydrogen is being attracted, that's going to weaken this oxygen-hydrogen bond. So hydrogen bonding weakens the oxygen-hydrogen bond, and we know if we're weakening the strength of a bond, that's like decreasing the force constant or decreasing the spring constant. And we saw in an earlier video, if you decrease k, you're going to decrease the frequency or decrease the wave number. And so the signal is going to change on your IR spectrum. And so at any moment in time, different alcohol molecules are going to have different amounts of hydrogen bonding. And so some molecules might have a little bit of hydrogen bonding, so k decreases a little bit, and the wave number decreases a little bit."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we saw in an earlier video, if you decrease k, you're going to decrease the frequency or decrease the wave number. And so the signal is going to change on your IR spectrum. And so at any moment in time, different alcohol molecules are going to have different amounts of hydrogen bonding. And so some molecules might have a little bit of hydrogen bonding, so k decreases a little bit, and the wave number decreases a little bit. But other molecules might have a lot of hydrogen bonding, and so we could decrease k even more, and so therefore we're going to decrease the wave number even more. And so you get a range of wave numbers, and since you get a range of wave numbers for the OH bond when hydrogen bonding is present, you get a very broad signal on your IR spectrum. So if we go over here in this region, so we're talking about the IR spectrum for 1-hexanol, this is the region for bonds to hydrogen."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so some molecules might have a little bit of hydrogen bonding, so k decreases a little bit, and the wave number decreases a little bit. But other molecules might have a lot of hydrogen bonding, and so we could decrease k even more, and so therefore we're going to decrease the wave number even more. And so you get a range of wave numbers, and since you get a range of wave numbers for the OH bond when hydrogen bonding is present, you get a very broad signal on your IR spectrum. So if we go over here in this region, so we're talking about the IR spectrum for 1-hexanol, this is the region for bonds to hydrogen. So we draw a line at 3,000 here, and we know that just below 3,000, we're talking, in this area, we're talking about a carbon-hydrogen bond stretch, where the carbon is sp3 hybridized. But this over here, this very broad signal right here, this is due to the OH. So let me go ahead and highlight that."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we go over here in this region, so we're talking about the IR spectrum for 1-hexanol, this is the region for bonds to hydrogen. So we draw a line at 3,000 here, and we know that just below 3,000, we're talking, in this area, we're talking about a carbon-hydrogen bond stretch, where the carbon is sp3 hybridized. But this over here, this very broad signal right here, this is due to the OH. So let me go ahead and highlight that. So this bond right here, this oxygen-hydrogen bond, gives us a very broad signal on our IR spectrum because of hydrogen bonding. So we get this very broad signal here because of the different wave numbers. And usually you're going to see this somewhere around 3,500 to 2,900."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight that. So this bond right here, this oxygen-hydrogen bond, gives us a very broad signal on our IR spectrum because of hydrogen bonding. So we get this very broad signal here because of the different wave numbers. And usually you're going to see this somewhere around 3,500 to 2,900. So if I find this is 31, 32, 33, 34, 35. So usually in this range, maybe even a little bit higher than that, you're going to find this very broad signal, in this case, the oxygen-hydrogen bond. And so you know immediately to think about the possibility of an alcohol functional group in your molecule."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And usually you're going to see this somewhere around 3,500 to 2,900. So if I find this is 31, 32, 33, 34, 35. So usually in this range, maybe even a little bit higher than that, you're going to find this very broad signal, in this case, the oxygen-hydrogen bond. And so you know immediately to think about the possibility of an alcohol functional group in your molecule. So also we can draw a line at 1,500 here, and this signal, actually, so somewhere around 1,100 wave numbers, this is actually the carbon-oxygen single bond. So let me go ahead and highlight that. So we have a carbon-oxygen single bond, and this is the single bond region in here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so you know immediately to think about the possibility of an alcohol functional group in your molecule. So also we can draw a line at 1,500 here, and this signal, actually, so somewhere around 1,100 wave numbers, this is actually the carbon-oxygen single bond. So let me go ahead and highlight that. So we have a carbon-oxygen single bond, and this is the single bond region in here. And that's this stretch right here. So not always going to be super useful to you, but it's just thinking about what we talked about in an earlier video. I think we calculated the approximate wave number for a carbon-oxygen single bond."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have a carbon-oxygen single bond, and this is the single bond region in here. And that's this stretch right here. So not always going to be super useful to you, but it's just thinking about what we talked about in an earlier video. I think we calculated the approximate wave number for a carbon-oxygen single bond. And so that's what the typical spectrum for an alcohol is going to look like. Look for that broad signal there. All right, let's compare this alcohol to another one here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I think we calculated the approximate wave number for a carbon-oxygen single bond. And so that's what the typical spectrum for an alcohol is going to look like. Look for that broad signal there. All right, let's compare this alcohol to another one here. So this molecule is a butylated hydroxy toluene, or BHT, and I drew two BHT molecules in there for a reason. And let's think about why. So you might think at first, oh, okay, I have another opportunity for hydrogen bonding."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's compare this alcohol to another one here. So this molecule is a butylated hydroxy toluene, or BHT, and I drew two BHT molecules in there for a reason. And let's think about why. So you might think at first, oh, okay, I have another opportunity for hydrogen bonding. So here's an opportunity for hydrogen bonding, and so I'm going to get a broad signal for this OH bond. So let me go ahead and highlight it here. So I might expect, since I have hydrogen bonding, that weakens this OH bond."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So you might think at first, oh, okay, I have another opportunity for hydrogen bonding. So here's an opportunity for hydrogen bonding, and so I'm going to get a broad signal for this OH bond. So let me go ahead and highlight it here. So I might expect, since I have hydrogen bonding, that weakens this OH bond. And so I might get another broad signal for my OH. But in this case, we have so much steric hindrance from these tert-butyl groups. So there's tons of steric hindrance here, and then we have these big ones over here, too."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I might expect, since I have hydrogen bonding, that weakens this OH bond. And so I might get another broad signal for my OH. But in this case, we have so much steric hindrance from these tert-butyl groups. So there's tons of steric hindrance here, and then we have these big ones over here, too. And that's going to prevent the hydrogen bonding from taking place. And so because of steric hindrance, these molecules can't get close enough to each other for hydrogen bonding to occur, so we don't get any hydrogen bonding. If we go over here to the IR spectrum for BHT, as usual, it helps to draw a line around 3,000 or so."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So there's tons of steric hindrance here, and then we have these big ones over here, too. And that's going to prevent the hydrogen bonding from taking place. And so because of steric hindrance, these molecules can't get close enough to each other for hydrogen bonding to occur, so we don't get any hydrogen bonding. If we go over here to the IR spectrum for BHT, as usual, it helps to draw a line around 3,000 or so. And then we don't see this broad signal. So this broad signal up here is missing down here, but what we do have is we have a sharp signal. So let me go ahead and highlight that here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we go over here to the IR spectrum for BHT, as usual, it helps to draw a line around 3,000 or so. And then we don't see this broad signal. So this broad signal up here is missing down here, but what we do have is we have a sharp signal. So let me go ahead and highlight that here. So we have a sharp signal right about, let's drop down and see where we are for our wave numbers. This would be 31, 32, 33, 34, 35, 36. And so somewhere around 3,600, we see this sharp signal, and this is that OH bond."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight that here. So we have a sharp signal right about, let's drop down and see where we are for our wave numbers. This would be 31, 32, 33, 34, 35, 36. And so somewhere around 3,600, we see this sharp signal, and this is that OH bond. So this is that OH bond, and we don't see a broad signal because we don't have hydrogen bonding to worry about. And so we don't see that broad shape. We see a sharp signal for the OH."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so somewhere around 3,600, we see this sharp signal, and this is that OH bond. So this is that OH bond, and we don't see a broad signal because we don't have hydrogen bonding to worry about. And so we don't see that broad shape. We see a sharp signal for the OH. So this allows you to think about where you would find this signal here. So if you have an oxygen-hydrogen bond with no hydrogen bonding, you expect to see it around 3,600. If you have an oxygen-hydrogen bond stretch and there is hydrogen bonding, look for this broad signal here over this large area."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We see a sharp signal for the OH. So this allows you to think about where you would find this signal here. So if you have an oxygen-hydrogen bond with no hydrogen bonding, you expect to see it around 3,600. If you have an oxygen-hydrogen bond stretch and there is hydrogen bonding, look for this broad signal here over this large area. All right, let's do one more. Let's do one more molecule where we're thinking about hydrogen bonding. And this time we're talking about a carboxylic acid."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you have an oxygen-hydrogen bond stretch and there is hydrogen bonding, look for this broad signal here over this large area. All right, let's do one more. Let's do one more molecule where we're thinking about hydrogen bonding. And this time we're talking about a carboxylic acid. So here's our carboxylic acid over here. And let's analyze the IR spectrum for this carboxylic acid. So we see this OH here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this time we're talking about a carboxylic acid. So here's our carboxylic acid over here. And let's analyze the IR spectrum for this carboxylic acid. So we see this OH here. We see this OH, and so we think to ourselves, ah, hydrogen bonding can occur. So where would that signal be? And before I said it would be somewhere around 3,500 to 2,900, somewhere in that range."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we see this OH here. We see this OH, and so we think to ourselves, ah, hydrogen bonding can occur. So where would that signal be? And before I said it would be somewhere around 3,500 to 2,900, somewhere in that range. And so if we look, we see an even broader signal here. So an even broader signal, a little bit even broader than the range we talked about before. And that's because carboxylic acids have more hydrogen bonding."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And before I said it would be somewhere around 3,500 to 2,900, somewhere in that range. And so if we look, we see an even broader signal here. So an even broader signal, a little bit even broader than the range we talked about before. And that's because carboxylic acids have more hydrogen bonding. So if I go down here, let me show you. Here's some hydrogen bonding for a carboxylic acid. So we have an opportunity for a hydrogen bond here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that's because carboxylic acids have more hydrogen bonding. So if I go down here, let me show you. Here's some hydrogen bonding for a carboxylic acid. So we have an opportunity for a hydrogen bond here. And we have an opportunity for hydrogen bonding here as well. So a large amount of hydrogen bonding makes the signal even broader when you're talking about the OH on a carboxylic acid. And so this very broad signal is talking about this oxygen-hydrogen bond stretch."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have an opportunity for a hydrogen bond here. And we have an opportunity for hydrogen bonding here as well. So a large amount of hydrogen bonding makes the signal even broader when you're talking about the OH on a carboxylic acid. And so this very broad signal is talking about this oxygen-hydrogen bond stretch. And once again, if we draw a line at 3,000, so we draw a line right here, we can see this little signal right in here. And that's actually the carbon-hydrogen stretch, where the carbon, I'm talking about sp3 hybridized carbon here. So the broad signal is often centered around 3,000, and so that partially obscures that carbon-hydrogen bond stretch that we've talked about before."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this very broad signal is talking about this oxygen-hydrogen bond stretch. And once again, if we draw a line at 3,000, so we draw a line right here, we can see this little signal right in here. And that's actually the carbon-hydrogen stretch, where the carbon, I'm talking about sp3 hybridized carbon here. So the broad signal is often centered around 3,000, and so that partially obscures that carbon-hydrogen bond stretch that we've talked about before. And so there's another hint that you're talking about a carboxylic acid. But of course the biggest hint is when you also see the very strong signal for the carbonyl occurring somewhere around 1,700. So let's go ahead and identify that."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the broad signal is often centered around 3,000, and so that partially obscures that carbon-hydrogen bond stretch that we've talked about before. And so there's another hint that you're talking about a carboxylic acid. But of course the biggest hint is when you also see the very strong signal for the carbonyl occurring somewhere around 1,700. So let's go ahead and identify that. So we draw a line right here, and then in the double bond region, we see this really intense signal right here, approximately around 1,700. So usually I think it's a tiny bit higher than that. But this is due to our carbon-oxygen double bond stretch."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and identify that. So we draw a line right here, and then in the double bond region, we see this really intense signal right here, approximately around 1,700. So usually I think it's a tiny bit higher than that. But this is due to our carbon-oxygen double bond stretch. So we're talking about our carbonyl here. So if you see this really broad signal, which tells you OH, and then this really strong signal, which tells you carbonyl, put an OH and a carbonyl together, and you get a carboxylic acid. And so it's pretty easy to identify a carboxylic acid using an IR spectrum here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But this is due to our carbon-oxygen double bond stretch. So we're talking about our carbonyl here. So if you see this really broad signal, which tells you OH, and then this really strong signal, which tells you carbonyl, put an OH and a carbonyl together, and you get a carboxylic acid. And so it's pretty easy to identify a carboxylic acid using an IR spectrum here. And just a quick note about hydrogen bonding. We talked about hydrogen bonding weakening a carbon-oxygen bond above. A similar thing happens here."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so it's pretty easy to identify a carboxylic acid using an IR spectrum here. And just a quick note about hydrogen bonding. We talked about hydrogen bonding weakening a carbon-oxygen bond above. A similar thing happens here. So let me go ahead and highlight that. So if we have hydrogen bonding right in here, that's going to weaken our carbonyl. So it's going to decrease the double bond character a little bit."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "A similar thing happens here. So let me go ahead and highlight that. So if we have hydrogen bonding right in here, that's going to weaken our carbonyl. So it's going to decrease the double bond character a little bit. And so that's actually going to change where we find the signal. So it actually changes the signal. If it's weakening the bond a little bit, so you're going to decrease the wave number."}, {"video_title": "Signal characteristics - shape Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to decrease the double bond character a little bit. And so that's actually going to change where we find the signal. So it actually changes the signal. If it's weakening the bond a little bit, so you're going to decrease the wave number. And so this carbonyl is a tiny bit lower in terms of wave number where you find the signal than what you would expect. And I'll briefly mention that in a later video when we talk about carbonyl chemistry and more about IR spectrum. So for the shape of the signal, remember broad."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And you'll hear many different ways to pronounce this functional group. So I usually say amide, but I've heard amids or amide, or however you want to say it. To me it's not really that big of a deal. It seems to be more where you're from is in terms of how you pronounce it. And so if our goal is to name this two carbon amide over here on the right, let's start with the two carbon carboxylic acid over here on the left. So we have two carbons. And if we think about the IUPAC name, so two carbons, this would be ethanoic acid."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "It seems to be more where you're from is in terms of how you pronounce it. And so if our goal is to name this two carbon amide over here on the right, let's start with the two carbon carboxylic acid over here on the left. So we have two carbons. And if we think about the IUPAC name, so two carbons, this would be ethanoic acid. So ethanoic acid. To name the corresponding amide, we're going to drop the ending. So we're going to drop the oic acid and add on amide."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And if we think about the IUPAC name, so two carbons, this would be ethanoic acid. So ethanoic acid. To name the corresponding amide, we're going to drop the ending. So we're going to drop the oic acid and add on amide. So let's go ahead and write that. So we would drop the oic acid and add on amide. So ethanamide or ethanamid is what you could call this molecule over here on the right."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to drop the oic acid and add on amide. So let's go ahead and write that. So we would drop the oic acid and add on amide. So ethanamide or ethanamid is what you could call this molecule over here on the right. Of course most people don't call this ethanoic acid. Most people call this acetic acid. So if we think about naming this derived from acetic acid, let's go ahead and write that down."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So ethanamide or ethanamid is what you could call this molecule over here on the right. Of course most people don't call this ethanoic acid. Most people call this acetic acid. So if we think about naming this derived from acetic acid, let's go ahead and write that down. So acetic acid, like that. If we drop the ending this time, we don't have an extra o, right? So we just have this ic."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about naming this derived from acetic acid, let's go ahead and write that down. So acetic acid, like that. If we drop the ending this time, we don't have an extra o, right? So we just have this ic. So we're going to drop the ic acid and add amide onto that. So we would make acetamide, right? If we think of it over here, or acetamid, however you want to say it."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we just have this ic. So we're going to drop the ic acid and add amide onto that. So we would make acetamide, right? If we think of it over here, or acetamid, however you want to say it. So once again, think about dropping the ending and adding amide on. So acetamide is an example of a primary amide. So this is a primary amide."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "If we think of it over here, or acetamid, however you want to say it. So once again, think about dropping the ending and adding amide on. So acetamide is an example of a primary amide. So this is a primary amide. And the way I like to think about this is this nitrogen here is bonded to only one carbon. And so that makes the carbon right here. So that makes this primary."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So this is a primary amide. And the way I like to think about this is this nitrogen here is bonded to only one carbon. And so that makes the carbon right here. So that makes this primary. If we want to think about naming this one down here, this time the nitrogen is bonded to two carbons, right? So the carbon in our carbonyl, and also this carbon over here. And so this one is an example of a secondary amide."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So that makes this primary. If we want to think about naming this one down here, this time the nitrogen is bonded to two carbons, right? So the carbon in our carbonyl, and also this carbon over here. And so this one is an example of a secondary amide. So a nitrogen bonded to two carbons. If we think about the parent carboxylic acids over here, one, two, three carboxylic acids, three carbons in this carboxylic acid. So the IUPAC name would be propanoic acid."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so this one is an example of a secondary amide. So a nitrogen bonded to two carbons. If we think about the parent carboxylic acids over here, one, two, three carboxylic acids, three carbons in this carboxylic acid. So the IUPAC name would be propanoic acid. So let's go ahead and write out propanoic acid. We're going to use this as the base for the amide on the right. So once again, we're going to drop the oic acid ending, just like we did up here."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So the IUPAC name would be propanoic acid. So let's go ahead and write out propanoic acid. We're going to use this as the base for the amide on the right. So once again, we're going to drop the oic acid ending, just like we did up here. So we drop the oic acid ending and add on amides. So we have propanamide. So let's go ahead and write that."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we're going to drop the oic acid ending, just like we did up here. So we drop the oic acid ending and add on amides. So we have propanamide. So let's go ahead and write that. So over here on the right, we would have propanamide. Propanamide or propanamid, like that. And then that takes care of, if you will, this portion of the molecule."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that. So over here on the right, we would have propanamide. Propanamide or propanamid, like that. And then that takes care of, if you will, this portion of the molecule. So now we still have this methyl group right here to worry about. And that methyl group is coming off of this nitrogen here. So we have a methyl group coming off of this nitrogen."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And then that takes care of, if you will, this portion of the molecule. So now we still have this methyl group right here to worry about. And that methyl group is coming off of this nitrogen here. So we have a methyl group coming off of this nitrogen. And the way we show that is to put an N here and then a methyl. So N-methylpropanamide tells us that we have a methyl group coming off of the nitrogen in our amide. And so that's how it's done."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we have a methyl group coming off of this nitrogen. And the way we show that is to put an N here and then a methyl. So N-methylpropanamide tells us that we have a methyl group coming off of the nitrogen in our amide. And so that's how it's done. All right, let's look at some more examples. So two more examples for amides right here. And let's see how to name them."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so that's how it's done. All right, let's look at some more examples. So two more examples for amides right here. And let's see how to name them. So let's start with the one on the left. So for this one, if I think about the carboxylic acid, it'd be one, it'd be a one, two, three, four. So four-carbon carboxylic acid, the IUPAC name is butanoic acid."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And let's see how to name them. So let's start with the one on the left. So for this one, if I think about the carboxylic acid, it'd be one, it'd be a one, two, three, four. So four-carbon carboxylic acid, the IUPAC name is butanoic acid. So we drop the oic acid part and add on amides. So it'd be butanamide. So let's go ahead and write that in."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So four-carbon carboxylic acid, the IUPAC name is butanoic acid. So we drop the oic acid part and add on amides. So it'd be butanamide. So let's go ahead and write that in. So this would be butanamide. And then we have to think about what else is attached to this nitrogen here. So the nitrogen in our amide right here has a methyl group here and another methyl group here."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that in. So this would be butanamide. And then we have to think about what else is attached to this nitrogen here. So the nitrogen in our amide right here has a methyl group here and another methyl group here. So two methyl groups. So this time we're gonna write N-N-dimethyl. So we have two methyl groups."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So the nitrogen in our amide right here has a methyl group here and another methyl group here. So two methyl groups. So this time we're gonna write N-N-dimethyl. So we have two methyl groups. Each one of those methyl groups is attached to that nitrogen. So N-N-dimethylbutanamide would be the name for this molecule. In terms of classifying it, so in terms of classifying this amide here, this would actually be a tertiary amide because this nitrogen is attached to one, two, three other carbons, and so it's a tertiary amide."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we have two methyl groups. Each one of those methyl groups is attached to that nitrogen. So N-N-dimethylbutanamide would be the name for this molecule. In terms of classifying it, so in terms of classifying this amide here, this would actually be a tertiary amide because this nitrogen is attached to one, two, three other carbons, and so it's a tertiary amide. Let's look at naming one more. So this one over here on the right. And if I think about the carboxylic acid from which this was derived, so you have to use your imagination a little bit."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "In terms of classifying it, so in terms of classifying this amide here, this would actually be a tertiary amide because this nitrogen is attached to one, two, three other carbons, and so it's a tertiary amide. Let's look at naming one more. So this one over here on the right. And if I think about the carboxylic acid from which this was derived, so you have to use your imagination a little bit. So if I think about this portion, instead of the nitrogen here, if I had an OH, that would be benzoic acid. So I'm just gonna go ahead and write that right here. So benzoic acid."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And if I think about the carboxylic acid from which this was derived, so you have to use your imagination a little bit. So if I think about this portion, instead of the nitrogen here, if I had an OH, that would be benzoic acid. So I'm just gonna go ahead and write that right here. So benzoic acid. We know we're gonna drop the ending, we're gonna drop the oic acid part and add amide. So it'd be benzamide. So let's go ahead and write that."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So benzoic acid. We know we're gonna drop the ending, we're gonna drop the oic acid part and add amide. So it'd be benzamide. So let's go ahead and write that. So as our parent name here, so we would have benzamide here. And then let's think about what else we have. We have a nitrogen, we have a methyl group, one methyl group bonded to that nitrogen."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that. So as our parent name here, so we would have benzamide here. And then let's think about what else we have. We have a nitrogen, we have a methyl group, one methyl group bonded to that nitrogen. And then we have another methyl group over here on the ring. So let's go ahead and number the ring and see where our other methyl group is. So our other methyl group, if we number our ring, this would be carbon one, if we think about this as being like benzoic acid."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "We have a nitrogen, we have a methyl group, one methyl group bonded to that nitrogen. And then we have another methyl group over here on the ring. So let's go ahead and number the ring and see where our other methyl group is. So our other methyl group, if we number our ring, this would be carbon one, if we think about this as being like benzoic acid. So if we're numbering our ring, this would be carbon one. And then carbon two, carbon three, and carbon four. And so we have a methyl group at carbon four, and we also have a methyl group coming off of our nitrogen."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So our other methyl group, if we number our ring, this would be carbon one, if we think about this as being like benzoic acid. So if we're numbering our ring, this would be carbon one. And then carbon two, carbon three, and carbon four. And so we have a methyl group at carbon four, and we also have a methyl group coming off of our nitrogen. So this time, we're going to write, let me go ahead and write right here, this would be N4-dimethyl. So we have N4-dimethylbenzamide as our name for this molecule. So notice the difference."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so we have a methyl group at carbon four, and we also have a methyl group coming off of our nitrogen. So this time, we're going to write, let me go ahead and write right here, this would be N4-dimethyl. So we have N4-dimethylbenzamide as our name for this molecule. So notice the difference. So here we have a methyl group on our nitrogen, right here, and we have a methyl group coming off of carbon four, so that's where our dimethyl comes from. For this one, we have N and another N, because both of our methyl groups are coming off of the nitrogen in the example on the left. And so it's just kind of useful to contrast these two molecules when we're thinking about nomenclature."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So notice the difference. So here we have a methyl group on our nitrogen, right here, and we have a methyl group coming off of carbon four, so that's where our dimethyl comes from. For this one, we have N and another N, because both of our methyl groups are coming off of the nitrogen in the example on the left. And so it's just kind of useful to contrast these two molecules when we're thinking about nomenclature. In terms of physical properties of amides, let's look at a little diagram here showing acetamide. So we have our amide right here. And physical properties in terms of what state of matter is this?"}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And so it's just kind of useful to contrast these two molecules when we're thinking about nomenclature. In terms of physical properties of amides, let's look at a little diagram here showing acetamide. So we have our amide right here. And physical properties in terms of what state of matter is this? Acetamide is actually a solid at room temperature and pressure. And so this is a solid, it has a lot of hydrogen bonding. So we could think about some hydrogen bonding here, we could think about some hydrogen bonding here."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "And physical properties in terms of what state of matter is this? Acetamide is actually a solid at room temperature and pressure. And so this is a solid, it has a lot of hydrogen bonding. So we could think about some hydrogen bonding here, we could think about some hydrogen bonding here. There are lots of opportunities for hydrogen bonding. So partial negative oxygen, partial positive hydrogen, bonded to this nitrogen, which is withdrawing some electron density from this hydrogen. So lots of opportunities for hydrogen bonding."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So we could think about some hydrogen bonding here, we could think about some hydrogen bonding here. There are lots of opportunities for hydrogen bonding. So partial negative oxygen, partial positive hydrogen, bonded to this nitrogen, which is withdrawing some electron density from this hydrogen. So lots of opportunities for hydrogen bonding. Makes acetamide a solid at room temperature, and so its melting point turns out to be approximately 82 degrees Celsius. So that's its melting point. Its boiling point is actually much higher."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So lots of opportunities for hydrogen bonding. Makes acetamide a solid at room temperature, and so its melting point turns out to be approximately 82 degrees Celsius. So that's its melting point. Its boiling point is actually much higher. So somewhere around 221 degrees, so higher. So this is actually the melting point, and the boiling point being much higher due to the very strong intermolecular forces that are present between amide molecules. In terms of solubility of amides in water, small amides are definitely soluble."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "Its boiling point is actually much higher. So somewhere around 221 degrees, so higher. So this is actually the melting point, and the boiling point being much higher due to the very strong intermolecular forces that are present between amide molecules. In terms of solubility of amides in water, small amides are definitely soluble. So if I think about water coming along, let's go ahead and draw water in here. Water is a polar molecule, so partial negative, partial positive. So like dissolves like."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "In terms of solubility of amides in water, small amides are definitely soluble. So if I think about water coming along, let's go ahead and draw water in here. Water is a polar molecule, so partial negative, partial positive. So like dissolves like. This polar amide, this will dissolve in polar water. You could think about some hydrogen bonding going on right here. So like dissolves like."}, {"video_title": "Nomenclature and properties of amides Organic chemistry Khan Academy.mp3", "Sentence": "So like dissolves like. This polar amide, this will dissolve in polar water. You could think about some hydrogen bonding going on right here. So like dissolves like. And there's also a resonance property to amides. So I won't get too much into resonance, but there's a lone pair of electrons on this nitrogen which can move into here, pushing these electrons off onto the oxygen, giving you a plus one formal charge on your nitrogen, and a negative one formal charge on your oxygen. And so amides can be polar as well, and so this also can explain the solubility of acetamide in water here."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we're gonna look at the stereochemistry of the dienophile, but first a quick review of the Diels-Alder reaction. On the left we have our diene, on the right is our dienophile. We know that our Diels-Alder reaction involves a concerted movement of six pi electrons. So these pi electrons move into here to form a bond, these pi electrons move into here to form a bond, and these pi electrons move down. So that gives us our product. So the electrons in red move into here to form this bond, the electrons, the pi electrons in blue form this bond, and finally the pi electrons in magenta are these electrons in our product. But let's think about the stereochemistry of the dienophile."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons move into here to form a bond, these pi electrons move into here to form a bond, and these pi electrons move down. So that gives us our product. So the electrons in red move into here to form this bond, the electrons, the pi electrons in blue form this bond, and finally the pi electrons in magenta are these electrons in our product. But let's think about the stereochemistry of the dienophile. So here is our dienophile, and we have our two R groups cis to each other, so they're on the same side. So what does that do for the product in terms of the stereochemistry? How do those two R groups end up?"}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "But let's think about the stereochemistry of the dienophile. So here is our dienophile, and we have our two R groups cis to each other, so they're on the same side. So what does that do for the product in terms of the stereochemistry? How do those two R groups end up? Well down here, let's look at the picture of the diene and the dienophile. So up top here is the diene, which you could think about as being one plane, and down here is the dienophile, which would be another plane. So the two R groups I made red right here, and the planes approach each other."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "How do those two R groups end up? Well down here, let's look at the picture of the diene and the dienophile. So up top here is the diene, which you could think about as being one plane, and down here is the dienophile, which would be another plane. So the two R groups I made red right here, and the planes approach each other. So the diene and the dienophile approach each other, and we know a bond forms between this carbon and this carbon, so think about a bond forming here, and a bond forms between this carbon and this carbon, so a bond forms here. And notice what happens when we go to our product over here on the right, when we're thinking about our dienophile. This carbon goes from being sp2 hybridized to being sp3 hybridized, and the same thing with this carbon, from sp2 to sp3 hybridized."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So the two R groups I made red right here, and the planes approach each other. So the diene and the dienophile approach each other, and we know a bond forms between this carbon and this carbon, so think about a bond forming here, and a bond forms between this carbon and this carbon, so a bond forms here. And notice what happens when we go to our product over here on the right, when we're thinking about our dienophile. This carbon goes from being sp2 hybridized to being sp3 hybridized, and the same thing with this carbon, from sp2 to sp3 hybridized. So we need to think about those R groups. Since we have a concerted movement of electrons, these two R groups end up on the same side, and if you're staring this way, down at your cyclohexene ring, these hydrogens will be going away from you in space, so these R groups are actually coming out at you in space. So when we're drawing in our stereochemistry, we draw in our cyclohexene ring, and we put in our R groups coming out at us in space."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "This carbon goes from being sp2 hybridized to being sp3 hybridized, and the same thing with this carbon, from sp2 to sp3 hybridized. So we need to think about those R groups. Since we have a concerted movement of electrons, these two R groups end up on the same side, and if you're staring this way, down at your cyclohexene ring, these hydrogens will be going away from you in space, so these R groups are actually coming out at you in space. So when we're drawing in our stereochemistry, we draw in our cyclohexene ring, and we put in our R groups coming out at us in space. So let's go back to this drawing over here on the left, and let's think about the stereochemistry. So I like to draw in the hydrogens over here sometimes, and think about the groups on the left and the right of this line. So if I'm drawing the dienophile approaching the diene in this fashion, the two R groups are on the right side, and those end up as both wedges coming out at us in space."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So when we're drawing in our stereochemistry, we draw in our cyclohexene ring, and we put in our R groups coming out at us in space. So let's go back to this drawing over here on the left, and let's think about the stereochemistry. So I like to draw in the hydrogens over here sometimes, and think about the groups on the left and the right of this line. So if I'm drawing the dienophile approaching the diene in this fashion, the two R groups are on the right side, and those end up as both wedges coming out at us in space. So the stuff on the right side of the line ends up as a wedge, and for the cis, right, we have a cis alkene on the left, and those two R groups end up cis on our ring as well. The two groups are on the same side of the ring. The stuff on the left side of the line that I drew, in this case two hydrogens, that ends up going down in our product."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm drawing the dienophile approaching the diene in this fashion, the two R groups are on the right side, and those end up as both wedges coming out at us in space. So the stuff on the right side of the line ends up as a wedge, and for the cis, right, we have a cis alkene on the left, and those two R groups end up cis on our ring as well. The two groups are on the same side of the ring. The stuff on the left side of the line that I drew, in this case two hydrogens, that ends up going down in our product. So here are those two hydrogens. Those would be dashes if we put them in on our product. What happens if we show our dienophile approaching our diene in a different way?"}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "The stuff on the left side of the line that I drew, in this case two hydrogens, that ends up going down in our product. So here are those two hydrogens. Those would be dashes if we put them in on our product. What happens if we show our dienophile approaching our diene in a different way? In this case, if we think about our double bond, the two R groups are on the left side of the double bond. So let me go ahead and make those red. So these two R groups on the model, you can see the R groups are right here."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "What happens if we show our dienophile approaching our diene in a different way? In this case, if we think about our double bond, the two R groups are on the left side of the double bond. So let me go ahead and make those red. So these two R groups on the model, you can see the R groups are right here. And on the right side of our double bond, in terms of how I've drawn it, we have two hydrogens, and here are the hydrogens on the model. We know that a bond forms between this carbon and this one, and a bond forms between this carbon and this one. So when we look at the product, here are the two bonds that I just pointed out, and we can see that our two R groups are on the same side, and if we're staring down in this direction at our cyclohexene ring, let me go ahead and draw in our cyclohexene ring like that, these two R groups are going away from us in space."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So these two R groups on the model, you can see the R groups are right here. And on the right side of our double bond, in terms of how I've drawn it, we have two hydrogens, and here are the hydrogens on the model. We know that a bond forms between this carbon and this one, and a bond forms between this carbon and this one. So when we look at the product, here are the two bonds that I just pointed out, and we can see that our two R groups are on the same side, and if we're staring down in this direction at our cyclohexene ring, let me go ahead and draw in our cyclohexene ring like that, these two R groups are going away from us in space. So those two R groups should go on dashes. So let me go ahead and put these dashes in. And so the stuff on the left side of our double bond, let me go ahead and go back to the dienophile over here, stuff on the left side, these two R groups, would end up going down."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So when we look at the product, here are the two bonds that I just pointed out, and we can see that our two R groups are on the same side, and if we're staring down in this direction at our cyclohexene ring, let me go ahead and draw in our cyclohexene ring like that, these two R groups are going away from us in space. So those two R groups should go on dashes. So let me go ahead and put these dashes in. And so the stuff on the left side of our double bond, let me go ahead and go back to the dienophile over here, stuff on the left side, these two R groups, would end up going down. So I'm just very consistent in how I draw my dienophile and thinking about my product. And I think a systematic way helps you when you're doing Diels-Alder reactions. The stuff on the right side, these hydrogens, these end up, so they are up relative to our plane if we're staring down in this direction."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And so the stuff on the left side of our double bond, let me go ahead and go back to the dienophile over here, stuff on the left side, these two R groups, would end up going down. So I'm just very consistent in how I draw my dienophile and thinking about my product. And I think a systematic way helps you when you're doing Diels-Alder reactions. The stuff on the right side, these hydrogens, these end up, so they are up relative to our plane if we're staring down in this direction. So we'd put those hydrogens as wedges if we put them in for the product. Now if these R groups are the same, it doesn't matter how you would represent the product. So let's look at an example of what I mean."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "The stuff on the right side, these hydrogens, these end up, so they are up relative to our plane if we're staring down in this direction. So we'd put those hydrogens as wedges if we put them in for the product. Now if these R groups are the same, it doesn't matter how you would represent the product. So let's look at an example of what I mean. So let's draw the product for this Diels-Alder reaction. We know these electrons move into here, so we form a bond between these two carbons. These electrons move into here, and these electrons move down."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at an example of what I mean. So let's draw the product for this Diels-Alder reaction. We know these electrons move into here, so we form a bond between these two carbons. These electrons move into here, and these electrons move down. So we get a cyclohexene ring, so let me sketch that in first. And then we think about the stereochemistry of our dienophile. We know that the stereochemistry is cis, so we need to have those two groups on the same side of the ring."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "These electrons move into here, and these electrons move down. So we get a cyclohexene ring, so let me sketch that in first. And then we think about the stereochemistry of our dienophile. We know that the stereochemistry is cis, so we need to have those two groups on the same side of the ring. We could use either wedges or dashes here, so I'm just gonna use two wedges. So let me put those wedges in, and then we draw in an ester up here, and we draw the same ester in down there. And there's our product of our Diels-Alder reaction."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "We know that the stereochemistry is cis, so we need to have those two groups on the same side of the ring. We could use either wedges or dashes here, so I'm just gonna use two wedges. So let me put those wedges in, and then we draw in an ester up here, and we draw the same ester in down there. And there's our product of our Diels-Alder reaction. What if we have our two R groups trans to each other? So let me go ahead and draw a line here. We look at the R group on the left."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And there's our product of our Diels-Alder reaction. What if we have our two R groups trans to each other? So let me go ahead and draw a line here. We look at the R group on the left. That's the one going towards the diene. So on the model, that's this R group, the one that's pointing towards the diene. The one to the right of the line, this R group is going away from the diene, so that's this one on the model."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "We look at the R group on the left. That's the one going towards the diene. So on the model, that's this R group, the one that's pointing towards the diene. The one to the right of the line, this R group is going away from the diene, so that's this one on the model. We know that a bond forms between these two carbons, and a bond forms between these two carbons, so we get the product on the right. So here are the bonds that I just pointed out. And if we're staring down at our cyclohexene ring, let me draw that in, this R group is going away from us in space."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "The one to the right of the line, this R group is going away from the diene, so that's this one on the model. We know that a bond forms between these two carbons, and a bond forms between these two carbons, so we get the product on the right. So here are the bonds that I just pointed out. And if we're staring down at our cyclohexene ring, let me draw that in, this R group is going away from us in space. So if we work backwards, that's this R group. That's the one on the left here. So that should be a dash for our product."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And if we're staring down at our cyclohexene ring, let me draw that in, this R group is going away from us in space. So if we work backwards, that's this R group. That's the one on the left here. So that should be a dash for our product. So we put that R group in as a dash. This R group is coming out at us in space, and that was this one, the one going away from our diene. So that's the one on the right."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So that should be a dash for our product. So we put that R group in as a dash. This R group is coming out at us in space, and that was this one, the one going away from our diene. So that's the one on the right. If that's coming out at us in space, that should be a wedge. So let me put that R group in. What if our dienophile approaches our diene in a different way?"}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So that's the one on the right. If that's coming out at us in space, that should be a wedge. So let me put that R group in. What if our dienophile approaches our diene in a different way? So now when we look at this line here, this is the R group that's going towards the diene, the one on the left. So that's this one back here. And this R group is going away from the diene."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "What if our dienophile approaches our diene in a different way? So now when we look at this line here, this is the R group that's going towards the diene, the one on the left. So that's this one back here. And this R group is going away from the diene. So that must be this R group. So if we draw in our bonds that we know are forming, a bond between those two carbons and a bond between these two carbons, we look at our product on the right with those bonds in there, and also the double bond, which I'm just not really focusing on at the moment. And we look down at our cyclohexene ring."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And this R group is going away from the diene. So that must be this R group. So if we draw in our bonds that we know are forming, a bond between those two carbons and a bond between these two carbons, we look at our product on the right with those bonds in there, and also the double bond, which I'm just not really focusing on at the moment. And we look down at our cyclohexene ring. So I draw that in. Now when we look down, let's look at this R group first. That R group is going away from us in space, and that was this one."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And we look down at our cyclohexene ring. So I draw that in. Now when we look down, let's look at this R group first. That R group is going away from us in space, and that was this one. That's the one going towards the diene. So that's this one right here. So to the left, the R group on the left is gonna get a dash."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "That R group is going away from us in space, and that was this one. That's the one going towards the diene. So that's this one right here. So to the left, the R group on the left is gonna get a dash. So that's this one. So I put in an R group on a dash. And then this R group is coming out at us."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So to the left, the R group on the left is gonna get a dash. So that's this one. So I put in an R group on a dash. And then this R group is coming out at us. It's going up, and that was this one, the one going away from the diene. And that's this R group. So if it's on the right, it gets a wedge."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And then this R group is coming out at us. It's going up, and that was this one, the one going away from the diene. And that's this R group. So if it's on the right, it gets a wedge. So this should be a wedge. Notice that if we have R groups that are the same, these two are enantiomers of each other. So we would get a pair of enantiomers for our reaction."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So if it's on the right, it gets a wedge. So this should be a wedge. Notice that if we have R groups that are the same, these two are enantiomers of each other. So we would get a pair of enantiomers for our reaction. Let's say we were given this Diels-Alder reaction on a test. On the left is our diene, and on the right is our dienophile. And we know a bond forms between these two carbons and between these two carbons."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So we would get a pair of enantiomers for our reaction. Let's say we were given this Diels-Alder reaction on a test. On the left is our diene, and on the right is our dienophile. And we know a bond forms between these two carbons and between these two carbons. So we could start by drawing in a cyclohexene ring. So let me put that in. Next, we think about the stereochemistry of our dienophile."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "And we know a bond forms between these two carbons and between these two carbons. So we could start by drawing in a cyclohexene ring. So let me put that in. Next, we think about the stereochemistry of our dienophile. And if we draw a line like that, the stuff to the right of the line ends up as a wedge. So this ester is going to be on a wedge. So let me put that."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "Next, we think about the stereochemistry of our dienophile. And if we draw a line like that, the stuff to the right of the line ends up as a wedge. So this ester is going to be on a wedge. So let me put that. Ester coming out at us in space. And the stuff to the left of our line should be on a dash. It's going away from us in space."}, {"video_title": "Diels-Alder stereochemistry of dienophile Organic chemistry Khan Academy.mp3", "Sentence": "So let me put that. Ester coming out at us in space. And the stuff to the left of our line should be on a dash. It's going away from us in space. So we put that ester in here. And we know from the pictures that we just saw, we're also going to get the enantiomer. So we get this compound plus its enantiomer as our products."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is just kind of the current thinking, the leading thinking on why plates are actually moving. Although we haven't seen the definitive evidence yet. And it's probably a combination of a bunch of things. Now before we even talk about plates, let's just talk about convection. And you might already be familiar with the term, but just in case you're not, let's do a little bit of review of convection. So let's say I have a pot over here. So that is my pot."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now before we even talk about plates, let's just talk about convection. And you might already be familiar with the term, but just in case you're not, let's do a little bit of review of convection. So let's say I have a pot over here. So that is my pot. And it contains some water. So I have water in my pot. And let's say I only heat one end of the pot."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that is my pot. And it contains some water. So I have water in my pot. And let's say I only heat one end of the pot. So I put a flame right over at that end of the pot. So what's going to happen? Well the water that's right over the flame is going to be warmed up more than any of the other water."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say I only heat one end of the pot. So I put a flame right over at that end of the pot. So what's going to happen? Well the water that's right over the flame is going to be warmed up more than any of the other water. So this water is going to get warm. But when it gets warm, it also becomes less dense. When you have a fluid, if you warm it up, the molecules are vibrating more."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well the water that's right over the flame is going to be warmed up more than any of the other water. So this water is going to get warm. But when it gets warm, it also becomes less dense. When you have a fluid, if you warm it up, the molecules are vibrating more. They have more kinetic energy. They're going to bounce further distances away from each other. They will become less dense."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When you have a fluid, if you warm it up, the molecules are vibrating more. They have more kinetic energy. They're going to bounce further distances away from each other. They will become less dense. And if you have something that's less dense and it's surrounded by things that are more dense, and we're dealing in kind of a fluid state right here, that warm, less dense water is going to move upwards. Well when it moves upwards, something has to replace it. So you're going to have cooler water from this side of the container kind of replacing where that water was."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They will become less dense. And if you have something that's less dense and it's surrounded by things that are more dense, and we're dealing in kind of a fluid state right here, that warm, less dense water is going to move upwards. Well when it moves upwards, something has to replace it. So you're going to have cooler water from this side of the container kind of replacing where that water was. Now this water, as it rises, what's going to happen to it? Well it's going to cool down. It's going to get further from the flame."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you're going to have cooler water from this side of the container kind of replacing where that water was. Now this water, as it rises, what's going to happen to it? Well it's going to cool down. It's going to get further from the flame. It's going to mix with maybe some of the other water, or transfer some of its kinetic energy to the neighboring water. So it'll cool down. But once it cools down, what's it going to want to do?"}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's going to get further from the flame. It's going to mix with maybe some of the other water, or transfer some of its kinetic energy to the neighboring water. So it'll cool down. But once it cools down, what's it going to want to do? Remember, in general, the closer you are to the flame. So the water closer to the flame in general is going to be warmer. So all of this stuff is going to be warmer."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But once it cools down, what's it going to want to do? Remember, in general, the closer you are to the flame. So the water closer to the flame in general is going to be warmer. So all of this stuff is going to be warmer. And all of this stuff up here, like the coldest water, is always going to be furthest from the flame. And so the coldest water is going to be over here. But remember, the coldest water is also the densest water."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So all of this stuff is going to be warmer. And all of this stuff up here, like the coldest water, is always going to be furthest from the flame. And so the coldest water is going to be over here. But remember, the coldest water is also the densest water. So this water over here is dense. And so it will sink. It's denser than the water around it."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But remember, the coldest water is also the densest water. So this water over here is dense. And so it will sink. It's denser than the water around it. And it also helps replace the water that's going here to get warmed up again. And so what you do is you have this cycle here. Warm water rises, moves over to the right down here, and then goes back down as it cools down."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's denser than the water around it. And it also helps replace the water that's going here to get warmed up again. And so what you do is you have this cycle here. Warm water rises, moves over to the right down here, and then goes back down as it cools down. And it's dense, and then it gets warmed up again. And so this process, essentially what it's doing is it's transferring the heat. It's allowing the heat to be transferred from this one spot throughout the fluid."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Warm water rises, moves over to the right down here, and then goes back down as it cools down. And it's dense, and then it gets warmed up again. And so this process, essentially what it's doing is it's transferring the heat. It's allowing the heat to be transferred from this one spot throughout the fluid. And so we call this process, this is convection. Now, the reason why we think the plates are moving is because we think that there are similar types of convection currents in the asthenosphere, in the mantle, in the more fluid part of the mantle. Remember, most of the mantle is kind of this mushy, spongy, not quite liquid, but not quite solid state, kind of plastic."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's allowing the heat to be transferred from this one spot throughout the fluid. And so we call this process, this is convection. Now, the reason why we think the plates are moving is because we think that there are similar types of convection currents in the asthenosphere, in the mantle, in the more fluid part of the mantle. Remember, most of the mantle is kind of this mushy, spongy, not quite liquid, but not quite solid state, kind of plastic. It can kind of mush past. It can kind of flow like super, super, super thick, I guess you could call them, like super viscous fluid. So not quite solid, not quite liquid."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, most of the mantle is kind of this mushy, spongy, not quite liquid, but not quite solid state, kind of plastic. It can kind of mush past. It can kind of flow like super, super, super thick, I guess you could call them, like super viscous fluid. So not quite solid, not quite liquid. But the same thing could be happening. You have certain areas in the mantle that are hotter than others, and it's particular in the asthenosphere. And those areas, that's where you're going to have the material in the mantle move up, because it's hotter, it's less dense, and it will move up."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So not quite solid, not quite liquid. But the same thing could be happening. You have certain areas in the mantle that are hotter than others, and it's particular in the asthenosphere. And those areas, that's where you're going to have the material in the mantle move up, because it's hotter, it's less dense, and it will move up. And maybe it'll cause one of these divergent rifts where kind of plate material and crustal material is forming. And then as it moves up, it cools down and eventually sinks, only to get heated up again. So it has these kind of circular motions, just like we saw with the boiling water."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And those areas, that's where you're going to have the material in the mantle move up, because it's hotter, it's less dense, and it will move up. And maybe it'll cause one of these divergent rifts where kind of plate material and crustal material is forming. And then as it moves up, it cools down and eventually sinks, only to get heated up again. So it has these kind of circular motions, just like we saw with the boiling water. And so that process, remember, this isn't completely liquid, so it is rocky. So it's going to potentially be able to take other things with it. It could maybe drag the crust along with it, and that would cause that."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it has these kind of circular motions, just like we saw with the boiling water. And so that process, remember, this isn't completely liquid, so it is rocky. So it's going to potentially be able to take other things with it. It could maybe drag the crust along with it, and that would cause that. Or I should say it could drag the lithosphere along with it, not just the crust. It could drag all this rigid rock up here along with it, causing it to move in that general direction. So you have the drag there."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It could maybe drag the crust along with it, and that would cause that. Or I should say it could drag the lithosphere along with it, not just the crust. It could drag all this rigid rock up here along with it, causing it to move in that general direction. So you have the drag there. You could also imagine that there's kind of a suction effect, where if you view it as a fluid, you have a bunch of fluid coming down here. So it would kind of pull the lithosphere down at those points, and it would kind of push the lithosphere up at those points. So you have these convection currents that are essentially driving."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have the drag there. You could also imagine that there's kind of a suction effect, where if you view it as a fluid, you have a bunch of fluid coming down here. So it would kind of pull the lithosphere down at those points, and it would kind of push the lithosphere up at those points. So you have these convection currents that are essentially driving. And these aren't going to be super fast-moving fluid convection currents like you would expect with boiling water or with heated water. These would be slow-moving convection currents, but they're moving enough, and they're able to kind of put enough drag on the lithosphere to take the lithosphere along with it. And so that's kind of, at a high level, the dominant theory as to why they're moving."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have these convection currents that are essentially driving. And these aren't going to be super fast-moving fluid convection currents like you would expect with boiling water or with heated water. These would be slow-moving convection currents, but they're moving enough, and they're able to kind of put enough drag on the lithosphere to take the lithosphere along with it. And so that's kind of, at a high level, the dominant theory as to why they're moving. There's other ones that talk about maybe the lithospheric plates. They thicken as they move further away from this area where the ridge is forming. And if we look at this oceanic crust right over here."}, {"video_title": "Plates moving due to convection in mantle Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that's kind of, at a high level, the dominant theory as to why they're moving. There's other ones that talk about maybe the lithospheric plates. They thicken as they move further away from this area where the ridge is forming. And if we look at this oceanic crust right over here. And so over time, they're denser over here because there's more cool-down material at these points. And it's already a little bit lower because this is where a lot of the land is being created, and so there's maybe some gravitational effects. But the dominant effect, we think, is due to this convection in the upper mantle."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we got that there should be 12.5 detectable civilizations in the Milky Way galaxy. And I talked about a bunch of reasons about why we aren't detecting them. But I left out one of the most obvious reasons that we're not detecting them. And it was rightfully pointed out in the comments below that video. And that's just the signal might be too weak. If there's 12.5, if there's on the order of 10, 11, 12 detectable civilizations in our galaxy, they could be quite far from us. This isn't the Milky Way, but this is a galaxy that probably doesn't look too different from our Milky Way."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it was rightfully pointed out in the comments below that video. And that's just the signal might be too weak. If there's 12.5, if there's on the order of 10, 11, 12 detectable civilizations in our galaxy, they could be quite far from us. This isn't the Milky Way, but this is a galaxy that probably doesn't look too different from our Milky Way. We could obviously never get this vantage point of our galaxy, at least not for a while, not unless we can travel quite far away from it. But let's say that we're over here. You could imagine if the 10 civilizations or the 12 civilizations are here."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This isn't the Milky Way, but this is a galaxy that probably doesn't look too different from our Milky Way. We could obviously never get this vantage point of our galaxy, at least not for a while, not unless we can travel quite far away from it. But let's say that we're over here. You could imagine if the 10 civilizations or the 12 civilizations are here. 1, 2, 3, 4. There's probably a lot more in the center, actually, because that's where our density is higher. So let me put it here."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You could imagine if the 10 civilizations or the 12 civilizations are here. 1, 2, 3, 4. There's probably a lot more in the center, actually, because that's where our density is higher. So let me put it here. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. The closest of them might be tens of thousands of light years away from us. And there might be a lot of stuff in between, all sorts of crazy things happening, stars exploding, all sorts of signals that we're receiving."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me put it here. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. The closest of them might be tens of thousands of light years away from us. And there might be a lot of stuff in between, all sorts of crazy things happening, stars exploding, all sorts of signals that we're receiving. And it might just be that the signals from those civilizations are too weak to reach us, or that there's somehow too much interference from all of the other craziness that's happening around the galaxy. There's also these other reasons that I talked about in the last video. Maybe they've gone beyond using radio as a form of communication, and that's why they never use it to begin with."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there might be a lot of stuff in between, all sorts of crazy things happening, stars exploding, all sorts of signals that we're receiving. And it might just be that the signals from those civilizations are too weak to reach us, or that there's somehow too much interference from all of the other craziness that's happening around the galaxy. There's also these other reasons that I talked about in the last video. Maybe they've gone beyond using radio as a form of communication, and that's why they never use it to begin with. And that's why we don't even see them ever using it. Or they use it for a very short period, kind of a transition period, and maybe in 100 years we'll discover the next best thing. The other idea behind why we're probably, or maybe why we might not be able to detect civilizations is that, well, there might be a lot fewer than 10."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe they've gone beyond using radio as a form of communication, and that's why they never use it to begin with. And that's why we don't even see them ever using it. Or they use it for a very short period, kind of a transition period, and maybe in 100 years we'll discover the next best thing. The other idea behind why we're probably, or maybe why we might not be able to detect civilizations is that, well, there might be a lot fewer than 10. When I did the Drake equation right over here, I just made a bunch of assumptions. None of these seemed crazy. But I kind of assumed a reality where you didn't have these kind of cataclysmic events in the galaxy at regular intervals."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The other idea behind why we're probably, or maybe why we might not be able to detect civilizations is that, well, there might be a lot fewer than 10. When I did the Drake equation right over here, I just made a bunch of assumptions. None of these seemed crazy. But I kind of assumed a reality where you didn't have these kind of cataclysmic events in the galaxy at regular intervals. But we know that there are cataclysmic events that happen in our galaxy, in other galaxies. The one that we know the most about, although there are probably all sorts of things that we don't know much about, are gamma ray bursts. And these are still kind of trying to be understood."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I kind of assumed a reality where you didn't have these kind of cataclysmic events in the galaxy at regular intervals. But we know that there are cataclysmic events that happen in our galaxy, in other galaxies. The one that we know the most about, although there are probably all sorts of things that we don't know much about, are gamma ray bursts. And these are still kind of trying to be understood. And you could watch the video on quasars. Those are essentially a lot of highly energetic rays being released when all of this material is being absorbed into supermassive black holes at the centers of galaxies that tend to be very, very, very often billions of light years away. And gamma rays are one of the things that get emitted from those."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these are still kind of trying to be understood. And you could watch the video on quasars. Those are essentially a lot of highly energetic rays being released when all of this material is being absorbed into supermassive black holes at the centers of galaxies that tend to be very, very, very often billions of light years away. And gamma rays are one of the things that get emitted from those. But you can also have gamma ray bursts within galaxies. We believe maybe certain types of stars, when they collapse into black holes, you have this burst of gamma rays. There might be certain types of neutron stars with the right properties that might every now and then release gamma rays."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And gamma rays are one of the things that get emitted from those. But you can also have gamma ray bursts within galaxies. We believe maybe certain types of stars, when they collapse into black holes, you have this burst of gamma rays. There might be certain types of neutron stars with the right properties that might every now and then release gamma rays. And the view is that if there is a civilization that is within a few thousand light years near one of these gamma ray bursts, and it's in the wrong place, it's kind of in the path of the burst, then it's a good chance that those civilizations will be completely wiped out, that those planets will be sterilized, because there's so much radiation coming out from that gamma ray. And there's even some theories that some of the extinction events that have happened in Earth's history, so we're not talking about the dinosaurs, we're talking about billions of years ago, maybe a billion years ago or two billion years ago, that these might have been caused by relatively local gamma ray bursts. The theory is that these might hit Earth on the order of once every billion years."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There might be certain types of neutron stars with the right properties that might every now and then release gamma rays. And the view is that if there is a civilization that is within a few thousand light years near one of these gamma ray bursts, and it's in the wrong place, it's kind of in the path of the burst, then it's a good chance that those civilizations will be completely wiped out, that those planets will be sterilized, because there's so much radiation coming out from that gamma ray. And there's even some theories that some of the extinction events that have happened in Earth's history, so we're not talking about the dinosaurs, we're talking about billions of years ago, maybe a billion years ago or two billion years ago, that these might have been caused by relatively local gamma ray bursts. The theory is that these might hit Earth on the order of once every billion years. And if you think about the galaxy as a whole, we're kind of in where our solar system is orbiting the galaxy. It's kind of a nice distance from the center of the galaxy. The closer you get into the center, the higher densities of stars you have."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The theory is that these might hit Earth on the order of once every billion years. And if you think about the galaxy as a whole, we're kind of in where our solar system is orbiting the galaxy. It's kind of a nice distance from the center of the galaxy. The closer you get into the center, the higher densities of stars you have. So you can imagine if Earth gets hit with one of these gamma ray bursts every couple of billion years, you could imagine something closer to the center of the galaxy gets hit with these gamma ray bursts much, much, much more frequently, just because there's more activity there. There's more stars that are closer by, more stars that are aging, more stars that might be collapsing into black holes. So the simple answer is we don't know."}, {"video_title": "Detectable civilizations in our galaxy 5 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The closer you get into the center, the higher densities of stars you have. So you can imagine if Earth gets hit with one of these gamma ray bursts every couple of billion years, you could imagine something closer to the center of the galaxy gets hit with these gamma ray bursts much, much, much more frequently, just because there's more activity there. There's more stars that are closer by, more stars that are aging, more stars that might be collapsing into black holes. So the simple answer is we don't know. There could be 1,000 civilizations out there. We're just not sophisticated enough to notice them just yet. Or there might be very, very few because of all of this craziness that happens in the galaxy."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Before we get into the physical properties of aldehydes and ketones, I just wanted to cover where the names for those functional groups come from. So, one way to make aldehydes and ketones is to oxidize alcohol. So if we start over here on the left, and we have methanol, we can oxidize that to methanol over here on the right, also called formaldehyde. And if we analyze the atoms here, one carbon on the left, one carbon on the right, one oxygen on the left, one oxygen on the right, four hydrogens on the left, and only two on the right. So, a loss of two hydrogens can convert methanol to methanol. And so the name of aldehyde comes from these words here. So if I write alcohol, and then dehydrogenatum, which refers to the fact that we are losing hydrogens, dehydrogenatum, if you look closely, you can see the name for aldehyde."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And if we analyze the atoms here, one carbon on the left, one carbon on the right, one oxygen on the left, one oxygen on the right, four hydrogens on the left, and only two on the right. So, a loss of two hydrogens can convert methanol to methanol. And so the name of aldehyde comes from these words here. So if I write alcohol, and then dehydrogenatum, which refers to the fact that we are losing hydrogens, dehydrogenatum, if you look closely, you can see the name for aldehyde. If you take the AL from alcohol, and then this portion of this word, and then add an E on, you get the name aldehyde. So that's the idea. You can also make ketones."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So if I write alcohol, and then dehydrogenatum, which refers to the fact that we are losing hydrogens, dehydrogenatum, if you look closely, you can see the name for aldehyde. If you take the AL from alcohol, and then this portion of this word, and then add an E on, you get the name aldehyde. So that's the idea. You can also make ketones. So if I oxidize this alcohol on the left to propanol, also called isopropanol or isopropyl alcohol, and then finally rubbing alcohol, if you oxidize this molecule, then you get this molecule over here on the right. So there are three carbons, so a three-carbon ketone is called a propanone. And of course, no one usually calls this propanone."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "You can also make ketones. So if I oxidize this alcohol on the left to propanol, also called isopropanol or isopropyl alcohol, and then finally rubbing alcohol, if you oxidize this molecule, then you get this molecule over here on the right. So there are three carbons, so a three-carbon ketone is called a propanone. And of course, no one usually calls this propanone. This is a famous molecule. This is acetone. And the old German word for acetone, if you spell out the old German word for acetone, it's easy to see where the word ketone comes from, right?"}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And of course, no one usually calls this propanone. This is a famous molecule. This is acetone. And the old German word for acetone, if you spell out the old German word for acetone, it's easy to see where the word ketone comes from, right? Because if I take this right here, and add an E on, I get ketone. So just a little bit of insight into those names, which I think is pretty interesting. In terms of physical properties, let's use these last two molecules here to describe boiling points of aldehydes and ketones."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And the old German word for acetone, if you spell out the old German word for acetone, it's easy to see where the word ketone comes from, right? Because if I take this right here, and add an E on, I get ketone. So just a little bit of insight into those names, which I think is pretty interesting. In terms of physical properties, let's use these last two molecules here to describe boiling points of aldehydes and ketones. So let's take 2-propanol over here on the left, and let's compare the boiling point of 2-propanol to acetone. So when you're talking about boiling point, you need to think about intermolecular forces, so the forces between molecules. So let's draw out two molecules of isopropanol here."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "In terms of physical properties, let's use these last two molecules here to describe boiling points of aldehydes and ketones. So let's take 2-propanol over here on the left, and let's compare the boiling point of 2-propanol to acetone. So when you're talking about boiling point, you need to think about intermolecular forces, so the forces between molecules. So let's draw out two molecules of isopropanol here. So let's go ahead and draw one. So we have our oxygen, we have our hydrogen right here. Now we know that oxygen is more electronegative than hydrogen, so the electrons in this bond are going to be pulled closer to the oxygen, giving the oxygen a partial negative charge, and giving this hydrogen a partial positive charge."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw out two molecules of isopropanol here. So let's go ahead and draw one. So we have our oxygen, we have our hydrogen right here. Now we know that oxygen is more electronegative than hydrogen, so the electrons in this bond are going to be pulled closer to the oxygen, giving the oxygen a partial negative charge, and giving this hydrogen a partial positive charge. If another molecule of isopropanol comes along, let's go ahead and show that. It has the same situation, right? The oxygen is partially negative, and the hydrogen is partially positive."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now we know that oxygen is more electronegative than hydrogen, so the electrons in this bond are going to be pulled closer to the oxygen, giving the oxygen a partial negative charge, and giving this hydrogen a partial positive charge. If another molecule of isopropanol comes along, let's go ahead and show that. It has the same situation, right? The oxygen is partially negative, and the hydrogen is partially positive. We know that opposite charges attract, right? So this partial positive charge is attracted to this partial negative charge, and this intermolecular force is called hydrogen bonding. So this is an example of hydrogen bonding, which we know is between hydrogen and a very electronegative atom like fluorine, oxygen, or nitrogen."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen is partially negative, and the hydrogen is partially positive. We know that opposite charges attract, right? So this partial positive charge is attracted to this partial negative charge, and this intermolecular force is called hydrogen bonding. So this is an example of hydrogen bonding, which we know is between hydrogen and a very electronegative atom like fluorine, oxygen, or nitrogen. And also, this hydrogen has to be bonded to another electronegative atom, so here we have oxygen. So this is an example of hydrogen bonding, the strongest type of intermolecular force. And so it takes a lot of energy to pull these molecules apart, so it takes a lot of heat."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is an example of hydrogen bonding, which we know is between hydrogen and a very electronegative atom like fluorine, oxygen, or nitrogen. And also, this hydrogen has to be bonded to another electronegative atom, so here we have oxygen. So this is an example of hydrogen bonding, the strongest type of intermolecular force. And so it takes a lot of energy to pull these molecules apart, so it takes a lot of heat. And so the boiling point of isopropanol is relatively high. The boiling point is approximately 83 degrees Celsius. All right, so let's compare that situation with acetone."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so it takes a lot of energy to pull these molecules apart, so it takes a lot of heat. And so the boiling point of isopropanol is relatively high. The boiling point is approximately 83 degrees Celsius. All right, so let's compare that situation with acetone. So let's go ahead and draw out acetone here. And so here's one molecule of acetone. If we think about oxygen compared to this carbonyl carbon here, oxygen is more electronegative, and so there is going to be a polarization."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's compare that situation with acetone. So let's go ahead and draw out acetone here. And so here's one molecule of acetone. If we think about oxygen compared to this carbonyl carbon here, oxygen is more electronegative, and so there is going to be a polarization. So the oxygen is going to withdraw electron density, making the oxygen partially negative. It's taking electron density away from this carbon, so this carbon right here, this carbonyl carbon is partially positive. And so we have a dipole situation."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we think about oxygen compared to this carbonyl carbon here, oxygen is more electronegative, and so there is going to be a polarization. So the oxygen is going to withdraw electron density, making the oxygen partially negative. It's taking electron density away from this carbon, so this carbon right here, this carbonyl carbon is partially positive. And so we have a dipole situation. So this molecule has a dipole moment. And if we think about another molecule of acetone, so another one has the exact same situation. The oxygen is partially negative."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we have a dipole situation. So this molecule has a dipole moment. And if we think about another molecule of acetone, so another one has the exact same situation. The oxygen is partially negative. This carbonyl carbon is partially positive. And so we have an attraction between this partially negative oxygen and this partially positive carbon. So there is an attraction between these two dipoles."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen is partially negative. This carbonyl carbon is partially positive. And so we have an attraction between this partially negative oxygen and this partially positive carbon. So there is an attraction between these two dipoles. So we call this dipole-dipole interaction, which is another type of intermolecular force. So actually hydrogen bonding is just an example of a very strong dipole-dipole interaction. And so dipole-dipole interactions are not as strong as hydrogen bonding."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So there is an attraction between these two dipoles. So we call this dipole-dipole interaction, which is another type of intermolecular force. So actually hydrogen bonding is just an example of a very strong dipole-dipole interaction. And so dipole-dipole interactions are not as strong as hydrogen bonding. And so molecules of acetone aren't held together as, aren't attracted to each other as much as molecules of isopropanol. So it doesn't take as much energy to pull apart molecules of acetone. And so therefore the boiling point is lower."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so dipole-dipole interactions are not as strong as hydrogen bonding. And so molecules of acetone aren't held together as, aren't attracted to each other as much as molecules of isopropanol. So it doesn't take as much energy to pull apart molecules of acetone. And so therefore the boiling point is lower. The boiling point of acetone is approximately 56 degrees Celsius. And so both of these temperatures are above room temperature, right? But both these boiling points are above room temperature."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore the boiling point is lower. The boiling point of acetone is approximately 56 degrees Celsius. And so both of these temperatures are above room temperature, right? But both these boiling points are above room temperature. So at room temperature and pressure, 2-propanol and acetone are both liquids. All right, let's look at some other molecules and let's compare them here. So we have all molecules with three carbons."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But both these boiling points are above room temperature. So at room temperature and pressure, 2-propanol and acetone are both liquids. All right, let's look at some other molecules and let's compare them here. So we have all molecules with three carbons. So over here on the left, this is propane. And the boiling point for propane is approximately negative 42 degrees Celsius. So that's well below room temperature."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have all molecules with three carbons. So over here on the left, this is propane. And the boiling point for propane is approximately negative 42 degrees Celsius. So that's well below room temperature. Room temperature is approximately between 20 and 25 degrees Celsius. And so since the boiling point for propane is well below room temp, the propane is already a gas. And so the state of matter for propane is a gas here."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So that's well below room temperature. Room temperature is approximately between 20 and 25 degrees Celsius. And so since the boiling point for propane is well below room temp, the propane is already a gas. And so the state of matter for propane is a gas here. In terms of intermolecular forces, the only intermolecular forces holding together alkanes are London dispersion forces. So let's go ahead and write that up here. So London dispersion forces."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so the state of matter for propane is a gas here. In terms of intermolecular forces, the only intermolecular forces holding together alkanes are London dispersion forces. So let's go ahead and write that up here. So London dispersion forces. Next, let's analyze an aldehyde. So three carbon aldehyde, right? One, two, three."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So London dispersion forces. Next, let's analyze an aldehyde. So three carbon aldehyde, right? One, two, three. So this must be propanol. And the boiling point for propanol is approximately 50 degrees Celsius. Once again, higher than room temperature."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three. So this must be propanol. And the boiling point for propanol is approximately 50 degrees Celsius. Once again, higher than room temperature. So propanol is a liquid. We've just analyzed acetone. The boiling point is approximately 56 degrees."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Once again, higher than room temperature. So propanol is a liquid. We've just analyzed acetone. The boiling point is approximately 56 degrees. And for both propanol and for acetone, you have the dipole-dipole interaction between molecules. So we already covered acetone. The same situation exists for this aldehyde."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The boiling point is approximately 56 degrees. And for both propanol and for acetone, you have the dipole-dipole interaction between molecules. So we already covered acetone. The same situation exists for this aldehyde. So we have a partial negative here and a partial positive right here. And so there's going to be dipole-dipole interaction between molecules of propanol. So we have London dispersion for our alkane."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The same situation exists for this aldehyde. So we have a partial negative here and a partial positive right here. And so there's going to be dipole-dipole interaction between molecules of propanol. So we have London dispersion for our alkane. And then for our aldehyde and our ketone, we have dipole-dipole interaction. And then finally, we have another alcohol. So instead of two propanol, this is one propanol, which has a boiling point of approximately 97 degrees."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have London dispersion for our alkane. And then for our aldehyde and our ketone, we have dipole-dipole interaction. And then finally, we have another alcohol. So instead of two propanol, this is one propanol, which has a boiling point of approximately 97 degrees. And one propanol also has, of course, hydrogen bonding. And so we can see that the boiling points reflect the type of intermolecular force. Hydrogen bonding is stronger than dipole-dipole interaction."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So instead of two propanol, this is one propanol, which has a boiling point of approximately 97 degrees. And one propanol also has, of course, hydrogen bonding. And so we can see that the boiling points reflect the type of intermolecular force. Hydrogen bonding is stronger than dipole-dipole interaction. And so therefore, the boiling point for alcohols are higher than the boiling point for aldehydes or ketones. But aldehydes and ketones have a higher boiling point than alkanes because dipole-dipole interactions are stronger than London dispersion forces. All right, so let's look at solubility next."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Hydrogen bonding is stronger than dipole-dipole interaction. And so therefore, the boiling point for alcohols are higher than the boiling point for aldehydes or ketones. But aldehydes and ketones have a higher boiling point than alkanes because dipole-dipole interactions are stronger than London dispersion forces. All right, so let's look at solubility next. So we just did boiling point. And now let's think about solubility in water. So let's go ahead and write that."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's look at solubility next. So we just did boiling point. And now let's think about solubility in water. So let's go ahead and write that. And once again, let's think about acetone as our example. And so if we draw this out, here's one molecule of acetone. And I can go ahead and put my lone pairs of electrons in there on my oxygen."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that. And once again, let's think about acetone as our example. And so if we draw this out, here's one molecule of acetone. And I can go ahead and put my lone pairs of electrons in there on my oxygen. Once again, the oxygen gets a partial negative charge. So the oxygen withdraws some electron density. So it gets a little bit more negative."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And I can go ahead and put my lone pairs of electrons in there on my oxygen. Once again, the oxygen gets a partial negative charge. So the oxygen withdraws some electron density. So it gets a little bit more negative. And this carbonyl carbon gets a little bit positive. And so we have this polarized situation in our acetone molecule. If we think about solubility in water, we'll go ahead and draw the dot structure for water."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it gets a little bit more negative. And this carbonyl carbon gets a little bit positive. And so we have this polarized situation in our acetone molecule. If we think about solubility in water, we'll go ahead and draw the dot structure for water. We know that water is also polarized here. So these electrons in this bond are pulled closer to the oxygen. So these electrons are pulled closer to the oxygen, giving the oxygen a partial negative and giving this hydrogen a partial positive."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we think about solubility in water, we'll go ahead and draw the dot structure for water. We know that water is also polarized here. So these electrons in this bond are pulled closer to the oxygen. So these electrons are pulled closer to the oxygen, giving the oxygen a partial negative and giving this hydrogen a partial positive. So we can see there's going to be an attractive force between this partial negative and this partial positive. And in terms of intermolecular forces, we should recognize that as hydrogen bonding. So this is hydrogen bonding right here."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons are pulled closer to the oxygen, giving the oxygen a partial negative and giving this hydrogen a partial positive. So we can see there's going to be an attractive force between this partial negative and this partial positive. And in terms of intermolecular forces, we should recognize that as hydrogen bonding. So this is hydrogen bonding right here. And because of that, we know that acetone is going to be soluble in water. So we have hydrogen bonding. Now, one quick point that I forgot to mention in the previous example, some people are confused as to why molecules of acetone can't hydrogen bond with themselves."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is hydrogen bonding right here. And because of that, we know that acetone is going to be soluble in water. So we have hydrogen bonding. Now, one quick point that I forgot to mention in the previous example, some people are confused as to why molecules of acetone can't hydrogen bond with themselves. So let's go back up here and look at those two molecules again. So if I think about the possibility of hydrogen bonding here, there is a hydrogen connected to this carbon. But that's the point."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Now, one quick point that I forgot to mention in the previous example, some people are confused as to why molecules of acetone can't hydrogen bond with themselves. So let's go back up here and look at those two molecules again. So if I think about the possibility of hydrogen bonding here, there is a hydrogen connected to this carbon. But that's the point. This hydrogen is connected to a carbon. It's not connected to something like oxygen, which is what we had over here. So hydrogen bonding between molecules of acetone is not possible because the hydrogen is bonded to a carbon and not to something like an oxygen."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But that's the point. This hydrogen is connected to a carbon. It's not connected to something like oxygen, which is what we had over here. So hydrogen bonding between molecules of acetone is not possible because the hydrogen is bonded to a carbon and not to something like an oxygen. So even though hydrogen bonding between molecules of acetone is not possible, hydrogen bonding between acetone and water is possible. And so acetone is going to be soluble in water. So same idea for other small aldehydes and ketones."}, {"video_title": "Physical properties of aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So hydrogen bonding between molecules of acetone is not possible because the hydrogen is bonded to a carbon and not to something like an oxygen. So even though hydrogen bonding between molecules of acetone is not possible, hydrogen bonding between acetone and water is possible. And so acetone is going to be soluble in water. So same idea for other small aldehydes and ketones. Small aldehydes and ketones are going to be relatively soluble in water. However, as you increase the chain length, so if we think about the alkyl groups attached to either a ketone or an aldehyde, so let's just look at the alkyl groups here. As you increase the number of carbons that are bonded to an aldehyde or a ketone, that increases the nonpolar character of the molecule."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the general formula for an alcohol we saw is some type of group or chain of carbons bonded to an oxygen, bonded to a hydrogen, and of course the oxygen will have two lone pairs just like that. Let's compare this to water. So water just looks like this. You have a hydrogen bonded to an oxygen, bonded to another hydrogen with two lone pairs. Now in the case of water, the oxygen is much more electronegative than the hydrogen, so it hogs the electrons towards it. So you have a partial negative charge at the oxygen end, and then you have partial positive charges at the hydrogen ends. And that's what allows water to kind of bond to itself, and for it to kind of have not a ridiculously low boiling point."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "You have a hydrogen bonded to an oxygen, bonded to another hydrogen with two lone pairs. Now in the case of water, the oxygen is much more electronegative than the hydrogen, so it hogs the electrons towards it. So you have a partial negative charge at the oxygen end, and then you have partial positive charges at the hydrogen ends. And that's what allows water to kind of bond to itself, and for it to kind of have not a ridiculously low boiling point. So let me show this. Let me copy and paste this. And we've seen all this before in regular chemistry."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And that's what allows water to kind of bond to itself, and for it to kind of have not a ridiculously low boiling point. So let me show this. Let me copy and paste this. And we've seen all this before in regular chemistry. So copy and paste. So let me draw some more water molecules here. And let me draw another water molecule here."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we've seen all this before in regular chemistry. So copy and paste. So let me draw some more water molecules here. And let me draw another water molecule here. So you see water, because the oxygen end has a partial negative charge and the hydrogen ends have partial positive charges, the oxygen of one water molecule will be attracted to the hydrogen of another water molecule. And we've seen this before. This we call hydrogen bonding."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And let me draw another water molecule here. So you see water, because the oxygen end has a partial negative charge and the hydrogen ends have partial positive charges, the oxygen of one water molecule will be attracted to the hydrogen of another water molecule. And we've seen this before. This we call hydrogen bonding. So that right there is hydrogen bonding. The exact same thing can happen with alcohols, although alcohols really only have the partial positive charge on the hydrogen. We don't know exactly what's going on here."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This we call hydrogen bonding. So that right there is hydrogen bonding. The exact same thing can happen with alcohols, although alcohols really only have the partial positive charge on the hydrogen. We don't know exactly what's going on here. We probably have carbons bonded to the oxygen. And the carbons, they're reasonably electronegative. They're not going to have their electrons hogged as much as a hydrogen would."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We don't know exactly what's going on here. We probably have carbons bonded to the oxygen. And the carbons, they're reasonably electronegative. They're not going to have their electrons hogged as much as a hydrogen would. So in the case of an alcohol, actually let me draw, instead of having this R for radical there, let me make it a little bit more concrete. Let me draw an actual alcohol. So an actual alcohol, maybe we have methanol."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "They're not going to have their electrons hogged as much as a hydrogen would. So in the case of an alcohol, actually let me draw, instead of having this R for radical there, let me make it a little bit more concrete. Let me draw an actual alcohol. So an actual alcohol, maybe we have methanol. It would look like that. It has a hydrogen right over here. Oxygen is much more electronegative than the hydrogen."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So an actual alcohol, maybe we have methanol. It would look like that. It has a hydrogen right over here. Oxygen is much more electronegative than the hydrogen. So you have a partial negative charge there, and then you have a partial positive charge there. So it too, because of these hydrogen bonds, it will have a reasonable boiling. It won't just turn immediately into the gaseous state."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen is much more electronegative than the hydrogen. So you have a partial negative charge there, and then you have a partial positive charge there. So it too, because of these hydrogen bonds, it will have a reasonable boiling. It won't just turn immediately into the gaseous state. It would actually try to bond to each other. So let me copy and paste that. So it can also form the hydrogen bonds, but they won't be quite as strong as what you see in water."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It won't just turn immediately into the gaseous state. It would actually try to bond to each other. So let me copy and paste that. So it can also form the hydrogen bonds, but they won't be quite as strong as what you see in water. And that's why something like methanol actually has a lower boiling point than water. It's easy to make it boil. It's easier to make these bonds break apart because you don't have as much of the hydrogen bonding."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it can also form the hydrogen bonds, but they won't be quite as strong as what you see in water. And that's why something like methanol actually has a lower boiling point than water. It's easy to make it boil. It's easier to make these bonds break apart because you don't have as much of the hydrogen bonding. So this is an example of hydrogen bonding with methanol. Now, because methanol can have hydrogen bonding and it has this slight polarity to it, and water obviously has hydrogen bonding, methanol is actually miscible in water. And all that means is that it's soluble in water in any proportion, no matter how much methanol or how much water you have, it is soluble."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It's easier to make these bonds break apart because you don't have as much of the hydrogen bonding. So this is an example of hydrogen bonding with methanol. Now, because methanol can have hydrogen bonding and it has this slight polarity to it, and water obviously has hydrogen bonding, methanol is actually miscible in water. And all that means is that it's soluble in water in any proportion, no matter how much methanol or how much water you have, it is soluble. So if I were to draw some methanol molecules, actually, maybe this is the water right here. So if you drop a methanol molecule right there, that would have a hydrogen bond right over there. If I were to draw another methanol molecule, maybe right over here, you would have another hydrogen bond right over there."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And all that means is that it's soluble in water in any proportion, no matter how much methanol or how much water you have, it is soluble. So if I were to draw some methanol molecules, actually, maybe this is the water right here. So if you drop a methanol molecule right there, that would have a hydrogen bond right over there. If I were to draw another methanol molecule, maybe right over here, you would have another hydrogen bond right over there. And that's what allows methanol to be soluble in water. Now, as this chain grows, or if you have alcohols with longer radical chains, then they become less and less soluble in water. But their boiling points actually do go up, and let's think about why that is."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If I were to draw another methanol molecule, maybe right over here, you would have another hydrogen bond right over there. And that's what allows methanol to be soluble in water. Now, as this chain grows, or if you have alcohols with longer radical chains, then they become less and less soluble in water. But their boiling points actually do go up, and let's think about why that is. So if I have something like, let me do butanol. So butanol is going to have four carbons. It's going to be H3C, let me just draw it like this, H3C, CH2, CH2, CH, let me do it like this, H2C, and then that carbon, that last carbon right there is going to be bonded to the oxygen, it's going to be bonded to an oxygen, which is bonded to a hydrogen."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But their boiling points actually do go up, and let's think about why that is. So if I have something like, let me do butanol. So butanol is going to have four carbons. It's going to be H3C, let me just draw it like this, H3C, CH2, CH2, CH, let me do it like this, H2C, and then that carbon, that last carbon right there is going to be bonded to the oxygen, it's going to be bonded to an oxygen, which is bonded to a hydrogen. Now, when you have a situation like this, the oxygen will have a partial negative charge, the hydrogen will still have a partial positive charge, just like we saw up here with both the water and the methanol. But now you have this big thing here that has no polarity. So this part of the alcohol is not going to be soluble in water, and it's going to make it harder for this part to be soluble over here."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be H3C, let me just draw it like this, H3C, CH2, CH2, CH, let me do it like this, H2C, and then that carbon, that last carbon right there is going to be bonded to the oxygen, it's going to be bonded to an oxygen, which is bonded to a hydrogen. Now, when you have a situation like this, the oxygen will have a partial negative charge, the hydrogen will still have a partial positive charge, just like we saw up here with both the water and the methanol. But now you have this big thing here that has no polarity. So this part of the alcohol is not going to be soluble in water, and it's going to make it harder for this part to be soluble over here. So this right here is less soluble. This is less soluble. It'll still be a little bit soluble, so if you have some oxygen here, you will still have a little bit of the hydrogen bonding."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this part of the alcohol is not going to be soluble in water, and it's going to make it harder for this part to be soluble over here. So this right here is less soluble. This is less soluble. It'll still be a little bit soluble, so if you have some oxygen here, you will still have a little bit of the hydrogen bonding. You still will have a little bit of the hydrogen bonding going on, but this part is kind of, you can imagine, it doesn't want to dissolve with the water. It is nonpolar. And so this is, you could actually, for example, butanol in particular, it actually is soluble in water, but not in any proportion."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It'll still be a little bit soluble, so if you have some oxygen here, you will still have a little bit of the hydrogen bonding. You still will have a little bit of the hydrogen bonding going on, but this part is kind of, you can imagine, it doesn't want to dissolve with the water. It is nonpolar. And so this is, you could actually, for example, butanol in particular, it actually is soluble in water, but not in any proportion. So methanol is miscible. Let me write this. This is a new word."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so this is, you could actually, for example, butanol in particular, it actually is soluble in water, but not in any proportion. So methanol is miscible. Let me write this. This is a new word. I don't think I've ever used it before in the context of the organic chemistry videos. So methanol is, let me write that in a brighter color since it's a new word. Methanol is miscible, which just means soluble in any proportion."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This is a new word. I don't think I've ever used it before in the context of the organic chemistry videos. So methanol is, let me write that in a brighter color since it's a new word. Methanol is miscible, which just means soluble in any proportion. So I don't care what percent is methanol, what percent is water, the methanol will dissolve into the water in any proportion. If you look at butanol, it is soluble, but not in any proportion. If you had a ton of butanol, some of it would not dissolve in the water."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Methanol is miscible, which just means soluble in any proportion. So I don't care what percent is methanol, what percent is water, the methanol will dissolve into the water in any proportion. If you look at butanol, it is soluble, but not in any proportion. If you had a ton of butanol, some of it would not dissolve in the water. So this is soluble. So the butanol right here is soluble, but not miscible in water. If you have too much of the butanol, all of a sudden some of it will not actually be able to be dissolved."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If you had a ton of butanol, some of it would not dissolve in the water. So this is soluble. So the butanol right here is soluble, but not miscible in water. If you have too much of the butanol, all of a sudden some of it will not actually be able to be dissolved. And if this chain, if this was a decanol or something with a really long carbon chain, then of course it's going to be very non-soluble. I mean, you might be able to get a couple of the molecules in the water, but most of them will not dissolve. Now the other reason I hinted, look, the reason why the alcohols have a reasonable, not too low of a boiling point, is that they're able to do this hydrogen bonding."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If you have too much of the butanol, all of a sudden some of it will not actually be able to be dissolved. And if this chain, if this was a decanol or something with a really long carbon chain, then of course it's going to be very non-soluble. I mean, you might be able to get a couple of the molecules in the water, but most of them will not dissolve. Now the other reason I hinted, look, the reason why the alcohols have a reasonable, not too low of a boiling point, is that they're able to do this hydrogen bonding. But you would say, well look, these longer carbon chains, these are going to have less of the hydrogen bonding going on. Maybe these would have lower boiling points. But actually the longer the chain gets, these actually have higher boiling points."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now the other reason I hinted, look, the reason why the alcohols have a reasonable, not too low of a boiling point, is that they're able to do this hydrogen bonding. But you would say, well look, these longer carbon chains, these are going to have less of the hydrogen bonding going on. Maybe these would have lower boiling points. But actually the longer the chain gets, these actually have higher boiling points. And that's because these chains can interact with each other. So the longer the chain, so the longer R, or the longer R chain, I guess I could say, we could say the higher the boiling point in an alcohol. Higher boiling point."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But actually the longer the chain gets, these actually have higher boiling points. And that's because these chains can interact with each other. So the longer the chain, so the longer R, or the longer R chain, I guess I could say, we could say the higher the boiling point in an alcohol. Higher boiling point. It's harder, you have to put more heat into the system, or the temperature has to be higher for the things to break apart. And that's because this is one decanol molecule here, another decanol molecule might look like this, maybe might look like this. You have an oxygen and a hydrogen, and then you have your carbons."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Higher boiling point. It's harder, you have to put more heat into the system, or the temperature has to be higher for the things to break apart. And that's because this is one decanol molecule here, another decanol molecule might look like this, maybe might look like this. You have an oxygen and a hydrogen, and then you have your carbons. So you have your CH, your CH2, CH2, H3C. So you have this other butanol here. And what the interaction between these two chains are, these are the van der Waals forces."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "You have an oxygen and a hydrogen, and then you have your carbons. So you have your CH, your CH2, CH2, H3C. So you have this other butanol here. And what the interaction between these two chains are, these are the van der Waals forces. So even though they have no polar, so these guys are going to have some polar interactions, they're going to have the hydrogen bonding, I've been saying that multiple times already. But these long chains, they're going to have the London dispersion forces, which are a subset of van der Waals forces, where even though they're neutral, every now and then, one of these might become slightly negative on one side. So you might have a temporary partial negative charge."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And what the interaction between these two chains are, these are the van der Waals forces. So even though they have no polar, so these guys are going to have some polar interactions, they're going to have the hydrogen bonding, I've been saying that multiple times already. But these long chains, they're going to have the London dispersion forces, which are a subset of van der Waals forces, where even though they're neutral, every now and then, one of these might become slightly negative on one side. So you might have a temporary partial negative charge. And that's just because of the randomness of how electrons move. On this side of the molecule, all of a sudden, you might have more electrons over there, and a partial negative charge. And because of that, you're going to have the electrons over here, they're not going to want to be there, so you're going to want to have a partial positive charge."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So you might have a temporary partial negative charge. And that's just because of the randomness of how electrons move. On this side of the molecule, all of a sudden, you might have more electrons over there, and a partial negative charge. And because of that, you're going to have the electrons over here, they're not going to want to be there, so you're going to want to have a partial positive charge. Then you're going to have a very temporary interaction. That's a very weak force, much weaker than hydrogen bonds. But as these chains get longer and longer, as they possibly even get intertwined with each other and get close to each other, these London dispersion forces, or van der Waals forces, are going to keep propagating."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And because of that, you're going to have the electrons over here, they're not going to want to be there, so you're going to want to have a partial positive charge. Then you're going to have a very temporary interaction. That's a very weak force, much weaker than hydrogen bonds. But as these chains get longer and longer, as they possibly even get intertwined with each other and get close to each other, these London dispersion forces, or van der Waals forces, are going to keep propagating. So all of a sudden, maybe these guys are going to be attracted to each other, and that's going to disappear. Then these guys are going to be attracted to each other, and that's going to disappear. And then these guys are going to be attracted to each other, and then that's going to disappear."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But as these chains get longer and longer, as they possibly even get intertwined with each other and get close to each other, these London dispersion forces, or van der Waals forces, are going to keep propagating. So all of a sudden, maybe these guys are going to be attracted to each other, and that's going to disappear. Then these guys are going to be attracted to each other, and that's going to disappear. And then these guys are going to be attracted to each other, and then that's going to disappear. So you can imagine, the longer the chain, the more of these types of interactions you're going to have, the more attracted they're going to be to each other, and it's going to be harder to break them apart, higher boiling point. So those are just kind of the two big takeaways on the properties of alcohols. Especially smaller-chained alcohols are soluble in water."}, {"video_title": "Alcohol properties Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then these guys are going to be attracted to each other, and then that's going to disappear. So you can imagine, the longer the chain, the more of these types of interactions you're going to have, the more attracted they're going to be to each other, and it's going to be harder to break them apart, higher boiling point. So those are just kind of the two big takeaways on the properties of alcohols. Especially smaller-chained alcohols are soluble in water. The very small ones are completely miscible. And the longer the chain you have, the harder it is to dissolve in water, but also the higher the boiling point. The harder it is to break them apart, because you have these London dispersion forces."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just as a reminder, in this drawing right here, this depiction right here, this circle right here, this solar system circle, it's not the size of the sun, it's not the size of the orbits of the Earth or Pluto or the Kuiper Belt. This is close to the size of the Oort Cloud. And the actual orbit of Earth is about 1... Well, the diameter of the orbit of Earth is about 1 50,000th of this. So you wouldn't even see it on this. It would not even make up a pixel on this screen right here, much less the actual size of the sun or something much smaller. And just remember, that orbit of the Earth, that was that huge distance. It took 8 minutes for light to get from the sun to the Earth."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you wouldn't even see it on this. It would not even make up a pixel on this screen right here, much less the actual size of the sun or something much smaller. And just remember, that orbit of the Earth, that was that huge distance. It took 8 minutes for light to get from the sun to the Earth. This super long distance. If you shot a bullet at the sun from Earth, it would take you that 17 years to actually get to the sun. So once again, this huge distance wouldn't even show up on this picture."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It took 8 minutes for light to get from the sun to the Earth. This super long distance. If you shot a bullet at the sun from Earth, it would take you that 17 years to actually get to the sun. So once again, this huge distance wouldn't even show up on this picture. Now, what we saw in the last video is if you travel at unimaginably fast speeds, if you travel at 60,000 kilometers per hour, you would have picked that speed because that's how fast Voyager 1 actually is traveling. That's one of the, I think, the fastest object we have out there in space right here. And it's actually kind of leaving the solar system as we speak."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So once again, this huge distance wouldn't even show up on this picture. Now, what we saw in the last video is if you travel at unimaginably fast speeds, if you travel at 60,000 kilometers per hour, you would have picked that speed because that's how fast Voyager 1 actually is traveling. That's one of the, I think, the fastest object we have out there in space right here. And it's actually kind of leaving the solar system as we speak. But even if you were able to get that fast, it would still take 80,000 years, 75 or 80,000 years, to travel the 4.2 light years to the Alpha Centauri cluster of stars, to the nearest star. It would take 80,000 years. And that scale of time is already amount of time that I have trouble comprehending."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's actually kind of leaving the solar system as we speak. But even if you were able to get that fast, it would still take 80,000 years, 75 or 80,000 years, to travel the 4.2 light years to the Alpha Centauri cluster of stars, to the nearest star. It would take 80,000 years. And that scale of time is already amount of time that I have trouble comprehending. As you can imagine, all of modern civilization has occurred, definitely in the last 10,000 years, but most of recorded history is in the last 4 or 5,000 years. So this is 80,000 years to travel to the nearest star. So it's a huge distance."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that scale of time is already amount of time that I have trouble comprehending. As you can imagine, all of modern civilization has occurred, definitely in the last 10,000 years, but most of recorded history is in the last 4 or 5,000 years. So this is 80,000 years to travel to the nearest star. So it's a huge distance. Another way to think about it is if the sun, if the sun were the size of a basketball and you put that basketball in London, if you wanted to do it in scale, the next closest star, which is actually a smaller basketball, right over here, Proxima Centauri, that smaller basketball, you would have to put in Kiev, Ukraine, in order to have a similar scale. So these are basketballs sitting in these cities and you would have to travel about 1,200 miles to place the next basketball. And these basketballs are representing these super huge things that we saw in the first video."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's a huge distance. Another way to think about it is if the sun, if the sun were the size of a basketball and you put that basketball in London, if you wanted to do it in scale, the next closest star, which is actually a smaller basketball, right over here, Proxima Centauri, that smaller basketball, you would have to put in Kiev, Ukraine, in order to have a similar scale. So these are basketballs sitting in these cities and you would have to travel about 1,200 miles to place the next basketball. And these basketballs are representing these super huge things that we saw in the first video. The sun, if you actually made the Earth relative to these basketballs, these would be little grains of sand. So if there are any little small planets over here, they would have to be grains of sand in Kiev, Ukraine, versus the grain of sand in London. So this is a massive, massive distance, already, at least in my mind, unimaginable."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these basketballs are representing these super huge things that we saw in the first video. The sun, if you actually made the Earth relative to these basketballs, these would be little grains of sand. So if there are any little small planets over here, they would have to be grains of sand in Kiev, Ukraine, versus the grain of sand in London. So this is a massive, massive distance, already, at least in my mind, unimaginable. And when it gets really racky is when you start realizing that even this is a super, super small distance relative to the galactic scale. So this whole depiction of kind of our neighborhood of stars, this thing over here is about, give or take, and we're doing rough estimates right here, it's about 30 light years. I'll just do LY for short."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is a massive, massive distance, already, at least in my mind, unimaginable. And when it gets really racky is when you start realizing that even this is a super, super small distance relative to the galactic scale. So this whole depiction of kind of our neighborhood of stars, this thing over here is about, give or take, and we're doing rough estimates right here, it's about 30 light years. I'll just do LY for short. So that's about 30 light years. And once again, you can take pictures of our galaxy from our point of view, but you actually can't take a picture of the whole galaxy from above it, so these are going to be artist depictions. But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly, these are all approximations, this is about, let me do this in a darker color, this is about 1,000 light years."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll just do LY for short. So that's about 30 light years. And once again, you can take pictures of our galaxy from our point of view, but you actually can't take a picture of the whole galaxy from above it, so these are going to be artist depictions. But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly, these are all approximations, this is about, let me do this in a darker color, this is about 1,000 light years. And this is the 1,000 light years of our sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here, and it would take forever to get anywhere over here, it would be 1 30th of this, so it would be about that big, this whole drawing."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if this is 30 light years, this drawing right here of kind of our local neighborhood of the galaxy, this right here is roughly, these are all approximations, this is about, let me do this in a darker color, this is about 1,000 light years. And this is the 1,000 light years of our sun's neighborhood, if you can even call it a neighborhood anymore. Even this isn't really a neighborhood if it takes you 80,000 years to get to your nearest neighbor. But this whole drawing over here, and it would take forever to get anywhere over here, it would be 1 30th of this, so it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here that spans 1,000 light years. But then when you start to really put it into perspective, so now let's zoom out a little bit, so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the sun, and once again the word local is used in a very liberal way at this point."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this whole drawing over here, and it would take forever to get anywhere over here, it would be 1 30th of this, so it would be about that big, this whole drawing. And what's really going to blow your mind is this would be roughly a little bit more than a pixel on this drawing right here that spans 1,000 light years. But then when you start to really put it into perspective, so now let's zoom out a little bit, so this drawing right here, this 1,000 light years is now this 1,000 light years over here. So this is the local vicinity of the sun, and once again the word local is used in a very liberal way at this point. So this right here is 1,000 light years. If you're sitting here and you're looking at an object that's sitting, let me do this in a darker color, if we're sitting here on earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago, because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the local vicinity of the sun, and once again the word local is used in a very liberal way at this point. So this right here is 1,000 light years. If you're sitting here and you're looking at an object that's sitting, let me do this in a darker color, if we're sitting here on earth and we're looking at an object out here that's 500 light years away, we're looking at it as it was 500 years ago, because the light that is reaching our eyeballs right now, or our telescopes right now, left this guy over here 500 years ago. In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're already talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, he's not going to even be there anymore. He probably has moved around a little bit. So just even on this scale, we're already talking about these unimaginably huge distances. And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly what the actual details of the actual shape of the Milky Way galaxy, the galaxy that we're in. But we're pretty sure, we're very sure we have these spiral arms and we have these spurs off of them, but it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy because it's kind of on the other side, on the other side of the center."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then when we zoom out, this is kind of our local part of the galaxy right over here. This piece right here, this is called the Orion Spur. And people are still trying to work out exactly what the actual details of the actual shape of the Milky Way galaxy, the galaxy that we're in. But we're pretty sure, we're very sure we have these spiral arms and we have these spurs off of them, but it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least, I mean it blows my mind if you really think about what it's saying. These unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we're pretty sure, we're very sure we have these spiral arms and we have these spurs off of them, but it's actually very hard to come up with the actual shape, especially because you can't see a lot of the galaxy because it's kind of on the other side, on the other side of the center. But really just to get a sense of something that at least, I mean it blows my mind if you really think about what it's saying. These unbelievable distances show up as a little dot here. This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This whole drawing shows up as a dot here. Now when we zoom out, over here that dot would no longer even show up. It wouldn't even register a pixel on this drawing right over here. And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully that gives you a sense of how small, how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion, billion stars."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then this whole drawing, this whole thing right over here, this whole picture is this grid right over here. It is this right over here. So hopefully that gives you a sense of how small, how small even our local neighborhood is relative to the galaxy as a whole. And the galaxy as a whole, just to give you a sense, has 200 to 400 billion, billion stars. Billion stars, or maybe I should say solar systems, just to give you a sense that when we saw the solar system, it's not just the sun, there's all this neat dynamic stuff in there, planets and asteroids and solar winds. So there's 200 to 400 billion stars and for the most part, 200 to 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the galaxy as a whole, just to give you a sense, has 200 to 400 billion, billion stars. Billion stars, or maybe I should say solar systems, just to give you a sense that when we saw the solar system, it's not just the sun, there's all this neat dynamic stuff in there, planets and asteroids and solar winds. So there's 200 to 400 billion stars and for the most part, 200 to 400 billion solar systems. So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, this isn't one star, this isn't two stars, these are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars, we're starting to talk in the millions of stars when you look at certain blotches here and there. Maybe it might be one star that's closer to you, or it might be a million stars that are far apart and that are just relatively close together and everything has to be used in kind of loose terms here."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's an unimaginably, I guess, complex or huge place. And just to make it clear, even when we zoom in to this picture right here, and I think it was obvious based on telling you about this, that these little white pockets right here, this isn't one star, this isn't two stars, these are thousands of stars here. So when you go over here, each little blotch of white that you see, that's not a star, that's not a thousand stars, we're starting to talk in the millions of stars when you look at certain blotches here and there. Maybe it might be one star that's closer to you, or it might be a million stars that are far apart and that are just relatively close together and everything has to be used in kind of loose terms here. And we'll talk more about other galaxies, but even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe it might be one star that's closer to you, or it might be a million stars that are far apart and that are just relatively close together and everything has to be used in kind of loose terms here. And we'll talk more about other galaxies, but even this isn't the upper bound of galaxies. People believe the Andromeda Galaxy has a trillion stars in it, a trillion solar systems. We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable distances. But if you really want to get the sense relative to the whole galaxy, this is an artist's depiction."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking about these huge, huge, immense distances. And so just to give you a sense of where we fit in the picture, this is a rough location of our sun. And remember, that little dot I drew just now is including millions of stars, millions of solar systems, already unimaginable distances. But if you really want to get the sense relative to the whole galaxy, this is an artist's depiction. Again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if you really want to get the sense relative to the whole galaxy, this is an artist's depiction. Again, we could never obviously get this perspective on the galaxy. It would take us forever to travel this far so that you could see the galaxy from above. But this is our best guess looking at things from our vantage point. We actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. And so it's very hard to see things on the other side. We think, or actually there's a super massive black hole at the center of the galaxy, and we think that they're at the center of all or most galaxies."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is our best guess looking at things from our vantage point. We actually can't even see this whole area over here because it's on the other side of the center of the galaxy, which is super, super dense and super bright. And so it's very hard to see things on the other side. We think, or actually there's a super massive black hole at the center of the galaxy, and we think that they're at the center of all or most galaxies. But the whole point of this video, actually this whole series of videos, this is just kind of, I don't know, put you in awe a little bit of just how huge this is. Because when you really think about the scales, it's, I don't know, no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We think, or actually there's a super massive black hole at the center of the galaxy, and we think that they're at the center of all or most galaxies. But the whole point of this video, actually this whole series of videos, this is just kind of, I don't know, put you in awe a little bit of just how huge this is. Because when you really think about the scales, it's, I don't know, no words can really describe it. But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean, when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But just to give you a sense, we're about 25,000 light years from the center of the galaxy. So even when we look at things in the center of the galaxy, that's as they were 25,000 years ago. It took 25,000 years for that light to get to us. I mean, when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances, and the whole galaxy over here, and once again, like solar systems, it's hard to say the edge of the galaxy because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out. But it gets less dense with stars. But the main density, the main disk, is about 100,000 light years."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I mean, when that light left the center of the galaxy, I won't even guess to think what humanity was like at that point in time. So it's these huge distances, and the whole galaxy over here, and once again, like solar systems, it's hard to say the edge of the galaxy because there's always going to be a few more stars and other things orbiting around the galaxy as you go further and further out. But it gets less dense with stars. But the main density, the main disk, is about 100,000 light years. Is the diameter, roughly, of the main part of the galaxy. And it's about 1,000 light years thick. So you kind of imagine it as this disk, this thing that's fairly flat."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the main density, the main disk, is about 100,000 light years. Is the diameter, roughly, of the main part of the galaxy. And it's about 1,000 light years thick. So you kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat, but it's still immensely, immensely thick."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you kind of imagine it as this disk, this thing that's fairly flat. But it's 1,000 light years thick. You would have to do this distance 250 times just to go from the top part of the galaxy to the bottom part, much less going across the galaxy. So it might seem relatively flat, but it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort cloud, roughly a light year in diameter, is a grain of sand, a millimeter in diameter grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might say, okay, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it might seem relatively flat, but it's still immensely, immensely thick. And just as another way to visualize it, if this thing right over here that includes the Oort cloud, roughly a light year in diameter, is a grain of sand, a millimeter in diameter grain of sand, then the universe as a whole is going to be the diameter of a football field. And that might say, okay, those are two tractable things. I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, remember that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that, or sorry, 15 years or 17 years. I forgot the exact number, but 15, 16, 17 years to even cover half of that distance, 30 years to cover an entire diameter of Earth."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I can imagine a grain of sand, a millimeter wide grain of sand in a football field. But remember, remember that grain of sand is still 50,000 or 60,000 times the diameter of Earth's orbit. And Earth's orbit, it would still take a bullet or something traveling as fast as a jet plane 15 hours to just go half of that, or sorry, 15 years or 17 years. I forgot the exact number, but 15, 16, 17 years to even cover half of that distance, 30 years to cover an entire diameter of Earth. So 30 years just to cover the diameter of Earth's orbit, that's 1 60,000th of our little grain of sand in the football field. And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy, so we're seeing, and this is looking towards the center."}, {"video_title": "Scale of the galaxy Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I forgot the exact number, but 15, 16, 17 years to even cover half of that distance, 30 years to cover an entire diameter of Earth. So 30 years just to cover the diameter of Earth's orbit, that's 1 60,000th of our little grain of sand in the football field. And just to kind of really, I don't know, have an appreciation for how mind-blowing this really is, this is actually a picture of the Milky Way galaxy, our galaxy, from our vantage point. As you can see, we're in the galaxy, so we're seeing, and this is looking towards the center. And even this picture, you start to appreciate the complexity of what 100 billion stars are. But what I really want to point out is even in this picture, when you're looking at these things, some of these things that look like stars, those aren't stars, those are thousands of stars or millions of stars. Maybe it could be one star closer up, but when we're starting to approach the center of the galaxy, these are thousands and thousands and millions of stars or solar systems that we're actually looking at."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Carboxylic acids. They have the general form. They still have that carbonyl group, just like we've seen in aldehydes and ketones. And they will be part of a longer carbon chain. But instead of having a hydrogen here, as is the case with an aldehyde, or having another carbon chain here, with the case of a ketone, we have an OH group, just like that. And you're probably saying, hey, why is this called an acid? It must be called an acid because it is an acid."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And they will be part of a longer carbon chain. But instead of having a hydrogen here, as is the case with an aldehyde, or having another carbon chain here, with the case of a ketone, we have an OH group, just like that. And you're probably saying, hey, why is this called an acid? It must be called an acid because it is an acid. And you'd be right. And the reason why it is an acid, and why it is more acidic than just something with an OH group, so the reason why it is more acidic, it's actually a good bit more acidic than alcohols, is because once this thing loses its hydrogen, remember, acids, depending on what type of acid you want to think about, acid in the Lewis sense could be an electron taker, and this oxygen can take the electron of this hydrogen. Or if you think about the Arrhenius definition of an acid, it is a proton donor, and this OH group can donate a proton."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It must be called an acid because it is an acid. And you'd be right. And the reason why it is an acid, and why it is more acidic than just something with an OH group, so the reason why it is more acidic, it's actually a good bit more acidic than alcohols, is because once this thing loses its hydrogen, remember, acids, depending on what type of acid you want to think about, acid in the Lewis sense could be an electron taker, and this oxygen can take the electron of this hydrogen. Or if you think about the Arrhenius definition of an acid, it is a proton donor, and this OH group can donate a proton. It takes away the electron of this hydrogen, gives away the proton. Either way. But the reason why this is more acidic than an alcohol is, once it gives away this proton, it is actually resident stabilized."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Or if you think about the Arrhenius definition of an acid, it is a proton donor, and this OH group can donate a proton. It takes away the electron of this hydrogen, gives away the proton. Either way. But the reason why this is more acidic than an alcohol is, once it gives away this proton, it is actually resident stabilized. So let me show you what that means. And to do that, let me actually show you the bond between this oxygen and this hydrogen. So the oxygen and the hydrogen, see the oxygen has this pink electron, and the hydrogen has that purple or that magenta electron right there."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But the reason why this is more acidic than an alcohol is, once it gives away this proton, it is actually resident stabilized. So let me show you what that means. And to do that, let me actually show you the bond between this oxygen and this hydrogen. So the oxygen and the hydrogen, see the oxygen has this pink electron, and the hydrogen has that purple or that magenta electron right there. If you put this in a solution of water, so you have some H2O over here, you have some H2O, this oxygen right here really wants to take back this magenta electron. So it really wants to take back that magenta electron, and as it takes back that magenta electron, it would essentially donate the hydrogen proton to a water molecule. So the water molecule would give one of its electrons to the hydrogen proton and then become positive."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the oxygen and the hydrogen, see the oxygen has this pink electron, and the hydrogen has that purple or that magenta electron right there. If you put this in a solution of water, so you have some H2O over here, you have some H2O, this oxygen right here really wants to take back this magenta electron. So it really wants to take back that magenta electron, and as it takes back that magenta electron, it would essentially donate the hydrogen proton to a water molecule. So the water molecule would give one of its electrons to the hydrogen proton and then become positive. So once this happens, the next step would look like this. What was our carboxylic acid will now turn into the carboxylate ion. So it will now look like this."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the water molecule would give one of its electrons to the hydrogen proton and then become positive. So once this happens, the next step would look like this. What was our carboxylic acid will now turn into the carboxylate ion. So it will now look like this. So it has our carbonyl group. Now this oxygen just took an extra electron. It just took an extra electron."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it will now look like this. So it has our carbonyl group. Now this oxygen just took an extra electron. It just took an extra electron. So it has, if I want to draw it, I have that one. Actually let me draw, so this is the oxygen. To start off with, the oxygen had two lone pairs."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It just took an extra electron. So it has, if I want to draw it, I have that one. Actually let me draw, so this is the oxygen. To start off with, the oxygen had two lone pairs. I want to draw those two lone pairs first. So to start off with, it had those two lone pairs. And now it had this pink electron from the get-go, and now it took this magenta electron."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "To start off with, the oxygen had two lone pairs. I want to draw those two lone pairs first. So to start off with, it had those two lone pairs. And now it had this pink electron from the get-go, and now it took this magenta electron. So now it has one extra valence electron. We can even draw them here. We can count them."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And now it had this pink electron from the get-go, and now it took this magenta electron. So now it has one extra valence electron. We can even draw them here. We can count them. One, two, three, four, five, six. Six is just a neutral oxygen, but now it gained another one. It has seven."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We can count them. One, two, three, four, five, six. Six is just a neutral oxygen, but now it gained another one. It has seven. It now has a negative charge. So the water has now become a hydronium ion. So you have the water here."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It has seven. It now has a negative charge. So the water has now become a hydronium ion. So you have the water here. We've increased the proton, or the hydronium concentration, in the water. This one water molecule is now a hydronium molecule. So this is now bonded with this hydrogen proton, just like this."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you have the water here. We've increased the proton, or the hydronium concentration, in the water. This one water molecule is now a hydronium molecule. So this is now bonded with this hydrogen proton, just like this. This oxygen gave away an electron to this proton, so now it has a positive charge. And this right here, this carboxylate anion, the reason why this thing was a stronger acid than something that just had an OH group is because the conjugate base, the carboxylate ion, is actually resonance stabilized. It is more stable than the conjugate base of an alcohol."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this is now bonded with this hydrogen proton, just like this. This oxygen gave away an electron to this proton, so now it has a positive charge. And this right here, this carboxylate anion, the reason why this thing was a stronger acid than something that just had an OH group is because the conjugate base, the carboxylate ion, is actually resonance stabilized. It is more stable than the conjugate base of an alcohol. Let me show you that. This thing can share its negative charge. It can give an electron."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It is more stable than the conjugate base of an alcohol. Let me show you that. This thing can share its negative charge. It can give an electron. Let me draw it. It can take this magenta electron, give it to this carbon, then this carbon will have an extra electron, so then it can give back an electron to this top oxygen. So then it can give back an electron to this top oxygen right there."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It can give an electron. Let me draw it. It can take this magenta electron, give it to this carbon, then this carbon will have an extra electron, so then it can give back an electron to this top oxygen. So then it can give back an electron to this top oxygen right there. So it is resonance stabilized with this structure right over here. Let me draw the same so it could look like this. That's too big."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So then it can give back an electron to this top oxygen right there. So it is resonance stabilized with this structure right over here. Let me draw the same so it could look like this. That's too big. Let me scroll down a little bit. It could look like this. Carbon, and now this took back this blue electron, so now that one of the bonds is gone and it now has, it started off with two lone pairs, so I want to draw that there, and now it has another lone pair."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "That's too big. Let me scroll down a little bit. It could look like this. Carbon, and now this took back this blue electron, so now that one of the bonds is gone and it now has, it started off with two lone pairs, so I want to draw that there, and now it has another lone pair. It has this electron, this electron, and now it has that blue electron right over there. And now this oxygen, this top oxygen, has a negative charge, and now the carbon has a double bond with this side oxygen. So now the carbon, let me go back to the yellow, that's the first bond with that oxygen."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Carbon, and now this took back this blue electron, so now that one of the bonds is gone and it now has, it started off with two lone pairs, so I want to draw that there, and now it has another lone pair. It has this electron, this electron, and now it has that blue electron right over there. And now this oxygen, this top oxygen, has a negative charge, and now the carbon has a double bond with this side oxygen. So now the carbon, let me go back to the yellow, that's the first bond with that oxygen. It had one, two lone pairs to begin with, and now it has this magenta bond. So this pink electron is at this end, and now this purple electron is at the other end, or this magenta electron, and now it has a double bond with this oxygen. And we know that when we have resonance stabilization, it's not like you're going back and forth."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So now the carbon, let me go back to the yellow, that's the first bond with that oxygen. It had one, two lone pairs to begin with, and now it has this magenta bond. So this pink electron is at this end, and now this purple electron is at the other end, or this magenta electron, and now it has a double bond with this oxygen. And we know that when we have resonance stabilization, it's not like you're going back and forth. The reality is that you kind of have a half double bond between both oxygens, that the electrons are just flowing across the whole place, and that stabilizes the molecule. And so to show that this is a resonance structure, let me put some brackets around it. And in general, if this R group right here is actually even better at withdrawing electrons, so if you put something that was really electronegative here, something that likes to hog electrons, it would make the carboxylate ion even more stable."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we know that when we have resonance stabilization, it's not like you're going back and forth. The reality is that you kind of have a half double bond between both oxygens, that the electrons are just flowing across the whole place, and that stabilizes the molecule. And so to show that this is a resonance structure, let me put some brackets around it. And in general, if this R group right here is actually even better at withdrawing electrons, so if you put something that was really electronegative here, something that likes to hog electrons, it would make the carboxylate ion even more stable. It would make the carboxylic acid even a better acid. So if you put something electronegative here, then you could imagine that some of this negative charge that we drew in these two resonance structures can be sucked to that R group, and then that would make it even more stable and would make the carboxylic acid even more acidic. Now, like in everything we've looked at, there are some common carboxylic acids that are not systematically named that it's probably a good idea to know."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And in general, if this R group right here is actually even better at withdrawing electrons, so if you put something that was really electronegative here, something that likes to hog electrons, it would make the carboxylate ion even more stable. It would make the carboxylic acid even a better acid. So if you put something electronegative here, then you could imagine that some of this negative charge that we drew in these two resonance structures can be sucked to that R group, and then that would make it even more stable and would make the carboxylic acid even more acidic. Now, like in everything we've looked at, there are some common carboxylic acids that are not systematically named that it's probably a good idea to know. And I'll start with one, and just to see the pattern that we've seen in other things. I mean, we've seen that this thing over here, if we have this, we call this acetaldehyde. This was the very simple aldehyde we studied."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now, like in everything we've looked at, there are some common carboxylic acids that are not systematically named that it's probably a good idea to know. And I'll start with one, and just to see the pattern that we've seen in other things. I mean, we've seen that this thing over here, if we have this, we call this acetaldehyde. This was the very simple aldehyde we studied. We saw that if we have something like this, which is a ketone, we call this acetone. So you could imagine what we're probably going to name, let me do this in a different color, what we're probably going to name this molecule. This molecule right over here."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This was the very simple aldehyde we studied. We saw that if we have something like this, which is a ketone, we call this acetone. So you could imagine what we're probably going to name, let me do this in a different color, what we're probably going to name this molecule. This molecule right over here. This is a carboxylic acid, clearly, and it has just that one methyl group, just like the acetaldehyde, just like the acetone. And so this is acetic acid. So the acetaldehyde, acetone, and acetic acid, for me, are fairly easy to memorize just because they all have the acete part."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This molecule right over here. This is a carboxylic acid, clearly, and it has just that one methyl group, just like the acetaldehyde, just like the acetone. And so this is acetic acid. So the acetaldehyde, acetone, and acetic acid, for me, are fairly easy to memorize just because they all have the acete part. They all have that as their prefix, and this part of the molecule is identical in every case. And the difference is the hydrogen, the methyl group over here, or the OH group, making this one a carboxylic acid, and the acetone, and all of that. Now, a couple of other ones that it wouldn't hurt for you to know is this one right over here."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the acetaldehyde, acetone, and acetic acid, for me, are fairly easy to memorize just because they all have the acete part. They all have that as their prefix, and this part of the molecule is identical in every case. And the difference is the hydrogen, the methyl group over here, or the OH group, making this one a carboxylic acid, and the acetone, and all of that. Now, a couple of other ones that it wouldn't hurt for you to know is this one right over here. This is, you could argue, even simpler than acetic acid. And this is formic acid. And then another one, and actually I recently did a general chemistry video with this, where we did a titration example, but this is essentially two carboxyl groups attached to each other."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now, a couple of other ones that it wouldn't hurt for you to know is this one right over here. This is, you could argue, even simpler than acetic acid. And this is formic acid. And then another one, and actually I recently did a general chemistry video with this, where we did a titration example, but this is essentially two carboxyl groups attached to each other. It looks like this. Two carboxyl groups attached to each other, and you can see them. This is one carboxyl group right over here, and then you have another carboxyl group right over here."}, {"video_title": "Carboxylic acid introduction Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then another one, and actually I recently did a general chemistry video with this, where we did a titration example, but this is essentially two carboxyl groups attached to each other. It looks like this. Two carboxyl groups attached to each other, and you can see them. This is one carboxyl group right over here, and then you have another carboxyl group right over here. And so this actually can be deprotonated twice. This hydrogen can be lost, and that hydrogen can be lost, and this is oxalic acid. I'll leave you there, and in the next video, we'll learn how to systematically name carboxylic acids."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Clearly a carboxylic acid. We have a carboxyl group right over here. Now to name it systematically, we do it just the way we've named our simple alkenes. When we first learned how to name any organic molecule, you look for the longest carbon chain. And the longest carbon chain is 1, 2, 3, 4 carbons. So our prefix will be bute. So it's butane."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "When we first learned how to name any organic molecule, you look for the longest carbon chain. And the longest carbon chain is 1, 2, 3, 4 carbons. So our prefix will be bute. So it's butane. And instead of calling it butane, instead of writing this e here, we know this is a carboxylic acid. It has this carboxyl group. So we call it butanoic acid."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it's butane. And instead of calling it butane, instead of writing this e here, we know this is a carboxylic acid. It has this carboxyl group. So we call it butanoic acid. And now you might wonder, hey, don't we have to specify where the carboxyl group is? And if you look at how carboxylic acids are arranged, you can tell that the carboxyl group is always going to be at one end of a carbon chain. So you don't have to specify."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we call it butanoic acid. And now you might wonder, hey, don't we have to specify where the carboxyl group is? And if you look at how carboxylic acids are arranged, you can tell that the carboxyl group is always going to be at one end of a carbon chain. So you don't have to specify. And in fact, you always want to start numbering at wherever the carboxyl carbon is. So if you had to number these, this would be the 1 carbon, the 2, the 3, and the 4. So you don't have to specify a number for the carboxyl group."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you don't have to specify. And in fact, you always want to start numbering at wherever the carboxyl carbon is. So if you had to number these, this would be the 1 carbon, the 2, the 3, and the 4. So you don't have to specify a number for the carboxyl group. Let's do another one. Let's say we had something that looked like this. Let's say we had something that looked like this."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you don't have to specify a number for the carboxyl group. Let's do another one. Let's say we had something that looked like this. Let's say we had something that looked like this. Let me put another carbon on there, just like that. And let's say that there's a methyl group right over there. Now, clearly a carboxylic acid, but to name it systematically, we just want to find the longest carbon chain."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we had something that looked like this. Let me put another carbon on there, just like that. And let's say that there's a methyl group right over there. Now, clearly a carboxylic acid, but to name it systematically, we just want to find the longest carbon chain. So we have 1, 2, 3, 4, 5, 6 carbons. So our prefix will be hex. So it's hexane."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now, clearly a carboxylic acid, but to name it systematically, we just want to find the longest carbon chain. So we have 1, 2, 3, 4, 5, 6 carbons. So our prefix will be hex. So it's hexane. It's clearly not just a hexane. It's a hexanoic acid. It has this carboxyl group right here."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it's hexane. It's clearly not just a hexane. It's a hexanoic acid. It has this carboxyl group right here. This is hexanoic acid. And we're not done, because we still have this methyl carbon right over here. And it is on the, we always want to start numbering at this carbonyl carbon, 1, 2, 3, 4, 5, 6."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It has this carboxyl group right here. This is hexanoic acid. And we're not done, because we still have this methyl carbon right over here. And it is on the, we always want to start numbering at this carbonyl carbon, 1, 2, 3, 4, 5, 6. It is at the number 3 carbon. So this is 3-methylhexanoic acid. Let's do one more."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And it is on the, we always want to start numbering at this carbonyl carbon, 1, 2, 3, 4, 5, 6. It is at the number 3 carbon. So this is 3-methylhexanoic acid. Let's do one more. Let's say we had a molecule that looked like this. Let's see, that's 1, 2, 3, 4, 5, 6, 7 carbons. And then we have our carboxyl group, just like that."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more. Let's say we had a molecule that looked like this. Let's see, that's 1, 2, 3, 4, 5, 6, 7 carbons. And then we have our carboxyl group, just like that. And let's say that we had a double bond right over there. What would we call this? Well, once again, look for the longest carbon chain."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our carboxyl group, just like that. And let's say that we had a double bond right over there. What would we call this? Well, once again, look for the longest carbon chain. We have 1, 2, 3, 4, 5, 6, 7 carbons. So the prefix is hept. Actually, let me be careful."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Well, once again, look for the longest carbon chain. We have 1, 2, 3, 4, 5, 6, 7 carbons. So the prefix is hept. Actually, let me be careful. This isn't an alkane. This has a double bond right here. This has a double bond right over here."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me be careful. This isn't an alkane. This has a double bond right here. This has a double bond right over here. So it's heptene. If this was just an alkene, we would just call it heptene. But we're not going to put this last e here, because this is a carboxylic acid."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This has a double bond right over here. So it's heptene. If this was just an alkene, we would just call it heptene. But we're not going to put this last e here, because this is a carboxylic acid. And to specify where that double bond is, we need to start numbering. And we start numbering at the carbonyl carbon, 1, 2, 3, 4, 5, 6, 7. So you could either name this 3-heptene."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But we're not going to put this last e here, because this is a carboxylic acid. And to specify where that double bond is, we need to start numbering. And we start numbering at the carbonyl carbon, 1, 2, 3, 4, 5, 6, 7. So you could either name this 3-heptene. And well, I haven't finished it yet. I haven't put this final e over here. Or you could name it hept-3-ene, just like that."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you could either name this 3-heptene. And well, I haven't finished it yet. I haven't put this final e over here. Or you could name it hept-3-ene, just like that. This is the more typical one that you would see, because it tells you, OK, we have a double bond. And it starts at the number 3 carbon, goes from the 3 to the 4 carbon. But this isn't just a regular alkene."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Or you could name it hept-3-ene, just like that. This is the more typical one that you would see, because it tells you, OK, we have a double bond. And it starts at the number 3 carbon, goes from the 3 to the 4 carbon. But this isn't just a regular alkene. This is a carboxylic acid. So instead of writing that final e here for an alkene, we write it as, we have a carboxyl group right here. So this is 3-heptenoic acid."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But this isn't just a regular alkene. This is a carboxylic acid. So instead of writing that final e here for an alkene, we write it as, we have a carboxyl group right here. So this is 3-heptenoic acid. And we are done. And actually, if you wanted to get really fancy on this one right over here, you could see that these two carbons that are on the double bond, so this carbon and this carbon, it's kind of arranged like this. If you view this as, let me draw it like this."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this is 3-heptenoic acid. And we are done. And actually, if you wanted to get really fancy on this one right over here, you could see that these two carbons that are on the double bond, so this carbon and this carbon, it's kind of arranged like this. If you view this as, let me draw it like this. They both have other hydrogens out there that we didn't draw. They're implicitly there. But if you wanted to rewrite or redraw this molecule, you could draw it like this."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "If you view this as, let me draw it like this. They both have other hydrogens out there that we didn't draw. They're implicitly there. But if you wanted to rewrite or redraw this molecule, you could draw it like this. You have two carbons just like this. This one has a hydrogen popping up like that. That one has a hydrogen popping down like that."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But if you wanted to rewrite or redraw this molecule, you could draw it like this. You have two carbons just like this. This one has a hydrogen popping up like that. That one has a hydrogen popping down like that. And then this carbon over here has this big functional group over here. We'll call that R. So this is R. And then this one over here, I'll do it in green, has this other functional group, has these three carbons. We could call that R prime."}, {"video_title": "Carboxylic acid naming Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "That one has a hydrogen popping down like that. And then this carbon over here has this big functional group over here. We'll call that R. So this is R. And then this one over here, I'll do it in green, has this other functional group, has these three carbons. We could call that R prime. These three carbons are R prime. And if you look at it this way, you're like, hey, look, the functional groups are on opposite sides of the double bond, they're away from each other. So if you wanted to, you could also call this trans-3-heptanoic acid, and this will specify that these guys are on opposite ends."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "What I want to do in this video is explore what happens when we get to really, really, really small scales. Before we even think about it, I want to familiarize ourselves with the units here. So we're all familiar with what a meter looks like. The average adult male is a little under 2 meters. If you were to divide a meter into a thousand units, you would get a millimeter. I think we probably know what a millimeter is if you've ever looked at a meter stick. It's the smallest measurement on that meter stick."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "The average adult male is a little under 2 meters. If you were to divide a meter into a thousand units, you would get a millimeter. I think we probably know what a millimeter is if you've ever looked at a meter stick. It's the smallest measurement on that meter stick. So it's already pretty hard to look at. If you were to divide each of those millimeters into a thousand sections, you'd get a micrometer. Or another way to think about a micrometer is it's one millionth of a meter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's the smallest measurement on that meter stick. So it's already pretty hard to look at. If you were to divide each of those millimeters into a thousand sections, you'd get a micrometer. Or another way to think about a micrometer is it's one millionth of a meter. So this is kind of beyond what we're capable of really perceiving. If you were to take each of those micrometers and divide them into a thousand sections, you would get a nanometer. So now we're at one billionth of a meter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Or another way to think about a micrometer is it's one millionth of a meter. So this is kind of beyond what we're capable of really perceiving. If you were to take each of those micrometers and divide them into a thousand sections, you would get a nanometer. So now we're at one billionth of a meter. You divide that by a thousand. You get a picometer. So a picometer is one thousand billionth of a meter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So now we're at one billionth of a meter. You divide that by a thousand. You get a picometer. So a picometer is one thousand billionth of a meter. Or you could say a trillionth of a meter. You divide one of those by a thousand and you would get a femtometer. So these are unimaginably small things."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So a picometer is one thousand billionth of a meter. Or you could say a trillionth of a meter. You divide one of those by a thousand and you would get a femtometer. So these are unimaginably small things. Now once you're familiar with the units, let's explore what types of things we can expect to find at these different scales. I'll start over here, and I've written them on the left as well, but it's more compelling when you see the pictures. We'll start over here with the b. I've arbitrarily picked something of this scale."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So these are unimaginably small things. Now once you're familiar with the units, let's explore what types of things we can expect to find at these different scales. I'll start over here, and I've written them on the left as well, but it's more compelling when you see the pictures. We'll start over here with the b. I've arbitrarily picked something of this scale. There's many, many, many, almost an infinite number of things I could have picked at this scale. But the average b is about two centimeters long. This b right over here."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "We'll start over here with the b. I've arbitrarily picked something of this scale. There's many, many, many, almost an infinite number of things I could have picked at this scale. But the average b is about two centimeters long. This b right over here. It's about, give or take, one hundredth the length of the average adult human being. But once again, a honey bee, not too exciting, although it is pretty exciting to see it zoomed in like this. But a honey bee is something that we can relate to."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This b right over here. It's about, give or take, one hundredth the length of the average adult human being. But once again, a honey bee, not too exciting, although it is pretty exciting to see it zoomed in like this. But a honey bee is something that we can relate to. We've all seen honey bees. Now, what I want to do is zoom in or look at something that's fifty times smaller than a honey bee. So something that if I were to kind of show how big it is relative to this honey bee, it would look something like this."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But a honey bee is something that we can relate to. We've all seen honey bees. Now, what I want to do is zoom in or look at something that's fifty times smaller than a honey bee. So something that if I were to kind of show how big it is relative to this honey bee, it would look something like this. I'm doing it very rough. And that is a dust mite. And this right here, these are both pictures of dust mites."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So something that if I were to kind of show how big it is relative to this honey bee, it would look something like this. I'm doing it very rough. And that is a dust mite. And this right here, these are both pictures of dust mites. Now, dust mites look like these strange and alien creatures. But what's amazing about them is that they are everywhere. They're all around us."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And this right here, these are both pictures of dust mites. Now, dust mites look like these strange and alien creatures. But what's amazing about them is that they are everywhere. They're all around us. You probably have many of them lying on your skin or wherever right now, which is kind of a creepy idea. But we're talking about scale here. And the average dust mite, so we were talking about centimeters before."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "They're all around us. You probably have many of them lying on your skin or wherever right now, which is kind of a creepy idea. But we're talking about scale here. And the average dust mite, so we were talking about centimeters before. Now we'll talk about millimeters. The average dust mite is less than half of a millimeter. Or if you want to talk in micrometers, it's about 400 micrometers long."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And the average dust mite, so we were talking about centimeters before. Now we'll talk about millimeters. The average dust mite is less than half of a millimeter. Or if you want to talk in micrometers, it's about 400 micrometers long. So this length right over here is about 400 micrometers, so about 150th the length. Remember, this huge thing that I'm showing right here, this is a honey bee. It's about 150th the length of a honey bee."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Or if you want to talk in micrometers, it's about 400 micrometers long. So this length right over here is about 400 micrometers, so about 150th the length. Remember, this huge thing that I'm showing right here, this is a honey bee. It's about 150th the length of a honey bee. Or maybe to put it in other terms that you might be familiar with, this is a zoomed in picture of human hair. And you might say, oh my god, this person has horrible hair. But no, if you were to look at your own hair under an electron microscope, you'd be lucky if it looked this good."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's about 150th the length of a honey bee. Or maybe to put it in other terms that you might be familiar with, this is a zoomed in picture of human hair. And you might say, oh my god, this person has horrible hair. But no, if you were to look at your own hair under an electron microscope, you'd be lucky if it looked this good. This person actually I've seen pictures of more damaged hair than this. This is probably smooth and silky hair right here. But the diameter of human hair, and this is on average."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But no, if you were to look at your own hair under an electron microscope, you'd be lucky if it looked this good. This person actually I've seen pictures of more damaged hair than this. This is probably smooth and silky hair right here. But the diameter of human hair, and this is on average. It depends on whose hair you're talking about. The diameter of human hair is about 100 micrometers thick. That's the diameter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But the diameter of human hair, and this is on average. It depends on whose hair you're talking about. The diameter of human hair is about 100 micrometers thick. That's the diameter. So it's about a fourth the length of a dust mite. Or if I were to draw some human hair relative to this honey bee, it would look something like this. And I'm drawing the whole hair, so its width would be the width of this thing that I just drew."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "That's the diameter. So it's about a fourth the length of a dust mite. Or if I were to draw some human hair relative to this honey bee, it would look something like this. And I'm drawing the whole hair, so its width would be the width of this thing that I just drew. Remember, we're looking at a honey bee here. It looks like some type of giant, but it is a honey bee. Let's zoom in even more."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And I'm drawing the whole hair, so its width would be the width of this thing that I just drew. Remember, we're looking at a honey bee here. It looks like some type of giant, but it is a honey bee. Let's zoom in even more. So we started with the honey bee. We zoomed in by 50 to get the dust mite. We zoomed in by another factor of 4 to get the width of human hair."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Let's zoom in even more. So we started with the honey bee. We zoomed in by 50 to get the dust mite. We zoomed in by another factor of 4 to get the width of human hair. If we zoom in, we're in the micrometer range now. If we zoom in by roughly another factor of 10, we get to the scale of cells. And this right here is a red blood cell."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "We zoomed in by another factor of 4 to get the width of human hair. If we zoom in, we're in the micrometer range now. If we zoom in by roughly another factor of 10, we get to the scale of cells. And this right here is a red blood cell. I think this is a white blood cell right over here. About 6 to 8 micrometers. So once again, if I were to draw a cell relative to this human hair, it would probably look something like this."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And this right here is a red blood cell. I think this is a white blood cell right over here. About 6 to 8 micrometers. So once again, if I were to draw a cell relative to this human hair, it would probably look something like this. Something on a similar scale that we can still kind of relate to is the width of spider silk. It's about 3 to 8 micrometers. So if I were to draw some spider silk on the same diagram, it would look something like this."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So once again, if I were to draw a cell relative to this human hair, it would probably look something like this. Something on a similar scale that we can still kind of relate to is the width of spider silk. It's about 3 to 8 micrometers. So if I were to draw some spider silk on the same diagram, it would look something like this. This is an actual image of spider silk. So once again, something that we can kind of perceive. You can bump into it."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So if I were to draw some spider silk on the same diagram, it would look something like this. This is an actual image of spider silk. So once again, something that we can kind of perceive. You can bump into it. You can touch spider silk. You can see it if the sun is reflecting just right or if it has a little bit of moisture on it. But it's about the thinnest thing that humans can perceive."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "You can bump into it. You can touch spider silk. You can see it if the sun is reflecting just right or if it has a little bit of moisture on it. But it's about the thinnest thing that humans can perceive. And this is in the ones of micrometer range. At that same range, you start to have some of your larger bacteria. Bacteria can be anywhere from, and I'm speaking very roughly, 1 to 10 micrometers."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But it's about the thinnest thing that humans can perceive. And this is in the ones of micrometer range. At that same range, you start to have some of your larger bacteria. Bacteria can be anywhere from, and I'm speaking very roughly, 1 to 10 micrometers. So in general, they're smaller than cells. Most bacteria are smaller than most cells. And just to figure out where we sit on our scale, have it over here."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Bacteria can be anywhere from, and I'm speaking very roughly, 1 to 10 micrometers. So in general, they're smaller than cells. Most bacteria are smaller than most cells. And just to figure out where we sit on our scale, have it over here. So we started off, I want to keep reminding ourselves, humans, you divide by 100, you get to the B. So each of these slashes right here are dividing by 10. So this is divide by 10."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And just to figure out where we sit on our scale, have it over here. So we started off, I want to keep reminding ourselves, humans, you divide by 100, you get to the B. So each of these slashes right here are dividing by 10. So this is divide by 10. Divide by 10 again, you're divided in size by 100. Divide by 10 again, you get to millimeter. You've divided by 1,000."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this is divide by 10. Divide by 10 again, you're divided in size by 100. Divide by 10 again, you get to millimeter. You've divided by 1,000. Divide by 10 again, you are doing tenths of millimeters, which is about the size of the human hair. You divide again by 10, you're going into tens of micrometers. By 10 again, you get into the micrometer range."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "You've divided by 1,000. Divide by 10 again, you are doing tenths of millimeters, which is about the size of the human hair. You divide again by 10, you're going into tens of micrometers. By 10 again, you get into the micrometer range. So now we're talking about cells, we're talking about bacteria. Now things are going to get really crazy. This was in the ones of micrometer range."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "By 10 again, you get into the micrometer range. So now we're talking about cells, we're talking about bacteria. Now things are going to get really crazy. This was in the ones of micrometer range. Now we're going to start getting into the hundreds of nanometer range. Just to get a sense of things. So remember, a nanometer is a thousandth of a micrometer."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This was in the ones of micrometer range. Now we're going to start getting into the hundreds of nanometer range. Just to get a sense of things. So remember, a nanometer is a thousandth of a micrometer. Or 100 nanometers would be a tenth of a micrometer. And this picture right here, this big, enormous planet or asteroid looking thing, this is a white blood cell. The enormous blue thing in this picture."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So remember, a nanometer is a thousandth of a micrometer. Or 100 nanometers would be a tenth of a micrometer. And this picture right here, this big, enormous planet or asteroid looking thing, this is a white blood cell. The enormous blue thing in this picture. And so if I were to zoom out, it might look something like this right over here. But what's really fascinating about this picture for multiple reasons are these little green things that are emerging, that are essentially reproducing, emerging from the surface of this white blood cell. And these things right here, these are AIDS viruses."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "The enormous blue thing in this picture. And so if I were to zoom out, it might look something like this right over here. But what's really fascinating about this picture for multiple reasons are these little green things that are emerging, that are essentially reproducing, emerging from the surface of this white blood cell. And these things right here, these are AIDS viruses. So now if we zoom in, roughly another factor of about 100 to 1000 from the size of a cell, you're now getting to the size of a virus. And all of the genetic material necessary to replicate that virus is right inside each of these little capsids, right inside each of these little green containers. So now going back to our scale, we are down to the scale of a virus."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And these things right here, these are AIDS viruses. So now if we zoom in, roughly another factor of about 100 to 1000 from the size of a cell, you're now getting to the size of a virus. And all of the genetic material necessary to replicate that virus is right inside each of these little capsids, right inside each of these little green containers. So now going back to our scale, we are down to the scale of a virus. So we're in the hundreds of nanometer range. If we divide by 10 and then divide by 10, you get to the nanometer range. And right in the ones of nanometer range, you get to the width of the double helix of a DNA molecule."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So now going back to our scale, we are down to the scale of a virus. So we're in the hundreds of nanometer range. If we divide by 10 and then divide by 10, you get to the nanometer range. And right in the ones of nanometer range, you get to the width of the double helix of a DNA molecule. So this right here is, if you were to zoom in, and this is an artist's depiction of it, obviously this is not a picture, so to speak, of a DNA molecule. But the width of this double helix is about 2 nanometers, or another way to think about it, about 160th the diameter of one of these viral capsids. It would have to be, because it's going to have to get all wound up and fit into one of these viral capsids."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And right in the ones of nanometer range, you get to the width of the double helix of a DNA molecule. So this right here is, if you were to zoom in, and this is an artist's depiction of it, obviously this is not a picture, so to speak, of a DNA molecule. But the width of this double helix is about 2 nanometers, or another way to think about it, about 160th the diameter of one of these viral capsids. It would have to be, because it's going to have to get all wound up and fit into one of these viral capsids. And DNA, just to make it clear, this is just the width of DNA. It's much, much, much, much, much, much longer, and we can talk about that in future videos. So once again, we're at a very, very small scale."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It would have to be, because it's going to have to get all wound up and fit into one of these viral capsids. And DNA, just to make it clear, this is just the width of DNA. It's much, much, much, much, much, much longer, and we can talk about that in future videos. So once again, we're at a very, very small scale. If you want to think of it in terms of meters, we're at 2 billionths of a meter. You could put 500 million of these side by side to get to a meter. Or you could even think of it this way."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So once again, we're at a very, very small scale. If you want to think of it in terms of meters, we're at 2 billionths of a meter. You could put 500 million of these side by side to get to a meter. Or you could even think of it this way. This is 2 millionths of a millimeter. So once again, super small. You could put these side by side, one DNA, another DNA."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Or you could even think of it this way. This is 2 millionths of a millimeter. So once again, super small. You could put these side by side, one DNA, another DNA. If you made them touch, you could put 500,000 next to each other in a millimeter. So this is an unbelievably small amount of space. It's going to fit into another unit that's not kind of in the conventional prefix followed by meters, and this is an angstrom."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "You could put these side by side, one DNA, another DNA. If you made them touch, you could put 500,000 next to each other in a millimeter. So this is an unbelievably small amount of space. It's going to fit into another unit that's not kind of in the conventional prefix followed by meters, and this is an angstrom. And 10 angstroms equal 1 nanometer. So the width of this DNA double helix would be 2 nanometers or 20 angstroms. Now, if we were to divide again by 10, you get to something that's 2 angstroms or 0.2 nanometers wide, and that is a water molecule."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's going to fit into another unit that's not kind of in the conventional prefix followed by meters, and this is an angstrom. And 10 angstroms equal 1 nanometer. So the width of this DNA double helix would be 2 nanometers or 20 angstroms. Now, if we were to divide again by 10, you get to something that's 2 angstroms or 0.2 nanometers wide, and that is a water molecule. Maybe instead of using red, I should have used blue or something. But this right here is the oxygen, and it is bonded to the two hydrogens right over here. So this is beyond, frankly, human perception, or even really stuff that we can conceptualize, not to even speak of perception."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Now, if we were to divide again by 10, you get to something that's 2 angstroms or 0.2 nanometers wide, and that is a water molecule. Maybe instead of using red, I should have used blue or something. But this right here is the oxygen, and it is bonded to the two hydrogens right over here. So this is beyond, frankly, human perception, or even really stuff that we can conceptualize, not to even speak of perception. We're still imagining how small we're dealing with right over here. Remember, we're dealing with less than a fifth of a billionth of a meter, or a fifth of a millionth of a millimeter, something that I really can't fathom. But we're going to get even smaller than that."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this is beyond, frankly, human perception, or even really stuff that we can conceptualize, not to even speak of perception. We're still imagining how small we're dealing with right over here. Remember, we're dealing with less than a fifth of a billionth of a meter, or a fifth of a millionth of a millimeter, something that I really can't fathom. But we're going to get even smaller than that. If we were to zoom in on one of these hydrogen atoms, and now things start to get kind of abstract, and we start dealing in the quantum realm, and it's hard to define where one thing ends and one thing begins, and what is real and what is not real, and all of that silliness. But if we try our best to do it, if we were to zoom in and we were to put some boundary on a hydrogen atom, because the electrons actually could jump around anywhere, but if we set some boundary of where the electrons are most likely to be found, the diameter of a hydrogen atom is roughly one angstrom, which makes sense from this diagram, too. It's about half of the diameter of this water molecule."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But we're going to get even smaller than that. If we were to zoom in on one of these hydrogen atoms, and now things start to get kind of abstract, and we start dealing in the quantum realm, and it's hard to define where one thing ends and one thing begins, and what is real and what is not real, and all of that silliness. But if we try our best to do it, if we were to zoom in and we were to put some boundary on a hydrogen atom, because the electrons actually could jump around anywhere, but if we set some boundary of where the electrons are most likely to be found, the diameter of a hydrogen atom is roughly one angstrom, which makes sense from this diagram, too. It's about half of the diameter of this water molecule. What's extra crazy is, one, this atom is super, super, duper small, something that we can't, you know, this is one ten billionth of a meter, or one ten millionth of a millimeter, so something we really, really can't fathom. But what's crazier than that is that it's mostly free space. We've gotten this small."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's about half of the diameter of this water molecule. What's extra crazy is, one, this atom is super, super, duper small, something that we can't, you know, this is one ten billionth of a meter, or one ten millionth of a millimeter, so something we really, really can't fathom. But what's crazier than that is that it's mostly free space. We've gotten this small. We're trying to get to these fundamental units, and this thing right here is mostly free space. And that's because if you look at an electron, and when we say radius here, it's really hard to define where it starts and ends, and you have to do some things related to the charge, and we're not even thinking about quantum effects and all of that. An electron has a radius of 3 times 10 to the negative 5th angstroms, and the nucleus of a hydrogen atom, which is really just a proton, has a radius a little bit, and, you know, don't even worry about this number right here."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "We've gotten this small. We're trying to get to these fundamental units, and this thing right here is mostly free space. And that's because if you look at an electron, and when we say radius here, it's really hard to define where it starts and ends, and you have to do some things related to the charge, and we're not even thinking about quantum effects and all of that. An electron has a radius of 3 times 10 to the negative 5th angstroms, and the nucleus of a hydrogen atom, which is really just a proton, has a radius a little bit, and, you know, don't even worry about this number right here. The general idea is it's the same order of magnitude. It's about one ten thousandth of an angstrom. And just to give a sense of what it's like, if you view the entire atomic radius to be about an angstrom, kind of just have a conception for scale of the atom and how much free space there is in an atom, if we even want to think what is free space."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "An electron has a radius of 3 times 10 to the negative 5th angstroms, and the nucleus of a hydrogen atom, which is really just a proton, has a radius a little bit, and, you know, don't even worry about this number right here. The general idea is it's the same order of magnitude. It's about one ten thousandth of an angstrom. And just to give a sense of what it's like, if you view the entire atomic radius to be about an angstrom, kind of just have a conception for scale of the atom and how much free space there is in an atom, if we even want to think what is free space. Imagine a nucleus being maybe a marble at the center of a football stadium, of a domed football stadium, and imagine an electron being a honeybee just randomly jumping around random parts of that entire volume inside of that football stadium. And obviously it's a quantum honeybee, so it can jump around from spot to spot, and it's not easy to predict where it's going to go next and all of the rest. But that will give you a sense of the scale of the electron and the proton relative to the atom as a whole."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we start with our benzene ring. And to benzene, we're going to add an alkyl chloride. And our catalyst is aluminum chloride. And the end result is to substitute an R group, the R group that was on the alkyl chloride, for a proton on the aromatic ring. Let's look at the mechanism for Friedel-Crafts alkylation. We start with our alkyl chloride. And we add our aluminum chloride, which we've already seen can function as a Lewis acid."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the end result is to substitute an R group, the R group that was on the alkyl chloride, for a proton on the aromatic ring. Let's look at the mechanism for Friedel-Crafts alkylation. We start with our alkyl chloride. And we add our aluminum chloride, which we've already seen can function as a Lewis acid. So it can be a Lewis acid because it can accept a pair of electrons. And so our Lewis base over here is going to be the alkyl chloride. It's going to donate an electron pair."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we add our aluminum chloride, which we've already seen can function as a Lewis acid. So it can be a Lewis acid because it can accept a pair of electrons. And so our Lewis base over here is going to be the alkyl chloride. It's going to donate an electron pair. So I'm going to say this electron pair is going to be donated to that aluminum. So we go ahead and draw the result of our Lewis acid base reaction here. So now the chlorine is bonded to the aluminum."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's going to donate an electron pair. So I'm going to say this electron pair is going to be donated to that aluminum. So we go ahead and draw the result of our Lewis acid base reaction here. So now the chlorine is bonded to the aluminum. And the aluminum is bonded to these other chlorines here. And I'm going to leave off the lone pairs of electrons on those just to save time. In terms of formal charges, aluminum now has a negative 1 formal charge."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So now the chlorine is bonded to the aluminum. And the aluminum is bonded to these other chlorines here. And I'm going to leave off the lone pairs of electrons on those just to save time. In terms of formal charges, aluminum now has a negative 1 formal charge. And this chlorine now has a plus 1 formal charge like that. So we know that halogens are relatively electronegative. And so this chlorine here could take these electrons."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In terms of formal charges, aluminum now has a negative 1 formal charge. And this chlorine now has a plus 1 formal charge like that. So we know that halogens are relatively electronegative. And so this chlorine here could take these electrons. We could show these electrons in here moving off onto that chlorine. And when we do that, we're taking a bond away from our alkyl group. And so therefore, our alkyl group is left with a positive 1 formal charge."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this chlorine here could take these electrons. We could show these electrons in here moving off onto that chlorine. And when we do that, we're taking a bond away from our alkyl group. And so therefore, our alkyl group is left with a positive 1 formal charge. And this is a carbocation. And this is our electrophile in our mechanism for electrophilic aromatic substitution. And since we actually form a carbocation in this mechanism, a carbocation intermediate, rearrangement is actually possible."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, our alkyl group is left with a positive 1 formal charge. And this is a carbocation. And this is our electrophile in our mechanism for electrophilic aromatic substitution. And since we actually form a carbocation in this mechanism, a carbocation intermediate, rearrangement is actually possible. So we have to be careful when we're looking at a Friedel-Crafts alkylation reaction in terms of predicting the products. Our other product here, we would have our aluminum still bonded to these chlorines. And of course, it's bonded to this other chlorine too."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And since we actually form a carbocation in this mechanism, a carbocation intermediate, rearrangement is actually possible. So we have to be careful when we're looking at a Friedel-Crafts alkylation reaction in terms of predicting the products. Our other product here, we would have our aluminum still bonded to these chlorines. And of course, it's bonded to this other chlorine too. And this chlorine now has three lone pairs of electrons around it. So these electrons in here, the ones that were in the bond between our R group and our chlorine left. And they are now on our chlorine like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, it's bonded to this other chlorine too. And this chlorine now has three lone pairs of electrons around it. So these electrons in here, the ones that were in the bond between our R group and our chlorine left. And they are now on our chlorine like that. And so our aluminum still has a negative 1 formal charge in there like that. All right, so now we have generated our electrophile, which is our carbocation. And we know, of course, that our benzene ring is going to react with our electrophile."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And they are now on our chlorine like that. And so our aluminum still has a negative 1 formal charge in there like that. All right, so now we have generated our electrophile, which is our carbocation. And we know, of course, that our benzene ring is going to react with our electrophile. So here's our benzene ring. And here is our carbocation, so positively charged. So we know that the pi electrons in our benzene ring are going to function as a nucleophile and attack the electrophile."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we know, of course, that our benzene ring is going to react with our electrophile. So here's our benzene ring. And here is our carbocation, so positively charged. So we know that the pi electrons in our benzene ring are going to function as a nucleophile and attack the electrophile. So negative charges are attracted to positive charges like that. And so we can go ahead and draw the results of our nucleophilic attack. So we now have our hydrogen still bonded here."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we know that the pi electrons in our benzene ring are going to function as a nucleophile and attack the electrophile. So negative charges are attracted to positive charges like that. And so we can go ahead and draw the results of our nucleophilic attack. So we now have our hydrogen still bonded here. And once again, I could add the alkyl group to either of those two carbons. Let me just highlight those again. So either one of these two."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we now have our hydrogen still bonded here. And once again, I could add the alkyl group to either of those two carbons. Let me just highlight those again. So either one of these two. But just to be consistent with how I've been doing it in the previous videos, I'm going to add the R group to the top carbon here, which means that I took a bond away from the bottom carbon. So let me go ahead and highlight that one. So this, of course, lost a bond, which means that is where our positive 1 formal charge would go."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So either one of these two. But just to be consistent with how I've been doing it in the previous videos, I'm going to add the R group to the top carbon here, which means that I took a bond away from the bottom carbon. So let me go ahead and highlight that one. So this, of course, lost a bond, which means that is where our positive 1 formal charge would go. So positive 1 formal charge, a carbocation. And just to save time, I'm not going to draw the other resonance structures for this ion. So remember, this will represent our sigma complex."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this, of course, lost a bond, which means that is where our positive 1 formal charge would go. So positive 1 formal charge, a carbocation. And just to save time, I'm not going to draw the other resonance structures for this ion. So remember, this will represent our sigma complex. And you can watch the previous videos for how to draw resonance structures for this carbocation. And so in the last step of our mechanism here, we're going to reform our aromatic ring. And so we need to deprotonate our sigma complex in order to do that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So remember, this will represent our sigma complex. And you can watch the previous videos for how to draw resonance structures for this carbocation. And so in the last step of our mechanism here, we're going to reform our aromatic ring. And so we need to deprotonate our sigma complex in order to do that. So we could think about these electrons in here moving out to take that proton and these electrons moving in here to take away the plus 1 formal charge and reform our aromatic ring. So now we have our aromatic ring reformed. Our R group is on our ring like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we need to deprotonate our sigma complex in order to do that. So we could think about these electrons in here moving out to take that proton and these electrons moving in here to take away the plus 1 formal charge and reform our aromatic ring. So now we have our aromatic ring reformed. Our R group is on our ring like that. Our other products would be HCl. And also we would regenerate our catalyst, aluminum chloride. Like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Our R group is on our ring like that. Our other products would be HCl. And also we would regenerate our catalyst, aluminum chloride. Like that. So let's go ahead and follow some electrons. So these electrons in magenta right here, when the proton leaves, those electrons in magenta reform this right here. And I can say that these electrons in here are forming a bond between the chlorine and the proton like that to form HCl."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Like that. So let's go ahead and follow some electrons. So these electrons in magenta right here, when the proton leaves, those electrons in magenta reform this right here. And I can say that these electrons in here are forming a bond between the chlorine and the proton like that to form HCl. So that's our mechanism for Friedel-Crafts alkylation. Let's look at a few examples of this reaction. So we'll start with benzene."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I can say that these electrons in here are forming a bond between the chlorine and the proton like that to form HCl. So that's our mechanism for Friedel-Crafts alkylation. Let's look at a few examples of this reaction. So we'll start with benzene. And to benzene, we are going to add 2-chloropropane. So go ahead and draw 2-chloropropane. That's going to be our alkyl chloride."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start with benzene. And to benzene, we are going to add 2-chloropropane. So go ahead and draw 2-chloropropane. That's going to be our alkyl chloride. And once again, we need our aluminum chloride as our catalyst. And so the way I do Friedel-Crafts alkylation reactions is I think about what sort of a carbocation am I going to get. And so without drawing the entire mechanism, I know that this chlorine is going to leave during the mechanism."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That's going to be our alkyl chloride. And once again, we need our aluminum chloride as our catalyst. And so the way I do Friedel-Crafts alkylation reactions is I think about what sort of a carbocation am I going to get. And so without drawing the entire mechanism, I know that this chlorine is going to leave during the mechanism. And if the chlorine leaves during the mechanism, well, that means that we took a bond away from this carbon right here. And so that's where our plus 1 formal charge is going to be. So we have a plus 1 formal charge on that top carbon right there."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so without drawing the entire mechanism, I know that this chlorine is going to leave during the mechanism. And if the chlorine leaves during the mechanism, well, that means that we took a bond away from this carbon right here. And so that's where our plus 1 formal charge is going to be. So we have a plus 1 formal charge on that top carbon right there. If we think about a possible rearrangement, this is a secondary carbocation. And there's no possible way that this carbocation could rearrange to form a tertiary carbocation. And so secondary carbocations are relatively stable."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a plus 1 formal charge on that top carbon right there. If we think about a possible rearrangement, this is a secondary carbocation. And there's no possible way that this carbocation could rearrange to form a tertiary carbocation. And so secondary carbocations are relatively stable. And so this will be the carbocation that reacts with our benzene ring. And so you can think about, once again, these pi electrons forming a bond with that carbon. So let's go ahead and draw the results of our nucleophilic attack."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so secondary carbocations are relatively stable. And so this will be the carbocation that reacts with our benzene ring. And so you can think about, once again, these pi electrons forming a bond with that carbon. So let's go ahead and draw the results of our nucleophilic attack. So we have our ring. We have our other pi electrons. We have our hydrogen that's already on our ring."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the results of our nucleophilic attack. So we have our ring. We have our other pi electrons. We have our hydrogen that's already on our ring. And now we formed a bond to three carbons. So let's go ahead and, once again, highlight some electrons here. So these electrons in magenta have now formed a bond between those two carbons right there."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our hydrogen that's already on our ring. And now we formed a bond to three carbons. So let's go ahead and, once again, highlight some electrons here. So these electrons in magenta have now formed a bond between those two carbons right there. And always double check yourself and count your carbons. So we have one, two, three carbons on that carbocation. So we have to make sure we have the same number over here, one, two, and three."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in magenta have now formed a bond between those two carbons right there. And always double check yourself and count your carbons. So we have one, two, three carbons on that carbocation. So we have to make sure we have the same number over here, one, two, and three. So in our last step, we know that deprotonation of our sigma complex, so this is a positive 1-formyl charge here, deprotonation of our sigma complex, we get rid of that positive 1-formyl charge and reform our aromatic ring. So we've seen that these electrons would move in here. And we have our product."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have to make sure we have the same number over here, one, two, and three. So in our last step, we know that deprotonation of our sigma complex, so this is a positive 1-formyl charge here, deprotonation of our sigma complex, we get rid of that positive 1-formyl charge and reform our aromatic ring. So we've seen that these electrons would move in here. And we have our product. So we have our aromatic ring like that. And then we have this isopropyl group coming off of our ring. So we form isopropyl benzene as our final product."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have our product. So we have our aromatic ring like that. And then we have this isopropyl group coming off of our ring. So we form isopropyl benzene as our final product. So let's do one more Fudelkraft's alkylation. This one will have a possible rearrangement. So let's go ahead and start with benzene again."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we form isopropyl benzene as our final product. So let's do one more Fudelkraft's alkylation. This one will have a possible rearrangement. So let's go ahead and start with benzene again. So we have our benzene ring. And to benzene, we are going to add butyl chloride here. So one, two, three, four carbons, and then a chlorine like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and start with benzene again. So we have our benzene ring. And to benzene, we are going to add butyl chloride here. So one, two, three, four carbons, and then a chlorine like that. And then aluminum chloride once again as our catalyst. And so if you're approaching this problem, once again, I like to think about what sort of a carbocation is going to form in the mechanism. So I know that this chlorine would kick off during the mechanism."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So one, two, three, four carbons, and then a chlorine like that. And then aluminum chloride once again as our catalyst. And so if you're approaching this problem, once again, I like to think about what sort of a carbocation is going to form in the mechanism. So I know that this chlorine would kick off during the mechanism. And so that would take away a bond from this carbon right here. So we're going to get a plus 1-formyl charge on that carbon. So if I go ahead and draw out my four carbons here, and I know there's going to be a plus 1-formyl charge on that carbon on the end, which we, of course, know this is a primary carbocation."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I know that this chlorine would kick off during the mechanism. And so that would take away a bond from this carbon right here. So we're going to get a plus 1-formyl charge on that carbon. So if I go ahead and draw out my four carbons here, and I know there's going to be a plus 1-formyl charge on that carbon on the end, which we, of course, know this is a primary carbocation. And primary carbocations are not very stable. So there could be some rearrangement here. And if you think about what's the best way to rearrange our carbocation to see if we can form a more stable one, you know that there are hydrogens attached to the carbon next door to that positive 1-formyl charge."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I go ahead and draw out my four carbons here, and I know there's going to be a plus 1-formyl charge on that carbon on the end, which we, of course, know this is a primary carbocation. And primary carbocations are not very stable. So there could be some rearrangement here. And if you think about what's the best way to rearrange our carbocation to see if we can form a more stable one, you know that there are hydrogens attached to the carbon next door to that positive 1-formyl charge. And so we get a hydride shift here. So we've seen how to do that in earlier videos. So you could think about the hydrogen and these two electrons."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if you think about what's the best way to rearrange our carbocation to see if we can form a more stable one, you know that there are hydrogens attached to the carbon next door to that positive 1-formyl charge. And so we get a hydride shift here. So we've seen how to do that in earlier videos. So you could think about the hydrogen and these two electrons. All of them are going to move over here. So if we go ahead and show the result of that hydride shift, so that hydrogen and those two electrons are going to move over there. That takes away the plus 1-formyl charge of the carbon on the end."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about the hydrogen and these two electrons. All of them are going to move over here. So if we go ahead and show the result of that hydride shift, so that hydrogen and those two electrons are going to move over there. That takes away the plus 1-formyl charge of the carbon on the end. And we took a bond away from this carbon now. So that's this carbon over here. So that is where the plus 1-formyl charge is going to go."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That takes away the plus 1-formyl charge of the carbon on the end. And we took a bond away from this carbon now. So that's this carbon over here. So that is where the plus 1-formyl charge is going to go. And so now we have a secondary carbocation. So just to refresh your memory, this is a secondary carbocation because the carbon in magenta is bonded to two other carbons. So we have a secondary carbocation, which we know is more stable than a primary carbocation."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that is where the plus 1-formyl charge is going to go. And so now we have a secondary carbocation. So just to refresh your memory, this is a secondary carbocation because the carbon in magenta is bonded to two other carbons. So we have a secondary carbocation, which we know is more stable than a primary carbocation. And so if we can think about two possible products here. So our benzene ring could react with our primary carbocation. Or more likely, it's going to react with our secondary carbocation over here."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a secondary carbocation, which we know is more stable than a primary carbocation. And so if we can think about two possible products here. So our benzene ring could react with our primary carbocation. Or more likely, it's going to react with our secondary carbocation over here. So let's go ahead and draw the result if it reacts with a secondary carbocation. So I could think about my electrons in here, in my pi bonds, going all the way over here to react with that carbocation, that secondary carbocation. And so we can go ahead and show the result of our nucleophilic attack."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Or more likely, it's going to react with our secondary carbocation over here. So let's go ahead and draw the result if it reacts with a secondary carbocation. So I could think about my electrons in here, in my pi bonds, going all the way over here to react with that carbocation, that secondary carbocation. And so we can go ahead and show the result of our nucleophilic attack. So once again, we have our pi electrons. We had a hydrogen already bonded to that ring there. And we're going to form a bond to what used to be our carbocation."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we can go ahead and show the result of our nucleophilic attack. So once again, we have our pi electrons. We had a hydrogen already bonded to that ring there. And we're going to form a bond to what used to be our carbocation. So we have four carbons. And one of them is going to be branching like that. So let's go ahead and analyze what we've done so far."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to form a bond to what used to be our carbocation. So we have four carbons. And one of them is going to be branching like that. So let's go ahead and analyze what we've done so far. So plus 1 formal charge on our sigma complex. So our electrons, our pi electrons in here, are going to be these electrons right here. So forming our carbon-carbon bond like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and analyze what we've done so far. So plus 1 formal charge on our sigma complex. So our electrons, our pi electrons in here, are going to be these electrons right here. So forming our carbon-carbon bond like that. And when we look at what's attached to that, let me go ahead and highlight some of these carbons here. So I'll start in red. So this carbon right here, I'm going to say is this carbon like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So forming our carbon-carbon bond like that. And when we look at what's attached to that, let me go ahead and highlight some of these carbons here. So I'll start in red. So this carbon right here, I'm going to say is this carbon like that. So in magenta, this carbon down here in magenta is, of course, this carbon right here in magenta, the one that's bonded to our ring. And then, of course, we could follow a few more carbons here. So I'm going to say that this carbon in green, that's, of course, this one right here."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here, I'm going to say is this carbon like that. So in magenta, this carbon down here in magenta is, of course, this carbon right here in magenta, the one that's bonded to our ring. And then, of course, we could follow a few more carbons here. So I'm going to say that this carbon in green, that's, of course, this one right here. And then, finally, this carbon right here in blue is this carbon. So always count your carbons and make sure that you have the right branching for something like this. So in the last step, of course, deprotonation of our sigma complex, these electrons would move into here like that."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to say that this carbon in green, that's, of course, this one right here. And then, finally, this carbon right here in blue is this carbon. So always count your carbons and make sure that you have the right branching for something like this. So in the last step, of course, deprotonation of our sigma complex, these electrons would move into here like that. And we have one of our final products here. So let's go ahead and draw this one. So we would have that as our major product for this reaction."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So in the last step, of course, deprotonation of our sigma complex, these electrons would move into here like that. And we have one of our final products here. So let's go ahead and draw this one. So we would have that as our major product for this reaction. It is possible for the primary carbocation to react as well. So let's go ahead and draw the result of that. So if these pi electrons instead went for the primary carbocation, they'd form a bond with that carbon right there."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we would have that as our major product for this reaction. It is possible for the primary carbocation to react as well. So let's go ahead and draw the result of that. So if these pi electrons instead went for the primary carbocation, they'd form a bond with that carbon right there. And so we could go ahead and draw the result of that. So we would have our ring once again. And once again, we have our hydrogen."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if these pi electrons instead went for the primary carbocation, they'd form a bond with that carbon right there. And so we could go ahead and draw the result of that. So we would have our ring once again. And once again, we have our hydrogen. And this time, we would have four carbons coming off of our ring. So one, two, three, four, like that. And once again, positive one formal charge on our sigma complex."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we have our hydrogen. And this time, we would have four carbons coming off of our ring. So one, two, three, four, like that. And once again, positive one formal charge on our sigma complex. So let's go ahead and, once again, highlight those electrons in magenta. So the electrons in magenta right here, they formed this carbon-carbon bond. And they're bonding to this carbon this time, so the one on the end."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, positive one formal charge on our sigma complex. So let's go ahead and, once again, highlight those electrons in magenta. So the electrons in magenta right here, they formed this carbon-carbon bond. And they're bonding to this carbon this time, so the one on the end. So I'm going to say it's that one right there. So next, we have the carbon in magenta right here. And so that would be this one."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And they're bonding to this carbon this time, so the one on the end. So I'm going to say it's that one right there. So next, we have the carbon in magenta right here. And so that would be this one. And just to be consistent with our colors, I'll go with green next. So this carbon is green. And so that one is green right here."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that would be this one. And just to be consistent with our colors, I'll go with green next. So this carbon is green. And so that one is green right here. And then finally, this carbon in blue is this carbon in blue. The reason I'm taking the time to show all these carbons is I find that this is one of the areas where students will make the most mistakes. So you have to be careful about following your electrons and counting your carbons when you're drawing your final products here."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that one is green right here. And then finally, this carbon in blue is this carbon in blue. The reason I'm taking the time to show all these carbons is I find that this is one of the areas where students will make the most mistakes. So you have to be careful about following your electrons and counting your carbons when you're drawing your final products here. So deprotonation of our sigma complex, these electrons move into there. And we would form butylbenzene as the minor product in this reaction. So let's go ahead and draw that out."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you have to be careful about following your electrons and counting your carbons when you're drawing your final products here. So deprotonation of our sigma complex, these electrons move into there. And we would form butylbenzene as the minor product in this reaction. So let's go ahead and draw that out. So we have four carbons coming off of our benzene ring. And this would be the minor product, since it comes from the less stable carbocation. The primary carbocation is not as stable as the secondary carbocation."}, {"video_title": "Friedel-Crafts alkylation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that out. So we have four carbons coming off of our benzene ring. And this would be the minor product, since it comes from the less stable carbocation. The primary carbocation is not as stable as the secondary carbocation. And so this shows one of the limitations to the Friedel Crafts alkylation reaction. You can't always control the alkyl group that you're putting onto your ring because of the possible rearrangement, because there's a carbocation present in the mechanism. So if your goal was to make butylbenzene, you wouldn't be able to make it in an extremely high yield using a Friedel Crafts alkylation."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even this, even if each of these dots were a star, this is a huge amount of stars, but a lot of these dots are thousands of stars, are thousands of stars. So this, you know, our mind was already blown, but what we're gonna see in this video is that in some ways this is kind of just the beginning. And to some degree, I'm gonna stop doing these particles of sand in a football field analogy, because at some point the particles of sand become so vast that our minds can't even grasp it to begin with. But let's just start with our Milky Way. And we saw in the last video, the Milky Way right here, we're sitting here about 25,000 light years from the center. It's roughly 100,000 light years in diameter. And then let's put it in perspective of its local neighborhood."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's just start with our Milky Way. And we saw in the last video, the Milky Way right here, we're sitting here about 25,000 light years from the center. It's roughly 100,000 light years in diameter. And then let's put it in perspective of its local neighborhood. So let's look at the local group. And when we talk about local group, we're talking about the local group of galaxies. Of galaxies."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then let's put it in perspective of its local neighborhood. So let's look at the local group. And when we talk about local group, we're talking about the local group of galaxies. Of galaxies. So this right here is the Milky Way's local group. That's us right there, sitting right over here, about 25,000 light years from the center of the Milky Way. You have some of these small, and I use the word, I use small in quotation marks, because these are also vast entities, also unimaginable entities."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Of galaxies. So this right here is the Milky Way's local group. That's us right there, sitting right over here, about 25,000 light years from the center of the Milky Way. You have some of these small, and I use the word, I use small in quotation marks, because these are also vast entities, also unimaginable entities. But we have these satellite galaxies around, under the gravitational influence, some of them, of the Milky Way. But the nearest large galaxy to us is Andromeda right over here. And this distance right over here, and now we're gonna start talking about the millions of light years."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have some of these small, and I use the word, I use small in quotation marks, because these are also vast entities, also unimaginable entities. But we have these satellite galaxies around, under the gravitational influence, some of them, of the Milky Way. But the nearest large galaxy to us is Andromeda right over here. And this distance right over here, and now we're gonna start talking about the millions of light years. So this distance right here is 2.5 million. 2.5 million light years. And just as a bit of a reference, if that's any reference at all, one light year is roughly the radius of the Oort Cloud."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this distance right over here, and now we're gonna start talking about the millions of light years. So this distance right here is 2.5 million. 2.5 million light years. And just as a bit of a reference, if that's any reference at all, one light year is roughly the radius of the Oort Cloud. And the Oort Cloud was, or another way to think about it, the Oort Cloud, or one radius of the Oort Cloud, is about 50 or 60,000 atomic, not atomic, astronomical units. And that's the distance from the sun to the Earth. So you could view this as 2.5 million times 60,000 or so, times the distance from the sun to the Earth."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just as a bit of a reference, if that's any reference at all, one light year is roughly the radius of the Oort Cloud. And the Oort Cloud was, or another way to think about it, the Oort Cloud, or one radius of the Oort Cloud, is about 50 or 60,000 atomic, not atomic, astronomical units. And that's the distance from the sun to the Earth. So you could view this as 2.5 million times 60,000 or so, times the distance from the sun to the Earth. So this is an unbelievably large distance we're talking about here, and that's to get to the next big galaxy over here. But even these things are huge things with many, I mean, just unfathomably many stars. But Andromeda in particular, you know, we said that the Milky Way has 200 to 400 billion stars."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you could view this as 2.5 million times 60,000 or so, times the distance from the sun to the Earth. So this is an unbelievably large distance we're talking about here, and that's to get to the next big galaxy over here. But even these things are huge things with many, I mean, just unfathomably many stars. But Andromeda in particular, you know, we said that the Milky Way has 200 to 400 billion stars. Andromeda, people believe, has on the order of one trillion. One trillion stars. So even, even, even, you know, these just start to become numbers."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But Andromeda in particular, you know, we said that the Milky Way has 200 to 400 billion stars. Andromeda, people believe, has on the order of one trillion. One trillion stars. So even, even, even, you know, these just start to become numbers. It's hard to grasp. But we're not gonna stop here. So in this over here, this whole diagram right here, it's about four light years across if you go from point to point."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So even, even, even, you know, these just start to become numbers. It's hard to grasp. But we're not gonna stop here. So in this over here, this whole diagram right here, it's about four light years across if you go from point to point. If you go from one side to the other side, this is about, not four light years, sorry, this is four million light years. Four million. Four million light years."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in this over here, this whole diagram right here, it's about four light years across if you go from point to point. If you go from one side to the other side, this is about, not four light years, sorry, this is four million light years. Four million. Four million light years. Four light years is just the distance from us to Alpha Centauri. So that was nothing. That would only take a Voyager 1 80,000 years to get, this is four million light years."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Four million light years. Four light years is just the distance from us to Alpha Centauri. So that was nothing. That would only take a Voyager 1 80,000 years to get, this is four million light years. So four million times the distance to the nearest star. But let's, but even this, even this is, I mean, I'm starting to stumble on my words because there's really no words to describe it. Even this is small on a, on an intergalactic scale."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That would only take a Voyager 1 80,000 years to get, this is four million light years. So four million times the distance to the nearest star. But let's, but even this, even this is, I mean, I'm starting to stumble on my words because there's really no words to describe it. Even this is small on a, on an intergalactic scale. Because when you zoom out more, you can see our local group, our local group is right over here, is right over here. And this right over here is the Virgo supercluster. And each dot here is at least one galaxy, but it might be more than one galaxy."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Even this is small on a, on an intergalactic scale. Because when you zoom out more, you can see our local group, our local group is right over here, is right over here. And this right over here is the Virgo supercluster. And each dot here is at least one galaxy, but it might be more than one galaxy. And this, and more than one galaxies. And the diameter here, the diameter here is 150 million. 150 million."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And each dot here is at least one galaxy, but it might be more than one galaxy. And this, and more than one galaxies. And the diameter here, the diameter here is 150 million. 150 million. 150 million light years. Light years. So what we saw in the local group in the last diagram, the distance from the Milky Way to Andromeda, which was two and a half million light years, which would be just a little dot, just like that."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "150 million. 150 million light years. Light years. So what we saw in the local group in the last diagram, the distance from the Milky Way to Andromeda, which was two and a half million light years, which would be just a little dot, just like that. That would be the distance between the Milky Way and Andromeda. And now we're looking at the Virgo supercluster that is 150 million light years. But we're not done yet."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what we saw in the local group in the last diagram, the distance from the Milky Way to Andromeda, which was two and a half million light years, which would be just a little dot, just like that. That would be the distance between the Milky Way and Andromeda. And now we're looking at the Virgo supercluster that is 150 million light years. But we're not done yet. We can zoom out even more. We can zoom out even more. And over here, so you had your Virgo supercluster, 150 million light years was that last diagram."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we're not done yet. We can zoom out even more. We can zoom out even more. And over here, so you had your Virgo supercluster, 150 million light years was that last diagram. This diagram right over here, I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And over here, so you had your Virgo supercluster, 150 million light years was that last diagram. This diagram right over here, I want to keep both of them on the screen if I can. This diagram right here, 150 million light years across. That would fit right about here on this diagram. So this is all of the superclusters that are near us. And once again, near has to be used very, very, very loosely. Here, this distance is about 150 million light years."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That would fit right about here on this diagram. So this is all of the superclusters that are near us. And once again, near has to be used very, very, very loosely. Here, this distance is about 150 million light years. A billion light years is two, three, four, five. A billion light years is about from here to there. So we're starting to talk on a fairly massive, I guess we've always been talking on a massive scale, but now it's an even more massive scale."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Here, this distance is about 150 million light years. A billion light years is two, three, four, five. A billion light years is about from here to there. So we're starting to talk on a fairly massive, I guess we've always been talking on a massive scale, but now it's an even more massive scale. But we're still not done because this whole diagram, now these dots that you're seeing now, I want to make it very clear, these aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're starting to talk on a fairly massive, I guess we've always been talking on a massive scale, but now it's an even more massive scale. But we're still not done because this whole diagram, now these dots that you're seeing now, I want to make it very clear, these aren't stars. These aren't even clusters of stars. These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies. Each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point, but we're still not done."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These aren't even clusters of millions or even billions of stars. Each of these dots are clusters of galaxies. Each of those galaxies having hundreds of billions to trillions of stars. So we're just on an unbelievably massive scale at this point, but we're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're just on an unbelievably massive scale at this point, but we're still not done. This is roughly about a billion light years across. But right here is actually the best estimate of the visible universe. And in future videos, we're going to talk a lot more about what the visible universe means. So if you were to zoom out enough, this entire diagram right here, about a billion light years, would fit right over, would fit just like that. We're talking about a super, super small amount of this part right here. And this is just the visible universe."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And in future videos, we're going to talk a lot more about what the visible universe means. So if you were to zoom out enough, this entire diagram right here, about a billion light years, would fit right over, would fit just like that. We're talking about a super, super small amount of this part right here. And this is just the visible universe. I want to make it clear, this is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about a point out here, and we're observing it, and that's, let's say, 13 billion light years away."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is just the visible universe. I want to make it clear, this is not the entire universe. And we say it's the visible universe because think about what's happening. When we think about a point out here, and we're observing it, and that's, let's say, 13 billion light years away. Let's say that point, 13 billion. We're going to talk more about this in future videos. 13 billion light years."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When we think about a point out here, and we're observing it, and that's, let's say, 13 billion light years away. Let's say that point, 13 billion. We're going to talk more about this in future videos. 13 billion light years. And I feel almost, it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, what we're observing, the light is just getting to us. This light left some point 13 billion light years ago."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "13 billion light years. And I feel almost, it's almost a sacrilege to be writing on this because this complexity that we're seeing here is just mind boggling. But this 13 billion light year away object, what we're observing, the light is just getting to us. This light left some point 13 billion light years ago. So what we're actually doing is observing that object close to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe its light hasn't reached us yet."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This light left some point 13 billion light years ago. So what we're actually doing is observing that object close to the beginning of the actual universe. And the reason why it's the visible universe is there might have been something a little bit further out. Maybe its light hasn't reached us yet. Or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark. It's actually a huge question mark on how big the actual universe is."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe its light hasn't reached us yet. Or maybe the universe itself, and we'll talk more about this, it's expanding so fast that the light will never, ever reach us. So it's actually a huge question mark. It's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter because this by itself is a huge distance. And I want to make it clear. You might say, OK, if this light over here, this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's actually a huge question mark on how big the actual universe is. And then some people might say, well, does it even matter because this by itself is a huge distance. And I want to make it clear. You might say, OK, if this light over here, this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful. Because remember, the universe is expanding. When this light was emitted, and I'll do a whole video on this because the geometry of it is kind of hard to visualize."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You might say, OK, if this light over here, this is coming from 13 billion light years away, or if this is 13 billion light years away, then you could say, hey, so everything that we can observe or that we can even observe the past of, the radius is about 26 billion light years. But even there, we have to be careful. Because remember, the universe is expanding. When this light was emitted, and I'll do a whole video on this because the geometry of it is kind of hard to visualize. When this light was emitted, where we are in the Virgo supercluster inside of the Milky Way galaxy, where we are was much closer to that point. It was on the order of, and I want to make sure I get this right, 36 million light years. So we were super close by, I guess, astronomical scales."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When this light was emitted, and I'll do a whole video on this because the geometry of it is kind of hard to visualize. When this light was emitted, where we are in the Virgo supercluster inside of the Milky Way galaxy, where we are was much closer to that point. It was on the order of, and I want to make sure I get this right, 36 million light years. So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object when that light was released. But that light was coming to us, and the whole time the universe expanding. We were also moving away from it."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we were super close by, I guess, astronomical scales. We were super close, only 36 million light years, to this object when that light was released. But that light was coming to us, and the whole time the universe expanding. We were also moving away from it. If you just think about all of the spaces, everything is expanding away from each other. And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We were also moving away from it. If you just think about all of the spaces, everything is expanding away from each other. And only 13 billion years later did it finally catch up with us. But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order of 40 or 45 billion light years away. We're just observing where that light was emitted 13 billion years ago."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the whole time that that was happening, this object has also been moving. This object has also been moving away from us. And so our best estimate of where this object is now, based on how space is expanding, is on the order of 40 or 45 billion light years away. We're just observing where that light was emitted 13 billion years ago. And I want to be very clear, what we're observing, this light is coming from something very, very, very primitive. That object, or that area of space where that light was emitted from, has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're just observing where that light was emitted 13 billion years ago. And I want to be very clear, what we're observing, this light is coming from something very, very, very primitive. That object, or that area of space where that light was emitted from, has now condensed into way more, I guess, mature astronomical structures. If you take it from the other point of view, people sitting in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. When I use words like shortly, I use that also loosely."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you take it from the other point of view, people sitting in this point of space now, and they've now moved 46 billion light years out, when they observe our region of space, they're not going to see us. They're not going to see Earth as it is now. They're going to see the region of space where Earth is at a super primitive stage, shortly after the Big Bang. When I use words like shortly, I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When I use words like shortly, I use that also loosely. We're still talking about hundreds of thousands or even millions of years. So we'll talk more about that in a future video. But the whole point of this video is it's beyond mind numbing. I would say the last video about the Milky Way, that alone was mind numbing. But now we're in a reality where just the Milky Way becomes something that's almost unbelievably insignificant when you think about this picture right here. And the really mind numbing thing is if someone told me that this is the entire universe, this by itself would certainly put things in perspective."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the whole point of this video is it's beyond mind numbing. I would say the last video about the Milky Way, that alone was mind numbing. But now we're in a reality where just the Milky Way becomes something that's almost unbelievably insignificant when you think about this picture right here. And the really mind numbing thing is if someone told me that this is the entire universe, this by itself would certainly put things in perspective. But it's unknown what's beyond it. There are some estimates that this might only be 1 times 10 to the 23rd of the entire universe. And we could even, we'll talk, it might even be the reality that the entire universe is smaller than this."}, {"video_title": "Intergalactic scale Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the really mind numbing thing is if someone told me that this is the entire universe, this by itself would certainly put things in perspective. But it's unknown what's beyond it. There are some estimates that this might only be 1 times 10 to the 23rd of the entire universe. And we could even, we'll talk, it might even be the reality that the entire universe is smaller than this. And that's an interesting thing to think about. But I'll leave you there because I think no matter how you think about it, it's just, I don't know. Actually, before doing this video, I stared at some of these photos for like half an hour."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And obviously, cyclohexane is just one ring. So it would be a monocyclic compound. The official IUPAC way of telling how many rings you have is how many cuts does it take to make an open-chain alkane? So if I just made a cut between those two carbons, I have an open-chain alkane. And it took me one cut to get there. So obviously, cyclohexane is monocyclic. What about if I had a cyclohexane ring, and then I had another ring fused to it like that?"}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So if I just made a cut between those two carbons, I have an open-chain alkane. And it took me one cut to get there. So obviously, cyclohexane is monocyclic. What about if I had a cyclohexane ring, and then I had another ring fused to it like that? Obviously, there are two rings here. So this would be a bicyclic compound. Once again, if you were to use the IUPAC way of figuring out how many rings you have, you would make a cut here, and a cut here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "What about if I had a cyclohexane ring, and then I had another ring fused to it like that? Obviously, there are two rings here. So this would be a bicyclic compound. Once again, if you were to use the IUPAC way of figuring out how many rings you have, you would make a cut here, and a cut here. And now you have an open-chain alkane. So it took you two cuts to get there. So you would classify this as a bicyclic molecule."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Once again, if you were to use the IUPAC way of figuring out how many rings you have, you would make a cut here, and a cut here. And now you have an open-chain alkane. So it took you two cuts to get there. So you would classify this as a bicyclic molecule. Now, these are very easy, simple compounds. But for more complicated compounds, you'll need to remember that IUPAC rule to figure out how many rings you actually have. Let's go ahead and redraw that second molecule here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So you would classify this as a bicyclic molecule. Now, these are very easy, simple compounds. But for more complicated compounds, you'll need to remember that IUPAC rule to figure out how many rings you actually have. Let's go ahead and redraw that second molecule here. And let's look at it in some more detail. So I had a cyclohexane ring. And then I had another ring fused to it like that."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and redraw that second molecule here. And let's look at it in some more detail. So I had a cyclohexane ring. And then I had another ring fused to it like that. Now, the carbons that are shared by both of these rings, so this carbon and this carbon, those are called your bridgehead carbons. Let's go ahead and write that. The bridgehead carbons."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then I had another ring fused to it like that. Now, the carbons that are shared by both of these rings, so this carbon and this carbon, those are called your bridgehead carbons. Let's go ahead and write that. The bridgehead carbons. And when you're numbering your bicyclic compounds, you want to start at the bridgehead carbon, then number along the longest path, then the second longest path, and then finally the shortest path. So let's go ahead and do that. Let's start with the top bridgehead carbon."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The bridgehead carbons. And when you're numbering your bicyclic compounds, you want to start at the bridgehead carbon, then number along the longest path, then the second longest path, and then finally the shortest path. So let's go ahead and do that. Let's start with the top bridgehead carbon. We'll make that number 1. And we're going to number along the longest path. So the longest path would be going left."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with the top bridgehead carbon. We'll make that number 1. And we're going to number along the longest path. So the longest path would be going left. So that's 1, 2, 3, 4, 5, and 6. And now the second longest path. Well, we can just continue."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So the longest path would be going left. So that's 1, 2, 3, 4, 5, and 6. And now the second longest path. Well, we can just continue. And that would be 7 and 8, since that's the second longest path. The third path, the shortest path, is usually the one between your bridgehead carbons. And there are no carbons between our bridgehead carbons."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well, we can just continue. And that would be 7 and 8, since that's the second longest path. The third path, the shortest path, is usually the one between your bridgehead carbons. And there are no carbons between our bridgehead carbons. So we're done in terms of numbering our compound. The general formula for naming a bicyclo compound. We've already determined this is a bicyclo compound."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And there are no carbons between our bridgehead carbons. So we're done in terms of numbering our compound. The general formula for naming a bicyclo compound. We've already determined this is a bicyclo compound. So you would have bicyclo here. And then in brackets, you would put the number of carbons in the longest path excluding the bridgehead carbon, which I will represent here with x. And then y is the number of carbons of the second longest path, excluding the bridgehead carbon."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We've already determined this is a bicyclo compound. So you would have bicyclo here. And then in brackets, you would put the number of carbons in the longest path excluding the bridgehead carbon, which I will represent here with x. And then y is the number of carbons of the second longest path, excluding the bridgehead carbon. And then z is the number of carbons in the shortest path, excluding the bridgehead carbon. And then you finish everything off with the alkane. So how many carbons are there in this molecule?"}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then y is the number of carbons of the second longest path, excluding the bridgehead carbon. And then z is the number of carbons in the shortest path, excluding the bridgehead carbon. And then you finish everything off with the alkane. So how many carbons are there in this molecule? So for this example, let's go ahead and name it. We've already determined that it's a bicyclic compound. So we'll go ahead and write bicyclo here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So how many carbons are there in this molecule? So for this example, let's go ahead and name it. We've already determined that it's a bicyclic compound. So we'll go ahead and write bicyclo here. And x, again, is the number of carbons in the longest path, excluding the bridgehead carbon. So we're going to find the longest path, which is on the left here. And we're going to ignore the bridgehead carbon."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we'll go ahead and write bicyclo here. And x, again, is the number of carbons in the longest path, excluding the bridgehead carbon. So we're going to find the longest path, which is on the left here. And we're going to ignore the bridgehead carbon. How many carbons were there? One, two, three, and four. So we'll go ahead and put a four here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to ignore the bridgehead carbon. How many carbons were there? One, two, three, and four. So we'll go ahead and put a four here. Why was the number of carbons in the second longest path? Well, that was over here, where we had this carbon and this carbon, again, excluding the bridgehead carbon. So that would get a two."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we'll go ahead and put a four here. Why was the number of carbons in the second longest path? Well, that was over here, where we had this carbon and this carbon, again, excluding the bridgehead carbon. So that would get a two. And finally, the shortest path, which was how many carbons are there between my bridgehead carbons? And there are, of course, zero carbons between my bridgehead carbons. So if there are zero carbons between your bridgehead carbons, obviously your bridgehead carbons are bonded directly together, which we can see in the dot structure."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So that would get a two. And finally, the shortest path, which was how many carbons are there between my bridgehead carbons? And there are, of course, zero carbons between my bridgehead carbons. So if there are zero carbons between your bridgehead carbons, obviously your bridgehead carbons are bonded directly together, which we can see in the dot structure. Finally, how many carbons are there total? There are eight. And we know that means octane."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So if there are zero carbons between your bridgehead carbons, obviously your bridgehead carbons are bonded directly together, which we can see in the dot structure. Finally, how many carbons are there total? There are eight. And we know that means octane. So the final name for this molecule is bicyclo 4, 2, 0 octane. Let's do another one following our general formula here for naming a bicyclo compound. So let's take a cyclohexane ring."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And we know that means octane. So the final name for this molecule is bicyclo 4, 2, 0 octane. Let's do another one following our general formula here for naming a bicyclo compound. So let's take a cyclohexane ring. And let's make this a bridgehead carbon. And we're actually going to put a carbon between our bridgehead carbons like that. Now, this is kind of hard to see when it's drawn this way."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a cyclohexane ring. And let's make this a bridgehead carbon. And we're actually going to put a carbon between our bridgehead carbons like that. Now, this is kind of hard to see when it's drawn this way. So usually, you'll see it drawn a little bit differently. Usually, you'll see it drawn with a little bit more three-dimensionality to it. So hopefully, we can do that here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now, this is kind of hard to see when it's drawn this way. So usually, you'll see it drawn a little bit differently. Usually, you'll see it drawn with a little bit more three-dimensionality to it. So hopefully, we can do that here. So it would look something like this. All right. So let's find our bridgehead carbons here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So hopefully, we can do that here. So it would look something like this. All right. So let's find our bridgehead carbons here. So we know that this is a bridgehead carbon. This is a bridgehead carbon. Those correspond to these guys over here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's find our bridgehead carbons here. So we know that this is a bridgehead carbon. This is a bridgehead carbon. Those correspond to these guys over here. And how many cuts would it take to turn this into an open-chain alkane? Well, let's go ahead and look at the drawing on the left. And if we cut it right here and we cut it here, we can see it's now an open-chain compound."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Those correspond to these guys over here. And how many cuts would it take to turn this into an open-chain alkane? Well, let's go ahead and look at the drawing on the left. And if we cut it right here and we cut it here, we can see it's now an open-chain compound. So it took two cuts, so it's bicyclo. So let's go ahead and put those bonds back in there like that. So it's a bicyclo compound."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And if we cut it right here and we cut it here, we can see it's now an open-chain compound. So it took two cuts, so it's bicyclo. So let's go ahead and put those bonds back in there like that. So it's a bicyclo compound. So we can go ahead and start naming it as a bicyclo compound. So this is bicyclo. We need to go ahead and number it."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So it's a bicyclo compound. So we can go ahead and start naming it as a bicyclo compound. So this is bicyclo. We need to go ahead and number it. So let's start with our bridgehead carbon. So we'll start with this as our bridgehead carbon. Find the longest path."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We need to go ahead and number it. So let's start with our bridgehead carbon. So we'll start with this as our bridgehead carbon. Find the longest path. Well, it doesn't matter if I go left or right here since it's the same length each way around. So all of a sudden, I'm going to have to go each way around. So I'll just go to the right."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Find the longest path. Well, it doesn't matter if I go left or right here since it's the same length each way around. So all of a sudden, I'm going to have to go each way around. So I'll just go to the right. So 1, 2, 3, 4. Second longest path. So I'm going to keep on going this way."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'll just go to the right. So 1, 2, 3, 4. Second longest path. So I'm going to keep on going this way. So 5 and 6. And then finally, the shortest path, which is the one carbon in between my two bridgehead carbons, which would then get a 7. So 7 total carbons."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to keep on going this way. So 5 and 6. And then finally, the shortest path, which is the one carbon in between my two bridgehead carbons, which would then get a 7. So 7 total carbons. All right, so when I'm naming this, I need to put my brackets in here. And I put the number of carbons in the longest path, excluding the bridgehead carbon here. So how many carbons were in my longest path, excluding the bridgehead?"}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So 7 total carbons. All right, so when I'm naming this, I need to put my brackets in here. And I put the number of carbons in the longest path, excluding the bridgehead carbon here. So how many carbons were in my longest path, excluding the bridgehead? Here's 1 and here's 2. So I'm going to go ahead and put a 2 in here. And then I go to my second longest path."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So how many carbons were in my longest path, excluding the bridgehead? Here's 1 and here's 2. So I'm going to go ahead and put a 2 in here. And then I go to my second longest path. Well, that was over on this side. And there are also two carbons on my second longest path. Obviously, it's the same length as the other one."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then I go to my second longest path. Well, that was over on this side. And there are also two carbons on my second longest path. Obviously, it's the same length as the other one. So it's 2, 2. And then my shortest path. How many carbons are there in my shortest path?"}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Obviously, it's the same length as the other one. So it's 2, 2. And then my shortest path. How many carbons are there in my shortest path? Not counting my bridgehead carbon, there's 1. So I go ahead and put a 1 here. So I have bicyclo-2, 2, 1, 0."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "How many carbons are there in my shortest path? Not counting my bridgehead carbon, there's 1. So I go ahead and put a 1 here. So I have bicyclo-2, 2, 1, 0. And then my total number of carbons is 7. So this is a bicycloheptane molecule. So bicyclo-2, 2, 1, heptane would be the official IUPAC name for this molecule."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I have bicyclo-2, 2, 1, 0. And then my total number of carbons is 7. So this is a bicycloheptane molecule. So bicyclo-2, 2, 1, heptane would be the official IUPAC name for this molecule. This molecule turns up a lot in nature, so much so that it actually has its own common name. This is also called norborane. So let's go ahead and write that here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So bicyclo-2, 2, 1, heptane would be the official IUPAC name for this molecule. This molecule turns up a lot in nature, so much so that it actually has its own common name. This is also called norborane. So let's go ahead and write that here. So this is norborane. Again, a structure found often in nature. All right, let's do one more example here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that here. So this is norborane. Again, a structure found often in nature. All right, let's do one more example here. And let's make it look similar to that norborane molecule. So we'll have it go like this. So definitely it's a little bit tricky to draw these molecules here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's do one more example here. And let's make it look similar to that norborane molecule. So we'll have it go like this. So definitely it's a little bit tricky to draw these molecules here. So let's put some substituents on this molecule. So I'm going to attempt to draw the same molecule over here. And we're going to take a look at two different ways to number it and see which one turns out to be correct."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So definitely it's a little bit tricky to draw these molecules here. So let's put some substituents on this molecule. So I'm going to attempt to draw the same molecule over here. And we're going to take a look at two different ways to number it and see which one turns out to be correct. So here I have the exact same molecule on both sides here. And let's see how to do this. So let's first find our bridgehead carbons."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to take a look at two different ways to number it and see which one turns out to be correct. So here I have the exact same molecule on both sides here. And let's see how to do this. So let's first find our bridgehead carbons. So here are my bridgehead carbons right here. Now, when I start numbering, I could start at either one of those bridgehead carbons. So if I identify my bridgehead carbons over on this one, I could start numbering with either of these being number one."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's first find our bridgehead carbons. So here are my bridgehead carbons right here. Now, when I start numbering, I could start at either one of those bridgehead carbons. So if I identify my bridgehead carbons over on this one, I could start numbering with either of these being number one. So let's just make this number one here. So if that's number one, I next go to my longest path. So that would be going to the right here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So if I identify my bridgehead carbons over on this one, I could start numbering with either of these being number one. So let's just make this number one here. So if that's number one, I next go to my longest path. So that would be going to the right here. So this would give me a 2 for this carbon, a 3 for this carbon, a 4 for this carbon, a 5 for this carbon. Next, the second longest path. Well, I could just keep going, make that a 6, make this a 7."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So that would be going to the right here. So this would give me a 2 for this carbon, a 3 for this carbon, a 4 for this carbon, a 5 for this carbon. Next, the second longest path. Well, I could just keep going, make that a 6, make this a 7. And then finally, my shortest path, which is up here at the top. So that would be eight total carbons for my parent name. On the dot structure on the left, again, the exact same dot structure."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well, I could just keep going, make that a 6, make this a 7. And then finally, my shortest path, which is up here at the top. So that would be eight total carbons for my parent name. On the dot structure on the left, again, the exact same dot structure. This time I'm going to start with the opposite bridgehead carbon. So I'm going to start with this bridgehead carbon and then follow the same rules. So longest path, excluding your, so longest path would be to the right here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "On the dot structure on the left, again, the exact same dot structure. This time I'm going to start with the opposite bridgehead carbon. So I'm going to start with this bridgehead carbon and then follow the same rules. So longest path, excluding your, so longest path would be to the right here. So 2, 3, 4, 5, 6, 7, and then 8. So which one of these numbering systems is the correct one? Remember, our goal is to get the lowest number possible for our substituents."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So longest path, excluding your, so longest path would be to the right here. So 2, 3, 4, 5, 6, 7, and then 8. So which one of these numbering systems is the correct one? Remember, our goal is to get the lowest number possible for our substituents. And if I look at the example on the right, I have a methyl group in the 7 position and a methyl group in the 8 position. The example on the left has a methyl group in the 6 position and a methyl group in the 8 position. Since I want to give my lowest number possible to my substituent, I'm going to not number it the way on the right."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Remember, our goal is to get the lowest number possible for our substituents. And if I look at the example on the right, I have a methyl group in the 7 position and a methyl group in the 8 position. The example on the left has a methyl group in the 6 position and a methyl group in the 8 position. Since I want to give my lowest number possible to my substituent, I'm going to not number it the way on the right. I'm going to go with the way on the left here. So let's go ahead and start naming it. We already know this is a bicyclo compound."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Since I want to give my lowest number possible to my substituent, I'm going to not number it the way on the right. I'm going to go with the way on the left here. So let's go ahead and start naming it. We already know this is a bicyclo compound. And we have two methyl groups, one at 6 and one at 8. So we can start naming it by saying 6, 8-dimethyl. And then we know this is a bicyclo compound, so bicyclo."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We already know this is a bicyclo compound. And we have two methyl groups, one at 6 and one at 8. So we can start naming it by saying 6, 8-dimethyl. And then we know this is a bicyclo compound, so bicyclo. And then in the brackets, we're going to do the longest path first, excluding the bridgehead carbon. So the longest path is on the right. How many carbons are there in the longest path?"}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then we know this is a bicyclo compound, so bicyclo. And then in the brackets, we're going to do the longest path first, excluding the bridgehead carbon. So the longest path is on the right. How many carbons are there in the longest path? There are three, excluding the bridgehead carbon. So we'll go ahead and put a 3 in here. Next, the second longest path, excluding the bridgehead carbon, well, that was these two right here."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "How many carbons are there in the longest path? There are three, excluding the bridgehead carbon. So we'll go ahead and put a 3 in here. Next, the second longest path, excluding the bridgehead carbon, well, that was these two right here. So next, there will be a 2. And then finally, the shortest path, excluding our bridgehead carbon. There is only one carbon in our shortest path."}, {"video_title": "Bicyclic compounds Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Next, the second longest path, excluding the bridgehead carbon, well, that was these two right here. So next, there will be a 2. And then finally, the shortest path, excluding our bridgehead carbon. There is only one carbon in our shortest path. So we put that in there. And then there were eight total carbons, so it is octane. So the final IUPAC name for this molecule is 6, 8-dimethylbicyclo-3, 2, 1-octane."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this is ethanol. It's a primary alcohol because the carbon bonded to the OH is bonded to one other carbon. And primary alcohols react with HBr to form an alkyl bromide via an SN2 process. And so we have HBr, a strong acid, and we have ethanol, which is gonna function as a base. So actually, the first step is to protonate the oxygen. So we get a proton transfer here. So a lone pair of electrons in the oxygen pick up this proton, and these electrons are left behind on the bromine."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so we have HBr, a strong acid, and we have ethanol, which is gonna function as a base. So actually, the first step is to protonate the oxygen. So we get a proton transfer here. So a lone pair of electrons in the oxygen pick up this proton, and these electrons are left behind on the bromine. So we form our bromide anion here. So we get a bromide anion, negative one formal charge. So these electrons in here in blue, let's say that these are these electrons right here on our bromide anion."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons in the oxygen pick up this proton, and these electrons are left behind on the bromine. So we form our bromide anion here. So we get a bromide anion, negative one formal charge. So these electrons in here in blue, let's say that these are these electrons right here on our bromide anion. So we're going to protonate the oxygen. So let's go ahead and show that oxygen being protonated here. So there's still one lone pair of electrons on this oxygen, which gives this oxygen a plus one formal charge."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here in blue, let's say that these are these electrons right here on our bromide anion. So we're going to protonate the oxygen. So let's go ahead and show that oxygen being protonated here. So there's still one lone pair of electrons on this oxygen, which gives this oxygen a plus one formal charge. So these electrons in magenta right here on this oxygen pick up this proton to form this bond right here. And so now we need to think about our SN2 type mechanism. So in an SN2 type mechanism, we need a nucleophile, and we need an electrophile, right?"}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So there's still one lone pair of electrons on this oxygen, which gives this oxygen a plus one formal charge. So these electrons in magenta right here on this oxygen pick up this proton to form this bond right here. And so now we need to think about our SN2 type mechanism. So in an SN2 type mechanism, we need a nucleophile, and we need an electrophile, right? And we're also going to have our leaving group here. And the reason why we protonate the oxygen is to form a much better leaving group. So if we just had a nucleophile attack ethanol right here, we'd have hydroxide as a leaving group."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So in an SN2 type mechanism, we need a nucleophile, and we need an electrophile, right? And we're also going to have our leaving group here. And the reason why we protonate the oxygen is to form a much better leaving group. So if we just had a nucleophile attack ethanol right here, we'd have hydroxide as a leaving group. That is not a good leaving group. Here we would have water as our leaving group, which is what we're going to see in a second. And so that helps this process occur."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So if we just had a nucleophile attack ethanol right here, we'd have hydroxide as a leaving group. That is not a good leaving group. Here we would have water as our leaving group, which is what we're going to see in a second. And so that helps this process occur. So let's think about this oxygen here, withdrawing some electron density from this carbon that I just marked in red. So this carbon in red is the electrophilic portion. So the nucleophile is going to attack the electrophile."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so that helps this process occur. So let's think about this oxygen here, withdrawing some electron density from this carbon that I just marked in red. So this carbon in red is the electrophilic portion. So the nucleophile is going to attack the electrophile. So the bromide anion is going to function as a nucleophile, and it's going to attack our electrophile. And it's going to form a bond between the bromine and the carbon. At the same time, these electrons in here are going to come off onto the oxygen."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So the nucleophile is going to attack the electrophile. So the bromide anion is going to function as a nucleophile, and it's going to attack our electrophile. And it's going to form a bond between the bromine and the carbon. At the same time, these electrons in here are going to come off onto the oxygen. So an SN2 type mechanism, you remember, is concerted. So the bond that forms, this happens at the same time that this bond breaks and the electrons come off onto our leaving group. So when we draw our final product, we would form ethyl bromide or bromoethane."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "At the same time, these electrons in here are going to come off onto the oxygen. So an SN2 type mechanism, you remember, is concerted. So the bond that forms, this happens at the same time that this bond breaks and the electrons come off onto our leaving group. So when we draw our final product, we would form ethyl bromide or bromoethane. Let's put in our electrons right here. And so these electrons in blue formed a bond between the carbon and the bromine. So this is the carbon in red down here."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So when we draw our final product, we would form ethyl bromide or bromoethane. Let's put in our electrons right here. And so these electrons in blue formed a bond between the carbon and the bromine. So this is the carbon in red down here. And so we form bromoethane. We would also form water. So let me go ahead and draw in water."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this is the carbon in red down here. And so we form bromoethane. We would also form water. So let me go ahead and draw in water. So H2O, which is a good leaving group because water is so stable. So we go ahead and think about these electrons in green here coming off onto the oxygen. And that gives us water."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw in water. So H2O, which is a good leaving group because water is so stable. So we go ahead and think about these electrons in green here coming off onto the oxygen. And that gives us water. So in an SN2 type mechanism, sometimes you have to think about stereochemistry, but not for this example. Because this carbon in red right here, so this carbon in red right here, it's not a chiral center. There are two hydrogens attached to that carbon in red."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And that gives us water. So in an SN2 type mechanism, sometimes you have to think about stereochemistry, but not for this example. Because this carbon in red right here, so this carbon in red right here, it's not a chiral center. There are two hydrogens attached to that carbon in red. And so this is our only product. This is the only thing that we have to think about. So bromoethane as our final result."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "There are two hydrogens attached to that carbon in red. And so this is our only product. This is the only thing that we have to think about. So bromoethane as our final result. Let's look at another SN2 reaction. This time we're doing a secondary alcohol. And so here we have our secondary alcohol."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So bromoethane as our final result. Let's look at another SN2 reaction. This time we're doing a secondary alcohol. And so here we have our secondary alcohol. The carbon bonded to the OH is bonded to two other carbons, so a secondary alcohol. And we saw in an earlier video that alcohols will react with tosyl chloride and pyridine to form a tosylate. So let's go ahead and draw the product."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so here we have our secondary alcohol. The carbon bonded to the OH is bonded to two other carbons, so a secondary alcohol. And we saw in an earlier video that alcohols will react with tosyl chloride and pyridine to form a tosylate. So let's go ahead and draw the product. At the first step, so we form a tosylate. We retain the stereochemistry. So we had a wedge over here, and we have a wedge down here as well."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the product. At the first step, so we form a tosylate. We retain the stereochemistry. So we had a wedge over here, and we have a wedge down here as well. So we have an oxygen, and then we'll just go ahead and put TS here for our abbreviation. So we talked in great detail about this in an earlier video. So once again, this gives us a better leaving group."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So we had a wedge over here, and we have a wedge down here as well. So we have an oxygen, and then we'll just go ahead and put TS here for our abbreviation. So we talked in great detail about this in an earlier video. So once again, this gives us a better leaving group. So in the previous reaction, we protonated the oxygen to give us a much better leaving group. And here we have an excellent leaving group. So that's one of the reasons for using tosylates here."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So once again, this gives us a better leaving group. So in the previous reaction, we protonated the oxygen to give us a much better leaving group. And here we have an excellent leaving group. So that's one of the reasons for using tosylates here. So in the second step, we're going to add sodium bromide, and we're going to get, again, an SN2 type mechanism. So a nucleophile is going to attack our electrophile. So we can identify our electrophile."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So that's one of the reasons for using tosylates here. So in the second step, we're going to add sodium bromide, and we're going to get, again, an SN2 type mechanism. So a nucleophile is going to attack our electrophile. So we can identify our electrophile. It's the carbon bonded to the oxygen. So the oxygen's withdrawing some electron density from this carbon. So this is the electrophilic portion of the molecule."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So we can identify our electrophile. It's the carbon bonded to the oxygen. So the oxygen's withdrawing some electron density from this carbon. So this is the electrophilic portion of the molecule. And then once again, our bromide anion is going to function as a nucleophile. So here's our bromide anion. I'm going to highlight this lone pair of electrons right here in blue."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this is the electrophilic portion of the molecule. And then once again, our bromide anion is going to function as a nucleophile. So here's our bromide anion. I'm going to highlight this lone pair of electrons right here in blue. And so our bromide anion attacks our electrophilic carbon and forms a bond. At the same time, these electrons come off onto the oxygen. So once again, a concerted SN2 type mechanism."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "I'm going to highlight this lone pair of electrons right here in blue. And so our bromide anion attacks our electrophilic carbon and forms a bond. At the same time, these electrons come off onto the oxygen. So once again, a concerted SN2 type mechanism. And this time, we do have to worry about stereochemistry. So we have this wedge in here. We have this wedge."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So once again, a concerted SN2 type mechanism. And this time, we do have to worry about stereochemistry. So we have this wedge in here. We have this wedge. So this part is coming out at us in space. And so the bromide anion has to attack from the opposite side. And so if it's attacking from the opposite side, when you draw the final product, you'd have to show this as a dash."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "We have this wedge. So this part is coming out at us in space. And so the bromide anion has to attack from the opposite side. And so if it's attacking from the opposite side, when you draw the final product, you'd have to show this as a dash. So the bromine, it had to attack from the opposite side, which gives us inversion of configuration, inversion of absolute configuration here. So when you assign your absolute configuration for your starting alcohol, this would be R. So our chiral center would be right here. So this carbon is chiral."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so if it's attacking from the opposite side, when you draw the final product, you'd have to show this as a dash. So the bromine, it had to attack from the opposite side, which gives us inversion of configuration, inversion of absolute configuration here. So when you assign your absolute configuration for your starting alcohol, this would be R. So our chiral center would be right here. So this carbon is chiral. And so this carbon is chiral for our products. And for our product, we would form the S in the antimer here. So we form the S in the antimer, and we start with the R in the antimer."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this carbon is chiral. And so this carbon is chiral for our products. And for our product, we would form the S in the antimer here. So we form the S in the antimer, and we start with the R in the antimer. So an SN2 type mechanism, inversion of configuration because the nucleophile has to attack from the opposite side. So let's look at one more example here. Let's do an SN1 mechanism using a tertiary alcohol."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So we form the S in the antimer, and we start with the R in the antimer. So an SN2 type mechanism, inversion of configuration because the nucleophile has to attack from the opposite side. So let's look at one more example here. Let's do an SN1 mechanism using a tertiary alcohol. So let's do that. So here we have tert-butyl chloride, or tert-butanol, reacting with concentrated hydrochloric acid. So concentrated hydrochloric acid is going to function as an acid."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "Let's do an SN1 mechanism using a tertiary alcohol. So let's do that. So here we have tert-butyl chloride, or tert-butanol, reacting with concentrated hydrochloric acid. So concentrated hydrochloric acid is going to function as an acid. Our alcohol is going to function as a base. And let me just go ahead and highlight the fact this is going to be an SN1 type mechanism because we have a tertiary alcohol. This carbon bonds to the OH."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So concentrated hydrochloric acid is going to function as an acid. Our alcohol is going to function as a base. And let me just go ahead and highlight the fact this is going to be an SN1 type mechanism because we have a tertiary alcohol. This carbon bonds to the OH. It's bonded to three other carbons. And so the alcohol functions as a base and is protonated. And we would form the chloride anion here."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "This carbon bonds to the OH. It's bonded to three other carbons. And so the alcohol functions as a base and is protonated. And we would form the chloride anion here. So let's go ahead and draw that in. So we'd have the chloride anion, so negative 1 formal charge. And let's go ahead and put those electrons in blue like we did before."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And we would form the chloride anion here. So let's go ahead and draw that in. So we'd have the chloride anion, so negative 1 formal charge. And let's go ahead and put those electrons in blue like we did before. So these electrons in here, let's say that those are these electrons, so the chloride anion. We protonate the oxygen. So let's go ahead and draw that."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and put those electrons in blue like we did before. So these electrons in here, let's say that those are these electrons, so the chloride anion. We protonate the oxygen. So let's go ahead and draw that. So if we protonate the oxygen, we have our tert-butyl group over here. Protonate the oxygen, so a plus 1 formal charge on this oxygen. So let's go ahead and draw that in."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that. So if we protonate the oxygen, we have our tert-butyl group over here. Protonate the oxygen, so a plus 1 formal charge on this oxygen. So let's go ahead and draw that in. So plus 1 formal charge on the oxygen. Electrons in magenta right here picked up this proton. Let's say it's this bond right in here."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that in. So plus 1 formal charge on the oxygen. Electrons in magenta right here picked up this proton. Let's say it's this bond right in here. So once again, we have an excellent leafing group like we talked about before. Water is a good leafing group. And so in an SN1 type mechanism, these electrons are going to come off onto the oxygen, which gives us water and also gives us a carbocation."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "Let's say it's this bond right in here. So once again, we have an excellent leafing group like we talked about before. Water is a good leafing group. And so in an SN1 type mechanism, these electrons are going to come off onto the oxygen, which gives us water and also gives us a carbocation. So let's go ahead and sketch in this carbocation, a plus 1 formal charge on this carbon right here. This carbon in red gets a plus 1 formal charge. That's this carbon."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so in an SN1 type mechanism, these electrons are going to come off onto the oxygen, which gives us water and also gives us a carbocation. So let's go ahead and sketch in this carbocation, a plus 1 formal charge on this carbon right here. This carbon in red gets a plus 1 formal charge. That's this carbon. Because it's losing a bond. So let me go ahead and highlight the electrons that it's losing. So these electrons in here are coming off onto the oxygen to form water."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "That's this carbon. Because it's losing a bond. So let me go ahead and highlight the electrons that it's losing. So these electrons in here are coming off onto the oxygen to form water. So we could go ahead and draw water over here. So we lose water at this point, so H2O. And the electrons in green come off onto this oxygen right here."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here are coming off onto the oxygen to form water. So we could go ahead and draw water over here. So we lose water at this point, so H2O. And the electrons in green come off onto this oxygen right here. So we form H2O. Taking a bond away from the carbon in red, so we form a carbocation. So this is a stable carbocation."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And the electrons in green come off onto this oxygen right here. So we form H2O. Taking a bond away from the carbon in red, so we form a carbocation. So this is a stable carbocation. This is a tertiary carbocation. So that's why this tertiary alcohol reacts via an SN1 type mechanism, the stability of the carbocation. And so in our final step, we have the nucleophile is going to attack our electrophile."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this is a stable carbocation. This is a tertiary carbocation. So that's why this tertiary alcohol reacts via an SN1 type mechanism, the stability of the carbocation. And so in our final step, we have the nucleophile is going to attack our electrophile. So the nucleophile attacks our electrophile. The chloride anion attacks our carbocation, attacks that carbon there. And so we form our final product, which is tert-butyl chloride."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so in our final step, we have the nucleophile is going to attack our electrophile. So the nucleophile attacks our electrophile. The chloride anion attacks our carbocation, attacks that carbon there. And so we form our final product, which is tert-butyl chloride. So let me go ahead and draw in these electrons. And let's highlight some again. The electrons in blue, these electrons form the bond."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so we form our final product, which is tert-butyl chloride. So let me go ahead and draw in these electrons. And let's highlight some again. The electrons in blue, these electrons form the bond. So they bonded right here. And so let me go ahead and highlight this carbon in red. So this carbon in red is this one right here."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "The electrons in blue, these electrons form the bond. So they bonded right here. And so let me go ahead and highlight this carbon in red. So this carbon in red is this one right here. So we form tert-butyl chloride. And we lost water in the process. So this is a very easy reaction to do."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this carbon in red is this one right here. So we form tert-butyl chloride. And we lost water in the process. So this is a very easy reaction to do. It occurs at room temperature. And you can just take tert-butanol and add some hydrochloric acid and just shake them together. And you could form your final product this way."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this is a very easy reaction to do. It occurs at room temperature. And you can just take tert-butanol and add some hydrochloric acid and just shake them together. And you could form your final product this way. So thinking about stereochemistry in an SN1 type mechanism, the carbon in red right here is not a chiral center. And so we don't have to worry about what kind of stereochemical outcome that we would predict for the product. So this is our final product."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And you could form your final product this way. So thinking about stereochemistry in an SN1 type mechanism, the carbon in red right here is not a chiral center. And so we don't have to worry about what kind of stereochemical outcome that we would predict for the product. So this is our final product. There's no stereochemistry. But just to refresh your memory, for an SN1 type mechanism, because of this formation of this carbocation, this carbon in the carbocation is sp2 hybridized. And so it's planar."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So this is our final product. There's no stereochemistry. But just to refresh your memory, for an SN1 type mechanism, because of this formation of this carbocation, this carbon in the carbocation is sp2 hybridized. And so it's planar. And so when we draw out that carbon in red here, let's say that's that carbon in red, it's sp2 hybridized, which means that the carbons that are directly bonded to it line the same plane. So these carbons line the same plane. sp2 hybridized with a p orbital."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "And so it's planar. And so when we draw out that carbon in red here, let's say that's that carbon in red, it's sp2 hybridized, which means that the carbons that are directly bonded to it line the same plane. So these carbons line the same plane. sp2 hybridized with a p orbital. So there's a p orbital. Let's sketch that in. So there's a plus 1 formal charge in this carbon."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "sp2 hybridized with a p orbital. So there's a p orbital. Let's sketch that in. So there's a plus 1 formal charge in this carbon. So when your nucleophile attacks, your nucleophile could attack from either side of that plane. So the nucleophile could attack from this side, or it could attack from this side. And so if your final product has a chiral center, you need to think about stereochemistry."}, {"video_title": "SN1 and SN2 reactions of alcohols AOrganic chemistry Khan Academy.mp3", "Sentence": "So there's a plus 1 formal charge in this carbon. So when your nucleophile attacks, your nucleophile could attack from either side of that plane. So the nucleophile could attack from this side, or it could attack from this side. And so if your final product has a chiral center, you need to think about stereochemistry. But not in this case. In this case, we don't have one. So we lucked out."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Here's a flow sheet created by Dr. Schwartz, who was my organic chemistry professor. And what's nice about this flow sheet is it shows you all of the reactions, or most of the reactions, that you've studied in the first semester of organic chemistry. And it shows you how all of those reactions are connected to each other. And this is very, very helpful when you're trying to do a synthesis problem. When you're trying to synthesize one molecule from another molecule, you have to know the reactions that connect those different functional groups together. And a flow sheet is one of the best ways of doing it, to show you all of those different connections here. And synthesis is what organic chemistry is all about."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this is very, very helpful when you're trying to do a synthesis problem. When you're trying to synthesize one molecule from another molecule, you have to know the reactions that connect those different functional groups together. And a flow sheet is one of the best ways of doing it, to show you all of those different connections here. And synthesis is what organic chemistry is all about. How do you synthesize this molecule from that molecule? How do you synthesize a molecule that's important to humanity, that's used in medicine to help to save lives, or help to make everyone's life better in society? So synthesis is really what organic chemistry is all about."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And synthesis is what organic chemistry is all about. How do you synthesize this molecule from that molecule? How do you synthesize a molecule that's important to humanity, that's used in medicine to help to save lives, or help to make everyone's life better in society? So synthesis is really what organic chemistry is all about. And it can be very difficult, which is why you'll see a lot of these problems on organic tests. And that's why you need to do practice problems using flow sheets. So let's check out our problem here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So synthesis is really what organic chemistry is all about. And it can be very difficult, which is why you'll see a lot of these problems on organic tests. And that's why you need to do practice problems using flow sheets. So let's check out our problem here. Synthesize the following molecules using only acetylene, methyl bromide, and any inorganic reagents or solvents. So we'll start with the 2-propanol molecule. And we'll think to ourselves, OK, how do I make this alcohol from acetylene?"}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's check out our problem here. Synthesize the following molecules using only acetylene, methyl bromide, and any inorganic reagents or solvents. So we'll start with the 2-propanol molecule. And we'll think to ourselves, OK, how do I make this alcohol from acetylene? And at first, it's not so obvious what to do. One approach is to think backwards. And this is called retro synthesis."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we'll think to ourselves, OK, how do I make this alcohol from acetylene? And at first, it's not so obvious what to do. One approach is to think backwards. And this is called retro synthesis. So I'm going to draw a retro synthesis arrow here. So it looks like that. I think, well, OK, what can I make an alcohol from?"}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this is called retro synthesis. So I'm going to draw a retro synthesis arrow here. So it looks like that. I think, well, OK, what can I make an alcohol from? So I can make an alcohol from an alkene. That's one of the reactions that we've seen this semester in organic chemistry. So I draw an alkene like that."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I think, well, OK, what can I make an alcohol from? So I can make an alcohol from an alkene. That's one of the reactions that we've seen this semester in organic chemistry. So I draw an alkene like that. And I go back up to my flow sheet. And let's see if I can find that reaction. So I'm going from an alkene to an alcohol."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I draw an alkene like that. And I go back up to my flow sheet. And let's see if I can find that reaction. So I'm going from an alkene to an alcohol. So I look at my flow sheet here. And here is an alkene right here. And here I'm going from an alkene to an alcohol."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going from an alkene to an alcohol. So I look at my flow sheet here. And here is an alkene right here. And here I'm going from an alkene to an alcohol. And so I'm using this arrow right here. So I have two choices of reagents. I can add water and sulfuric acid, which would be a Markovnikov addition of the OH."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And here I'm going from an alkene to an alcohol. And so I'm using this arrow right here. So I have two choices of reagents. I can add water and sulfuric acid, which would be a Markovnikov addition of the OH. Or I could do a hydroboration oxidation, which would be an anti-Markovnikov addition of the OH. So I look down here and say, OK, which one do I want? I want a Markovnikov addition."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I can add water and sulfuric acid, which would be a Markovnikov addition of the OH. Or I could do a hydroboration oxidation, which would be an anti-Markovnikov addition of the OH. So I look down here and say, OK, which one do I want? I want a Markovnikov addition. I want to add the OH to the most substituted carbon, like this. So it's going to be water and sulfuric acid for this transition. So water and sulfuric acid like that."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I want a Markovnikov addition. I want to add the OH to the most substituted carbon, like this. So it's going to be water and sulfuric acid for this transition. So water and sulfuric acid like that. Now, I have an alkene. But I want to get back to acetylene. So I go back up here to my flow sheet."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So water and sulfuric acid like that. Now, I have an alkene. But I want to get back to acetylene. So I go back up here to my flow sheet. And I try to find acetylene on my flow sheet. So acetylene is all the way over here on the top right here. That's acetylene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I go back up here to my flow sheet. And I try to find acetylene on my flow sheet. So acetylene is all the way over here on the top right here. That's acetylene. And I have an alkene. So I need to think about retrosynthesis. I need to work backwards."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's acetylene. And I have an alkene. So I need to think about retrosynthesis. I need to work backwards. So what can I make an alkene from? I can make it from this alkyne. So I can make it from this alkyne right here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I need to work backwards. So what can I make an alkene from? I can make it from this alkyne. So I can make it from this alkyne right here. And how do I turn that alkyne into an alkene? Well, I can either use hydrogen gas and linlar palladium, which would give me a cis product. Or I can use the sodium, metal, and ammonia, which would give me the trans product here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can make it from this alkyne right here. And how do I turn that alkyne into an alkene? Well, I can either use hydrogen gas and linlar palladium, which would give me a cis product. Or I can use the sodium, metal, and ammonia, which would give me the trans product here. So let's think about retrosynthesis. So I have an alkene. I'm going to make that alkene from an alkyne."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Or I can use the sodium, metal, and ammonia, which would give me the trans product here. So let's think about retrosynthesis. So I have an alkene. I'm going to make that alkene from an alkyne. I need to think about what reagents do I want to use here. So what would the alkyne look like? Well, if I'm going to add hydrogen across a triple bond, I can just think about turning this molecule on the left into an alkyne."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to make that alkene from an alkyne. I need to think about what reagents do I want to use here. So what would the alkyne look like? Well, if I'm going to add hydrogen across a triple bond, I can just think about turning this molecule on the left into an alkyne. So all I have to do is go like that. Just add a triple bond to the molecule. I made it linear because that's, of course, what alkynes are."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, if I'm going to add hydrogen across a triple bond, I can just think about turning this molecule on the left into an alkyne. So all I have to do is go like that. Just add a triple bond to the molecule. I made it linear because that's, of course, what alkynes are. So how did I do that? Well, in this case, you don't have to worry about cis or trans. So it's pretty easy just to use hydrogen gas and then linlar palladium for this transition here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I made it linear because that's, of course, what alkynes are. So how did I do that? Well, in this case, you don't have to worry about cis or trans. So it's pretty easy just to use hydrogen gas and then linlar palladium for this transition here. So now I have this as my alkyne. And I want to get back to acetylene. So I draw my retrosynthesis arrow here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's pretty easy just to use hydrogen gas and then linlar palladium for this transition here. So now I have this as my alkyne. And I want to get back to acetylene. So I draw my retrosynthesis arrow here. And I think, OK, how can I make that molecule from acetylene like that? So I go back up to my flow sheet. So how do I make that terminal alkyne from acetylene?"}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I draw my retrosynthesis arrow here. And I think, OK, how can I make that molecule from acetylene like that? So I go back up to my flow sheet. So how do I make that terminal alkyne from acetylene? So I'm right here. I have my alkyne. And I want to make it from acetylene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So how do I make that terminal alkyne from acetylene? So I'm right here. I have my alkyne. And I want to make it from acetylene. So I'm going to do an alkylation reaction. And because I only need one alkyl group, I'm only going to do it one time. So I need a one-time alkylation reaction."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I want to make it from acetylene. So I'm going to do an alkylation reaction. And because I only need one alkyl group, I'm only going to do it one time. So I need a one-time alkylation reaction. And the first step you add is a very strong base, sodium amide. And the second step, you add a primary alkyl halide. So let's go ahead and draw that down here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I need a one-time alkylation reaction. And the first step you add is a very strong base, sodium amide. And the second step, you add a primary alkyl halide. So let's go ahead and draw that down here. So first step, I want to add sodium amide like that. And my second step, I want to add an alkyl halide. So what alkyl halide do I want to add?"}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that down here. So first step, I want to add sodium amide like that. And my second step, I want to add an alkyl halide. So what alkyl halide do I want to add? Well, I want to add a methyl group onto my alkyne. I want to add a methyl group. And up here, it says I can use methyl bromide."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what alkyl halide do I want to add? Well, I want to add a methyl group onto my alkyne. I want to add a methyl group. And up here, it says I can use methyl bromide. So that will be the alkyl halide that I will use for this alkylation reaction, so CH3Br like that. And so now we've done it. So we used retrosynthesis."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And up here, it says I can use methyl bromide. So that will be the alkyl halide that I will use for this alkylation reaction, so CH3Br like that. And so now we've done it. So we used retrosynthesis. And if you were to write this on a test, you'd probably write it in the reverse order here. So you would start with acetylene and then use the regular arrow. And the first step, add sodium amides and then add methyl bromide for an alkylation reaction to put this alkyl group onto acetylene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we used retrosynthesis. And if you were to write this on a test, you'd probably write it in the reverse order here. So you would start with acetylene and then use the regular arrow. And the first step, add sodium amides and then add methyl bromide for an alkylation reaction to put this alkyl group onto acetylene. Now you have a terminal alkyne, which you can turn into an alkene by the addition of hydrogen gas and a poison catalyst, which stops the hydrogenation at the alkene form. And then you can use a Markovnikov addition of OH using water and sulfuric acid to add OH to your alkene. And then you're finally done."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the first step, add sodium amides and then add methyl bromide for an alkylation reaction to put this alkyl group onto acetylene. Now you have a terminal alkyne, which you can turn into an alkene by the addition of hydrogen gas and a poison catalyst, which stops the hydrogenation at the alkene form. And then you can use a Markovnikov addition of OH using water and sulfuric acid to add OH to your alkene. And then you're finally done. So that's the approach that you should take when you're doing a synthesis reaction. All right, so we have time to do one more here. Let's do the synthesis of the molecule on the right."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then you're finally done. So that's the approach that you should take when you're doing a synthesis reaction. All right, so we have time to do one more here. Let's do the synthesis of the molecule on the right. So you look at the molecule on the right. And you think to yourself, all right, so let's go ahead and redraw that molecule on the right down here. We have some more room."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's do the synthesis of the molecule on the right. So you look at the molecule on the right. And you think to yourself, all right, so let's go ahead and redraw that molecule on the right down here. We have some more room. So here I have my molecule on the right. Let's go ahead and use the yellow for our synthesis problems here. So this is my molecule."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have some more room. So here I have my molecule on the right. Let's go ahead and use the yellow for our synthesis problems here. So this is my molecule. I have two bromines trans from each other. And immediately that should make you think about a reaction we just did in the last video. Halogenation will add these guys on trans to an alkyne."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is my molecule. I have two bromines trans from each other. And immediately that should make you think about a reaction we just did in the last video. Halogenation will add these guys on trans to an alkyne. So I can add bromine. And my solvent would be carbon tetrachloride, so CCl4. We just did this in the last video."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Halogenation will add these guys on trans to an alkyne. So I can add bromine. And my solvent would be carbon tetrachloride, so CCl4. We just did this in the last video. So watch the last video for halogenation of alkynes. And what would that give us? What would that give us for our alkyne here?"}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We just did this in the last video. So watch the last video for halogenation of alkynes. And what would that give us? What would that give us for our alkyne here? So we have our alkyne. And then we'd have to have an ethyl group on this side. And that would be the reaction."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What would that give us for our alkyne here? So we have our alkyne. And then we'd have to have an ethyl group on this side. And that would be the reaction. So now we have an alkyne. And I need to make this alkyne from acetylene. So I'm once again at this stage."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that would be the reaction. So now we have an alkyne. And I need to make this alkyne from acetylene. So I'm once again at this stage. So I'm once again right here. I have this alkyne. I want to make it from acetylene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm once again at this stage. So I'm once again right here. I have this alkyne. I want to make it from acetylene. So I'm going to do my alkylation again. So I'm going to add sodium amide, my first step. And an alkyl halide in my second step."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I want to make it from acetylene. So I'm going to do my alkylation again. So I'm going to add sodium amide, my first step. And an alkyl halide in my second step. So let's go ahead and draw that. So let's go ahead and draw my retrosynthesis arrow here. So I'm trying to synthesize that alkyne from acetylene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And an alkyl halide in my second step. So let's go ahead and draw that. So let's go ahead and draw my retrosynthesis arrow here. So I'm trying to synthesize that alkyne from acetylene. So let's go ahead and draw acetylene in here. And I know I can do that in two steps. First step, add my sodium amide like that."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm trying to synthesize that alkyne from acetylene. So let's go ahead and draw acetylene in here. And I know I can do that in two steps. First step, add my sodium amide like that. And in my second step, I need to add an alkyl halide. What kind of alkyl halide do I need to add? I need to add an ethyl group onto acetylene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "First step, add my sodium amide like that. And in my second step, I need to add an alkyl halide. What kind of alkyl halide do I need to add? I need to add an ethyl group onto acetylene. So I need something like ethyl bromide would work. So ethyl bromide like that. And that would give me my product."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I need to add an ethyl group onto acetylene. So I need something like ethyl bromide would work. So ethyl bromide like that. And that would give me my product. And you might think I'm done. I'm done with my synthesis. But in reality, you are not done."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that would give me my product. And you might think I'm done. I'm done with my synthesis. But in reality, you are not done. Because if you go back and you read the question, the question says, synthesize the following molecules using only acetylene or methyl bromide, not ethyl bromide. So you actually can't use ethyl bromide in terms of stopping right here. You need to figure out a way to make ethyl bromide."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But in reality, you are not done. Because if you go back and you read the question, the question says, synthesize the following molecules using only acetylene or methyl bromide, not ethyl bromide. So you actually can't use ethyl bromide in terms of stopping right here. You need to figure out a way to make ethyl bromide. Ethyl bromide contains two carbons. So let's see if we can think about a way to make ethyl bromide from acetylene right here. So let's think to ourselves, OK, so this way works."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You need to figure out a way to make ethyl bromide. Ethyl bromide contains two carbons. So let's see if we can think about a way to make ethyl bromide from acetylene right here. So let's think to ourselves, OK, so this way works. Now I have to figure a way to make ethyl bromide from acetylene. So once again, we use retrosynthesis. So retrosynthesis here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's think to ourselves, OK, so this way works. Now I have to figure a way to make ethyl bromide from acetylene. So once again, we use retrosynthesis. So retrosynthesis here. And I think, OK, I have an alkyl halide. How do I make an alkyl halide from an alkyne? And once again, it's helpful to look at your flow sheet."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So retrosynthesis here. And I think, OK, I have an alkyl halide. How do I make an alkyl halide from an alkyne? And once again, it's helpful to look at your flow sheet. So let's go back up and look at our flow sheet. So we want to make an alkyl halide from an alkyne. So let's find our alkyl halide."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And once again, it's helpful to look at your flow sheet. So let's go back up and look at our flow sheet. So we want to make an alkyl halide from an alkyne. So let's find our alkyl halide. And it's all the way over here. So our alkyl halide is all the way over here. So if we're doing retrosynthesis, I want to get back to an alkyne."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's find our alkyl halide. And it's all the way over here. So our alkyl halide is all the way over here. So if we're doing retrosynthesis, I want to get back to an alkyne. So I can make an alkyl halide from an alkene if I add a hydrogen halide, so HX. So retrosynthesis, I'm going to make my alkyl halide from an alkene using hydrogen halide here. So I'm going to go ahead and let's see."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if we're doing retrosynthesis, I want to get back to an alkyne. So I can make an alkyl halide from an alkene if I add a hydrogen halide, so HX. So retrosynthesis, I'm going to make my alkyl halide from an alkene using hydrogen halide here. So I'm going to go ahead and let's see. What am I going to write? I'm going to have it start with an alkene. So that's my alkene."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and let's see. What am I going to write? I'm going to have it start with an alkene. So that's my alkene. It's just ethene or ethylene. And my halogen is bromine here. So my hydrogen halide would have to be HBr."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's my alkene. It's just ethene or ethylene. And my halogen is bromine here. So my hydrogen halide would have to be HBr. So if I add HBr to ethylene, I will make ethyl bromide. And I'm getting there. I have an alkene now."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So my hydrogen halide would have to be HBr. So if I add HBr to ethylene, I will make ethyl bromide. And I'm getting there. I have an alkene now. But I have to start with acetylene. So how do I make an alkene from an alkyne? Once again, just go back to refresh everyone's memory on the flow sheet."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I have an alkene now. But I have to start with acetylene. So how do I make an alkene from an alkyne? Once again, just go back to refresh everyone's memory on the flow sheet. So now I am right here on the flow sheet. I have my alkene. And I know how to make an alkene from an alkyne."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Once again, just go back to refresh everyone's memory on the flow sheet. So now I am right here on the flow sheet. I have my alkene. And I know how to make an alkene from an alkyne. All I have to do is hydrogenate it using my poison catalyst, using my linlar palladium. So let's go ahead and draw that in. So I can make my alkene from an alkyne if I add hydrogen gas and my poisoned catalyst, linlar palladium here."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I know how to make an alkene from an alkyne. All I have to do is hydrogenate it using my poison catalyst, using my linlar palladium. So let's go ahead and draw that in. So I can make my alkene from an alkyne if I add hydrogen gas and my poisoned catalyst, linlar palladium here. And that'll be acetylene, actually. So if I take acetylene here. And let's go ahead and check our synthesis."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can make my alkene from an alkyne if I add hydrogen gas and my poisoned catalyst, linlar palladium here. And that'll be acetylene, actually. So if I take acetylene here. And let's go ahead and check our synthesis. I meet all my qualifications. And again, if I start with acetylene, and let's just walk back through here. And I add hydrogen gas and poison catalyst."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and check our synthesis. I meet all my qualifications. And again, if I start with acetylene, and let's just walk back through here. And I add hydrogen gas and poison catalyst. I'm going to hydrogenate my alkyne to form my alkene right here. I take my alkene and I add HBr to it. And that's going to add H plus and Br minus across my double bond to give me ethyl bromide as my product."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I add hydrogen gas and poison catalyst. I'm going to hydrogenate my alkyne to form my alkene right here. I take my alkene and I add HBr to it. And that's going to add H plus and Br minus across my double bond to give me ethyl bromide as my product. And I take another molecule of acetylene. And to that, I add sodium amide. And I add the ethyl bromide that I just created from acetylene to alkylate my alkyne, put an ethyl group on there."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that's going to add H plus and Br minus across my double bond to give me ethyl bromide as my product. And I take another molecule of acetylene. And to that, I add sodium amide. And I add the ethyl bromide that I just created from acetylene to alkylate my alkyne, put an ethyl group on there. And then finally, I do a halogenation reaction of alkynes to add on my bromines anti to each other like that. And so we're done. And so once again, if you're writing this on a test, you'd probably make sense to write it the other way."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I add the ethyl bromide that I just created from acetylene to alkylate my alkyne, put an ethyl group on there. And then finally, I do a halogenation reaction of alkynes to add on my bromines anti to each other like that. And so we're done. And so once again, if you're writing this on a test, you'd probably make sense to write it the other way. But this is just a good way of thinking about it with the flow sheet. Let's go back up here, take one more look at the flow sheet. So I encourage you to use this flow sheet."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, if you're writing this on a test, you'd probably make sense to write it the other way. But this is just a good way of thinking about it with the flow sheet. Let's go back up here, take one more look at the flow sheet. So I encourage you to use this flow sheet. Make your own. Do lots of practice problems. Do lots of synthesis practice problems."}, {"video_title": "Synthesis using alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I encourage you to use this flow sheet. Make your own. Do lots of practice problems. Do lots of synthesis practice problems. You can have your friends make them. And then you can solve their synthesis problems. You can make synthesis problems yourself."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "We call these divergent boundaries. And the example we showed of this was the Mid-Atlantic Ridge, where essentially new crustal material is being created. Now on the other side of the equation, you have areas where plates are ramming into each other. We see that over here, where the Nazca plate is running into the South American plate. We see it over here, where the Pacific plate is running into the Filipino plate. They're running into each other. So what happens over there?"}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "We see that over here, where the Nazca plate is running into the South American plate. We see it over here, where the Pacific plate is running into the Filipino plate. They're running into each other. So what happens over there? So what we're going to do is just go through the different scenarios. The general idea is that one plate is going to get subducted under another. They're ramming into each other."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "So what happens over there? So what we're going to do is just go through the different scenarios. The general idea is that one plate is going to get subducted under another. They're ramming into each other. One is going to get essentially pushed under the other one. This diagram shows some subduction over here. This is essentially an oceanic plate being subducted under another oceanic plate."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "They're ramming into each other. One is going to get essentially pushed under the other one. This diagram shows some subduction over here. This is essentially an oceanic plate being subducted under another oceanic plate. So not too different than what might happen where the Pacific plate runs into the Filipino plate right over here. And then on this side of the diagram, we see an oceanic plate and the oceanic crust getting subducted under a continental plate right over here. And this is what's happening when the Nazca plate is getting subducted under the South American plate."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "This is essentially an oceanic plate being subducted under another oceanic plate. So not too different than what might happen where the Pacific plate runs into the Filipino plate right over here. And then on this side of the diagram, we see an oceanic plate and the oceanic crust getting subducted under a continental plate right over here. And this is what's happening when the Nazca plate is getting subducted under the South American plate. And when that happens, a couple of things. So you have the oceanic plate being pushed under. And what happens at the same time, the continental plate gets pushed upwards, causing mountain ranges like the Andes, and that's exactly what has created the Andes."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And this is what's happening when the Nazca plate is getting subducted under the South American plate. And when that happens, a couple of things. So you have the oceanic plate being pushed under. And what happens at the same time, the continental plate gets pushed upwards, causing mountain ranges like the Andes, and that's exactly what has created the Andes. It's that upward force from the Nazca plate being pushed under the South American plate at that coastline. And what you're also going to see is, and you can imagine, you have these huge plates grinding past each other. And it's not a very smooth process."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And what happens at the same time, the continental plate gets pushed upwards, causing mountain ranges like the Andes, and that's exactly what has created the Andes. It's that upward force from the Nazca plate being pushed under the South American plate at that coastline. And what you're also going to see is, and you can imagine, you have these huge plates grinding past each other. And it's not a very smooth process. Every now and then, you kind of reach a breaking point and huge amounts of energy get released. So you're also going to see a lot of earthquakes in those areas. And we know that Chile has a lot of earthquakes."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And it's not a very smooth process. Every now and then, you kind of reach a breaking point and huge amounts of energy get released. So you're also going to see a lot of earthquakes in those areas. And we know that Chile has a lot of earthquakes. And then on top of that, this is going to result in a lot of heat and a lot of the friction of the plates grinding past each other, essentially allowing magma to form at that part of the rock. Because it's getting so heated, and so you'll also have volcanoes in these areas, where essentially something is being subducted underneath a continental plate. Now, we also talked about what's happening in the Pacific, where we have the Pacific plate being subducted under the Filipino plate."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And we know that Chile has a lot of earthquakes. And then on top of that, this is going to result in a lot of heat and a lot of the friction of the plates grinding past each other, essentially allowing magma to form at that part of the rock. Because it's getting so heated, and so you'll also have volcanoes in these areas, where essentially something is being subducted underneath a continental plate. Now, we also talked about what's happening in the Pacific, where we have the Pacific plate being subducted under the Filipino plate. That's what we kept referring to over here. And that's doing a couple of interesting things. Whenever you have subduction, you have trenches."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "Now, we also talked about what's happening in the Pacific, where we have the Pacific plate being subducted under the Filipino plate. That's what we kept referring to over here. And that's doing a couple of interesting things. Whenever you have subduction, you have trenches. But it's most interesting, or at least in my mind, the deepest trenches have been created where you have an oceanic plate being subducted under another oceanic plate. So a couple of things are going to happen. You're going to have a very deep, you're going to have trenches form."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "Whenever you have subduction, you have trenches. But it's most interesting, or at least in my mind, the deepest trenches have been created where you have an oceanic plate being subducted under another oceanic plate. So a couple of things are going to happen. You're going to have a very deep, you're going to have trenches form. Over here, we see in this diagram, we also have a trench in the first example. But you have trenches form where one oceanic plate is being subducted under another. And then you have that same type of friction that you saw over here create volcanoes."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "You're going to have a very deep, you're going to have trenches form. Over here, we see in this diagram, we also have a trench in the first example. But you have trenches form where one oceanic plate is being subducted under another. And then you have that same type of friction that you saw over here create volcanoes. And those volcanoes will initially be underwater volcanoes, since these are both oceanic plates. Or we're dealing with oceanic crust at that point of the plate. It doesn't have to be entirely an oceanic plate."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And then you have that same type of friction that you saw over here create volcanoes. And those volcanoes will initially be underwater volcanoes, since these are both oceanic plates. Or we're dealing with oceanic crust at that point of the plate. It doesn't have to be entirely an oceanic plate. And they'll first be underwater volcanoes. But as the lava piles up and hardens, it'll eventually turn into a group of islands. And we have that happening where the Pacific plate runs into the Filipino plate."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "It doesn't have to be entirely an oceanic plate. And they'll first be underwater volcanoes. But as the lava piles up and hardens, it'll eventually turn into a group of islands. And we have that happening where the Pacific plate runs into the Filipino plate. And first, we have the trench. So let me just draw everything right here. So this is the boundary, roughly speaking, between the two plates."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And we have that happening where the Pacific plate runs into the Filipino plate. And first, we have the trench. So let me just draw everything right here. So this is the boundary, roughly speaking, between the two plates. This is the Pacific plate. And this is the Filipino plate, right over here. And so where it's being subducted, you have the Mariana Trench, which is the deepest trench in the world."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "So this is the boundary, roughly speaking, between the two plates. This is the Pacific plate. And this is the Filipino plate, right over here. And so where it's being subducted, you have the Mariana Trench, which is the deepest trench in the world. It goes down 11 kilometers, 11,000 meters. That's deeper than Mount Everest is high. Mount Everest is about 9,000 meters high."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And so where it's being subducted, you have the Mariana Trench, which is the deepest trench in the world. It goes down 11 kilometers, 11,000 meters. That's deeper than Mount Everest is high. Mount Everest is about 9,000 meters high. And we'll see that's also due to another convergent plate boundary, another place where plates are running into each other. So not only do you see the Mariana Trench here, because one plate is being subducted under the other, you see the formation of the Mariana Islands, which are essentially created from underwater volcanoes because of all of the energy being released. And this is actually a depiction of what's the subduction that's happening at the Mariana Trench."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "Mount Everest is about 9,000 meters high. And we'll see that's also due to another convergent plate boundary, another place where plates are running into each other. So not only do you see the Mariana Trench here, because one plate is being subducted under the other, you see the formation of the Mariana Islands, which are essentially created from underwater volcanoes because of all of the energy being released. And this is actually a depiction of what's the subduction that's happening at the Mariana Trench. You have this subduction over here, and then you have the Mariana Islands being created by essentially the energy causing a magma and lava, essentially magma before it surfaces, to flow to the top. And as lava just goes and starts building these islands. Now the last type of convergent boundary is when you have two parts of continental crust running into each other."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "And this is actually a depiction of what's the subduction that's happening at the Mariana Trench. You have this subduction over here, and then you have the Mariana Islands being created by essentially the energy causing a magma and lava, essentially magma before it surfaces, to flow to the top. And as lava just goes and starts building these islands. Now the last type of convergent boundary is when you have two parts of continental crust running into each other. So that's the situation that we have where the Indian plate is running into the Eurasian plate. And I think you might already guess what's going to happen there. When you have two pieces of continental crust running into each other, one isn't more or less dense than the other."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "Now the last type of convergent boundary is when you have two parts of continental crust running into each other. So that's the situation that we have where the Indian plate is running into the Eurasian plate. And I think you might already guess what's going to happen there. When you have two pieces of continental crust running into each other, one isn't more or less dense than the other. And so at least the crustal portions of them are just going to keep jamming into each other. And so they're just going to push things upward. This is a depiction right here that I got from the USGS."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "When you have two pieces of continental crust running into each other, one isn't more or less dense than the other. And so at least the crustal portions of them are just going to keep jamming into each other. And so they're just going to push things upward. This is a depiction right here that I got from the USGS. And what's kind of depicting is this is the Indian plate, this is the Eurasian plate. This is if you rewind a good bit before they've really had a chance to jam into each other. But as they're jamming into each other, the Indian plate is kind of digging in a little bit, not being fully subducted, and it's causing the land to rise."}, {"video_title": "Plate Tectonics-- Geological features of Convergent Plate Boundaries.mp3", "Sentence": "This is a depiction right here that I got from the USGS. And what's kind of depicting is this is the Indian plate, this is the Eurasian plate. This is if you rewind a good bit before they've really had a chance to jam into each other. But as they're jamming into each other, the Indian plate is kind of digging in a little bit, not being fully subducted, and it's causing the land to rise. And what that essentially ends up with is you end up with something like the Himalayas. And this right here is a picture of Mount Everest, which is almost 9,000 meters high, 9,000 meters above sea level. So it's almost as high as the Mariana Trench is deep."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It has enormous pressure, enormous inward pressure on this core. Because as we form heavier and heavier elements in the core, the core gets denser and denser and denser. And so we keep fusing more and more elements into iron. This iron core becomes more and more massive, more and more dense. It's squeezing in on itself. And it's not fusing. That is not exothermic anymore."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This iron core becomes more and more massive, more and more dense. It's squeezing in on itself. And it's not fusing. That is not exothermic anymore. If iron were to fuse, it would not even be an exothermic process. It would require energy. So it wouldn't be even something that could be helped to fend off this squeezing, to fend off this increasing density of the core."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is not exothermic anymore. If iron were to fuse, it would not even be an exothermic process. It would require energy. So it wouldn't be even something that could be helped to fend off this squeezing, to fend off this increasing density of the core. So we have this iron here, and it just gets more and more massive, more and more dense. And so at some mass, already a reasonably high mass, the only thing that's keeping this from just completely collapsing is what we could call electron degeneracy pressure. So let me write this here."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it wouldn't be even something that could be helped to fend off this squeezing, to fend off this increasing density of the core. So we have this iron here, and it just gets more and more massive, more and more dense. And so at some mass, already a reasonably high mass, the only thing that's keeping this from just completely collapsing is what we could call electron degeneracy pressure. So let me write this here. Electron degeneracy pressure. And all this means is we have all of these iron atoms. We have all of these iron atoms getting really, really, really close to each other."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me write this here. Electron degeneracy pressure. And all this means is we have all of these iron atoms. We have all of these iron atoms getting really, really, really close to each other. The only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether is that they have these electrons. You have these electrons, and these are being squeezed together now. We're talking about unbelievably dense states of matter."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We have all of these iron atoms getting really, really, really close to each other. The only thing that keeps it from collapsing at this earlier stage, the only thing that keeps it from collapsing altogether is that they have these electrons. You have these electrons, and these are being squeezed together now. We're talking about unbelievably dense states of matter. And electron degeneracy pressure is essentially what it's saying. These electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it, but they cannot be squeezed into each other anymore."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking about unbelievably dense states of matter. And electron degeneracy pressure is essentially what it's saying. These electrons don't want to be in the same place at the same time. I won't go into the quantum mechanics of it, but they cannot be squeezed into each other anymore. So that, at least temporarily, holds this thing from collapsing even further. And in the case of a less massive star, in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure, so this is our core now, even more gravitational pressure, eventually even this electron degeneracy, I guess we could call it force or pressure, this outward pressure, this thing that keeps it from collapsing, even that gives in."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I won't go into the quantum mechanics of it, but they cannot be squeezed into each other anymore. So that, at least temporarily, holds this thing from collapsing even further. And in the case of a less massive star, in the case of a white dwarf, that's how a white dwarf actually maintains its shape, because of the electron degeneracy pressure. But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure, so this is our core now, even more gravitational pressure, eventually even this electron degeneracy, I guess we could call it force or pressure, this outward pressure, this thing that keeps it from collapsing, even that gives in. And then we have something called electron capture. Which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleus."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But as this iron core gets even more massive, more dense, and we get more and more gravitational pressure, so this is our core now, even more gravitational pressure, eventually even this electron degeneracy, I guess we could call it force or pressure, this outward pressure, this thing that keeps it from collapsing, even that gives in. And then we have something called electron capture. Which is essentially the electrons get captured by protons in the nucleus. They start collapsing into the nucleus. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons, you have neutrinos being released, but you can imagine, an enormous amount of energy is also being released. So this is kind of a temporary, and then all of a sudden this collapses. This collapses even more, until all you have, and all the protons are turning into neutrons, because they're capturing electrons."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They start collapsing into the nucleus. It's kind of the opposite of beta negative decay, where you have the electrons get captured, protons get turned into neutrons, you have neutrinos being released, but you can imagine, an enormous amount of energy is also being released. So this is kind of a temporary, and then all of a sudden this collapses. This collapses even more, until all you have, and all the protons are turning into neutrons, because they're capturing electrons. So then what you eventually have is this entire core is collapsing into a dense ball of neutrons. So dense neutrons. You can kind of view them as just one really, really, really, really, really massive atom, because it's just a dense ball of neutrons."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This collapses even more, until all you have, and all the protons are turning into neutrons, because they're capturing electrons. So then what you eventually have is this entire core is collapsing into a dense ball of neutrons. So dense neutrons. You can kind of view them as just one really, really, really, really, really massive atom, because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can kind of view them as just one really, really, really, really, really massive atom, because it's just a dense ball of neutrons. At the same time, when this collapse happens, you have an enormous amount of energy being released in the form of neutrinos. Did I say that neutrons are being released? No, no, no. The electrons are being captured by the protons, protons turning into neutrons, this dense ball of neutrons right here, and in the process neutrinos get released, these fundamental particles. We won't go into the details here. But it's an enormous amount of energy."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "No, no, no. The electrons are being captured by the protons, protons turning into neutrons, this dense ball of neutrons right here, and in the process neutrinos get released, these fundamental particles. We won't go into the details here. But it's an enormous amount of energy. And this actually is not really, really well understood of all of the dynamics here, because at the same time, that this iron core is undergoing through this, it first kind of pauses due to the electron degeneracy pressure, and then it finally gives in because it's so massive, and then it collapses into this dense ball of neutrons. But when it does it, all of this energy is released. It's not clear how, because it has to be a lot of energy, because remember, this is a massive star, so you have a lot of mass in this area over here."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's an enormous amount of energy. And this actually is not really, really well understood of all of the dynamics here, because at the same time, that this iron core is undergoing through this, it first kind of pauses due to the electron degeneracy pressure, and then it finally gives in because it's so massive, and then it collapses into this dense ball of neutrons. But when it does it, all of this energy is released. It's not clear how, because it has to be a lot of energy, because remember, this is a massive star, so you have a lot of mass in this area over here. But it's so much energy that it causes the rest of the star to explode outward in an unbelievable, I guess, unbelievably bright or energetic explosion, and that's called a supernova. And the reason why it's called nova, it comes from, I believe, I'm not an expert here, Latin for new. And the first time people observed a nova, they thought it was a new star, because all of a sudden, something they didn't see before, all of a sudden, it looks like a star appeared, because maybe it wasn't bright enough for us to observe it before, but then when the nova occurred, it did become bright enough, so it comes from the idea of new."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's not clear how, because it has to be a lot of energy, because remember, this is a massive star, so you have a lot of mass in this area over here. But it's so much energy that it causes the rest of the star to explode outward in an unbelievable, I guess, unbelievably bright or energetic explosion, and that's called a supernova. And the reason why it's called nova, it comes from, I believe, I'm not an expert here, Latin for new. And the first time people observed a nova, they thought it was a new star, because all of a sudden, something they didn't see before, all of a sudden, it looks like a star appeared, because maybe it wasn't bright enough for us to observe it before, but then when the nova occurred, it did become bright enough, so it comes from the idea of new. But a supernova is when you have a pretty massive star's core collapsing, and that energy is being released to explode the rest of the star out at unbelievable velocities. So just to kind of fathom the amount of energy that's being released in a supernova, it can temporarily outshine an entire galaxy, and in a galaxy, we're talking about hundreds of billions of stars. Or another way to think about it, in that very short period of time, it can release as much energy as the sun will in its entire lifetime."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the first time people observed a nova, they thought it was a new star, because all of a sudden, something they didn't see before, all of a sudden, it looks like a star appeared, because maybe it wasn't bright enough for us to observe it before, but then when the nova occurred, it did become bright enough, so it comes from the idea of new. But a supernova is when you have a pretty massive star's core collapsing, and that energy is being released to explode the rest of the star out at unbelievable velocities. So just to kind of fathom the amount of energy that's being released in a supernova, it can temporarily outshine an entire galaxy, and in a galaxy, we're talking about hundreds of billions of stars. Or another way to think about it, in that very short period of time, it can release as much energy as the sun will in its entire lifetime. So these are unbelievably energetic events. And so you actually have the material that's not in the core being shot out of the star at appreciable percentages of the actual speed of light. So we're talking about things being shot out at up to 10%, 10% the speed of light."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or another way to think about it, in that very short period of time, it can release as much energy as the sun will in its entire lifetime. So these are unbelievably energetic events. And so you actually have the material that's not in the core being shot out of the star at appreciable percentages of the actual speed of light. So we're talking about things being shot out at up to 10%, 10% the speed of light. That's 30,000 kilometers per second. That's almost circumnavigating the Earth every second. So this is unbelievably energetic events that we're talking about here."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're talking about things being shot out at up to 10%, 10% the speed of light. That's 30,000 kilometers per second. That's almost circumnavigating the Earth every second. So this is unbelievably energetic events that we're talking about here. And so if the star, if the original star was, and these are rough estimates, people don't have kind of a hard limit here. If the original star is 9 to 20, I'll say approximately 9 to 20 times the mass of the sun, then it will supernova, and the core will turn into what's called a neutron star. This is a neutron star."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is unbelievably energetic events that we're talking about here. And so if the star, if the original star was, and these are rough estimates, people don't have kind of a hard limit here. If the original star is 9 to 20, I'll say approximately 9 to 20 times the mass of the sun, then it will supernova, and the core will turn into what's called a neutron star. This is a neutron star. Which you can imagine is just this dense ball, it's this dense ball of neutrons. And just to give you a sense of it, it'll be something about maybe 2 times the mass of the sun, give or take, 1.5 to 3 times the mass of the sun. So this is 1.5 to 3 times the mass of the sun in a volume that has a diameter of about, on the order of tens of kilometers."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is a neutron star. Which you can imagine is just this dense ball, it's this dense ball of neutrons. And just to give you a sense of it, it'll be something about maybe 2 times the mass of the sun, give or take, 1.5 to 3 times the mass of the sun. So this is 1.5 to 3 times the mass of the sun in a volume that has a diameter of about, on the order of tens of kilometers. So roughly the size of a city, in a diameter of a city. So this is unbelievably dense, diameter of a city. I mean, we know how much larger the sun is relative to the earth, and we know how much larger the earth is relative to a city, but this is something more mass than the sun being squeezed into the density or into the size of the city."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is 1.5 to 3 times the mass of the sun in a volume that has a diameter of about, on the order of tens of kilometers. So roughly the size of a city, in a diameter of a city. So this is unbelievably dense, diameter of a city. I mean, we know how much larger the sun is relative to the earth, and we know how much larger the earth is relative to a city, but this is something more mass than the sun being squeezed into the density or into the size of the city. So unbelievably dense. Now, if the original star is even more massive, if it's more than 20 times the sun, so let me write it over here, let me scroll up. If it's greater than 20 times the sun, then even the neutron degeneracy pressure, even the pressure, even the neutrons' inability to squeeze further will give up and it'll turn into a black hole."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I mean, we know how much larger the sun is relative to the earth, and we know how much larger the earth is relative to a city, but this is something more mass than the sun being squeezed into the density or into the size of the city. So unbelievably dense. Now, if the original star is even more massive, if it's more than 20 times the sun, so let me write it over here, let me scroll up. If it's greater than 20 times the sun, then even the neutron degeneracy pressure, even the pressure, even the neutrons' inability to squeeze further will give up and it'll turn into a black hole. I can do many videos on that, and that's actually an open area of research still on exactly what's going on inside of a black hole. But then you turn into a black hole where essentially all of the mass gets condensed into an infinitely small and dense point, so something unbelievably hard to imagine. And just to give you a sense of it, this will be more mass than even 3 times the mass of the sun, so we're talking about an incredibly high amount of mass."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If it's greater than 20 times the sun, then even the neutron degeneracy pressure, even the pressure, even the neutrons' inability to squeeze further will give up and it'll turn into a black hole. I can do many videos on that, and that's actually an open area of research still on exactly what's going on inside of a black hole. But then you turn into a black hole where essentially all of the mass gets condensed into an infinitely small and dense point, so something unbelievably hard to imagine. And just to give you a sense of it, this will be more mass than even 3 times the mass of the sun, so we're talking about an incredibly high amount of mass. So just to kind of visualize things, here's actually a remnant of a supernova. This is the Crab Nebula. This right here is the Crab Nebula."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just to give you a sense of it, this will be more mass than even 3 times the mass of the sun, so we're talking about an incredibly high amount of mass. So just to kind of visualize things, here's actually a remnant of a supernova. This is the Crab Nebula. This right here is the Crab Nebula. And it's about 6,500 light years away. So it's still, you know, from galactic scale, if you think of our galaxy as being 100,000 light years in diameter, it's still not too far from us on those scales, but it's an enormous distance. The closest star to us is 4 light years away, and it would take Voyager traveling at 60,000 kilometers an hour 80,000 years to get there."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right here is the Crab Nebula. And it's about 6,500 light years away. So it's still, you know, from galactic scale, if you think of our galaxy as being 100,000 light years in diameter, it's still not too far from us on those scales, but it's an enormous distance. The closest star to us is 4 light years away, and it would take Voyager traveling at 60,000 kilometers an hour 80,000 years to get there. So this is a very, very... That's only 4 light years. That is 6,500 light years. So this supernova, it's believed happened 1,000 years ago, right at the center."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The closest star to us is 4 light years away, and it would take Voyager traveling at 60,000 kilometers an hour 80,000 years to get there. So this is a very, very... That's only 4 light years. That is 6,500 light years. So this supernova, it's believed happened 1,000 years ago, right at the center. And so at the center here, we should have a neutron star. And this cloud, the shock wave that you see here, this is still the material traveling outward from that supernova over 1,000 years. This shock wave, or the diameter of the sphere of material, is 6 light years."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this supernova, it's believed happened 1,000 years ago, right at the center. And so at the center here, we should have a neutron star. And this cloud, the shock wave that you see here, this is still the material traveling outward from that supernova over 1,000 years. This shock wave, or the diameter of the sphere of material, is 6 light years. So we could say this distance right here is 6 light years. So this is an enormously big shock wave cloud. And actually, we believe that our solar system started to form, it started to condense because of a shock wave created by a supernova relatively near to us."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This shock wave, or the diameter of the sphere of material, is 6 light years. So we could say this distance right here is 6 light years. So this is an enormously big shock wave cloud. And actually, we believe that our solar system started to form, it started to condense because of a shock wave created by a supernova relatively near to us. And just to answer another question that was kind of jumping up probably in the last video, and this is still not really, really well understood, we talk about how elements up to iron or maybe nickel can be formed inside of the cores of massive stars. So you can imagine when the star explodes, a lot of that material is released into the universe. And so that's why we have a lot of these materials in our own bodies."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And actually, we believe that our solar system started to form, it started to condense because of a shock wave created by a supernova relatively near to us. And just to answer another question that was kind of jumping up probably in the last video, and this is still not really, really well understood, we talk about how elements up to iron or maybe nickel can be formed inside of the cores of massive stars. So you can imagine when the star explodes, a lot of that material is released into the universe. And so that's why we have a lot of these materials in our own bodies. In fact, we could not exist if these heavier elements were not formed inside of the cores of primitive stars, stars that have supernovaed a long time ago. Now, the question is, how do these heavier elements form? How do we get all of this other stuff on the periodic table?"}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that's why we have a lot of these materials in our own bodies. In fact, we could not exist if these heavier elements were not formed inside of the cores of primitive stars, stars that have supernovaed a long time ago. Now, the question is, how do these heavier elements form? How do we get all of this other stuff on the periodic table? How do we get all these other heavier elements? And they're formed during the supernova itself. It's so energetic."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "How do we get all of this other stuff on the periodic table? How do we get all these other heavier elements? And they're formed during the supernova itself. It's so energetic. You have all sorts of particles streaming out and streaming in, streaming out because of the force of the shock wave, streaming in because of the gravity. You have all sorts of kind of a mishmash of elements forming, and that's actually where you have your heavier elements forming. And because, and I'll talk more about this in future videos, most of the uranium, or actually all of the uranium on Earth right now must have been formed in some type of a supernova explosion, at least based on our current understanding."}, {"video_title": "Supernova (supernovae) Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's so energetic. You have all sorts of particles streaming out and streaming in, streaming out because of the force of the shock wave, streaming in because of the gravity. You have all sorts of kind of a mishmash of elements forming, and that's actually where you have your heavier elements forming. And because, and I'll talk more about this in future videos, most of the uranium, or actually all of the uranium on Earth right now must have been formed in some type of a supernova explosion, at least based on our current understanding. And it looks to be about 4.6 billion years old. So given that it looks to be about 4.6 billion years old, based on how fast it's decayed, and I'll do a whole video on that, that's why we think that Earth was probably, that our solar system was first formed from some type of supernova explosion because that uranium would have been formed right at about the birth of our solar system. Anyway, hopefully you found that interesting."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And a carbocation has a carbon that's positively charged, which we call a cation in chemistry. So if you take these words, carbon and cation, and combine them, you get carbocation. Let's look at the picture in the middle first. So this carbon is positively charged. Normally, carbon has four bonds to it, but here it has only three. So here's one, here's two, and here's three. And because it has only three bonds, it has a plus one formal charge."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So this carbon is positively charged. Normally, carbon has four bonds to it, but here it has only three. So here's one, here's two, and here's three. And because it has only three bonds, it has a plus one formal charge. And carbocations are very reactive, because carbon likes to form four bonds. Let's look at the same carbocation over here on the left. And the carbon in magenta, the one with the plus one formal charge, is this one."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And because it has only three bonds, it has a plus one formal charge. And carbocations are very reactive, because carbon likes to form four bonds. Let's look at the same carbocation over here on the left. And the carbon in magenta, the one with the plus one formal charge, is this one. And since we aren't drawing in our atoms on this bond line structure, sometimes students forget that because this carbon in magenta has a plus one formal charge, that means it must have a hydrogen bonded to it. So don't forget about that when you're looking at bond line structures. Finally, let's look at this same carbocation on the right."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And the carbon in magenta, the one with the plus one formal charge, is this one. And since we aren't drawing in our atoms on this bond line structure, sometimes students forget that because this carbon in magenta has a plus one formal charge, that means it must have a hydrogen bonded to it. So don't forget about that when you're looking at bond line structures. Finally, let's look at this same carbocation on the right. The carbon in magenta is right here. That's the one with the plus one formal charge on it. Because the carbon in magenta has only three bonds to it, this carbon is sp2 hybridized."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "Finally, let's look at this same carbocation on the right. The carbon in magenta is right here. That's the one with the plus one formal charge on it. Because the carbon in magenta has only three bonds to it, this carbon is sp2 hybridized. And we know from earlier videos, an sp2 hybridized carbon is going to have an unhybridized p orbital. So here is our unhybridized p orbital. And also the geometry around that sp2 hybridized carbon is planar."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "Because the carbon in magenta has only three bonds to it, this carbon is sp2 hybridized. And we know from earlier videos, an sp2 hybridized carbon is going to have an unhybridized p orbital. So here is our unhybridized p orbital. And also the geometry around that sp2 hybridized carbon is planar. So let me see if I can sketch in a plane here, indicating the atoms that are directly bonded to that carbon are in a plane around that carbon here. Now let's look at a model of the same carbocation. Here is our sp2 hybridized carbon, and here is our unfilled p orbital."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And also the geometry around that sp2 hybridized carbon is planar. So let me see if I can sketch in a plane here, indicating the atoms that are directly bonded to that carbon are in a plane around that carbon here. Now let's look at a model of the same carbocation. Here is our sp2 hybridized carbon, and here is our unfilled p orbital. On the left, a methyl group is directly bonded to that positively charged carbon, and on the right, we have another methyl group. So two alkyl groups are bonded to that positively charged carbon, and we call this a secondary carbocation, since it has two alkyl groups directly attached to the positive charge. And these alkyl groups can help to stabilize our carbocation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "Here is our sp2 hybridized carbon, and here is our unfilled p orbital. On the left, a methyl group is directly bonded to that positively charged carbon, and on the right, we have another methyl group. So two alkyl groups are bonded to that positively charged carbon, and we call this a secondary carbocation, since it has two alkyl groups directly attached to the positive charge. And these alkyl groups can help to stabilize our carbocation. So let me go ahead and show that. So some of this electron density in here in this bond can be donated to this empty p orbital, and opposites attract. So donating some electron density helps to stabilize the carbocation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And these alkyl groups can help to stabilize our carbocation. So let me go ahead and show that. So some of this electron density in here in this bond can be donated to this empty p orbital, and opposites attract. So donating some electron density helps to stabilize the carbocation. So alkyl groups stabilize carbocations. This alkyl group on the right can do the same thing. So some electron density from in here can help to stabilize our carbocation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So donating some electron density helps to stabilize the carbocation. So alkyl groups stabilize carbocations. This alkyl group on the right can do the same thing. So some electron density from in here can help to stabilize our carbocation. But notice in the back, our hydrogen, the electron density in this bond can't be donated into the p orbital. So the geometry isn't right. So alkyl groups stabilize carbocations, and hydrogens do not."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So some electron density from in here can help to stabilize our carbocation. But notice in the back, our hydrogen, the electron density in this bond can't be donated into the p orbital. So the geometry isn't right. So alkyl groups stabilize carbocations, and hydrogens do not. So it makes sense that the more alkyl groups you have, the more stable your carbocation. We just saw that alkyl groups stabilize a carbocation by donating electron density to the empty p orbital. This effect is called hyperconjugation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So alkyl groups stabilize carbocations, and hydrogens do not. So it makes sense that the more alkyl groups you have, the more stable your carbocation. We just saw that alkyl groups stabilize a carbocation by donating electron density to the empty p orbital. This effect is called hyperconjugation. So as you increase in the number of alkyl groups, you should increase in the stabilization of your carbocation. So let's look at this carbocation on the left. There's only one alkyl group directly bonded to this positively charged carbon."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "This effect is called hyperconjugation. So as you increase in the number of alkyl groups, you should increase in the stabilization of your carbocation. So let's look at this carbocation on the left. There's only one alkyl group directly bonded to this positively charged carbon. So we would call this a primary carbocation. In the middle, we have two alkyl groups directly bonded to this positively charged carbon. That would be a secondary carbocation, like the example in the picture above."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "There's only one alkyl group directly bonded to this positively charged carbon. So we would call this a primary carbocation. In the middle, we have two alkyl groups directly bonded to this positively charged carbon. That would be a secondary carbocation, like the example in the picture above. And finally, if we have three alkyl groups directly bonded to our positively charged carbon, that would be a tertiary carbocation. The more alkyl groups you have, the more you stabilize your carbocation. So a tertiary carbocation is more stable than a secondary carbocation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "That would be a secondary carbocation, like the example in the picture above. And finally, if we have three alkyl groups directly bonded to our positively charged carbon, that would be a tertiary carbocation. The more alkyl groups you have, the more you stabilize your carbocation. So a tertiary carbocation is more stable than a secondary carbocation. And a secondary carbocation is much more stable than a primary carbocation. So these are so unstable, they might not even exist. So we'll focus on secondary and tertiary carbocations."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So a tertiary carbocation is more stable than a secondary carbocation. And a secondary carbocation is much more stable than a primary carbocation. So these are so unstable, they might not even exist. So we'll focus on secondary and tertiary carbocations. Now that we understand carbocation stability, let's look at an introduction to carbocation rearrangements. One possible rearrangement is something called a hydride shift. So first, let's study what a hydride ion is."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So we'll focus on secondary and tertiary carbocations. Now that we understand carbocation stability, let's look at an introduction to carbocation rearrangements. One possible rearrangement is something called a hydride shift. So first, let's study what a hydride ion is. We know that hydrogen has one valence electron. And if you take away the one valence electron from hydrogen, you would be left with H+, which we know is a proton. But if you added an electron to a neutral hydrogen atom, you would now have two valence electrons."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So first, let's study what a hydride ion is. We know that hydrogen has one valence electron. And if you take away the one valence electron from hydrogen, you would be left with H+, which we know is a proton. But if you added an electron to a neutral hydrogen atom, you would now have two valence electrons. Let me draw that in there, which would give this a negative one formal charge. And that is a hydride ion. So hydrogen with two electrons and a negative one formal charge is what's called a hydride."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "But if you added an electron to a neutral hydrogen atom, you would now have two valence electrons. Let me draw that in there, which would give this a negative one formal charge. And that is a hydride ion. So hydrogen with two electrons and a negative one formal charge is what's called a hydride. So a hydride shift could occur to form a more stable carbocation. So let's look at this carbocation here. We know that this carbon has the plus one formal charge."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So hydrogen with two electrons and a negative one formal charge is what's called a hydride. So a hydride shift could occur to form a more stable carbocation. So let's look at this carbocation here. We know that this carbon has the plus one formal charge. Here's one alkyl group, and here's a second alkyl group. So this is a secondary carbocation. And we can do a hydride shift."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "We know that this carbon has the plus one formal charge. Here's one alkyl group, and here's a second alkyl group. So this is a secondary carbocation. And we can do a hydride shift. Before I show how to draw a hydride shift, let's go to the video so we can see it with the model set. So here is that secondary carbocation. You can see the geometry around our sp2 hybridized carbon is planar."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And we can do a hydride shift. Before I show how to draw a hydride shift, let's go to the video so we can see it with the model set. So here is that secondary carbocation. You can see the geometry around our sp2 hybridized carbon is planar. You can also see our empty p orbita here with the paddles. And we have two alkyl groups directly bonded to this carbon. On the right, there's a methyl group, and on the left, we have this big alkyl group."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "You can see the geometry around our sp2 hybridized carbon is planar. You can also see our empty p orbita here with the paddles. And we have two alkyl groups directly bonded to this carbon. On the right, there's a methyl group, and on the left, we have this big alkyl group. And in the back, we have a hydrogen. So notice that we are donating some electron density from this bond into our empty p orbital. We know that helps to stabilize our carbocation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "On the right, there's a methyl group, and on the left, we have this big alkyl group. And in the back, we have a hydrogen. So notice that we are donating some electron density from this bond into our empty p orbital. We know that helps to stabilize our carbocation. But let me just take these paddles off here, and let's show a hydride shift. So remember, a hydride is a hydrogen and its two electrons. So I'm gonna take this hydrogen and these two electrons in this bond, and I'm going to show a shift from this carbon to the carbon on the right."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "We know that helps to stabilize our carbocation. But let me just take these paddles off here, and let's show a hydride shift. So remember, a hydride is a hydrogen and its two electrons. So I'm gonna take this hydrogen and these two electrons in this bond, and I'm going to show a shift from this carbon to the carbon on the right. And that's a hydride shift. Now, on the left, we took a bond away from this carbon, so that should be positively charged, and it should be planar or flat. But it's not because of the model set."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So I'm gonna take this hydrogen and these two electrons in this bond, and I'm going to show a shift from this carbon to the carbon on the right. And that's a hydride shift. Now, on the left, we took a bond away from this carbon, so that should be positively charged, and it should be planar or flat. But it's not because of the model set. I had to use a tetrahedral carbon here. So I'll show a new model set in a minute to show that it actually is planar. And on the right, this should be tetrahedral."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "But it's not because of the model set. I had to use a tetrahedral carbon here. So I'll show a new model set in a minute to show that it actually is planar. And on the right, this should be tetrahedral. This carbon went from being sp2 hybridized to sp3. So let me get out the new model set here so we can better visualize what the carbocation actually looks like. So this carbon is our sp2 hybridized carbon now, and you can see it's planar around that carbon."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And on the right, this should be tetrahedral. This carbon went from being sp2 hybridized to sp3. So let me get out the new model set here so we can better visualize what the carbocation actually looks like. So this carbon is our sp2 hybridized carbon now, and you can see it's planar around that carbon. And this is a tertiary carbocation. We have a methyl group here, a methyl group here, and then an alkyl group over here. So a tertiary carbocation is more stable than a secondary."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So this carbon is our sp2 hybridized carbon now, and you can see it's planar around that carbon. And this is a tertiary carbocation. We have a methyl group here, a methyl group here, and then an alkyl group over here. So a tertiary carbocation is more stable than a secondary. You can also see at this carbon, now we have an sp3 hybridized carbon here. So the geometry around that carbon is tetrahedral. It went from being planar to tetrahedral geometry."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So a tertiary carbocation is more stable than a secondary. You can also see at this carbon, now we have an sp3 hybridized carbon here. So the geometry around that carbon is tetrahedral. It went from being planar to tetrahedral geometry. Let's draw the hydride shift that we saw in the video. So attached to this carbon, we know there is a hydrogen, and this hydrogen and these two electrons can move over to this carbon on the right to form a more stable carbocation. So let's draw in that new carbocation."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "It went from being planar to tetrahedral geometry. Let's draw the hydride shift that we saw in the video. So attached to this carbon, we know there is a hydrogen, and this hydrogen and these two electrons can move over to this carbon on the right to form a more stable carbocation. So let's draw in that new carbocation. So let me draw this in here. I'm gonna draw in that hydrogen in red. Let me go ahead and highlight it in red here like that."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So let's draw in that new carbocation. So let me draw this in here. I'm gonna draw in that hydrogen in red. Let me go ahead and highlight it in red here like that. And notice we're taking a bond away from this carbon on the left. So that's this carbon here. And that's a tertiary carbocation as we saw in the video."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "Let me go ahead and highlight it in red here like that. And notice we're taking a bond away from this carbon on the left. So that's this carbon here. And that's a tertiary carbocation as we saw in the video. So I'm gonna put a plus one formal charge on this carbon. Notice that there was a hydrogen on this carbon to start with, and it's still there in our tertiary carbocation. You can see it better in the video."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "And that's a tertiary carbocation as we saw in the video. So I'm gonna put a plus one formal charge on this carbon. Notice that there was a hydrogen on this carbon to start with, and it's still there in our tertiary carbocation. You can see it better in the video. But the goal is to form a more stable carbocation in a rearrangement, and we go from a secondary carbocation on the left to a tertiary carbocation on the right, which we know is more stable. Finally, let's do one more kind of carbocation rearrangement. This one's called a methyl shift."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "You can see it better in the video. But the goal is to form a more stable carbocation in a rearrangement, and we go from a secondary carbocation on the left to a tertiary carbocation on the right, which we know is more stable. Finally, let's do one more kind of carbocation rearrangement. This one's called a methyl shift. So this carbocation is secondary. The carbon with the plus one formal charge is directly bonded to a methyl group and this alkyl group over here. So for a methyl shift, we could take this methyl group here and we could show this methyl group moving from this carbon on the left to this carbon on the right."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "This one's called a methyl shift. So this carbocation is secondary. The carbon with the plus one formal charge is directly bonded to a methyl group and this alkyl group over here. So for a methyl shift, we could take this methyl group here and we could show this methyl group moving from this carbon on the left to this carbon on the right. Let me go ahead and color code. So I'm gonna say the carbon on the left I'm referring to is red, and the carbon on the right that I'm referring to is blue. So the methyl group shifts from the carbon in red to the carbon in blue, and let's show the result of that methyl shift."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So for a methyl shift, we could take this methyl group here and we could show this methyl group moving from this carbon on the left to this carbon on the right. Let me go ahead and color code. So I'm gonna say the carbon on the left I'm referring to is red, and the carbon on the right that I'm referring to is blue. So the methyl group shifts from the carbon in red to the carbon in blue, and let's show the result of that methyl shift. So now we would have a carbocation that looks like this. So the carbon in red loses a bond. So here's the carbon in red."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So the methyl group shifts from the carbon in red to the carbon in blue, and let's show the result of that methyl shift. So now we would have a carbocation that looks like this. So the carbon in red loses a bond. So here's the carbon in red. It's sp3 hybridized, it's tetrahedral, but when it loses a bond, here's the carbon in red, now it's sp2 hybridized and has planar geometry with a plus one formal charge. The carbon in blue, let me circle that here, it was sp2 hybridized and planar, but it's gaining a methyl group, right? So here's the methyl group that it's gaining and this is the carbon in blue, and it goes from being sp2 hybridized to now being sp3 hybridized."}, {"video_title": "Carbocation stability and rearrangements.mp3", "Sentence": "So here's the carbon in red. It's sp3 hybridized, it's tetrahedral, but when it loses a bond, here's the carbon in red, now it's sp2 hybridized and has planar geometry with a plus one formal charge. The carbon in blue, let me circle that here, it was sp2 hybridized and planar, but it's gaining a methyl group, right? So here's the methyl group that it's gaining and this is the carbon in blue, and it goes from being sp2 hybridized to now being sp3 hybridized. So remember, there was a hydrogen on this carbon in blue to begin with and it's still there for our carbocation. So this is a tertiary carbocation, which we know is more stable than a secondary. So a methyl shift resulted in a more stable carbocation."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if current is moving in this direction in a loop of wire, so I represents current, a magnetic field is created, and at the very center of this loop the magnetic field is pointing straight down. As you move away from the center, I can draw in some more magnetic field lines, so we didn't do this in the earlier video. And as you get closer to the edge of this loop, inside of the loop the magnetic field will be pointing down, but outside of the loop of wire the magnetic field will be pointing up. Same thing on this side, so pointing down inside, and the magnetic field points up on the outside of the loop. So when you're talking about current, you're thinking about positive charges moving, but that's not what's happening. We know that electrons are really what are moving, and moving charges create a magnetic field. The electrons are moving in a direction opposite to how we define current."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Same thing on this side, so pointing down inside, and the magnetic field points up on the outside of the loop. So when you're talking about current, you're thinking about positive charges moving, but that's not what's happening. We know that electrons are really what are moving, and moving charges create a magnetic field. The electrons are moving in a direction opposite to how we define current. So we have electron density moving this way, and we get a magnetic field. If we think about benzene, benzene has six pi electrons, so up here is benzene, so let's go ahead and identify the pi electrons, two, four, and six. And if we put benzene in an applied magnetic field, so here is our applied magnetic field, B naught, so it's pointing up, those six pi electrons of benzene are going to circulate to create an induced magnetic field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The electrons are moving in a direction opposite to how we define current. So we have electron density moving this way, and we get a magnetic field. If we think about benzene, benzene has six pi electrons, so up here is benzene, so let's go ahead and identify the pi electrons, two, four, and six. And if we put benzene in an applied magnetic field, so here is our applied magnetic field, B naught, so it's pointing up, those six pi electrons of benzene are going to circulate to create an induced magnetic field. So let me go ahead and draw a picture of the pi electrons in benzene circulating. So the pi electrons are going in this direction. If the pi electrons are going in that direction, then we know the induced magnetic field will be pointing down here."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if we put benzene in an applied magnetic field, so here is our applied magnetic field, B naught, so it's pointing up, those six pi electrons of benzene are going to circulate to create an induced magnetic field. So let me go ahead and draw a picture of the pi electrons in benzene circulating. So the pi electrons are going in this direction. If the pi electrons are going in that direction, then we know the induced magnetic field will be pointing down here. So at the very center, the induced magnetic field will be pointing down. So induced magnetic field points down. As you move away from the center, once again, we can draw in some more magnetic field lines."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If the pi electrons are going in that direction, then we know the induced magnetic field will be pointing down here. So at the very center, the induced magnetic field will be pointing down. So induced magnetic field points down. As you move away from the center, once again, we can draw in some more magnetic field lines. And then as you get to the edge of the ring, edge of the benzene ring here, once again, inside of the ring, the magnetic field points down, but outside of the ring, the magnetic field is going to point up. So same thing on this side, right? Inside it points down, outside of the ring, the magnetic field points up."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "As you move away from the center, once again, we can draw in some more magnetic field lines. And then as you get to the edge of the ring, edge of the benzene ring here, once again, inside of the ring, the magnetic field points down, but outside of the ring, the magnetic field is going to point up. So same thing on this side, right? Inside it points down, outside of the ring, the magnetic field points up. So let's think about the magnetic field experienced by this proton. So that proton experiences the applied magnetic field, B naught, but it also feels this induced magnetic field, which is in the same direction as the external magnetic field. So this is the direction of the induced magnetic field outside of the ring."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Inside it points down, outside of the ring, the magnetic field points up. So let's think about the magnetic field experienced by this proton. So that proton experiences the applied magnetic field, B naught, but it also feels this induced magnetic field, which is in the same direction as the external magnetic field. So this is the direction of the induced magnetic field outside of the ring. So the effective magnetic field felt by this proton, you'd have to add the induced magnetic field to the applied magnetic field to find the effective magnetic field. So outside of the ring, we get a larger magnetic field. So we get a large magnetic field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is the direction of the induced magnetic field outside of the ring. So the effective magnetic field felt by this proton, you'd have to add the induced magnetic field to the applied magnetic field to find the effective magnetic field. So outside of the ring, we get a larger magnetic field. So we get a large magnetic field. We get a large difference between the alpha and the beta spin states in terms of energy. And a greater difference in terms of energy means a higher frequency absorbed, and therefore you get a higher chemical shift. And so the proton on benzene has a chemical shift of approximately 7.27 parts per million."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we get a large magnetic field. We get a large difference between the alpha and the beta spin states in terms of energy. And a greater difference in terms of energy means a higher frequency absorbed, and therefore you get a higher chemical shift. And so the proton on benzene has a chemical shift of approximately 7.27 parts per million. So this is just for any proton on any kind of benzene ring here, your general range is going to be 6.528. And so there are several molecules that demonstrate this effect very dramatically. And let's take a look at one of them."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the proton on benzene has a chemical shift of approximately 7.27 parts per million. So this is just for any proton on any kind of benzene ring here, your general range is going to be 6.528. And so there are several molecules that demonstrate this effect very dramatically. And let's take a look at one of them. So how do we know that this effect is even true? So if I look at this molecule, we would have a giant ring here. So let me go around so you can see the outline of this giant ring."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's take a look at one of them. So how do we know that this effect is even true? So if I look at this molecule, we would have a giant ring here. So let me go around so you can see the outline of this giant ring. So a much bigger ring than benzene. We have a lot of pi electrons, so more pi electrons than benzene. So I'll just highlight some of them, 2, 4, 6, 8, and so on."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go around so you can see the outline of this giant ring. So a much bigger ring than benzene. We have a lot of pi electrons, so more pi electrons than benzene. So I'll just highlight some of them, 2, 4, 6, 8, and so on. You can see we have alternating single double bonds here in this molecule. And so if you put this molecule into an external magnetic field, you're going to get the same situation as benzene. So let's think about these inner protons here."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I'll just highlight some of them, 2, 4, 6, 8, and so on. You can see we have alternating single double bonds here in this molecule. And so if you put this molecule into an external magnetic field, you're going to get the same situation as benzene. So let's think about these inner protons here. So we have six inner protons. If we look at the diagram for benzene, if you have an applied external field B0, in the center of the ring, the inner protons experience an induced magnetic field that's down. It opposes the external magnetic field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about these inner protons here. So we have six inner protons. If we look at the diagram for benzene, if you have an applied external field B0, in the center of the ring, the inner protons experience an induced magnetic field that's down. It opposes the external magnetic field. So let me go ahead and draw that out here. So if we apply an external magnetic field B0, the inner protons have an induced magnetic field caused by the movement of those pi electrons. The induced magnetic field opposes the applied field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It opposes the external magnetic field. So let me go ahead and draw that out here. So if we apply an external magnetic field B0, the inner protons have an induced magnetic field caused by the movement of those pi electrons. The induced magnetic field opposes the applied field. And so the effective magnetic field felt by those inner protons is smaller. So we get a smaller effective magnetic field. Smaller effective magnetic field means a smaller energy difference between the alpha and the beta spin states."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The induced magnetic field opposes the applied field. And so the effective magnetic field felt by those inner protons is smaller. So we get a smaller effective magnetic field. Smaller effective magnetic field means a smaller energy difference between the alpha and the beta spin states. Therefore, we get a lower frequency signal and a lower chemical shift. And the chemical shift for these six inner protons turns out to be negative 2 parts per million. So think about what that means."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Smaller effective magnetic field means a smaller energy difference between the alpha and the beta spin states. Therefore, we get a lower frequency signal and a lower chemical shift. And the chemical shift for these six inner protons turns out to be negative 2 parts per million. So think about what that means. Negative 2 is past TMS. So if I go back up here, TMS was at 0. So negative 2 would be to the right."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So think about what that means. Negative 2 is past TMS. So if I go back up here, TMS was at 0. So negative 2 would be to the right. I don't even have room to show it on this chemical shift right here, so way past TMS. So a pretty dramatic effect. We can look at the protons outside of the ring as well."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So negative 2 would be to the right. I don't even have room to show it on this chemical shift right here, so way past TMS. So a pretty dramatic effect. We can look at the protons outside of the ring as well. So let me go ahead and highlight those. So we have 12 protons outside of the ring. Since those protons are outside of the ring, the induced magnetic field is now in the same direction as the applied magnetic field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We can look at the protons outside of the ring as well. So let me go ahead and highlight those. So we have 12 protons outside of the ring. Since those protons are outside of the ring, the induced magnetic field is now in the same direction as the applied magnetic field. So therefore, we get a larger effective magnetic field felt by one of those protons. A larger magnetic field means a greater difference in energy between your alpha and your beta spin states. So you get a higher frequency signal and a higher chemical shift."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Since those protons are outside of the ring, the induced magnetic field is now in the same direction as the applied magnetic field. So therefore, we get a larger effective magnetic field felt by one of those protons. A larger magnetic field means a greater difference in energy between your alpha and your beta spin states. So you get a higher frequency signal and a higher chemical shift. The chemical shift is about 9 parts per million. So the dramatic difference between these chemical shifts for these inner and outer protons shows you how powerful this effect can be. Let's use this effect to explain the shift for a proton on a triple bond."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So you get a higher frequency signal and a higher chemical shift. The chemical shift is about 9 parts per million. So the dramatic difference between these chemical shifts for these inner and outer protons shows you how powerful this effect can be. Let's use this effect to explain the shift for a proton on a triple bond. So if we think about acetylene, so here's acetylene. And we're thinking about the signal for this proton. Let's think about the carbon it's attached to."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's use this effect to explain the shift for a proton on a triple bond. So if we think about acetylene, so here's acetylene. And we're thinking about the signal for this proton. Let's think about the carbon it's attached to. So this carbon right here is sp hybridized. And in the previous video, we talked about the fact that an sp hybrid orbital has more s character than an sp2 or sp3 hybrid orbital. And therefore, the electron density is going to be closer to that carbon."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about the carbon it's attached to. So this carbon right here is sp hybridized. And in the previous video, we talked about the fact that an sp hybrid orbital has more s character than an sp2 or sp3 hybrid orbital. And therefore, the electron density is going to be closer to that carbon. So you can think about an sp hybridized carbon as being more electronegative than an sp2 or sp3 hybridized carbon. So the electron density is closer to this carbon here, which you would think would deshield this proton and give you a higher chemical shift than a proton on a double bond. But that's not what we observed."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, the electron density is going to be closer to that carbon. So you can think about an sp hybridized carbon as being more electronegative than an sp2 or sp3 hybridized carbon. So the electron density is closer to this carbon here, which you would think would deshield this proton and give you a higher chemical shift than a proton on a double bond. But that's not what we observed. The shift for this proton turns out to be approximately 2 to 2.5. So it's actually a lower chemical shift than a proton on a double bond. And let's see if we can explain why."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But that's not what we observed. The shift for this proton turns out to be approximately 2 to 2.5. So it's actually a lower chemical shift than a proton on a double bond. And let's see if we can explain why. So if we apply an external magnetic field, so B0 is our applied external magnetic field, we know that causes pi electrons to circulate. And if we have an upright orientation of acetylene, so the orientation of the molecule matters. So if it's facing in this direction, the pi electrons are going to circulate like this."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's see if we can explain why. So if we apply an external magnetic field, so B0 is our applied external magnetic field, we know that causes pi electrons to circulate. And if we have an upright orientation of acetylene, so the orientation of the molecule matters. So if it's facing in this direction, the pi electrons are going to circulate like this. And just like we talked about in benzene, if the pi electrons circulate like that, we get an induced magnetic field down in this direction, like that. So we can draw a few more magnetic field lines like that. And think about the magnetic field experienced by, let's say, this proton."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if it's facing in this direction, the pi electrons are going to circulate like this. And just like we talked about in benzene, if the pi electrons circulate like that, we get an induced magnetic field down in this direction, like that. So we can draw a few more magnetic field lines like that. And think about the magnetic field experienced by, let's say, this proton. So this proton is feeling the applied magnetic field. It's also feeling the induced magnetic field. But the induced magnetic field is in the opposite direction of the applied magnetic field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And think about the magnetic field experienced by, let's say, this proton. So this proton is feeling the applied magnetic field. It's also feeling the induced magnetic field. But the induced magnetic field is in the opposite direction of the applied magnetic field. So we could draw the induced magnetic field opposing the applied magnetic field. So that proton that I circled there actually feels a smaller effective magnetic field here. So if you get a smaller effective magnetic field, you're decreasing the energy difference between your alpha and your beta spin states."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But the induced magnetic field is in the opposite direction of the applied magnetic field. So we could draw the induced magnetic field opposing the applied magnetic field. So that proton that I circled there actually feels a smaller effective magnetic field here. So if you get a smaller effective magnetic field, you're decreasing the energy difference between your alpha and your beta spin states. So you get a lower chemical shift than expected due to this effect. And so that's currently how we explain the chemical shift of somewhere around 2.5 for a proton on a triple bond. And so this effect holds true any time you have pi electrons that can circulate when you put a molecule in an applied magnetic field."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you get a smaller effective magnetic field, you're decreasing the energy difference between your alpha and your beta spin states. So you get a lower chemical shift than expected due to this effect. And so that's currently how we explain the chemical shift of somewhere around 2.5 for a proton on a triple bond. And so this effect holds true any time you have pi electrons that can circulate when you put a molecule in an applied magnetic field. And so we could also use this to explain, for example, the proton on a double bond. So here's some pi electrons. Or the proton here we have next to a carbonyl here."}, {"video_title": "Diamagnetic anisotropy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this effect holds true any time you have pi electrons that can circulate when you put a molecule in an applied magnetic field. And so we could also use this to explain, for example, the proton on a double bond. So here's some pi electrons. Or the proton here we have next to a carbonyl here. So we have pi electrons here. So any time you have pi electrons, this effect can be present. And as we've seen, it can be a very powerful effect and really affect the chemical shift."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "You can imagine I'm drawing it like this because this hydrogen is poking out of the page. Then maybe you have another hydrogen that's in the page. You have one above the carbon, and then you have one that's behind the page. So you can imagine it's like a tripod with a pole sticking out of the top of the tripod. Or if you were to imagine the shape another way, if you were to connect the hydrogens, you would have a four-sided pyramid with a triangle as each of the sides. So it would look something like this. It would look something like this if you could see through it."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So you can imagine it's like a tripod with a pole sticking out of the top of the tripod. Or if you were to imagine the shape another way, if you were to connect the hydrogens, you would have a four-sided pyramid with a triangle as each of the sides. So it would look something like this. It would look something like this if you could see through it. So this would be one side. Another side would be over here. And then the back side would be over here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "It would look something like this if you could see through it. So this would be one side. Another side would be over here. And then the back side would be over here. And then the fourth side is actually the side that's transparent out front. So the fourth side would be the actual kind of thing that we're looking through when we look at this pyramid. It would be this front side right over here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then the back side would be over here. And then the fourth side is actually the side that's transparent out front. So the fourth side would be the actual kind of thing that we're looking through when we look at this pyramid. It would be this front side right over here. So you could imagine it different ways. This was the case with methane. Now let's extend this into a slightly more complex molecule, and that's ethane."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "It would be this front side right over here. So you could imagine it different ways. This was the case with methane. Now let's extend this into a slightly more complex molecule, and that's ethane. So the way we've been drawing it so far, I guess the simplest way to draw ethane is just like that. This is ethane. By implication, you have a carbon there and a carbon there, and they'll each have three hydrogens bonded to it."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now let's extend this into a slightly more complex molecule, and that's ethane. So the way we've been drawing it so far, I guess the simplest way to draw ethane is just like that. This is ethane. By implication, you have a carbon there and a carbon there, and they'll each have three hydrogens bonded to it. And we've drawn it something like this. Three hydrogens bonded to each of these guys. But now we know that carbon has these sp3 hybridized orbitals that it likes to form more of a tetrahedral shape when it bonds."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "By implication, you have a carbon there and a carbon there, and they'll each have three hydrogens bonded to it. And we've drawn it something like this. Three hydrogens bonded to each of these guys. But now we know that carbon has these sp3 hybridized orbitals that it likes to form more of a tetrahedral shape when it bonds. So an ethane molecule would actually look more like this. Let me draw the carbons. So I'll do the carbons in orange."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But now we know that carbon has these sp3 hybridized orbitals that it likes to form more of a tetrahedral shape when it bonds. So an ethane molecule would actually look more like this. Let me draw the carbons. So I'll do the carbons in orange. So if that's the carbon and that's the carbon. So you could imagine you have a carbon molecule here. I'll draw it as this little circle."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'll do the carbons in orange. So if that's the carbon and that's the carbon. So you could imagine you have a carbon molecule here. I'll draw it as this little circle. And then if we have some perspective. So the carbon-carbon bond is going to look like that. And then you have another carbon molecule right over there, so that's that bond over here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "I'll draw it as this little circle. And then if we have some perspective. So the carbon-carbon bond is going to look like that. And then you have another carbon molecule right over there, so that's that bond over here. And we want both the carbons, all of their bonds, to be kind of in a tetrahedral shape. So then you could imagine this bond over here going like this, this bond going like that. And you have your hydrogen at the end."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then you have another carbon molecule right over there, so that's that bond over here. And we want both the carbons, all of their bonds, to be kind of in a tetrahedral shape. So then you could imagine this bond over here going like this, this bond going like that. And you have your hydrogen at the end. Let's make the green circles the hydrogens. So you have that hydrogen, or actually just the circles, we'll call them hydrogen. And then you could imagine this one, maybe it's coming out of the page a little bit."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And you have your hydrogen at the end. Let's make the green circles the hydrogens. So you have that hydrogen, or actually just the circles, we'll call them hydrogen. And then you could imagine this one, maybe it's coming out of the page a little bit. That is that hydrogen. Let me label the hydrogens actually. I'm doing it all in different colors so you can see what I'm talking about."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then you could imagine this one, maybe it's coming out of the page a little bit. That is that hydrogen. Let me label the hydrogens actually. I'm doing it all in different colors so you can see what I'm talking about. And then this hydrogen is going right below it, maybe pointed back a little bit. So that hydrogen is right over there. So you can see this carbon, its bonds have a tetrahedral shape, or if you just looked at this part of it, these would be the base of the tripod and this would be the thing sticking up."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "I'm doing it all in different colors so you can see what I'm talking about. And then this hydrogen is going right below it, maybe pointed back a little bit. So that hydrogen is right over there. So you can see this carbon, its bonds have a tetrahedral shape, or if you just looked at this part of it, these would be the base of the tripod and this would be the thing sticking up. Now for this carbon, it would be a very similar idea. This hydrogen right here might be sticking down like this. And I'll stop switching colors soon enough."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So you can see this carbon, its bonds have a tetrahedral shape, or if you just looked at this part of it, these would be the base of the tripod and this would be the thing sticking up. Now for this carbon, it would be a very similar idea. This hydrogen right here might be sticking down like this. And I'll stop switching colors soon enough. It takes a lot of time. That hydrogen over there is pointing out in that direction like that. And then you would have, let's see what colors do I have left."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And I'll stop switching colors soon enough. It takes a lot of time. That hydrogen over there is pointing out in that direction like that. And then you would have, let's see what colors do I have left. I'll just do yellow. That hydrogen right there, maybe it's pointing out like that. So this is a possible configuration for ethane."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then you would have, let's see what colors do I have left. I'll just do yellow. That hydrogen right there, maybe it's pointing out like that. So this is a possible configuration for ethane. And the way that I've drawn this right now, and you can actually have a model that has this where you have little wooden sticks with balls and this, that balls represent the actual atoms, this is called a ball and stick model. And this is a ball and stick model for ethane. Now a simpler way we could have drawn this, and this is called a horseshoe projection, or actually not, a saw horse projection, I always say horseshoe."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is a possible configuration for ethane. And the way that I've drawn this right now, and you can actually have a model that has this where you have little wooden sticks with balls and this, that balls represent the actual atoms, this is called a ball and stick model. And this is a ball and stick model for ethane. Now a simpler way we could have drawn this, and this is called a horseshoe projection, or actually not, a saw horse projection, I always say horseshoe. A saw horse projection, it would look like this. This exact same configuration of ethane in a saw horse projection, and you know what a saw horse looks like. Well, it looks like what I'm about to draw."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now a simpler way we could have drawn this, and this is called a horseshoe projection, or actually not, a saw horse projection, I always say horseshoe. A saw horse projection, it would look like this. This exact same configuration of ethane in a saw horse projection, and you know what a saw horse looks like. Well, it looks like what I'm about to draw. It looks like this. So you could, well I could draw it exactly the way I drew it here, so you have the carbon, carbon, and then you have the hydrogen, hydrogen, hydrogen, and then you have a, the way I've drawn it, well, the way I've drawn it up here is more like this, just so we see the parallel and then we can rotate things around. The way I drew it up here, you have a hydrogen, hydrogen, and hydrogen, and then over here you have a hydrogen, hydrogen, and hydrogen."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well, it looks like what I'm about to draw. It looks like this. So you could, well I could draw it exactly the way I drew it here, so you have the carbon, carbon, and then you have the hydrogen, hydrogen, hydrogen, and then you have a, the way I've drawn it, well, the way I've drawn it up here is more like this, just so we see the parallel and then we can rotate things around. The way I drew it up here, you have a hydrogen, hydrogen, and hydrogen, and then over here you have a hydrogen, hydrogen, and hydrogen. This is a saw horse projection. Now, either way you depict it, I mean these are really the same way, this is kind of just like the laser way of doing it on some level, you're not drawing all of these circles and all of that, and you're putting a little less care into actually showing the angle, how things are angled away from the carbon and showing the tetrahedral shape. But in either case, when you start visualizing the molecule in this way, you start to realize there's, well, there's actually an infinite ways that these things can be configured, and that all comes from the notion that this is just a sigma bond right here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The way I drew it up here, you have a hydrogen, hydrogen, and hydrogen, and then over here you have a hydrogen, hydrogen, and hydrogen. This is a saw horse projection. Now, either way you depict it, I mean these are really the same way, this is kind of just like the laser way of doing it on some level, you're not drawing all of these circles and all of that, and you're putting a little less care into actually showing the angle, how things are angled away from the carbon and showing the tetrahedral shape. But in either case, when you start visualizing the molecule in this way, you start to realize there's, well, there's actually an infinite ways that these things can be configured, and that all comes from the notion that this is just a sigma bond right here. We learned that in the video on sp3 hybridization and sigma and pi bonds. This is just a sigma bond. And so we can rotate around the bonds."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But in either case, when you start visualizing the molecule in this way, you start to realize there's, well, there's actually an infinite ways that these things can be configured, and that all comes from the notion that this is just a sigma bond right here. We learned that in the video on sp3 hybridization and sigma and pi bonds. This is just a sigma bond. And so we can rotate around the bonds. One of these carbons could rotate around the axis of that bond without the other carbon having to necessarily rotate with it. If this was a double bond, if this was a pi bond, they would have to rotate together. So you could have a situation like I've drawn here, or you could have a situation where they're kind of rotated the inverse of each other."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And so we can rotate around the bonds. One of these carbons could rotate around the axis of that bond without the other carbon having to necessarily rotate with it. If this was a double bond, if this was a pi bond, they would have to rotate together. So you could have a situation like I've drawn here, or you could have a situation where they're kind of rotated the inverse of each other. This is what I mean. So I'll do a ball and stick. So let's say this is our front carbon, that is our back carbon, and we'll compare it to this one over here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So you could have a situation like I've drawn here, or you could have a situation where they're kind of rotated the inverse of each other. This is what I mean. So I'll do a ball and stick. So let's say this is our front carbon, that is our back carbon, and we'll compare it to this one over here. So let me draw this guy the exact same way. So he's got a hydrogen down here, he's got a hydrogen down there, he's got a hydrogen up there, and then he's got a hydrogen up here. So that part of the ethane looks identical."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's say this is our front carbon, that is our back carbon, and we'll compare it to this one over here. So let me draw this guy the exact same way. So he's got a hydrogen down here, he's got a hydrogen down there, he's got a hydrogen up there, and then he's got a hydrogen up here. So that part of the ethane looks identical. Now what I'm going to do is I'm going to flip the other side of the ethane. And I want you to pay close attention, because hopefully you'll see the difference between the two. So instead of doing this blue hydrogen down here, I'm going to do it on top."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So that part of the ethane looks identical. Now what I'm going to do is I'm going to flip the other side of the ethane. And I want you to pay close attention, because hopefully you'll see the difference between the two. So instead of doing this blue hydrogen down here, I'm going to do it on top. So this blue one, I want to do that in blue. This blue hydrogen, put it on top. So I'm just rotating this around."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So instead of doing this blue hydrogen down here, I'm going to do it on top. So this blue one, I want to do that in blue. This blue hydrogen, put it on top. So I'm just rotating this around. So if the blue hydrogen's on top, so if I've rotated it so the blue one's on top now, and now the green one is going to go over here, so now the green hydrogen is now over here. And then this purple, or this magenta hydrogen, the way I've rotated it, is now going to go over here, is now going to go over there. So what's the difference between this configuration and this configuration right here?"}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just rotating this around. So if the blue hydrogen's on top, so if I've rotated it so the blue one's on top now, and now the green one is going to go over here, so now the green hydrogen is now over here. And then this purple, or this magenta hydrogen, the way I've rotated it, is now going to go over here, is now going to go over there. So what's the difference between this configuration and this configuration right here? And we could have had every other configuration in between. But what's the real difference here? Well here, the hydrogens are all, you can imagine that hydrogen is kind of, if you were looking from that direction, that hydrogen is directly behind that hydrogen."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So what's the difference between this configuration and this configuration right here? And we could have had every other configuration in between. But what's the real difference here? Well here, the hydrogens are all, you can imagine that hydrogen is kind of, if you were looking from that direction, that hydrogen is directly behind that hydrogen. That hydrogen is directly behind that hydrogen. That hydrogen is directly behind that. This is called an eclipsed configuration, or an eclipsed confirmation."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well here, the hydrogens are all, you can imagine that hydrogen is kind of, if you were looking from that direction, that hydrogen is directly behind that hydrogen. That hydrogen is directly behind that hydrogen. That hydrogen is directly behind that. This is called an eclipsed configuration, or an eclipsed confirmation. And this right here, nothing is behind anything. If you went straight back from this guy, you'd get to this point, and no one's behind it. And if you went straight forward from this guy, you'll get, so in no way are any of the guys in the back, if once again, if you're viewing from this direction, are they blocked by any of the people here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "This is called an eclipsed configuration, or an eclipsed confirmation. And this right here, nothing is behind anything. If you went straight back from this guy, you'd get to this point, and no one's behind it. And if you went straight forward from this guy, you'll get, so in no way are any of the guys in the back, if once again, if you're viewing from this direction, are they blocked by any of the people here. So we call this a staggered confirmation. Now why do we even care? You know, OK, I can twist around this back molecule."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And if you went straight forward from this guy, you'll get, so in no way are any of the guys in the back, if once again, if you're viewing from this direction, are they blocked by any of the people here. So we call this a staggered confirmation. Now why do we even care? You know, OK, I can twist around this back molecule. What's that even going to do for our actual, why does it even matter? Well one, it's just interesting that you can actually change, that this thing can twist around without changing the front, without the front molecule having to twist with it. But even more important, these have different energy levels."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "You know, OK, I can twist around this back molecule. What's that even going to do for our actual, why does it even matter? Well one, it's just interesting that you can actually change, that this thing can twist around without changing the front, without the front molecule having to twist with it. But even more important, these have different energy levels. So you can kind of think of them as, you're kind of twisting a spring, and the spring might want to go back to one confirmation or another. And to visualize it a little better, I'll draw what's called a Newman projection. So I'm going to draw this exact thing."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But even more important, these have different energy levels. So you can kind of think of them as, you're kind of twisting a spring, and the spring might want to go back to one confirmation or another. And to visualize it a little better, I'll draw what's called a Newman projection. So I'm going to draw this exact thing. But with a Newman projection, you draw the carbon molecules directly in front or directly behind each other. So in this situation, you would draw the carbon molecule in front would just be the intersection of these bonds. So for a Newman projection, let me draw that out."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to draw this exact thing. But with a Newman projection, you draw the carbon molecules directly in front or directly behind each other. So in this situation, you would draw the carbon molecule in front would just be the intersection of these bonds. So for a Newman projection, let me draw that out. So it's a Newman projection, and I'll start with the Newman projection for the staggered confirmation. So in the front, we'll consider this carbon the front carbon, we have our hydrogen pointing straight down, and we have a hydrogen coming out to the top left like that. And then we have this hydrogen over here coming up to the right, I want to do that in that same color."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So for a Newman projection, let me draw that out. So it's a Newman projection, and I'll start with the Newman projection for the staggered confirmation. So in the front, we'll consider this carbon the front carbon, we have our hydrogen pointing straight down, and we have a hydrogen coming out to the top left like that. And then we have this hydrogen over here coming up to the right, I want to do that in that same color. So the front carbon is implicitly at the intersection of the bonds of these three hydrogen. And then the back carbon, I said the front carbon is the intersection of the bonds of these three hydrogens. The back carbon, you represent it as a circle."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then we have this hydrogen over here coming up to the right, I want to do that in that same color. So the front carbon is implicitly at the intersection of the bonds of these three hydrogen. And then the back carbon, I said the front carbon is the intersection of the bonds of these three hydrogens. The back carbon, you represent it as a circle. So this circle represents the back carbon. The front carbon is kind of that point there, just as a way of visualizing it. But if we were to draw it this way, the back carbon now has that blue hydrogen popping off of it."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The back carbon, you represent it as a circle. So this circle represents the back carbon. The front carbon is kind of that point there, just as a way of visualizing it. But if we were to draw it this way, the back carbon now has that blue hydrogen popping off of it. So it has that blue hydrogen, this green hydrogen, and then this magenta hydrogen. And when you look at it like this, it is more clear that it's staggered. We're just looking straight on to this ethane molecule."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But if we were to draw it this way, the back carbon now has that blue hydrogen popping off of it. So it has that blue hydrogen, this green hydrogen, and then this magenta hydrogen. And when you look at it like this, it is more clear that it's staggered. We're just looking straight on to this ethane molecule. When we look straight on the front carbon, they're obviously blocking each other. But this way you can see the front carbon's hydrogens are staggered relative to the back one. So this is right here."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We're just looking straight on to this ethane molecule. When we look straight on the front carbon, they're obviously blocking each other. But this way you can see the front carbon's hydrogens are staggered relative to the back one. So this is right here. Once again, this is staggered. Now let's draw the eclipsed conformation as a Newman projection. So as a Newman projection, the front's going to look the same."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is right here. Once again, this is staggered. Now let's draw the eclipsed conformation as a Newman projection. So as a Newman projection, the front's going to look the same. You have a hydrogen there. You have this hydrogen. You have that hydrogen."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So as a Newman projection, the front's going to look the same. You have a hydrogen there. You have this hydrogen. You have that hydrogen. And then you have that purple hydrogen down here. So that's the front of it. But the back, they're right behind it."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "You have that hydrogen. And then you have that purple hydrogen down here. So that's the front of it. But the back, they're right behind it. So let me draw the back carbon. Remember, the front carbon is kind of represented by just that dot. The back carbon will represent like that."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But the back, they're right behind it. So let me draw the back carbon. Remember, the front carbon is kind of represented by just that dot. The back carbon will represent like that. Now the staggered conformation, if this guy's really behind that, you're going to have to draw it like right there, like right behind it. But since that's a little bit messy, normally when people draw a staggered, an eclipsed conformation as a Newman projection, instead of directly eclipsing that back hydrogen, they'll push it off a little bit. So that's that hydrogen."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "The back carbon will represent like that. Now the staggered conformation, if this guy's really behind that, you're going to have to draw it like right there, like right behind it. But since that's a little bit messy, normally when people draw a staggered, an eclipsed conformation as a Newman projection, instead of directly eclipsing that back hydrogen, they'll push it off a little bit. So that's that hydrogen. That magenta hydrogen is right there. It's really right behind the front one. But this is just so you can actually see it, that it's there, and then finally you have this blue one down there."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So that's that hydrogen. That magenta hydrogen is right there. It's really right behind the front one. But this is just so you can actually see it, that it's there, and then finally you have this blue one down there. So that blue one is going to be right over here. So this is the eclipsed conformation as a Newman projection. And as you can see, it's eclipsed."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But this is just so you can actually see it, that it's there, and then finally you have this blue one down there. So that blue one is going to be right over here. So this is the eclipsed conformation as a Newman projection. And as you can see, it's eclipsed. The back hydrogens are eclipsed by the front ones. If I were to draw it perfectly, they would be right behind it. Now there's one more idea I want to introduce you to, and that's the notion of the angle between the different hydrogens."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And as you can see, it's eclipsed. The back hydrogens are eclipsed by the front ones. If I were to draw it perfectly, they would be right behind it. Now there's one more idea I want to introduce you to, and that's the notion of the angle between the different hydrogens. So if you wanted to say, well, what is this angle between the blue hydrogen and this pink hydrogen right here? Now when you actually think of it in three dimensions, it's like, well, you can't really say the angle over here between the blue and the pink. But on a Newman projection, when you're just saying how much are they rotated away from each other, this angle right here is called a dihedral angle."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now there's one more idea I want to introduce you to, and that's the notion of the angle between the different hydrogens. So if you wanted to say, well, what is this angle between the blue hydrogen and this pink hydrogen right here? Now when you actually think of it in three dimensions, it's like, well, you can't really say the angle over here between the blue and the pink. But on a Newman projection, when you're just saying how much are they rotated away from each other, this angle right here is called a dihedral angle. Sometimes this just says this hydrogen relative to that hydrogen has a dA of, in this case, 60 degrees. In this case, this hydrogen relative to that hydrogen has a dihedral angle of 0 degrees. And it's a way of saying how staggered or how eclipsed you are."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But on a Newman projection, when you're just saying how much are they rotated away from each other, this angle right here is called a dihedral angle. Sometimes this just says this hydrogen relative to that hydrogen has a dA of, in this case, 60 degrees. In this case, this hydrogen relative to that hydrogen has a dihedral angle of 0 degrees. And it's a way of saying how staggered or how eclipsed you are. Now one last thing. I touched on the idea. It's like, why do we even care?"}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And it's a way of saying how staggered or how eclipsed you are. Now one last thing. I touched on the idea. It's like, why do we even care? Well, all of these hydrogens have electron clouds around them, and all of these bonds have electron clouds around them, and electron clouds, they're all negative. So they want to get as far away from each other as possible, and they're all stable now because they've bonded in ways that they have nice, stable structures. Everyone feels like they have full valence shells, full orbitals."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "It's like, why do we even care? Well, all of these hydrogens have electron clouds around them, and all of these bonds have electron clouds around them, and electron clouds, they're all negative. So they want to get as far away from each other as possible, and they're all stable now because they've bonded in ways that they have nice, stable structures. Everyone feels like they have full valence shells, full orbitals. And so the electron crowds want to get away from each other. Now in this situation, in the eclipsed conformation, this hydrogen and this hydrogen are closer to each other than when you go to the staggered conformation. The staggered conformation, the closest hydrogen to this guy, is going to be either that hydrogen or that hydrogen, but they're both further away than this hydrogen was in the eclipsed conformation."}, {"video_title": "Newman projections Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Everyone feels like they have full valence shells, full orbitals. And so the electron crowds want to get away from each other. Now in this situation, in the eclipsed conformation, this hydrogen and this hydrogen are closer to each other than when you go to the staggered conformation. The staggered conformation, the closest hydrogen to this guy, is going to be either that hydrogen or that hydrogen, but they're both further away than this hydrogen was in the eclipsed conformation. So in general, the staggered conformation is going to be more stable. It's going to have a lower potential energy. You can imagine that if you start with an eclipsed conformation, these guys all want to get away from each other, so it's kind of like this is the wound conformation."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In this video, I'm going to use words like eras, periods, and ages to refer to segments of time in the human or in the pre-human past. And what I want to clarify right from the get-go, because frankly this is something that's confused me in the past, is that archaeologists will refer to eras, periods, and ages in the human past, and they're usually referring to periods of tens of thousands of years or thousands of years. But these are different eras, periods, and ages than the ones that geologists would refer to when they're talking about geological time. In geological time, era means several hundred millions of years. Periods and ages mean millions of years. When archaeologists, when we're starting the human past, they're just generally talking about long segments of human time, but not in the millions of years, usually in the thousands or the 10,000s of years. So what I want to do with that out of the way is talk about what has happened in the distant human past or the distant pre-human past, and also touch on some of the classifications for these segments of time, because they actually tell us what were the interesting developments that happened to humanity over the 200,000 years that Homo sapiens have been on this planet, or that we believe that Homo sapiens have been on this planet."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In geological time, era means several hundred millions of years. Periods and ages mean millions of years. When archaeologists, when we're starting the human past, they're just generally talking about long segments of human time, but not in the millions of years, usually in the thousands or the 10,000s of years. So what I want to do with that out of the way is talk about what has happened in the distant human past or the distant pre-human past, and also touch on some of the classifications for these segments of time, because they actually tell us what were the interesting developments that happened to humanity over the 200,000 years that Homo sapiens have been on this planet, or that we believe that Homo sapiens have been on this planet. So the longest period of time in human past, or the category of human time, and there's different ways you can categorize it, is the Paleolithic era right over here, and what really makes that period of time. So this begins even in pre-history or pre-human history, so before Homo sapiens even existed, you have the beginning of the Paleolithic era that really began with the development of stone tools, and as we learned in the video on human evolution, there were pre- Homo sapiens species that were using stone tools. And so the Paleolithic era, it's really kind of signified by one, the stone tools, but even more that either the pre-humans, or once you go about 200,000 years ago, the humans showed up."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what I want to do with that out of the way is talk about what has happened in the distant human past or the distant pre-human past, and also touch on some of the classifications for these segments of time, because they actually tell us what were the interesting developments that happened to humanity over the 200,000 years that Homo sapiens have been on this planet, or that we believe that Homo sapiens have been on this planet. So the longest period of time in human past, or the category of human time, and there's different ways you can categorize it, is the Paleolithic era right over here, and what really makes that period of time. So this begins even in pre-history or pre-human history, so before Homo sapiens even existed, you have the beginning of the Paleolithic era that really began with the development of stone tools, and as we learned in the video on human evolution, there were pre- Homo sapiens species that were using stone tools. And so the Paleolithic era, it's really kind of signified by one, the stone tools, but even more that either the pre-humans, or once you go about 200,000 years ago, the humans showed up. It's kind of distinguished by humans being hunter-gatherers, which essentially means to survive, we used to walk around a lot, if we couldn't see something obvious to hunt, maybe a woolly mammoth or something, if we didn't see something obvious to hunt, we would look around for snails or mushrooms or whatever else, and that's how we would survive, that's how we would live. And because we were constantly adapting to our environment based on the seasons, we were maybe following animals as they migrated, hunter-gatherers were fundamentally nomadic, which means that they never settled in one place for a long time. They were always ready to kind of pick up, probably, their tents and follow the herd or follow whatever animals they were hunting or follow the seasons so they could go to warmer climates, maybe where there's more likely to find something to find on the ground to eat, maybe, during the winter, who knows."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the Paleolithic era, it's really kind of signified by one, the stone tools, but even more that either the pre-humans, or once you go about 200,000 years ago, the humans showed up. It's kind of distinguished by humans being hunter-gatherers, which essentially means to survive, we used to walk around a lot, if we couldn't see something obvious to hunt, maybe a woolly mammoth or something, if we didn't see something obvious to hunt, we would look around for snails or mushrooms or whatever else, and that's how we would survive, that's how we would live. And because we were constantly adapting to our environment based on the seasons, we were maybe following animals as they migrated, hunter-gatherers were fundamentally nomadic, which means that they never settled in one place for a long time. They were always ready to kind of pick up, probably, their tents and follow the herd or follow whatever animals they were hunting or follow the seasons so they could go to warmer climates, maybe where there's more likely to find something to find on the ground to eat, maybe, during the winter, who knows. So the Paleolithic era is really distinguished by that. It's a huge swath of time in human history, and it doesn't come to an end until you get to the advent of farming. So the Paleolithic era, I mean, we're literally talking about over 2 million years ago is when it starts, before Homo sapiens even existed as a species, and it goes all the way to the advent of farming that we believe first came about around 11,000 to 7,000 years ago."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They were always ready to kind of pick up, probably, their tents and follow the herd or follow whatever animals they were hunting or follow the seasons so they could go to warmer climates, maybe where there's more likely to find something to find on the ground to eat, maybe, during the winter, who knows. So the Paleolithic era is really distinguished by that. It's a huge swath of time in human history, and it doesn't come to an end until you get to the advent of farming. So the Paleolithic era, I mean, we're literally talking about over 2 million years ago is when it starts, before Homo sapiens even existed as a species, and it goes all the way to the advent of farming that we believe first came about around 11,000 to 7,000 years ago. And this abbreviation right here, this BP, this does not stand for British Petroleum. It stands for Before Present or Before the Present Time. So one more acronym to kind of have in your toolkit when you see things."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the Paleolithic era, I mean, we're literally talking about over 2 million years ago is when it starts, before Homo sapiens even existed as a species, and it goes all the way to the advent of farming that we believe first came about around 11,000 to 7,000 years ago. And this abbreviation right here, this BP, this does not stand for British Petroleum. It stands for Before Present or Before the Present Time. So one more acronym to kind of have in your toolkit when you see things. And obviously, if we're 11,000 years before the present, that's the same thing as 9,000 years before Christ or before the Common Era because Christ was, we believe, born 2,000 years ago. Now, it may or may not be obvious to you, but the advent of agriculture is a super big deal, arguably the biggest deal in the development of human civilization or in all of human history. And you might say, hey, what's the big deal about agriculture?"}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So one more acronym to kind of have in your toolkit when you see things. And obviously, if we're 11,000 years before the present, that's the same thing as 9,000 years before Christ or before the Common Era because Christ was, we believe, born 2,000 years ago. Now, it may or may not be obvious to you, but the advent of agriculture is a super big deal, arguably the biggest deal in the development of human civilization or in all of human history. And you might say, hey, what's the big deal about agriculture? These characters over here look pretty happy. They're able to walk around a lot. They're able to hunt."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you might say, hey, what's the big deal about agriculture? These characters over here look pretty happy. They're able to walk around a lot. They're able to hunt. What's the big deal of all of a sudden people plowing fields and domesticating cattle and having chickens to lay eggs and whatever else? And the big deal about that, besides the fact that it would change people's diet, is that for the first time, it allowed them to not be nomadic. It allowed them to, and you could have probably had some hunters who were somewhat settled, maybe living near the ocean, maybe they did some fishing and all the rest."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're able to hunt. What's the big deal of all of a sudden people plowing fields and domesticating cattle and having chickens to lay eggs and whatever else? And the big deal about that, besides the fact that it would change people's diet, is that for the first time, it allowed them to not be nomadic. It allowed them to, and you could have probably had some hunters who were somewhat settled, maybe living near the ocean, maybe they did some fishing and all the rest. But for the most part, with the development of agriculture, it forced people to stay in one place. So you have the Paleolithic era all the way to the advent of agriculture, which was about 11,000 to 7,000 years ago, and besides the fact that it changed people's diet, it allowed them to settle. So agriculture allowed human beings to settle down in one area."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It allowed them to, and you could have probably had some hunters who were somewhat settled, maybe living near the ocean, maybe they did some fishing and all the rest. But for the most part, with the development of agriculture, it forced people to stay in one place. So you have the Paleolithic era all the way to the advent of agriculture, which was about 11,000 to 7,000 years ago, and besides the fact that it changed people's diet, it allowed them to settle. So agriculture allowed human beings to settle down in one area. And it wasn't just that they were settling in one area, but because they were able to control their environment. They were able to increase the density of things, of crops that humans could consume, and animals that humans could consume, and lower the density of crops that humans can't consume, and animals that they can't consume, or that they don't want around, like pests of some type. What it allowed them to do is also settle in more dense environments."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So agriculture allowed human beings to settle down in one area. And it wasn't just that they were settling in one area, but because they were able to control their environment. They were able to increase the density of things, of crops that humans could consume, and animals that humans could consume, and lower the density of crops that humans can't consume, and animals that they can't consume, or that they don't want around, like pests of some type. What it allowed them to do is also settle in more dense environments. You can imagine, when you just have people walking around, you need a lot of land to support even the calorie requirements of one human being. But all of a sudden, if you are able to develop agriculture, you're able to domesticate animals, all of a sudden you can have, in the same amount of land, you can have more calories being generated. And because you have more calories being generated in a smaller amount of land, people can settle, and they can settle in a denser environment."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What it allowed them to do is also settle in more dense environments. You can imagine, when you just have people walking around, you need a lot of land to support even the calorie requirements of one human being. But all of a sudden, if you are able to develop agriculture, you're able to domesticate animals, all of a sudden you can have, in the same amount of land, you can have more calories being generated. And because you have more calories being generated in a smaller amount of land, people can settle, and they can settle in a denser environment. And so agriculture was really this necessary requirement for people to kind of develop civilization, or to develop villages, and cities. And maybe also giving them the free time to start thinking about, hey, maybe we want to think about how we can record what we know, how we can develop even more technologies. And so just to give us a sense of the categorization that an archaeologist would use for these different periods of time, I told you we start with the Paleolithic era, with the advent of stone tools, pre-humans, most of human time on this planet."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And because you have more calories being generated in a smaller amount of land, people can settle, and they can settle in a denser environment. And so agriculture was really this necessary requirement for people to kind of develop civilization, or to develop villages, and cities. And maybe also giving them the free time to start thinking about, hey, maybe we want to think about how we can record what we know, how we can develop even more technologies. And so just to give us a sense of the categorization that an archaeologist would use for these different periods of time, I told you we start with the Paleolithic era, with the advent of stone tools, pre-humans, most of human time on this planet. And then about 11,000 years ago, the development of agriculture. And it developed independently at different places around the world, which is by itself an interesting phenomenon. And people think that it might just be that the climate might have warmed up a little bit, so that people, maybe naturally there were some human edible crops that existed in a little bit of a denser environment, and humans learned to optimize that slowly, and they did that independently."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so just to give us a sense of the categorization that an archaeologist would use for these different periods of time, I told you we start with the Paleolithic era, with the advent of stone tools, pre-humans, most of human time on this planet. And then about 11,000 years ago, the development of agriculture. And it developed independently at different places around the world, which is by itself an interesting phenomenon. And people think that it might just be that the climate might have warmed up a little bit, so that people, maybe naturally there were some human edible crops that existed in a little bit of a denser environment, and humans learned to optimize that slowly, and they did that independently. But it's an interesting question of why did it develop just then, after 180,000, 190,000 years, why did agriculture all of a sudden happen? But just to get the terminology, the Paleolithic era is that period before agriculture. And then once agriculture starts developing, we are now in the Neolithic era."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And people think that it might just be that the climate might have warmed up a little bit, so that people, maybe naturally there were some human edible crops that existed in a little bit of a denser environment, and humans learned to optimize that slowly, and they did that independently. But it's an interesting question of why did it develop just then, after 180,000, 190,000 years, why did agriculture all of a sudden happen? But just to get the terminology, the Paleolithic era is that period before agriculture. And then once agriculture starts developing, we are now in the Neolithic era. And some archaeologists will describe a transition period between the Paleolithic and the Neolithic era, called the Mesolithic. And just so you know what these words mean, because they actually make sense when you know what they mean, paleo means old, and lithic means stone, or of stone. So they're really talking about the old stone age."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then once agriculture starts developing, we are now in the Neolithic era. And some archaeologists will describe a transition period between the Paleolithic and the Neolithic era, called the Mesolithic. And just so you know what these words mean, because they actually make sense when you know what they mean, paleo means old, and lithic means stone, or of stone. So they're really talking about the old stone age. Neolithic, as you can imagine, means new, new stone. So it's kind of the new stone age. And meso means middle."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they're really talking about the old stone age. Neolithic, as you can imagine, means new, new stone. So it's kind of the new stone age. And meso means middle. So it is the middle stone age. So another way of thinking about this whole period, from when people were hunter-gatherers all the way to about 11,000 to 7,000 years ago, when they developed agriculture, this whole period is called the stone age. And the stone age is kind of this biggest age."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And meso means middle. So it is the middle stone age. So another way of thinking about this whole period, from when people were hunter-gatherers all the way to about 11,000 to 7,000 years ago, when they developed agriculture, this whole period is called the stone age. And the stone age is kind of this biggest age. And there's just different ways of describing it. Because if you just call it the stone age, you're really making importance out of the actual tools that people could shape, they weren't able to use metal at this point. When you refer to Paleolithic and Neolithic, you're maybe referring a little bit more, and there's other ways to think about it, but you're referring a little bit more to the lifestyles of the human beings."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the stone age is kind of this biggest age. And there's just different ways of describing it. Because if you just call it the stone age, you're really making importance out of the actual tools that people could shape, they weren't able to use metal at this point. When you refer to Paleolithic and Neolithic, you're maybe referring a little bit more, and there's other ways to think about it, but you're referring a little bit more to the lifestyles of the human beings. Paleolithic being hunter-gatherers, Neolithic having actually settled, having actually started to develop kind of primitive villages and even cities. And then, of course, Mesolithic is in between. And just for kind of a pop culture reference, you might have heard of the Paleolithic diet that some people are going on now."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When you refer to Paleolithic and Neolithic, you're maybe referring a little bit more, and there's other ways to think about it, but you're referring a little bit more to the lifestyles of the human beings. Paleolithic being hunter-gatherers, Neolithic having actually settled, having actually started to develop kind of primitive villages and even cities. And then, of course, Mesolithic is in between. And just for kind of a pop culture reference, you might have heard of the Paleolithic diet that some people are going on now. And those are people who try to live like hunter-gatherers. Their belief is that most of human evolution occurred while we were hunter-gatherers. And so that's what our bodies are most accustomed to."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just for kind of a pop culture reference, you might have heard of the Paleolithic diet that some people are going on now. And those are people who try to live like hunter-gatherers. Their belief is that most of human evolution occurred while we were hunter-gatherers. And so that's what our bodies are most accustomed to. So they like to eat meat, and they like to eat a lot of nuts. And I even met a co-worker once who used to only eat raw meat. And I don't know if that is even justified or that's even somehow validated by the archaeological record."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that's what our bodies are most accustomed to. So they like to eat meat, and they like to eat a lot of nuts. And I even met a co-worker once who used to only eat raw meat. And I don't know if that is even justified or that's even somehow validated by the archaeological record. These people probably did cook their meat. Now, at the end of the Stone Age, we would have, I would say, the number two most significant development in human history. And now we're talking about 3,000 BC, which is about 5,000 years ago."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I don't know if that is even justified or that's even somehow validated by the archaeological record. These people probably did cook their meat. Now, at the end of the Stone Age, we would have, I would say, the number two most significant development in human history. And now we're talking about 3,000 BC, which is about 5,000 years ago. And this is the development of writing. So we were hunter-gatherers about 9,000 to 10,000, 11,000 years ago. People start developing agriculture."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now we're talking about 3,000 BC, which is about 5,000 years ago. And this is the development of writing. So we were hunter-gatherers about 9,000 to 10,000, 11,000 years ago. People start developing agriculture. It allows them to settle in more dense environments. It also gives them a little bit more free time because they don't have to hunt and gather all the time. And then you go."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "People start developing agriculture. It allows them to settle in more dense environments. It also gives them a little bit more free time because they don't have to hunt and gather all the time. And then you go. And once again, we'll probably discover things as we go forward in time that maybe these dates need to be pushed back or whatever else. Or we discover new civilizations or who knows. But our best sense is you have these villages, you have these civilizations developing."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then you go. And once again, we'll probably discover things as we go forward in time that maybe these dates need to be pushed back or whatever else. Or we discover new civilizations or who knows. But our best sense is you have these villages, you have these civilizations developing. And by about 5,000 years ago, so this would be 5,000 before the present or 3,000 BC, before Christ, you have people saying, hey, why don't we start trying to write down what we know so that when I tell someone orally, it doesn't actually lose information there. And so our descendants can slowly collect all of the knowledge we have and maybe accelerate. I don't know if they did it explicitly thinking of these, but let's just write down what we know."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But our best sense is you have these villages, you have these civilizations developing. And by about 5,000 years ago, so this would be 5,000 before the present or 3,000 BC, before Christ, you have people saying, hey, why don't we start trying to write down what we know so that when I tell someone orally, it doesn't actually lose information there. And so our descendants can slowly collect all of the knowledge we have and maybe accelerate. I don't know if they did it explicitly thinking of these, but let's just write down what we know. And so at about that period of time, you have, as far as we can tell, the first development of a pictogram based system of writing. And the earliest system of writing we know is cuneiform, which is from the Sumerian civilization, which is now in what present day Iraq. And what's the really big deal about this is that this is, on some level, the beginning of recorded history."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't know if they did it explicitly thinking of these, but let's just write down what we know. And so at about that period of time, you have, as far as we can tell, the first development of a pictogram based system of writing. And the earliest system of writing we know is cuneiform, which is from the Sumerian civilization, which is now in what present day Iraq. And what's the really big deal about this is that this is, on some level, the beginning of recorded history. We could talk about the word history. You could say that history is all of the past. And we could use the archaeological record to figure out stuff before people started to write things down."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's the really big deal about this is that this is, on some level, the beginning of recorded history. We could talk about the word history. You could say that history is all of the past. And we could use the archaeological record to figure out stuff before people started to write things down. But when they started to write things down, now it was recorded. Now we're actually getting actual accounts of what people know, of actual people's knowledge. And the reason why this is a big deal, I mean, agriculture, hopefully you now appreciate that it was a pretty big deal."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we could use the archaeological record to figure out stuff before people started to write things down. But when they started to write things down, now it was recorded. Now we're actually getting actual accounts of what people know, of actual people's knowledge. And the reason why this is a big deal, I mean, agriculture, hopefully you now appreciate that it was a pretty big deal. But the reason why writing was a big deal is that now civilization could collect its knowledge. And it could build upon it generation after generation without having to worry about people forgetting it or information getting distorted verbally from ancestor to descendant. And with that, you also have the beginning of the Bronze Age, and the Bronze Age is kind of known for this beginning of, even though it's referring to a material which comes from the first time that people started using bronze as a tool, or using bronze for their tools and for their weapons."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the reason why this is a big deal, I mean, agriculture, hopefully you now appreciate that it was a pretty big deal. But the reason why writing was a big deal is that now civilization could collect its knowledge. And it could build upon it generation after generation without having to worry about people forgetting it or information getting distorted verbally from ancestor to descendant. And with that, you also have the beginning of the Bronze Age, and the Bronze Age is kind of known for this beginning of, even though it's referring to a material which comes from the first time that people started using bronze as a tool, or using bronze for their tools and for their weapons. And bronze, just so you know, it's a mixture of mostly copper and a little bit of tin. But the Bronze Age, at least in my mind, the biggest deal of what started at the beginning of the Bronze Age really, really was the writing. So once again, just as a review, because I actually find this kind of confusing, our current understanding, most of human prehistory, and even pre-human prehistory, were spent as hunter-gatherers using stone tools until about 11,000 years ago."}, {"video_title": "Development of agriculture and writing Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And with that, you also have the beginning of the Bronze Age, and the Bronze Age is kind of known for this beginning of, even though it's referring to a material which comes from the first time that people started using bronze as a tool, or using bronze for their tools and for their weapons. And bronze, just so you know, it's a mixture of mostly copper and a little bit of tin. But the Bronze Age, at least in my mind, the biggest deal of what started at the beginning of the Bronze Age really, really was the writing. So once again, just as a review, because I actually find this kind of confusing, our current understanding, most of human prehistory, and even pre-human prehistory, were spent as hunter-gatherers using stone tools until about 11,000 years ago. And then we became a little bit more settled. We became farmers, essentially, using stone tools. And then you fast forward another about 5,000, 6,000 years, and then we started to become farmers who started to write down the things that we knew, and we started to use bronze tools."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And maybe an even more interesting question is, can we detect any of those civilizations? Have they gotten to the level of technological progress like us, that they're emitting electromagnetic waves into space that other civilizations like ours can detect and say, hey, there's someone else out there watching television or using radio or whatever else they might be doing. And so what I want to do in this video is not answer that question. It's a big, open question. We don't know the answer. We don't have anywhere near enough information to definitively answer that question. But what I want to do is come up with a framework for at least thinking about that question, a way of actually estimating how many detectable civilizations there are in just our galaxy."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's a big, open question. We don't know the answer. We don't have anywhere near enough information to definitively answer that question. But what I want to do is come up with a framework for at least thinking about that question, a way of actually estimating how many detectable civilizations there are in just our galaxy. And there's a formula that you may or may not have heard of called the Drake equation. And what we're going to do is independently derive our own version of the Drake equation. It's going to be slightly different, but it's the same thought process."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what I want to do is come up with a framework for at least thinking about that question, a way of actually estimating how many detectable civilizations there are in just our galaxy. And there's a formula that you may or may not have heard of called the Drake equation. And what we're going to do is independently derive our own version of the Drake equation. It's going to be slightly different, but it's the same thought process. And in a future video, I'm going to maybe reconcile what we come up with with the Drake equation. And just so you know, the Drake equation is named for Frank Drake, who is a professor at University of California, Santa Cruz. He first kind of put some structure around this problem, and that's why the formula or the equation has his name."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's going to be slightly different, but it's the same thought process. And in a future video, I'm going to maybe reconcile what we come up with with the Drake equation. And just so you know, the Drake equation is named for Frank Drake, who is a professor at University of California, Santa Cruz. He first kind of put some structure around this problem, and that's why the formula or the equation has his name. But the equation, it's not an equation that you can apply on a daily basis and get results that you can use to build things. But what it is, is it structures our thinking around this question of how many detectable civilizations are there in our galaxy. Now to answer this question, I'm going to start a little bit differently than Frank Drake did."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "He first kind of put some structure around this problem, and that's why the formula or the equation has his name. But the equation, it's not an equation that you can apply on a daily basis and get results that you can use to build things. But what it is, is it structures our thinking around this question of how many detectable civilizations are there in our galaxy. Now to answer this question, I'm going to start a little bit differently than Frank Drake did. He starts with the number of new stars that are born each year. And we'll see that our definitions are actually pretty close to each other. What I want to do is start with the total number of stars."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now to answer this question, I'm going to start a little bit differently than Frank Drake did. He starts with the number of new stars that are born each year. And we'll see that our definitions are actually pretty close to each other. What I want to do is start with the total number of stars. So what we're trying to come up with is, I'll call it n, and this is the number of detectable civilizations in the Milky Way, in our galaxy. And once again, there could be civilizations, looking back at this star field right over here. This star right over here, maybe it has a planet that's in the right place."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What I want to do is start with the total number of stars. So what we're trying to come up with is, I'll call it n, and this is the number of detectable civilizations in the Milky Way, in our galaxy. And once again, there could be civilizations, looking back at this star field right over here. This star right over here, maybe it has a planet that's in the right place. It has liquid water. And maybe there's intelligent life on that planet. But they might not be detectable because they aren't technologically advanced enough that they're using electromagnetic radiation."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This star right over here, maybe it has a planet that's in the right place. It has liquid water. And maybe there's intelligent life on that planet. But they might not be detectable because they aren't technologically advanced enough that they're using electromagnetic radiation. Or maybe they just figured out some other way to communicate. Or maybe they're beyond using electromagnetic radiation, radio waves and all the rest, to communicate. And so we'll never be able to detect them."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But they might not be detectable because they aren't technologically advanced enough that they're using electromagnetic radiation. Or maybe they just figured out some other way to communicate. Or maybe they're beyond using electromagnetic radiation, radio waves and all the rest, to communicate. And so we'll never be able to detect them. We're talking about civilizations like ours that are, to some degree, using technology not too different than our own. So that's what we mean by detectable. So let's think about that a little bit."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we'll never be able to detect them. We're talking about civilizations like ours that are, to some degree, using technology not too different than our own. So that's what we mean by detectable. So let's think about that a little bit. So I like to start with just the total number of stars in our solar system. So let's just start with, I'll call it n star and asterisk. And this is the number of stars in our, not in our solar system, number of stars in our galaxy."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's think about that a little bit. So I like to start with just the total number of stars in our solar system. So let's just start with, I'll call it n star and asterisk. And this is the number of stars in our, not in our solar system, number of stars in our galaxy. Number of stars in the galaxy. And our best guess, I said, is this is going to be 100 billion to 400 billion stars. We don't even know how many there are."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is the number of stars in our, not in our solar system, number of stars in our galaxy. Number of stars in the galaxy. And our best guess, I said, is this is going to be 100 billion to 400 billion stars. We don't even know how many there are. Some of them are undetectable. In the center of our galaxy, it's just a big blur to us. And we don't even know what's on the other side of that."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We don't even know how many there are. Some of them are undetectable. In the center of our galaxy, it's just a big blur to us. And we don't even know what's on the other side of that. And we can't even see all the stars that are packed into the center. So this is our best guess, 100 billion to 400 billion stars. Now obviously, it's going to be a subset of those stars that even have planets."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we don't even know what's on the other side of that. And we can't even see all the stars that are packed into the center. So this is our best guess, 100 billion to 400 billion stars. Now obviously, it's going to be a subset of those stars that even have planets. So let's multiply it times that subset. So let's multiply it times the frequency of having a planet. If you're a star, this is the percent chance or the frequency or the fraction of these stars that have planets."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now obviously, it's going to be a subset of those stars that even have planets. So let's multiply it times that subset. So let's multiply it times the frequency of having a planet. If you're a star, this is the percent chance or the frequency or the fraction of these stars that have planets. So I'll write it this way. Fraction that have planets. So if this is 100 billion, and let's say I'm making a guess here, and we're learning more about this every day, there are all these discoveries of exoplanets, planets outside of our solar system, maybe this is 1 4th."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you're a star, this is the percent chance or the frequency or the fraction of these stars that have planets. So I'll write it this way. Fraction that have planets. So if this is 100 billion, and let's say I'm making a guess here, and we're learning more about this every day, there are all these discoveries of exoplanets, planets outside of our solar system, maybe this is 1 4th. Then we could say, well, that means that 100 billion times 1 4th means that there are 25 billion stars that have planets around them. But that's still not enough to go to civilizations. We also need to think about planets."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if this is 100 billion, and let's say I'm making a guess here, and we're learning more about this every day, there are all these discoveries of exoplanets, planets outside of our solar system, maybe this is 1 4th. Then we could say, well, that means that 100 billion times 1 4th means that there are 25 billion stars that have planets around them. But that's still not enough to go to civilizations. We also need to think about planets. Because a planet could be a planet like Jupiter, and we don't know how life as we know it can survive on a planet like Jupiter or Neptune or Mercury. It has to have planets that are good for sustaining life. Preferably have a rocky core, liquid water on the outside."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We also need to think about planets. Because a planet could be a planet like Jupiter, and we don't know how life as we know it can survive on a planet like Jupiter or Neptune or Mercury. It has to have planets that are good for sustaining life. Preferably have a rocky core, liquid water on the outside. That's what we think are the ingredients that you need for life. Maybe we're just not being creative enough. That's what we know as life as being."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Preferably have a rocky core, liquid water on the outside. That's what we think are the ingredients that you need for life. Maybe we're just not being creative enough. That's what we know as life as being. So let's multiply this times the average number of life-sustaining or planets that could sustain life on them. So we don't necessarily that they're going to have life, but they seem like they're just the right distance from the star, not too hot, not too cold. They have the right amount of gravity, water, all the other stuff."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's what we know as life as being. So let's multiply this times the average number of life-sustaining or planets that could sustain life on them. So we don't necessarily that they're going to have life, but they seem like they're just the right distance from the star, not too hot, not too cold. They have the right amount of gravity, water, all the other stuff. And we still don't know exactly what this means. But this means average number. So given that there's a solar system with planets, what's the average number of planets that are capable of sustaining life?"}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They have the right amount of gravity, water, all the other stuff. And we still don't know exactly what this means. But this means average number. So given that there's a solar system with planets, what's the average number of planets that are capable of sustaining life? And once again, we don't know this answer. Maybe it's 0.1. It's probably less than 1."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So given that there's a solar system with planets, what's the average number of planets that are capable of sustaining life? And once again, we don't know this answer. Maybe it's 0.1. It's probably less than 1. For any given solar system that has planets, the average number capable of sustaining life, maybe it's 0.1. Maybe it's more than 1. I don't know."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's probably less than 1. For any given solar system that has planets, the average number capable of sustaining life, maybe it's 0.1. Maybe it's more than 1. I don't know. We don't know the exact answer here. But I'll throw out a guess. Maybe it is 0.1."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't know. We don't know the exact answer here. But I'll throw out a guess. Maybe it is 0.1. And here, the fraction that have planets, I don't know. I'll throw that out as. And once again, I'm just making up these numbers."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe it is 0.1. And here, the fraction that have planets, I don't know. I'll throw that out as. And once again, I'm just making up these numbers. We really don't know the right answer. This is 1 4th. But if we were to multiply this out, we would have the average number of planets in our solar system that are capable of sustaining life, that are around stars that have planets."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And once again, I'm just making up these numbers. We really don't know the right answer. This is 1 4th. But if we were to multiply this out, we would have the average number of planets in our solar system that are capable of sustaining life, that are around stars that have planets. And these planets are capable of sustaining life. Now, and this would give us the total number, because this is average per solar system that has planets. This is the total number of solar systems with planets."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if we were to multiply this out, we would have the average number of planets in our solar system that are capable of sustaining life, that are around stars that have planets. And these planets are capable of sustaining life. Now, and this would give us the total number, because this is average per solar system that has planets. This is the total number of solar systems with planets. You multiply it out. Total number of planets in our galaxy capable of sustaining life. Now, just because you have liquid water and the right temperature and all of the rest ingredients doesn't necessarily mean that you will actually have life happening on your planet."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the total number of solar systems with planets. You multiply it out. Total number of planets in our galaxy capable of sustaining life. Now, just because you have liquid water and the right temperature and all of the rest ingredients doesn't necessarily mean that you will actually have life happening on your planet. So let's multiply that times the fraction that actually generate life. So this is the fraction that actually have life. And this is actually a very, we don't know this answer."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, just because you have liquid water and the right temperature and all of the rest ingredients doesn't necessarily mean that you will actually have life happening on your planet. So let's multiply that times the fraction that actually generate life. So this is the fraction that actually have life. And this is actually a very, we don't know this answer. So this is a fraction that have life on them. And this is a really big open question. Maybe if you have the ingredients, almost every planet has life."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is actually a very, we don't know this answer. So this is a fraction that have life on them. And this is a really big open question. Maybe if you have the ingredients, almost every planet has life. Maybe it's a frequent thing that's happening in our galaxy and frankly our universe. Or maybe it's a very infrequent thing. Maybe it's just the right kind of freak set of circumstances that just have to happen."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe if you have the ingredients, almost every planet has life. Maybe it's a frequent thing that's happening in our galaxy and frankly our universe. Or maybe it's a very infrequent thing. Maybe it's just the right kind of freak set of circumstances that just have to happen. I'll throw out a number just for the sake of just to have a number there. Maybe it's 1 out of every 10 planets that have all of the right ingredients for life actually do generate life. My personal guess is probably higher than that, given that life seems such a robust and flexible thing that we've been seeing in all sorts of weird circumstances."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe it's just the right kind of freak set of circumstances that just have to happen. I'll throw out a number just for the sake of just to have a number there. Maybe it's 1 out of every 10 planets that have all of the right ingredients for life actually do generate life. My personal guess is probably higher than that, given that life seems such a robust and flexible thing that we've been seeing in all sorts of weird circumstances. Actually, let me make it even a higher number than that. So let me make it 1 half, assuming that we have all of the ingredients. So this should tell us essentially how many planets, if we were to multiply all of these, how many planets in our galaxy have had life on them at some point in those planets' lives."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "My personal guess is probably higher than that, given that life seems such a robust and flexible thing that we've been seeing in all sorts of weird circumstances. Actually, let me make it even a higher number than that. So let me make it 1 half, assuming that we have all of the ingredients. So this should tell us essentially how many planets, if we were to multiply all of these, how many planets in our galaxy have had life on them at some point in those planets' lives. The life might have come and gone. It maybe destroyed itself through nuclear war or whatever. But this would tell us the number of life planets in our galaxy that have had life on them at at least one point in their history."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this should tell us essentially how many planets, if we were to multiply all of these, how many planets in our galaxy have had life on them at some point in those planets' lives. The life might have come and gone. It maybe destroyed itself through nuclear war or whatever. But this would tell us the number of life planets in our galaxy that have had life on them at at least one point in their history. Now, we care about civilization. So let's multiply this times. If you make all of these, you even get to the point that you have life, we care about, well, do you get intelligent life?"}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this would tell us the number of life planets in our galaxy that have had life on them at at least one point in their history. Now, we care about civilization. So let's multiply this times. If you make all of these, you even get to the point that you have life, we care about, well, do you get intelligent life? So maybe if the asteroid never hit Earth, the dinosaurs would have stayed on Earth, and they would have never evolved to the point of generating radios and TVs and telephones and all the rest. And so it's kind of a freak circumstance that we were, because they were destroyed, these gaps in the ecosystem developed so that we could emerge and be intelligent and do all of these crazy things like make YouTube videos and all the rest. So let's multiply this times the fraction."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you make all of these, you even get to the point that you have life, we care about, well, do you get intelligent life? So maybe if the asteroid never hit Earth, the dinosaurs would have stayed on Earth, and they would have never evolved to the point of generating radios and TVs and telephones and all the rest. And so it's kind of a freak circumstance that we were, because they were destroyed, these gaps in the ecosystem developed so that we could emerge and be intelligent and do all of these crazy things like make YouTube videos and all the rest. So let's multiply this times the fraction. If you get all of this, the fraction that actually end up having intelligent life. And maybe this fraction is intelligent life over here. So intelligent life, this is maybe out to our number 1 10th."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's multiply this times the fraction. If you get all of this, the fraction that actually end up having intelligent life. And maybe this fraction is intelligent life over here. So intelligent life, this is maybe out to our number 1 10th. And probably in the next video, I'll calculate it all. And this is very important to realize, because once again, you could have life. These are all examples of life right over here."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So intelligent life, this is maybe out to our number 1 10th. And probably in the next video, I'll calculate it all. And this is very important to realize, because once again, you could have life. These are all examples of life right over here. This is actually life on our planet, even though this looks quite alien. This is a weevil that kind of looked very close up. But there's all sorts of forms of life, many of which we probably can't even begin to imagine."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These are all examples of life right over here. This is actually life on our planet, even though this looks quite alien. This is a weevil that kind of looked very close up. But there's all sorts of forms of life, many of which we probably can't even begin to imagine. But what we care is that intelligent life starts to emerge on the planet. Because only intelligent life has a chance, we believe, of being able to eventually communicate in ways that are detectable by us. Now, I said intelligent life, but maybe not all intelligent life will eventually get to the technological sophistication where they will be using radio waves and electromagnetic radiation to communicate with each other."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But there's all sorts of forms of life, many of which we probably can't even begin to imagine. But what we care is that intelligent life starts to emerge on the planet. Because only intelligent life has a chance, we believe, of being able to eventually communicate in ways that are detectable by us. Now, I said intelligent life, but maybe not all intelligent life will eventually get to the technological sophistication where they will be using radio waves and electromagnetic radiation to communicate with each other. Maybe we might have stagnated at this stage if the right things didn't happen. So what we need to do now is multiply this. So right here, we would have the number of planets in our galaxy that have had intelligent life on them at some point in their history, maybe not at a time that coincides with ours."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, I said intelligent life, but maybe not all intelligent life will eventually get to the technological sophistication where they will be using radio waves and electromagnetic radiation to communicate with each other. Maybe we might have stagnated at this stage if the right things didn't happen. So what we need to do now is multiply this. So right here, we would have the number of planets in our galaxy that have had intelligent life on them at some point in their history, maybe not at a time that coincides with ours. But what we want to do is whittle it down even more to the percentage that get to the point that they can develop technology that allows us to detect them. So let me multiply it times the fraction that are, I'll put a c here for maybe they're using communications, c for communications, that allow us to detect them. So this is detectable, the fraction that are detectable."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So right here, we would have the number of planets in our galaxy that have had intelligent life on them at some point in their history, maybe not at a time that coincides with ours. But what we want to do is whittle it down even more to the percentage that get to the point that they can develop technology that allows us to detect them. So let me multiply it times the fraction that are, I'll put a c here for maybe they're using communications, c for communications, that allow us to detect them. So this is detectable, the fraction that are detectable. Detectable. Now you might think that we're done. This would give you the total number of civilizations or life forms in our galaxy, or the planets that have life forms that developed detectable technologies at some point in their history."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is detectable, the fraction that are detectable. Detectable. Now you might think that we're done. This would give you the total number of civilizations or life forms in our galaxy, or the planets that have life forms that developed detectable technologies at some point in their history. Now it would be nice if civilizations did not kind of be born and then die. But the reality is they do die. They might destroy themselves or whatever."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This would give you the total number of civilizations or life forms in our galaxy, or the planets that have life forms that developed detectable technologies at some point in their history. Now it would be nice if civilizations did not kind of be born and then die. But the reality is they do die. They might destroy themselves or whatever. And they might exist for only a small period of time for the history of that planet or the history of that solar system. So in order to make it the number of civilizations that are in existence now, and I'll clarify what now means in the next video. Because it's really, if we're detecting something from a star that's 10,000 light years away, our now means we're just receiving their signals, which means that they released the signals 10,000 years ago."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They might destroy themselves or whatever. And they might exist for only a small period of time for the history of that planet or the history of that solar system. So in order to make it the number of civilizations that are in existence now, and I'll clarify what now means in the next video. Because it's really, if we're detecting something from a star that's 10,000 light years away, our now means we're just receiving their signals, which means that they released the signals 10,000 years ago. So what I want to do is, what is the fraction of these whose signals are achieving, are reaching us right now? And here I'm going to say, well, what's the average lifespan of a civilization? I'll put that L. Who knows what that is?"}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because it's really, if we're detecting something from a star that's 10,000 light years away, our now means we're just receiving their signals, which means that they released the signals 10,000 years ago. So what I want to do is, what is the fraction of these whose signals are achieving, are reaching us right now? And here I'm going to say, well, what's the average lifespan of a civilization? I'll put that L. Who knows what that is? Maybe 10,000 years. So civilization lifespan. And it's going to be that over the life of the star."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll put that L. Who knows what that is? Maybe 10,000 years. So civilization lifespan. And it's going to be that over the life of the star. So that is over, I'll put a T here. T for the star. So the average lifespan for the star."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's going to be that over the life of the star. So that is over, I'll put a T here. T for the star. So the average lifespan for the star. And I could say the average lifespan for the planet or whatever, but we're assuming that once our star supernovas, you're not going to have any chance for Earth to develop life on it anymore. So maybe this thing up here is 10,000 years, and this down here is maybe 10 billion years. And if you were to multiply all of this out, you should get the number of detectable civilizations in our galaxy right now."}, {"video_title": "Detectable civilizations in our galaxy 1 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the average lifespan for the star. And I could say the average lifespan for the planet or whatever, but we're assuming that once our star supernovas, you're not going to have any chance for Earth to develop life on it anymore. So maybe this thing up here is 10,000 years, and this down here is maybe 10 billion years. And if you were to multiply all of this out, you should get the number of detectable civilizations in our galaxy right now. I'll leave you there for this video. In the next video, we'll discuss it a little bit more and reconcile it with the more famous version of Drake's equation. And I'll also try to talk about this piece a little bit, because I think this might be a little bit confusing."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "Let's start with this carbocation. So we have a plus one formal charge on this carbon, and then we have the carbon with a positive charge bonded to two other carbons, so this is a secondary carbocation. Let's think about what could possibly shift here. So first, let's pick one of the two hydrogens on this carbon here, and let's say one of them is involved in a hydride shift. That'll be our first attempt. So we take this hydrogen and these two electrons in a hydride shift, and we move them over to this carbon. So let's draw what we would form."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So first, let's pick one of the two hydrogens on this carbon here, and let's say one of them is involved in a hydride shift. That'll be our first attempt. So we take this hydrogen and these two electrons in a hydride shift, and we move them over to this carbon. So let's draw what we would form. So I'll draw in the ring here, and let me go ahead and put in the methyl group over here on the right. We know that there was already a hydrogen, I should say, on this carbon that's marked in magenta, so let me draw in that hydrogen. And the hydride in red, that's going to move onto that carbon as well, so let me go ahead and mark this as being the red hydrogen."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So let's draw what we would form. So I'll draw in the ring here, and let me go ahead and put in the methyl group over here on the right. We know that there was already a hydrogen, I should say, on this carbon that's marked in magenta, so let me draw in that hydrogen. And the hydride in red, that's going to move onto that carbon as well, so let me go ahead and mark this as being the red hydrogen. That leaves one hydrogen, and I'll make this one in green, on this carbon in green. So let's say this is the carbon in green, and then there's one hydrogen left on that carbon. We took a bond away from the carbon in green, and so that carbon now has a plus one formal charge, so plus one formal charge on this carbon."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "And the hydride in red, that's going to move onto that carbon as well, so let me go ahead and mark this as being the red hydrogen. That leaves one hydrogen, and I'll make this one in green, on this carbon in green. So let's say this is the carbon in green, and then there's one hydrogen left on that carbon. We took a bond away from the carbon in green, and so that carbon now has a plus one formal charge, so plus one formal charge on this carbon. So we still have a secondary carbocation. The carbon in green is bonded to two other carbons, so this is a secondary carbocation, which is not an improvement upon our original secondary carbocation. Now, we could also draw it like this."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "We took a bond away from the carbon in green, and so that carbon now has a plus one formal charge, so plus one formal charge on this carbon. So we still have a secondary carbocation. The carbon in green is bonded to two other carbons, so this is a secondary carbocation, which is not an improvement upon our original secondary carbocation. Now, we could also draw it like this. So I draw out all the hydrogens, but when you do a lot of practice, you don't have to draw in all of the hydrogens. You could just think about a hydride shift, and you have your methyl group right here forming a carbocation at this carbon. So when you're first doing this, you need to draw in all the hydrogens just for practice."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "Now, we could also draw it like this. So I draw out all the hydrogens, but when you do a lot of practice, you don't have to draw in all of the hydrogens. You could just think about a hydride shift, and you have your methyl group right here forming a carbocation at this carbon. So when you're first doing this, you need to draw in all the hydrogens just for practice. All right, let's think about the possibility of a methyl shift, because that hydride shift didn't work out very well, so let's try to move this methyl group over to the carbon in magenta. So let's draw what we would form. So here is our ring, and we're moving our methyl group in red over to this carbon."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So when you're first doing this, you need to draw in all the hydrogens just for practice. All right, let's think about the possibility of a methyl shift, because that hydride shift didn't work out very well, so let's try to move this methyl group over to the carbon in magenta. So let's draw what we would form. So here is our ring, and we're moving our methyl group in red over to this carbon. So let me go ahead and draw in a CH3 in red. That carbon already had a hydrogen on it, so I'll go ahead and draw in the hydrogen that was already on that carbon. And we're taking a bond away from, let me make this carbon blue."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So here is our ring, and we're moving our methyl group in red over to this carbon. So let me go ahead and draw in a CH3 in red. That carbon already had a hydrogen on it, so I'll go ahead and draw in the hydrogen that was already on that carbon. And we're taking a bond away from, let me make this carbon blue. So this carbon in blue is losing that CH3, but there was a hydrogen on that carbon originally. So let me go ahead and draw it in. There was a hydrogen here, so I'll make that one in blue."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "And we're taking a bond away from, let me make this carbon blue. So this carbon in blue is losing that CH3, but there was a hydrogen on that carbon originally. So let me go ahead and draw it in. There was a hydrogen here, so I'll make that one in blue. And so it's still there, but we did take away a bond from the carbon in blue, which means that this carbon in blue has a plus one formal charge. This is still a secondary carbocation. The carbon in blue was directly bonded to two other carbons."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "There was a hydrogen here, so I'll make that one in blue. And so it's still there, but we did take away a bond from the carbon in blue, which means that this carbon in blue has a plus one formal charge. This is still a secondary carbocation. The carbon in blue was directly bonded to two other carbons. So this is a secondary carbocation, which again, is not an improvement upon our original carbocation. And also, instead of drawing your carbocation this way with all these bonds in here, you could just go ahead and show your methyl group as being here, and then a plus one formal charge on this carbon. So finally, what's left?"}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "The carbon in blue was directly bonded to two other carbons. So this is a secondary carbocation, which again, is not an improvement upon our original carbocation. And also, instead of drawing your carbocation this way with all these bonds in here, you could just go ahead and show your methyl group as being here, and then a plus one formal charge on this carbon. So finally, what's left? What possibility is left? Well, we could try a hydride shift again, but instead of trying one of the two hydrogens on the left, we could try this hydrogen on the right, the one in blue. So let's try a hydride shift with that one."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So finally, what's left? What possibility is left? Well, we could try a hydride shift again, but instead of trying one of the two hydrogens on the left, we could try this hydrogen on the right, the one in blue. So let's try a hydride shift with that one. So this hydrogen and these two electrons are gonna move over here to the carbon in magenta. So let's draw what we would form. We have our ring, and the methyl group stays there."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So let's try a hydride shift with that one. So this hydrogen and these two electrons are gonna move over here to the carbon in magenta. So let's draw what we would form. We have our ring, and the methyl group stays there. The hydrogen in blue just moved over to here. So actually, let me go ahead and make that blue so we can distinguish it. So here is the hydrogen and those two electrons that moved."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "We have our ring, and the methyl group stays there. The hydrogen in blue just moved over to here. So actually, let me go ahead and make that blue so we can distinguish it. So here is the hydrogen and those two electrons that moved. There was already a hydrogen on that carbon, so I will draw in the original hydrogen here. And we took a bond away from the carbon in blue. So here's the carbon in blue, which means that is where our carbocation is."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So here is the hydrogen and those two electrons that moved. There was already a hydrogen on that carbon, so I will draw in the original hydrogen here. And we took a bond away from the carbon in blue. So here's the carbon in blue, which means that is where our carbocation is. That is a plus one formal charge. Now, let's look at this resulting carbocation. The carbon that's in blue is directly bonded to one, two, three other carbons."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So here's the carbon in blue, which means that is where our carbocation is. That is a plus one formal charge. Now, let's look at this resulting carbocation. The carbon that's in blue is directly bonded to one, two, three other carbons. So this is a tertiary carbocation. And we know from the previous video that a tertiary carbocation is more stable than a secondary carbocation. So this is the rearrangement that we would see."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "The carbon that's in blue is directly bonded to one, two, three other carbons. So this is a tertiary carbocation. And we know from the previous video that a tertiary carbocation is more stable than a secondary carbocation. So this is the rearrangement that we would see. We're going from a secondary carbocation to a tertiary carbocation via a hydride shift. And just like the previous examples, we don't need to draw in the hydrogens. We could just show our tertiary carbocation leaving out the hydrogen."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So this is the rearrangement that we would see. We're going from a secondary carbocation to a tertiary carbocation via a hydride shift. And just like the previous examples, we don't need to draw in the hydrogens. We could just show our tertiary carbocation leaving out the hydrogen. So put a methyl group in here, plus one formal charge on this carbon. So this tertiary carbocation is more stable than the secondary ones. Let's do another carbocation rearrangement problem."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "We could just show our tertiary carbocation leaving out the hydrogen. So put a methyl group in here, plus one formal charge on this carbon. So this tertiary carbocation is more stable than the secondary ones. Let's do another carbocation rearrangement problem. So this one's actually a little bit easier than the previous one. So here's our carbocation, and the carbon with the plus one formal charge is directly bonded to two other carbons, which makes this a secondary carbocation. So let's think about what kind of shifts that we could possibly have."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "Let's do another carbocation rearrangement problem. So this one's actually a little bit easier than the previous one. So here's our carbocation, and the carbon with the plus one formal charge is directly bonded to two other carbons, which makes this a secondary carbocation. So let's think about what kind of shifts that we could possibly have. So first, we know that this carbon has two hydrogens on it. So we could try doing a hydride shift with one of those hydrogens. So one of these hydrogens and these two electrons could do a hydride shift and move over here to this carbon."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So let's think about what kind of shifts that we could possibly have. So first, we know that this carbon has two hydrogens on it. So we could try doing a hydride shift with one of those hydrogens. So one of these hydrogens and these two electrons could do a hydride shift and move over here to this carbon. So let's draw what we would make. Let's draw our ring in here, and let's put in these two methyl groups coming off of that carbon. So the hydrogen in red is now this hydrogen."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So one of these hydrogens and these two electrons could do a hydride shift and move over here to this carbon. So let's draw what we would make. Let's draw our ring in here, and let's put in these two methyl groups coming off of that carbon. So the hydrogen in red is now this hydrogen. Let me go ahead and make this red, and let me highlight these two electrons. So that's what moved in our hydride shift. And that moved to this top carbon here in magenta."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So the hydrogen in red is now this hydrogen. Let me go ahead and make this red, and let me highlight these two electrons. So that's what moved in our hydride shift. And that moved to this top carbon here in magenta. We know there was already a hydrogen bonded to that magenta carbon in the beginning, right? So let's go ahead and put in that original hydrogen on that magenta carbon. We took a bond away from this carbon, so I'll make this the carbon in green, but we still have a hydrogen attached to that carbon."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "And that moved to this top carbon here in magenta. We know there was already a hydrogen bonded to that magenta carbon in the beginning, right? So let's go ahead and put in that original hydrogen on that magenta carbon. We took a bond away from this carbon, so I'll make this the carbon in green, but we still have a hydrogen attached to that carbon. So the carbon in green is this one, and there's still a hydrogen attached to it, but since we took a bond away from it, that means that we have a plus one formal charge on the carbon in green. And the carbon in green is directly bonded to two other carbons. So we formed a secondary carbocation, which is not an improvement upon our original secondary carbocation, so we haven't increased our stability."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "We took a bond away from this carbon, so I'll make this the carbon in green, but we still have a hydrogen attached to that carbon. So the carbon in green is this one, and there's still a hydrogen attached to it, but since we took a bond away from it, that means that we have a plus one formal charge on the carbon in green. And the carbon in green is directly bonded to two other carbons. So we formed a secondary carbocation, which is not an improvement upon our original secondary carbocation, so we haven't increased our stability. Remember, we could also draw this without all those hydrogens in there, just a little bit easier to see. So we have our methyl groups with a plus one formal charge on this carbon. Let's try doing another kind of shift."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So we formed a secondary carbocation, which is not an improvement upon our original secondary carbocation, so we haven't increased our stability. Remember, we could also draw this without all those hydrogens in there, just a little bit easier to see. So we have our methyl groups with a plus one formal charge on this carbon. Let's try doing another kind of shift. So if we look over here to this carbon in blue, we have these two methyl groups. So we could do a methyl shift with one of these methyl groups. So let's take the CH3, we're gonna move it over here to the carbon in magenta."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "Let's try doing another kind of shift. So if we look over here to this carbon in blue, we have these two methyl groups. So we could do a methyl shift with one of these methyl groups. So let's take the CH3, we're gonna move it over here to the carbon in magenta. So let's try a methyl shift, and let's draw in our ring. Well, one of the methyl groups is going to remain on this carbon in blue, so let me highlight it here. So this is the carbon in blue, and this is one of the methyl groups that stayed behind."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So let's take the CH3, we're gonna move it over here to the carbon in magenta. So let's try a methyl shift, and let's draw in our ring. Well, one of the methyl groups is going to remain on this carbon in blue, so let me highlight it here. So this is the carbon in blue, and this is one of the methyl groups that stayed behind. Another one moved over here to this carbon, so let me go ahead and draw that in, CH3. And let me just go ahead and highlight that in blue. This was the methyl group that underwent our methyl shift, so it's this one right here."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So this is the carbon in blue, and this is one of the methyl groups that stayed behind. Another one moved over here to this carbon, so let me go ahead and draw that in, CH3. And let me just go ahead and highlight that in blue. This was the methyl group that underwent our methyl shift, so it's this one right here. And again, on that carbon in magenta, there was originally a hydrogen, so let's draw in that hydrogen. But notice that we took a bond away from the carbon in blue. So this carbon in blue gets a plus one formal charge."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "This was the methyl group that underwent our methyl shift, so it's this one right here. And again, on that carbon in magenta, there was originally a hydrogen, so let's draw in that hydrogen. But notice that we took a bond away from the carbon in blue. So this carbon in blue gets a plus one formal charge. And if we think about what kind of carbocation this is, the carbon in blue is directly bonded to one, two, three other carbons. So this is a tertiary carbocation, which we know is more stable than a secondary carbocation. So this is the rearrangement that's going to occur."}, {"video_title": "Carbocation rearrangement practice.mp3", "Sentence": "So this carbon in blue gets a plus one formal charge. And if we think about what kind of carbocation this is, the carbon in blue is directly bonded to one, two, three other carbons. So this is a tertiary carbocation, which we know is more stable than a secondary carbocation. So this is the rearrangement that's going to occur. We get a methyl shift to form a tertiary carbocation. Let's go ahead and draw it without the hydrogen in there so we can see what it looks like a little bit better. So we have those methyl groups, and we have a plus one formal charge on this carbon."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And to that alkene, we're going to add O3 in the first step, which is ozone. And the second step, we're going to add DMS, which is dimethyl sulfide. And be careful, because there are different reagents you could add in the second step. So make sure to learn the one that your professor wants you to use. If you use DMS, you're going to get a mixture of aldehydes and or ketones for your product, depending on what is attached to your initial alkene. So let's start by looking at a dot structure for ozone. So over here on the left is a possible dot structure for ozone."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So make sure to learn the one that your professor wants you to use. If you use DMS, you're going to get a mixture of aldehydes and or ketones for your product, depending on what is attached to your initial alkene. So let's start by looking at a dot structure for ozone. So over here on the left is a possible dot structure for ozone. And we can draw a resonance structure by taking these electrons and moving them in here to form a double bond between those two oxygens. And that would push these electrons in here off onto the oxygen on the left. So let's go ahead and draw the other resonance structure."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the left is a possible dot structure for ozone. And we can draw a resonance structure by taking these electrons and moving them in here to form a double bond between those two oxygens. And that would push these electrons in here off onto the oxygen on the left. So let's go ahead and draw the other resonance structure. So now I would have a double bond between my two oxygens on the right. The oxygen on the far right has two lone pairs of electrons around it now. The oxygen in the center still has a lone pair of electrons on it."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the other resonance structure. So now I would have a double bond between my two oxygens on the right. The oxygen on the far right has two lone pairs of electrons around it now. The oxygen in the center still has a lone pair of electrons on it. And the oxygen on the far left now has three lone pairs of electrons on it. So the oxygen on the far left now has a negative 1 formal charge. And the oxygen on the top here still has a plus 1 formal charge."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen in the center still has a lone pair of electrons on it. And the oxygen on the far left now has three lone pairs of electrons on it. So the oxygen on the far left now has a negative 1 formal charge. And the oxygen on the top here still has a plus 1 formal charge. And so those are our two resonance structures. Remember that the actual molecule is a hybrid of these two resonance structures. So let's go ahead and pick one of those resonance structures."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the oxygen on the top here still has a plus 1 formal charge. And so those are our two resonance structures. Remember that the actual molecule is a hybrid of these two resonance structures. So let's go ahead and pick one of those resonance structures. I'm just going to take the one on the right. So let me just go ahead and redraw the resonance structure on the right. And so we're going to do the one that has the negative charge on the oxygen on the far left."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and pick one of those resonance structures. I'm just going to take the one on the right. So let me just go ahead and redraw the resonance structure on the right. And so we're going to do the one that has the negative charge on the oxygen on the far left. And this top oxygen has a plus 1 formal charge. And the oxygen on the right has no charge in this resonance structure. So let's go ahead and draw in our alkene."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to do the one that has the negative charge on the oxygen on the far left. And this top oxygen has a plus 1 formal charge. And the oxygen on the right has no charge in this resonance structure. So let's go ahead and draw in our alkene. And so here is our alkene with unknown substituents at the moment. And we think about our mechanism, the negative charge on the oxygen. This lone pair of electrons here is going to attack this carbon, which would push these pi electrons off."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in our alkene. And so here is our alkene with unknown substituents at the moment. And we think about our mechanism, the negative charge on the oxygen. This lone pair of electrons here is going to attack this carbon, which would push these pi electrons off. And those pi electrons are actually going to go to this oxygen right here, which would push these pi electrons in here off onto this oxygen. So it's a concerted mechanism here. And so let's go ahead and draw the results of those electrons moving."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This lone pair of electrons here is going to attack this carbon, which would push these pi electrons off. And those pi electrons are actually going to go to this oxygen right here, which would push these pi electrons in here off onto this oxygen. So it's a concerted mechanism here. And so let's go ahead and draw the results of those electrons moving. So the oxygen on the left is now bonded to the carbon on the left. The carbon on the left now has a single bond to the carbon on the right. The carbon on the right is now bonded to this oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and draw the results of those electrons moving. So the oxygen on the left is now bonded to the carbon on the left. The carbon on the left now has a single bond to the carbon on the right. The carbon on the right is now bonded to this oxygen. And then these two oxygens are bonded to an oxygen in the center. For lone pairs of electrons, all of our oxygens are going to have two lone pairs of electrons like that. And so we had so many electrons moving, let's see if we can follow them."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the right is now bonded to this oxygen. And then these two oxygens are bonded to an oxygen in the center. For lone pairs of electrons, all of our oxygens are going to have two lone pairs of electrons like that. And so we had so many electrons moving, let's see if we can follow them. So let's color coordinate some electrons here. So I'm going to say that these electrons in blue, those are the ones that formed this bond between the oxygen and the carbon. And I'm going to say that these pi electrons here in red, those are the ones that formed this bond between this carbon and this oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we had so many electrons moving, let's see if we can follow them. So let's color coordinate some electrons here. So I'm going to say that these electrons in blue, those are the ones that formed this bond between the oxygen and the carbon. And I'm going to say that these pi electrons here in red, those are the ones that formed this bond between this carbon and this oxygen. And then finally, these electrons, I'm saying these are my pi electrons in here, are going to move off onto this top oxygen. So you could say that those magenta electrons would be right there. And so now we have this structure."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to say that these pi electrons here in red, those are the ones that formed this bond between this carbon and this oxygen. And then finally, these electrons, I'm saying these are my pi electrons in here, are going to move off onto this top oxygen. So you could say that those magenta electrons would be right there. And so now we have this structure. Now, oxygen-oxygen bonds are relatively weak. So they're unstable. And so we have two oxygen-oxygen bonds in this molecule."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so now we have this structure. Now, oxygen-oxygen bonds are relatively weak. So they're unstable. And so we have two oxygen-oxygen bonds in this molecule. And so one of those oxygen-oxygen bonds is going to break in the next step of the mechanism. And so I could pick either one since they're symmetrical. I'm just going to say that these electrons over here, I'm going to say that this oxygen-oxygen bond is going to break in the next part of our mechanism."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we have two oxygen-oxygen bonds in this molecule. And so one of those oxygen-oxygen bonds is going to break in the next step of the mechanism. And so I could pick either one since they're symmetrical. I'm just going to say that these electrons over here, I'm going to say that this oxygen-oxygen bond is going to break in the next part of our mechanism. And so if we think about these electrons and this oxygen moving in here, that would break this bond between the two carbons because there'd be too many bonds to the carbon on the right. So this bond is going to break, push those electrons into here. And then these electrons in green are going to come off onto the top oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm just going to say that these electrons over here, I'm going to say that this oxygen-oxygen bond is going to break in the next part of our mechanism. And so if we think about these electrons and this oxygen moving in here, that would break this bond between the two carbons because there'd be too many bonds to the carbon on the right. So this bond is going to break, push those electrons into here. And then these electrons in green are going to come off onto the top oxygen. So our unstable oxygen-oxygen bond breaks. And so let's go ahead and draw what we would get. Well, on the left side, we have a carbon."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons in green are going to come off onto the top oxygen. So our unstable oxygen-oxygen bond breaks. And so let's go ahead and draw what we would get. Well, on the left side, we have a carbon. And now that carbon is going to have two bonds to the oxygen. And the oxygen is going to have two lone pairs of electrons. So make a carbonyl compound."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, on the left side, we have a carbon. And now that carbon is going to have two bonds to the oxygen. And the oxygen is going to have two lone pairs of electrons. So make a carbonyl compound. On the right, the carbon on the right is bonded to two other things. And it now has a double bond to this oxygen here. And now this oxygen has only one lone pair of electrons on it."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So make a carbonyl compound. On the right, the carbon on the right is bonded to two other things. And it now has a double bond to this oxygen here. And now this oxygen has only one lone pair of electrons on it. And then this oxygen is bonded to the other oxygen. And that oxygen now has three lone pairs of electrons, which would give it a negative 1 formal charge. So we have a negative 1 formal charge here."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now this oxygen has only one lone pair of electrons on it. And then this oxygen is bonded to the other oxygen. And that oxygen now has three lone pairs of electrons, which would give it a negative 1 formal charge. So we have a negative 1 formal charge here. This oxygen right here gets a positive 1 formal charge. And we form a carbonyl oxide on the right. So this charged compound on the right is called a carbonyl oxide."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have a negative 1 formal charge here. This oxygen right here gets a positive 1 formal charge. And we form a carbonyl oxide on the right. So this charged compound on the right is called a carbonyl oxide. So in the next step of the mechanism, if we think about the carbonyl compound on the left, so this carbonyl compound on the left here, the double bond between oxygen and carbon, oxygen is more electronegative than carbon. So oxygen is going to pull those electrons closer to it to give the oxygen a partial negative charge. Carbon is going to lose some electron density."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this charged compound on the right is called a carbonyl oxide. So in the next step of the mechanism, if we think about the carbonyl compound on the left, so this carbonyl compound on the left here, the double bond between oxygen and carbon, oxygen is more electronegative than carbon. So oxygen is going to pull those electrons closer to it to give the oxygen a partial negative charge. Carbon is going to lose some electron density. So this carbon is going to have a partial positive charge. And you can see that you have this partially negatively charged oxygen right next to a negatively charged oxygen on the carbonyl oxide. And so those, of course, are going to repel."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Carbon is going to lose some electron density. So this carbon is going to have a partial positive charge. And you can see that you have this partially negatively charged oxygen right next to a negatively charged oxygen on the carbonyl oxide. And so those, of course, are going to repel. So we're going to flip over the carbonyl compound on the left. So we're going to flip over this compound on the left. And I'm going to draw the carbonyl oxide on the right the same way."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so those, of course, are going to repel. So we're going to flip over the carbonyl compound on the left. So we're going to flip over this compound on the left. And I'm going to draw the carbonyl oxide on the right the same way. So as we continue in our mechanism here, we're going to flip over the carbonyl compound on the left. And we're going to keep the carbonyl oxide the same, the same orientation like that. And so let's go ahead and put in all of our lone pairs of electrons and formal charges."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to draw the carbonyl oxide on the right the same way. So as we continue in our mechanism here, we're going to flip over the carbonyl compound on the left. And we're going to keep the carbonyl oxide the same, the same orientation like that. And so let's go ahead and put in all of our lone pairs of electrons and formal charges. So negative formal charge here, positive formal charge here, and two lone pairs of electrons on this oxygen. So let's get a little bit more room here. And once again, we can think about our carbonyl compound being polarized."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and put in all of our lone pairs of electrons and formal charges. So negative formal charge here, positive formal charge here, and two lone pairs of electrons on this oxygen. So let's get a little bit more room here. And once again, we can think about our carbonyl compound being polarized. The oxygen is partially negative. And this carbon right here is partially positive. And so in the next step of the mechanism, we're going to get a nucleophilic attack."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we can think about our carbonyl compound being polarized. The oxygen is partially negative. And this carbon right here is partially positive. And so in the next step of the mechanism, we're going to get a nucleophilic attack. We're going to get this negatively charged oxygen is going to attack this partially positive carbon right here. And if the oxygen forms a bond with this carbon, that would be too many bonds to this carbon. And so this lone pair of electrons, these pi electrons in here, are going to go after this carbon, which would push these electrons in here off onto that oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so in the next step of the mechanism, we're going to get a nucleophilic attack. We're going to get this negatively charged oxygen is going to attack this partially positive carbon right here. And if the oxygen forms a bond with this carbon, that would be too many bonds to this carbon. And so this lone pair of electrons, these pi electrons in here, are going to go after this carbon, which would push these electrons in here off onto that oxygen. So once again, we have a lot of electrons moving. So let's see if we can go ahead and draw the product and see if we can follow those electrons a little bit. So now we're going to have this carbon over here is bonded to an oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so this lone pair of electrons, these pi electrons in here, are going to go after this carbon, which would push these electrons in here off onto that oxygen. So once again, we have a lot of electrons moving. So let's see if we can go ahead and draw the product and see if we can follow those electrons a little bit. So now we're going to have this carbon over here is bonded to an oxygen. This oxygen is now bonded to what used to be our carbonyl oxide carbon. This carbon is bonded to two other things. It's also bonded to an oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now we're going to have this carbon over here is bonded to an oxygen. This oxygen is now bonded to what used to be our carbonyl oxide carbon. This carbon is bonded to two other things. It's also bonded to an oxygen. And that oxygen is bonded to an oxygen, which is now bonded to this carbon. So let me go ahead and put in our lone pairs of electrons, each of our oxygens is going to have two lone pairs of electrons like that. And once again, let's see if we can follow those electrons, because there was a lot that was happening."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's also bonded to an oxygen. And that oxygen is bonded to an oxygen, which is now bonded to this carbon. So let me go ahead and put in our lone pairs of electrons, each of our oxygens is going to have two lone pairs of electrons like that. And once again, let's see if we can follow those electrons, because there was a lot that was happening. So let's go ahead and follow these electrons right here in blue, so the ones on the negatively charged oxygen. Those are the ones that are going to nucleophilic attack that carbonyl carbon right here and form this bond. So this bond in blue represents those blue electrons."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And once again, let's see if we can follow those electrons, because there was a lot that was happening. So let's go ahead and follow these electrons right here in blue, so the ones on the negatively charged oxygen. Those are the ones that are going to nucleophilic attack that carbonyl carbon right here and form this bond. So this bond in blue represents those blue electrons. If we follow these pi electrons here, so these pi electrons in that carbonyl in red, those are going to form a bond between this oxygen and this carbon like that. And then finally, these pi electrons in here on the carbonyl oxide are going to move off onto that oxygen like that. And so we have this compound now."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this bond in blue represents those blue electrons. If we follow these pi electrons here, so these pi electrons in that carbonyl in red, those are going to form a bond between this oxygen and this carbon like that. And then finally, these pi electrons in here on the carbonyl oxide are going to move off onto that oxygen like that. And so we have this compound now. So that's all for the first step. That's adding ozone to our alkene. And now we can go ahead and add our dimethyl sulfide."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we have this compound now. So that's all for the first step. That's adding ozone to our alkene. And now we can go ahead and add our dimethyl sulfide. So dimethyl sulfide comes along. Let's go ahead and draw that in here. So we'd have a sulfur."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now we can go ahead and add our dimethyl sulfide. So dimethyl sulfide comes along. Let's go ahead and draw that in here. So we'd have a sulfur. And then that sulfur is bonded to two methyl groups. And the sulfur has two lone pairs of electrons like that. And so the one of those lone pairs on the sulfur is going to attack this oxygen right here."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we'd have a sulfur. And then that sulfur is bonded to two methyl groups. And the sulfur has two lone pairs of electrons like that. And so the one of those lone pairs on the sulfur is going to attack this oxygen right here. So one lone pair is going to attack this oxygen. And once again, we have a weak oxygen-oxygen bond. So let me see if I can highlight that weak oxygen-oxygen bond."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so the one of those lone pairs on the sulfur is going to attack this oxygen right here. So one lone pair is going to attack this oxygen. And once again, we have a weak oxygen-oxygen bond. So let me see if I can highlight that weak oxygen-oxygen bond. So right here is our weak oxygen-oxygen bond, which is going to break. And those electrons are going to move into here. And that would be too many bonds to this carbon."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me see if I can highlight that weak oxygen-oxygen bond. So right here is our weak oxygen-oxygen bond, which is going to break. And those electrons are going to move into here. And that would be too many bonds to this carbon. So these electrons in red are going to move into here. And then finally, the electrons in blue are going to come back off onto that top oxygen. So let's go ahead and draw that."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that would be too many bonds to this carbon. So these electrons in red are going to move into here. And then finally, the electrons in blue are going to come back off onto that top oxygen. So let's go ahead and draw that. So what happens now? Now we would have a carbon over here on the left. It's now double bonded to this oxygen."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that. So what happens now? Now we would have a carbon over here on the left. It's now double bonded to this oxygen. And this oxygen gets two lone pairs of electrons. The carbon over here on the right is now doubly bonded to an oxygen like that. And then this carbon is also bonded to two other things."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's now double bonded to this oxygen. And this oxygen gets two lone pairs of electrons. The carbon over here on the right is now doubly bonded to an oxygen like that. And then this carbon is also bonded to two other things. And the sulfur is now bonded to this oxygen here. The sulfur is bonded to this oxygen. This oxygen has three lone pairs of electrons, which gives it a negative 1 formal charge."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then this carbon is also bonded to two other things. And the sulfur is now bonded to this oxygen here. The sulfur is bonded to this oxygen. This oxygen has three lone pairs of electrons, which gives it a negative 1 formal charge. The sulfur is bonded to two methyl groups. And it still has a lone pair of electrons on it, because it had two lone pairs to start with, which gives the sulfur a plus 1 formal charge like that. And so if you just focus in on the sulfur compound up here, you could draw a resonance structure."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen has three lone pairs of electrons, which gives it a negative 1 formal charge. The sulfur is bonded to two methyl groups. And it still has a lone pair of electrons on it, because it had two lone pairs to start with, which gives the sulfur a plus 1 formal charge like that. And so if you just focus in on the sulfur compound up here, you could draw a resonance structure. You could move these electrons in here. And so if we drew a resonance structure for that, we would now have sulfur doubly bonded to an oxygen like that with two methyl groups and still having a lone pair of electrons on it. So we can now recognize it as DMSO or dimethyl sulfoxide."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so if you just focus in on the sulfur compound up here, you could draw a resonance structure. You could move these electrons in here. And so if we drew a resonance structure for that, we would now have sulfur doubly bonded to an oxygen like that with two methyl groups and still having a lone pair of electrons on it. So we can now recognize it as DMSO or dimethyl sulfoxide. And so we produce DMSO. But more importantly, we produce these two carbonyls down here. Once again, aldehydes or ketones, depending on what is attached."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we can now recognize it as DMSO or dimethyl sulfoxide. And so we produce DMSO. But more importantly, we produce these two carbonyls down here. Once again, aldehydes or ketones, depending on what is attached. So we produce these aldehydes or ketones here. And so if we follow some of our electrons, so these electrons in red over here on the left, those moved in here on our carbonyl compound. And the electrons in green over here on the left, those moved in here to form this carbonyl compound."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Once again, aldehydes or ketones, depending on what is attached. So we produce these aldehydes or ketones here. And so if we follow some of our electrons, so these electrons in red over here on the left, those moved in here on our carbonyl compound. And the electrons in green over here on the left, those moved in here to form this carbonyl compound. So after an extremely long mechanism, we finally get our mixture of aldehydes or ketones. Let's go ahead and do one example to make sure that you can apply this reaction. Let's look at an alkene."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons in green over here on the left, those moved in here to form this carbonyl compound. So after an extremely long mechanism, we finally get our mixture of aldehydes or ketones. Let's go ahead and do one example to make sure that you can apply this reaction. Let's look at an alkene. So here is our alkene. And to our alkene, in the first step, we're going to add ozone, so O3. And the second step, we're going to add dimethyl sulfide, so DMS."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an alkene. So here is our alkene. And to our alkene, in the first step, we're going to add ozone, so O3. And the second step, we're going to add dimethyl sulfide, so DMS. So the way to think about this reaction, or an easy way to think about it, let me go ahead and redraw it here. We know that we're going to cleave the double bond. We're going to break the double bond here."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the second step, we're going to add dimethyl sulfide, so DMS. So the way to think about this reaction, or an easy way to think about it, let me go ahead and redraw it here. We know that we're going to cleave the double bond. We're going to break the double bond here. We know that our double bond, this is a methyl group. And over here is a hydrogen. We didn't draw it in over here on the left side, but we know it's there."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to break the double bond here. We know that our double bond, this is a methyl group. And over here is a hydrogen. We didn't draw it in over here on the left side, but we know it's there. We're going to cleave this double bond. We're going to cleave it right here. And so one way to do this would be just to erase that part."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We didn't draw it in over here on the left side, but we know it's there. We're going to cleave this double bond. We're going to cleave it right here. And so one way to do this would be just to erase that part. So if we think about erasing this part right here and adding in an oxygen to either one of those carbons here, so we're going to add in an oxygen right here and add in an oxygen down here. And so it's a little bit hard to see, so let me go ahead and redraw it. Let me go ahead and redraw it here."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so one way to do this would be just to erase that part. So if we think about erasing this part right here and adding in an oxygen to either one of those carbons here, so we're going to add in an oxygen right here and add in an oxygen down here. And so it's a little bit hard to see, so let me go ahead and redraw it. Let me go ahead and redraw it here. So down at the bottom, we now have a carbonyl, and that's a ketone. And up here at the top, we're going to have an aldehyde. And so this is your product."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and redraw it here. So down at the bottom, we now have a carbonyl, and that's a ketone. And up here at the top, we're going to have an aldehyde. And so this is your product. Now, this is probably not the easiest way to draw it, so let's go ahead and redraw it in a different way here. So if we're going to draw it as a straight chain, let's see how many carbons we would have to deal with. So we have 1, 2, 3, 4, 5, 6 carbons."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so this is your product. Now, this is probably not the easiest way to draw it, so let's go ahead and redraw it in a different way here. So if we're going to draw it as a straight chain, let's see how many carbons we would have to deal with. So we have 1, 2, 3, 4, 5, 6 carbons. Let's go ahead and straighten out that molecule a little bit. So we would have a 1, and then we have a ketone here. So that's two carbons."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have 1, 2, 3, 4, 5, 6 carbons. Let's go ahead and straighten out that molecule a little bit. So we would have a 1, and then we have a ketone here. So that's two carbons. 3, 4, 5, 6 carbons, and then we have our aldehyde right here. And so there's our aldehyde, which would be this carbon right here. So a total of seven carbons."}, {"video_title": "Ozonolysis Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's two carbons. 3, 4, 5, 6 carbons, and then we have our aldehyde right here. And so there's our aldehyde, which would be this carbon right here. So a total of seven carbons. So 1, 2, 3, 4, 5, 6, 7. So this reaction produced an aldehyde and a ketone. So once again, be careful with the second step, and pay attention to what your professor wants to use."}, {"video_title": "Friedel crafts acylation addendum Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Its electron was allowed to go back into the benzene ring. So the aluminum lost a chlorine. This benzene ring lost a proton. It was able to take back that electron. And I forgot to draw the hydrogen and the chlorine. So then that results in hydrogen chloride. So you have the hydrogen proton bonded."}, {"video_title": "Friedel crafts acylation addendum Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It was able to take back that electron. And I forgot to draw the hydrogen and the chlorine. So then that results in hydrogen chloride. So you have the hydrogen proton bonded. It took this electron from the aluminum, and it is bonded to the chlorine. I didn't want to forget that in that last step of the Friedel-Crafts acylation. We also get some hydrogen chloride."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If we try the same reaction with benzene, we're not going to get anything for our products. So there's no reaction. And so benzene is more stable than cyclohexene. At first, you might think that the stability is due to the fact that benzene is conjugated. But numerous other experiments have shown that it is even more stable than we would expect, and that extra stability is called aromaticity, or aromatic stabilization. And so benzene is an aromatic molecule. Let's look at the criteria to determine if a compound is aromatic."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "At first, you might think that the stability is due to the fact that benzene is conjugated. But numerous other experiments have shown that it is even more stable than we would expect, and that extra stability is called aromaticity, or aromatic stabilization. And so benzene is an aromatic molecule. Let's look at the criteria to determine if a compound is aromatic. So a compound is aromatic if it contains a ring of continuously overlapping p orbitals. And so if the molecule is planar, that's what allows the p orbitals to overlap. It also has to have 4n plus 2 pi electrons in the ring, where n is equal to 0, 1, 2, or any other positive integer."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the criteria to determine if a compound is aromatic. So a compound is aromatic if it contains a ring of continuously overlapping p orbitals. And so if the molecule is planar, that's what allows the p orbitals to overlap. It also has to have 4n plus 2 pi electrons in the ring, where n is equal to 0, 1, 2, or any other positive integer. And this is called Huckel's rule. So let's go ahead and analyze benzene in a little bit more detail. So if I look at the dot structure, I can see that benzene has 2 pi electrons there, 2 here, and 2 more here, for a total of 6 pi electrons."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It also has to have 4n plus 2 pi electrons in the ring, where n is equal to 0, 1, 2, or any other positive integer. And this is called Huckel's rule. So let's go ahead and analyze benzene in a little bit more detail. So if I look at the dot structure, I can see that benzene has 2 pi electrons there, 2 here, and 2 more here, for a total of 6 pi electrons. If I look at the carbons of benzene, I can see that each carbon has a double bond to it. So each carbon is sp2 hybridized. And if each carbon is sp2 hybridized, that means that each carbon has a free p orbital."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at the dot structure, I can see that benzene has 2 pi electrons there, 2 here, and 2 more here, for a total of 6 pi electrons. If I look at the carbons of benzene, I can see that each carbon has a double bond to it. So each carbon is sp2 hybridized. And if each carbon is sp2 hybridized, that means that each carbon has a free p orbital. So I'm going to go ahead and sketch in the unhybridized free p orbital on each of the 6 carbons of benzene. Now, since benzene is a planar molecule, that's going to allow those p orbitals to overlap side by side. So you get some overlap side by side of those p orbitals."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if each carbon is sp2 hybridized, that means that each carbon has a free p orbital. So I'm going to go ahead and sketch in the unhybridized free p orbital on each of the 6 carbons of benzene. Now, since benzene is a planar molecule, that's going to allow those p orbitals to overlap side by side. So you get some overlap side by side of those p orbitals. And so benzene contains a ring of continuously overlapping p orbitals. So p orbitals are considered to be atomic orbitals. And so there are a total of 6 atomic orbitals in benzene."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you get some overlap side by side of those p orbitals. And so benzene contains a ring of continuously overlapping p orbitals. So p orbitals are considered to be atomic orbitals. And so there are a total of 6 atomic orbitals in benzene. According to MO theory, those 6 atomic orbitals are going to cease to exist. And we will get 6 molecular orbitals instead. And so benzene has 6 molecular orbitals."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so there are a total of 6 atomic orbitals in benzene. According to MO theory, those 6 atomic orbitals are going to cease to exist. And we will get 6 molecular orbitals instead. And so benzene has 6 molecular orbitals. Drawing out these molecular orbitals would be a little bit too complicated for this video. So check out your textbook for some nice diagrams of the 6 molecular orbitals of benzene. However, it is important to understand those 6 molecular orbitals in terms of their relative energy levels."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so benzene has 6 molecular orbitals. Drawing out these molecular orbitals would be a little bit too complicated for this video. So check out your textbook for some nice diagrams of the 6 molecular orbitals of benzene. However, it is important to understand those 6 molecular orbitals in terms of their relative energy levels. And the simplest way to do that is to draw a frost circle. And so here I have a circle already drawn. And inside the circle, we're going to inscribe a polygon."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "However, it is important to understand those 6 molecular orbitals in terms of their relative energy levels. And the simplest way to do that is to draw a frost circle. And so here I have a circle already drawn. And inside the circle, we're going to inscribe a polygon. And since benzene is a 6-membered ring, we're going to inscribe a hexagon in our frost circle. I'm going to go ahead and draw a center line through the circle just to help out with the drawing here. And when you're inscribing your polygon in your frost circle, you always start at the bottom."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And inside the circle, we're going to inscribe a polygon. And since benzene is a 6-membered ring, we're going to inscribe a hexagon in our frost circle. I'm going to go ahead and draw a center line through the circle just to help out with the drawing here. And when you're inscribing your polygon in your frost circle, you always start at the bottom. So we're going to start down here. So we're going to inscribe a hexagon. So let's see if we can put a hexagon in here."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And when you're inscribing your polygon in your frost circle, you always start at the bottom. So we're going to start down here. So we're going to inscribe a hexagon. So let's see if we can put a hexagon in here. So we have a 6-sided figure in our frost circle. The key point about a frost circle is everywhere your polygon intersects with your circle, that represents the energy level of a molecular orbital. And so this intersection right here, this intersection here, and then all the way around."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's see if we can put a hexagon in here. So we have a 6-sided figure in our frost circle. The key point about a frost circle is everywhere your polygon intersects with your circle, that represents the energy level of a molecular orbital. And so this intersection right here, this intersection here, and then all the way around. And so we have our 6 molecular orbitals. And we have the relative energy levels of those 6 molecular orbitals. So let me go ahead and draw them over here."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this intersection right here, this intersection here, and then all the way around. And so we have our 6 molecular orbitals. And we have the relative energy levels of those 6 molecular orbitals. So let me go ahead and draw them over here. So we have 3 molecular orbitals which are above the center line. And those are higher in energy. And we know that those are called anti-bonding molecular orbitals."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw them over here. So we have 3 molecular orbitals which are above the center line. And those are higher in energy. And we know that those are called anti-bonding molecular orbitals. So these are anti-bonding molecular orbitals, which are the highest in energy. But look down here. There are 3 molecular orbitals which are below the center line."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we know that those are called anti-bonding molecular orbitals. So these are anti-bonding molecular orbitals, which are the highest in energy. But look down here. There are 3 molecular orbitals which are below the center line. And those are our bonding molecular orbitals. So those are lower in energy. And if we had some molecular orbitals that were on the center line, those would be non-bonding molecular orbitals."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "There are 3 molecular orbitals which are below the center line. And those are our bonding molecular orbitals. So those are lower in energy. And if we had some molecular orbitals that were on the center line, those would be non-bonding molecular orbitals. We're going to go ahead and fill our molecular orbitals with our pi electrons. So if I go back over here, remember that benzene has 6 pi electrons. And so filling molecular orbitals is analogous to electron configurations."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if we had some molecular orbitals that were on the center line, those would be non-bonding molecular orbitals. We're going to go ahead and fill our molecular orbitals with our pi electrons. So if I go back over here, remember that benzene has 6 pi electrons. And so filling molecular orbitals is analogous to electron configurations. You're going to fill the lowest molecular orbital first. And each orbital can hold 2 electrons, like electron configurations. And so we're going to go ahead and put 2 electrons into the lowest bonding molecular orbital."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so filling molecular orbitals is analogous to electron configurations. You're going to fill the lowest molecular orbital first. And each orbital can hold 2 electrons, like electron configurations. And so we're going to go ahead and put 2 electrons into the lowest bonding molecular orbital. So I have 4 more pi electrons to worry about. So 4 more pi electrons. And I go ahead and put those in."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to go ahead and put 2 electrons into the lowest bonding molecular orbital. So I have 4 more pi electrons to worry about. So 4 more pi electrons. And I go ahead and put those in. And I have filled the bonding molecular orbitals of benzene. So I have represented all 6 pi electrons. If I think about Huckel's rule, 4n plus 2, I have 6 pi electrons."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I go ahead and put those in. And I have filled the bonding molecular orbitals of benzene. So I have represented all 6 pi electrons. If I think about Huckel's rule, 4n plus 2, I have 6 pi electrons. So if n is equal to 1, Huckel's rule is satisfied. Because I would do 4 times 1 plus 2. And so I would get a total of 6 pi electrons."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If I think about Huckel's rule, 4n plus 2, I have 6 pi electrons. So if n is equal to 1, Huckel's rule is satisfied. Because I would do 4 times 1 plus 2. And so I would get a total of 6 pi electrons. And so 6 pi electrons follows Huckel's rule. If we look at the frost circle and we look at the molecular orbitals, we can understand Huckel's rule a little bit better visually. So if I think about these 2 electrons down here, you can think about that's where the 2 comes from in Huckel's rule."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I would get a total of 6 pi electrons. And so 6 pi electrons follows Huckel's rule. If we look at the frost circle and we look at the molecular orbitals, we can understand Huckel's rule a little bit better visually. So if I think about these 2 electrons down here, you can think about that's where the 2 comes from in Huckel's rule. And if I think about these 4 electrons up here, that would be 4 electrons times our positive integer of 1. So 4 times 1 plus 2 gives us 6 pi electrons. And we have filled the bonding molecular orbitals of benzene, which confers the extra stability that we call aromaticity or aromatic stabilization."}, {"video_title": "Aromatic stability I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I think about these 2 electrons down here, you can think about that's where the 2 comes from in Huckel's rule. And if I think about these 4 electrons up here, that would be 4 electrons times our positive integer of 1. So 4 times 1 plus 2 gives us 6 pi electrons. And we have filled the bonding molecular orbitals of benzene, which confers the extra stability that we call aromaticity or aromatic stabilization. And so benzene is aromatic. It follows our different criteria. In the next few videos, we're going to look at several other examples of aromatic compounds and ions."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that I had a ketone that looked like this. That's my, draw my carbonyl group, just like that. And then it is bonded to a carbon that is bonded to two other CH3 groups. And just to make it clear, I mean there's three hydrogens off of this carbon there implicitly, but I'm going to draw the fourth bond here, which is to a hydrogen, because this hydrogen is going to be important for this reaction. Now we know that the oxygen has two lone pairs of electrons. Let me draw it up here. And let's just imagine it's floating around in some water, and we know that in water there is some concentration of hydronium."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And just to make it clear, I mean there's three hydrogens off of this carbon there implicitly, but I'm going to draw the fourth bond here, which is to a hydrogen, because this hydrogen is going to be important for this reaction. Now we know that the oxygen has two lone pairs of electrons. Let me draw it up here. And let's just imagine it's floating around in some water, and we know that in water there is some concentration of hydronium. And let's say that one of the hydroniums is right over here. So let's see, hydronium is just a positively charged, so this right here, let me make it clear, this, let me do it a different color, this is what water looks like. And if water gives away an electron to a proton, it looks like this."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's just imagine it's floating around in some water, and we know that in water there is some concentration of hydronium. And let's say that one of the hydroniums is right over here. So let's see, hydronium is just a positively charged, so this right here, let me make it clear, this, let me do it a different color, this is what water looks like. And if water gives away an electron to a proton, it looks like this. It is hydronium, and it only has one lone pair of electrons. It gave away one of the other electrons in its other lone pair to a proton. So you can imagine a reality where it's like, hey, I could grab that proton from this hydronium, and then this will turn back into water."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if water gives away an electron to a proton, it looks like this. It is hydronium, and it only has one lone pair of electrons. It gave away one of the other electrons in its other lone pair to a proton. So you can imagine a reality where it's like, hey, I could grab that proton from this hydronium, and then this will turn back into water. And in that situation, the mechanism would look like this. This electron, let me do it in a different color, this blue electron gets given to this proton, if they just bump into each other just right, and then the hydrogen's electron gets taken back by what will become a water molecule. So if that happens, what do our molecules now look like?"}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So you can imagine a reality where it's like, hey, I could grab that proton from this hydronium, and then this will turn back into water. And in that situation, the mechanism would look like this. This electron, let me do it in a different color, this blue electron gets given to this proton, if they just bump into each other just right, and then the hydrogen's electron gets taken back by what will become a water molecule. So if that happens, what do our molecules now look like? So now our, what was a ketone, looks a little bit different than a ketone. So it looks like this. I changed it to a slightly lighter color of green."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if that happens, what do our molecules now look like? So now our, what was a ketone, looks a little bit different than a ketone. So it looks like this. I changed it to a slightly lighter color of green. So it looks like that. We have our lone pair over here, but we no longer have this lone pair. At this end, we still have this magenta electron, but now it is in a covalent bond with the blue electron, which was now given to the hydrogen proton."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I changed it to a slightly lighter color of green. So it looks like that. We have our lone pair over here, but we no longer have this lone pair. At this end, we still have this magenta electron, but now it is in a covalent bond with the blue electron, which was now given to the hydrogen proton. Let me scroll up a little bit. It was given to this hydrogen proton up here. It was given to that hydrogen proton up here, and then this hydronium molecule, it took back an electron, and now it is just neutral water."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "At this end, we still have this magenta electron, but now it is in a covalent bond with the blue electron, which was now given to the hydrogen proton. Let me scroll up a little bit. It was given to this hydrogen proton up here. It was given to that hydrogen proton up here, and then this hydronium molecule, it took back an electron, and now it is just neutral water. It is now just neutral water. It took back that magenta electron, so now it has two lone pairs again. So it is just neutral water."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It was given to that hydrogen proton up here, and then this hydronium molecule, it took back an electron, and now it is just neutral water. It is now just neutral water. It took back that magenta electron, so now it has two lone pairs again. So it is just neutral water. Now this, since this oxygen up here in the carbonyl group gave away an electron, it now has a positive charge. But this is actually resonance stabilized. You could maybe see that this would be in resonance, or another resonance form of this would be if this guy is positive, so he wants to gain an electron."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it is just neutral water. Now this, since this oxygen up here in the carbonyl group gave away an electron, it now has a positive charge. But this is actually resonance stabilized. You could maybe see that this would be in resonance, or another resonance form of this would be if this guy is positive, so he wants to gain an electron. Maybe he takes an electron from this carbon, the carbon in the carbonyl group right over there. If he takes that electron, then the other resonance form would look like this. Let me do it in the same colors."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You could maybe see that this would be in resonance, or another resonance form of this would be if this guy is positive, so he wants to gain an electron. Maybe he takes an electron from this carbon, the carbon in the carbonyl group right over there. If he takes that electron, then the other resonance form would look like this. Let me do it in the same colors. You have now only a single bond with this oxygen up here. This carbon down here is still bonded to the same carbons, and then this carbon over here. We could call this an alpha carbon."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me do it in the same colors. You have now only a single bond with this oxygen up here. This carbon down here is still bonded to the same carbons, and then this carbon over here. We could call this an alpha carbon. This is an alpha carbon to the carbonyl group. It still has a hydrogen on it right over there. This oxygen, since it gained this magenta electron, so now it has two lone pairs again."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could call this an alpha carbon. This is an alpha carbon to the carbonyl group. It still has a hydrogen on it right over there. This oxygen, since it gained this magenta electron, so now it has two lone pairs again. It has this pair over there, and then it gained this electron and this electron, so it has another lone pair. Of course, it has the bond to the hydrogen. Since it gained an electron, it is now neutral."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen, since it gained this magenta electron, so now it has two lone pairs again. It has this pair over there, and then it gained this electron and this electron, so it has another lone pair. Of course, it has the bond to the hydrogen. Since it gained an electron, it is now neutral. This carbon lost an electron, so now it is positive. Now this carbon right over here is positive. These two are two different resonance forms, so they help stabilize each other."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Since it gained an electron, it is now neutral. This carbon lost an electron, so now it is positive. Now this carbon right over here is positive. These two are two different resonance forms, so they help stabilize each other. The reality is actually someplace in between. I could actually draw it in brackets to show that these are two resonance structures. Now you can imagine, just as likely, and actually I shouldn't just draw this as a one-way arrow because this guy could take a hydrogen from this hydronium, or a water could take a hydrogen from this guy."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These two are two different resonance forms, so they help stabilize each other. The reality is actually someplace in between. I could actually draw it in brackets to show that these are two resonance structures. Now you can imagine, just as likely, and actually I shouldn't just draw this as a one-way arrow because this guy could take a hydrogen from this hydronium, or a water could take a hydrogen from this guy. This actually could go in both directions. Let me make that clear. This could go in both directions."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now you can imagine, just as likely, and actually I shouldn't just draw this as a one-way arrow because this guy could take a hydrogen from this hydronium, or a water could take a hydrogen from this guy. This actually could go in both directions. Let me make that clear. This could go in both directions. You could say that they are in equilibrium with each other. You are just as likely to go in that direction as you really, for the most part, are to go in the other direction. But you can now imagine, okay, this is now turned in from a carbonyl group."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This could go in both directions. You could say that they are in equilibrium with each other. You are just as likely to go in that direction as you really, for the most part, are to go in the other direction. But you can now imagine, okay, this is now turned in from a carbonyl group. This is now an OH group. This is now turned into an alcohol, although we have this carbocation here. This does not like being positive."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But you can now imagine, okay, this is now turned in from a carbonyl group. This is now an OH group. This is now turned into an alcohol, although we have this carbocation here. This does not like being positive. You can imagine where this electron right here, being attracted on this hydrogen nucleus, might want to go really bad to this carbocation. It just needs something to nab the proton off for it to go there. The perfect candidate for that would just be a water molecule."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This does not like being positive. You can imagine where this electron right here, being attracted on this hydrogen nucleus, might want to go really bad to this carbocation. It just needs something to nab the proton off for it to go there. The perfect candidate for that would just be a water molecule. We have this water floating around, so I can draw another water molecule just like this. It has two lone pairs. It can act as a weak base."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The perfect candidate for that would just be a water molecule. We have this water floating around, so I can draw another water molecule just like this. It has two lone pairs. It can act as a weak base. It can give one of its electrons to this hydrogen proton. If it does that at the exact same time, bumps into it in the exact same way, this electron can then go to the carbocation. If that happened, you could go in either direction."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It can act as a weak base. It can give one of its electrons to this hydrogen proton. If it does that at the exact same time, bumps into it in the exact same way, this electron can then go to the carbocation. If that happened, you could go in either direction. This reaction is just as likely to happen as the reverse reaction. We could put this in equilibrium. If that were to happen, what started off as our ketone now looks like this."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If that happened, you could go in either direction. This reaction is just as likely to happen as the reverse reaction. We could put this in equilibrium. If that were to happen, what started off as our ketone now looks like this. We have a bond to an OH group, just like this. Over here, let me draw the rest of it. We had our molecule that looked like that."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If that were to happen, what started off as our ketone now looks like this. We have a bond to an OH group, just like this. Over here, let me draw the rest of it. We had our molecule that looked like that. This electron gets given back to this carbocation. We now have a double bond here between our carbonyl carbon, or what was a carbonyl carbon, and our alpha carbon. Now we have this double bond right over here."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We had our molecule that looked like that. This electron gets given back to this carbocation. We now have a double bond here between our carbonyl carbon, or what was a carbonyl carbon, and our alpha carbon. Now we have this double bond right over here. We have this double bond right over here. That hydrogen has been taken by the water. Now that is hydronium."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now we have this double bond right over here. We have this double bond right over here. That hydrogen has been taken by the water. Now that is hydronium. Let me draw the water, or the hydronium. That water had that one lone pair. The other lone pair got broken up because it gave one of the electrons to this hydrogen right over here, and it went back to being hydronium."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now that is hydronium. Let me draw the water, or the hydronium. That water had that one lone pair. The other lone pair got broken up because it gave one of the electrons to this hydrogen right over here, and it went back to being hydronium. What happened here? We started with a ketone. Sometimes we'll call this the keto form of the molecule."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The other lone pair got broken up because it gave one of the electrons to this hydrogen right over here, and it went back to being hydronium. What happened here? We started with a ketone. Sometimes we'll call this the keto form of the molecule. Then we ended up with something called an enol form. An enol comes from the fact that it is an alkene that is also an alcohol. We even call it an alkenol."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes we'll call this the keto form of the molecule. Then we ended up with something called an enol form. An enol comes from the fact that it is an alkene that is also an alcohol. We even call it an alkenol. It has a double bond, and on one of the carbons it has a double bond. It has an OH group. The whole reason I show you this mechanism is one, just to show you a mechanism that could happen with an aldehyde or a ketone."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We even call it an alkenol. It has a double bond, and on one of the carbons it has a double bond. It has an OH group. The whole reason I show you this mechanism is one, just to show you a mechanism that could happen with an aldehyde or a ketone. This was a ketone, but if this was a hydrogen right here, this would have been occurring with an aldehyde. But even more, this is a pretty common mechanism that you'll see in organic chemistry classes. It actually has a lot of functions in biology in general."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The whole reason I show you this mechanism is one, just to show you a mechanism that could happen with an aldehyde or a ketone. This was a ketone, but if this was a hydrogen right here, this would have been occurring with an aldehyde. But even more, this is a pretty common mechanism that you'll see in organic chemistry classes. It actually has a lot of functions in biology in general. These two molecules, this ketone and this enol form, these are called tautomers. These are called tautomers. The keto form is actually the much more stable form."}, {"video_title": "Keto-enol tautomerization (by Sal) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It actually has a lot of functions in biology in general. These two molecules, this ketone and this enol form, these are called tautomers. These are called tautomers. The keto form is actually the much more stable form. In a solution, you won't see much of the enol form, but these can occur. It can spontaneously, through equilibrium, get to the actual enol form. You could imagine these are tautomers, so this mechanism is actually called a tautomerization."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is an epoxide right here, which is where you have oxygen in a three-membered ring with those two carbons there. You can open up this ring using either acid or base catalyzed. And we're going to talk about acid catalyzed reaction in this video. And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you can assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right, let's look at the mechanism to form our epoxide."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And what ends up happening is you get two OH groups that add on anti, so anti to each other across from your double bond. So the net result is you end up oxidizing your alkene. So you can assign some oxidation numbers on an actual problem and find out that this is an oxidation reaction. All right, let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid, except it has an extra oxygen. And the bond between these two oxygen atoms is weak. So this bond is going to break in the mechanism."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at the mechanism to form our epoxide. So we start with our percarboxylic acid here, which looks a lot like a carboxylic acid, except it has an extra oxygen. And the bond between these two oxygen atoms is weak. So this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular conformation that it's in. So this hydrogen ends up being very close to this oxygen because there's a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this bond is going to break in the mechanism. The other important thing to note about the structure of our percarboxylic acid is the particular conformation that it's in. So this hydrogen ends up being very close to this oxygen because there's a source of attraction between those atoms. There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this conformation, the mechanism will begin. This is a concerted eight-electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "There's some intramolecular hydrogen bonding that keeps it in this conformation. When the percarboxylic acid approaches the alkene, when it gets close enough in this conformation, the mechanism will begin. This is a concerted eight-electron mechanism, which means that eight electrons are going to move at the same time. So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in this bond between oxygen and hydrogen are going to move down here to form a bond with this carbon. The electrons in this pi bond here are going to move out and grab this oxygen. That's going to break this weak oxygen-oxygen bond, and those electrons move into here. And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight-electron mechanism. So of course, at the bottom here, we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, the electrons in this pi bond are going to move to here to form an actual bond between that oxygen and that hydrogen. So let's see if we can draw the results of this concerted eight-electron mechanism. So of course, at the bottom here, we're going to form our epoxide. So we draw in our carbons, and then we can put in our oxygen here. And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we draw in our carbons, and then we can put in our oxygen here. And then we show the bond between those like that. And then up at the top here, here's my carbonyl carbon. So now there's only one bond between that carbon and this oxygen. There was a new bond that formed between that oxygen and that hydrogen. And there was our R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now there's only one bond between that carbon and this oxygen. There was a new bond that formed between that oxygen and that hydrogen. And there was our R group over here. And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then there used to be only one bond to this oxygen, but another lone pair of electrons moved in to form a carbonyl here. So this is our other product, which you can see is a carboxylic acid. Let's color code these electrons so we can follow them a little bit better. So let's make these electrons in here. So those electrons are going to form the bond on the left side between the carbon and the oxygen like that. All right, let's follow these electrons next. So now let's look at these electrons in here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's make these electrons in here. So those electrons are going to form the bond on the left side between the carbon and the oxygen like that. All right, let's follow these electrons next. So now let's look at these electrons in here. The electrons in this pi bond, those are the ones that are going to form this side of our epoxide ring like that. And let's make our oxygen-oxygen bond blue here. So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now let's look at these electrons in here. The electrons in this pi bond, those are the ones that are going to form this side of our epoxide ring like that. And let's make our oxygen-oxygen bond blue here. So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that. And then let's go ahead and make these green right here. The electrons in this bond right here, these are the ones that moved out here to form the bond between our oxygen and our hydrogen. So our end result is to form a carboxylic acid and our epoxide."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in this bond, those are the ones that moved in here to form our carbonyl like that. And then let's go ahead and make these green right here. The electrons in this bond right here, these are the ones that moved out here to form the bond between our oxygen and our hydrogen. So our end result is to form a carboxylic acid and our epoxide. Let's look at a reaction, an actual reaction, for the formation of epoxide. And then we'll talk about how to form a diol from that. So if we start with cyclohexene, let's go ahead and draw cyclohexene in here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So our end result is to form a carboxylic acid and our epoxide. Let's look at a reaction, an actual reaction, for the formation of epoxide. And then we'll talk about how to form a diol from that. So if we start with cyclohexene, let's go ahead and draw cyclohexene in here. Let's do another one. That one wasn't very good. So we draw our cyclohexene ring like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if we start with cyclohexene, let's go ahead and draw cyclohexene in here. Let's do another one. That one wasn't very good. So we draw our cyclohexene ring like that. And to cyclohexene, we're going to add peroxyacetic acid. So what does peroxyacetic acid look like? Well, it's based on acetic acid, but it has one extra oxygen in there."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we draw our cyclohexene ring like that. And to cyclohexene, we're going to add peroxyacetic acid. So what does peroxyacetic acid look like? Well, it's based on acetic acid, but it has one extra oxygen in there. So it looks like that. So that's our peroxyacetic acid. So we add cyclohexene to peroxyacetic acid."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, it's based on acetic acid, but it has one extra oxygen in there. So it looks like that. So that's our peroxyacetic acid. So we add cyclohexene to peroxyacetic acid. We're going to form an epoxide. So we're going to form a three-membered ring, including oxygen. I'm going to say the oxygen adds to the top face of our ring."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we add cyclohexene to peroxyacetic acid. We're going to form an epoxide. So we're going to form a three-membered ring, including oxygen. I'm going to say the oxygen adds to the top face of our ring. It doesn't really matter for this example. But we'll go ahead and put in our epoxide using wedges here. And that must mean that going away from us, those are hydrogens in space."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to say the oxygen adds to the top face of our ring. It doesn't really matter for this example. But we'll go ahead and put in our epoxide using wedges here. And that must mean that going away from us, those are hydrogens in space. So that's the epoxide that would form using the mechanism that we put above there. Let's go ahead and open this epoxide using acid. So just to refresh everyone's memory, go back up here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that must mean that going away from us, those are hydrogens in space. So that's the epoxide that would form using the mechanism that we put above there. Let's go ahead and open this epoxide using acid. So just to refresh everyone's memory, go back up here. Now we're going to look at the second part, where we add H3O plus to form our diol. So let's take a look at that now. So we're going to add H3O plus to this epoxide."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So just to refresh everyone's memory, go back up here. Now we're going to look at the second part, where we add H3O plus to form our diol. So let's take a look at that now. So we're going to add H3O plus to this epoxide. So adding H3O plus. And I'm going to redraw our epoxide to give us a better viewpoint here. So I'm going to put my oxygen right here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to add H3O plus to this epoxide. So adding H3O plus. And I'm going to redraw our epoxide to give us a better viewpoint here. So I'm going to put my oxygen right here. And then that's bonded to our two carbons like this. And then let me see if I can draw the rest of the ring. So in the back here, here is the rest of my cyclohexane ring like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to put my oxygen right here. And then that's bonded to our two carbons like this. And then let me see if I can draw the rest of the ring. So in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. And now I have my H3O plus in here like this."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in the back here, here is the rest of my cyclohexane ring like that. And we'll go ahead and put in our lone pairs of electrons. So this is the same exact drawing above here. And now I have my H3O plus in here like this. So my hydronium ion is present with a lone pair of electrons, giving us a plus 1 formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now I have my H3O plus in here like this. So my hydronium ion is present with a lone pair of electrons, giving us a plus 1 formal charge like that. So the oxygen on our epoxide is going to act as a base. It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to take a proton. So this lone pair of electrons is going to take this proton right here, which would kick these electrons in here off onto my oxygen. So let's draw the result of that acid-base reaction. So I'm going to make a protonated epoxide. So let's go ahead and draw our oxygen here. And it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to make a protonated epoxide. So let's go ahead and draw our oxygen here. And it's connected to those carbons down here. So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed a bond on that proton. So this is my structure now."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and draw the rest of my ring in the back here like that. And then one lone pair of electrons didn't do anything, so it's still there. One lone pair of electrons is the one that formed a bond on that proton. So this is my structure now. And this would give this oxygen a plus 1 formal charge. So it's positively charged now. So this is the same structure that we saw in the earlier videos with our cyclic halonium ion."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is my structure now. And this would give this oxygen a plus 1 formal charge. So it's positively charged now. So this is the same structure that we saw in the earlier videos with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos, check out the halohydrin video, you're going to get a partial carbocationic character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, the lone pair of electrons on water are going to be attracted to those carbons."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is the same structure that we saw in the earlier videos with our cyclic halonium ion. And just like the cyclic halonium ion in those earlier videos, check out the halohydrin video, you're going to get a partial carbocationic character with these carbons down here. So the resonance hybrid is going to give these carbons some partial positive character. So when water comes along as a nucleophile, the lone pair of electrons on water are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when water comes along as a nucleophile, the lone pair of electrons on water are going to be attracted to those carbons. So opposite charges attract. These two blue carbons are partially positive. The negative electrons are attracted to the partially positive carbon. And you're going to get nucleophilic attack. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The negative electrons are attracted to the partially positive carbon. And you're going to get nucleophilic attack. So let's say this lone pair of electrons attacks right here. Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the results of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance?"}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, that would kick the electrons in this bond off onto your oxygen. So let's go ahead and draw the results of an attack on the carbon on the left. So let's get some more room here. So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what would happen in that instance? Well, let's go ahead and draw our cyclohexane ring back here. So here is our cyclohexane ring. The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack. So it has two hydrogens on it. It has one lone pair of electrons now."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen attacked the carbon on the left. So there is the oxygen that did the nucleophilic attack. So it has two hydrogens on it. It has one lone pair of electrons now. And it formed a plus 1 formal charge. Our epoxide opened. The electrons kicked off onto the top oxygen."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It has one lone pair of electrons now. And it formed a plus 1 formal charge. Our epoxide opened. The electrons kicked off onto the top oxygen. And that means that the top oxygen moves over here, like that. So that would be your structure. Well, this lone pair of electrons could have attacked this carbon."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The electrons kicked off onto the top oxygen. And that means that the top oxygen moves over here, like that. So that would be your structure. Well, this lone pair of electrons could have attacked this carbon. This carbon could have been the partially positive one in the resonance hybrid, which would kick these electrons off onto this oxygen. So let's go ahead and draw the result of that nucleophilic attack. So I'll go ahead and put in my cyclohexane ring like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, this lone pair of electrons could have attacked this carbon. This carbon could have been the partially positive one in the resonance hybrid, which would kick these electrons off onto this oxygen. So let's go ahead and draw the result of that nucleophilic attack. So I'll go ahead and put in my cyclohexane ring like that. This time, our oxygen is going to bond with the carbon on the left. Two hydrogens attached to it. A lone pair of electrons."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and put in my cyclohexane ring like that. This time, our oxygen is going to bond with the carbon on the left. Two hydrogens attached to it. A lone pair of electrons. A positive 1 formal charge. And then this time, the oxygen on top is going to kick off onto this oxygen over here on the left, like that. So you're going to get an OH over there."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "A lone pair of electrons. A positive 1 formal charge. And then this time, the oxygen on top is going to kick off onto this oxygen over here on the left, like that. So you're going to get an OH over there. So in the next step of the mechanism, we're almost done. We've almost formed our diol. We're going to have water come along."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you're going to get an OH over there. So in the next step of the mechanism, we're almost done. We've almost formed our diol. We're going to have water come along. And this time, instead of water acting as a nucleophile, water is going to act as a base. It's going to take a proton. So let's look at the product on the left here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to have water come along. And this time, instead of water acting as a nucleophile, water is going to act as a base. It's going to take a proton. So let's look at the product on the left here. So this lone pair of electrons would take one of these protons, kick these electrons off onto your oxygen like that. So let's go ahead and draw the result of that acid-base reaction. So let's draw our cyclohexane ring like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at the product on the left here. So this lone pair of electrons would take one of these protons, kick these electrons off onto your oxygen like that. So let's go ahead and draw the result of that acid-base reaction. So let's draw our cyclohexane ring like that. And now we have an OH down here. So this is now an OH. And this was an OH."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw our cyclohexane ring like that. And now we have an OH down here. So this is now an OH. And this was an OH. So we've achieved our product. We've added two OHs anti to each other. Same thing could happen over here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this was an OH. So we've achieved our product. We've added two OHs anti to each other. Same thing could happen over here. You could grab this proton, kick these electrons off onto your oxygen like that. And so on the right, after we draw our cyclohexane ring, so there's our cyclohexane ring, we're going to have an OH right here. And then we're going to have an OH over here like that."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Same thing could happen over here. You could grab this proton, kick these electrons off onto your oxygen like that. And so on the right, after we draw our cyclohexane ring, so there's our cyclohexane ring, we're going to have an OH right here. And then we're going to have an OH over here like that. So we have two products. And if you look at them, they are mirror images of each other. If I were to put a mirror right here, you would see that they would be reflected in a mirror."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we're going to have an OH over here like that. So we have two products. And if you look at them, they are mirror images of each other. If I were to put a mirror right here, you would see that they would be reflected in a mirror. And they are non-superimposable. So non-superimposable mirror images and nanti-mirrors. So you're going to get two products for this reaction."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If I were to put a mirror right here, you would see that they would be reflected in a mirror. And they are non-superimposable. So non-superimposable mirror images and nanti-mirrors. So you're going to get two products for this reaction. So just to summarize this reaction, let's do it one more time. We'll start with cyclohexane. And in the first step of our mechanism, the first of our reaction here, we added peroxyacetic acid."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you're going to get two products for this reaction. So just to summarize this reaction, let's do it one more time. We'll start with cyclohexane. And in the first step of our mechanism, the first of our reaction here, we added peroxyacetic acid. So we added CH3CO3H. And the second step of our mechanism, in the second step of our reactions here, we added H3O+. And so that opened up the epoxide that formed to form our diol."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And in the first step of our mechanism, the first of our reaction here, we added peroxyacetic acid. So we added CH3CO3H. And the second step of our mechanism, in the second step of our reactions here, we added H3O+. And so that opened up the epoxide that formed to form our diol. And we get two products. So this product over here on the left, let's go ahead and redraw this product over here on the left in a way that's a little bit more familiar. So once again, I put my eye right here."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so that opened up the epoxide that formed to form our diol. And we get two products. So this product over here on the left, let's go ahead and redraw this product over here on the left in a way that's a little bit more familiar. So once again, I put my eye right here. I stare down. At this top carbon, if this is top of my head, this OH is coming out at me. So I would draw that product."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So once again, I put my eye right here. I stare down. At this top carbon, if this is top of my head, this OH is coming out at me. So I would draw that product. I would draw my cyclohexane ring. And at that top carbon, I would show the OH coming out at me. And then, of course, at this carbon down here, the OH would be going away from me."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I would draw that product. I would draw my cyclohexane ring. And at that top carbon, I would show the OH coming out at me. And then, of course, at this carbon down here, the OH would be going away from me. So I go ahead and draw my OH as a dash here, down here. And then I do the same thing with this one right here. So if I stare down at that molecule, once again, if I stare down here and look this way, this time at this top carbon, if my head's right here, top of my head, this OH would be down."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then, of course, at this carbon down here, the OH would be going away from me. So I go ahead and draw my OH as a dash here, down here. And then I do the same thing with this one right here. So if I stare down at that molecule, once again, if I stare down here and look this way, this time at this top carbon, if my head's right here, top of my head, this OH would be down. So I go ahead and put a dash right here. And put my OH. And then over here at this carbon, it would be going up."}, {"video_title": "Epoxide formation and anti dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I stare down at that molecule, once again, if I stare down here and look this way, this time at this top carbon, if my head's right here, top of my head, this OH would be down. So I go ahead and put a dash right here. And put my OH. And then over here at this carbon, it would be going up. So it might be easier to see that these are enantiomers when you look at them drawn like this, a different absolute configuration at both carbons. So this is how to form an epoxide and one way to make a diol. In the next video, we will see another way to make a diol, although it will add in a different way to give you a slightly different product."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So here's our general structure of an acyl halide. On the left side we have an acyl group. On the right side we have a halogen. You can also call these acid halides. They're derived from carboxylic acids. And so if we look at this carboxylic acid on the left here, so a two carbon carboxylic acid, we could convert that to a two carbon acyl halide over here on the right. And if we want to name our acyl halide, we have to think about the name of the carboxylic acid."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "You can also call these acid halides. They're derived from carboxylic acids. And so if we look at this carboxylic acid on the left here, so a two carbon carboxylic acid, we could convert that to a two carbon acyl halide over here on the right. And if we want to name our acyl halide, we have to think about the name of the carboxylic acid. And so this, of course, is acetic acid. So let's go ahead and write out acetic acid here. And if we want to name the corresponding acyl halide, we need to think about dropping the IC ending and then the acid."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And if we want to name our acyl halide, we have to think about the name of the carboxylic acid. And so this, of course, is acetic acid. So let's go ahead and write out acetic acid here. And if we want to name the corresponding acyl halide, we need to think about dropping the IC ending and then the acid. So we drop IC in acid, and we add YL and then the halide. So let's go ahead and write that out. So we drop the IC and we add YL, and then we add the halide."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And if we want to name the corresponding acyl halide, we need to think about dropping the IC ending and then the acid. So we drop IC in acid, and we add YL and then the halide. So let's go ahead and write that out. So we drop the IC and we add YL, and then we add the halide. So we have a chlorine here, so we're going to write chloride. And so we would call this acetyl chloride. So let's go ahead and show that right here."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So we drop the IC and we add YL, and then we add the halide. So we have a chlorine here, so we're going to write chloride. And so we would call this acetyl chloride. So let's go ahead and show that right here. So we add the YL, and then we add the halide portion. We could have also called this ethanoic acid. So ethanoic acid would be the IUPAC name, but everyone says acetic acid."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So let's go ahead and show that right here. So we add the YL, and then we add the halide portion. We could have also called this ethanoic acid. So ethanoic acid would be the IUPAC name, but everyone says acetic acid. So if we were to call this ethanoic acid, once again, think about drop your IC and then the acid part. So drop this portion, and then add YL and then chloride. So let's go ahead and write that in here."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So ethanoic acid would be the IUPAC name, but everyone says acetic acid. So if we were to call this ethanoic acid, once again, think about drop your IC and then the acid part. So drop this portion, and then add YL and then chloride. So let's go ahead and write that in here. So we go ahead and add the YL in and then the chloride, like that. So that would be ethanolyl chloride as our name. Let's go ahead and show this portion."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So let's go ahead and write that in here. So we go ahead and add the YL in and then the chloride, like that. So that would be ethanolyl chloride as our name. Let's go ahead and show this portion. Once again, the YL portion and then our halide. In terms of physical properties of acyl halides, we need to think about the interaction of two molecules here. So let me go ahead and draw in another molecule of acetyl chloride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "Let's go ahead and show this portion. Once again, the YL portion and then our halide. In terms of physical properties of acyl halides, we need to think about the interaction of two molecules here. So let me go ahead and draw in another molecule of acetyl chloride. Acetyl chloride has a boiling point of approximately 51 degrees Celsius. So let me go ahead and write that in. So approximately 51 degrees Celsius."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So let me go ahead and draw in another molecule of acetyl chloride. Acetyl chloride has a boiling point of approximately 51 degrees Celsius. So let me go ahead and write that in. So approximately 51 degrees Celsius. And we know that acetyl chloride is a polar molecule. The oxygen right here is more electronegative than this carbon. So we have a partial negative and we have a partial positive."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So approximately 51 degrees Celsius. And we know that acetyl chloride is a polar molecule. The oxygen right here is more electronegative than this carbon. So we have a partial negative and we have a partial positive. This chlorine is also withdrawing electron density from our partially positive carbon. And so we have a polar molecule. So acetyl chloride is polar right here."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So we have a partial negative and we have a partial positive. This chlorine is also withdrawing electron density from our partially positive carbon. And so we have a polar molecule. So acetyl chloride is polar right here. So this is polar. Same molecule. So this is polar."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So acetyl chloride is polar right here. So this is polar. Same molecule. So this is polar. We have a partial negative, partial positive. Once again, this chlorine is also withdrawing electron density this way. And so we have two polar molecules interacting, which we know is a dipole-dipole intermolecular force."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So this is polar. We have a partial negative, partial positive. Once again, this chlorine is also withdrawing electron density this way. And so we have two polar molecules interacting, which we know is a dipole-dipole intermolecular force. So there's an attractive force between these molecules, which is dipole-dipole. So let me go ahead and write that. So it's a dipole-dipole interaction with molecules of acetyl chloride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And so we have two polar molecules interacting, which we know is a dipole-dipole intermolecular force. So there's an attractive force between these molecules, which is dipole-dipole. So let me go ahead and write that. So it's a dipole-dipole interaction with molecules of acetyl chloride. And we know that dipole-dipole interactions are stronger than London dispersion forces. And so acetyl chloride has a higher boiling point than, say, a two-carbon alkane, so like ethane. So it's a little bit harder to pull these molecules apart than it is to pull molecules of ethane apart."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So it's a dipole-dipole interaction with molecules of acetyl chloride. And we know that dipole-dipole interactions are stronger than London dispersion forces. And so acetyl chloride has a higher boiling point than, say, a two-carbon alkane, so like ethane. So it's a little bit harder to pull these molecules apart than it is to pull molecules of ethane apart. And so therefore, this boiling point is higher than that for a two-carbon alkane. However, this boiling point is lower than that of acetic acid. And to think about that, we need to draw on another molecule of acetic acid."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So it's a little bit harder to pull these molecules apart than it is to pull molecules of ethane apart. And so therefore, this boiling point is higher than that for a two-carbon alkane. However, this boiling point is lower than that of acetic acid. And to think about that, we need to draw on another molecule of acetic acid. So let me go ahead and do that. So drawing in another molecule of acetic acid, we can see that there's opportunities for hydrogen bonding. So there's a hydrogen bond here and a hydrogen bond here."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And to think about that, we need to draw on another molecule of acetic acid. So let me go ahead and do that. So drawing in another molecule of acetic acid, we can see that there's opportunities for hydrogen bonding. So there's a hydrogen bond here and a hydrogen bond here. And hydrogen bonding, let me go ahead and write that. So hydrogen bonding is the strongest type of intermolecular force. And so therefore, the boiling point of acetic acid is going to be higher."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So there's a hydrogen bond here and a hydrogen bond here. And hydrogen bonding, let me go ahead and write that. So hydrogen bonding is the strongest type of intermolecular force. And so therefore, the boiling point of acetic acid is going to be higher. So it's somewhere around 118 degrees Celsius. And so it's harder to pull these two molecules apart, because hydrogen bonding is a stronger intermolecular force than dipole-dipole. So that just gives you some idea of the boiling point of acyl halides."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And so therefore, the boiling point of acetic acid is going to be higher. So it's somewhere around 118 degrees Celsius. And so it's harder to pull these two molecules apart, because hydrogen bonding is a stronger intermolecular force than dipole-dipole. So that just gives you some idea of the boiling point of acyl halides. In terms of solubility in water, you can't really say that something like acetyl chloride is soluble in water, because it reacts so violently with it. So acetyl chloride is extremely reactive. And it reacts very quickly and often violently with water."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So that just gives you some idea of the boiling point of acyl halides. In terms of solubility in water, you can't really say that something like acetyl chloride is soluble in water, because it reacts so violently with it. So acetyl chloride is extremely reactive. And it reacts very quickly and often violently with water. And so we can't really say that it dissolves in water. Let's move on to acid anhydrides. And so let's look at how to name an acid anhydride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And it reacts very quickly and often violently with water. And so we can't really say that it dissolves in water. Let's move on to acid anhydrides. And so let's look at how to name an acid anhydride. Acid anhydrides can be thought of as being derived from carboxylic acids, too. And so if we look over here on the left, once again, we have acetic acid here. So this is acetic acid."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And so let's look at how to name an acid anhydride. Acid anhydrides can be thought of as being derived from carboxylic acids, too. And so if we look over here on the left, once again, we have acetic acid here. So this is acetic acid. And if we take two molecules of acetic acid and combine them, we can form an acid anhydride. And let's go ahead and think about what happens. We're going to lose water here."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So this is acetic acid. And if we take two molecules of acetic acid and combine them, we can form an acid anhydride. And let's go ahead and think about what happens. We're going to lose water here. And the word anhydride means without water. So if we take off the water and take this portion, take this acyl group and this over here and stick them together, then we form our anhydride over here on the right. And because our anhydride was formed from acetic acid, we call this acetic anhydride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "We're going to lose water here. And the word anhydride means without water. So if we take off the water and take this portion, take this acyl group and this over here and stick them together, then we form our anhydride over here on the right. And because our anhydride was formed from acetic acid, we call this acetic anhydride. So these are pretty simple names. You keep the acetic part, and then you drop the acid and just add anhydride. So this is acetic anhydride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And because our anhydride was formed from acetic acid, we call this acetic anhydride. So these are pretty simple names. You keep the acetic part, and then you drop the acid and just add anhydride. So this is acetic anhydride. If you thought of acetic acid as ethanoic acid, if you prefer to use the IUPAC name, so ethanoic acid. If you write ethanoic acid here. Once again, just drop the acid part and add anhydride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So this is acetic anhydride. If you thought of acetic acid as ethanoic acid, if you prefer to use the IUPAC name, so ethanoic acid. If you write ethanoic acid here. Once again, just drop the acid part and add anhydride. So you could call this ethanoic anhydride. So ethanoic anhydride. So once again, anhydride meaning without water."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "Once again, just drop the acid part and add anhydride. So you could call this ethanoic anhydride. So ethanoic anhydride. So once again, anhydride meaning without water. Let's look at how to name another anhydride. So let me go down here and get some more room. So we're trying to name this anhydride over here on the right."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So once again, anhydride meaning without water. Let's look at how to name another anhydride. So let me go down here and get some more room. So we're trying to name this anhydride over here on the right. And to do that, we need to think about the carboxylic acid from which it can be thought of as being derived. So here we have two molecules of benzoic acid. So let's go ahead and write benzoic acid here."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So we're trying to name this anhydride over here on the right. And to do that, we need to think about the carboxylic acid from which it can be thought of as being derived. So here we have two molecules of benzoic acid. So let's go ahead and write benzoic acid here. And I'm not talking about exact chemical reactions. I'm just thinking about the acid anhydride and how to put it into the different carboxylic acids. But if we do the same thing we did before, we think about the term anhydride being loss of water."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So let's go ahead and write benzoic acid here. And I'm not talking about exact chemical reactions. I'm just thinking about the acid anhydride and how to put it into the different carboxylic acids. But if we do the same thing we did before, we think about the term anhydride being loss of water. And we take out water here and stick those together. Once again, you can see we form the anhydride on the right. So this portion plus this portion gives us our acid anhydride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "But if we do the same thing we did before, we think about the term anhydride being loss of water. And we take out water here and stick those together. Once again, you can see we form the anhydride on the right. So this portion plus this portion gives us our acid anhydride. And once again, we're not doing exact chemical reactions here. So just for the sake of nomenclature, we can just drop the acid and add anhydride. So this would be benzoic anhydride."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So this portion plus this portion gives us our acid anhydride. And once again, we're not doing exact chemical reactions here. So just for the sake of nomenclature, we can just drop the acid and add anhydride. So this would be benzoic anhydride. So this would be benzoic anhydride, like that. Let's look at another example. This time, we don't have symmetry."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So this would be benzoic anhydride. So this would be benzoic anhydride, like that. Let's look at another example. This time, we don't have symmetry. So when I'm thinking about some carboxylic acids for this one, over here on the left, I recognize benzoic acids. Let me go ahead and write that down. So benzoic acid is being present."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "This time, we don't have symmetry. So when I'm thinking about some carboxylic acids for this one, over here on the left, I recognize benzoic acids. Let me go ahead and write that down. So benzoic acid is being present. And if I think about over on the right side, so this portion, if I think about a carboxylic acid this way, I see that's acetic acid. So I have benzoic acid and acetic acid. To name our anhydride, we drop the acid part."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So benzoic acid is being present. And if I think about over on the right side, so this portion, if I think about a carboxylic acid this way, I see that's acetic acid. So I have benzoic acid and acetic acid. To name our anhydride, we drop the acid part. And we're going to add anhydride. And we have to think about using the alphabet here. So A comes before B."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "To name our anhydride, we drop the acid part. And we're going to add anhydride. And we have to think about using the alphabet here. So A comes before B. So to write the name of our anhydride, we would write acetic benzoic anhydride. So acetic benzoic anhydride. So once again, when you see an anhydride and you're trying to name it, just think about the carboxylic acids."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So A comes before B. So to write the name of our anhydride, we would write acetic benzoic anhydride. So acetic benzoic anhydride. So once again, when you see an anhydride and you're trying to name it, just think about the carboxylic acids. And that will help you figure out the name. In terms of physical properties of acid anhydrides, let's look at an example here. So over here on the left, we have acetic anhydride, which is a polar molecule."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So once again, when you see an anhydride and you're trying to name it, just think about the carboxylic acids. And that will help you figure out the name. In terms of physical properties of acid anhydrides, let's look at an example here. So over here on the left, we have acetic anhydride, which is a polar molecule. So it's moderately polar because we have these carbonyls here. So the oxygen is partially negative. This carbon down here is partially positive."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So over here on the left, we have acetic anhydride, which is a polar molecule. So it's moderately polar because we have these carbonyls here. So the oxygen is partially negative. This carbon down here is partially positive. Same thing for all of these carbonyls. And so it's a moderately polar molecule. And so there's going to be, that's a negative sign right there, so there's going to be some attraction between these molecules."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "This carbon down here is partially positive. Same thing for all of these carbonyls. And so it's a moderately polar molecule. And so there's going to be, that's a negative sign right there, so there's going to be some attraction between these molecules. So there's going to be some attraction between the negative and the positive charges. So we have a fairly polar molecule and a fairly polar molecule. So we can say that there's some dipole-dipole interaction presence."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And so there's going to be, that's a negative sign right there, so there's going to be some attraction between these molecules. So there's going to be some attraction between the negative and the positive charges. So we have a fairly polar molecule and a fairly polar molecule. So we can say that there's some dipole-dipole interaction presence. So between molecules of acetic anhydride, there's some dipole-dipole interaction. And there's also, of course, London dispersion forces as well. And so the boiling point for acetic anhydride turns out to be approximately 140 degrees Celsius."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So we can say that there's some dipole-dipole interaction presence. So between molecules of acetic anhydride, there's some dipole-dipole interaction. And there's also, of course, London dispersion forces as well. And so the boiling point for acetic anhydride turns out to be approximately 140 degrees Celsius. So let's go ahead and write that in here. So approximately 140 degrees Celsius. And we can compare that to a carboxylic acid that's similar in terms of number of carbons and oxygens."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And so the boiling point for acetic anhydride turns out to be approximately 140 degrees Celsius. So let's go ahead and write that in here. So approximately 140 degrees Celsius. And we can compare that to a carboxylic acid that's similar in terms of number of carbons and oxygens. So for acetic anhydride, we had one, two, three, four carbons. And so for over here on the right, this is butanoic acid. So we have one, two, three, four carbons."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And we can compare that to a carboxylic acid that's similar in terms of number of carbons and oxygens. So for acetic anhydride, we had one, two, three, four carbons. And so for over here on the right, this is butanoic acid. So we have one, two, three, four carbons. And then we have two oxygens for butanoic acid. And we have three oxygens for acetic anhydride. So they're similar in terms of sizes."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So we have one, two, three, four carbons. And then we have two oxygens for butanoic acid. And we have three oxygens for acetic anhydride. So they're similar in terms of sizes. But when we think about comparing their boiling points, so over here on the right, butanoic acid has a boiling point of approximately 164 degrees Celsius. And so it has a higher boiling point. Because once again, there's some hydrogen bonding present."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So they're similar in terms of sizes. But when we think about comparing their boiling points, so over here on the right, butanoic acid has a boiling point of approximately 164 degrees Celsius. And so it has a higher boiling point. Because once again, there's some hydrogen bonding present. So there's some hydrogen bonding present because we are talking about a carboxylic acid here. And once again, hydrogen bonding is a stronger intermolecular force than dipole-dipole. So it's harder to pull apart molecules of butanoic acid."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "Because once again, there's some hydrogen bonding present. So there's some hydrogen bonding present because we are talking about a carboxylic acid here. And once again, hydrogen bonding is a stronger intermolecular force than dipole-dipole. So it's harder to pull apart molecules of butanoic acid. Therefore, it takes more energy. It takes a higher temperature to pull these molecules apart to turn them into a gas. And so once again, H bonding is a stronger intermolecular force than dipole-dipole."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "So it's harder to pull apart molecules of butanoic acid. Therefore, it takes more energy. It takes a higher temperature to pull these molecules apart to turn them into a gas. And so once again, H bonding is a stronger intermolecular force than dipole-dipole. When we think about the solubility of an acid anhydride in water, once again, it's kind of difficult. Because something like acetic anhydride is going to react with the water. So acetic anhydride is also fairly reactive, not quite as reactive as an acyl halide."}, {"video_title": "Nomenclature and properties of acyl (acid) halides and acid anhydrides Khan Academy.mp3", "Sentence": "And so once again, H bonding is a stronger intermolecular force than dipole-dipole. When we think about the solubility of an acid anhydride in water, once again, it's kind of difficult. Because something like acetic anhydride is going to react with the water. So acetic anhydride is also fairly reactive, not quite as reactive as an acyl halide. But it does react with water. So we can't really say that it dissolves very well in it. And so we'll talk much more about the reactivity of these carboxylic acid derivatives in a later video."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "Intermolecular forces are the forces that are between molecules. And so that's different from an intramolecular force, which is the force within a molecule. So a force within a molecule would be something like the covalent bond. An intermolecular force would be the force that are between molecules. And so let's look at the first intermolecular force. It's called a dipole-dipole interaction. And let's analyze why it has that name."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "An intermolecular force would be the force that are between molecules. And so let's look at the first intermolecular force. It's called a dipole-dipole interaction. And let's analyze why it has that name. If I look at one of these molecules of acetone here, and I focus in on the carbon that's double bonded to the oxygen, I know that oxygen is more electronegative than carbon. And so we have four electrons in this double bond between the carbon and the oxygen. So I'll try to highlight them right here."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And let's analyze why it has that name. If I look at one of these molecules of acetone here, and I focus in on the carbon that's double bonded to the oxygen, I know that oxygen is more electronegative than carbon. And so we have four electrons in this double bond between the carbon and the oxygen. So I'll try to highlight them right here. And since oxygen is more electronegative, oxygen is going to pull those electrons closer to it, therefore giving oxygen a partial negative charge. Those electrons in yellow are moving away from this carbon. So the carbon's losing a little bit of electron density."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So I'll try to highlight them right here. And since oxygen is more electronegative, oxygen is going to pull those electrons closer to it, therefore giving oxygen a partial negative charge. Those electrons in yellow are moving away from this carbon. So the carbon's losing a little bit of electron density. And this carbon is becoming partially positive, like that. And so for this molecule, we're going to get a separation of charge, a positive and a negative charge. So we have a polarized double bond situation here."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So the carbon's losing a little bit of electron density. And this carbon is becoming partially positive, like that. And so for this molecule, we're going to get a separation of charge, a positive and a negative charge. So we have a polarized double bond situation here. We also have a polarized molecule. And so there's two different poles, a negative and a positive pole here. And so we say that this is a polar molecule."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So we have a polarized double bond situation here. We also have a polarized molecule. And so there's two different poles, a negative and a positive pole here. And so we say that this is a polar molecule. So acetone is a relatively polar molecule. The same thing happens to this acetone molecule down here. So we get a partial negative, and we get a partial positive."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And so we say that this is a polar molecule. So acetone is a relatively polar molecule. The same thing happens to this acetone molecule down here. So we get a partial negative, and we get a partial positive. So this is a polar molecule as well. It has two poles. So we call this a dipole."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So we get a partial negative, and we get a partial positive. So this is a polar molecule as well. It has two poles. So we call this a dipole. So each molecule has a dipole moment. And because each molecule is polar and has a separation of positive and negative charge, in organic chemistry, we know that opposite charges attract. So this negatively charged oxygen is going to be attracted to this positively charged carbon."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So we call this a dipole. So each molecule has a dipole moment. And because each molecule is polar and has a separation of positive and negative charge, in organic chemistry, we know that opposite charges attract. So this negatively charged oxygen is going to be attracted to this positively charged carbon. And so there's going to be an electrostatic attraction between those two molecules. And that's what's going to hold these two molecules together. And you would therefore need energy if you were to try to pull them apart."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So this negatively charged oxygen is going to be attracted to this positively charged carbon. And so there's going to be an electrostatic attraction between those two molecules. And that's what's going to hold these two molecules together. And you would therefore need energy if you were to try to pull them apart. And so the boiling point of acetone turns out to be approximately 56 degrees Celsius. And since room temperature is between 20 and 25, at room temperature, we have not reached the boiling point of acetone. And therefore, acetone is still a liquid."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And you would therefore need energy if you were to try to pull them apart. And so the boiling point of acetone turns out to be approximately 56 degrees Celsius. And since room temperature is between 20 and 25, at room temperature, we have not reached the boiling point of acetone. And therefore, acetone is still a liquid. So at room temperature and pressure, acetone is liquid. And it has to do with the intermolecular force of dipole-dipole interactions holding those molecules together. And the intermolecular force, in turn, depends on the electronegativity."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And therefore, acetone is still a liquid. So at room temperature and pressure, acetone is liquid. And it has to do with the intermolecular force of dipole-dipole interactions holding those molecules together. And the intermolecular force, in turn, depends on the electronegativity. Let's look at another intermolecular force. And this one's called hydrogen bonding. So here we have two water molecules."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And the intermolecular force, in turn, depends on the electronegativity. Let's look at another intermolecular force. And this one's called hydrogen bonding. So here we have two water molecules. And once again, if I think about these electrons here, which are between the oxygen and the hydrogen, I know oxygen's more electronegative than hydrogen. So oxygen's going to pull those electrons closer to it, giving the oxygen a partial negative charge like that. The hydrogen is losing a little bit of electron density, therefore becoming partially positive."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So here we have two water molecules. And once again, if I think about these electrons here, which are between the oxygen and the hydrogen, I know oxygen's more electronegative than hydrogen. So oxygen's going to pull those electrons closer to it, giving the oxygen a partial negative charge like that. The hydrogen is losing a little bit of electron density, therefore becoming partially positive. And the same situation exists in the water molecule down here. So we have a partial negative, and we have a partial positive. And so like the last example, we can see there's going to be some sort of electrostatic attraction between those opposite charges, between the negatively partially charged oxygen and the partially positive hydrogen, like that."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "The hydrogen is losing a little bit of electron density, therefore becoming partially positive. And the same situation exists in the water molecule down here. So we have a partial negative, and we have a partial positive. And so like the last example, we can see there's going to be some sort of electrostatic attraction between those opposite charges, between the negatively partially charged oxygen and the partially positive hydrogen, like that. And so this is a polar molecule. Of course, water is a polar molecule. And so you would think that this would be an example of dipole-dipole interaction."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And so like the last example, we can see there's going to be some sort of electrostatic attraction between those opposite charges, between the negatively partially charged oxygen and the partially positive hydrogen, like that. And so this is a polar molecule. Of course, water is a polar molecule. And so you would think that this would be an example of dipole-dipole interaction. And it is, except in this case, it's an even stronger version of dipole-dipole interaction that we call hydrogen bonding. So at one time, it was thought that it was possible for hydrogen to form an extra bond. And that's where the term originally comes from."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And so you would think that this would be an example of dipole-dipole interaction. And it is, except in this case, it's an even stronger version of dipole-dipole interaction that we call hydrogen bonding. So at one time, it was thought that it was possible for hydrogen to form an extra bond. And that's where the term originally comes from. But of course, it's not an actual intramolecular force. We're talking about an intermolecular force. But it is the strongest intermolecular force."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And that's where the term originally comes from. But of course, it's not an actual intramolecular force. We're talking about an intermolecular force. But it is the strongest intermolecular force. The way to recognize when hydrogen bonding is present, as opposed to just dipole-dipole, is to see what the hydrogen is bonded to. And so in this case, we have a very electronegative atom, hydrogen, bonded, oxygen I should say, bonded to hydrogen. And then that hydrogen is interacting with another electronegative atom like that."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "But it is the strongest intermolecular force. The way to recognize when hydrogen bonding is present, as opposed to just dipole-dipole, is to see what the hydrogen is bonded to. And so in this case, we have a very electronegative atom, hydrogen, bonded, oxygen I should say, bonded to hydrogen. And then that hydrogen is interacting with another electronegative atom like that. So we have a partial negative, and we have a partial positive, and then we have another partial negative over here. And this is the situation that you need to have when you have hydrogen bonding. Here's your hydrogen, showing intermolecular force here."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And then that hydrogen is interacting with another electronegative atom like that. So we have a partial negative, and we have a partial positive, and then we have another partial negative over here. And this is the situation that you need to have when you have hydrogen bonding. Here's your hydrogen, showing intermolecular force here. And what some students forget is that this hydrogen actually has to be bonded to another electronegative atom in order for there to be a big enough difference in electronegativity for there to be a little bit extra attraction. And so the three electronegative elements that you should remember for hydrogen bonding are fluorine, oxygen, and nitrogen. And so the mnemonic that students use is phone."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "Here's your hydrogen, showing intermolecular force here. And what some students forget is that this hydrogen actually has to be bonded to another electronegative atom in order for there to be a big enough difference in electronegativity for there to be a little bit extra attraction. And so the three electronegative elements that you should remember for hydrogen bonding are fluorine, oxygen, and nitrogen. And so the mnemonic that students use is phone. So if you remember phone as the electronegative atoms that could participate in hydrogen bonding, you should be able to remember this intermolecular force. The boiling point of water is, of course, about 100 degrees Celsius, so higher than what we saw for acetone. And this just is due to the fact that hydrogen bonding is a stronger version of dipole-dipole interaction."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And so the mnemonic that students use is phone. So if you remember phone as the electronegative atoms that could participate in hydrogen bonding, you should be able to remember this intermolecular force. The boiling point of water is, of course, about 100 degrees Celsius, so higher than what we saw for acetone. And this just is due to the fact that hydrogen bonding is a stronger version of dipole-dipole interaction. And therefore, it takes more energy or more heat to pull these water molecules apart in order to turn them into a gas. And so, of course, water is a liquid at room temperature. Let's look at another intermolecular force."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And this just is due to the fact that hydrogen bonding is a stronger version of dipole-dipole interaction. And therefore, it takes more energy or more heat to pull these water molecules apart in order to turn them into a gas. And so, of course, water is a liquid at room temperature. Let's look at another intermolecular force. And this one is called London dispersion forces. So these are the weakest intermolecular forces. And they have to do with the electrons that are always moving around in orbitals."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "Let's look at another intermolecular force. And this one is called London dispersion forces. So these are the weakest intermolecular forces. And they have to do with the electrons that are always moving around in orbitals. And even though the methane molecule here, if we look at it, we have a carbon surrounded by four hydrogens for methane. It's hard to tell in how I've drawn the structure here. But if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon and they're equivalent in all directions."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And they have to do with the electrons that are always moving around in orbitals. And even though the methane molecule here, if we look at it, we have a carbon surrounded by four hydrogens for methane. It's hard to tell in how I've drawn the structure here. But if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon and they're equivalent in all directions. And there's a very small difference in electronegativity between the carbon and the hydrogen. And that small difference is canceled out in three dimensions. So the methane molecule becomes nonpolar as a result of that."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "But if you go back and you look at the video for the tetrahedral bond angle proof, you can see that in three dimensions, these hydrogens are coming off of the carbon and they're equivalent in all directions. And there's a very small difference in electronegativity between the carbon and the hydrogen. And that small difference is canceled out in three dimensions. So the methane molecule becomes nonpolar as a result of that. So this one's nonpolar. Of course, this one's nonpolar. And so there's no dipole-dipole interaction."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "So the methane molecule becomes nonpolar as a result of that. So this one's nonpolar. Of course, this one's nonpolar. And so there's no dipole-dipole interaction. There's no hydrogen bonding. The only intermolecular force that's holding two methane molecules together would be London dispersion forces. And so once again, you could think about the electrons that are in these bonds moving in those orbitals."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And so there's no dipole-dipole interaction. There's no hydrogen bonding. The only intermolecular force that's holding two methane molecules together would be London dispersion forces. And so once again, you could think about the electrons that are in these bonds moving in those orbitals. And let's say for the molecule on the left, for a brief transient moment in time, you get a little bit of negative charge on this side of the molecule. So it might turn out to be those electrons have a net negative charge on this side. And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And so once again, you could think about the electrons that are in these bonds moving in those orbitals. And let's say for the molecule on the left, for a brief transient moment in time, you get a little bit of negative charge on this side of the molecule. So it might turn out to be those electrons have a net negative charge on this side. And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive. And so there could be a very, very small bit of attraction between these two methane molecules. It's very weak, which is why London dispersion forces are the weakest intermolecular forces. But it is there."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And then for this molecule, the electrons could be moving the opposite direction, giving this a partial positive. And so there could be a very, very small bit of attraction between these two methane molecules. It's very weak, which is why London dispersion forces are the weakest intermolecular forces. But it is there. And that's the only thing that's holding together these methane molecules. And since it's weak, we would expect the boiling point for methane to be extremely low. And of course, it is."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "But it is there. And that's the only thing that's holding together these methane molecules. And since it's weak, we would expect the boiling point for methane to be extremely low. And of course, it is. So the boiling point for methane is somewhere around negative, 164 degrees Celsius. And so since room temperature is somewhere around 20 to 25, obviously methane has already boiled, if you will, and turned into a gas. So methane is obviously a gas at room temperature and pressure."}, {"video_title": "Intermolecular forces Chemistry of life Biology Khan Academy.mp3", "Sentence": "And of course, it is. So the boiling point for methane is somewhere around negative, 164 degrees Celsius. And so since room temperature is somewhere around 20 to 25, obviously methane has already boiled, if you will, and turned into a gas. So methane is obviously a gas at room temperature and pressure. So methane is a gas. Now, if you increase the number of carbons, you're going to increase the number of attractive forces that are possible. And if you do that, you can actually increase the boiling point of other hydrocarbons dramatically."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Here's the general reaction for the nitration of benzene. So you start off with benzene, and to it you add concentrated nitric and concentrated sulfuric acids. And that puts a nitro group onto your benzene ring in place of this proton. Let's look at the mechanism for the nitration of benzene. So we start over here with the dot structure for nitric acid. And here's the dot structure for sulfuric acid. And sulfuric acid is actually a stronger acid than nitric acid."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the mechanism for the nitration of benzene. So we start over here with the dot structure for nitric acid. And here's the dot structure for sulfuric acid. And sulfuric acid is actually a stronger acid than nitric acid. So the first step is sulfuric acid is going to function as a Bronsted-Lowry acid and donate a proton. And so I'm going to say it's this proton right here. And nitric acid is actually going to function as a base and accept that proton."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And sulfuric acid is actually a stronger acid than nitric acid. So the first step is sulfuric acid is going to function as a Bronsted-Lowry acid and donate a proton. And so I'm going to say it's this proton right here. And nitric acid is actually going to function as a base and accept that proton. So we can go ahead and show this lone pair of electrons picking up this proton and these two electrons in here remaining behind on that oxygen. So let's go ahead and show that acid-base reaction. So we would have our compound over here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And nitric acid is actually going to function as a base and accept that proton. So we can go ahead and show this lone pair of electrons picking up this proton and these two electrons in here remaining behind on that oxygen. So let's go ahead and show that acid-base reaction. So we would have our compound over here. Let's go ahead and draw in these atoms. So this oxygen is still going to have a negative 1 formal charge on it. And then over here on the right, this oxygen already had one hydrogen bonded to it."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our compound over here. Let's go ahead and draw in these atoms. So this oxygen is still going to have a negative 1 formal charge on it. And then over here on the right, this oxygen already had one hydrogen bonded to it. It just picked up another proton. So it still has one lone pair of electrons, which gives that a plus 1 formal charge. So let me go ahead and highlight these electrons here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then over here on the right, this oxygen already had one hydrogen bonded to it. It just picked up another proton. So it still has one lone pair of electrons, which gives that a plus 1 formal charge. So let me go ahead and highlight these electrons here. So I'm saying that this lone pair of electrons right here picked up a proton right like that, which gives that oxygen a plus 1 formal charge. And notice that forms water as a leaving group. And so if these electrons were to move in here, that would kick these electrons in here off onto the oxygen."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight these electrons here. So I'm saying that this lone pair of electrons right here picked up a proton right like that, which gives that oxygen a plus 1 formal charge. And notice that forms water as a leaving group. And so if these electrons were to move in here, that would kick these electrons in here off onto the oxygen. So let's go ahead and draw the result of that. So we would have water that left. So let's go ahead and show H2O over here on the right."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if these electrons were to move in here, that would kick these electrons in here off onto the oxygen. So let's go ahead and draw the result of that. So we would have water that left. So let's go ahead and show H2O over here on the right. So go ahead and draw in those electrons like that. So I'm saying that these electrons in here in magenta are the ones that came off onto the water molecule, so water as our leaving group. And if water leaves, that leaves what's called the nitronium ion behind."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show H2O over here on the right. So go ahead and draw in those electrons like that. So I'm saying that these electrons in here in magenta are the ones that came off onto the water molecule, so water as our leaving group. And if water leaves, that leaves what's called the nitronium ion behind. So let me go ahead and draw the nitronium ion here so it looks like this. Now let me go ahead and highlight those electrons. So I'm saying that these electrons in here moved in to form a pi bond."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if water leaves, that leaves what's called the nitronium ion behind. So let me go ahead and draw the nitronium ion here so it looks like this. Now let me go ahead and highlight those electrons. So I'm saying that these electrons in here moved in to form a pi bond. So I'm saying it could be represented by those electrons down here on my structure. Now for formal charges, this nitrogen has a plus 1 formal charge. And that nitrogen is actually sp hybridized, which makes the nitronium ion linear."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'm saying that these electrons in here moved in to form a pi bond. So I'm saying it could be represented by those electrons down here on my structure. Now for formal charges, this nitrogen has a plus 1 formal charge. And that nitrogen is actually sp hybridized, which makes the nitronium ion linear. And the nitronium ion is positively charged. This is going to be the electrophile in our mechanism. So this is our electrophile, our positively charged ion."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that nitrogen is actually sp hybridized, which makes the nitronium ion linear. And the nitronium ion is positively charged. This is going to be the electrophile in our mechanism. So this is our electrophile, our positively charged ion. So up here, if we think about it, we would also create the conjugate base to sulfuric acid. So HSO4 minus would be over here. All right, so let's go ahead and show what happens now that we've formed our electrophile."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is our electrophile, our positively charged ion. So up here, if we think about it, we would also create the conjugate base to sulfuric acid. So HSO4 minus would be over here. All right, so let's go ahead and show what happens now that we've formed our electrophile. So the electrophile is going to add to the benzene ring. So we're going to have a nucleophile electrophile reaction. So let's go ahead and draw our benzene ring over here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's go ahead and show what happens now that we've formed our electrophile. So the electrophile is going to add to the benzene ring. So we're going to have a nucleophile electrophile reaction. So let's go ahead and draw our benzene ring over here. And I draw in my hydrogen on my benzene ring. We've now formed our nitronium ion. So the point of the catalyst was to produce our electrophile here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw our benzene ring over here. And I draw in my hydrogen on my benzene ring. We've now formed our nitronium ion. So the point of the catalyst was to produce our electrophile here. So we have a positively charged nitronium ion like that. So now we have a nucleophile electrophile situation, where once again, the pi electrons in our benzene ring are going to function as a nucleophile and attack our electrophile. So those electrons are going to attack that positively charged nitrogen, which would kick these electrons in here off onto that oxygen."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the point of the catalyst was to produce our electrophile here. So we have a positively charged nitronium ion like that. So now we have a nucleophile electrophile situation, where once again, the pi electrons in our benzene ring are going to function as a nucleophile and attack our electrophile. So those electrons are going to attack that positively charged nitrogen, which would kick these electrons in here off onto that oxygen. So let's go ahead and draw the result of our nucleophilic attack. So we have our ring right here. We have a hydrogen."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So those electrons are going to attack that positively charged nitrogen, which would kick these electrons in here off onto that oxygen. So let's go ahead and draw the result of our nucleophilic attack. So we have our ring right here. We have a hydrogen. And once again, just for convention's sake, I'm going to show our electrophile adding to that top carbon of the double bond there, of what used to be the double bond. And so now we have a nitrogen there. We have a nitrogen double bonded to our top oxygen."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have a hydrogen. And once again, just for convention's sake, I'm going to show our electrophile adding to that top carbon of the double bond there, of what used to be the double bond. And so now we have a nitrogen there. We have a nitrogen double bonded to our top oxygen. And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have a nitrogen double bonded to our top oxygen. And then over here, we would have an oxygen with three lone pairs of electrons, giving that a negative 1 formal charge. And the nitrogen, of course, is still going to have a plus 1 formal charge like that. All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge. Nucleophile, electrophile, and those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "All right, let me go ahead and highlight those electrons. So once again, these pi electrons are going to be attracted to the positive charge. Nucleophile, electrophile, and those pi electrons are going to form this bond right here to our nitro group. Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Well, once again, as we've seen several times before, we took away a bond from this carbon. So that's where our plus 1 formal charge is going to go like that. And so we can draw some resonance structures. So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show a resonance structure for this. We could move these pi electrons in over here. So let's go ahead and draw that. So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. These pi electrons over here are still there."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we had a hydrogen up here. And you could just show a nitro group as NO2. So I'm just going to go ahead and do that to save some time. These pi electrons over here are still there. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "These pi electrons over here are still there. And I'm saying that those pi electrons moved over to here. So let me go ahead and highlight those. So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons in blue move over to here, took a bond away from that carbon. So now we can put a plus 1 formal charge at that carbon like that. We can draw yet another resonance structure. So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could show these electrons over here moving to here. So let me go ahead and draw that. So we have our ring. We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our nitro group already on our ring. We have some pi electrons right here. And we have some more pi electrons moving from here to here, which, of course, takes a bond away from this top carbon. So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, so deprotonation of our sigma complex."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's where our positive 1 formal charge is now. So now we have our three resonance structures. And remember, once again, that the sigma complex is a hybrid of these three. And we're now ready for our last step, so deprotonation of our sigma complex. So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're now ready for our last step, so deprotonation of our sigma complex. So if we go back up to here, we think, what could function as a base? Well, the water molecule here could function as a base. So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on our water molecule are going to take that proton, which would cause these electrons to move in here to reform your aromatic ring. So let's go ahead and show that. So we're going to reform our benzene ring here. And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we took off the proton. So deprotonation of the sigma complex yields our product with a nitro group substituted in. So let me go ahead and highlight those electrons again. So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring. And we have created our product. We have added in our nitro group."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this time I'll use green. So these electrons right in here, when that sigma complex is deprotonated, those electrons are going to move in here to restore the aromatic ring. And we have created our product. We have added in our nitro group. So that's the mechanism for nitration. Now, once you form a benzene ring with a nitro group on it, sometimes when you're doing synthesis, it's helpful to turn that nitro group into an amine. So let's go ahead and real quickly look at another useful reaction here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have added in our nitro group. So that's the mechanism for nitration. Now, once you form a benzene ring with a nitro group on it, sometimes when you're doing synthesis, it's helpful to turn that nitro group into an amine. So let's go ahead and real quickly look at another useful reaction here. So once you form your nitro group like that, you can turn that nitro group into an amine a couple of different ways. So one of the classic ways to reduce a nitro group would be in the first step to use either iron or tin. And a source of protons, something like HCl, will work well."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and real quickly look at another useful reaction here. So once you form your nitro group like that, you can turn that nitro group into an amine a couple of different ways. So one of the classic ways to reduce a nitro group would be in the first step to use either iron or tin. And a source of protons, something like HCl, will work well. And since you're doing this in an acidic environment, in the second step you would need to neutralize it with something like sodium hydroxide. And those steps will reduce the nitro group to an amine. So let me go ahead and show the product here."}, {"video_title": "Nitration Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And a source of protons, something like HCl, will work well. And since you're doing this in an acidic environment, in the second step you would need to neutralize it with something like sodium hydroxide. And those steps will reduce the nitro group to an amine. So let me go ahead and show the product here. So instead of an NO2 on the ring, now you have an NH2. So of course, this would be the aniline molecule. And there are other ways to reduce a nitro group to an amine."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The n plus one rule allows us to predict how many peaks we would expect to see for a signal in an NMR spectrum. So if we think about the signal for one proton, if that proton has n neighboring protons, we would expect to see n plus one peaks on the NMR spectrum. The n plus one rule only applies when the neighboring protons are chemically equivalent to each other. So in the last video, we looked at this molecule and we focused in on this proton right here in red, and we said that this signal right here was due to the proton in red, and we also talked about this proton right here in blue, and this signal's due to the proton in blue. Let's look at the proton in red. All right, so let's see how many neighboring protons we have. So if you think about this carbon, right, and you think about the carbon next door, so this carbon right here, so this proton is a neighbor, right?"}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in the last video, we looked at this molecule and we focused in on this proton right here in red, and we said that this signal right here was due to the proton in red, and we also talked about this proton right here in blue, and this signal's due to the proton in blue. Let's look at the proton in red. All right, so let's see how many neighboring protons we have. So if you think about this carbon, right, and you think about the carbon next door, so this carbon right here, so this proton is a neighbor, right? So we have one neighbor so far. We go over here to this carbon, this next door carbon here. There are no protons on that carbon."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you think about this carbon, right, and you think about the carbon next door, so this carbon right here, so this proton is a neighbor, right? So we have one neighbor so far. We go over here to this carbon, this next door carbon here. There are no protons on that carbon. So we have a total of only one neighboring proton for the red proton, so one neighboring proton, so n is equal to one. So we're gonna see one plus one peaks, so one plus one is equal to two. We expect to see two peaks for the signal for the red proton, all right?"}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "There are no protons on that carbon. So we have a total of only one neighboring proton for the red proton, so one neighboring proton, so n is equal to one. So we're gonna see one plus one peaks, so one plus one is equal to two. We expect to see two peaks for the signal for the red proton, all right? So here's the signal, and we see our two peaks, so this is called a doublet. So the signal for the red proton is split into two peaks because of the presence of the neighboring blue proton. All right, let's do the same thing for the blue proton."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We expect to see two peaks for the signal for the red proton, all right? So here's the signal, and we see our two peaks, so this is called a doublet. So the signal for the red proton is split into two peaks because of the presence of the neighboring blue proton. All right, let's do the same thing for the blue proton. All right, so we think about the signal for the blue proton. How many neighboring protons do we have? Well, if we go to the carbon next door, so this carbon, there are no protons on this carbon, and on this carbon, we of course have one proton, so only one neighbor, so n is equal to one."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's do the same thing for the blue proton. All right, so we think about the signal for the blue proton. How many neighboring protons do we have? Well, if we go to the carbon next door, so this carbon, there are no protons on this carbon, and on this carbon, we of course have one proton, so only one neighbor, so n is equal to one. We're gonna see n plus one peaks, so one plus one is equal to two. We expect to see two peaks, so we go over here to the signal for the blue proton, and we see our two peaks, so we get a doublet. One neighbor gives us a doublet here."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, if we go to the carbon next door, so this carbon, there are no protons on this carbon, and on this carbon, we of course have one proton, so only one neighbor, so n is equal to one. We're gonna see n plus one peaks, so one plus one is equal to two. We expect to see two peaks, so we go over here to the signal for the blue proton, and we see our two peaks, so we get a doublet. One neighbor gives us a doublet here. What about the protons over here, these protons in magenta, all right? So how many neighbors do we have for those protons? Go over here to the oxygen, all right?"}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One neighbor gives us a doublet here. What about the protons over here, these protons in magenta, all right? So how many neighbors do we have for those protons? Go over here to the oxygen, all right? No protons on that, no neighbors, so n is equal to zero, n plus one peaks, so zero plus one is equal to one. We'd expect only one peak for the signal for these three equivalent protons, and of course, this is the signal, right? Only one peak, and we call this a singlet."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Go over here to the oxygen, all right? No protons on that, no neighbors, so n is equal to zero, n plus one peaks, so zero plus one is equal to one. We'd expect only one peak for the signal for these three equivalent protons, and of course, this is the signal, right? Only one peak, and we call this a singlet. All right, let's do another one. So we also saw this molecule in the last video, right? So over here, and let's look at our protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Only one peak, and we call this a singlet. All right, let's do another one. So we also saw this molecule in the last video, right? So over here, and let's look at our protons. We expect one signal for the blue proton, and we would expect one signal for these red protons here. Let's think about the red protons first. So how many neighboring protons do we have, right?"}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So over here, and let's look at our protons. We expect one signal for the blue proton, and we would expect one signal for these red protons here. Let's think about the red protons first. So how many neighboring protons do we have, right? So we go to the carbon next door, and then we have one proton, so one neighbor. So n is equal to one, so we expect one plus one peaks, so we expect two peaks. So the signal for the red protons needs to have two peaks, so this one right here, so there's one peak, and there's the second peak, so we get a doublet."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So how many neighboring protons do we have, right? So we go to the carbon next door, and then we have one proton, so one neighbor. So n is equal to one, so we expect one plus one peaks, so we expect two peaks. So the signal for the red protons needs to have two peaks, so this one right here, so there's one peak, and there's the second peak, so we get a doublet. All right, what about the signal for the blue proton, right? So we get a signal for the blue proton. How many neighboring protons do we have?"}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the signal for the red protons needs to have two peaks, so this one right here, so there's one peak, and there's the second peak, so we get a doublet. All right, what about the signal for the blue proton, right? So we get a signal for the blue proton. How many neighboring protons do we have? So we go to the carbon next door, and we have one, two neighbors. So for the blue proton, we have two neighbors. N is equal to two."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "How many neighboring protons do we have? So we go to the carbon next door, and we have one, two neighbors. So for the blue proton, we have two neighbors. N is equal to two. Expect n plus one peaks, so two plus one is equal to three. So three peaks for this signal. So this signal is right here, and we get one, two, and three."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "N is equal to two. Expect n plus one peaks, so two plus one is equal to three. So three peaks for this signal. So this signal is right here, and we get one, two, and three. Three peaks, which is called a triplet. All right, let's go to the next one. So let's look at this one right here."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this signal is right here, and we get one, two, and three. Three peaks, which is called a triplet. All right, let's go to the next one. So let's look at this one right here. Okay, so we have bromoethane. So let's first draw in the protons. So on this carbon, we have two protons, and on this carbon, we have three protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this one right here. Okay, so we have bromoethane. So let's first draw in the protons. So on this carbon, we have two protons, and on this carbon, we have three protons. All right, so let's start with the signal, let's start with the signal for these protons right here. So these are equivalent. We would expect one signal for those protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So on this carbon, we have two protons, and on this carbon, we have three protons. All right, so let's start with the signal, let's start with the signal for these protons right here. So these are equivalent. We would expect one signal for those protons. How many neighbors do those protons have? All right, we go to the carbon next door, and we have one, two. We have two neighbors."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We would expect one signal for those protons. How many neighbors do those protons have? All right, we go to the carbon next door, and we have one, two. We have two neighbors. So n is equal to two. We would expect n plus one peaks. So two plus one is equal to three."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have two neighbors. So n is equal to two. We would expect n plus one peaks. So two plus one is equal to three. We would expect a triplet for this signal, and here is our triplet, right? One, two, and three peaks for that one. Next, let's think about the signal for these two protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So two plus one is equal to three. We would expect a triplet for this signal, and here is our triplet, right? One, two, and three peaks for that one. Next, let's think about the signal for these two protons. All right, so how many neighbors do those two protons have? So we go to the carbon next door. One, two, three neighbors."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's think about the signal for these two protons. All right, so how many neighbors do those two protons have? So we go to the carbon next door. One, two, three neighbors. So n is equal to three. So three plus one is equal to four. We would expect a signal with four peaks for the protons in blue, and here's our signal with four peaks."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three neighbors. So n is equal to three. So three plus one is equal to four. We would expect a signal with four peaks for the protons in blue, and here's our signal with four peaks. One, two, three, and four. So we call this a quartet. All right, so four peaks, we call it a quartet."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We would expect a signal with four peaks for the protons in blue, and here's our signal with four peaks. One, two, three, and four. So we call this a quartet. All right, so four peaks, we call it a quartet. Let's do another one. Let's look at this molecule right here. All right, and let's first draw in the protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so four peaks, we call it a quartet. Let's do another one. Let's look at this molecule right here. All right, and let's first draw in the protons. All right, so on this carbon, there are three protons. So here are the three protons. On this carbon, there are three protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, and let's first draw in the protons. All right, so on this carbon, there are three protons. So here are the three protons. On this carbon, there are three protons. And on this carbon, there would only be one proton. So here's this one. Let's think about how many signals we would expect to see."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "On this carbon, there are three protons. And on this carbon, there would only be one proton. So here's this one. Let's think about how many signals we would expect to see. So not thinking about the n plus one rule or spin-spin splitting here. So a signal for these protons. These protons are in the same environment as these protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about how many signals we would expect to see. So not thinking about the n plus one rule or spin-spin splitting here. So a signal for these protons. These protons are in the same environment as these protons. So that's one signal for those. And then we would expect a signal for this proton right here in a different environment. All right, let's think about the signal for the red protons first."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These protons are in the same environment as these protons. So that's one signal for those. And then we would expect a signal for this proton right here in a different environment. All right, let's think about the signal for the red protons first. How many neighboring protons do the red protons have? We go to the carbon next door, and this is the carbon next door for both of those red protons. And there's one neighbor."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's think about the signal for the red protons first. How many neighboring protons do the red protons have? We go to the carbon next door, and this is the carbon next door for both of those red protons. And there's one neighbor. So one neighbor, n is equal to one. So one plus one is equal to two. We would expect a doublet for this signal."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And there's one neighbor. So one neighbor, n is equal to one. So one plus one is equal to two. We would expect a doublet for this signal. And so that must be this. So this is kind of like a zoom-in. And it's not the exact same drawing."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We would expect a doublet for this signal. And so that must be this. So this is kind of like a zoom-in. And it's not the exact same drawing. I hand-drew all this stuff, so it's not exactly perfect. But you can see there are two peaks here. Two peaks, so this is a zoom-in."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it's not the exact same drawing. I hand-drew all this stuff, so it's not exactly perfect. But you can see there are two peaks here. Two peaks, so this is a zoom-in. This is supposed to represent a zoom-in of those two peaks there. So that's that signal. What about the blue proton?"}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Two peaks, so this is a zoom-in. This is supposed to represent a zoom-in of those two peaks there. So that's that signal. What about the blue proton? So how many neighbors does the blue proton have? So we look at this carbon, and we look at the next door carbon. So this is the carbon next door, one, two, three."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What about the blue proton? So how many neighbors does the blue proton have? So we look at this carbon, and we look at the next door carbon. So this is the carbon next door, one, two, three. This carbon is also next door to this carbon. So one, two, three, so a total of six. So n is equal to six, so expect n plus one peak."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is the carbon next door, one, two, three. This carbon is also next door to this carbon. So one, two, three, so a total of six. So n is equal to six, so expect n plus one peak. So six plus one is equal to seven. So we'd expect seven peaks, called a septet. Let me go ahead and rewrite that here."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So n is equal to six, so expect n plus one peak. So six plus one is equal to seven. So we'd expect seven peaks, called a septet. Let me go ahead and rewrite that here. So let me go ahead and write septet here. And let's look at this signal. It must be this right here."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and rewrite that here. So let me go ahead and write septet here. And let's look at this signal. It must be this right here. And this is pretty hard to draw. And once again, this is a blow-up of what you're looking at. So we'd expect seven peaks."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It must be this right here. And this is pretty hard to draw. And once again, this is a blow-up of what you're looking at. So we'd expect seven peaks. So one, two, three, four, five, six, and seven. So that's the idea of the n plus one rule. And let's talk a little bit more about spin-spin splitting and when to expect a signal."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we'd expect seven peaks. So one, two, three, four, five, six, and seven. So that's the idea of the n plus one rule. And let's talk a little bit more about spin-spin splitting and when to expect a signal. So if you have chemically equivalent protons, they don't show spin-spin splitting. So if I look at these two protons right here, these two protons are in the same environment as these two protons, right? Think about symmetry."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's talk a little bit more about spin-spin splitting and when to expect a signal. So if you have chemically equivalent protons, they don't show spin-spin splitting. So if I look at these two protons right here, these two protons are in the same environment as these two protons, right? Think about symmetry. So we'd expect only one signal. We'd expect only one signal on the NMR spectrum. And these protons are not going to split these protons even though they're next door because they're chemically equivalent to each other."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Think about symmetry. So we'd expect only one signal. We'd expect only one signal on the NMR spectrum. And these protons are not going to split these protons even though they're next door because they're chemically equivalent to each other. So chemically equivalent protons do not show spin-spin splitting. All right, let's look at these examples right here. So what we've been talking about is next door protons."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And these protons are not going to split these protons even though they're next door because they're chemically equivalent to each other. So chemically equivalent protons do not show spin-spin splitting. All right, let's look at these examples right here. So what we've been talking about is next door protons. So this proton and this proton will split each other's signal if they're in different environments. So splitting is observed. We're talking about these next door protons here."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So what we've been talking about is next door protons. So this proton and this proton will split each other's signal if they're in different environments. So splitting is observed. We're talking about these next door protons here. But if you have this situation, so let's just once again make this the red proton and make this the blue proton, there's an extra carbon in between. So splitting is generally not observed for this situation. So the red and the blue protons won't split each other generally."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're talking about these next door protons here. But if you have this situation, so let's just once again make this the red proton and make this the blue proton, there's an extra carbon in between. So splitting is generally not observed for this situation. So the red and the blue protons won't split each other generally. Again, there are exceptions. But for simple NMR spectra, you really have to think about the situation on the left here. So these protons are too far apart for them to feel any effect."}, {"video_title": "Multiplicity n + 1 rule Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the red and the blue protons won't split each other generally. Again, there are exceptions. But for simple NMR spectra, you really have to think about the situation on the left here. So these protons are too far apart for them to feel any effect. And then finally, it is possible to have splitting from protons on the same carbon. So if these are in different environments, so if this proton is in a different environment from this proton, it's possible for splitting to be observed. And we'll talk more about that in the next video."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so chemists knew that benzene contained six carbons and six hydrogens, but they weren't sure exactly how those atoms were connected. And so there were several different proposals for structures for benzene. And the winning structure was proposed by Auguste Kekul\u00e9, who said that benzene contains six carbons in a ring, and you have alternating single and double bonds in that ring. And the story goes that Kekul\u00e9 came up with this dot structure when he had a dream, and he saw some snakes bite each other's tails, and the snakes then whirled around in a circle. And that gave Kekul\u00e9 the idea for the ring. So this is one possible dot structure for benzene, but I didn't have to draw my double bonds in this place. I can actually show a resonance structure for benzene."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the story goes that Kekul\u00e9 came up with this dot structure when he had a dream, and he saw some snakes bite each other's tails, and the snakes then whirled around in a circle. And that gave Kekul\u00e9 the idea for the ring. So this is one possible dot structure for benzene, but I didn't have to draw my double bonds in this place. I can actually show a resonance structure for benzene. I could take these electrons, move them over here, push these electrons over here, and then these electrons would be over here. And so a resonance structure for benzene. I could have my pi electrons over here, over here, and over here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I can actually show a resonance structure for benzene. I could take these electrons, move them over here, push these electrons over here, and then these electrons would be over here. And so a resonance structure for benzene. I could have my pi electrons over here, over here, and over here. So either one of these Kekul\u00e9 structures is an acceptable dot structure for benzene. And remember, in reality, since these are resonance structures, the actual molecule is more of a hybrid of these two molecules. And with that in mind, sometimes chemists will prefer to represent benzene with the six carbons in a ring and with a circle here in the center to represent the delocalization of those pi electrons."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I could have my pi electrons over here, over here, and over here. So either one of these Kekul\u00e9 structures is an acceptable dot structure for benzene. And remember, in reality, since these are resonance structures, the actual molecule is more of a hybrid of these two molecules. And with that in mind, sometimes chemists will prefer to represent benzene with the six carbons in a ring and with a circle here in the center to represent the delocalization of those pi electrons. And so maybe this is the lazy way to represent benzene. And this is called a Robinson circle after the great synthetic organic chemist Sir Robert Robinson. And so sometimes the Robinson circle is a useful way to represent benzene."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And with that in mind, sometimes chemists will prefer to represent benzene with the six carbons in a ring and with a circle here in the center to represent the delocalization of those pi electrons. And so maybe this is the lazy way to represent benzene. And this is called a Robinson circle after the great synthetic organic chemist Sir Robert Robinson. And so sometimes the Robinson circle is a useful way to represent benzene. And sometimes a Kekul\u00e9 structure is over here on the left. And you would use a Kekul\u00e9 structure if you're trying to show the mechanism of reactions that benzene does. All right, let's look at how to name derivatives of benzene."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so sometimes the Robinson circle is a useful way to represent benzene. And sometimes a Kekul\u00e9 structure is over here on the left. And you would use a Kekul\u00e9 structure if you're trying to show the mechanism of reactions that benzene does. All right, let's look at how to name derivatives of benzene. So here we have benzene with a methyl group coming off of it. And so you could just call this molecule methylbenzene. So that's one possible name for it."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at how to name derivatives of benzene. So here we have benzene with a methyl group coming off of it. And so you could just call this molecule methylbenzene. So that's one possible name for it. But most people don't call it methylbenzene. It's called toluene. So it's such a common molecule in organic chemistry that toluene is an acceptable IUPAC name."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's one possible name for it. But most people don't call it methylbenzene. It's called toluene. So it's such a common molecule in organic chemistry that toluene is an acceptable IUPAC name. And in this example, we have an alkyl substituent that has only one carbon. This methyl group has only one carbon versus the six in the benzene ring. Well, what happened if your alkyl group had more carbons than your benzene ring, which is the situation over here on the right?"}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's such a common molecule in organic chemistry that toluene is an acceptable IUPAC name. And in this example, we have an alkyl substituent that has only one carbon. This methyl group has only one carbon versus the six in the benzene ring. Well, what happened if your alkyl group had more carbons than your benzene ring, which is the situation over here on the right? So if we count up how many carbons we have, we have a total of seven carbons. And so in this case, we're actually going to name this as an alkane and name the benzene ring as a substituent coming off of our alkane. So a seven-carbon alkane would be called heptane."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Well, what happened if your alkyl group had more carbons than your benzene ring, which is the situation over here on the right? So if we count up how many carbons we have, we have a total of seven carbons. And so in this case, we're actually going to name this as an alkane and name the benzene ring as a substituent coming off of our alkane. So a seven-carbon alkane would be called heptane. So I can go ahead and write heptane here. Then I have a benzene ring coming off of carbon 4. And when you're naming a benzene ring as a substituent, so it would be C6H5, substituent coming off of my ring here, we call it a phenyl group."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So a seven-carbon alkane would be called heptane. So I can go ahead and write heptane here. Then I have a benzene ring coming off of carbon 4. And when you're naming a benzene ring as a substituent, so it would be C6H5, substituent coming off of my ring here, we call it a phenyl group. So I have a phenyl group coming off of carbon 4. So this would be 4-phenylheptane as the IUPAC name for this molecule. Let's look at some other examples where we see benzene with one group on the ring."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And when you're naming a benzene ring as a substituent, so it would be C6H5, substituent coming off of my ring here, we call it a phenyl group. So I have a phenyl group coming off of carbon 4. So this would be 4-phenylheptane as the IUPAC name for this molecule. Let's look at some other examples where we see benzene with one group on the ring. And these are all very famous monosubstituted benzenes. And because they're so famous, their common name is acceptable in IUPAC nomenclature. So a benzene ring with an OH group on it is called phenol."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at some other examples where we see benzene with one group on the ring. And these are all very famous monosubstituted benzenes. And because they're so famous, their common name is acceptable in IUPAC nomenclature. So a benzene ring with an OH group on it is called phenol. And we can use that when we're naming molecules. Benzaldehyde would be an aldehyde coming off of a benzene ring. And benzaldehyde is, of course, famous for having the smell of almonds."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So a benzene ring with an OH group on it is called phenol. And we can use that when we're naming molecules. Benzaldehyde would be an aldehyde coming off of a benzene ring. And benzaldehyde is, of course, famous for having the smell of almonds. It's a really, really wonderful smell, which is also why you'll hear some of these molecules referred to as aromatic compounds. And it originally is because of the smell. And we're going to see in future videos what aromatic means in a chemistry sense."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And benzaldehyde is, of course, famous for having the smell of almonds. It's a really, really wonderful smell, which is also why you'll hear some of these molecules referred to as aromatic compounds. And it originally is because of the smell. And we're going to see in future videos what aromatic means in a chemistry sense. Over here we have benzoic acid, a carboxylic acid functional group coming off of our benzene ring here like that. And so here I have seven of the most famous and most common monosubstituted benzene derivatives. And so these are molecules that most professors will have you memorize because you can use these names when you're trying to name more complicated benzene derivatives."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to see in future videos what aromatic means in a chemistry sense. Over here we have benzoic acid, a carboxylic acid functional group coming off of our benzene ring here like that. And so here I have seven of the most famous and most common monosubstituted benzene derivatives. And so these are molecules that most professors will have you memorize because you can use these names when you're trying to name more complicated benzene derivatives. So commit these to memory. All right, let's look at some disubstituted benzene derivatives. So this molecule over here on the left has two methyl groups coming off of the benzene ring."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so these are molecules that most professors will have you memorize because you can use these names when you're trying to name more complicated benzene derivatives. So commit these to memory. All right, let's look at some disubstituted benzene derivatives. So this molecule over here on the left has two methyl groups coming off of the benzene ring. And so this first molecule has a methyl group coming off of carbon 1 and coming off of carbon 2. So we could call this 1, 2-dimethylbenzene. That would be an acceptable IUPAC name."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this molecule over here on the left has two methyl groups coming off of the benzene ring. And so this first molecule has a methyl group coming off of carbon 1 and coming off of carbon 2. So we could call this 1, 2-dimethylbenzene. That would be an acceptable IUPAC name. But whenever you have a benzene ring with two methyl groups on it, the common name for that is xylene. And so if we wanted to call this molecule xylene, technically all three of these molecules would be xylene. They're all benzene ring with two methyl groups coming off of it."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That would be an acceptable IUPAC name. But whenever you have a benzene ring with two methyl groups on it, the common name for that is xylene. And so if we wanted to call this molecule xylene, technically all three of these molecules would be xylene. They're all benzene ring with two methyl groups coming off of it. And so we have to distinguish these xylenes from each other. And so when you have two groups that are right next to each other on a benzene ring, in this case my methyl groups, my methyl groups are right next to each other on my benzene ring, we say that that relationship is ortho. So I could call this molecule ortho xylene, and that would be another acceptable name."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "They're all benzene ring with two methyl groups coming off of it. And so we have to distinguish these xylenes from each other. And so when you have two groups that are right next to each other on a benzene ring, in this case my methyl groups, my methyl groups are right next to each other on my benzene ring, we say that that relationship is ortho. So I could call this molecule ortho xylene, and that would be another acceptable name. And sometimes you'll just see an O there. So you call it O xylene, and that's fine too. We go over here to this molecule."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could call this molecule ortho xylene, and that would be another acceptable name. And sometimes you'll just see an O there. So you call it O xylene, and that's fine too. We go over here to this molecule. This is also xylene, but we can see the methyl groups are in slightly different positions. Right now we would have 1, 3-dimethylbenzene. So 1, 3-dimethylbenzene would be an acceptable name for this molecule."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We go over here to this molecule. This is also xylene, but we can see the methyl groups are in slightly different positions. Right now we would have 1, 3-dimethylbenzene. So 1, 3-dimethylbenzene would be an acceptable name for this molecule. And when you have two groups that are carbon away from each other, so this relationship is said to be meta in organic chemistry. So you could call this meta. Let me take off that M here, so I'm running out of room here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 3-dimethylbenzene would be an acceptable name for this molecule. And when you have two groups that are carbon away from each other, so this relationship is said to be meta in organic chemistry. So you could call this meta. Let me take off that M here, so I'm running out of room here. You could call this meta xylene. So this is meta xylene, or just M xylene. And then finally, another xylene molecule."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me take off that M here, so I'm running out of room here. You could call this meta xylene. So this is meta xylene, or just M xylene. And then finally, another xylene molecule. This time our two methyl groups would be at carbons 1 and 4. So you call this 1, 4-dimethylbenzene. So let me go ahead and write that."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, another xylene molecule. This time our two methyl groups would be at carbons 1 and 4. So you call this 1, 4-dimethylbenzene. So let me go ahead and write that. And 1, 4-dimethylbenzene would be one IUPAC name for it. But again, most people would name this as a xylene derivative. So it's a xylene derivative."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write that. And 1, 4-dimethylbenzene would be one IUPAC name for it. But again, most people would name this as a xylene derivative. So it's a xylene derivative. This time my two groups are opposite each other. So they're opposite each other on the ring, and we call this relationship para in organic chemistry. So you could say it's para xylene or also p xylene."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's a xylene derivative. This time my two groups are opposite each other. So they're opposite each other on the ring, and we call this relationship para in organic chemistry. So you could say it's para xylene or also p xylene. Let's look at some more examples of di-substituted benzene rings. And so here we go. We're actually going to use the monosubstituted derivatives that we talked about above."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could say it's para xylene or also p xylene. Let's look at some more examples of di-substituted benzene rings. And so here we go. We're actually going to use the monosubstituted derivatives that we talked about above. So if I look at this molecule over here on the left, I can see that this is the phenol portion of the molecule. So I can go ahead and say that this is phenol. And then in terms of identifying the bromine, I have two options here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We're actually going to use the monosubstituted derivatives that we talked about above. So if I look at this molecule over here on the left, I can see that this is the phenol portion of the molecule. So I can go ahead and say that this is phenol. And then in terms of identifying the bromine, I have two options here. I could use a number. I could say, OK, that bromine is at carbon 4. So I could call this 4-bromophenol, and that's an acceptable IUPAC name."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then in terms of identifying the bromine, I have two options here. I could use a number. I could say, OK, that bromine is at carbon 4. So I could call this 4-bromophenol, and that's an acceptable IUPAC name. Or I could use the OMP system that we talked about above. And when you have two groups that are opposite on the ring, we call that para. So I could also call this molecule para-bromophenol."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could call this 4-bromophenol, and that's an acceptable IUPAC name. Or I could use the OMP system that we talked about above. And when you have two groups that are opposite on the ring, we call that para. So I could also call this molecule para-bromophenol. So para-bromophenol, and that's an acceptable name as well. Or I could even shorten it to P-bromophenol. Let's do this di-substituted benzene ring over here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could also call this molecule para-bromophenol. So para-bromophenol, and that's an acceptable name as well. Or I could even shorten it to P-bromophenol. Let's do this di-substituted benzene ring over here. And if I look at it, I can see that is benzoic acid. That's one of the ones that we memorized above. So I can go ahead and write the parent name as being benzoic acid right here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's do this di-substituted benzene ring over here. And if I look at it, I can see that is benzoic acid. That's one of the ones that we memorized above. So I can go ahead and write the parent name as being benzoic acid right here. And I now have to identify my substituent coming off of my benzene ring. So benzoic acid would make this carbon 1, and then my substituent is coming off of carbon 3, and my substituent is a nitro group. So I could call this 3-nitrobenzoic acid."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and write the parent name as being benzoic acid right here. And I now have to identify my substituent coming off of my benzene ring. So benzoic acid would make this carbon 1, and then my substituent is coming off of carbon 3, and my substituent is a nitro group. So I could call this 3-nitrobenzoic acid. Or I could say that the relationship between those two things coming off my benzene ring would be meta. So I could call this meta-nitrobenzoic acid. So let me see if I have room here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could call this 3-nitrobenzoic acid. Or I could say that the relationship between those two things coming off my benzene ring would be meta. So I could call this meta-nitrobenzoic acid. So let me see if I have room here. So meta-nitrobenzoic acid. Or I could just say M-nitrobenzoic acid. So all of those would be correct IUPAC names."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me see if I have room here. So meta-nitrobenzoic acid. Or I could just say M-nitrobenzoic acid. So all of those would be correct IUPAC names. Let's look at poly-substituted benzene derivatives now. And so these are actually two of the more famous examples that we could do. Once again, we're going to try to find a monosubstituted parent name here."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So all of those would be correct IUPAC names. Let's look at poly-substituted benzene derivatives now. And so these are actually two of the more famous examples that we could do. Once again, we're going to try to find a monosubstituted parent name here. And if I look, I can see that right here. This would be benzaldehyde. And so usually aldehydes have precedence over alcohols."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we're going to try to find a monosubstituted parent name here. And if I look, I can see that right here. This would be benzaldehyde. And so usually aldehydes have precedence over alcohols. And so that's why we're going to name this as a benzaldehyde derivative. So I'm going to go ahead and, let's see, just to make sure I have enough room, I'm going to start naming it by saying the parent name is benzaldehyde right here. And since the aldehyde gets precedence, this would get a carbon 1."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so usually aldehydes have precedence over alcohols. And so that's why we're going to name this as a benzaldehyde derivative. So I'm going to go ahead and, let's see, just to make sure I have enough room, I'm going to start naming it by saying the parent name is benzaldehyde right here. And since the aldehyde gets precedence, this would get a carbon 1. We want to give the lowest number possible to our substituents coming off my ring. So I'm going to go this way. And I can see that I have a group coming off of carbon 3."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And since the aldehyde gets precedence, this would get a carbon 1. We want to give the lowest number possible to our substituents coming off my ring. So I'm going to go this way. And I can see that I have a group coming off of carbon 3. And it's an ether group. So we talked about how to name ethers as substituents. This would be a methoxy substituent."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I can see that I have a group coming off of carbon 3. And it's an ether group. So we talked about how to name ethers as substituents. This would be a methoxy substituent. So this would be coming off of carbon 3. So I'm going to write 3 and methoxy like that. And then I go over here to carbon 4."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This would be a methoxy substituent. So this would be coming off of carbon 3. So I'm going to write 3 and methoxy like that. And then I go over here to carbon 4. And I can see I have an OH group. And if I'm naming OH and alcohol as a substituent, I would call this a hydroxy or a hydroxyl group. And I'm going to say that this is a hydroxy group at carbon 4."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then I go over here to carbon 4. And I can see I have an OH group. And if I'm naming OH and alcohol as a substituent, I would call this a hydroxy or a hydroxyl group. And I'm going to say that this is a hydroxy group at carbon 4. So 4-hydroxy-3-methoxybenzaldehyde would be the IUPAC name for this molecule. This molecule is better known as vanillin. So the smell of vanilla."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to say that this is a hydroxy group at carbon 4. So 4-hydroxy-3-methoxybenzaldehyde would be the IUPAC name for this molecule. This molecule is better known as vanillin. So the smell of vanilla. It's probably my favorite smell. And so I really enjoy doing labs that involve the vanillin molecule. Let's do one more, another extremely famous example."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the smell of vanilla. It's probably my favorite smell. And so I really enjoy doing labs that involve the vanillin molecule. Let's do one more, another extremely famous example. I can see that if I'm trying to find a monosubstituted benzene derivative, that would be toluene right here. So I'm going to go ahead and put toluene as my parent name. And that would make this carbon 1."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more, another extremely famous example. I can see that if I'm trying to find a monosubstituted benzene derivative, that would be toluene right here. So I'm going to go ahead and put toluene as my parent name. And that would make this carbon 1. The methyl group would become carbon 1. So I go ahead and number to give the lowest number possible. And I can see that I have three nitro substituents located at 2, 4, and 6."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that would make this carbon 1. The methyl group would become carbon 1. So I go ahead and number to give the lowest number possible. And I can see that I have three nitro substituents located at 2, 4, and 6. So to finish my IUPAC name, it'd be 2, 4, 6. I have three nitros, so that'd be tri as my prefix. So trinitrotoluene."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I can see that I have three nitro substituents located at 2, 4, and 6. So to finish my IUPAC name, it'd be 2, 4, 6. I have three nitros, so that'd be tri as my prefix. So trinitrotoluene. And so this is also a very famous molecule. It's not normally called 2, 4, 6-trinitrotoluene. Most people in the general public would, of course, know this as TNT, so the famous explosive."}, {"video_title": "Naming benzene derivatives Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So trinitrotoluene. And so this is also a very famous molecule. It's not normally called 2, 4, 6-trinitrotoluene. Most people in the general public would, of course, know this as TNT, so the famous explosive. So this would be TNT. And we just named it. Of course, the name comes from the nomenclature for polysubstituted benzene derivatives."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And even though this idea isn't perfect, it does allow you to predict the solubility of compounds. For example, a polar solvent will dissolve a polar compound in general, so like dissolves like. I also have here, a polar solvent will dissolve an ionic solute, because you don't usually describe ionic compounds as being polar. Next, a non-polar solvent will dissolve a non-polar compound, so like dissolves like. But, a polar solvent will not dissolve a non-polar compound. So this would be like and unlike here. An example of a polar solvent is water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, a non-polar solvent will dissolve a non-polar compound, so like dissolves like. But, a polar solvent will not dissolve a non-polar compound. So this would be like and unlike here. An example of a polar solvent is water. An example of a non-polar compound could be something like oil. And we know that water will not dissolve oil. Let's go back to this first idea of a polar solvent being able to dissolve a polar compound, or a polar solvent dissolving an ionic compound like sodium chloride."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "An example of a polar solvent is water. An example of a non-polar compound could be something like oil. And we know that water will not dissolve oil. Let's go back to this first idea of a polar solvent being able to dissolve a polar compound, or a polar solvent dissolving an ionic compound like sodium chloride. We know from experience that sodium chloride, or salt, is soluble in water. So over here on the left, we have part of a salt crystal. We know that crystals are held together by attractive forces."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's go back to this first idea of a polar solvent being able to dissolve a polar compound, or a polar solvent dissolving an ionic compound like sodium chloride. We know from experience that sodium chloride, or salt, is soluble in water. So over here on the left, we have part of a salt crystal. We know that crystals are held together by attractive forces. The positively charged sodium cation is attracted to the negatively charged chloride anion. So opposite charges attract, and our crystal is held together by these attractive forces. If we get some water molecules to come along, we know that water is a polar solvent."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We know that crystals are held together by attractive forces. The positively charged sodium cation is attracted to the negatively charged chloride anion. So opposite charges attract, and our crystal is held together by these attractive forces. If we get some water molecules to come along, we know that water is a polar solvent. Water is a polar molecule. The oxygen is more electronegative than this hydrogen, so the oxygen pulls some of the electron density in this bond closer to it, giving it a partial negative charge. If we are withdrawing electron density from this hydrogen, this hydrogen gets a partial positive charge."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If we get some water molecules to come along, we know that water is a polar solvent. Water is a polar molecule. The oxygen is more electronegative than this hydrogen, so the oxygen pulls some of the electron density in this bond closer to it, giving it a partial negative charge. If we are withdrawing electron density from this hydrogen, this hydrogen gets a partial positive charge. Since opposite charges attract, the partially positive hydrogen on water is attracted to the negatively charged chloride anion. So there's an interaction here. If we get a bunch of water molecules, here's another one right here, so partially negative oxygen, partially positive hydrogen, so there's another attractive force, we can pull off these chloride anions from the solid and bring the anion into solution."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If we are withdrawing electron density from this hydrogen, this hydrogen gets a partial positive charge. Since opposite charges attract, the partially positive hydrogen on water is attracted to the negatively charged chloride anion. So there's an interaction here. If we get a bunch of water molecules, here's another one right here, so partially negative oxygen, partially positive hydrogen, so there's another attractive force, we can pull off these chloride anions from the solid and bring the anion into solution. So on the right here, we have our chloride anion in solution, surrounded by a bunch of water molecules, and we have all these partially positive hydrogens interacting with our negatively charged chloride anion. For the sodium cations, let's go back to our solid on the left. Since the sodium cation is positively charged, that's going to interact with the partially negatively charged oxygen in the water molecule."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If we get a bunch of water molecules, here's another one right here, so partially negative oxygen, partially positive hydrogen, so there's another attractive force, we can pull off these chloride anions from the solid and bring the anion into solution. So on the right here, we have our chloride anion in solution, surrounded by a bunch of water molecules, and we have all these partially positive hydrogens interacting with our negatively charged chloride anion. For the sodium cations, let's go back to our solid on the left. Since the sodium cation is positively charged, that's going to interact with the partially negatively charged oxygen in the water molecule. So opposite charges attract, and if you get enough water molecules, you can pull off these sodium cations and bring the sodium cations into solution. So we have the partially negative oxygens on water interacting with our positively charged sodium cations in our solution. So our polar solvent water needs to be able to interact with our solutes, and in this case, the polar solvent attacks the solid over here on the left, and it replaces these ion-ion interactions of our crystal with ion-dipole interactions in our solution."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Since the sodium cation is positively charged, that's going to interact with the partially negatively charged oxygen in the water molecule. So opposite charges attract, and if you get enough water molecules, you can pull off these sodium cations and bring the sodium cations into solution. So we have the partially negative oxygens on water interacting with our positively charged sodium cations in our solution. So our polar solvent water needs to be able to interact with our solutes, and in this case, the polar solvent attacks the solid over here on the left, and it replaces these ion-ion interactions of our crystal with ion-dipole interactions in our solution. And by ion-dipole, I mean we have a cation right here, so that's our ion, and then our dipole would be water. Water is a polar molecule. It has a dipole moment."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So our polar solvent water needs to be able to interact with our solutes, and in this case, the polar solvent attacks the solid over here on the left, and it replaces these ion-ion interactions of our crystal with ion-dipole interactions in our solution. And by ion-dipole, I mean we have a cation right here, so that's our ion, and then our dipole would be water. Water is a polar molecule. It has a dipole moment. So we have all of these ion-dipole interactions. So ionic solutes that are able to participate in these interactions will dissolve in water. If you have a polar compound, a similar idea, you have attractive forces that allow the polar compounds to be dissolved in a polar solvent like water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It has a dipole moment. So we have all of these ion-dipole interactions. So ionic solutes that are able to participate in these interactions will dissolve in water. If you have a polar compound, a similar idea, you have attractive forces that allow the polar compounds to be dissolved in a polar solvent like water. Let's move on to a nonpolar compound. So a nonpolar compound is something like this molecule on the left here, and this molecule is called naphthalene. Naphthalene is a solid with a very distinctive smell to it."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If you have a polar compound, a similar idea, you have attractive forces that allow the polar compounds to be dissolved in a polar solvent like water. Let's move on to a nonpolar compound. So a nonpolar compound is something like this molecule on the left here, and this molecule is called naphthalene. Naphthalene is a solid with a very distinctive smell to it. So the first time I smelled naphthalene in the lab, it reminded me of my grandparents' house because my grandparents, when I was a kid, had mothballs that were made of naphthalene, so it's a very distinctive smell. Naphthalene is nonpolar because it's composed of only carbons and hydrogens. It's a hydrocarbon."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Naphthalene is a solid with a very distinctive smell to it. So the first time I smelled naphthalene in the lab, it reminded me of my grandparents' house because my grandparents, when I was a kid, had mothballs that were made of naphthalene, so it's a very distinctive smell. Naphthalene is nonpolar because it's composed of only carbons and hydrogens. It's a hydrocarbon. So naphthalene is nonpolar, and you would need a nonpolar solvent to get it to dissolve. Toluene is a nonpolar solvent. Again, this is a hydrocarbon."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It's a hydrocarbon. So naphthalene is nonpolar, and you would need a nonpolar solvent to get it to dissolve. Toluene is a nonpolar solvent. Again, this is a hydrocarbon. So if you take solid naphthalene and liquid toluene, naphthalene will dissolve in toluene. So like dissolves like. Our nonpolar solvent will dissolve our nonpolar compound."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Again, this is a hydrocarbon. So if you take solid naphthalene and liquid toluene, naphthalene will dissolve in toluene. So like dissolves like. Our nonpolar solvent will dissolve our nonpolar compound. But finally, let's look at this last idea here. So a polar solvent, something like water, should not dissolve a nonpolar compound, something like naphthalene. And that's true."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Our nonpolar solvent will dissolve our nonpolar compound. But finally, let's look at this last idea here. So a polar solvent, something like water, should not dissolve a nonpolar compound, something like naphthalene. And that's true. Naphthalene will not dissolve in water. So water doesn't interact well enough with the naphthalene molecules to get them to dissolve and form a solution. So this concept of like dissolves like is important because it allows you to predict whether or not a compound will be soluble in water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And that's true. Naphthalene will not dissolve in water. So water doesn't interact well enough with the naphthalene molecules to get them to dissolve and form a solution. So this concept of like dissolves like is important because it allows you to predict whether or not a compound will be soluble in water. Let's look at several organic compounds and determine whether or not those compounds are soluble in water. And we'll start with ethanol. Ethanol has a polar oxygen-hydrogen bond."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this concept of like dissolves like is important because it allows you to predict whether or not a compound will be soluble in water. Let's look at several organic compounds and determine whether or not those compounds are soluble in water. And we'll start with ethanol. Ethanol has a polar oxygen-hydrogen bond. The oxygen is more electronegative than hydrogen, so the oxygen withdraws some electron density, making the oxygen partially negative and leaving the hydrogen partially positive. If water comes along, I'll draw in a water molecule here, and we know that water is a polar solvent. Water is a polar molecule."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Ethanol has a polar oxygen-hydrogen bond. The oxygen is more electronegative than hydrogen, so the oxygen withdraws some electron density, making the oxygen partially negative and leaving the hydrogen partially positive. If water comes along, I'll draw in a water molecule here, and we know that water is a polar solvent. Water is a polar molecule. The oxygen has a partial negative, and the hydrogens have partial positive charges. We can see that there's an opportunity for an attractive force. Opposite charges attract, so the partially positive hydrogen on ethanol is attracted to the partially negatively charged oxygen on water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Water is a polar molecule. The oxygen has a partial negative, and the hydrogens have partial positive charges. We can see that there's an opportunity for an attractive force. Opposite charges attract, so the partially positive hydrogen on ethanol is attracted to the partially negatively charged oxygen on water. This is an example of hydrogen bonding. So if you remember hydrogen bonding from earlier videos, here is a good example of that. We could even have some more hydrogen bonding."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Opposite charges attract, so the partially positive hydrogen on ethanol is attracted to the partially negatively charged oxygen on water. This is an example of hydrogen bonding. So if you remember hydrogen bonding from earlier videos, here is a good example of that. We could even have some more hydrogen bonding. I could draw in another water molecule down here, so let me go ahead and do that. We know that the oxygen is partially negative. The hydrogens are partially positive."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We could even have some more hydrogen bonding. I could draw in another water molecule down here, so let me go ahead and do that. We know that the oxygen is partially negative. The hydrogens are partially positive. Here's another opportunity for hydrogen bonding between the partially negative oxygen on ethanol and the partially positive hydrogen on water. This portion of the ethanol molecule is polar and loves water. This is the polar region, and this portion loves water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogens are partially positive. Here's another opportunity for hydrogen bonding between the partially negative oxygen on ethanol and the partially positive hydrogen on water. This portion of the ethanol molecule is polar and loves water. This is the polar region, and this portion loves water. We call this hydrophilic. Let me write that down here. This portion of the molecule is hydrophilic, or water loving."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This is the polar region, and this portion loves water. We call this hydrophilic. Let me write that down here. This portion of the molecule is hydrophilic, or water loving. Let's look at the other portion of the ethanol molecule. This portion on the left, we have a CH2 here and a CH3 here. Carbons and hydrogens, which we know are nonpolar."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This portion of the molecule is hydrophilic, or water loving. Let's look at the other portion of the ethanol molecule. This portion on the left, we have a CH2 here and a CH3 here. Carbons and hydrogens, which we know are nonpolar. This region is nonpolar. This region doesn't like water. It's scared of water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Carbons and hydrogens, which we know are nonpolar. This region is nonpolar. This region doesn't like water. It's scared of water. We call this hydrophobic, or water fearing. We know that ethanol is soluble in water just by experience. That must mean that this hydrophobic region doesn't overcome the hydrophilic region."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It's scared of water. We call this hydrophobic, or water fearing. We know that ethanol is soluble in water just by experience. That must mean that this hydrophobic region doesn't overcome the hydrophilic region. The hydrophilic region, this polar region of the ethanol molecule, is enough to make ethanol soluble in water. If we think about that same concept and we look at a different molecule, on the right here's 1-octanol. 1-octanol has an opportunity for hydrogen bonding."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That must mean that this hydrophobic region doesn't overcome the hydrophilic region. The hydrophilic region, this polar region of the ethanol molecule, is enough to make ethanol soluble in water. If we think about that same concept and we look at a different molecule, on the right here's 1-octanol. 1-octanol has an opportunity for hydrogen bonding. We have this OH here. It's the same situation as the ethanol on the left. We have a polar or hydrophilic region of the molecule."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "1-octanol has an opportunity for hydrogen bonding. We have this OH here. It's the same situation as the ethanol on the left. We have a polar or hydrophilic region of the molecule. However, the difference is this time we have an extremely large, nonpolar, hydrophobic portion of the molecule. This nonpolar region overcomes the slightly polar region, making the 1-octanol molecule nonpolar overall. 1-octanol will not dissolve in water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have a polar or hydrophilic region of the molecule. However, the difference is this time we have an extremely large, nonpolar, hydrophobic portion of the molecule. This nonpolar region overcomes the slightly polar region, making the 1-octanol molecule nonpolar overall. 1-octanol will not dissolve in water. This one is a no, and this one over here was a yes. Ethanol is a yes. Next, let's look at cinnamaldehyde."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "1-octanol will not dissolve in water. This one is a no, and this one over here was a yes. Ethanol is a yes. Next, let's look at cinnamaldehyde. Down here on the left is cinnamaldehyde. Let's focus in on this carbon-oxygen double bond first. We know that oxygen is more electronegative than this carbon here."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at cinnamaldehyde. Down here on the left is cinnamaldehyde. Let's focus in on this carbon-oxygen double bond first. We know that oxygen is more electronegative than this carbon here. The oxygen withdraws some electron density, making it partially negative. This carbon would be, therefore, partially positive. This very small portion of the molecule is polar."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We know that oxygen is more electronegative than this carbon here. The oxygen withdraws some electron density, making it partially negative. This carbon would be, therefore, partially positive. This very small portion of the molecule is polar. This small portion could interact with water. However, we have an extremely large, nonpolar region of the molecule. All these carbons and hydrogens over here on the left."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This very small portion of the molecule is polar. This small portion could interact with water. However, we have an extremely large, nonpolar region of the molecule. All these carbons and hydrogens over here on the left. This very hydrophobic region, or nonpolar region, overcomes the small polar region, making cinnamaldehyde overall nonpolar. Since it's overall nonpolar, cinnamaldehyde will not dissolve in water. If it's nonpolar, you would need a more nonpolar solvent to get cinnamaldehyde to dissolve."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "All these carbons and hydrogens over here on the left. This very hydrophobic region, or nonpolar region, overcomes the small polar region, making cinnamaldehyde overall nonpolar. Since it's overall nonpolar, cinnamaldehyde will not dissolve in water. If it's nonpolar, you would need a more nonpolar solvent to get cinnamaldehyde to dissolve. There are several examples of nonpolar organic solvents that will do that. Next, let's look at sucrose. Over here on the right is sucrose, or one way to draw or represent the sucrose compound."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If it's nonpolar, you would need a more nonpolar solvent to get cinnamaldehyde to dissolve. There are several examples of nonpolar organic solvents that will do that. Next, let's look at sucrose. Over here on the right is sucrose, or one way to draw or represent the sucrose compound. We see lots of carbons and hydrogens. All of these right here, let me just go ahead and highlight all these carbons in this ring. All these carbons in these rings, they're all these hydrogens."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Over here on the right is sucrose, or one way to draw or represent the sucrose compound. We see lots of carbons and hydrogens. All of these right here, let me just go ahead and highlight all these carbons in this ring. All these carbons in these rings, they're all these hydrogens. At first, you might think, okay, there's lots of carbons and hydrogens. This might be nonpolar. Of course, we have lots of these OH groups."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "All these carbons in these rings, they're all these hydrogens. At first, you might think, okay, there's lots of carbons and hydrogens. This might be nonpolar. Of course, we have lots of these OH groups. Let me go ahead and circle a few of them. We have all of these OH groups in the sucrose molecule, so lots of them. That means opportunities for hydrogen bonding."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Of course, we have lots of these OH groups. Let me go ahead and circle a few of them. We have all of these OH groups in the sucrose molecule, so lots of them. That means opportunities for hydrogen bonding. Because of all those opportunities for hydrogen bonding, sucrose is soluble in water, which we know from experience, of course. Sucrose, or sugar, sugar will dissolve in water. The opportunity for hydrogen bonding is the reason for that."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That means opportunities for hydrogen bonding. Because of all those opportunities for hydrogen bonding, sucrose is soluble in water, which we know from experience, of course. Sucrose, or sugar, sugar will dissolve in water. The opportunity for hydrogen bonding is the reason for that. Benzoic acid is a solid at room temperature. If you take some benzoic acid crystals and you put them in some room temperature water, the crystals won't dissolve. We can explain that by looking at the structure for benzoic acid."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The opportunity for hydrogen bonding is the reason for that. Benzoic acid is a solid at room temperature. If you take some benzoic acid crystals and you put them in some room temperature water, the crystals won't dissolve. We can explain that by looking at the structure for benzoic acid. While we do have this portion of the compound, which we know is polar and hydrophilic due to the presence of the electronegative oxygens, we also have this portion of the compound on the left, which is nonpolar and hydrophobic due to the presence of all the carbons and hydrogens. Since the benzoic acid crystals don't dissolve at room temperature water, the hydrophobic portion of the compound must overcome the hydrophilic portion of the compound. You actually can get benzoic acid crystals to dissolve in water if you heat up the water, if you increase the solubility of the compound by increasing the temperature of the solvent."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We can explain that by looking at the structure for benzoic acid. While we do have this portion of the compound, which we know is polar and hydrophilic due to the presence of the electronegative oxygens, we also have this portion of the compound on the left, which is nonpolar and hydrophobic due to the presence of all the carbons and hydrogens. Since the benzoic acid crystals don't dissolve at room temperature water, the hydrophobic portion of the compound must overcome the hydrophilic portion of the compound. You actually can get benzoic acid crystals to dissolve in water if you heat up the water, if you increase the solubility of the compound by increasing the temperature of the solvent. Let's think about benzoic acid crystals in room temperature water. Let's add a base. Let's add sodium hydroxide."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "You actually can get benzoic acid crystals to dissolve in water if you heat up the water, if you increase the solubility of the compound by increasing the temperature of the solvent. Let's think about benzoic acid crystals in room temperature water. Let's add a base. Let's add sodium hydroxide. Sodium hydroxide is going to react with the most acidic proton on benzoic acid. Benzoic acid is acidic. It will donate this proton right here."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's add sodium hydroxide. Sodium hydroxide is going to react with the most acidic proton on benzoic acid. Benzoic acid is acidic. It will donate this proton right here. That means the electrons in red in this bond are left behind on the oxygen. I'll show those electrons in red over here. That gives this oxygen a negative charge."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It will donate this proton right here. That means the electrons in red in this bond are left behind on the oxygen. I'll show those electrons in red over here. That gives this oxygen a negative charge. We form sodium benzoate. I won't get too much into acid-base chemistry, but we took the most acidic proton off of benzoic acid to give us the conjugate base sodium benzoate. Sodium benzoate is highly soluble in room temperature water."}, {"video_title": "Solubility of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That gives this oxygen a negative charge. We form sodium benzoate. I won't get too much into acid-base chemistry, but we took the most acidic proton off of benzoic acid to give us the conjugate base sodium benzoate. Sodium benzoate is highly soluble in room temperature water. That must mean we increase this hydrophilic portion because now we have a negative charge. The hydrophilic portion now is able to overcome the hydrophobic portion. Sodium benzoate is soluble."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "If we have ingredients very similar to what we saw in the last video, but instead of our nucleophile or our base being methoxide, it's going to be something slightly more involved. So it's still going to have the O minus, but it's going to be bonded to a carbon, which is then bonded to three methyl groups. CH3, CH3, CH3, just like that. So we don't have methoxide anymore. We don't have methoxide anymore. We have this thing right over here. So just like before, we have the exact same solvent."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "So we don't have methoxide anymore. We don't have methoxide anymore. We have this thing right over here. So just like before, we have the exact same solvent. We have dimethylformamide. It's an aprotic solvent. That by itself would put us in the SN2 or E2 direction."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "So just like before, we have the exact same solvent. We have dimethylformamide. It's an aprotic solvent. That by itself would put us in the SN2 or E2 direction. But now we don't have methoxide anymore. Methoxide was both a strong base, very strong base, and it's also a very small molecule. And so it can really get in there and react with the substrate."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "That by itself would put us in the SN2 or E2 direction. But now we don't have methoxide anymore. Methoxide was both a strong base, very strong base, and it's also a very small molecule. And so it can really get in there and react with the substrate. So it was also a strong nucleophile. Now, this more bulky molecule, it is still a strong base. It is still an extremely strong base."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "And so it can really get in there and react with the substrate. So it was also a strong nucleophile. Now, this more bulky molecule, it is still a strong base. It is still an extremely strong base. But now it's this big bulky molecule. It would actually have trouble getting in to react with your substrate, so it is no longer a good nucleophile. This is not a good nucleophile."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "It is still an extremely strong base. But now it's this big bulky molecule. It would actually have trouble getting in to react with your substrate, so it is no longer a good nucleophile. This is not a good nucleophile. So by making the base more, I guess, bulky, it's now an, or I guess you could also call it the nucleophile, or the thing that would act as a nucleophile, more bulky, it is no longer a strong nucleophile. So it would no longer be good for an SN2 reaction. So just by changing the base a little bit or the nucleophile a little bit, now this one would go strictly in the E2 direction."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "This is not a good nucleophile. So by making the base more, I guess, bulky, it's now an, or I guess you could also call it the nucleophile, or the thing that would act as a nucleophile, more bulky, it is no longer a strong nucleophile. So it would no longer be good for an SN2 reaction. So just by changing the base a little bit or the nucleophile a little bit, now this one would go strictly in the E2 direction. So we wouldn't see anything like this in the last video. We would only see something like this. But obviously, the base in this example is no longer just the methoxide."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "So just by changing the base a little bit or the nucleophile a little bit, now this one would go strictly in the E2 direction. So we wouldn't see anything like this in the last video. We would only see something like this. But obviously, the base in this example is no longer just the methoxide. It looks like this. Let me clear it. Do my best to clear it."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "But obviously, the base in this example is no longer just the methoxide. It looks like this. Let me clear it. Do my best to clear it. Edit, clear. Let me clear it over here as well. So now instead of just being bonded to a methyl group, it's bonded to a carbon that's bonded to three methyl groups."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "Do my best to clear it. Edit, clear. Let me clear it over here as well. So now instead of just being bonded to a methyl group, it's bonded to a carbon that's bonded to three methyl groups. So CH3, CH3, CH3. Or you could call this a tert-butyl group, this whole thing over here. So that's a carbon bonded to a CH3, a CH3, and a CH3."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "So now instead of just being bonded to a methyl group, it's bonded to a carbon that's bonded to three methyl groups. So CH3, CH3, CH3. Or you could call this a tert-butyl group, this whole thing over here. So that's a carbon bonded to a CH3, a CH3, and a CH3. So the reaction occurs just like what we saw in the last video, except this base is this big old bulky thing. But it can still act as a strong base. So it still nabs the hydrogen, or really just the proton."}, {"video_title": "E2 E1 Sn2 Sn1 Reactions Example 2.mp3", "Sentence": "So that's a carbon bonded to a CH3, a CH3, and a CH3. So the reaction occurs just like what we saw in the last video, except this base is this big old bulky thing. But it can still act as a strong base. So it still nabs the hydrogen, or really just the proton. The hydrogen's electron that was bonded now goes to the alpha carbon. The alpha carbon will then lose an electron to the bromo group, and that becomes bromide. So the same exact mechanism, different base."}, {"video_title": "Triple bonds cause linear configurations Organic chemistry Khan Academy.mp3", "Sentence": "I want to do a quick clarification on the video on alcohols. In that video, I gave this example of this alkanol right over here. It has a triple bond between the five and six carbons. And I just want to clarify that in reality, it would not ever be drawn this way. That this was an error to actually draw it this way. The way I should have drawn it is this right over here. And you see the only difference between these two pictures, starting with the one carbon, two carbon, three carbon, everything looks the same."}, {"video_title": "Triple bonds cause linear configurations Organic chemistry Khan Academy.mp3", "Sentence": "And I just want to clarify that in reality, it would not ever be drawn this way. That this was an error to actually draw it this way. The way I should have drawn it is this right over here. And you see the only difference between these two pictures, starting with the one carbon, two carbon, three carbon, everything looks the same. Four carbon attached to the hydroxyl group, everything looks the same. And then we get to the five carbon. But instead of bending back up, we just keep going straight to the six carbon."}, {"video_title": "Triple bonds cause linear configurations Organic chemistry Khan Academy.mp3", "Sentence": "And you see the only difference between these two pictures, starting with the one carbon, two carbon, three carbon, everything looks the same. Four carbon attached to the hydroxyl group, everything looks the same. And then we get to the five carbon. But instead of bending back up, we just keep going straight to the six carbon. That's where we have the triple bond between the five and six carbon. And then we go straight again to the seven carbon that's attached to the two bromo groups. And then we get to the eighth carbon."}, {"video_title": "Triple bonds cause linear configurations Organic chemistry Khan Academy.mp3", "Sentence": "But instead of bending back up, we just keep going straight to the six carbon. That's where we have the triple bond between the five and six carbon. And then we go straight again to the seven carbon that's attached to the two bromo groups. And then we get to the eighth carbon. And the whole point why this right over here, that this is the correct way to draw it, is that triple bonds, this triple bond right here, it forces a linear configuration. So on both sides of that triple bond, you would go straight out. So whenever you see an alkyne drawn, the triple bonds should essentially straighten out."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Over here on the left, we have the ethanol molecule. So this is our two-carbon alcohol. And the carbon that we're most concerned with is this carbon right here, which has one bond to this oxygen atom. And in the liver, ethanol is oxidized to ethanol. So over here on the right is the ethanol molecule, a two-carbon aldehyde. And once again, we're concerned with that carbon in yellow. And so one easy way to tell that ethanol was oxidized to ethanol is to see that on the left, we have one bond of that carbon to oxygen."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And in the liver, ethanol is oxidized to ethanol. So over here on the right is the ethanol molecule, a two-carbon aldehyde. And once again, we're concerned with that carbon in yellow. And so one easy way to tell that ethanol was oxidized to ethanol is to see that on the left, we have one bond of that carbon to oxygen. And over here on the right, we now have two bonds of that carbon to oxygen. So an increase in the number of bonds to oxygen is oxidation. You could also assign oxidation states to this carbon."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so one easy way to tell that ethanol was oxidized to ethanol is to see that on the left, we have one bond of that carbon to oxygen. And over here on the right, we now have two bonds of that carbon to oxygen. So an increase in the number of bonds to oxygen is oxidation. You could also assign oxidation states to this carbon. And you will see that there's an increase in the oxidation state of that carbon. And then you could also think about electrons. So Leo, the line goes ger."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "You could also assign oxidation states to this carbon. And you will see that there's an increase in the oxidation state of that carbon. And then you could also think about electrons. So Leo, the line goes ger. Loss of electrons is oxidation. Gain of electrons is reduction. And so if I think about these electrons here in magenta, you can see that those electrons are lost from the ethanol molecule."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So Leo, the line goes ger. Loss of electrons is oxidation. Gain of electrons is reduction. And so if I think about these electrons here in magenta, you can see that those electrons are lost from the ethanol molecule. So loss of electrons is oxidation. Ethanol is oxidized. If ethanol is oxidized, something else must be reduced."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so if I think about these electrons here in magenta, you can see that those electrons are lost from the ethanol molecule. So loss of electrons is oxidation. Ethanol is oxidized. If ethanol is oxidized, something else must be reduced. That's how redox reactions work. And what's reduced is NAD plus over here on the left. So this is NAD plus, which stands for nicotinamide adenine dinucleotide."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If ethanol is oxidized, something else must be reduced. That's how redox reactions work. And what's reduced is NAD plus over here on the left. So this is NAD plus, which stands for nicotinamide adenine dinucleotide. The adenine is hiding in this R portion. And we have a nitrogenous-based ring with an amide functional group over here on the right for the nicotinamide portion of the molecule. Plus one formal charge on this nitrogen gives us NAD plus."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this is NAD plus, which stands for nicotinamide adenine dinucleotide. The adenine is hiding in this R portion. And we have a nitrogenous-based ring with an amide functional group over here on the right for the nicotinamide portion of the molecule. Plus one formal charge on this nitrogen gives us NAD plus. So this is nicotinamide adenine dinucleotide, NAD plus. And since ethanol is oxidized, NAD plus must be reduced. So reduction means gaining of electrons."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Plus one formal charge on this nitrogen gives us NAD plus. So this is nicotinamide adenine dinucleotide, NAD plus. And since ethanol is oxidized, NAD plus must be reduced. So reduction means gaining of electrons. So NAD plus is going to gain those electrons in magenta from ethanol. So if we think about a possible mechanism, if I took these electrons between the oxygen and the hydrogen and moved them in here, that would form our double bond between the carbon and the oxygen. But there would be too many bonds to this carbon right here."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So reduction means gaining of electrons. So NAD plus is going to gain those electrons in magenta from ethanol. So if we think about a possible mechanism, if I took these electrons between the oxygen and the hydrogen and moved them in here, that would form our double bond between the carbon and the oxygen. But there would be too many bonds to this carbon right here. So the electrons in magenta are going to move to this carbon down here on NAD plus, to this carbon. That would push these electrons over here. And now it would push these electrons here off onto the nitrogen."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But there would be too many bonds to this carbon right here. So the electrons in magenta are going to move to this carbon down here on NAD plus, to this carbon. That would push these electrons over here. And now it would push these electrons here off onto the nitrogen. So if we showed what happened with the movement of all of those electrons over here on the right, this carbon right here at the top already had a hydrogen bonded to it. And it gained another hydrogen with two electrons. The two electrons were the ones in magenta right here."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And now it would push these electrons here off onto the nitrogen. So if we showed what happened with the movement of all of those electrons over here on the right, this carbon right here at the top already had a hydrogen bonded to it. And it gained another hydrogen with two electrons. The two electrons were the ones in magenta right here. So this hydrogen right here is this hydrogen. And the electrons in magenta move over there to our ring. And then we would also have pi electrons moved over here."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "The two electrons were the ones in magenta right here. So this hydrogen right here is this hydrogen. And the electrons in magenta move over there to our ring. And then we would also have pi electrons moved over here. And then we had a lone pair of electrons move off onto the nitrogen like that. And then we still had some pi electrons over here on the right. And so this molecule is called NADH."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then we would also have pi electrons moved over here. And then we had a lone pair of electrons move off onto the nitrogen like that. And then we still had some pi electrons over here on the right. And so this molecule is called NADH. So it's gained the equivalent of a hydride, hydrogen with two electrons. And so we can see that NAD plus gains two electrons. And gaining electrons is reduction."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so this molecule is called NADH. So it's gained the equivalent of a hydride, hydrogen with two electrons. And so we can see that NAD plus gains two electrons. And gaining electrons is reduction. So NAD plus is reduced to NADH. Since NAD plus is reduced, it allows ethanol to be oxidized. And so we would refer to NAD plus as an oxidizing agent."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And gaining electrons is reduction. So NAD plus is reduced to NADH. Since NAD plus is reduced, it allows ethanol to be oxidized. And so we would refer to NAD plus as an oxidizing agent. It is the oxidizing agent for ethanol, even though it itself is being reduced. So that's something that confuses some general chemistry students sometimes. So now over here we have the NADH molecule."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so we would refer to NAD plus as an oxidizing agent. It is the oxidizing agent for ethanol, even though it itself is being reduced. So that's something that confuses some general chemistry students sometimes. So now over here we have the NADH molecule. And this reaction is catalyzed by an enzyme. And the enzyme is alcohol dehydrogenase. So this is catalyzed by the alcohol dehydrogenase enzyme like that."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So now over here we have the NADH molecule. And this reaction is catalyzed by an enzyme. And the enzyme is alcohol dehydrogenase. So this is catalyzed by the alcohol dehydrogenase enzyme like that. And this reaction is reversible. So if we think about the reverse reaction, we think about ethanol being reduced to ethanol. And so if ethanol is reduced to ethanol, NADH would be oxidized to NAD plus."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this is catalyzed by the alcohol dehydrogenase enzyme like that. And this reaction is reversible. So if we think about the reverse reaction, we think about ethanol being reduced to ethanol. And so if ethanol is reduced to ethanol, NADH would be oxidized to NAD plus. And so let's think about a mechanism where we could oxidize NADH and reduce the ethanol. If I took this lone pair of electrons in the nitrogen and move it back in here, that would push these electrons off over here. And now the electrons in magenta on this bond right here would attack this carbon right here."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so if ethanol is reduced to ethanol, NADH would be oxidized to NAD plus. And so let's think about a mechanism where we could oxidize NADH and reduce the ethanol. If I took this lone pair of electrons in the nitrogen and move it back in here, that would push these electrons off over here. And now the electrons in magenta on this bond right here would attack this carbon right here. So the electrons in magenta, we could think about the electrons in magenta as being right here. And you could think about that as being a hydride. So a hydrogen with two electrons giving it a negative 1 formal charge."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And now the electrons in magenta on this bond right here would attack this carbon right here. So the electrons in magenta, we could think about the electrons in magenta as being right here. And you could think about that as being a hydride. So a hydrogen with two electrons giving it a negative 1 formal charge. And even though we've seen in some earlier videos that hydride isn't necessarily the best nucleophile, you could think about this as being a nucleophilic attack if it makes it easier for you. Because this carbon right here would be partially positive. So the negatively charged electrons would attack that carbon."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So a hydrogen with two electrons giving it a negative 1 formal charge. And even though we've seen in some earlier videos that hydride isn't necessarily the best nucleophile, you could think about this as being a nucleophilic attack if it makes it easier for you. Because this carbon right here would be partially positive. So the negatively charged electrons would attack that carbon. And in doing so, that would push these pi electrons off to then grab this proton here. And that would give you your ethanol molecule. And that would convert NADH back into NAD plus."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the negatively charged electrons would attack that carbon. And in doing so, that would push these pi electrons off to then grab this proton here. And that would give you your ethanol molecule. And that would convert NADH back into NAD plus. So you could think about NADH as being oxidized. It is losing two electrons, the electrons in magenta. Loss of electrons is oxidation."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And that would convert NADH back into NAD plus. So you could think about NADH as being oxidized. It is losing two electrons, the electrons in magenta. Loss of electrons is oxidation. And since NADH is the agent for the reduction of ethanol to ethanol, you would say that NADH would be the reducing agent for this example. And the best way to remember that NADH is the reducing agent is it is the one that has the hydrogen on it. So it has the hydride, which is capable of being the agent for the reduction."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Loss of electrons is oxidation. And since NADH is the agent for the reduction of ethanol to ethanol, you would say that NADH would be the reducing agent for this example. And the best way to remember that NADH is the reducing agent is it is the one that has the hydrogen on it. So it has the hydride, which is capable of being the agent for the reduction. So therefore, NADH is the reducing agent. This NAD plus, NADH conversion, and vice versa, is extremely important in biochemistry. This happens in numerous biochemical reactions."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it has the hydride, which is capable of being the agent for the reduction. So therefore, NADH is the reducing agent. This NAD plus, NADH conversion, and vice versa, is extremely important in biochemistry. This happens in numerous biochemical reactions. And so it's important to understand what's happening with those electrons on these molecules. Let's look at another biochemical example of redox. And here we have, on the left, phenol."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This happens in numerous biochemical reactions. And so it's important to understand what's happening with those electrons on these molecules. Let's look at another biochemical example of redox. And here we have, on the left, phenol. So this is our phenol molecule. And once again, we're most concerned about this carbon, the one that's attached to this oxygen. And there are many ways to oxidize phenols."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And here we have, on the left, phenol. So this is our phenol molecule. And once again, we're most concerned about this carbon, the one that's attached to this oxygen. And there are many ways to oxidize phenols. So if we oxidize phenol, something like the Jones reagent with sodium dichromate, sulfuric acid, and water would be capable of oxidizing phenol to this molecule over here on the right, which we call benzoquinone. So this right here is a benzoquinone molecule. And just real fast, you could see that this carbon right now has two bonds of carbon to oxygen."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And there are many ways to oxidize phenols. So if we oxidize phenol, something like the Jones reagent with sodium dichromate, sulfuric acid, and water would be capable of oxidizing phenol to this molecule over here on the right, which we call benzoquinone. So this right here is a benzoquinone molecule. And just real fast, you could see that this carbon right now has two bonds of carbon to oxygen. So it has been oxidized. So phenol can be oxidized to benzoquinone using numerous organic reagents. Once you make benzoquinone, you could reduce that to this molecule over here on the right, which is called hydroquinone."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And just real fast, you could see that this carbon right now has two bonds of carbon to oxygen. So it has been oxidized. So phenol can be oxidized to benzoquinone using numerous organic reagents. Once you make benzoquinone, you could reduce that to this molecule over here on the right, which is called hydroquinone. So there are several, again, organic reagents that can reduce benzoquinone to hydroquinone. Hydroquinone, let me change that spelling there. And then from hydroquinone, you could oxidize hydroquinone back to benzoquinone pretty easily."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Once you make benzoquinone, you could reduce that to this molecule over here on the right, which is called hydroquinone. So there are several, again, organic reagents that can reduce benzoquinone to hydroquinone. Hydroquinone, let me change that spelling there. And then from hydroquinone, you could oxidize hydroquinone back to benzoquinone pretty easily. And so once again, in organic chemistry, there are lots of reagents that can do these redox conversions. And in the body, you're usually talking about the NAD plus NADH system. So we've just studied that."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then from hydroquinone, you could oxidize hydroquinone back to benzoquinone pretty easily. And so once again, in organic chemistry, there are lots of reagents that can do these redox conversions. And in the body, you're usually talking about the NAD plus NADH system. So we've just studied that. And if we look here at this molecule, you can see it's a quinone. So you can see the benzoquinone portion of this molecule. And this is called ubiquinone."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we've just studied that. And if we look here at this molecule, you can see it's a quinone. So you can see the benzoquinone portion of this molecule. And this is called ubiquinone. You'd be referring to the fact that this is ubiquitous. This compound is found everywhere. It's found in all the cells in nature."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And this is called ubiquinone. You'd be referring to the fact that this is ubiquitous. This compound is found everywhere. It's found in all the cells in nature. And the other name for this would be coenzyme Q. And so this is a very important part of the electron transport chain. And if we look at ubiquinone, going to this molecule over here on the right, you can see this is like a hydroquinone analog here."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It's found in all the cells in nature. And the other name for this would be coenzyme Q. And so this is a very important part of the electron transport chain. And if we look at ubiquinone, going to this molecule over here on the right, you can see this is like a hydroquinone analog here. So this is ubiquinol. And so these carbons are being reduced from this chemical reaction that I've drawn here. So ubiquinone is being reduced to ubiquinol."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And if we look at ubiquinone, going to this molecule over here on the right, you can see this is like a hydroquinone analog here. So this is ubiquinol. And so these carbons are being reduced from this chemical reaction that I've drawn here. So ubiquinone is being reduced to ubiquinol. And so if ubiquinone is being reduced, something else must be oxidized. And so the NADH is being oxidized to NAD plus. And so the NADH, it's the one that has this hydride on here, which can serve as the reducing agent."}, {"video_title": "Biological redox reactions Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So ubiquinone is being reduced to ubiquinol. And so if ubiquinone is being reduced, something else must be oxidized. And so the NADH is being oxidized to NAD plus. And so the NADH, it's the one that has this hydride on here, which can serve as the reducing agent. So here, NADH is acting as the reducing agent. The agent for the reduction of ubiquinone to the ubiquinol molecule over here on the right. And so this is just an oversimplification of part of the electron transport chain, where you're transporting electrons, which eventually leads to oxidative phosphorylation and also ATP synthesis, which, of course, gives us energy."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We've learned in previous videos that relative to the orbital plane around the sun, or the plane of Earth's orbit around the sun, the Earth has a certain tilt. So let me draw the Earth's tilt relative to that orbital plane right over here. So if this is the orbital plane right over here, so we're looking right directly sideways on this orbital plane, right sideways along this orbital plane that I've drawn in orange, and maybe at the point in Earth's orbit right now, maybe the sun is to the left, and so the rays from the sun are coming in this general direction, we've learned that Earth has a certain tilt. Earth has a tilt, and what I mean by that, it means if you think about the axis around which it's rotating, it's not straight up from the orbital plane, it is at an angle. Let me draw that. So if I were to draw an arrow that's coming out of the North Pole, it would look like that, and I'll draw an arrow coming out of the South Pole, and the Earth is rotating in that direction right over here, and you notice this axis that I've drawn this arrow on, it is not straight up and down, and right now it is an angle of, it is at an angle of 23.4 degrees with the vertical, with being straight up and down. And we've learned how this is what is the primary cause of our seasons, in that when the Northern Hemisphere is pointed towards the sun, it's getting a disproportionate amount of the solar radiation, whatever is going through the atmosphere has to go through less atmosphere, and the things in the Northern Hemisphere are getting more daylight."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Earth has a tilt, and what I mean by that, it means if you think about the axis around which it's rotating, it's not straight up from the orbital plane, it is at an angle. Let me draw that. So if I were to draw an arrow that's coming out of the North Pole, it would look like that, and I'll draw an arrow coming out of the South Pole, and the Earth is rotating in that direction right over here, and you notice this axis that I've drawn this arrow on, it is not straight up and down, and right now it is an angle of, it is at an angle of 23.4 degrees with the vertical, with being straight up and down. And we've learned how this is what is the primary cause of our seasons, in that when the Northern Hemisphere is pointed towards the sun, it's getting a disproportionate amount of the solar radiation, whatever is going through the atmosphere has to go through less atmosphere, and the things in the Northern Hemisphere are getting more daylight. And when the Earth is on the other side of the sun, and the Northern Hemisphere is pointed away from the sun, then the opposite is going to happen, and the reverse is true for the Southern Hemisphere. But in that video when we talk about how tilt can affect the seasons, I also kind of hinted a little bit that this is the current tilt right now, and over long periods of time that this tilt will change. And in particular, it will vary, and even the boundaries for this varying are different for the past million years than they will be for the next million years, but it varies roughly between 22.1 degrees and 24.5 degrees."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we've learned how this is what is the primary cause of our seasons, in that when the Northern Hemisphere is pointed towards the sun, it's getting a disproportionate amount of the solar radiation, whatever is going through the atmosphere has to go through less atmosphere, and the things in the Northern Hemisphere are getting more daylight. And when the Earth is on the other side of the sun, and the Northern Hemisphere is pointed away from the sun, then the opposite is going to happen, and the reverse is true for the Southern Hemisphere. But in that video when we talk about how tilt can affect the seasons, I also kind of hinted a little bit that this is the current tilt right now, and over long periods of time that this tilt will change. And in particular, it will vary, and even the boundaries for this varying are different for the past million years than they will be for the next million years, but it varies roughly between 22.1 degrees and 24.5 degrees. And just to make it clear that it's not wobbling back and forth like this, and just to visualize 22.1 versus 24.5, it's not a huge difference. So if this is 23.4, and I'm not measuring exactly, maybe pointing in this direction, maybe pointing in that, maybe 22.1 would look something like that. In fact, I've exaggerated it."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And in particular, it will vary, and even the boundaries for this varying are different for the past million years than they will be for the next million years, but it varies roughly between 22.1 degrees and 24.5 degrees. And just to make it clear that it's not wobbling back and forth like this, and just to visualize 22.1 versus 24.5, it's not a huge difference. So if this is 23.4, and I'm not measuring exactly, maybe pointing in this direction, maybe pointing in that, maybe 22.1 would look something like that. In fact, I've exaggerated it. And maybe 24.5 would look something like that. And so it's not a huge difference, but it is enough of a difference, so we believe, to actually have a significant impact on what the climate is like, or what the seasons are like, especially in terms of how much of a chance different parts of our planet have a chance to freeze over, or not freeze over, and all the rest, or how much sunlight they get, and all the rest. So it has some impact, but I want to make it clear that it takes a long period of time."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, I've exaggerated it. And maybe 24.5 would look something like that. And so it's not a huge difference, but it is enough of a difference, so we believe, to actually have a significant impact on what the climate is like, or what the seasons are like, especially in terms of how much of a chance different parts of our planet have a chance to freeze over, or not freeze over, and all the rest, or how much sunlight they get, and all the rest. So it has some impact, but I want to make it clear that it takes a long period of time. It actually takes 41,000 years to go from a minimum tilt to a maximum tilt, and then back to a minimum tilt. 41,000 years. And right now, at a tilt of 23.4 degrees, we're someplace right smack in between."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it has some impact, but I want to make it clear that it takes a long period of time. It actually takes 41,000 years to go from a minimum tilt to a maximum tilt, and then back to a minimum tilt. 41,000 years. And right now, at a tilt of 23.4 degrees, we're someplace right smack in between. And we think the last maximum was at 8,700 BCE, before the common era, or you could say before Christ, and that the next minimum, when our tilt has been minimized, the next time our tilt will be minimized, will be at 11,800. So this isn't something that's happening overnight, but it is something that could affect our climate over long periods of time. And this is just one factor, and sometimes this changing of the tilt, a fancier word for tilt is sometimes given, is obliquity."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And right now, at a tilt of 23.4 degrees, we're someplace right smack in between. And we think the last maximum was at 8,700 BCE, before the common era, or you could say before Christ, and that the next minimum, when our tilt has been minimized, the next time our tilt will be minimized, will be at 11,800. So this isn't something that's happening overnight, but it is something that could affect our climate over long periods of time. And this is just one factor, and sometimes this changing of the tilt, a fancier word for tilt is sometimes given, is obliquity. But this is really just a fancy word for tilt. This changing of the obliquity, or changing of the tilt, is one of these changes in Earth's rotation, or Earth's orbit around the sun, that might have long-term cycles or effects on Earth's climate, and maybe they do help cause certain ice ages when they act together with each other over certain cycles. And broadly, this entire class of cycles are called Milankovitch cycles."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is just one factor, and sometimes this changing of the tilt, a fancier word for tilt is sometimes given, is obliquity. But this is really just a fancy word for tilt. This changing of the obliquity, or changing of the tilt, is one of these changes in Earth's rotation, or Earth's orbit around the sun, that might have long-term cycles or effects on Earth's climate, and maybe they do help cause certain ice ages when they act together with each other over certain cycles. And broadly, this entire class of cycles are called Milankovitch cycles. Milankovitch, he was a Serbian scientist who was the guy who theorized that these changes in Earth's orbit might be responsible for long-term climate change, or maybe some cycles where we enter ice ages and get out of ice ages, or we have more extreme or less extreme weather. So these are Milankovitch cycles. And changes in the tilt, or the obliquity, are just one of the possible factors playing into Milankovitch cycles."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And broadly, this entire class of cycles are called Milankovitch cycles. Milankovitch, he was a Serbian scientist who was the guy who theorized that these changes in Earth's orbit might be responsible for long-term climate change, or maybe some cycles where we enter ice ages and get out of ice ages, or we have more extreme or less extreme weather. So these are Milankovitch cycles. And changes in the tilt, or the obliquity, are just one of the possible factors playing into Milankovitch cycles. And what I want to do in this video and the next few is talk about all of the different factors, or at least summarize all of the different factors. Now another one, this one is pretty intuitive for me, that this tilt can change. One that's a little bit less intuitive when you first think about it is something called precession."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And changes in the tilt, or the obliquity, are just one of the possible factors playing into Milankovitch cycles. And what I want to do in this video and the next few is talk about all of the different factors, or at least summarize all of the different factors. Now another one, this one is pretty intuitive for me, that this tilt can change. One that's a little bit less intuitive when you first think about it is something called precession. And the idea behind precession, I guess the best analogy I can think of, is if you imagine a top, or maybe you can imagine Earth as a top right over here. The top is spinning in this direction. And obliquity tells you essentially how much it's wobbling."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "One that's a little bit less intuitive when you first think about it is something called precession. And the idea behind precession, I guess the best analogy I can think of, is if you imagine a top, or maybe you can imagine Earth as a top right over here. The top is spinning in this direction. And obliquity tells you essentially how much it's wobbling. Well, actually let me think of it this way. Imagine a wobbling top. So it's rotating like this, it's tilted, and then it's also, if you imagine that this was a pole up here that's coming out of the pole, if this was actually a physical arrow, that that arrow itself would be rotating."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And obliquity tells you essentially how much it's wobbling. Well, actually let me think of it this way. Imagine a wobbling top. So it's rotating like this, it's tilted, and then it's also, if you imagine that this was a pole up here that's coming out of the pole, if this was actually a physical arrow, that that arrow itself would be rotating. So the best way to think about it is a wobbling top. After some point of time, this thing would wobble, so it would look like this. So now the arrow is pointing that way."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's rotating like this, it's tilted, and then it's also, if you imagine that this was a pole up here that's coming out of the pole, if this was actually a physical arrow, that that arrow itself would be rotating. So the best way to think about it is a wobbling top. After some point of time, this thing would wobble, so it would look like this. So now the arrow is pointing that way. And if you wait a few more seconds, now maybe the arrow is pointing a little bit out of the page. And then you wait a few more seconds, then it's pointing in this direction, then it's pointing into the page. And so this whole time the obliquity isn't changing."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now the arrow is pointing that way. And if you wait a few more seconds, now maybe the arrow is pointing a little bit out of the page. And then you wait a few more seconds, then it's pointing in this direction, then it's pointing into the page. And so this whole time the obliquity isn't changing. The obliquity you can kind of view it as how far is that wobble? You can imagine how far from vertical is that wobble. And no matter where we are in that rotation, it hasn't changed."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this whole time the obliquity isn't changing. The obliquity you can kind of view it as how far is that wobble? You can imagine how far from vertical is that wobble. And no matter where we are in that rotation, it hasn't changed. And you can imagine it as a procession as where we are in the wobble. And I want to, this is a little bit hard to visualize, and hopefully as we think about it in different ways and I draw different diagrams, it'll make it a little bit clearer. But I want to make it clear, just as it takes a long time for the inclination to change from a minimum value to a maximum value and back, it takes a huge amount of time for Earth's precession to change in a significant way."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And no matter where we are in that rotation, it hasn't changed. And you can imagine it as a procession as where we are in the wobble. And I want to, this is a little bit hard to visualize, and hopefully as we think about it in different ways and I draw different diagrams, it'll make it a little bit clearer. But I want to make it clear, just as it takes a long time for the inclination to change from a minimum value to a maximum value and back, it takes a huge amount of time for Earth's precession to change in a significant way. So for this top to kind of, if you imagined this arrow popping out, for this arrow to actually trace out an entire loop, it takes 26,000 years. So 26,000 years to have an entire cycle of precession. Now what I want to do is think about, given that this precession is occurring, I want to think about how that would affect our seasons or how it would actually affect how we think about the year, the calendar."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I want to make it clear, just as it takes a long time for the inclination to change from a minimum value to a maximum value and back, it takes a huge amount of time for Earth's precession to change in a significant way. So for this top to kind of, if you imagined this arrow popping out, for this arrow to actually trace out an entire loop, it takes 26,000 years. So 26,000 years to have an entire cycle of precession. Now what I want to do is think about, given that this precession is occurring, I want to think about how that would affect our seasons or how it would actually affect how we think about the year, the calendar. So let's draw the orbit of Earth around the sun. So here is my sun right over here, and here is the orbit of Earth. And I'm not going to think too much."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now what I want to do is think about, given that this precession is occurring, I want to think about how that would affect our seasons or how it would actually affect how we think about the year, the calendar. So let's draw the orbit of Earth around the sun. So here is my sun right over here, and here is the orbit of Earth. And I'm not going to think too much. I'm going to assume that it's almost circular for the sake of this video. In future videos, we'll talk about how the eccentricity or how elliptical the orbit is can also affect the Milankovitch cycles or play into the Milankovitch cycles. But let's just draw the orbit of Earth around the sun over here."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm not going to think too much. I'm going to assume that it's almost circular for the sake of this video. In future videos, we'll talk about how the eccentricity or how elliptical the orbit is can also affect the Milankovitch cycles or play into the Milankovitch cycles. But let's just draw the orbit of Earth around the sun over here. And so you can imagine this is at one point in time. This is the Earth. Let's say it is tilted towards the sun right now."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's just draw the orbit of Earth around the sun over here. And so you can imagine this is at one point in time. This is the Earth. Let's say it is tilted towards the sun right now. So in the northern hemisphere, and I'm assuming this arrow is coming out of the North Pole, this would be the summer in the northern hemisphere. And then if you had no precession, absolutely no precession, when you go to this time of year, you still have the same direction of tilt. Let me do that in blue."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say it is tilted towards the sun right now. So in the northern hemisphere, and I'm assuming this arrow is coming out of the North Pole, this would be the summer in the northern hemisphere. And then if you had no precession, absolutely no precession, when you go to this time of year, you still have the same direction of tilt. Let me do that in blue. You still have the same direction of tilt. We're still pointing to the same part of the universe. We still have the same north star."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me do that in blue. You still have the same direction of tilt. We're still pointing to the same part of the universe. We still have the same north star. At this time, we're still tilting in the same direction relative to the universe, but we're not tilting away from the sun. And now this would be the winter in the northern hemisphere. And we'd keep going around."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We still have the same north star. At this time, we're still tilting in the same direction relative to the universe, but we're not tilting away from the sun. And now this would be the winter in the northern hemisphere. And we'd keep going around. If you had no precession, when you get back to this point over here, we'd be tilting in the exact same direction. If your obliquity or if your tilt changed a little bit, you might move up or down, away or towards the sun a little bit. But this is all assuming no precession."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we'd keep going around. If you had no precession, when you get back to this point over here, we'd be tilting in the exact same direction. If your obliquity or if your tilt changed a little bit, you might move up or down, away or towards the sun a little bit. But this is all assuming no precession. Now I'm going to think about what happens if you do have precession. So what's happening with precession is when you go around one time around the sun, by the time you get to this point again, you're not pointing at exactly the same direction. You're now pointing a little bit further."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is all assuming no precession. Now I'm going to think about what happens if you do have precession. So what's happening with precession is when you go around one time around the sun, by the time you get to this point again, you're not pointing at exactly the same direction. You're now pointing a little bit further. So this arrow, let me draw it a little bit bigger. So this is the earth, and this is that arrow. And this is hard to visualize, or at least it's hard for me to visualize."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You're now pointing a little bit further. So this arrow, let me draw it a little bit bigger. So this is the earth, and this is that arrow. And this is hard to visualize, or at least it's hard for me to visualize. Well, once you get it, it's easier to visualize. But the first time I tried to understand it, it was hard for me to understand how precession was different than obliquity or different than tilt. Obliquity is how much we're going from vertical."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is hard to visualize, or at least it's hard for me to visualize. Well, once you get it, it's easier to visualize. But the first time I tried to understand it, it was hard for me to understand how precession was different than obliquity or different than tilt. Obliquity is how much we're going from vertical. And so if we had no precession, we would be exactly pointing in that same direction every year. Now with just precession alone, what happens is every year, this arrow is slowly tracing out a circle that goes like this. So I'm going to exaggerate how much it's happening, just so that you can visualize it."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Obliquity is how much we're going from vertical. And so if we had no precession, we would be exactly pointing in that same direction every year. Now with just precession alone, what happens is every year, this arrow is slowly tracing out a circle that goes like this. So I'm going to exaggerate how much it's happening, just so that you can visualize it. So maybe after several years, that arrow is not, when you're at that same point relative to the sun, that same point in the solar system, that arrow is no longer pointing in that direction. It is now traced out a little bit of that circle. So it is now pointing in this direction."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I'm going to exaggerate how much it's happening, just so that you can visualize it. So maybe after several years, that arrow is not, when you're at that same point relative to the sun, that same point in the solar system, that arrow is no longer pointing in that direction. It is now traced out a little bit of that circle. So it is now pointing in this direction. So if it is now pointing in this direction, will that same point in the solar system, that same point relative to the sun, that same exact point in the orbit, will it still be the summer in the northern hemisphere? Well, it won't, because we're now not pointing directly, or we're not most inclined to the sun at that point. Now we would have been most inclined to the sun a little bit earlier in the year or a little bit earlier in the orbit."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it is now pointing in this direction. So if it is now pointing in this direction, will that same point in the solar system, that same point relative to the sun, that same exact point in the orbit, will it still be the summer in the northern hemisphere? Well, it won't, because we're now not pointing directly, or we're not most inclined to the sun at that point. Now we would have been most inclined to the sun a little bit earlier in the year or a little bit earlier in the orbit. So we would have been most inclined to the sun maybe over here. And it would take many, many, many actual thousands of years for the precession to change this much. But then over here, this is where, at this point in that year, when we would be pointed most towards the sun."}, {"video_title": "Milankovitch cycles precession and obliquity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now we would have been most inclined to the sun a little bit earlier in the year or a little bit earlier in the orbit. So we would have been most inclined to the sun maybe over here. And it would take many, many, many actual thousands of years for the precession to change this much. But then over here, this is where, at this point in that year, when we would be pointed most towards the sun. So what the real effect of precession is doing to our seasons and doing to what our sense of what our year is, is that every year, relative to our orbit on Earth, because Earth is kind of a top that's slowly circling, slowly tracing out this circle with, I guess you could say, with its pole, what it's doing is it's making it tilt towards the sun or away from the sun a little bit earlier each year. I know it's hard to visualize, but you could even take a top out and have a basketball as the sun. And if you play with it, you'll see how that works."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Let's practice identifying functional groups in different compounds. So this molecule on the left is found in perfumes. And let's look for some of the functional groups that we've talked about in the previous videos. Well here is a carbon-carbon double bond. And we know that a carbon-carbon double bond is an alkene. So here is an alkene functional group. Here's another alkene, right?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well here is a carbon-carbon double bond. And we know that a carbon-carbon double bond is an alkene. So here is an alkene functional group. Here's another alkene, right? Here's another carbon-carbon double bond. What is this functional group? We have an OH, and then we have the rest of the molecule."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Here's another alkene, right? Here's another carbon-carbon double bond. What is this functional group? We have an OH, and then we have the rest of the molecule. So we have ROH. ROH is an alcohol. So there's also an alcohol present in this compound."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have an OH, and then we have the rest of the molecule. So we have ROH. ROH is an alcohol. So there's also an alcohol present in this compound. Next let's look at aspirin. So what functional groups can we find in aspirin? Well here is an aromatic ring, right?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So there's also an alcohol present in this compound. Next let's look at aspirin. So what functional groups can we find in aspirin? Well here is an aromatic ring, right? So this is an arene. So there's an arene functional group present in aspirin. What about this one up here?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well here is an aromatic ring, right? So this is an arene. So there's an arene functional group present in aspirin. What about this one up here? We have an OH, and the oxygen is directly bonded to a carbonyl. So let's go ahead and write that out. We have an OH, where the oxygen is directly bonded to a carbon, double bonded to an oxygen."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "What about this one up here? We have an OH, and the oxygen is directly bonded to a carbonyl. So let's go ahead and write that out. We have an OH, where the oxygen is directly bonded to a carbon, double bonded to an oxygen. And then we have the rest of the molecule. So hopefully you recognize this as being a carboxylic acid. So let me go ahead and write that out here."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have an OH, where the oxygen is directly bonded to a carbon, double bonded to an oxygen. And then we have the rest of the molecule. So hopefully you recognize this as being a carboxylic acid. So let me go ahead and write that out here. So this is a carboxylic acid. All right, our next functional group. We have an oxygen, and that oxygen is directly bonded to a carbonyl, right?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write that out here. So this is a carboxylic acid. All right, our next functional group. We have an oxygen, and that oxygen is directly bonded to a carbonyl, right? So here's a carbon double bonded to an oxygen. So let's write this out. We have an oxygen directly bonded to a carbonyl."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have an oxygen, and that oxygen is directly bonded to a carbonyl, right? So here's a carbon double bonded to an oxygen. So let's write this out. We have an oxygen directly bonded to a carbonyl. And then for this oxygen, we'd have the rest of the molecule. So that's all of this stuff over here. And then on the other side of the carbonyl, we have another R group."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have an oxygen directly bonded to a carbonyl. And then for this oxygen, we'd have the rest of the molecule. So that's all of this stuff over here. And then on the other side of the carbonyl, we have another R group. So I'll go ahead and write that in. So that is an ester. ROC double bond O, R is an ester."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then on the other side of the carbonyl, we have another R group. So I'll go ahead and write that in. So that is an ester. ROC double bond O, R is an ester. So there's an ester functional group present in the aspirin molecule. Let's look at some of the common mistakes that students make. One of them is students will say a carboxylic acid is an alcohol."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "ROC double bond O, R is an ester. So there's an ester functional group present in the aspirin molecule. Let's look at some of the common mistakes that students make. One of them is students will say a carboxylic acid is an alcohol. So let me write out here a carboxylic acid so we can talk about that. So sometimes students will look at that and say, oh, well I see an OH, and then I see the rest of the molecule. So isn't that an alcohol?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "One of them is students will say a carboxylic acid is an alcohol. So let me write out here a carboxylic acid so we can talk about that. So sometimes students will look at that and say, oh, well I see an OH, and then I see the rest of the molecule. So isn't that an alcohol? But since this oxygen is right next to this carbonyl, this is a carboxylic acid. So this is an example of a carboxylic acid. If we move the OH further away from the carbonyl, let's go ahead and draw one out like that."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So isn't that an alcohol? But since this oxygen is right next to this carbonyl, this is a carboxylic acid. So this is an example of a carboxylic acid. If we move the OH further away from the carbonyl, let's go ahead and draw one out like that. So here is our carbonyl. And now the OH is moved further away. Now we do have an alcohol."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "If we move the OH further away from the carbonyl, let's go ahead and draw one out like that. So here is our carbonyl. And now the OH is moved further away. Now we do have an alcohol. Now we have an OH and then the rest of the molecule. So this would be, let me go ahead and use a different color here. So now we are talking about an alcohol."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now we do have an alcohol. Now we have an OH and then the rest of the molecule. So this would be, let me go ahead and use a different color here. So now we are talking about an alcohol. So this is an alcohol. And what would this one be? We have a carbonyl and then we have an R group on one side and R group on the other side."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So now we are talking about an alcohol. So this is an alcohol. And what would this one be? We have a carbonyl and then we have an R group on one side and R group on the other side. That is a ketone. Let me draw this out. So when you have a carbonyl and an R group on one side and R group on the other side, they could be the same R group, they could be a different R group."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have a carbonyl and then we have an R group on one side and R group on the other side. That is a ketone. Let me draw this out. So when you have a carbonyl and an R group on one side and R group on the other side, they could be the same R group, they could be a different R group. Sometimes you'll see like R prime drawn for that. So this is a ketone. So now we have a ketone and an alcohol, so two functional groups present in the same compound."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So when you have a carbonyl and an R group on one side and R group on the other side, they could be the same R group, they could be a different R group. Sometimes you'll see like R prime drawn for that. So this is a ketone. So now we have a ketone and an alcohol, so two functional groups present in the same compound. So hopefully you can see the difference between this compound and this compound. This one is a carboxylic acid and this one is a ketone and an alcohol. Another common mistake that I've seen a lot is on this functional group right here on aspirin."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So now we have a ketone and an alcohol, so two functional groups present in the same compound. So hopefully you can see the difference between this compound and this compound. This one is a carboxylic acid and this one is a ketone and an alcohol. Another common mistake that I've seen a lot is on this functional group right here on aspirin. Students will look at this oxygen here and say, okay, I have an oxygen and I have an R group on one side and I have an R group on the other side. So an R group on one side of the oxygen and R group on the other side of the oxygen. Isn't that an ether?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Another common mistake that I've seen a lot is on this functional group right here on aspirin. Students will look at this oxygen here and say, okay, I have an oxygen and I have an R group on one side and I have an R group on the other side. So an R group on one side of the oxygen and R group on the other side of the oxygen. Isn't that an ether? Well, this is, ROR would represent an ether. However, we have this carbonyl here. So this carbonyl right next to this oxygen is what makes this an ester."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Isn't that an ether? Well, this is, ROR would represent an ether. However, we have this carbonyl here. So this carbonyl right next to this oxygen is what makes this an ester. How could we turn that into an ether? Let me go ahead and redraw this molecule here. So I'll first put in our ring."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this carbonyl right next to this oxygen is what makes this an ester. How could we turn that into an ether? Let me go ahead and redraw this molecule here. So I'll first put in our ring. So I drew the double bonds a little bit differently from how I drew it up here, but it doesn't really matter. And then I'll put in our carboxylic acid up here and now when I draw in this oxygen, I'm gonna take out the carbonyl. So now the carbonyl is gone and now we do have an ether."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'll first put in our ring. So I drew the double bonds a little bit differently from how I drew it up here, but it doesn't really matter. And then I'll put in our carboxylic acid up here and now when I draw in this oxygen, I'm gonna take out the carbonyl. So now the carbonyl is gone and now we do have an ether. So this actually is an ether now. We have an oxygen, we have an R group on one side and we have the rest of the molecule over here on the other side. So now this is an ether."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So now the carbonyl is gone and now we do have an ether. So this actually is an ether now. We have an oxygen, we have an R group on one side and we have the rest of the molecule over here on the other side. So now this is an ether. So hopefully you see the difference there. Look for the carbonyl right next to the oxygen. That makes it an ester."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So now this is an ether. So hopefully you see the difference there. Look for the carbonyl right next to the oxygen. That makes it an ester. All right, some more common mistakes that students make is they mix up these two functional groups. So let's look at the functional groups in these two molecules here. We start with benzaldehyde and the name is a dead giveaway as to the functional group."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "That makes it an ester. All right, some more common mistakes that students make is they mix up these two functional groups. So let's look at the functional groups in these two molecules here. We start with benzaldehyde and the name is a dead giveaway as to the functional group. We're talking about an aldehyde here. So first we have our aromatic ring, our arene, and then we have an aldehyde. We have a carbonyl and we have a hydrogen that's directly bonded to the carbonyl carbon."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We start with benzaldehyde and the name is a dead giveaway as to the functional group. We're talking about an aldehyde here. So first we have our aromatic ring, our arene, and then we have an aldehyde. We have a carbonyl and we have a hydrogen that's directly bonded to the carbonyl carbon. So we have an R group and then we have a carbonyl and then we have a hydrogen directly bonded to our carbonyl carbon. That is an aldehyde. If we took off that hydrogen and we put a CH3 instead, that would be the compound on the right."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We have a carbonyl and we have a hydrogen that's directly bonded to the carbonyl carbon. So we have an R group and then we have a carbonyl and then we have a hydrogen directly bonded to our carbonyl carbon. That is an aldehyde. If we took off that hydrogen and we put a CH3 instead, that would be the compound on the right. So now we have a CH3 directly bonded to this carbonyl carbon. So now we have an R group on one side, a carbonyl and then another R group. So we have R, C double bond, O, R. And that is a ketone."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "If we took off that hydrogen and we put a CH3 instead, that would be the compound on the right. So now we have a CH3 directly bonded to this carbonyl carbon. So now we have an R group on one side, a carbonyl and then another R group. So we have R, C double bond, O, R. And that is a ketone. And you can tell by the ending of our name here that we have a ketone present in this compound. So again, this difference is subtle but it's important and a lot of students mess this up. An aldehyde has a hydrogen directly bonded to this carbonyl carbon."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we have R, C double bond, O, R. And that is a ketone. And you can tell by the ending of our name here that we have a ketone present in this compound. So again, this difference is subtle but it's important and a lot of students mess this up. An aldehyde has a hydrogen directly bonded to this carbonyl carbon. But if there's no hydrogen, we're talking about a ketone here. So R, C double bond, O, R is a ketone. Finally, let's look at one giant compound with lots of different functional groups and let's see if we can identify all the functional groups present in this molecule."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "An aldehyde has a hydrogen directly bonded to this carbonyl carbon. But if there's no hydrogen, we're talking about a ketone here. So R, C double bond, O, R is a ketone. Finally, let's look at one giant compound with lots of different functional groups and let's see if we can identify all the functional groups present in this molecule. This molecule is called atenolol. This is a beta blocker, so this is a heart medication. Alright, let's look for some functional groups we've seen before."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Finally, let's look at one giant compound with lots of different functional groups and let's see if we can identify all the functional groups present in this molecule. This molecule is called atenolol. This is a beta blocker, so this is a heart medication. Alright, let's look for some functional groups we've seen before. Here is that aromatic ring. So we know that an arene is present in atenolol. So let me go ahead and write this in here."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Alright, let's look for some functional groups we've seen before. Here is that aromatic ring. So we know that an arene is present in atenolol. So let me go ahead and write this in here. Next, we have an oxygen and there's an R group on one side of the oxygen and R group on the other side of the oxygen. So R, O, R, we know that's an ether. So there's an ether present in this compound."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write this in here. Next, we have an oxygen and there's an R group on one side of the oxygen and R group on the other side of the oxygen. So R, O, R, we know that's an ether. So there's an ether present in this compound. Next, we have an OH and then the rest of the molecule. So R, O, H would be an alcohol. So there's an alcohol present."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So there's an ether present in this compound. Next, we have an OH and then the rest of the molecule. So R, O, H would be an alcohol. So there's an alcohol present. Alright, next we have a nitrogen with a lone pair of electrons. There's an R group on one side, there's an R group on the other side. So this is an amine."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So there's an alcohol present. Alright, next we have a nitrogen with a lone pair of electrons. There's an R group on one side, there's an R group on the other side. So this is an amine. So we have an amine. And finally, over here on the left, so this is one that is messed up a lot. We do have a nitrogen with a lone pair of electrons on it, so it's tempting to say we have an amine here."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is an amine. So we have an amine. And finally, over here on the left, so this is one that is messed up a lot. We do have a nitrogen with a lone pair of electrons on it, so it's tempting to say we have an amine here. But this nitrogen is right next to a carbonyl, so it's not an amine, it's an amide or amid. So this is an amide. So a lot of people pronounce this amid."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We do have a nitrogen with a lone pair of electrons on it, so it's tempting to say we have an amine here. But this nitrogen is right next to a carbonyl, so it's not an amine, it's an amide or amid. So this is an amide. So a lot of people pronounce this amid. So it's not an amine. So let's talk more about the difference between an amide and an amine. So let me go ahead and draw out another compound here."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So a lot of people pronounce this amid. So it's not an amine. So let's talk more about the difference between an amide and an amine. So let me go ahead and draw out another compound here. So we can see we have our NH2, and then we have our carbonyl. So for this one, we have our nitrogen directly bonded to the carbonyl carbon. And that's what makes this an amide."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw out another compound here. So we can see we have our NH2, and then we have our carbonyl. So for this one, we have our nitrogen directly bonded to the carbonyl carbon. And that's what makes this an amide. We can move these electrons into here and push these electrons off onto the oxygen. So resonance is possible with this compound. So this is an amide or an amid."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And that's what makes this an amide. We can move these electrons into here and push these electrons off onto the oxygen. So resonance is possible with this compound. So this is an amide or an amid. If we move the nitrogen further away from the carbonyl, let's go ahead and do that over here. So we have our carbonyl, and now our nitrogen is further away. Now we don't have any more resonance, right?"}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is an amide or an amid. If we move the nitrogen further away from the carbonyl, let's go ahead and do that over here. So we have our carbonyl, and now our nitrogen is further away. Now we don't have any more resonance, right? You can't draw a resonance structure showing the delocalization of the lone pair of electrons on the nitrogen. So now we do have an amine. So this over here, this would be an amine."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Now we don't have any more resonance, right? You can't draw a resonance structure showing the delocalization of the lone pair of electrons on the nitrogen. So now we do have an amine. So this over here, this would be an amine. Let me change colors. Let me do blue. This is an amine."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this over here, this would be an amine. Let me change colors. Let me do blue. This is an amine. And then what would this functional group be? We have a carbonyl, and then we have an R group on one side, R group on the other side. That is a ketone."}, {"video_title": "Identifying functional groups Organic chemistry Khan Academy.mp3", "Sentence": "This is an amine. And then what would this functional group be? We have a carbonyl, and then we have an R group on one side, R group on the other side. That is a ketone. So this is a ketone and an amine. And then over here, we have an amide or an amid. So make sure to know the difference between these."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when it is closer to the sun, so let's say that this is the time of the orbit when it's closer to the sun, this is the perihelion. And when it's furthest from the sun, and I'm exaggerating the difference, this is aphelion. This is the aphelion in our orbit when we are furthest from the sun. Maybe our orbit looks something like this. And what I point out in the first video where we discussed this is that this is not the cause of the seasons. Even though we are 3% closer right now, the way our orbit is set up, and we'll see in future videos that the difference or the eccentricity or how elliptical the orbit is does change over time, how much it deviates from being circular, that's one way to think about eccentricity, that does change over time. But right now, when we are closest to the sun, we are 3% closer than when we are furthest from the sun."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe our orbit looks something like this. And what I point out in the first video where we discussed this is that this is not the cause of the seasons. Even though we are 3% closer right now, the way our orbit is set up, and we'll see in future videos that the difference or the eccentricity or how elliptical the orbit is does change over time, how much it deviates from being circular, that's one way to think about eccentricity, that does change over time. But right now, when we are closest to the sun, we are 3% closer than when we are furthest from the sun. So 3% closer than at aphelion. And we point out in the first video when we discussed this, that this is not the cause of the seasons. And in particular, perihelion, when we are closest to the sun, when we actually have the most radiation from the sun, that's actually when we have the northern hemisphere winter."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But right now, when we are closest to the sun, we are 3% closer than when we are furthest from the sun. So 3% closer than at aphelion. And we point out in the first video when we discussed this, that this is not the cause of the seasons. And in particular, perihelion, when we are closest to the sun, when we actually have the most radiation from the sun, that's actually when we have the northern hemisphere winter. So this occurs right over here. This occurs in January. And aphelion occurs in July."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And in particular, perihelion, when we are closest to the sun, when we actually have the most radiation from the sun, that's actually when we have the northern hemisphere winter. So this occurs right over here. This occurs in January. And aphelion occurs in July. Now, based on this, this might lead to an interesting question. Because, so let's think about January when we're at perihelion, and let's think about July when we're at aphelion. And let me draw a quick globe right over here."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And aphelion occurs in July. Now, based on this, this might lead to an interesting question. Because, so let's think about January when we're at perihelion, and let's think about July when we're at aphelion. And let me draw a quick globe right over here. And let's make that the equator. And I'll draw it in both situations. So January is obviously when we have the northern hemisphere winter."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me draw a quick globe right over here. And let's make that the equator. And I'll draw it in both situations. So January is obviously when we have the northern hemisphere winter. So I'll paint it in blue right over here. It is winter. And July is when we have the northern hemisphere summer or the southern hemisphere winter."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So January is obviously when we have the northern hemisphere winter. So I'll paint it in blue right over here. It is winter. And July is when we have the northern hemisphere summer or the southern hemisphere winter. So then we have winter during July in the southern hemisphere. And let me put summer in a more summery color. I guess that orange is a pretty good color."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And July is when we have the northern hemisphere summer or the southern hemisphere winter. So then we have winter during July in the southern hemisphere. And let me put summer in a more summery color. I guess that orange is a pretty good color. That's not orange. Here's orange. All right."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I guess that orange is a pretty good color. That's not orange. Here's orange. All right. That's orange. And that's orange. So these are summer."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "All right. That's orange. And that's orange. So these are summer. So that's the summer in the southern hemisphere, which occurs during the winter in the northern hemisphere, and vice versa. Summer in the northern hemisphere occurs during winter in the southern hemisphere. And so the question might be rising in your head."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So these are summer. So that's the summer in the southern hemisphere, which occurs during the winter in the northern hemisphere, and vice versa. Summer in the northern hemisphere occurs during winter in the southern hemisphere. And so the question might be rising in your head. And I did see a few comments on that first video asking this question. And it's a good one. If we are closer to the sun in January, or we are really closest to the sun in January, and so we're getting more solar radiation in January, does that moderate the winter in the northern hemisphere?"}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the question might be rising in your head. And I did see a few comments on that first video asking this question. And it's a good one. If we are closer to the sun in January, or we are really closest to the sun in January, and so we're getting more solar radiation in January, does that moderate the winter in the northern hemisphere? Or I guess another way to think about it, does it make the summer in the southern hemisphere when we are closer to the sun, does it make it more extreme or hotter? And vice versa, in July, when we are furthest from the sun, does that moderate the northern hemisphere winter? Because it's hot up there, but hey, we're a little bit further from the sun."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we are closer to the sun in January, or we are really closest to the sun in January, and so we're getting more solar radiation in January, does that moderate the winter in the northern hemisphere? Or I guess another way to think about it, does it make the summer in the southern hemisphere when we are closer to the sun, does it make it more extreme or hotter? And vice versa, in July, when we are furthest from the sun, does that moderate the northern hemisphere winter? Because it's hot up there, but hey, we're a little bit further from the sun. And does it make the southern hemisphere winter colder? So once again, does it make this more extreme? Because it's already winter and we're further from the sun, so maybe we're also getting less radiation."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because it's hot up there, but hey, we're a little bit further from the sun. And does it make the southern hemisphere winter colder? So once again, does it make this more extreme? Because it's already winter and we're further from the sun, so maybe we're also getting less radiation. And so there's a couple ways to think about it. One, it is true that when we're further, we are getting a little bit less radiation from the sun, or we're getting heated up a little bit less. But the one reality is that the southern hemisphere climate as a whole is not more extreme, despite getting heated up, getting more solar energy in the summer, and getting less solar energy in the winter."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because it's already winter and we're further from the sun, so maybe we're also getting less radiation. And so there's a couple ways to think about it. One, it is true that when we're further, we are getting a little bit less radiation from the sun, or we're getting heated up a little bit less. But the one reality is that the southern hemisphere climate as a whole is not more extreme, despite getting heated up, getting more solar energy in the summer, and getting less solar energy in the winter. And the reason why it is not as extreme, let me draw the equator here, just so that we can separate our hemispheres. The main reason, it is believed, why it is not more extreme is that the southern hemisphere has a lot more water in it. So just if you look at the surface of the southern hemisphere, you're looking at a lot more water than the surface of the northern hemisphere."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the one reality is that the southern hemisphere climate as a whole is not more extreme, despite getting heated up, getting more solar energy in the summer, and getting less solar energy in the winter. And the reason why it is not as extreme, let me draw the equator here, just so that we can separate our hemispheres. The main reason, it is believed, why it is not more extreme is that the southern hemisphere has a lot more water in it. So just if you look at the surface of the southern hemisphere, you're looking at a lot more water than the surface of the northern hemisphere. This is, of course, a Mercator projection, and so it distorts things so that things near the poles get really kind of built up to look really huge, even though they really aren't that big. Greenland really isn't larger than South America. It just spreads them out so that you can flatten out the map, so to speak."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So just if you look at the surface of the southern hemisphere, you're looking at a lot more water than the surface of the northern hemisphere. This is, of course, a Mercator projection, and so it distorts things so that things near the poles get really kind of built up to look really huge, even though they really aren't that big. Greenland really isn't larger than South America. It just spreads them out so that you can flatten out the map, so to speak. But the southern hemisphere has more water, and as you may have learned in chemistry class, water has a higher specific heat, higher specific energy. It takes more energy, more heat, to raise water a degree than it does to raise, say, land a degree. And so water can absorb more energy."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It just spreads them out so that you can flatten out the map, so to speak. But the southern hemisphere has more water, and as you may have learned in chemistry class, water has a higher specific heat, higher specific energy. It takes more energy, more heat, to raise water a degree than it does to raise, say, land a degree. And so water can absorb more energy. Or when there's less energy, water will release more energy without dropping as much of a temperature. So water has a moderating influence on the climate. So even though the summers in the southern hemisphere actually are getting more radiation than the summers in the northern hemisphere, it's moderated on the actual temperature because the water has the ability to absorb more of that heat without changing the temperature as dramatically."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so water can absorb more energy. Or when there's less energy, water will release more energy without dropping as much of a temperature. So water has a moderating influence on the climate. So even though the summers in the southern hemisphere actually are getting more radiation than the summers in the northern hemisphere, it's moderated on the actual temperature because the water has the ability to absorb more of that heat without changing the temperature as dramatically. Now with that said, it is true that in general, Antarctica is colder than the North Pole. But the main reason why Antarctica is colder, besides the fact that it's on land as opposed to the North Pole being in the center of the Arctic Ocean, is that it's actually a huge, very high-altitude ice sheet. And so the altitude for most of Antarctica is around 8,000 feet."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So even though the summers in the southern hemisphere actually are getting more radiation than the summers in the northern hemisphere, it's moderated on the actual temperature because the water has the ability to absorb more of that heat without changing the temperature as dramatically. Now with that said, it is true that in general, Antarctica is colder than the North Pole. But the main reason why Antarctica is colder, besides the fact that it's on land as opposed to the North Pole being in the center of the Arctic Ocean, is that it's actually a huge, very high-altitude ice sheet. And so the altitude for most of Antarctica is around 8,000 feet. So it's kind of like an alpine altitude. So the main reason why it's colder is possibly being further away from the sun in winter might play some role there. But the main reason why it's colder is that it's just at a much higher altitude."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the altitude for most of Antarctica is around 8,000 feet. So it's kind of like an alpine altitude. So the main reason why it's colder is possibly being further away from the sun in winter might play some role there. But the main reason why it's colder is that it's just at a much higher altitude. And it's to some degree insulated from the water. Or I guess you could say it's on the land. So especially during the long winters, it's going to get that much colder."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the main reason why it's colder is that it's just at a much higher altitude. And it's to some degree insulated from the water. Or I guess you could say it's on the land. So especially during the long winters, it's going to get that much colder. But I'll leave you there. And to some degree, and this is the other aspect of it, during the summers, and all of this stuff is super complicated, so you can't just throw out one rule of thumb and say this is the reason. But these are all influences."}, {"video_title": "Are southern hemisphere seasons more severe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So especially during the long winters, it's going to get that much colder. But I'll leave you there. And to some degree, and this is the other aspect of it, during the summers, and all of this stuff is super complicated, so you can't just throw out one rule of thumb and say this is the reason. But these are all influences. Is that if you have a large, super large ice sheet, it's also more likely to reflect more of the energy. Because it's white, as opposed to a darker color like the ocean or the land. And so you can think about all of those factors."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The whole point of this video is really just to look at what, in my mind, is one of the coolest pictures ever taken by anything. And this was actually taken by the Hubble telescope. Let me get my pen tool in place. And what they did is they pointed the telescope at this area of our night sky. And obviously, the Hubble telescope, it's out in orbit, so it doesn't have to worry about all of the interference from our actual atmosphere. So it gets a nice, good look at things. But it's right over here, relative to the moon."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what they did is they pointed the telescope at this area of our night sky. And obviously, the Hubble telescope, it's out in orbit, so it doesn't have to worry about all of the interference from our actual atmosphere. So it gets a nice, good look at things. But it's right over here, relative to the moon. Obviously, the moon's moving around. But on that day, it was here, relative to the moon. And they picked this location right over here because there weren't that many nearby stars there."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's right over here, relative to the moon. Obviously, the moon's moving around. But on that day, it was here, relative to the moon. And they picked this location right over here because there weren't that many nearby stars there. So it really allowed the telescope, because if there were nearby stars, that light would have outshone things that are behind it, further away, maybe perhaps galaxies. So just keep in mind, everything you're going to see is in this little patch of the night sky. And I think the main point for showing the moon here, obviously, the moon's moving around."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they picked this location right over here because there weren't that many nearby stars there. So it really allowed the telescope, because if there were nearby stars, that light would have outshone things that are behind it, further away, maybe perhaps galaxies. So just keep in mind, everything you're going to see is in this little patch of the night sky. And I think the main point for showing the moon here, obviously, the moon's moving around. I'm not telling you which exact patch of sky this is. But to really give you an idea of how small of a patch of sky that really is. But you really could have done this with any patch of sky."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I think the main point for showing the moon here, obviously, the moon's moving around. I'm not telling you which exact patch of sky this is. But to really give you an idea of how small of a patch of sky that really is. But you really could have done this with any patch of sky. But in other patches of sky, there would have been other nearby stars that would have blocked things. But the galaxies are there, beyond that, in the clusters of galaxies, in the super clusters of galaxies. So with that said, just remember, everything we're talking about in this video is in this little patch of sky right over here."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But you really could have done this with any patch of sky. But in other patches of sky, there would have been other nearby stars that would have blocked things. But the galaxies are there, beyond that, in the clusters of galaxies, in the super clusters of galaxies. So with that said, just remember, everything we're talking about in this video is in this little patch of sky right over here. And the whole point of this, once again, like all of these videos, is really to kind of just blow your mind. So this right here is what the Hubble telescope saw in that patch. Everything."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So with that said, just remember, everything we're talking about in this video is in this little patch of sky right over here. And the whole point of this, once again, like all of these videos, is really to kind of just blow your mind. So this right here is what the Hubble telescope saw in that patch. Everything. Everything that I'm showing you right here. I just want to be clear. Some of these things are nearby stars."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Everything. Everything that I'm showing you right here. I just want to be clear. Some of these things are nearby stars. But most of these things are galaxies. That's a galaxy. That is a galaxy."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some of these things are nearby stars. But most of these things are galaxies. That's a galaxy. That is a galaxy. That is a galaxy. That's a galaxy. And that's a galaxy."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is a galaxy. That is a galaxy. That's a galaxy. And that's a galaxy. And the reason I wanted to do this, obviously, in the last video, I showed you our local group. I showed you the Virgo super cluster. I showed you the kind of clusters of clusters."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's a galaxy. And the reason I wanted to do this, obviously, in the last video, I showed you our local group. I showed you the Virgo super cluster. I showed you the kind of clusters of clusters. And I even showed you a depiction of the observable universe. But what just is amazing about this, this is an actual picture. This is actually an image of a galaxy."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I showed you the kind of clusters of clusters. And I even showed you a depiction of the observable universe. But what just is amazing about this, this is an actual picture. This is actually an image of a galaxy. Hey, there's another galaxy. Oh look, there's another galaxy up here. So you could keep doing that forever."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is actually an image of a galaxy. Hey, there's another galaxy. Oh look, there's another galaxy up here. So you could keep doing that forever. And this is just in that little patch of sky. And this is obviously not all of the galaxies in the universe. These are just the ones that we could see."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you could keep doing that forever. And this is just in that little patch of sky. And this is obviously not all of the galaxies in the universe. These are just the ones that we could see. There are ones that might be even further. Or there definitely are ones that are further back. And their light is just even more."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These are just the ones that we could see. There are ones that might be even further. Or there definitely are ones that are further back. And their light is just even more. We could probably even focus even on a patch of sky like that and see that many galaxies again. So you could kind of keep, keep zooming in. But I just, this thing, and I encourage you."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And their light is just even more. We could probably even focus even on a patch of sky like that and see that many galaxies again. So you could kind of keep, keep zooming in. But I just, this thing, and I encourage you. I mean, there's so many unbelievable images. You could look up, they're all on the NASA website. A lot of these are on Wikipedia."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I just, this thing, and I encourage you. I mean, there's so many unbelievable images. You could look up, they're all on the NASA website. A lot of these are on Wikipedia. But these images are just, I mean, unbelievable. I mean, you see a galaxy, another galaxy, another galaxy, another galaxy. I suspect some of this stuff might actually be clusters of galaxies."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "A lot of these are on Wikipedia. But these images are just, I mean, unbelievable. I mean, you see a galaxy, another galaxy, another galaxy, another galaxy. I suspect some of this stuff might actually be clusters of galaxies. A galaxy, another galaxy. And remember, where I'm just circling these galaxies just left and right, you kind of lose sight of what each galaxy actually is. These galaxies have hundreds of billions of stars."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I suspect some of this stuff might actually be clusters of galaxies. A galaxy, another galaxy. And remember, where I'm just circling these galaxies just left and right, you kind of lose sight of what each galaxy actually is. These galaxies have hundreds of billions of stars. That even from one pixel on these galaxies is an unimaginable distance. Something that we could never, based with current technology or even current science, we could ever hope to traverse in the lifetime of humanity. Much less the lifetime of one individual."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These galaxies have hundreds of billions of stars. That even from one pixel on these galaxies is an unimaginable distance. Something that we could never, based with current technology or even current science, we could ever hope to traverse in the lifetime of humanity. Much less the lifetime of one individual. So these are just enormous things. And they're just an infinite number of them. So I just wanted to highlight this with you."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Much less the lifetime of one individual. So these are just enormous things. And they're just an infinite number of them. So I just wanted to highlight this with you. And really just, I mean, just look at it. I mean, it's kind of breathtaking. There's a galaxy right over there."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I just wanted to highlight this with you. And really just, I mean, just look at it. I mean, it's kind of breathtaking. There's a galaxy right over there. Another galaxy. I mean, I could keep doing it all day. But it's really, and I feel bad about drawing on it."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's a galaxy right over there. Another galaxy. I mean, I could keep doing it all day. But it's really, and I feel bad about drawing on it. It kind of ruins the picture. But look at that. That's a nice looking galaxy right there."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's really, and I feel bad about drawing on it. It kind of ruins the picture. But look at that. That's a nice looking galaxy right there. You got a bunch right over here. So I mean, it really is just, you know. This video, it's probably the least dense with actual instruction."}, {"video_title": "Hubble image of galaxies Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's a nice looking galaxy right there. You got a bunch right over here. So I mean, it really is just, you know. This video, it's probably the least dense with actual instruction. But hopefully it's one of the most dense with inspiration. Anyway, hopefully you enjoyed that. Let me just finish looking at the entire picture."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video, I gave a little bit of a hand-wavy explanation about why S waves don't travel in liquid or air. And what I want to do in this video is give you a little bit more intuitive understanding of that and really go down to the molecular level. So let's draw a solid. And it has nice covalent bonds, strong bonds between the different molecules. And the bonds are drawn by these lines in between. So if I were to hit this solid, I have this really small hammer where I just hit it at a molecular level. But if I were to hit these molecules hard enough that they move, but not so hard enough that it breaks the bonds, then essentially what it's going to look like is this kind of row of molecules are going to move to the left."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it has nice covalent bonds, strong bonds between the different molecules. And the bonds are drawn by these lines in between. So if I were to hit this solid, I have this really small hammer where I just hit it at a molecular level. But if I were to hit these molecules hard enough that they move, but not so hard enough that it breaks the bonds, then essentially what it's going to look like is this kind of row of molecules are going to move to the left. So you're going to have that row of molecules moving to the left. And then the row above it won't fully move to the left just yet, but it will start to get pulled. So let me just draw all of the bonds."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if I were to hit these molecules hard enough that they move, but not so hard enough that it breaks the bonds, then essentially what it's going to look like is this kind of row of molecules are going to move to the left. So you're going to have that row of molecules moving to the left. And then the row above it won't fully move to the left just yet, but it will start to get pulled. So let me just draw all of the bonds. I'm just drawing all of the same bonds. Because these are strong bonds that we have in a solid, actually they could be ionic bonds as well. Because they are strong bonds that we have in this solid, they'll essentially be pulled."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me just draw all of the bonds. I'm just drawing all of the same bonds. Because these are strong bonds that we have in a solid, actually they could be ionic bonds as well. Because they are strong bonds that we have in this solid, they'll essentially be pulled. They'll be pulled in the direction, the top row will be pulled in the direction of the bottom row. And so they'll start kind of moving in that direction. And then the bottom row will essentially recoil back."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because they are strong bonds that we have in this solid, they'll essentially be pulled. They'll be pulled in the direction, the top row will be pulled in the direction of the bottom row. And so they'll start kind of moving in that direction. And then the bottom row will essentially recoil back. And then you fast forward a little bit. And so then the top row will have moved to the left. And now the bottom row will start to move back."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then the bottom row will essentially recoil back. And then you fast forward a little bit. And so then the top row will have moved to the left. And now the bottom row will start to move back. And then the bottom row will start to kind of move back. Especially because remember, it's bonded to other things down here. It's bonded to more of the solid down here."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now the bottom row will start to move back. And then the bottom row will start to kind of move back. Especially because remember, it's bonded to other things down here. It's bonded to more of the solid down here. So it'll move back. And you can see this transverse wave. You can see this S wave propagating."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's bonded to more of the solid down here. So it'll move back. And you can see this transverse wave. You can see this S wave propagating. Essentially right over here, the kind of peak of the S wave is here. Now it has moved up. Now let's think about the exact same situation with the liquids."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can see this S wave propagating. Essentially right over here, the kind of peak of the S wave is here. Now it has moved up. Now let's think about the exact same situation with the liquids. In liquids, you don't have these strong ionic or covalent bonds between the different molecules. You just have these weak kind of bonds, usually formed due to polarity. So in a liquid, water is a good example."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now let's think about the exact same situation with the liquids. In liquids, you don't have these strong ionic or covalent bonds between the different molecules. You just have these weak kind of bonds, usually formed due to polarity. So in a liquid, water is a good example. You just have these kind of weaker bonds formed because water is a polar molecule. So the kind of halfway polar sides or the halfway positive sides are somewhat attracted to the halfway negative sides. So they kind of flow past each other."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in a liquid, water is a good example. You just have these kind of weaker bonds formed because water is a polar molecule. So the kind of halfway polar sides or the halfway positive sides are somewhat attracted to the halfway negative sides. So they kind of flow past each other. But if I were to hit these water molecules right here with my hammer, what would happen? Well, they're going to start moving to the left. And this one's going to bump into that one, which is going to bump into that one, which is going to bump into that one."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they kind of flow past each other. But if I were to hit these water molecules right here with my hammer, what would happen? Well, they're going to start moving to the left. And this one's going to bump into that one, which is going to bump into that one, which is going to bump into that one. And they're going to move to the left. So they're going to move to the left. But these molecules aren't going to move with them."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this one's going to bump into that one, which is going to bump into that one, which is going to bump into that one. And they're going to move to the left. So they're going to move to the left. But these molecules aren't going to move with them. You could view it as going to break that very weak bond due to polarity. They're going to move away from each other. Let me draw these top molecules in green."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But these molecules aren't going to move with them. You could view it as going to break that very weak bond due to polarity. They're going to move away from each other. Let me draw these top molecules in green. They're essentially just going to flow past each other. And this guy might have had also weak bonds with stuff below it, too. I should draw just dotted lines with stuff below it, too."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me draw these top molecules in green. They're essentially just going to flow past each other. And this guy might have had also weak bonds with stuff below it, too. I should draw just dotted lines with stuff below it, too. But because of the impact here, these guys are just going to flow. They're actually going to compress in this direction. You're going to have a P wave, a compression wave, go in this direction, where this one bumps into that one and then goes back."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I should draw just dotted lines with stuff below it, too. But because of the impact here, these guys are just going to flow. They're actually going to compress in this direction. You're going to have a P wave, a compression wave, go in this direction, where this one bumps into that one and then goes back. And then this one bumps into that one and goes back. And then this one bumps into that one. But the bonds aren't strong enough."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You're going to have a P wave, a compression wave, go in this direction, where this one bumps into that one and then goes back. And then this one bumps into that one and goes back. And then this one bumps into that one. But the bonds aren't strong enough. And it's even more the case with air. But the bonds aren't strong enough for these blue guys to take these green guys for a ride. And the bonds are also not strong enough for the adjacent molecules to kind of help these blue guys to retract to their original position."}, {"video_title": "Why S-waves only travel in solids Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the bonds aren't strong enough. And it's even more the case with air. But the bonds aren't strong enough for these blue guys to take these green guys for a ride. And the bonds are also not strong enough for the adjacent molecules to kind of help these blue guys to retract to their original position. So when I talked about the elasticity in the last video, that's what I was talking about. The bonds aren't strong enough to cause the things that have deformed to kind of move back to where they were. And also, their bonds aren't strong enough to allow the things that are deformed to pull other things with it."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "Here's another SN1 carbocation rearrangement, but this one's pretty challenging. So if we look on the left, this is our starting alkyl halide, and we're heating this alkyl halide with water to produce this tertiary alcohol on the right. So the first step of this mechanism should be loss of a leaving group. So these electrons come off onto the bromine to form the bromide anion. So when we do that, we're taking a bond away from this carbon in red, so that gets a plus one formal charge. So if I draw in my six-membered ring, the carbon in red is this one, and it has a plus one formal charge, so we have a carbocation, and we put in these two methyl groups here. This is a secondary carbocation, because the carbon in red is directly bonded to two other carbons, so this is secondary."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So these electrons come off onto the bromine to form the bromide anion. So when we do that, we're taking a bond away from this carbon in red, so that gets a plus one formal charge. So if I draw in my six-membered ring, the carbon in red is this one, and it has a plus one formal charge, so we have a carbocation, and we put in these two methyl groups here. This is a secondary carbocation, because the carbon in red is directly bonded to two other carbons, so this is secondary. If we look at our product, look at our product here, we have a five-membered ring, and not a six-membered ring. And we know our nucleophile would have to be water in this reaction, and so the oxygen and water forms a bond in our product with this carbon. So let me mark that carbon in blue."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "This is a secondary carbocation, because the carbon in red is directly bonded to two other carbons, so this is secondary. If we look at our product, look at our product here, we have a five-membered ring, and not a six-membered ring. And we know our nucleophile would have to be water in this reaction, and so the oxygen and water forms a bond in our product with this carbon. So let me mark that carbon in blue. So this carbon in blue, which means that must be the plus one formal charge. That must be the carbocation of how we form the product, just based on what we know from earlier examples. So let's sketch that in."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So let me mark that carbon in blue. So this carbon in blue, which means that must be the plus one formal charge. That must be the carbocation of how we form the product, just based on what we know from earlier examples. So let's sketch that in. So we have our five-membered ring, and the carbon in blue is this carbon right here, and that one must have our plus one formal charge, because our nucleophile would attack that carbon in blue. So let me go ahead and just draw in the water molecule here, our nucleophile attacking that. So here's our water molecule."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So let's sketch that in. So we have our five-membered ring, and the carbon in blue is this carbon right here, and that one must have our plus one formal charge, because our nucleophile would attack that carbon in blue. So let me go ahead and just draw in the water molecule here, our nucleophile attacking that. So here's our water molecule. Two lone pairs of electrons on the oxygen, and our nucleophile attacks our electrophile to form a bond between the oxygen and that carbon. So that would form, let's draw that in here next. We have our five-membered ring."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So here's our water molecule. Two lone pairs of electrons on the oxygen, and our nucleophile attacks our electrophile to form a bond between the oxygen and that carbon. So that would form, let's draw that in here next. We have our five-membered ring. We have our carbon in blue, which is right here. And let's draw in a bond to our oxygen. Our oxygen is bonded to two other hydrogens."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "We have our five-membered ring. We have our carbon in blue, which is right here. And let's draw in a bond to our oxygen. Our oxygen is bonded to two other hydrogens. We still have a lone pair of electrons on this oxygen, which gives this oxygen a plus one formal charge. So the last step of this mechanism is just loss of a proton. So we have a proton transfer, so I'm just gonna write here minus H plus."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "Our oxygen is bonded to two other hydrogens. We still have a lone pair of electrons on this oxygen, which gives this oxygen a plus one formal charge. So the last step of this mechanism is just loss of a proton. So we have a proton transfer, so I'm just gonna write here minus H plus. And a base, like a water molecule, would come along and take one of those protons to give us our final product. So we've seen that in lots of earlier videos. But now let's think to ourselves, how do we go from the carbocation on the left to the carbocation on the right?"}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So we have a proton transfer, so I'm just gonna write here minus H plus. And a base, like a water molecule, would come along and take one of those protons to give us our final product. So we've seen that in lots of earlier videos. But now let's think to ourselves, how do we go from the carbocation on the left to the carbocation on the right? Notice the carbocation on the right is a tertiary carbocation. The carbon in blue is directly bonded to three other carbons so this one is tertiary. So we must get some sort of rearrangement going from a secondary carbocation to a tertiary carbocation."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "But now let's think to ourselves, how do we go from the carbocation on the left to the carbocation on the right? Notice the carbocation on the right is a tertiary carbocation. The carbon in blue is directly bonded to three other carbons so this one is tertiary. So we must get some sort of rearrangement going from a secondary carbocation to a tertiary carbocation. But this one's different from any rearrangement we've seen so far. So let's go to the video to make this a little bit more clear. Here's a model of our carbocation."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So we must get some sort of rearrangement going from a secondary carbocation to a tertiary carbocation. But this one's different from any rearrangement we've seen so far. So let's go to the video to make this a little bit more clear. Here's a model of our carbocation. The carbon, the plus one formal charge is this carbon. I left the hydrogen in on that carbon only just to make it easier to see. So for our carbocation rearrangement, I'm gonna take these electrons in the rings."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "Here's a model of our carbocation. The carbon, the plus one formal charge is this carbon. I left the hydrogen in on that carbon only just to make it easier to see. So for our carbocation rearrangement, I'm gonna take these electrons in the rings. We're gonna break the ring, go from a six-membered ring to a five-membered ring with this carbocation rearrangement. And we change the hybridization states of two carbons. So let's get a different model so we can see better what the carbocation looks like."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So for our carbocation rearrangement, I'm gonna take these electrons in the rings. We're gonna break the ring, go from a six-membered ring to a five-membered ring with this carbocation rearrangement. And we change the hybridization states of two carbons. So let's get a different model so we can see better what the carbocation looks like. So the carbon with the hydrogen is now sp3 hybridized and this carbon is sp2 hybridized and planar. I took some images from the video to help us understand this tricky rearrangement. So let's start with this picture."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So let's get a different model so we can see better what the carbocation looks like. So the carbon with the hydrogen is now sp3 hybridized and this carbon is sp2 hybridized and planar. I took some images from the video to help us understand this tricky rearrangement. So let's start with this picture. The carbon in red that I marked above with this carbon, let me circle it one more time here, that is this carbon. Let me highlight another carbon on here and I'll do this one in green. So this carbon in green is this one over here."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So let's start with this picture. The carbon in red that I marked above with this carbon, let me circle it one more time here, that is this carbon. Let me highlight another carbon on here and I'll do this one in green. So this carbon in green is this one over here. So in the video, we took these electrons and we moved them over to this carbon. So when we're drawing our mechanism up here, we take these electrons and we move them over to the carbon in red. And that forms a bond between the carbon in green and the carbon in red."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So this carbon in green is this one over here. So in the video, we took these electrons and we moved them over to this carbon. So when we're drawing our mechanism up here, we take these electrons and we move them over to the carbon in red. And that forms a bond between the carbon in green and the carbon in red. So let's show that up here on the drawing. So that moves us to a five-membered ring and the carbon in red is this one and the carbon in green is this one. And let's go ahead and make this carbon right here blue."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "And that forms a bond between the carbon in green and the carbon in red. So let's show that up here on the drawing. So that moves us to a five-membered ring and the carbon in red is this one and the carbon in green is this one. And let's go ahead and make this carbon right here blue. So this carbon in blue is still attached to the carbon in red. So let me just sketch that in here and then we have two methyl groups coming off that carbon. So the carbon in blue is this one now."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "And let's go ahead and make this carbon right here blue. So this carbon in blue is still attached to the carbon in red. So let me just sketch that in here and then we have two methyl groups coming off that carbon. So the carbon in blue is this one now. When we move to, actually let's highlight the carbon in blue over here. So this is the carbon in blue on this picture. So now let's move to this central picture here."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So the carbon in blue is this one now. When we move to, actually let's highlight the carbon in blue over here. So this is the carbon in blue on this picture. So now let's move to this central picture here. So the carbon in green is this one. The carbon in red is this one. And the carbon in blue is this one down here."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So now let's move to this central picture here. So the carbon in green is this one. The carbon in red is this one. And the carbon in blue is this one down here. So this forms our tertiary carbocation. So there's a plus one formal charge on the carbon in blue. And this is the same thing, right?"}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "And the carbon in blue is this one down here. So this forms our tertiary carbocation. So there's a plus one formal charge on the carbon in blue. And this is the same thing, right? This is just two different ways of drawing our carbocation. The one on the right is a little bit better in terms of how to draw it. But it's really the same picture."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "And this is the same thing, right? This is just two different ways of drawing our carbocation. The one on the right is a little bit better in terms of how to draw it. But it's really the same picture. And let's identify those carbons here. So the carbon in red is this one and on our picture it is this one here. The carbon in green is this one, which is this carbon."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "But it's really the same picture. And let's identify those carbons here. So the carbon in red is this one and on our picture it is this one here. The carbon in green is this one, which is this carbon. And finally the carbon in blue right here is this carbon. And the carbon in red goes, let's go back to the picture all the way over here on the left. The carbon in red starting off with a plus one formal charge is sp2 hybridized."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "The carbon in green is this one, which is this carbon. And finally the carbon in blue right here is this carbon. And the carbon in red goes, let's go back to the picture all the way over here on the left. The carbon in red starting off with a plus one formal charge is sp2 hybridized. But notice when we move over here, we're moving to sp3 hybridization. So tetrahedral geometry around the carbon in red. The carbon in blue is going from sp3 hybridization over here to sp2 hybridization."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "The carbon in red starting off with a plus one formal charge is sp2 hybridized. But notice when we move over here, we're moving to sp3 hybridization. So tetrahedral geometry around the carbon in red. The carbon in blue is going from sp3 hybridization over here to sp2 hybridization. And that's why I switched the model sets because this carbon in blue is now sp2 hybridized when it's a carbocation. So here's our carbon in blue. So it has planar geometry around it."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "The carbon in blue is going from sp3 hybridization over here to sp2 hybridization. And that's why I switched the model sets because this carbon in blue is now sp2 hybridized when it's a carbocation. So here's our carbon in blue. So it has planar geometry around it. So hopefully the models helped clear up this strange carbocation rearrangement or challenging I should say. So let's go back here and let's look at the entire mechanism. So the first step is loss of our leaving group."}, {"video_title": "Sn1 carbocation rearrangement challenge problem.mp3", "Sentence": "So it has planar geometry around it. So hopefully the models helped clear up this strange carbocation rearrangement or challenging I should say. So let's go back here and let's look at the entire mechanism. So the first step is loss of our leaving group. Then we get our carbocation rearrangement going from a secondary carbocation to a tertiary carbocation. The next step is nucleophilic attack where the water molecule attacks our positive charge. And then finally we have an acid-base reaction."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at an acid-base reaction. On the left, acetic acid is gonna function as our Bronsted-Lowry acid. It's gonna be a proton donor. On the right, we have sodium hydroxide, and hydroxide is going to accept a proton. It is going to be a Bronsted-Lowry base. So when you're drawing the mechanism, you use curved arrows to show the flow of electrons, and these two electrons on hydroxide, let's say, are the two that are going to grab the acidic proton on acetic acid. So there's my curved arrow."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the right, we have sodium hydroxide, and hydroxide is going to accept a proton. It is going to be a Bronsted-Lowry base. So when you're drawing the mechanism, you use curved arrows to show the flow of electrons, and these two electrons on hydroxide, let's say, are the two that are going to grab the acidic proton on acetic acid. So there's my curved arrow. Only the protons, so these two electrons, are left behind on this oxygen. So let's draw the products, I should say, of this acid-base reaction. So on the left, we would have our carbon double-bonded to our oxygen, and then now this oxygen would have three lone pairs of electrons around it, which gives this oxygen a negative one formal charge."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there's my curved arrow. Only the protons, so these two electrons, are left behind on this oxygen. So let's draw the products, I should say, of this acid-base reaction. So on the left, we would have our carbon double-bonded to our oxygen, and then now this oxygen would have three lone pairs of electrons around it, which gives this oxygen a negative one formal charge. We'd also have a sodium cation here, so we could think about that, right, an ionic bond. And then what do you get if you add an H plus to OH minus? You would get H2O, or water."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, we would have our carbon double-bonded to our oxygen, and then now this oxygen would have three lone pairs of electrons around it, which gives this oxygen a negative one formal charge. We'd also have a sodium cation here, so we could think about that, right, an ionic bond. And then what do you get if you add an H plus to OH minus? You would get H2O, or water. So let me go ahead and draw water in here, and I'll put in my lone pairs of electrons. Let's follow our electrons along. So the two electrons, right, this lone pair right here on the hydroxide anion, picked up this proton."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You would get H2O, or water. So let me go ahead and draw water in here, and I'll put in my lone pairs of electrons. Let's follow our electrons along. So the two electrons, right, this lone pair right here on the hydroxide anion, picked up this proton. So let's say those two electrons in magenta are these two electrons, and this was the proton that they picked up. And then we also need to follow the electrons in here. I'll make them blue."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the two electrons, right, this lone pair right here on the hydroxide anion, picked up this proton. So let's say those two electrons in magenta are these two electrons, and this was the proton that they picked up. And then we also need to follow the electrons in here. I'll make them blue. So these electrons in blue come off onto the oxygen. So let's say those electrons in blue are right here, which gives the oxygen a negative one formal charge. So this is an acid-base reaction, and we could even identify conjugate acid-base pairs here."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll make them blue. So these electrons in blue come off onto the oxygen. So let's say those electrons in blue are right here, which gives the oxygen a negative one formal charge. So this is an acid-base reaction, and we could even identify conjugate acid-base pairs here. So on the left, right, on the left, this was acetic acid. This was our Bronsted-Lowry acid. What is the conjugate base to acetic acid?"}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is an acid-base reaction, and we could even identify conjugate acid-base pairs here. So on the left, right, on the left, this was acetic acid. This was our Bronsted-Lowry acid. What is the conjugate base to acetic acid? Well, that would be over here, right? Just take away a proton, and this would be the conjugate base. So let me identify this as being the conjugate base."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "What is the conjugate base to acetic acid? Well, that would be over here, right? Just take away a proton, and this would be the conjugate base. So let me identify this as being the conjugate base. This is the acetate anion, right? So this is our conjugate base. For hydroxide, hydroxide on the left side functioned as a base, right?"}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me identify this as being the conjugate base. This is the acetate anion, right? So this is our conjugate base. For hydroxide, hydroxide on the left side functioned as a base, right? So the conjugate acid must be on the right side. So if you add a proton to OH-, you get H2O. So water is the conjugate, whoops, I'm writing conjugate base here, but it's really the conjugate acid, right?"}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "For hydroxide, hydroxide on the left side functioned as a base, right? So the conjugate acid must be on the right side. So if you add a proton to OH-, you get H2O. So water is the conjugate, whoops, I'm writing conjugate base here, but it's really the conjugate acid, right? So we've identified our conjugate acid-base pairs. All right, the biggest mistake that I see when students are drawing acid-base mechanisms is they mess up their curved arrows. So the biggest mistake I see, and I'll do this in red so it'll remind you not to do it, is they show this proton right here moving to the hydroxide anion, and that is incorrect, right?"}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So water is the conjugate, whoops, I'm writing conjugate base here, but it's really the conjugate acid, right? So we've identified our conjugate acid-base pairs. All right, the biggest mistake that I see when students are drawing acid-base mechanisms is they mess up their curved arrows. So the biggest mistake I see, and I'll do this in red so it'll remind you not to do it, is they show this proton right here moving to the hydroxide anion, and that is incorrect, right? That is a very common mistake because curved arrows show the movement of electrons, right? And that's not, this is not what's happening here. These two electrons up here in magenta are the ones that are taking that acidic proton."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the biggest mistake I see, and I'll do this in red so it'll remind you not to do it, is they show this proton right here moving to the hydroxide anion, and that is incorrect, right? That is a very common mistake because curved arrows show the movement of electrons, right? And that's not, this is not what's happening here. These two electrons up here in magenta are the ones that are taking that acidic proton. So this is incorrect. Don't draw your acid-base mechanisms like this. Let's do one more acid-base mechanism for some extra practice here."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These two electrons up here in magenta are the ones that are taking that acidic proton. So this is incorrect. Don't draw your acid-base mechanisms like this. Let's do one more acid-base mechanism for some extra practice here. So on the left we have acetone, and on the right we have the hydronium ion, H3O+. So the hydronium ion is gonna function as our Bronsted-Lowry acid. It's going to donate a proton to acetone, which is going to be our Bronsted-Lowry base."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more acid-base mechanism for some extra practice here. So on the left we have acetone, and on the right we have the hydronium ion, H3O+. So the hydronium ion is gonna function as our Bronsted-Lowry acid. It's going to donate a proton to acetone, which is going to be our Bronsted-Lowry base. Remember, when you're drawing an acid-base mechanism, your curved arrows show the movement of electrons. So if acetone functions as our base, a lone pair of electrons on this oxygen could take this proton right here and leave these electrons behind on this oxygen. So let's show the result of our acid-base mechanism."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's going to donate a proton to acetone, which is going to be our Bronsted-Lowry base. Remember, when you're drawing an acid-base mechanism, your curved arrows show the movement of electrons. So if acetone functions as our base, a lone pair of electrons on this oxygen could take this proton right here and leave these electrons behind on this oxygen. So let's show the result of our acid-base mechanism. So on the left, right, the lone pair on the left of the oxygen didn't do anything. The lone pair on the right of the oxygen picked up a proton, formed a bond. And so we get this with a plus one formal charge on the oxygen."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's show the result of our acid-base mechanism. So on the left, right, the lone pair on the left of the oxygen didn't do anything. The lone pair on the right of the oxygen picked up a proton, formed a bond. And so we get this with a plus one formal charge on the oxygen. We would also form water here. So H2O, let me draw that in. And show our lone pairs of electrons."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we get this with a plus one formal charge on the oxygen. We would also form water here. So H2O, let me draw that in. And show our lone pairs of electrons. And let's follow our electrons again. So the electrons in magenta right here on the oxygen picked up this proton, forming this bond. So this bond right here are the two electrons in magenta."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And show our lone pairs of electrons. And let's follow our electrons again. So the electrons in magenta right here on the oxygen picked up this proton, forming this bond. So this bond right here are the two electrons in magenta. And then the electrons in blue here move off onto the oxygen to add another lone pair of electrons onto that oxygen, giving us water. So identifying our conjugate acid-base pairs again. On the left, hydronium, H3O+, is functioning as our bronzed Lowry acid."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this bond right here are the two electrons in magenta. And then the electrons in blue here move off onto the oxygen to add another lone pair of electrons onto that oxygen, giving us water. So identifying our conjugate acid-base pairs again. On the left, hydronium, H3O+, is functioning as our bronzed Lowry acid. And you take a proton away from that and you're left with the conjugate base. So on the right would be water, which is our conjugate base. And on the left, acetone is functioning as a bronzed Lowry base."}, {"video_title": "Organic acid-base mechanisms Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left, hydronium, H3O+, is functioning as our bronzed Lowry acid. And you take a proton away from that and you're left with the conjugate base. So on the right would be water, which is our conjugate base. And on the left, acetone is functioning as a bronzed Lowry base. So on the right, this right here must be the conjugate acid. So this is the conjugate acid on the right. We've identified our conjugate acid-base pairs and we've shown the movement of electrons using curved arrows."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "In an E2 mechanism, we need an alkyl halide and we need a strong base. On our alkyl halide, the carbon that's directly bonded to the halogen is the alpha carbon, and the carbon next to the alpha carbon is the beta carbon, and we need a beta hydrogen for this reaction to occur. And all four of those atoms must be in the same plane, so this hydrogen, this carbon, this carbon, and this halogen are all in the same plane, which is why I have the bonds drawn in as straight lines here. The hydrogen and the halogen must be on opposite sides if I draw a little line here through the carbon-carbon bond, and that's said to be anti. So for this mechanism, we need anti-periplanar hydrogen and halogen. The hydrogen and halogen are anti to each other, and they are in the same plane. The mechanism for this reaction is a concerted mechanism, so the strong base comes along and takes this proton, which leaves these electrons."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "The hydrogen and the halogen must be on opposite sides if I draw a little line here through the carbon-carbon bond, and that's said to be anti. So for this mechanism, we need anti-periplanar hydrogen and halogen. The hydrogen and halogen are anti to each other, and they are in the same plane. The mechanism for this reaction is a concerted mechanism, so the strong base comes along and takes this proton, which leaves these electrons. These electrons move into here to form our double bond at the same time that these electrons are coming off onto our halogen, so it's concerted. And let's follow those electrons in red here. The electrons in the carbon-hydrogen bond move in here to form our double bond, so our product is an alkene."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "The mechanism for this reaction is a concerted mechanism, so the strong base comes along and takes this proton, which leaves these electrons. These electrons move into here to form our double bond at the same time that these electrons are coming off onto our halogen, so it's concerted. And let's follow those electrons in red here. The electrons in the carbon-hydrogen bond move in here to form our double bond, so our product is an alkene. When you look at the kinetics for an E2 mechanism, the overall rate of the reaction is equal to the rate constant times the concentration of the substrate to the first power, and the substrate is your alkyl halides, times the concentration of the base to the first power, so our strong base. And so this is an elimination reaction that depends on the concentration of both the substrate and the base, and that's why we call this an E2 reaction, E2 mechanism. The E stands for elimination, and let me go ahead and write that in here, so the E is for elimination."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "The electrons in the carbon-hydrogen bond move in here to form our double bond, so our product is an alkene. When you look at the kinetics for an E2 mechanism, the overall rate of the reaction is equal to the rate constant times the concentration of the substrate to the first power, and the substrate is your alkyl halides, times the concentration of the base to the first power, so our strong base. And so this is an elimination reaction that depends on the concentration of both the substrate and the base, and that's why we call this an E2 reaction, E2 mechanism. The E stands for elimination, and let me go ahead and write that in here, so the E is for elimination. This is an elimination reaction, and the two is because this mechanism is bimolecular, meaning it's dependent on the concentration of two things, the concentration of the substrate and the base. So let's say you increase the concentration of your substrate, which is your alkyl halides, by a factor of two, so increase the concentration of that by a factor of two. Let's say you also increase the concentration of your base by a factor of two, so increase that by a factor of two."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "The E stands for elimination, and let me go ahead and write that in here, so the E is for elimination. This is an elimination reaction, and the two is because this mechanism is bimolecular, meaning it's dependent on the concentration of two things, the concentration of the substrate and the base. So let's say you increase the concentration of your substrate, which is your alkyl halides, by a factor of two, so increase the concentration of that by a factor of two. Let's say you also increase the concentration of your base by a factor of two, so increase that by a factor of two. What happens to the overall rate of the reaction? So the rate would be equal to, now you're thinking about two to the first, times two to the first, because both of these are to the first power, and we're doubling the concentration of both. So two times two is, of course, four, so we're gonna increase the rate by a factor of four if you double the concentration of both."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "Let's say you also increase the concentration of your base by a factor of two, so increase that by a factor of two. What happens to the overall rate of the reaction? So the rate would be equal to, now you're thinking about two to the first, times two to the first, because both of these are to the first power, and we're doubling the concentration of both. So two times two is, of course, four, so we're gonna increase the rate by a factor of four if you double the concentration of both. So this reaction is similar to an SN2 reaction in terms of the kinetics. Remember, in SN2, the two is there because it's also a bimolecular mechanism. So here's an E2 reaction."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So two times two is, of course, four, so we're gonna increase the rate by a factor of four if you double the concentration of both. So this reaction is similar to an SN2 reaction in terms of the kinetics. Remember, in SN2, the two is there because it's also a bimolecular mechanism. So here's an E2 reaction. On the left is our alkyl halide. The carbon that's bonded to our halogen is our alpha carbon, and then all of these carbons around here would be beta carbons, but all of our beta carbons are equivalent in this case. Our strong base is the hydroxide anion, so we need a source of the hydroxide anion here, and the mechanism would be E2."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So here's an E2 reaction. On the left is our alkyl halide. The carbon that's bonded to our halogen is our alpha carbon, and then all of these carbons around here would be beta carbons, but all of our beta carbons are equivalent in this case. Our strong base is the hydroxide anion, so we need a source of the hydroxide anion here, and the mechanism would be E2. So let me go ahead and redraw our alkyl halide, so I'll put in lone pairs of electrons on the bromine. We know the mechanism is a concerted mechanism. It happens in one step."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "Our strong base is the hydroxide anion, so we need a source of the hydroxide anion here, and the mechanism would be E2. So let me go ahead and redraw our alkyl halide, so I'll put in lone pairs of electrons on the bromine. We know the mechanism is a concerted mechanism. It happens in one step. So let me draw in our hydroxide over here, and we know that hydroxide is going to take a proton from a beta carbon. So I'll draw in a hydrogen here on that beta carbon, and as our base takes this beta proton, the electrons in this bond would move in here to form a double bond. At the same time, these electrons come off onto the bromine."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "It happens in one step. So let me draw in our hydroxide over here, and we know that hydroxide is going to take a proton from a beta carbon. So I'll draw in a hydrogen here on that beta carbon, and as our base takes this beta proton, the electrons in this bond would move in here to form a double bond. At the same time, these electrons come off onto the bromine. So for our product, we would have an alkene. So let me draw in the alkene that would form, and let me highlight our electron. So these electrons in magenta moved in here to form our double bond, and this is our alkene."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "At the same time, these electrons come off onto the bromine. So for our product, we would have an alkene. So let me draw in the alkene that would form, and let me highlight our electron. So these electrons in magenta moved in here to form our double bond, and this is our alkene. So we would also have the bromide anion, so let me just go ahead and draw this in here. So we would have our bromide anion with a negative one formal charge. So we could say that these electrons in here in blue came off onto bromine to form the bromide anion, and if you take hydroxide and you add a proton onto hydroxide, you would also have water."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So these electrons in magenta moved in here to form our double bond, and this is our alkene. So we would also have the bromide anion, so let me just go ahead and draw this in here. So we would have our bromide anion with a negative one formal charge. So we could say that these electrons in here in blue came off onto bromine to form the bromide anion, and if you take hydroxide and you add a proton onto hydroxide, you would also have water. So let me draw that in here as well. So your bromide anion, and you would have water. But we're really only thinking about this alkene that forms here."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So we could say that these electrons in here in blue came off onto bromine to form the bromide anion, and if you take hydroxide and you add a proton onto hydroxide, you would also have water. So let me draw that in here as well. So your bromide anion, and you would have water. But we're really only thinking about this alkene that forms here. So when we talked about kinetics, we said that our E2 mechanism was similar to an SN2 mechanism. They're both bimolecular. But when you're thinking about the structure of the substrate, E2 is very different from SN2."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "But we're really only thinking about this alkene that forms here. So when we talked about kinetics, we said that our E2 mechanism was similar to an SN2 mechanism. They're both bimolecular. But when you're thinking about the structure of the substrate, E2 is very different from SN2. SN2 reactions did not occur when you had a tertiary substrate, and that was because of steric hindrance. There was too much steric hindrance for the nucleophile to attack. So if you think about hydroxide trying to attack this alkyl halide, there's too much steric hindrance because of these methyl groups here."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "But when you're thinking about the structure of the substrate, E2 is very different from SN2. SN2 reactions did not occur when you had a tertiary substrate, and that was because of steric hindrance. There was too much steric hindrance for the nucleophile to attack. So if you think about hydroxide trying to attack this alkyl halide, there's too much steric hindrance because of these methyl groups here. But when you're thinking about an E2 mechanism, it's different. The hydroxide anion doesn't have to get close to this carbon. It's only taking a proton away, and there's enough room for it to do that."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So if you think about hydroxide trying to attack this alkyl halide, there's too much steric hindrance because of these methyl groups here. But when you're thinking about an E2 mechanism, it's different. The hydroxide anion doesn't have to get close to this carbon. It's only taking a proton away, and there's enough room for it to do that. And I think it'll be a little bit more clear if I show you a video so we can see the difference between those two reactions. Here we have our tertiary alkyl halide with yellow being the halogen, and that's directly bonded to our alpha carbon. And we have three methyl groups around that, so one, two, and three."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "It's only taking a proton away, and there's enough room for it to do that. And I think it'll be a little bit more clear if I show you a video so we can see the difference between those two reactions. Here we have our tertiary alkyl halide with yellow being the halogen, and that's directly bonded to our alpha carbon. And we have three methyl groups around that, so one, two, and three. If we think about this tertiary alkyl halide trying to participate in an SN2 mechanism, if the hydroxide anion tried to function as a nucleophile and attack this carbon, we have these bulky methyl groups in the way which block the hydroxide anion. So there's too much steric hindrance from these relatively bulky methyl groups to attack this carbon. If we think about an E2 mechanism, remember in an E2 mechanism, we're trying to take a proton from the beta carbon."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "And we have three methyl groups around that, so one, two, and three. If we think about this tertiary alkyl halide trying to participate in an SN2 mechanism, if the hydroxide anion tried to function as a nucleophile and attack this carbon, we have these bulky methyl groups in the way which block the hydroxide anion. So there's too much steric hindrance from these relatively bulky methyl groups to attack this carbon. If we think about an E2 mechanism, remember in an E2 mechanism, we're trying to take a proton from the beta carbon. So here's a beta carbon right here. Let's say we're trying to take this proton. We need to get these four atoms in the same plane."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "If we think about an E2 mechanism, remember in an E2 mechanism, we're trying to take a proton from the beta carbon. So here's a beta carbon right here. Let's say we're trying to take this proton. We need to get these four atoms in the same plane. It's easiest to see that in a Newman projection. So if I rotate a little bit and I turn this, we have a different confirmation. Now we can see that these four atoms are all in the same plane, and that's what we need for an E2 mechanism."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "We need to get these four atoms in the same plane. It's easiest to see that in a Newman projection. So if I rotate a little bit and I turn this, we have a different confirmation. Now we can see that these four atoms are all in the same plane, and that's what we need for an E2 mechanism. And it's pretty easy for hydroxide to come along and take this beta proton right here. There's not much steric hindrance at all. So an E2 mechanism is possible."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "Now we can see that these four atoms are all in the same plane, and that's what we need for an E2 mechanism. And it's pretty easy for hydroxide to come along and take this beta proton right here. There's not much steric hindrance at all. So an E2 mechanism is possible. We've just seen that it's possible for a tertiary alkyl halide to undergo an E2 reaction. And actually, as you go from primary to secondary to tertiary, you increase in the rate of reactivity in an E2 reaction. To explain why, let's look at a mechanism for a secondary and a tertiary substrate, and let's analyze the products and see if we can explain why this is the case."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So an E2 mechanism is possible. We've just seen that it's possible for a tertiary alkyl halide to undergo an E2 reaction. And actually, as you go from primary to secondary to tertiary, you increase in the rate of reactivity in an E2 reaction. To explain why, let's look at a mechanism for a secondary and a tertiary substrate, and let's analyze the products and see if we can explain why this is the case. So I'll draw in a beta proton here on my tertiary substrate and think about our strong base taking this proton. These electrons move in here. These electrons come off onto our halogen."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "To explain why, let's look at a mechanism for a secondary and a tertiary substrate, and let's analyze the products and see if we can explain why this is the case. So I'll draw in a beta proton here on my tertiary substrate and think about our strong base taking this proton. These electrons move in here. These electrons come off onto our halogen. So our product would be this alkene. So our electrons in magenta formed our double bond here. So the same thing for our secondary substrate."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "These electrons come off onto our halogen. So our product would be this alkene. So our electrons in magenta formed our double bond here. So the same thing for our secondary substrate. I'll draw in a beta proton, and we have our strong base that's going to take that beta proton. These electrons move in here, and these electrons come off onto our leaving group. And so our product for a secondary substrate would look like this."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So the same thing for our secondary substrate. I'll draw in a beta proton, and we have our strong base that's going to take that beta proton. These electrons move in here, and these electrons come off onto our leaving group. And so our product for a secondary substrate would look like this. The electrons in magenta formed our double bonds. Now let's analyze our two products. So on the left, this alkene is a monosubstituted alkene."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "And so our product for a secondary substrate would look like this. The electrons in magenta formed our double bonds. Now let's analyze our two products. So on the left, this alkene is a monosubstituted alkene. We have one alkyl group bonded to this carbon. On the right, we have a disubstituted alkene. So we have two alkyl groups bonded to this carbon."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So on the left, this alkene is a monosubstituted alkene. We have one alkyl group bonded to this carbon. On the right, we have a disubstituted alkene. So we have two alkyl groups bonded to this carbon. And as the double bond forms in this concerted mechanism, it is stabilized by the presence of these alkyl groups. So remember from the alkene stability video, the increased substitution of the double bond means increased stability. So this is more stable here, so increased stability."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So we have two alkyl groups bonded to this carbon. And as the double bond forms in this concerted mechanism, it is stabilized by the presence of these alkyl groups. So remember from the alkene stability video, the increased substitution of the double bond means increased stability. So this is more stable here, so increased stability. And this alkene has two alkyl groups to stabilize the forming double bond. The alkene on the left has only one alkyl group to stabilize the forming double bond. So that's one way to think about why a tertiary alkyl halide would actually react the fastest."}, {"video_title": "E2 mechanism kinetics and substrate.mp3", "Sentence": "So this is more stable here, so increased stability. And this alkene has two alkyl groups to stabilize the forming double bond. The alkene on the left has only one alkyl group to stabilize the forming double bond. So that's one way to think about why a tertiary alkyl halide would actually react the fastest. Now it is possible for a primary alkyl halide to undergo an E2 mechanism. So don't think that a primary won't. As a matter of fact, it will."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a relatively large separation of charge means a relatively large change in the dipole moment of the carbonyl when it stretches. And so therefore, we get a very strong signal on our IR spectrum. And the signal for the ketone carbonyl shows up at a wave number of approximately 1715 or 1720. It's a little different for a conjugated ketone. So down here, we have the dot structure for a conjugated ketone. Conjugation lowers the signal for the wave number. And let's see why."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's a little different for a conjugated ketone. So down here, we have the dot structure for a conjugated ketone. Conjugation lowers the signal for the wave number. And let's see why. So we could think about resonance here. So if I move these electrons into here, push those electrons off onto oxygen, I could draw a resonance structure. So over here, we would now have a double bond and here, we would now have a single bond of carbon to oxygen and three lone pairs of electrons around the oxygen."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's see why. So we could think about resonance here. So if I move these electrons into here, push those electrons off onto oxygen, I could draw a resonance structure. So over here, we would now have a double bond and here, we would now have a single bond of carbon to oxygen and three lone pairs of electrons around the oxygen. Formal charge is negative one formal charge for the oxygen and this carbon right here has a plus one formal charge. So if the electrons in magenta move over to here like that, then the electrons in blue right here could move off onto your oxygen. So remember, the actual structure of the molecule is a hybrid of your resonance structures."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So over here, we would now have a double bond and here, we would now have a single bond of carbon to oxygen and three lone pairs of electrons around the oxygen. Formal charge is negative one formal charge for the oxygen and this carbon right here has a plus one formal charge. So if the electrons in magenta move over to here like that, then the electrons in blue right here could move off onto your oxygen. So remember, the actual structure of the molecule is a hybrid of your resonance structures. And so the resonance structure on the left, this looks like a carbon-oxygen double bond. The one on the right looks like a carbon-oxygen single bond. So in reality, it's more of a hybrid of those two."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So remember, the actual structure of the molecule is a hybrid of your resonance structures. And so the resonance structure on the left, this looks like a carbon-oxygen double bond. The one on the right looks like a carbon-oxygen single bond. So in reality, it's more of a hybrid of those two. And so over here, I'm going to draw a carbon bonded to an oxygen. I'm gonna give a partial bond right here. So I'm saying it's stronger than a single bond and not quite as strong as a double bond."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in reality, it's more of a hybrid of those two. And so over here, I'm going to draw a carbon bonded to an oxygen. I'm gonna give a partial bond right here. So I'm saying it's stronger than a single bond and not quite as strong as a double bond. The idea is resonance lowers the double bond character of your carbonyl and so that's weakening, that's weakening the carbonyl. And so if you're weakening the bonds, right, remember the value for K, the spring or the force constant goes down. And from earlier videos, we've seen if you decrease K, you're going to decrease the frequency."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I'm saying it's stronger than a single bond and not quite as strong as a double bond. The idea is resonance lowers the double bond character of your carbonyl and so that's weakening, that's weakening the carbonyl. And so if you're weakening the bonds, right, remember the value for K, the spring or the force constant goes down. And from earlier videos, we've seen if you decrease K, you're going to decrease the frequency. You're going to decrease the wave number where you find the signal. And so the signal for the carbonyl, right, let me go ahead and use green here. So the signal for our carbonyl moves down to approximately 1680 or so."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And from earlier videos, we've seen if you decrease K, you're going to decrease the frequency. You're going to decrease the wave number where you find the signal. And so the signal for the carbonyl, right, let me go ahead and use green here. So the signal for our carbonyl moves down to approximately 1680 or so. So definitely less than 1700. And once again, that's because resonance, right, that's weakening, you can think about that decreasing the bond strength. All right, what happens when you add in an electronegative atom, right?"}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the signal for our carbonyl moves down to approximately 1680 or so. So definitely less than 1700. And once again, that's because resonance, right, that's weakening, you can think about that decreasing the bond strength. All right, what happens when you add in an electronegative atom, right? So now we're talking about a generic carboxylic acid derivative here. So why is some electronegative atom, like oxygen for example, right, we could also have resonance here, right? So we could draw a resonance structure."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, what happens when you add in an electronegative atom, right? So now we're talking about a generic carboxylic acid derivative here. So why is some electronegative atom, like oxygen for example, right, we could also have resonance here, right? So we could draw a resonance structure. We could move the lone pair of electrons in here and push those electrons onto our oxygen and we could go ahead and draw a resonance structure. So now once again, the top oxygen would have a negative one formal charge. And now the Y would have a positive one formal charge like that."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we could draw a resonance structure. We could move the lone pair of electrons in here and push those electrons onto our oxygen and we could go ahead and draw a resonance structure. So now once again, the top oxygen would have a negative one formal charge. And now the Y would have a positive one formal charge like that. So if we're keeping with our colors here, so the lone pair of electrons moves in here, right, and then the pi electrons in here in our carbonyl, right, move off onto our oxygen. So once again, for a carboxylic acid derivative, we have resonance. And when you're thinking about the carbonyl, what is that doing to the carbonyl, right?"}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And now the Y would have a positive one formal charge like that. So if we're keeping with our colors here, so the lone pair of electrons moves in here, right, and then the pi electrons in here in our carbonyl, right, move off onto our oxygen. So once again, for a carboxylic acid derivative, we have resonance. And when you're thinking about the carbonyl, what is that doing to the carbonyl, right? In your hybrid, right, it's decreasing the double bond character of the carbonyl. So you're decreasing the strength of the carbonyl and we've already seen, right, decrease K, right, decreasing the strength of the carbonyl, you decrease K, you decrease the signal where you find the wave number, you decrease the frequency of bond vibration. All right, so that's the idea of resonance."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And when you're thinking about the carbonyl, what is that doing to the carbonyl, right? In your hybrid, right, it's decreasing the double bond character of the carbonyl. So you're decreasing the strength of the carbonyl and we've already seen, right, decrease K, right, decreasing the strength of the carbonyl, you decrease K, you decrease the signal where you find the wave number, you decrease the frequency of bond vibration. All right, so that's the idea of resonance. All right, so resonance is present. But when we're talking about a carboxylic acid, you also have an inductive effect. All right, so we talked about this in the video on carboxylic acid derivatives."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so that's the idea of resonance. All right, so resonance is present. But when we're talking about a carboxylic acid, you also have an inductive effect. All right, so we talked about this in the video on carboxylic acid derivatives. There's a competition between resonance and induction. So let's think about induction here. And if we have our group, our carbon, our oxygen, and our electronegative atom, Y, like this, if we are thinking about the double bond as being decreased by resonance, right, so decreased double bond character because of resonance, what is induction?"}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we talked about this in the video on carboxylic acid derivatives. There's a competition between resonance and induction. So let's think about induction here. And if we have our group, our carbon, our oxygen, and our electronegative atom, Y, like this, if we are thinking about the double bond as being decreased by resonance, right, so decreased double bond character because of resonance, what is induction? Induction, of course, refers to the electronegativity of this atom here. So if we have a very electronegative atom relative to carbon, like oxygen, right, this is going to withdraw electron density in this direction. So we can think about, all right, we can think about a lone pair of electrons and this oxygen, right, some electron density, I should say, moving in here to strengthen the carbonyl."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if we have our group, our carbon, our oxygen, and our electronegative atom, Y, like this, if we are thinking about the double bond as being decreased by resonance, right, so decreased double bond character because of resonance, what is induction? Induction, of course, refers to the electronegativity of this atom here. So if we have a very electronegative atom relative to carbon, like oxygen, right, this is going to withdraw electron density in this direction. So we can think about, all right, we can think about a lone pair of electrons and this oxygen, right, some electron density, I should say, moving in here to strengthen the carbonyl. All right, so if you're withdrawing electron density, you can think about some electron density from the oxygen moving in to increase the strength of the carbonyl. And so this has the effect of giving us more of a double bond again, right? So that's just how I like to think about it here."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we can think about, all right, we can think about a lone pair of electrons and this oxygen, right, some electron density, I should say, moving in here to strengthen the carbonyl. All right, so if you're withdrawing electron density, you can think about some electron density from the oxygen moving in to increase the strength of the carbonyl. And so this has the effect of giving us more of a double bond again, right? So that's just how I like to think about it here. And so induction, all right, let me go ahead and write this here, induction, by withdrawing electron density, we're increasing the strength of the carbonyl, so we're increasing the spring constant, or K. So we're increasing the frequency of bond vibration. We expect to find the signal at a higher wave number. And so, once again, these two competing effects are something we have to think about for carboxylic acid derivatives, right?"}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's just how I like to think about it here. And so induction, all right, let me go ahead and write this here, induction, by withdrawing electron density, we're increasing the strength of the carbonyl, so we're increasing the spring constant, or K. So we're increasing the frequency of bond vibration. We expect to find the signal at a higher wave number. And so, once again, these two competing effects are something we have to think about for carboxylic acid derivatives, right? Resonance is going to lower the signal and induction is going to increase the signal. So let's look at a whole bunch of carbonyl compounds here, and let's look at the different IR signals and see if we can explain the signals. Here we have a bunch of compounds that contain the carbonyl, and we would expect to find the signal for the carbonyl at a wave number in a range of 1850 to 1650."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so, once again, these two competing effects are something we have to think about for carboxylic acid derivatives, right? Resonance is going to lower the signal and induction is going to increase the signal. So let's look at a whole bunch of carbonyl compounds here, and let's look at the different IR signals and see if we can explain the signals. Here we have a bunch of compounds that contain the carbonyl, and we would expect to find the signal for the carbonyl at a wave number in a range of 1850 to 1650. So somewhere in that range, we would expect to find the signal. And the way I like to think about this is to kind of divide that in half. So 1850 and 1650, so somewhere in the middle would be about 1750."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Here we have a bunch of compounds that contain the carbonyl, and we would expect to find the signal for the carbonyl at a wave number in a range of 1850 to 1650. So somewhere in that range, we would expect to find the signal. And the way I like to think about this is to kind of divide that in half. So 1850 and 1650, so somewhere in the middle would be about 1750. And if you go a little bit lower than that, 1740, I like to think about that as being the average signal. So therefore, an average value for K, or an average carbonyl strength, right? So if we go a little bit higher than that, so we're looking at an ester now."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 1850 and 1650, so somewhere in the middle would be about 1750. And if you go a little bit lower than that, 1740, I like to think about that as being the average signal. So therefore, an average value for K, or an average carbonyl strength, right? So if we go a little bit higher than that, so we're looking at an ester now. In the video on the reactivity of carboxylic acid derivatives, I told you how to think about the competing effects of resonance and induction. And for an ester, the inductive effect is a little bit stronger than resonance. So if the inductive effect is a little bit stronger than resonance, we've just seen that has the effect of increasing the strength of the carbonyl."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we go a little bit higher than that, so we're looking at an ester now. In the video on the reactivity of carboxylic acid derivatives, I told you how to think about the competing effects of resonance and induction. And for an ester, the inductive effect is a little bit stronger than resonance. So if the inductive effect is a little bit stronger than resonance, we've just seen that has the effect of increasing the strength of the carbonyl. So we're going to increase the value for K a little bit. And if you increase the value for K, you increase the wave number where you find the signal. So if you increase K, you increase the frequency of bond vibration."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if the inductive effect is a little bit stronger than resonance, we've just seen that has the effect of increasing the strength of the carbonyl. So we're going to increase the value for K a little bit. And if you increase the value for K, you increase the wave number where you find the signal. So if you increase K, you increase the frequency of bond vibration. So you increase the wave number. And so the wave number will go up a little bit from 1740 to approximately 1745. So again, this is approximately where you'd find the signal for the carbonyl for an ester."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you increase K, you increase the frequency of bond vibration. So you increase the wave number. And so the wave number will go up a little bit from 1740 to approximately 1745. So again, this is approximately where you'd find the signal for the carbonyl for an ester. Next, let's do an acid anhydride. So for an acid anhydride, the inductive effect is even more important. And if the inductive effect is even more important, so if you think about this oxygen here as being very electronegative, that means that we're going to get an even higher value for K. So an even higher value for K, an even stronger carbonyl."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So again, this is approximately where you'd find the signal for the carbonyl for an ester. Next, let's do an acid anhydride. So for an acid anhydride, the inductive effect is even more important. And if the inductive effect is even more important, so if you think about this oxygen here as being very electronegative, that means that we're going to get an even higher value for K. So an even higher value for K, an even stronger carbonyl. And so if we have a higher value for K, we have a higher frequency of vibration. We expect to find the signal at a higher wave number. And so the wave number goes up to approximately 1760 to 1790."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if the inductive effect is even more important, so if you think about this oxygen here as being very electronegative, that means that we're going to get an even higher value for K. So an even higher value for K, an even stronger carbonyl. And so if we have a higher value for K, we have a higher frequency of vibration. We expect to find the signal at a higher wave number. And so the wave number goes up to approximately 1760 to 1790. And this is the first signal for an acid anhydride. So this is actually the symmetrical stretch. So we have two carbonyls, right?"}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the wave number goes up to approximately 1760 to 1790. And this is the first signal for an acid anhydride. So this is actually the symmetrical stretch. So we have two carbonyls, right? And if they're stretching in phase with each other, that's the signal that we would expect to find. So we also have an asymmetric stretch. So we're actually going to see a second signal for an acid anhydride."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have two carbonyls, right? And if they're stretching in phase with each other, that's the signal that we would expect to find. So we also have an asymmetric stretch. So we're actually going to see a second signal for an acid anhydride. We talked about this in the previous video. It takes a little bit more energy for the asymmetric stretch. And so you find that at a higher wave number."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're actually going to see a second signal for an acid anhydride. We talked about this in the previous video. It takes a little bit more energy for the asymmetric stretch. And so you find that at a higher wave number. So approximately 1810 for the asymmetric stretch. So two signals for the acid anhydride. For an acyl or acid chloride, the inductive effect completely dominates the resonance effect."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so you find that at a higher wave number. So approximately 1810 for the asymmetric stretch. So two signals for the acid anhydride. For an acyl or acid chloride, the inductive effect completely dominates the resonance effect. So even more strongly than the previous two examples. So the chlorine is withdrawing some electron density. We're strengthening the carbonyl even more."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For an acyl or acid chloride, the inductive effect completely dominates the resonance effect. So even more strongly than the previous two examples. So the chlorine is withdrawing some electron density. We're strengthening the carbonyl even more. So the force constant goes up even more. We have a higher frequency of vibration. We get a higher wave number."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're strengthening the carbonyl even more. So the force constant goes up even more. We have a higher frequency of vibration. We get a higher wave number. So the signal for this carbonyl appears at a higher wave number. So approximately 1800 or even a little bit higher than that. So 1815 or so."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We get a higher wave number. So the signal for this carbonyl appears at a higher wave number. So approximately 1800 or even a little bit higher than that. So 1815 or so. So once again, that's the approximate value for the signal of the carbonyl for an acyl or acid chloride. Let's go the other direction. So if this is our average spring constant and pretty close to our average signal for a wave number, let's look at an aldehyde next."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 1815 or so. So once again, that's the approximate value for the signal of the carbonyl for an acyl or acid chloride. Let's go the other direction. So if this is our average spring constant and pretty close to our average signal for a wave number, let's look at an aldehyde next. So here we have an aldehyde. We have a hydrogen here, so we don't have an electronegative atom to think about in the previous examples. We had oxygen, we had oxygen, we had chlorine to think about."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if this is our average spring constant and pretty close to our average signal for a wave number, let's look at an aldehyde next. So here we have an aldehyde. We have a hydrogen here, so we don't have an electronegative atom to think about in the previous examples. We had oxygen, we had oxygen, we had chlorine to think about. But here we have hydrogen. So we're not really worried about the inductive effect here. We're more concerned with the electron donating of the alkyl group."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We had oxygen, we had oxygen, we had chlorine to think about. But here we have hydrogen. So we're not really worried about the inductive effect here. We're more concerned with the electron donating of the alkyl group. So we have an alkyl group right here and alkyl groups are electron donating. So if we're donating some electron density in this direction, we're going to lose a little bit of electron density from our carbonyl. So like charges repel here."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're more concerned with the electron donating of the alkyl group. So we have an alkyl group right here and alkyl groups are electron donating. So if we're donating some electron density in this direction, we're going to lose a little bit of electron density from our carbonyl. So like charges repel here. And therefore, we're decreasing the strength of the carbonyl. So we're decreasing the force constant K. And if we decrease the force constant K, we would expect the signal to be at a lower wave number here. So the average value would go down to approximately 1725."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So like charges repel here. And therefore, we're decreasing the strength of the carbonyl. So we're decreasing the force constant K. And if we decrease the force constant K, we would expect the signal to be at a lower wave number here. So the average value would go down to approximately 1725. And again, all these are just approximate values. I'm not saying it has to be exactly that. It's just somewhere around that."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the average value would go down to approximately 1725. And again, all these are just approximate values. I'm not saying it has to be exactly that. It's just somewhere around that. We would expect to find the signal for the carbonyl for an aldehyde. Comparing an aldehyde to a ketone, a ketone has two alkyl groups. We have even more electron donating."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's just somewhere around that. We would expect to find the signal for the carbonyl for an aldehyde. Comparing an aldehyde to a ketone, a ketone has two alkyl groups. We have even more electron donating. And so we're going to weaken our carbonyl even more. And so if we're weakening our carbonyl, we can think about K decreasing even more than before. And so we should expect the signal for the ketone to be at a lower wave number."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have even more electron donating. And so we're going to weaken our carbonyl even more. And so if we're weakening our carbonyl, we can think about K decreasing even more than before. And so we should expect the signal for the ketone to be at a lower wave number. And it does go down to approximately 1715 or 1720. And so once again, this is pretty close to where you'd find the signal for a ketone here. All right, let's talk about carboxylic acids next."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so we should expect the signal for the ketone to be at a lower wave number. And it does go down to approximately 1715 or 1720. And so once again, this is pretty close to where you'd find the signal for a ketone here. All right, let's talk about carboxylic acids next. And so if we have a carboxylic acid, let's think about it, let's draw another one here. If we have a carboxylic acid in the dimeric form, so we have another carboxylic acid here, we can get some pretty strong hydrogen bonding. So we get some pretty strong hydrogen bonding here."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's talk about carboxylic acids next. And so if we have a carboxylic acid, let's think about it, let's draw another one here. If we have a carboxylic acid in the dimeric form, so we have another carboxylic acid here, we can get some pretty strong hydrogen bonding. So we get some pretty strong hydrogen bonding here. And that's going to have the effect of weakening the carbonyl. So you can think about some electron density moving into here because of your strong hydrogen bonding. And so that turns out to decrease the value for the force constant even more compared to our average value."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we get some pretty strong hydrogen bonding here. And that's going to have the effect of weakening the carbonyl. So you can think about some electron density moving into here because of your strong hydrogen bonding. And so that turns out to decrease the value for the force constant even more compared to our average value. And so for the carboxylic acid in the dimeric form, we would expect to see it somewhere around 1710, although this signal can, the signal can definitely vary for a carboxylic acid, so you might not see this exact value somewhere else because everything depends on things like, you know, what form is it in. Here we're talking about the dimeric form. Finally, we have our amide here."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that turns out to decrease the value for the force constant even more compared to our average value. And so for the carboxylic acid in the dimeric form, we would expect to see it somewhere around 1710, although this signal can, the signal can definitely vary for a carboxylic acid, so you might not see this exact value somewhere else because everything depends on things like, you know, what form is it in. Here we're talking about the dimeric form. Finally, we have our amide here. And a carboxylic acid derivative, so we think about resonance versus induction. And this is the example where resonance dominates. So resonance dominates induction, and we have to think about a resonance structure."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Finally, we have our amide here. And a carboxylic acid derivative, so we think about resonance versus induction. And this is the example where resonance dominates. So resonance dominates induction, and we have to think about a resonance structure. We move these electrons in here, we push those electrons off, so we have the effect of weakening the carbonyl. So we're decreasing the strength of the carbonyl and resonance dominates. So we're gonna decrease the value for K a lot here."}, {"video_title": "IR signals for carbonyl compounds Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So resonance dominates induction, and we have to think about a resonance structure. We move these electrons in here, we push those electrons off, so we have the effect of weakening the carbonyl. So we're decreasing the strength of the carbonyl and resonance dominates. So we're gonna decrease the value for K a lot here. So we're gonna decrease the value for K a lot compared to our made-up average value. And so we're gonna decrease the signal for the wave number. So we're gonna get a much lower wave number here, so approximately 1650 to 1690."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video, we started with a star in its main sequence, like the sun. And inside the core of that star, you had hydrogen fusion going on. So that is hydrogen fusion. And then outside of the core, you just had hydrogen. You just had hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like the soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then outside of the core, you just had hydrogen. You just had hydrogen plasma. And when we say plasma, it's the electrons and protons of the individual atoms have been disassociated because the temperatures and pressures are so high. So they're really just kind of like the soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. This is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they're really just kind of like the soup of electrons and protons, as opposed to proper atoms that we associate with at lower temperatures. So this is a main sequence star right over here. This is a main sequence star right over here. And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense because helium is a more massive atom. It is able to pack more mass in a smaller volume."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we saw in the last video that this hydrogen is fusing into helium. So we start having more and more helium here. And as we have more and more helium, the core becomes more and more dense because helium is a more massive atom. It is able to pack more mass in a smaller volume. So this gets more and more dense. So core becomes more dense until at some point. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is able to pack more mass in a smaller volume. So this gets more and more dense. So core becomes more dense until at some point. And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing. So it starts to fuse hotter. So let me write this. So the fusion."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so while the core is becoming more and more dense, that actually makes the fusion happen faster and faster because it's more dense, more gravitational pressure, more mass wanting to get to it, more pressure on the hydrogen that's fusing. So it starts to fuse hotter. So let me write this. So the fusion. So hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the fusion. So hydrogen fuses faster. And actually, we even see this in our sun. Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually, you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Our sun today is brighter and hotter. It's fusing faster than it was when it was born 4.5 or 4.6 billion years ago. But eventually, you're going to get to the point so that the core, you only have helium. So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. Not all of it."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So there's going to be some point where the entire core is all helium. And it's going to be way denser than this core over here. All of that mass over there has now been turned into helium. Not all of it. A lot of it has been turned into energy. But most of it is now in helium. And it's going to be in a much, much smaller volume."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Not all of it. A lot of it has been turned into energy. But most of it is now in helium. And it's going to be in a much, much smaller volume. And the whole time, the temperature is increasing. The fusion is getting faster and faster. And now that it's in this dense volume of helium that's not fusing, you do have, and we saw this in this video, a shell around it of hydrogen that is fusing."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's going to be in a much, much smaller volume. And the whole time, the temperature is increasing. The fusion is getting faster and faster. And now that it's in this dense volume of helium that's not fusing, you do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, was what's going on in the core is that the core is getting more and more dense."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now that it's in this dense volume of helium that's not fusing, you do have, and we saw this in this video, a shell around it of hydrogen that is fusing. So this right here is hydrogen fusion going on. And then this over here is just hydrogen plasma. Now the unintuitive thing, or at least this was unintuitive to me at first, was what's going on in the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now the unintuitive thing, or at least this was unintuitive to me at first, was what's going on in the core is that the core is getting more and more dense. It's fusing at a faster rate. And so it's getting hotter and hotter. So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the core is hotter, fusing faster, getting more and more dense. I kind of imagine it's starting to collapse. Every time it collapses, it's getting hotter and more dense. But at the same time that's happening, the star itself is getting bigger. The star itself is getting bigger. And this is actually not drawn to scale. Red giants are much, much larger than main sequence stars, but the whole time that this is getting more dense, the rest of the stars, you can kind of view it as getting less dense."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But at the same time that's happening, the star itself is getting bigger. The star itself is getting bigger. And this is actually not drawn to scale. Red giants are much, much larger than main sequence stars, but the whole time that this is getting more dense, the rest of the stars, you can kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset, the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler than the surface of a main sequence star, this right here is hotter."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Red giants are much, much larger than main sequence stars, but the whole time that this is getting more dense, the rest of the stars, you can kind of view it as getting less dense. And that's because this is generating so much energy that it's able to more than offset, or better offset, the gravitational pull into it. So even though this is hotter, it's able to disperse the rest of the material in the sun over a larger volume. And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler than the surface of a main sequence star, this right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today. Or another way to put it, it will have the same diameter as the Earth's orbit around the current sun. Or another way to view it is, where we are right now will be on the surface, or near the surface, or maybe even inside of that future sun."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that volume is so big that the surface, and we saw this in the last video, the surface of the red giant is actually cooler than the surface of a main sequence star, this right here is hotter. And just to put things in perspective, when the sun becomes a red giant, and it will become a red giant, its diameter will be 100 times the diameter that it is today. Or another way to put it, it will have the same diameter as the Earth's orbit around the current sun. Or another way to view it is, where we are right now will be on the surface, or near the surface, or maybe even inside of that future sun. Or another way to put it, when the sun becomes a red giant, the Earth's going to be not even a speck out here, and it will be liquefied and vaporized at that point in time, so this is super, super huge. And we've even thought about it. Just for light to reach the current sun to our point in orbit, it takes 8 minutes."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or another way to view it is, where we are right now will be on the surface, or near the surface, or maybe even inside of that future sun. Or another way to put it, when the sun becomes a red giant, the Earth's going to be not even a speck out here, and it will be liquefied and vaporized at that point in time, so this is super, super huge. And we've even thought about it. Just for light to reach the current sun to our point in orbit, it takes 8 minutes. So that's how big one of these stars are. To get from one side of the star to another side of the star, it'll take 16 minutes for light to travel, if it was traveling that diameter. And even slightly longer, if it was to travel in a circumference."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just for light to reach the current sun to our point in orbit, it takes 8 minutes. So that's how big one of these stars are. To get from one side of the star to another side of the star, it'll take 16 minutes for light to travel, if it was traveling that diameter. And even slightly longer, if it was to travel in a circumference. So these are huge, huge, huge stars. And we'll talk about other stars in the future that are even bigger than this when they become super giants. But anyway, we have the helium in the center."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even slightly longer, if it was to travel in a circumference. So these are huge, huge, huge stars. And we'll talk about other stars in the future that are even bigger than this when they become super giants. But anyway, we have the helium in the center. Let me write this down. We have a helium core in the center. We're fusing faster and faster and faster."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But anyway, we have the helium in the center. Let me write this down. We have a helium core in the center. We're fusing faster and faster and faster. We're now a red giant. The core is getting hotter and hotter and hotter, until it gets to the temperature for ignition of helium. So until it gets to 100 million Kelvin."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're fusing faster and faster and faster. We're now a red giant. The core is getting hotter and hotter and hotter, until it gets to the temperature for ignition of helium. So until it gets to 100 million Kelvin. Remember, the ignition temperature for hydrogen was 10 million Kelvin. So now we're at 100 million Kelvin, factor of 10. And now, all of a sudden, in the core, you actually start to have helium fusion."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So until it gets to 100 million Kelvin. Remember, the ignition temperature for hydrogen was 10 million Kelvin. So now we're at 100 million Kelvin, factor of 10. And now, all of a sudden, in the core, you actually start to have helium fusion. And we touched on this in the last video, but the helium is fusing into heavier elements. And some of those heavier elements, and predominantly it will be carbon and oxygen. And you may suspect, this is how heavier and heavier elements form in the universe."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now, all of a sudden, in the core, you actually start to have helium fusion. And we touched on this in the last video, but the helium is fusing into heavier elements. And some of those heavier elements, and predominantly it will be carbon and oxygen. And you may suspect, this is how heavier and heavier elements form in the universe. They form literally due to fusion in the core of stars. Especially when we're talking about elements up to iron. But anyway, the core is now experiencing helium fusion."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you may suspect, this is how heavier and heavier elements form in the universe. They form literally due to fusion in the core of stars. Especially when we're talking about elements up to iron. But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there. Does not quite have the pressures and temperatures to fuse yet. So just regular helium."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But anyway, the core is now experiencing helium fusion. It has a shell around it of helium that is not quite there. Does not quite have the pressures and temperatures to fuse yet. So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So just regular helium. But then outside of that, we do have the pressures and temperatures for hydrogen to continue to fuse. So out here, you do have hydrogen fusion. And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden, now this is once again providing some type of outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then outside over here, you just have the regular hydrogen plasma. So what just happened here? When you have helium fusion all of a sudden, now this is once again providing some type of outward support for the core. So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense. Because now we have energy going outward. Energy pushing things outward. But at the same time that that is happening, more and more hydrogen in this layer is turning into helium."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's going to counteract the ever-increasing contraction of the core as it gets more and more dense. Because now we have energy going outward. Energy pushing things outward. But at the same time that that is happening, more and more hydrogen in this layer is turning into helium. Is fusing into helium. So it's making this inert part of the helium core even larger and larger denser. And even larger and larger."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But at the same time that that is happening, more and more hydrogen in this layer is turning into helium. Is fusing into helium. So it's making this inert part of the helium core even larger and larger denser. And even larger and larger. And putting even more pressure on this inside part. And so what's actually going to happen within a few moments from, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super, I shouldn't use igniting or fusing, in a super hot level. But it's contained to do all of this pressure."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even larger and larger. And putting even more pressure on this inside part. And so what's actually going to happen within a few moments from, I guess, especially from a cosmological point of view, this helium fusion is going to be burning super, I shouldn't use igniting or fusing, in a super hot level. But it's contained to do all of this pressure. But at some point, that pressure won't be able to contain it and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's contained to do all of this pressure. But at some point, that pressure won't be able to contain it and the core is going to explode. But it's not going to be one of these catastrophic explosions where the star is going to be destroyed. It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. A helium flash. But once that happens all of a sudden, then now the star is going to be more stable."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's just going to release a lot of energy all of a sudden into the star. And that's called a helium flash. A helium flash. But once that happens all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down. Because red giants in general are already getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But once that happens all of a sudden, then now the star is going to be more stable. And I'll use that in quotes without writing it down. Because red giants in general are already getting to be less stable than a main sequence star. But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But once that happens, you now will have a slightly larger volume. So it's not being contained in as small of a tight volume. That helium flash kind of took care of that. So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now you have helium fusing into carbon and oxygen. And there's all sorts of other combinations of things. Obviously, there's many elements in between helium and carbon and oxygen. But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But these are the ones that dominate. And then outside of that, you have helium forming. You have helium that is not fusing. And then outside of that, you have your fusing hydrogen. You have over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages?"}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then outside of that, you have your fusing hydrogen. You have over here, you have hydrogen fusing into helium. And then out here in the rest of the radius of our super huge red giant, you just have your hydrogen plasma out here. Now what's going to happen as this star ages? Well, if we fast forward this a bunch. And remember, as the star gets denser and denser in the core and the reactions happen faster and faster in this core, it's expelling more and more energy outward. The star keeps growing."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now what's going to happen as this star ages? Well, if we fast forward this a bunch. And remember, as the star gets denser and denser in the core and the reactions happen faster and faster in this core, it's expelling more and more energy outward. The star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun. If it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The star keeps growing. And the surface gets cooler and cooler. So if we fast forward a bunch, and this is what's going to happen to something the mass of our sun. If it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually, you will have a core of carbon and oxygen, or mainly carbon and oxygen, surrounded by fusing helium, surrounded by non-fusing helium, surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just a hydrogen plasma of the sun. But eventually, all of this fuel will run out."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If it's more massive, then at some point, the core of carbon and oxygen that's forming can start to fuse into even heavier elements. But in the case of the sun, it will never get to that 600 million Kelvin to actually fuse the carbon and the oxygen. And so eventually, you will have a core of carbon and oxygen, or mainly carbon and oxygen, surrounded by fusing helium, surrounded by non-fusing helium, surrounded by fusing hydrogen, which is surrounded by non-fusing hydrogen, or just a hydrogen plasma of the sun. But eventually, all of this fuel will run out. All of the hydrogen will run out in the stars. All of this fusing hydrogen will run out. All of this fusing helium will run out."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But eventually, all of this fuel will run out. All of the hydrogen will run out in the stars. All of this fusing hydrogen will run out. All of this fusing helium will run out. All of this is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "All of this fusing helium will run out. All of this is the fusing hydrogen. This is the inert helium, which will run out. It'll be used in kind of this core being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super dense. This whole time, it'll be getting more and more dense as heavier and heavier elements show up in the core. So it gets denser and denser and denser."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It'll be used in kind of this core being fused into the carbon and oxygen, until you get to a point where you literally just have a really hot core of carbon and oxygen. And it's super dense. This whole time, it'll be getting more and more dense as heavier and heavier elements show up in the core. So it gets denser and denser and denser. But this super dense thing will not, in the case of the sun, if it was a more massive star, it would get there. But in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this dense ball of carbon, super dense ball of carbon and oxygen."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it gets denser and denser and denser. But this super dense thing will not, in the case of the sun, if it was a more massive star, it would get there. But in the case of the sun, it will not get hot enough for the carbon and the oxygen to form. So it really will just be this dense ball of carbon, super dense ball of carbon and oxygen. And all of the other material in the sun, remember, was super energetic. It was releasing tons and tons of energy. The more that we progressed down this, the more energy was releasing outward."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it really will just be this dense ball of carbon, super dense ball of carbon and oxygen. And all of the other material in the sun, remember, was super energetic. It was releasing tons and tons of energy. The more that we progressed down this, the more energy was releasing outward. And the larger the radius of the star gained. And the cooler the outside of the star became. Until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The more that we progressed down this, the more energy was releasing outward. And the larger the radius of the star gained. And the cooler the outside of the star became. Until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center, so I could just draw it as this huge, this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a super dense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Until the outside just becomes this kind of cloud, this huge cloud of gas around what once was the star. And in the center, so I could just draw it as this huge, this is now way far away from the star, much even bigger than the radius or the diameter of a red giant. And all we'll have left is a mass, a super dense mass of, I would call it, inert carbon or oxygen. This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot, we'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many years until it becomes, when it's no longer, when it's completely cooled down, lost all of its energy, it'll just be this super dense ball of carbon and oxygen."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is in the case of the sun. And at first, when it's hot, and it will be releasing radiation because it's so hot, we'll call this a white dwarf. This right here is called a white dwarf. And it'll cool down over many, many, many, many, many, many, many years until it becomes, when it's no longer, when it's completely cooled down, lost all of its energy, it'll just be this super dense ball of carbon and oxygen. At which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it."}, {"video_title": "White and black dwarfs Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it'll cool down over many, many, many, many, many, many, many years until it becomes, when it's no longer, when it's completely cooled down, lost all of its energy, it'll just be this super dense ball of carbon and oxygen. At which point, we would call it a black dwarf. And these are obviously very hard to observe because they're not emitting light. And they don't have quite the mass of something like a black hole that isn't even emitting light, but you can see how it's affecting things around it. So that's what's going to happen to the sun. In the next few videos, we're going to talk about what would happen to things less massive than the sun and what would happen to things more massive than the sun. Although I think you can imagine the more massive, then there would be so much pressure on these things because you have so much mass around it that these would begin to fuse into heavier and heavier elements until we get to iron."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we'll start with pyrrole right down here. So the pyrrole molecule, as you can see, five atoms in the ring. And if we take a look at the carbons in the ring, we can see that those carbons all have a double bond to them. Therefore, each of those carbons is sp2 hybridized, meaning each of the carbons has a free p orbital. So I can go ahead and draw a free p orbital on each of those four carbons like that. In terms of the nitrogen on the ring, I need to know the hybridization state of this nitrogen. So the best way to do that is to calculate the steric number."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, each of those carbons is sp2 hybridized, meaning each of the carbons has a free p orbital. So I can go ahead and draw a free p orbital on each of those four carbons like that. In terms of the nitrogen on the ring, I need to know the hybridization state of this nitrogen. So the best way to do that is to calculate the steric number. So we know the steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. And so I can see that here is a sigma bond, here is a sigma bond, and here is a sigma bond. So three sigma bonds plus lone pairs of electrons."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the best way to do that is to calculate the steric number. So we know the steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. And so I can see that here is a sigma bond, here is a sigma bond, and here is a sigma bond. So three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen there. So the steric number should be equal to 4, which would imply an sp3 hybridization state for pyrrole. So we know that's not the case because pyrrole is an aromatic molecule."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So three sigma bonds plus lone pairs of electrons. There's one lone pair of electrons on that nitrogen there. So the steric number should be equal to 4, which would imply an sp3 hybridization state for pyrrole. So we know that's not the case because pyrrole is an aromatic molecule. An sp3 hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp2 hybridized. And of course, we saw how to do that in the end of the last video."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we know that's not the case because pyrrole is an aromatic molecule. An sp3 hybridized nitrogen would mean no p orbitals at that nitrogen, which would violate the first criterion for this compound to be aromatic. And so there must be some way to get that nitrogen to be sp2 hybridized. And of course, we saw how to do that in the end of the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. So if those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, we saw how to do that in the end of the last video. This lone pair of electrons on this nitrogen is actually not localized to this nitrogen. We can take this lone pair of electrons and move them in here so that lone pair of electrons can participate in resonance. So if those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon. So the resonance structure will have nitrogen with a pi bond here now. And a lone pair of electrons on this carbon, which would give this carbon a negative 1 formal charge. We still have a pi bond over here like that."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if those lone pairs of electrons move into there to form a pi bond, that would kick these electrons off onto this carbon. So the resonance structure will have nitrogen with a pi bond here now. And a lone pair of electrons on this carbon, which would give this carbon a negative 1 formal charge. We still have a pi bond over here like that. And this nitrogen now would have a plus 1 formal charge. Now when we analyze the hybridization state of this nitrogen, we can see that, once again, we're going for sigma bonds. So 1 sigma bond, 2 sigma bonds, 3 sigma bonds."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We still have a pi bond over here like that. And this nitrogen now would have a plus 1 formal charge. Now when we analyze the hybridization state of this nitrogen, we can see that, once again, we're going for sigma bonds. So 1 sigma bond, 2 sigma bonds, 3 sigma bonds. So 3 sigma bonds. This time, no lone pairs of electrons because that lone pair of electrons is now delocalized in resonance. And so 3 plus 0 is, of course, 3, meaning that this nitrogen is now sp2 hybridized."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So 1 sigma bond, 2 sigma bonds, 3 sigma bonds. So 3 sigma bonds. This time, no lone pairs of electrons because that lone pair of electrons is now delocalized in resonance. And so 3 plus 0 is, of course, 3, meaning that this nitrogen is now sp2 hybridized. Since that nitrogen is sp2 hybridized, it has a free p orbital. So we can go ahead and draw the p orbital on that nitrogen. And so you could think about in terms of dot structure."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so 3 plus 0 is, of course, 3, meaning that this nitrogen is now sp2 hybridized. Since that nitrogen is sp2 hybridized, it has a free p orbital. So we can go ahead and draw the p orbital on that nitrogen. And so you could think about in terms of dot structure. So these two electrons over here, these two electrons are actually delocalized and participate in resonance. So that lone pair of electrons you could think about as occupying a p orbital here. And they're actually delocalized."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so you could think about in terms of dot structure. So these two electrons over here, these two electrons are actually delocalized and participate in resonance. So that lone pair of electrons you could think about as occupying a p orbital here. And they're actually delocalized. So we have all these pi electrons delocalized throughout our ring. And so let's go ahead and check the criteria for aromaticity. So pyrrole does contain a ring of continuously overlapping p orbitals."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And they're actually delocalized. So we have all these pi electrons delocalized throughout our ring. And so let's go ahead and check the criteria for aromaticity. So pyrrole does contain a ring of continuously overlapping p orbitals. And it does have 4n plus 2 pi electrons in that ring. So let's go ahead and highlight those. So we had these pi electrons."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So pyrrole does contain a ring of continuously overlapping p orbitals. And it does have 4n plus 2 pi electrons in that ring. So let's go ahead and highlight those. So we had these pi electrons. So that's 2. These pi electrons. So that's 4."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we had these pi electrons. So that's 2. These pi electrons. So that's 4. And then these pi electrons here in magenta are actually delocalized in the ring. So that gives us 6 pi electrons. So if n is equal to 1, 4 times 1 plus 2 gives me 6 pi electrons."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's 4. And then these pi electrons here in magenta are actually delocalized in the ring. So that gives us 6 pi electrons. So if n is equal to 1, 4 times 1 plus 2 gives me 6 pi electrons. So pyrrole has 6 pi electrons. And it also has a ring of continuously overlapping p orbitals. So we can say that it is aromatic."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if n is equal to 1, 4 times 1 plus 2 gives me 6 pi electrons. So pyrrole has 6 pi electrons. And it also has a ring of continuously overlapping p orbitals. So we can say that it is aromatic. Let's go ahead and look at another molecule here. So similar to it, this is imidazole. And so for imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we can say that it is aromatic. Let's go ahead and look at another molecule here. So similar to it, this is imidazole. And so for imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here. So at first, it looks like that nitrogen might be sp3 hybridized. But we can draw a resonance structure for it. So we can take these electrons in here and move them in here, which would kick these electrons off onto that top nitrogen."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so for imidazole, once again, we have the same sort of situation that we had for pyrrole with this nitrogen right here. So at first, it looks like that nitrogen might be sp3 hybridized. But we can draw a resonance structure for it. So we can take these electrons in here and move them in here, which would kick these electrons off onto that top nitrogen. So let's go ahead and draw the resonance structure for that. So we would have some pi electrons here. We would have a double bond here."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we can take these electrons in here and move them in here, which would kick these electrons off onto that top nitrogen. So let's go ahead and draw the resonance structure for that. So we would have some pi electrons here. We would have a double bond here. We would now have a plus 1 formal charge on this nitrogen. And we would have a negative 1 formal charge on this top nitrogen right here like that. So once again, we've seen that this nitrogen in blue, it's actually sp2 hybridized."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We would have a double bond here. We would now have a plus 1 formal charge on this nitrogen. And we would have a negative 1 formal charge on this top nitrogen right here like that. So once again, we've seen that this nitrogen in blue, it's actually sp2 hybridized. So let me go ahead and highlight it in blue over here on the right. This nitrogen is actually sp2 hybridized. So it's the exact same situation that we saw in pyrrole, which at first, it looks like those electrons might be localized to that nitrogen."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we've seen that this nitrogen in blue, it's actually sp2 hybridized. So let me go ahead and highlight it in blue over here on the right. This nitrogen is actually sp2 hybridized. So it's the exact same situation that we saw in pyrrole, which at first, it looks like those electrons might be localized to that nitrogen. Those electrons are actually not. Those electrons are delocalized in the ring because of the possible resonance structure. And so we've determined that that nitrogen in blue that's bonded to that hydrogen is sp2 hybridized."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's the exact same situation that we saw in pyrrole, which at first, it looks like those electrons might be localized to that nitrogen. Those electrons are actually not. Those electrons are delocalized in the ring because of the possible resonance structure. And so we've determined that that nitrogen in blue that's bonded to that hydrogen is sp2 hybridized. So over here, I can go ahead and identify that nitrogen in blue that's bonded to the hydrogen. It's sp2 hybridized. Therefore, it has a p orbital like that."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we've determined that that nitrogen in blue that's bonded to that hydrogen is sp2 hybridized. So over here, I can go ahead and identify that nitrogen in blue that's bonded to the hydrogen. It's sp2 hybridized. Therefore, it has a p orbital like that. So I can go ahead and draw the p orbital on that nitrogen. If I look at the carbons in this molecule, so I'm going to go ahead and highlight these carbons in red here. So these three carbons of my original dot structure, we can see they have a double bond to them."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, it has a p orbital like that. So I can go ahead and draw the p orbital on that nitrogen. If I look at the carbons in this molecule, so I'm going to go ahead and highlight these carbons in red here. So these three carbons of my original dot structure, we can see they have a double bond to them. So those are all sp2 hybridized. So I can go ahead and draw in the p orbital on those carbons as well. And then once again, going back to the original dot structure, and this time looking at the nitrogen, so I'm going to go ahead and draw this one in magenta."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these three carbons of my original dot structure, we can see they have a double bond to them. So those are all sp2 hybridized. So I can go ahead and draw in the p orbital on those carbons as well. And then once again, going back to the original dot structure, and this time looking at the nitrogen, so I'm going to go ahead and draw this one in magenta. So this nitrogen right here, if I look at this first dot structure, it's just like the example in pyridine that we saw in the last video. So it's actually sp2 hybridized. And since it's sp2 hybridized, we can think about these electrons in here in magenta as participating resonance and the electrons in blue here as being localized to that nitrogen atom, localized to an sp2 hybridized orbital."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then once again, going back to the original dot structure, and this time looking at the nitrogen, so I'm going to go ahead and draw this one in magenta. So this nitrogen right here, if I look at this first dot structure, it's just like the example in pyridine that we saw in the last video. So it's actually sp2 hybridized. And since it's sp2 hybridized, we can think about these electrons in here in magenta as participating resonance and the electrons in blue here as being localized to that nitrogen atom, localized to an sp2 hybridized orbital. And so I can go ahead and identify this nitrogen in magenta over here. So I'm saying this nitrogen is in magenta. I can say that it is sp2 hybridized."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And since it's sp2 hybridized, we can think about these electrons in here in magenta as participating resonance and the electrons in blue here as being localized to that nitrogen atom, localized to an sp2 hybridized orbital. And so I can go ahead and identify this nitrogen in magenta over here. So I'm saying this nitrogen is in magenta. I can say that it is sp2 hybridized. It's the same situation as pyridine. So I can go ahead and draw a p orbital on it. And that lone pair of electrons in blue that's on that nitrogen, that lone pair of electrons occupies, I'm going to go ahead and put them in blue, that lone pair of electrons is going to occupy an sp2 hybridized orbital on that nitrogen."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I can say that it is sp2 hybridized. It's the same situation as pyridine. So I can go ahead and draw a p orbital on it. And that lone pair of electrons in blue that's on that nitrogen, that lone pair of electrons occupies, I'm going to go ahead and put them in blue, that lone pair of electrons is going to occupy an sp2 hybridized orbital on that nitrogen. So those electrons are not involved in resonance. Those electrons are localized to that nitrogen. And so when we go ahead and think about the criteria for aromaticity, our first criteria are we have overlapping p orbitals here."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that lone pair of electrons in blue that's on that nitrogen, that lone pair of electrons occupies, I'm going to go ahead and put them in blue, that lone pair of electrons is going to occupy an sp2 hybridized orbital on that nitrogen. So those electrons are not involved in resonance. Those electrons are localized to that nitrogen. And so when we go ahead and think about the criteria for aromaticity, our first criteria are we have overlapping p orbitals here. And so I can see that that is the case. We have overlapping p orbitals. Everything is sp2 hybridized in our ring."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so when we go ahead and think about the criteria for aromaticity, our first criteria are we have overlapping p orbitals here. And so I can see that that is the case. We have overlapping p orbitals. Everything is sp2 hybridized in our ring. And of course, we also need 4n plus 2 pi electrons. And so let me go ahead and label those in magenta. So my pi electrons, going back to my original dot structure, here's 2, here's 4."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Everything is sp2 hybridized in our ring. And of course, we also need 4n plus 2 pi electrons. And so let me go ahead and label those in magenta. So my pi electrons, going back to my original dot structure, here's 2, here's 4. And then this lone pair on our nitrogen, those are actually pi electrons. When we think about this resonance structure here, I'm going to go ahead and label them in magenta again. So there actually are 6 pi electrons in the imidazole molecule."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So my pi electrons, going back to my original dot structure, here's 2, here's 4. And then this lone pair on our nitrogen, those are actually pi electrons. When we think about this resonance structure here, I'm going to go ahead and label them in magenta again. So there actually are 6 pi electrons in the imidazole molecule. And those pi electrons are delocalized around this ring, around these overlapping p orbitals. And so the imidazole molecule is aromatic as well. So the imidazole molecule is actually extremely important in biochemistry."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there actually are 6 pi electrons in the imidazole molecule. And those pi electrons are delocalized around this ring, around these overlapping p orbitals. And so the imidazole molecule is aromatic as well. So the imidazole molecule is actually extremely important in biochemistry. So let's take a look at a famous molecule that contains the imidazole ring. And this molecule is called histamine, which anyone who has allergies has heard of histamine. And you can see the imidazole ring over here in the left on the histamine molecule."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the imidazole molecule is actually extremely important in biochemistry. So let's take a look at a famous molecule that contains the imidazole ring. And this molecule is called histamine, which anyone who has allergies has heard of histamine. And you can see the imidazole ring over here in the left on the histamine molecule. So if you want to understand biochemistry, it's very useful to understand these concepts found in organic chemistry. And so histamine would be an example of a biological aromatic heterocycle. This portion of the molecule, the imidazole ring, is aromatic."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you can see the imidazole ring over here in the left on the histamine molecule. So if you want to understand biochemistry, it's very useful to understand these concepts found in organic chemistry. And so histamine would be an example of a biological aromatic heterocycle. This portion of the molecule, the imidazole ring, is aromatic. It satisfies the two criteria as we have seen. So let's do one more example of an aromatic heterocycle here. This time, let's look at an example that has sulfur in it."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This portion of the molecule, the imidazole ring, is aromatic. It satisfies the two criteria as we have seen. So let's do one more example of an aromatic heterocycle here. This time, let's look at an example that has sulfur in it. So this molecule is called thiophene. And this is really the sulfur analog to the pyrrole molecule that we studied. So let's start by analyzing this sulfur in terms of its steric number."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This time, let's look at an example that has sulfur in it. So this molecule is called thiophene. And this is really the sulfur analog to the pyrrole molecule that we studied. So let's start by analyzing this sulfur in terms of its steric number. So I'm going to look at this sulfur right here. I'm going to calculate the steric number of that sulfur. So number of sigma bonds to that atom."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's start by analyzing this sulfur in terms of its steric number. So I'm going to look at this sulfur right here. I'm going to calculate the steric number of that sulfur. So number of sigma bonds to that atom. So here's a sigma bond, and here's a sigma bond. So we have steric numbers equal to number of sigma bonds plus number of lone pairs of electrons. So we have two lone pairs of electrons around that sulfur like that."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So number of sigma bonds to that atom. So here's a sigma bond, and here's a sigma bond. So we have steric numbers equal to number of sigma bonds plus number of lone pairs of electrons. So we have two lone pairs of electrons around that sulfur like that. So that would be 2 plus 2, which is equal to 4, which implies that that sulfur is sp3 hybridized. But that doesn't work for our concept of aromaticity. Because in order for something to be aromatic, our atom needs to be sp2 hybridized so it has a free p orbital."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have two lone pairs of electrons around that sulfur like that. So that would be 2 plus 2, which is equal to 4, which implies that that sulfur is sp3 hybridized. But that doesn't work for our concept of aromaticity. Because in order for something to be aromatic, our atom needs to be sp2 hybridized so it has a free p orbital. Just like these carbons right here. So once again, these carbons all have a double bond to them. So those carbons all have a p orbital."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Because in order for something to be aromatic, our atom needs to be sp2 hybridized so it has a free p orbital. Just like these carbons right here. So once again, these carbons all have a double bond to them. So those carbons all have a p orbital. So I can go ahead and sketch in those p orbitals on those carbons like that. So this sulfur looks like it's sp3 hybridized. But of course, it's going to be a similar example to the pyrrole molecule."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So those carbons all have a p orbital. So I can go ahead and sketch in those p orbitals on those carbons like that. So this sulfur looks like it's sp3 hybridized. But of course, it's going to be a similar example to the pyrrole molecule. I can show a resonance structure for this thiophene molecule. I can take one of these lone pairs of electrons. I'm going to say it's the lone pair on the right here."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But of course, it's going to be a similar example to the pyrrole molecule. I can show a resonance structure for this thiophene molecule. I can take one of these lone pairs of electrons. I'm going to say it's the lone pair on the right here. I could show them moving in to form a pi bond between the sulfur and this carbon, which would, of course, kick these electrons off onto that carbon. And so when I go ahead and draw this resonance structure, so now there's a pi bond between the sulfur and this carbon. Lone pair of electrons moved out onto this carbon, giving that carbon a negative 1 formal charge."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to say it's the lone pair on the right here. I could show them moving in to form a pi bond between the sulfur and this carbon, which would, of course, kick these electrons off onto that carbon. And so when I go ahead and draw this resonance structure, so now there's a pi bond between the sulfur and this carbon. Lone pair of electrons moved out onto this carbon, giving that carbon a negative 1 formal charge. There was still a pi bond over here. And there was still a lone pair of electrons left on this sulfur. So now we can also give the sulfur a plus 1 formal charge."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Lone pair of electrons moved out onto this carbon, giving that carbon a negative 1 formal charge. There was still a pi bond over here. And there was still a lone pair of electrons left on this sulfur. So now we can also give the sulfur a plus 1 formal charge. And we can analyze it in terms of its steric number. So once again, number of sigma bonds. So this is a sigma bond."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So now we can also give the sulfur a plus 1 formal charge. And we can analyze it in terms of its steric number. So once again, number of sigma bonds. So this is a sigma bond. This is a sigma bond. And so the steric number would be equal to 2 plus how many lone pairs of electrons? Now there's only 1."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is a sigma bond. This is a sigma bond. And so the steric number would be equal to 2 plus how many lone pairs of electrons? Now there's only 1. So 2 plus 1 is equal to 3. So now we can see that it's actually sp2 hybridized. So it has 3 sp2 hybrid orbitals."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now there's only 1. So 2 plus 1 is equal to 3. So now we can see that it's actually sp2 hybridized. So it has 3 sp2 hybrid orbitals. And one of those sp2 hybrid orbitals is going to contain that lone pair of electrons in blue there. So that lone pair of electrons. And again, it's sp2 hybridized, which means it has two other sp2 hybrid orbitals."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it has 3 sp2 hybrid orbitals. And one of those sp2 hybrid orbitals is going to contain that lone pair of electrons in blue there. So that lone pair of electrons. And again, it's sp2 hybridized, which means it has two other sp2 hybrid orbitals. And those hybrid orbitals are forming bonds with those carbons there. Since it's sp2 hybridized, it also has an unhybridized p orbital. So I can go ahead and draw in that unhybridized p orbital."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And again, it's sp2 hybridized, which means it has two other sp2 hybrid orbitals. And those hybrid orbitals are forming bonds with those carbons there. Since it's sp2 hybridized, it also has an unhybridized p orbital. So I can go ahead and draw in that unhybridized p orbital. And so I could think about one of those lone pairs of electrons on my original dot structure occupying a p orbital. So let's go back here and look at the thiophene original dot structure I drew. I could think about one of these lone pair of electrons."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and draw in that unhybridized p orbital. And so I could think about one of those lone pairs of electrons on my original dot structure occupying a p orbital. So let's go back here and look at the thiophene original dot structure I drew. I could think about one of these lone pair of electrons. I'm going to mark it in magenta as occupying a p orbital. So I can go ahead and put it in there like that. And I could think about the other lone pair of electrons on that dot structure as occupying an sp2 hybridized orbital out to the side."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I could think about one of these lone pair of electrons. I'm going to mark it in magenta as occupying a p orbital. So I can go ahead and put it in there like that. And I could think about the other lone pair of electrons on that dot structure as occupying an sp2 hybridized orbital out to the side. So whenever you see this sort of situation, think about where those lone pairs of electrons actually are. And so I also have these as being pi electrons and these as being pi electrons. So 2, 4, 6, a total of 6 pi electrons, which is Huckel's number, delocalized throughout our ring, throughout our overlapping p orbitals."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I could think about the other lone pair of electrons on that dot structure as occupying an sp2 hybridized orbital out to the side. So whenever you see this sort of situation, think about where those lone pairs of electrons actually are. And so I also have these as being pi electrons and these as being pi electrons. So 2, 4, 6, a total of 6 pi electrons, which is Huckel's number, delocalized throughout our ring, throughout our overlapping p orbitals. And so we can say that the thiophene molecule is aromatic. It satisfies both of the criteria for this. And so there are, of course, analogs to thiophene using oxygen."}, {"video_title": "Aromatic heterocycles II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So 2, 4, 6, a total of 6 pi electrons, which is Huckel's number, delocalized throughout our ring, throughout our overlapping p orbitals. And so we can say that the thiophene molecule is aromatic. It satisfies both of the criteria for this. And so there are, of course, analogs to thiophene using oxygen. And that's, of course, a similar situation. And you can draw other resonance structures. But I just wanted to draw this one to show that one of those lone pairs is actually delocalized and involved in resonance."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Most organic molecules don't have any color at all. An example of that would be ethene or ethylene. So ethene has two carbons, and each of those carbons is sp2 hybridized, so each of those carbons has a p orbital. So we have a total of two p orbitals, or two atomic orbitals, and those two atomic orbitals are going to recombine to form two molecular orbitals. So one bonding molecular orbital, and one anti-bonding molecular orbital. The bonding molecular orbital has a lower energy, and so this represents the bonding molecular orbital down here. The anti-bonding molecular orbital is higher in energy, so this is the anti-bonding molecular orbital."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have a total of two p orbitals, or two atomic orbitals, and those two atomic orbitals are going to recombine to form two molecular orbitals. So one bonding molecular orbital, and one anti-bonding molecular orbital. The bonding molecular orbital has a lower energy, and so this represents the bonding molecular orbital down here. The anti-bonding molecular orbital is higher in energy, so this is the anti-bonding molecular orbital. Ethene has two pi electrons, and we can highlight those two pi electrons in magenta here, and those two pi electrons go into the bonding molecular orbital, and so that orbital is occupied. We could also call this orbital the highest occupied molecular orbital, or the HOMO, and this would be the lowest unoccupied molecular orbital, or the LUMO, and there's a difference in energy between the HOMO and the LUMO, and that difference in energy is very important, because that difference in energy corresponds to a wavelength of light, and so energy is equal to Planck's constant times the speed of light divided by the wavelength, and so energy and wavelength are inversely proportional to each other, and a certain amount of energy corresponds to a certain wavelength of light, and so this energy difference, this energy difference between the HOMO and the LUMO corresponds to a certain wavelength of light, and that wavelength of light turns out to be approximately 171 nanometers, and so when ethene absorbs light at a wavelength of 171 nanometers, that corresponds to the proper amount of energy between the HOMO and the LUMO, and that's enough energy for one of those pi electrons to jump from the HOMO to the LUMO in a pi to pi star transition, and so we talked a lot about pi to pi star transitions in the first video on UV-Vis spectroscopy, so make sure to watch that video before you watch this one, and so that's the idea of what's happening with ethene. It absorbs light at a wavelength of 171 nanometers."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The anti-bonding molecular orbital is higher in energy, so this is the anti-bonding molecular orbital. Ethene has two pi electrons, and we can highlight those two pi electrons in magenta here, and those two pi electrons go into the bonding molecular orbital, and so that orbital is occupied. We could also call this orbital the highest occupied molecular orbital, or the HOMO, and this would be the lowest unoccupied molecular orbital, or the LUMO, and there's a difference in energy between the HOMO and the LUMO, and that difference in energy is very important, because that difference in energy corresponds to a wavelength of light, and so energy is equal to Planck's constant times the speed of light divided by the wavelength, and so energy and wavelength are inversely proportional to each other, and a certain amount of energy corresponds to a certain wavelength of light, and so this energy difference, this energy difference between the HOMO and the LUMO corresponds to a certain wavelength of light, and that wavelength of light turns out to be approximately 171 nanometers, and so when ethene absorbs light at a wavelength of 171 nanometers, that corresponds to the proper amount of energy between the HOMO and the LUMO, and that's enough energy for one of those pi electrons to jump from the HOMO to the LUMO in a pi to pi star transition, and so we talked a lot about pi to pi star transitions in the first video on UV-Vis spectroscopy, so make sure to watch that video before you watch this one, and so that's the idea of what's happening with ethene. It absorbs light at a wavelength of 171 nanometers. Let's move on to 1,3-butadiene, which has four carbons. Each one of those carbons is sp2 hybridized, and so each one of those carbons has a p orbital, so we have four p orbitals, four atomic orbitals, which would recombine to form four molecular orbitals, two bonding and two antibonding, so the two bonding molecular orbitals are lower in energy than the two antibonding molecular orbitals, and we have a total of four pi electrons for butadiene, so here's two pi electrons, and here are the other two, so four pi electrons. Those pi electrons occupy the two bonding molecular orbitals."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It absorbs light at a wavelength of 171 nanometers. Let's move on to 1,3-butadiene, which has four carbons. Each one of those carbons is sp2 hybridized, and so each one of those carbons has a p orbital, so we have four p orbitals, four atomic orbitals, which would recombine to form four molecular orbitals, two bonding and two antibonding, so the two bonding molecular orbitals are lower in energy than the two antibonding molecular orbitals, and we have a total of four pi electrons for butadiene, so here's two pi electrons, and here are the other two, so four pi electrons. Those pi electrons occupy the two bonding molecular orbitals. Next, our job is to find the highest occupied molecular orbital, that's this one, and the lowest unoccupied molecular orbital, that's this one, and so the energy difference between the HOMO and the LUMO is what we're thinking about here, and notice that this energy difference, right, this difference in energy is smaller than the difference in energy in the previous example, and so if you think about the equation that relates energy and wavelength, if you decrease the energy, since they're inversely proportional to each other, you must increase the wavelength, and so we must have a higher wavelength than before, so instead of 171 nanometers, 1,3-butadiene's going to absorb light at approximately a wavelength of 217 nanometers. Finally, let's look at 1,3,5-hexatriene, so if I look at the pi electrons, we have two, four, and six pi electrons, so six pi electrons fill the three bonding molecular orbitals. Next, we find the highest occupied molecular orbital and the lowest unoccupied molecular orbital, and look at the energy difference between them."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Those pi electrons occupy the two bonding molecular orbitals. Next, our job is to find the highest occupied molecular orbital, that's this one, and the lowest unoccupied molecular orbital, that's this one, and so the energy difference between the HOMO and the LUMO is what we're thinking about here, and notice that this energy difference, right, this difference in energy is smaller than the difference in energy in the previous example, and so if you think about the equation that relates energy and wavelength, if you decrease the energy, since they're inversely proportional to each other, you must increase the wavelength, and so we must have a higher wavelength than before, so instead of 171 nanometers, 1,3-butadiene's going to absorb light at approximately a wavelength of 217 nanometers. Finally, let's look at 1,3,5-hexatriene, so if I look at the pi electrons, we have two, four, and six pi electrons, so six pi electrons fill the three bonding molecular orbitals. Next, we find the highest occupied molecular orbital and the lowest unoccupied molecular orbital, and look at the energy difference between them. Notice the energy difference has gotten even smaller, and if we, once again, decrease the energy, we increase the wavelength of light that's absorbed, so the wavelength of light must be even higher than this, and it goes up to approximately 258 nanometers, so hexatriene absorbs light at approximately 258 nanometers. That's still in the UV region of the electromagnetic spectrum, and so hexatriene doesn't have a color. In order for something to have a color, it has to absorb light in the visible region, and that is only accomplished by thinking about conjugation, so here we have hexatriene, which is conjugated, double bond, single bond, double bond, single bond, double bond, but it's still absorbing light in the UV region, but imagine a molecule that has much more conjugation than hexatriene."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Next, we find the highest occupied molecular orbital and the lowest unoccupied molecular orbital, and look at the energy difference between them. Notice the energy difference has gotten even smaller, and if we, once again, decrease the energy, we increase the wavelength of light that's absorbed, so the wavelength of light must be even higher than this, and it goes up to approximately 258 nanometers, so hexatriene absorbs light at approximately 258 nanometers. That's still in the UV region of the electromagnetic spectrum, and so hexatriene doesn't have a color. In order for something to have a color, it has to absorb light in the visible region, and that is only accomplished by thinking about conjugation, so here we have hexatriene, which is conjugated, double bond, single bond, double bond, single bond, double bond, but it's still absorbing light in the UV region, but imagine a molecule that has much more conjugation than hexatriene. Increased conjugation means an increased wavelength of light that's absorbed, so notice our trend here. As we increase in the amount of conjugation, we increase the wavelength of light that's absorbed, and so if we have an extensively conjugated molecule, we could absorb light at an even higher wavelength, and if we get past approximately 400 nanometers, we're into the visible region, and those molecules should have color, and so that's the idea. Extensive conjugation leads to color, and let's look at beta carotene, which is what we talked about in the previous video, and we said that beta carotene is orange, so we have an orange molecule here, and once again, the key is conjugation."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "In order for something to have a color, it has to absorb light in the visible region, and that is only accomplished by thinking about conjugation, so here we have hexatriene, which is conjugated, double bond, single bond, double bond, single bond, double bond, but it's still absorbing light in the UV region, but imagine a molecule that has much more conjugation than hexatriene. Increased conjugation means an increased wavelength of light that's absorbed, so notice our trend here. As we increase in the amount of conjugation, we increase the wavelength of light that's absorbed, and so if we have an extensively conjugated molecule, we could absorb light at an even higher wavelength, and if we get past approximately 400 nanometers, we're into the visible region, and those molecules should have color, and so that's the idea. Extensive conjugation leads to color, and let's look at beta carotene, which is what we talked about in the previous video, and we said that beta carotene is orange, so we have an orange molecule here, and once again, the key is conjugation. Look how conjugated beta carotene is. Look at all the alternating single and double bonds. The extensive conjugation means that this molecule can absorb light at a longer wavelength, and it absorbs light in the visible region of the electromagnetic spectrum, and beta carotene absorbs blue light and reflects orange, which is why we said carotene and the molecule looks orange here, so it's conjugation that you have to think about."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Extensive conjugation leads to color, and let's look at beta carotene, which is what we talked about in the previous video, and we said that beta carotene is orange, so we have an orange molecule here, and once again, the key is conjugation. Look how conjugated beta carotene is. Look at all the alternating single and double bonds. The extensive conjugation means that this molecule can absorb light at a longer wavelength, and it absorbs light in the visible region of the electromagnetic spectrum, and beta carotene absorbs blue light and reflects orange, which is why we said carotene and the molecule looks orange here, so it's conjugation that you have to think about. Let's move on to one more example, phenolphthalein. Alright, so phenolphthalein is probably the most famous acid-base indicator, so if you've taken a general chemistry lab, I'm sure you've used phenolphthalein as an indicator for your acid-base titrations, and so you know at a low pH in an acidic environment, phenolphthalein is colorless, but if you add base, so here we'll talk about adding base here, so you increase the pH, you see a pink or magenta color, and let's see if we can explain why. So first let's add some sodium hydroxide, so let me add some sodium hydroxide here, so there's a hydroxide anion, and I'm drawing in another hydroxide anion here, so a negative one charge on the oxygen."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The extensive conjugation means that this molecule can absorb light at a longer wavelength, and it absorbs light in the visible region of the electromagnetic spectrum, and beta carotene absorbs blue light and reflects orange, which is why we said carotene and the molecule looks orange here, so it's conjugation that you have to think about. Let's move on to one more example, phenolphthalein. Alright, so phenolphthalein is probably the most famous acid-base indicator, so if you've taken a general chemistry lab, I'm sure you've used phenolphthalein as an indicator for your acid-base titrations, and so you know at a low pH in an acidic environment, phenolphthalein is colorless, but if you add base, so here we'll talk about adding base here, so you increase the pH, you see a pink or magenta color, and let's see if we can explain why. So first let's add some sodium hydroxide, so let me add some sodium hydroxide here, so there's a hydroxide anion, and I'm drawing in another hydroxide anion here, so a negative one charge on the oxygen. Hydroxide anion is going to function as a base. It's going to take this proton, which we could take these electrons and move them into here, which pushes these electrons over to here. These electrons move in here, and these electrons come off onto the oxygen."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So first let's add some sodium hydroxide, so let me add some sodium hydroxide here, so there's a hydroxide anion, and I'm drawing in another hydroxide anion here, so a negative one charge on the oxygen. Hydroxide anion is going to function as a base. It's going to take this proton, which we could take these electrons and move them into here, which pushes these electrons over to here. These electrons move in here, and these electrons come off onto the oxygen. The other hydroxide would take this proton, and these electrons would end up on the oxygen here, so let's follow those electrons. Electrons in magenta move into here like that. Let's use blue for the next one here, so these electrons in blue move into here, and let's go for green, so these electrons in green move into here, and then finally these electrons in red move off onto this oxygen, so we give that oxygen a negative one formal charge."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These electrons move in here, and these electrons come off onto the oxygen. The other hydroxide would take this proton, and these electrons would end up on the oxygen here, so let's follow those electrons. Electrons in magenta move into here like that. Let's use blue for the next one here, so these electrons in blue move into here, and let's go for green, so these electrons in green move into here, and then finally these electrons in red move off onto this oxygen, so we give that oxygen a negative one formal charge. If this hydroxide anion takes this proton, then these electrons end up on this oxygen, giving that oxygen a negative one formal charge, and so we form an ion here, and this ion has a pink or magenta color to it. If we look closely at it, we can see why. Look at the conjugation that's present."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's use blue for the next one here, so these electrons in blue move into here, and let's go for green, so these electrons in green move into here, and then finally these electrons in red move off onto this oxygen, so we give that oxygen a negative one formal charge. If this hydroxide anion takes this proton, then these electrons end up on this oxygen, giving that oxygen a negative one formal charge, and so we form an ion here, and this ion has a pink or magenta color to it. If we look closely at it, we can see why. Look at the conjugation that's present. There's all kinds of alternating single and double bonds, so double bond, single bond, double bond, single bond, double bond, single bond, double bond, single bond, double bond, and so on. Pretty much the entire ion is conjugated, so we have this extensive conjugation which allows the ion to absorb in the visible region, and that's why it appears to be a pinkish color here. We could go back the other direction."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Look at the conjugation that's present. There's all kinds of alternating single and double bonds, so double bond, single bond, double bond, single bond, double bond, single bond, double bond, single bond, double bond, and so on. Pretty much the entire ion is conjugated, so we have this extensive conjugation which allows the ion to absorb in the visible region, and that's why it appears to be a pinkish color here. We could go back the other direction. If we add some acid, we could turn this back into the colorless form of phenolphthalein here. The reason why this is colorless is because this carbon, let me go ahead and highlight this carbon right here. This carbon is sp3 hybridized."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We could go back the other direction. If we add some acid, we could turn this back into the colorless form of phenolphthalein here. The reason why this is colorless is because this carbon, let me go ahead and highlight this carbon right here. This carbon is sp3 hybridized. It's sp3 hybridized, therefore it doesn't have a p orbital. We have a little bit of conjugation in the benzene rings, so we have these alternating single and double bonds here, but this conjugation is disrupted when we get to this central carbon, and so we don't have conjugation throughout the entire molecule here. We don't have enough conjugation, and so it doesn't absorb in the visible spectrum, and that's why it appears to be colorless."}, {"video_title": "Conjugation and color Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is sp3 hybridized. It's sp3 hybridized, therefore it doesn't have a p orbital. We have a little bit of conjugation in the benzene rings, so we have these alternating single and double bonds here, but this conjugation is disrupted when we get to this central carbon, and so we don't have conjugation throughout the entire molecule here. We don't have enough conjugation, and so it doesn't absorb in the visible spectrum, and that's why it appears to be colorless. If we look at that carbon over here on the right, let's use red here, so this carbon right here, this carbon is sp2 hybridized, so it has a p orbital which allows delocalization of these electrons. We have extensive conjugation, and so we get the color. Hopefully this just helps you understand how important conjugation is for a color."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we have nine carbons, the maximum number of hydrogens we can have is two times nine plus two. And two times nine plus two is equal to 20. So for nine carbons, 20 hydrogens is the maximum number. Here we have only 10 hydrogens. So we are missing 10 hydrogens, or we're missing five pairs of hydrogens. Therefore, the hydrogen deficiency index is equal to five. With an HDI of four or higher, you think benzene ring."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Here we have only 10 hydrogens. So we are missing 10 hydrogens, or we're missing five pairs of hydrogens. Therefore, the hydrogen deficiency index is equal to five. With an HDI of four or higher, you think benzene ring. So I'm going to go ahead and draw a benzene ring in here because I'm pretty sure there's one in our molecule. And if we look down here for this very complicated looking signal, it's really a bunch of overlapping signals from protons on the benzene ring. And I know that because we're in the aromatic proton region for our NMR."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "With an HDI of four or higher, you think benzene ring. So I'm going to go ahead and draw a benzene ring in here because I'm pretty sure there's one in our molecule. And if we look down here for this very complicated looking signal, it's really a bunch of overlapping signals from protons on the benzene ring. And I know that because we're in the aromatic proton region for our NMR. So right in here. So I have five aromatic protons. Let me go ahead and draw them in."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And I know that because we're in the aromatic proton region for our NMR. So right in here. So I have five aromatic protons. Let me go ahead and draw them in. So I put my five aromatic protons in slightly different environments, giving us overlapping signals, which give us this complicated looking one down here. That takes care of an HDI of four, but we have an HDI of five. So we have one more thing."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and draw them in. So I put my five aromatic protons in slightly different environments, giving us overlapping signals, which give us this complicated looking one down here. That takes care of an HDI of four, but we have an HDI of five. So we have one more thing. And it's, of course, going to be a double bond. And I know that because of this signal down here between nine and ten. Remember, a signal between nine and ten, that's the region for an aldehyde proton."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have one more thing. And it's, of course, going to be a double bond. And I know that because of this signal down here between nine and ten. Remember, a signal between nine and ten, that's the region for an aldehyde proton. So we have an aldehyde proton over here. So I draw in my carbonyl, I draw in my hydrogen. And then this is another piece of the puzzle here."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Remember, a signal between nine and ten, that's the region for an aldehyde proton. So we have an aldehyde proton over here. So I draw in my carbonyl, I draw in my hydrogen. And then this is another piece of the puzzle here. So this aldehyde is connected to something. Let's go to these other two signals down here. So this signal represents two protons."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And then this is another piece of the puzzle here. So this aldehyde is connected to something. Let's go to these other two signals down here. So this signal represents two protons. So I'm going to write a CH2 here. How many neighboring protons do we have for those two CH2 protons? Well, there's one, two, three peaks."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this signal represents two protons. So I'm going to write a CH2 here. How many neighboring protons do we have for those two CH2 protons? Well, there's one, two, three peaks. So if there's three peaks, just subtract one to figure out how many neighboring protons you have. So three minus one is equal to two. So this is a CH2 with two neighbors."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, there's one, two, three peaks. So if there's three peaks, just subtract one to figure out how many neighboring protons you have. So three minus one is equal to two. So this is a CH2 with two neighbors. It's the exact same thing for this signal. So this represents two protons, so this would be a CH2. And once again, we have three peaks."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is a CH2 with two neighbors. It's the exact same thing for this signal. So this represents two protons, so this would be a CH2. And once again, we have three peaks. One, two, three. So three minus one is two. So this is a CH2 with two neighbors."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we have three peaks. One, two, three. So three minus one is two. So this is a CH2 with two neighbors. And so these two CH2s must be right next to each other. So let me draw that out here. So if we have one CH2 next to another CH2, each of those CH2 protons have two neighbors."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is a CH2 with two neighbors. And so these two CH2s must be right next to each other. So let me draw that out here. So if we have one CH2 next to another CH2, each of those CH2 protons have two neighbors. For example, if I think about these right here, how many neighboring protons? Well, this is the carbon next door, and I have two neighboring protons. So that makes sense when we look at the signal on the NMR."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we have one CH2 next to another CH2, each of those CH2 protons have two neighbors. For example, if I think about these right here, how many neighboring protons? Well, this is the carbon next door, and I have two neighboring protons. So that makes sense when we look at the signal on the NMR. There's only one way to put together these different pieces of the puzzle. So we would have to put a CH2 coming off this place on our benzene ring, and then another CH2, and then finally our aldehyde. So I'll go ahead and draw in our aldehyde like that."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that makes sense when we look at the signal on the NMR. There's only one way to put together these different pieces of the puzzle. So we would have to put a CH2 coming off this place on our benzene ring, and then another CH2, and then finally our aldehyde. So I'll go ahead and draw in our aldehyde like that. And so that must be the structure of our molecule. So a little bit about this aldehyde proton. Let me go ahead and highlight it over here."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and draw in our aldehyde like that. And so that must be the structure of our molecule. So a little bit about this aldehyde proton. Let me go ahead and highlight it over here. So this aldehyde proton right here, we only see a singlet on the NMR spectrum, but it does have two neighbors. So let me go ahead and draw in the neighboring protons. So they're on this carbon right here."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and highlight it over here. So this aldehyde proton right here, we only see a singlet on the NMR spectrum, but it does have two neighbors. So let me go ahead and draw in the neighboring protons. So they're on this carbon right here. So if we're thinking about the aldehyde proton, this is the carbon next door, and so we have two neighboring protons. And with two neighbors, you might think we would get some splitting for the signal for this aldehyde proton here, but we don't notice any on the NMR. And normally you don't see any splitting because the coupling constant is usually very small, and so therefore the signal often looks like it's a singlet."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So they're on this carbon right here. So if we're thinking about the aldehyde proton, this is the carbon next door, and so we have two neighboring protons. And with two neighbors, you might think we would get some splitting for the signal for this aldehyde proton here, but we don't notice any on the NMR. And normally you don't see any splitting because the coupling constant is usually very small, and so therefore the signal often looks like it's a singlet. But sometimes if you zoom in, you can observe some splitting for the aldehyde proton. For this NMR, we have a signal for two protons, so that must be a CH2. And how many neighboring protons?"}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And normally you don't see any splitting because the coupling constant is usually very small, and so therefore the signal often looks like it's a singlet. But sometimes if you zoom in, you can observe some splitting for the aldehyde proton. For this NMR, we have a signal for two protons, so that must be a CH2. And how many neighboring protons? Well, for this signal, we have four peaks, one, two, three, four. So four minus one is three, so three neighboring protons for these two CH2 protons. What about the chemical shift for this signal?"}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And how many neighboring protons? Well, for this signal, we have four peaks, one, two, three, four. So four minus one is three, so three neighboring protons for these two CH2 protons. What about the chemical shift for this signal? So the chemical shift is getting close to four parts per million, and in that range, that makes us think about those protons being bonded to a carbon that's bonded to an electronegative atom. And if we look at our molecular formula, the only electronegative atom we see on here is oxygen. So that oxygen must be bonded to this carbon."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What about the chemical shift for this signal? So the chemical shift is getting close to four parts per million, and in that range, that makes us think about those protons being bonded to a carbon that's bonded to an electronegative atom. And if we look at our molecular formula, the only electronegative atom we see on here is oxygen. So that oxygen must be bonded to this carbon. So let's go ahead and draw that. So that oxygen is bonded to that carbon, and that carbon is bonded to two hydrogens. So that's what we have so far."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that oxygen must be bonded to this carbon. So let's go ahead and draw that. So that oxygen is bonded to that carbon, and that carbon is bonded to two hydrogens. So that's what we have so far. So the oxygen is more electronegative than carbon. The oxygen is withdrawing some electron density from these two protons, giving us a higher value for the chemical shift. So the signal for these two protons in magenta is right here."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's what we have so far. So the oxygen is more electronegative than carbon. The oxygen is withdrawing some electron density from these two protons, giving us a higher value for the chemical shift. So the signal for these two protons in magenta is right here. Next, let's look at this signal. So we have three protons, so a CH3. How many neighboring protons?"}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the signal for these two protons in magenta is right here. Next, let's look at this signal. So we have three protons, so a CH3. How many neighboring protons? Well, for our signal, we see one, two, three peaks. So three minus one is two, so two neighboring protons. And so two neighboring protons must be the ones in magenta, right?"}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "How many neighboring protons? Well, for our signal, we see one, two, three peaks. So three minus one is two, so two neighboring protons. And so two neighboring protons must be the ones in magenta, right? So we can go ahead and put our methyl group on here, and let's use red for the methyl protons. So these three methyl protons are giving us this signal. We predicted two neighbors, and those neighboring protons are the ones in magenta right here."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so two neighboring protons must be the ones in magenta, right? So we can go ahead and put our methyl group on here, and let's use red for the methyl protons. So these three methyl protons are giving us this signal. We predicted two neighbors, and those neighboring protons are the ones in magenta right here. So those are the two neighbors. For the ones in magenta, we predicted three neighbors. And so that's one, two, and three."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We predicted two neighbors, and those neighboring protons are the ones in magenta right here. So those are the two neighbors. For the ones in magenta, we predicted three neighbors. And so that's one, two, and three. Finally, let's look at the last signal. So we only have one more proton to think about. There's only one place to put it."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that's one, two, and three. Finally, let's look at the last signal. So we only have one more proton to think about. There's only one place to put it. It must go on the oxygen. So this represents, this is the NMR spectrum for an alcohol, for ethanol. So this, this proton in blue, is this signal on the NMR spectrum."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "There's only one place to put it. It must go on the oxygen. So this represents, this is the NMR spectrum for an alcohol, for ethanol. So this, this proton in blue, is this signal on the NMR spectrum. And the chemical shift is hard to predict for an alcoholic proton. Usually you see two to five parts per million, but it's really hard to predict exactly where this signal is going to appear. And also, let's think about how many neighboring protons this proton in blue has."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this, this proton in blue, is this signal on the NMR spectrum. And the chemical shift is hard to predict for an alcoholic proton. Usually you see two to five parts per million, but it's really hard to predict exactly where this signal is going to appear. And also, let's think about how many neighboring protons this proton in blue has. So the carbon next door has two neighbors. So you would think, you would expect two plus one peaks. If N is equal to two, two plus one gives us three."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And also, let's think about how many neighboring protons this proton in blue has. So the carbon next door has two neighbors. So you would think, you would expect two plus one peaks. If N is equal to two, two plus one gives us three. So we'd expect a triplet for this signal, but we only see a singlet. And that's because this proton, this alcoholic proton, rapidly passes from one molecule to another. And this proton transfer is so fast that the proton never stays in place long enough to interact with these neighboring protons."}, {"video_title": "Proton NMR practice 3 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If N is equal to two, two plus one gives us three. So we'd expect a triplet for this signal, but we only see a singlet. And that's because this proton, this alcoholic proton, rapidly passes from one molecule to another. And this proton transfer is so fast that the proton never stays in place long enough to interact with these neighboring protons. And so the NMR machine usually doesn't show any splitting. Under the right conditions, it is possible for splitting of the alcoholic proton to occur, and you might see a triplet here. But on most NMRs, you're only going to see a singlet, which is another clue on your NMR spectrum."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with methane, and let's find the oxidation state of carbon in methane. One approach is more of a general chemistry approach where we know that hydrogen usually has an oxidation state of plus one, and we have four hydrogens for a total of plus four. The sum has to be equal to zero, so we know that carbon's oxidation state must be minus four immediately, since we have only one carbon here. So let's go ahead and verify that with our dot structure. So remember, when we're using, when we're calculating the oxidation state using dot structures, we're thinking about bonding electrons and we know that each bond consists of two electrons, so we need to put in the bonding electrons for all of our bonds. Next, we think about the oxidation state for carbon, and we start with the number of valence electrons in the free atom, or the number of valence electrons that carbon is supposed to have. And we know carbon's supposed to have four valence electrons."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and verify that with our dot structure. So remember, when we're using, when we're calculating the oxidation state using dot structures, we're thinking about bonding electrons and we know that each bond consists of two electrons, so we need to put in the bonding electrons for all of our bonds. Next, we think about the oxidation state for carbon, and we start with the number of valence electrons in the free atom, or the number of valence electrons that carbon is supposed to have. And we know carbon's supposed to have four valence electrons. So from that number, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons carbon has in our drawing. But now we need to think about these covalent bonds as being ionic, and so the more electronegative atom is going to take all of the electrons in the bond. So we need to think about electronegativity differences, and we're comparing carbon to hydrogen."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we know carbon's supposed to have four valence electrons. So from that number, we subtract the number of valence electrons in the bonded atom, or the number of valence electrons carbon has in our drawing. But now we need to think about these covalent bonds as being ionic, and so the more electronegative atom is going to take all of the electrons in the bond. So we need to think about electronegativity differences, and we're comparing carbon to hydrogen. So which is more electronegative? We know that carbon is more electronegative than hydrogen, so the two electrons in this bond here, carbon's gonna take both of them. So it's winner takes all, carbon's gonna hog all of those electrons in this bond."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we need to think about electronegativity differences, and we're comparing carbon to hydrogen. So which is more electronegative? We know that carbon is more electronegative than hydrogen, so the two electrons in this bond here, carbon's gonna take both of them. So it's winner takes all, carbon's gonna hog all of those electrons in this bond. Same for this next carbon-hydrogen bond, carbon is more electronegative, so it takes those electrons, and all the way around. So we can see that carbon is now surrounded by eight electrons, let's count them up here. One, two, three, four, five, six, seven, eight."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's winner takes all, carbon's gonna hog all of those electrons in this bond. Same for this next carbon-hydrogen bond, carbon is more electronegative, so it takes those electrons, and all the way around. So we can see that carbon is now surrounded by eight electrons, let's count them up here. One, two, three, four, five, six, seven, eight. So four minus eight is equal to minus four. So we already knew that minus four was going to be the oxidation state for carbon. Let's move on to another molecule here, so C2H4, this is ethene or ethylene."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six, seven, eight. So four minus eight is equal to minus four. So we already knew that minus four was going to be the oxidation state for carbon. Let's move on to another molecule here, so C2H4, this is ethene or ethylene. What's the oxidation state of carbon in this molecule? Well, hydrogen should be plus one, and we have four of them for a total of plus four. So the total for carbon should be minus four, because that total has to sum to zero."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's move on to another molecule here, so C2H4, this is ethene or ethylene. What's the oxidation state of carbon in this molecule? Well, hydrogen should be plus one, and we have four of them for a total of plus four. So the total for carbon should be minus four, because that total has to sum to zero. But this time we have two carbons, so minus four divided by two gives us minus two. Each carbon should have an oxidation state of minus two. Let's verify that, let's put in our bonding electrons, and let's calculate the oxidation state of carbon by using our little formula here."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the total for carbon should be minus four, because that total has to sum to zero. But this time we have two carbons, so minus four divided by two gives us minus two. Each carbon should have an oxidation state of minus two. Let's verify that, let's put in our bonding electrons, and let's calculate the oxidation state of carbon by using our little formula here. So we put in our bonding electrons, and let's just pick one of the carbons to start off with. Let's say we're talking about the carbon on the right. So we think about electronegativity differences, and we know carbon's more electronegative than hydrogen, so carbon's going to steal those electrons."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's verify that, let's put in our bonding electrons, and let's calculate the oxidation state of carbon by using our little formula here. So we put in our bonding electrons, and let's just pick one of the carbons to start off with. Let's say we're talking about the carbon on the right. So we think about electronegativity differences, and we know carbon's more electronegative than hydrogen, so carbon's going to steal those electrons. And same thing over here, carbon's more electronegative, so carbon steals those electrons too. When we get to this double bond here between the two carbons, we have these four electrons. And now we're trying to compare the electronegativity of carbon to carbon, and obviously that's the same electronegativity."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we think about electronegativity differences, and we know carbon's more electronegative than hydrogen, so carbon's going to steal those electrons. And same thing over here, carbon's more electronegative, so carbon steals those electrons too. When we get to this double bond here between the two carbons, we have these four electrons. And now we're trying to compare the electronegativity of carbon to carbon, and obviously that's the same electronegativity. So with four electrons, we're going to divide those electrons equally, because both those carbons have the same value for the electronegativity. And so we take those four electrons, we divide them in half, so we give two electrons to one carbon and two electrons to the other carbon. So carbon normally has four valence electrons, so we're using our formula here for oxidation state."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And now we're trying to compare the electronegativity of carbon to carbon, and obviously that's the same electronegativity. So with four electrons, we're going to divide those electrons equally, because both those carbons have the same value for the electronegativity. And so we take those four electrons, we divide them in half, so we give two electrons to one carbon and two electrons to the other carbon. So carbon normally has four valence electrons, so we're using our formula here for oxidation state. Oxidation state is equal to the number of valence electrons that carbon is supposed to have minus the number of valence electrons around carbon in our drawing. So let's count them up after we've accounted for electronegativity. That's one, two, three, four, five, and six."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So carbon normally has four valence electrons, so we're using our formula here for oxidation state. Oxidation state is equal to the number of valence electrons that carbon is supposed to have minus the number of valence electrons around carbon in our drawing. So let's count them up after we've accounted for electronegativity. That's one, two, three, four, five, and six. So four minus six gives us an oxidation state for carbon of minus two, which is in agreement with what we calculated over here. Keep in mind that each carbon was supposed to have an oxidation state of minus two, so if we look over here at the carbon on the left, and we just assign those electrons really quickly, we can see that that would be the same calculation. Four minus six gives us minus two."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's one, two, three, four, five, and six. So four minus six gives us an oxidation state for carbon of minus two, which is in agreement with what we calculated over here. Keep in mind that each carbon was supposed to have an oxidation state of minus two, so if we look over here at the carbon on the left, and we just assign those electrons really quickly, we can see that that would be the same calculation. Four minus six gives us minus two. So each carbon has an oxidation state of minus two. We move on to another molecule, so CH2O. This is formaldehyde, and we know that oxygen usually has an oxidation state of minus two, and we have one oxygen for a total of minus two down here."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Four minus six gives us minus two. So each carbon has an oxidation state of minus two. We move on to another molecule, so CH2O. This is formaldehyde, and we know that oxygen usually has an oxidation state of minus two, and we have one oxygen for a total of minus two down here. Hydrogen usually has an oxidation state of plus one, and we have two of them for plus two. This total has to sum to zero, so carbon should have an oxidation state of zero in this molecule. Let's go ahead and look at the dot structure."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is formaldehyde, and we know that oxygen usually has an oxidation state of minus two, and we have one oxygen for a total of minus two down here. Hydrogen usually has an oxidation state of plus one, and we have two of them for plus two. This total has to sum to zero, so carbon should have an oxidation state of zero in this molecule. Let's go ahead and look at the dot structure. We put in all of our bonding electrons so we can find the oxidation state easier, and we think about electronegativity differences. We know carbon's more electronegative than hydrogen, so carbon takes those electrons. Same for this bond, carbon takes those electrons."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and look at the dot structure. We put in all of our bonding electrons so we can find the oxidation state easier, and we think about electronegativity differences. We know carbon's more electronegative than hydrogen, so carbon takes those electrons. Same for this bond, carbon takes those electrons. But now, now we're gonna compare carbon to oxygen, and we know oxygen is more electronegative than carbon. So oxygen's gonna take all of those electrons, so all four of those electrons in that double bond. So it's winner takes all when you're thinking about oxidation states."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Same for this bond, carbon takes those electrons. But now, now we're gonna compare carbon to oxygen, and we know oxygen is more electronegative than carbon. So oxygen's gonna take all of those electrons, so all four of those electrons in that double bond. So it's winner takes all when you're thinking about oxidation states. So because oxygen is more electronegative, it takes all of those electrons, and now we can see that carbon is surrounded by four electrons, so one, two, three, four. Carbon is supposed to have four valence electrons around it, and here we see it is surrounded by four once we take into account electronegativity. So four minus four gives us an oxidation state of zero."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's winner takes all when you're thinking about oxidation states. So because oxygen is more electronegative, it takes all of those electrons, and now we can see that carbon is surrounded by four electrons, so one, two, three, four. Carbon is supposed to have four valence electrons around it, and here we see it is surrounded by four once we take into account electronegativity. So four minus four gives us an oxidation state of zero. So in the formaldehyde molecule, carbon has an oxidation state of zero, which is what we predicted over here. Let's think about oxygen. We said that oxygen would have an oxidation state of minus two."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So four minus four gives us an oxidation state of zero. So in the formaldehyde molecule, carbon has an oxidation state of zero, which is what we predicted over here. Let's think about oxygen. We said that oxygen would have an oxidation state of minus two. So let's look at oxygen. Oxygen's right here, and let's think about how many valence electrons oxygen should have. Well, because of its position on the periodic table, oxygen should have six valence electrons."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We said that oxygen would have an oxidation state of minus two. So let's look at oxygen. Oxygen's right here, and let's think about how many valence electrons oxygen should have. Well, because of its position on the periodic table, oxygen should have six valence electrons. How many electrons are around it now once we've accounted for electronegativity? Well, here's one, two, three, four, five, six, seven, and eight. So six minus eight gives us an oxidation state for oxygen of minus two."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, because of its position on the periodic table, oxygen should have six valence electrons. How many electrons are around it now once we've accounted for electronegativity? Well, here's one, two, three, four, five, six, seven, and eight. So six minus eight gives us an oxidation state for oxygen of minus two. What about for hydrogen? So hydrogen should have one valence electron, but there's zero electrons around it here because carbon's more electronegative than hydrogen. So one minus zero gives us an oxidation state of plus one, which is what we predicted over here, and the same thing for this hydrogen."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So six minus eight gives us an oxidation state for oxygen of minus two. What about for hydrogen? So hydrogen should have one valence electron, but there's zero electrons around it here because carbon's more electronegative than hydrogen. So one minus zero gives us an oxidation state of plus one, which is what we predicted over here, and the same thing for this hydrogen. So this formula works for other atoms, too. All right, let's do another example, and this time we're doing formic acid, so CH2O2. Oxygen should be minus two, and we have two of them, so we have minus four for totaling our oxygens."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So one minus zero gives us an oxidation state of plus one, which is what we predicted over here, and the same thing for this hydrogen. So this formula works for other atoms, too. All right, let's do another example, and this time we're doing formic acid, so CH2O2. Oxygen should be minus two, and we have two of them, so we have minus four for totaling our oxygens. Hydrogen should be plus one, and we have two of them for a total of plus two. And now this total has to add up to equal zero, so we know that carbon's oxidation state should be plus two in the formic acid molecule here. So let's go over and put in our bonding electrons."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen should be minus two, and we have two of them, so we have minus four for totaling our oxygens. Hydrogen should be plus one, and we have two of them for a total of plus two. And now this total has to add up to equal zero, so we know that carbon's oxidation state should be plus two in the formic acid molecule here. So let's go over and put in our bonding electrons. So we put in all of our bonding electrons here, and we think about the oxidation state of carbon. We think about electronegativities. So carbon's more electronegative than hydrogen, so carbon gets those two electrons in that bond, but over here, oxygen is more electronegative than carbon, so oxygen takes those electrons, and the same thing above."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go over and put in our bonding electrons. So we put in all of our bonding electrons here, and we think about the oxidation state of carbon. We think about electronegativities. So carbon's more electronegative than hydrogen, so carbon gets those two electrons in that bond, but over here, oxygen is more electronegative than carbon, so oxygen takes those electrons, and the same thing above. Oxygen's more electronegative than carbon, so oxygen takes those electrons, too. So what's the oxidation state of carbon in formic acid? Well, carbon is supposed to have four valence electrons, and from that, we subtract the number of electrons around carbon once we've accounted for electronegativity, once we pretend like everything is ionic."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So carbon's more electronegative than hydrogen, so carbon gets those two electrons in that bond, but over here, oxygen is more electronegative than carbon, so oxygen takes those electrons, and the same thing above. Oxygen's more electronegative than carbon, so oxygen takes those electrons, too. So what's the oxidation state of carbon in formic acid? Well, carbon is supposed to have four valence electrons, and from that, we subtract the number of electrons around carbon once we've accounted for electronegativity, once we pretend like everything is ionic. And that's only two electrons this time, so four minus two gives us an oxidation state for carbon of plus two, just like we predicted over here. So you can see, every molecule we've done so far has had a different oxidation state for carbon. Carbon is unique, right?"}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, carbon is supposed to have four valence electrons, and from that, we subtract the number of electrons around carbon once we've accounted for electronegativity, once we pretend like everything is ionic. And that's only two electrons this time, so four minus two gives us an oxidation state for carbon of plus two, just like we predicted over here. So you can see, every molecule we've done so far has had a different oxidation state for carbon. Carbon is unique, right? It has all these different oxidation states. Let's finally look at carbon dioxide. So oxygen should have an oxidation state of minus two, and we have two of them for a total of minus four, so carbon must have an oxidation state of plus four."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Carbon is unique, right? It has all these different oxidation states. Let's finally look at carbon dioxide. So oxygen should have an oxidation state of minus two, and we have two of them for a total of minus four, so carbon must have an oxidation state of plus four. So we look at our dot structure over here, and we put in our bonding electrons, and we think about electronegativity. Oxygen is more electronegative than carbon, so oxygen's going to steal, each oxygen's gonna steal the electrons in those bonds. So let me redraw it here."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen should have an oxidation state of minus two, and we have two of them for a total of minus four, so carbon must have an oxidation state of plus four. So we look at our dot structure over here, and we put in our bonding electrons, and we think about electronegativity. Oxygen is more electronegative than carbon, so oxygen's going to steal, each oxygen's gonna steal the electrons in those bonds. So let me redraw it here. I'll show oxygen with all these electrons around it, so the four electrons in that double bond here are all going to be on this oxygen on the left. Same thing for the oxygen on the right. It has these lone pairs of electrons, and it's more electronegative than carbon, so it steals all of those electrons, which leaves carbon with zero electrons around it."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me redraw it here. I'll show oxygen with all these electrons around it, so the four electrons in that double bond here are all going to be on this oxygen on the left. Same thing for the oxygen on the right. It has these lone pairs of electrons, and it's more electronegative than carbon, so it steals all of those electrons, which leaves carbon with zero electrons around it. So to find carbon's oxidation state, we know carbon should have four valence electrons, and here, we have zero electrons around carbon, so four minus zero gives us an oxidation state of plus four, just as we predicted. And if you wanted to do each oxygen, oxygen should have six valence electrons, and here, we have one, two, three, four, five, six, seven, eight, so six minus eight gives us minus two for the oxidation state of oxygen. All right, let's go back up to the beginning here."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It has these lone pairs of electrons, and it's more electronegative than carbon, so it steals all of those electrons, which leaves carbon with zero electrons around it. So to find carbon's oxidation state, we know carbon should have four valence electrons, and here, we have zero electrons around carbon, so four minus zero gives us an oxidation state of plus four, just as we predicted. And if you wanted to do each oxygen, oxygen should have six valence electrons, and here, we have one, two, three, four, five, six, seven, eight, so six minus eight gives us minus two for the oxidation state of oxygen. All right, let's go back up to the beginning here. Let's look at all these different oxidation states that we've covered for carbon. So we started off with an oxidation state for carbon of minus four. We went to minus two, and then we went to zero, and then we went to plus two, and then finally, we ended up with plus four."}, {"video_title": "Oxidation states of carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's go back up to the beginning here. Let's look at all these different oxidation states that we've covered for carbon. So we started off with an oxidation state for carbon of minus four. We went to minus two, and then we went to zero, and then we went to plus two, and then finally, we ended up with plus four. So carbon can have a range of oxidation states from minus four to plus four when you're talking about carbon with four bonds. And you can also have in between, right? It's possible, of course, for carbon to have an oxidation state of plus three."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Most of these diagrams that we see in science textbooks, they don't give justice. In fact, when I showed this sun over here that was about 5 or 6 inches across, I said the earth would be just this little speck about 40 feet. It wouldn't be this distance, it would be about 40 feet to the left or the right. Its orbit would have a radius of about 40 feet. You wouldn't even notice it if you were looking at this thing over here. It would be this little speck orbiting at this huge, huge distance. If you look at this sun over here, if I were to draw the whole sun, it looks like it would have a diameter of about 20 inches."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Its orbit would have a radius of about 40 feet. You wouldn't even notice it if you were looking at this thing over here. It would be this little speck orbiting at this huge, huge distance. If you look at this sun over here, if I were to draw the whole sun, it looks like it would have a diameter of about 20 inches. In this situation, this earth right here, this is drawn to scale, this earth would not be anywhere near this close. It would be about 200 feet that way, or about 60 or 70 meters. You can imagine if the sun was this size, sitting on something like a football field, this little speck of an earth, this little thing right here would be sitting on the other 40 yard line, 60 meters away."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you look at this sun over here, if I were to draw the whole sun, it looks like it would have a diameter of about 20 inches. In this situation, this earth right here, this is drawn to scale, this earth would not be anywhere near this close. It would be about 200 feet that way, or about 60 or 70 meters. You can imagine if the sun was this size, sitting on something like a football field, this little speck of an earth, this little thing right here would be sitting on the other 40 yard line, 60 meters away. You wouldn't even notice it. You might notice this from a distance, but you wouldn't even see this thing over here. The other planets are even further."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can imagine if the sun was this size, sitting on something like a football field, this little speck of an earth, this little thing right here would be sitting on the other 40 yard line, 60 meters away. You wouldn't even notice it. You might notice this from a distance, but you wouldn't even see this thing over here. The other planets are even further. Well, not all of the other planets. Obviously, you have Mercury here. I think most of us are familiar with these, but I'll just list them here just in case."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The other planets are even further. Well, not all of the other planets. Obviously, you have Mercury here. I think most of us are familiar with these, but I'll just list them here just in case. That's Mercury. This is Venus. Mercury is the smallest of the planets where it's not debated whether it's a planet."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I think most of us are familiar with these, but I'll just list them here just in case. That's Mercury. This is Venus. Mercury is the smallest of the planets where it's not debated whether it's a planet. Pluto is the smallest, but some people debate whether it's really a planet or just a large solar body or a dwarf planet or any of those type of things. Then you have Venus, probably the closest in size to the earth, or it is the closest in size to the earth. Then you have Mars."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Mercury is the smallest of the planets where it's not debated whether it's a planet. Pluto is the smallest, but some people debate whether it's really a planet or just a large solar body or a dwarf planet or any of those type of things. Then you have Venus, probably the closest in size to the earth, or it is the closest in size to the earth. Then you have Mars. Then you have Jupiter. Just to give a sense of, once again, how far these things are, if I were to go back to the analogy of this being the size of the sun, then Jupiter is five times further than earth. If I were to actually do the scale distance, this would be 300 meters away."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then you have Mars. Then you have Jupiter. Just to give a sense of, once again, how far these things are, if I were to go back to the analogy of this being the size of the sun, then Jupiter is five times further than earth. If I were to actually do the scale distance, this would be 300 meters away. 300 meters. If I had a nice, big, maybe medicine-ball-sized sun right over here, maybe basketball-sized sun, a little bit bigger than a basketball, this looks on my screen, then I would put this little thing that's smaller than a ping-pong ball, I would put this three football fields away. That's how far Jupiter is."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If I were to actually do the scale distance, this would be 300 meters away. 300 meters. If I had a nice, big, maybe medicine-ball-sized sun right over here, maybe basketball-sized sun, a little bit bigger than a basketball, this looks on my screen, then I would put this little thing that's smaller than a ping-pong ball, I would put this three football fields away. That's how far Jupiter is. Then Saturn's about twice as far as that. Saturn is about nine times the distance. Let me make it clear."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's how far Jupiter is. Then Saturn's about twice as far as that. Saturn is about nine times the distance. Let me make it clear. The earth is one astronomical unit away from the sun, roughly. Its distance changes. It's not a perfectly circular orbit."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me make it clear. The earth is one astronomical unit away from the sun, roughly. Its distance changes. It's not a perfectly circular orbit. Jupiter is approximately a little bit five plus astronomical units, so a little bit more than five times the distance of the sun to the earth. Saturn is approximately nine astronomical units, or nine times the distance from the sun to the earth. Once again, this would be nine football fields away."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's not a perfectly circular orbit. Jupiter is approximately a little bit five plus astronomical units, so a little bit more than five times the distance of the sun to the earth. Saturn is approximately nine astronomical units, or nine times the distance from the sun to the earth. Once again, this would be nine football fields away. Another way to think about it, it would be essentially a kilometer away. If we had a medicine-ball-sized sun, this little, smaller-than-a-ping-pong-balled Saturn would be a kilometer away. I just want to really reiterate that because you never visualize it that way."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once again, this would be nine football fields away. Another way to think about it, it would be essentially a kilometer away. If we had a medicine-ball-sized sun, this little, smaller-than-a-ping-pong-balled Saturn would be a kilometer away. I just want to really reiterate that because you never visualize it that way. Just for the sake of being able to draw it on a page, you see diagrams that look like this. They really don't give you a sense of how small these planets are relative to the sun, and especially relative to their distance from the sun. After Saturn, you have Uranus and then Neptune."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I just want to really reiterate that because you never visualize it that way. Just for the sake of being able to draw it on a page, you see diagrams that look like this. They really don't give you a sense of how small these planets are relative to the sun, and especially relative to their distance from the sun. After Saturn, you have Uranus and then Neptune. Obviously, these guys are even further. Just to give you a sense, it's very easy to start talking about galaxies and universes or the universe. Already, what we've talked about, we're talking about huge distances, huge scale."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "After Saturn, you have Uranus and then Neptune. Obviously, these guys are even further. Just to give you a sense, it's very easy to start talking about galaxies and universes or the universe. Already, what we've talked about, we're talking about huge distances, huge scale. We already talked about that it would take a jet plane 17 years to travel from the earth to the sun. Multiply that by 5, about 100 years to go from Jupiter to the sun, 200 years to go from Saturn to the sun. You could have had Abraham Lincoln get into a jet plane, and he still wouldn't have gotten, if he left from Saturn, he still would not have gotten to the sun."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Already, what we've talked about, we're talking about huge distances, huge scale. We already talked about that it would take a jet plane 17 years to travel from the earth to the sun. Multiply that by 5, about 100 years to go from Jupiter to the sun, 200 years to go from Saturn to the sun. You could have had Abraham Lincoln get into a jet plane, and he still wouldn't have gotten, if he left from Saturn, he still would not have gotten to the sun. These are huge, huge distances, but we're not done with the solar system there. Just to give a sense of scale, this right here, that's the sun. Each of these planets are actually narrower than these orbits, so they just draw these orbits here, but you wouldn't actually even see the actual planets here at this type of a scale."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You could have had Abraham Lincoln get into a jet plane, and he still wouldn't have gotten, if he left from Saturn, he still would not have gotten to the sun. These are huge, huge distances, but we're not done with the solar system there. Just to give a sense of scale, this right here, that's the sun. Each of these planets are actually narrower than these orbits, so they just draw these orbits here, but you wouldn't actually even see the actual planets here at this type of a scale. This is one astronomical unit right over here, the distance from the sun to the earth. Then you have Mars. Then you have the asteroid belt."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Each of these planets are actually narrower than these orbits, so they just draw these orbits here, but you wouldn't actually even see the actual planets here at this type of a scale. This is one astronomical unit right over here, the distance from the sun to the earth. Then you have Mars. Then you have the asteroid belt. Mars also has some pretty big things in it itself, and it has these things that are kind of considered almost dwarf planets, things like Ceres. You could look those type things up. Then you have Jupiter out here."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then you have the asteroid belt. Mars also has some pretty big things in it itself, and it has these things that are kind of considered almost dwarf planets, things like Ceres. You could look those type things up. Then you have Jupiter out here. Once again, we said it would take 100 years, or roughly 100 years for a jet plane to get from Jupiter to the sun. Even if you take this whole box here, which is a huge amount of distance, roughly about 5 astronomical units, it would take about 40 minutes for light to get from the sun to Jupiter. This is a huge, huge distance, but even this huge distance, we can put it into this little box right over here."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then you have Jupiter out here. Once again, we said it would take 100 years, or roughly 100 years for a jet plane to get from Jupiter to the sun. Even if you take this whole box here, which is a huge amount of distance, roughly about 5 astronomical units, it would take about 40 minutes for light to get from the sun to Jupiter. This is a huge, huge distance, but even this huge distance, we can put it into this little box right over here. This whole box right over there can be fit into this box. You need to do that in order to appreciate the orbits of the outer planets. On this scale, the earth and Venus and Mercury and Mars, their orbits look pretty much, you can't even differentiate them from the sun."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is a huge, huge distance, but even this huge distance, we can put it into this little box right over here. This whole box right over there can be fit into this box. You need to do that in order to appreciate the orbits of the outer planets. On this scale, the earth and Venus and Mercury and Mars, their orbits look pretty much, you can't even differentiate them from the sun. They look so close. They almost look like they're part of the sun when you look at it on this scale. Then you have the outer planets, Saturn, Uranus, Neptune."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "On this scale, the earth and Venus and Mercury and Mars, their orbits look pretty much, you can't even differentiate them from the sun. They look so close. They almost look like they're part of the sun when you look at it on this scale. Then you have the outer planets, Saturn, Uranus, Neptune. Then we have a Kuiper belt. This is more asteroids, but these are kind of more frozen. When we think of ice, you always think of water ice."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then you have the outer planets, Saturn, Uranus, Neptune. Then we have a Kuiper belt. This is more asteroids, but these are kind of more frozen. When we think of ice, you always think of water ice. But out here, it's so cold and it's relatively getting dark now because we're pretty far from the sun that things that we normally associate as gases are going to be in their solid form out here. This isn't just rocky elements. This will also be things that we normally associate as gases like methane, frozen methane."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When we think of ice, you always think of water ice. But out here, it's so cold and it's relatively getting dark now because we're pretty far from the sun that things that we normally associate as gases are going to be in their solid form out here. This isn't just rocky elements. This will also be things that we normally associate as gases like methane, frozen methane. But even here, we're not done. We're not even out of the solar system yet. Actually, just to give you a sense of the scale we're operating right here, I have this chart right here from the Voyager mission."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This will also be things that we normally associate as gases like methane, frozen methane. But even here, we're not done. We're not even out of the solar system yet. Actually, just to give you a sense of the scale we're operating right here, I have this chart right here from the Voyager mission. The Voyager missions, Voyager 1 and 2, actually Voyager 2 left a little bit earlier, a month earlier. Voyager 1 is just traveling faster. They left about a year after I was born."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, just to give you a sense of the scale we're operating right here, I have this chart right here from the Voyager mission. The Voyager missions, Voyager 1 and 2, actually Voyager 2 left a little bit earlier, a month earlier. Voyager 1 is just traveling faster. They left about a year after I was born. Their current velocity, just to give you a sense of how fast, Voyager 1 right here, is right now traveling at 61,000 kilometers per hour. That's about 17 kilometers per second, the size of a city every second. It's going that fast."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They left about a year after I was born. Their current velocity, just to give you a sense of how fast, Voyager 1 right here, is right now traveling at 61,000 kilometers per hour. That's about 17 kilometers per second, the size of a city every second. It's going that fast. That's, at least in my mind, an unfathomably fast velocity. This thing has been traveling roughly that fast. It's been going around planets and gaining acceleration as it went around orbits."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's going that fast. That's, at least in my mind, an unfathomably fast velocity. This thing has been traveling roughly that fast. It's been going around planets and gaining acceleration as it went around orbits. But for most of the time, it's been going at a pretty fast speed. Just to translate it to people who don't relate to kilometers, that's about 38,000 miles per hour. This huge, huge, unfathomably fast speed."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's been going around planets and gaining acceleration as it went around orbits. But for most of the time, it's been going at a pretty fast speed. Just to translate it to people who don't relate to kilometers, that's about 38,000 miles per hour. This huge, huge, unfathomably fast speed. It's been doing it since 1977. I was learning to walk, and when I was learning to walk, it was traveling at this super fast speed. Then when I was learning to talk, our whole lives, when we're sleeping, everything, we're eating, I'm in elementary school, it's still rocketing out of the solar system at roughly this speed."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This huge, huge, unfathomably fast speed. It's been doing it since 1977. I was learning to walk, and when I was learning to walk, it was traveling at this super fast speed. Then when I was learning to talk, our whole lives, when we're sleeping, everything, we're eating, I'm in elementary school, it's still rocketing out of the solar system at roughly this speed. Its velocity has changed, but especially once it got outside of the planets, it's been roughly at this velocity. It's just been rocketing out, and I don't want to say only, but it's gotten this far. It's gotten this far."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then when I was learning to talk, our whole lives, when we're sleeping, everything, we're eating, I'm in elementary school, it's still rocketing out of the solar system at roughly this speed. Its velocity has changed, but especially once it got outside of the planets, it's been roughly at this velocity. It's just been rocketing out, and I don't want to say only, but it's gotten this far. It's gotten this far. If we look at it on this scale, it's gotten about that far right there. It's about 115, 116 astronomical units. To give a sense, there's two ways to think about it."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's gotten this far. If we look at it on this scale, it's gotten about that far right there. It's about 115, 116 astronomical units. To give a sense, there's two ways to think about it. One is, wow, that's really far, because we know that even on this scale, you can't even see Earth's orbit. This looks like it's a pretty, pretty far distance. Just to give you a sense of how far 116 astronomical units are, if 2,000 years ago, Jesus got on a plane, I actually pasted a copy of Jesus just for visualization purposes, but if he got on a jetliner at 1,000 kilometers per hour and went straight in that direction, in the direction of Voyager, Voyager would only just now be catching up to Jesus."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "To give a sense, there's two ways to think about it. One is, wow, that's really far, because we know that even on this scale, you can't even see Earth's orbit. This looks like it's a pretty, pretty far distance. Just to give you a sense of how far 116 astronomical units are, if 2,000 years ago, Jesus got on a plane, I actually pasted a copy of Jesus just for visualization purposes, but if he got on a jetliner at 1,000 kilometers per hour and went straight in that direction, in the direction of Voyager, Voyager would only just now be catching up to Jesus. This is a huge, huge, huge, huge distance. At the same time, even though it's a huge distance, especially relative to everything else we've talked about, relative to just even the outer reaches of the solar system, we're still talking in terms of a small scale. That's how far Voyager is."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just to give you a sense of how far 116 astronomical units are, if 2,000 years ago, Jesus got on a plane, I actually pasted a copy of Jesus just for visualization purposes, but if he got on a jetliner at 1,000 kilometers per hour and went straight in that direction, in the direction of Voyager, Voyager would only just now be catching up to Jesus. This is a huge, huge, huge, huge distance. At the same time, even though it's a huge distance, especially relative to everything else we've talked about, relative to just even the outer reaches of the solar system, we're still talking in terms of a small scale. That's how far Voyager is. Just to give a sense, on this scale, this whole box over here can be contained in this box. When you look at this box, Voyager's only gotten about that far. After traveling at this unbelievable velocity for over 30 years, for about 33 years."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's how far Voyager is. Just to give a sense, on this scale, this whole box over here can be contained in this box. When you look at this box, Voyager's only gotten about that far. After traveling at this unbelievable velocity for over 30 years, for about 33 years. Just to give you an idea of these other things, Sedna right here is one of the... It's a reasonably large-sized outer solar system object. It's one of the furthest objects that we know of in the solar system."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "After traveling at this unbelievable velocity for over 30 years, for about 33 years. Just to give you an idea of these other things, Sedna right here is one of the... It's a reasonably large-sized outer solar system object. It's one of the furthest objects that we know of in the solar system. It has this very eccentric orbit. It gets, I don't want to say relatively close, but not unreasonably far away. Then it gets really far away from the sun."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's one of the furthest objects that we know of in the solar system. It has this very eccentric orbit. It gets, I don't want to say relatively close, but not unreasonably far away. Then it gets really far away from the sun. Even Sedna's orbit, if I were to look at this, this whole box over here can be contained right over here. In this diagram right here, you wouldn't even be able to see, it would be like a speck how far Voyager has traveled in 33 years at 38,000 miles per hour. You would not even be able to notice that distance."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then it gets really far away from the sun. Even Sedna's orbit, if I were to look at this, this whole box over here can be contained right over here. In this diagram right here, you wouldn't even be able to see, it would be like a speck how far Voyager has traveled in 33 years at 38,000 miles per hour. You would not even be able to notice that distance. Even though you can't even notice that distance, we still have the sun's influence. The gravitational pull is still attracting things to it. This right here, we speculate that there's the Oort cloud."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You would not even be able to notice that distance. Even though you can't even notice that distance, we still have the sun's influence. The gravitational pull is still attracting things to it. This right here, we speculate that there's the Oort cloud. This is where the comets originate from. This is just a bunch of frozen gases and ice particles and things like that. We're starting to get to the outer reaches of the solar system."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right here, we speculate that there's the Oort cloud. This is where the comets originate from. This is just a bunch of frozen gases and ice particles and things like that. We're starting to get to the outer reaches of the solar system. This distance right here is about 50,000 astronomical units. Just to give a scale, because you hear a lot about light years and all of that, light years are about 63,000 astronomical units. If you go a light year out from the sun, you'll end up in the Oort cloud, the hypothesized Oort cloud."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're starting to get to the outer reaches of the solar system. This distance right here is about 50,000 astronomical units. Just to give a scale, because you hear a lot about light years and all of that, light years are about 63,000 astronomical units. If you go a light year out from the sun, you'll end up in the Oort cloud, the hypothesized Oort cloud. Just to give a sense, another scale, the Oort cloud is actually, most of the planets' orbits are roughly in the same plane. This right here is the orbit of the planets. Once again, these lines are drawn too thick."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you go a light year out from the sun, you'll end up in the Oort cloud, the hypothesized Oort cloud. Just to give a sense, another scale, the Oort cloud is actually, most of the planets' orbits are roughly in the same plane. This right here is the orbit of the planets. Once again, these lines are drawn too thick. They're just drawn the thinnest possible so that you can see them, but they're still drawn too thick. This gets us all the way to the Kuiper belt, but all of this over here, all the way out to the Kuiper belt, all the way out to all of the major planets, this is Pluto's orbit right over here. This whole diagram is only sitting in right over there."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once again, these lines are drawn too thick. They're just drawn the thinnest possible so that you can see them, but they're still drawn too thick. This gets us all the way to the Kuiper belt, but all of this over here, all the way out to the Kuiper belt, all the way out to all of the major planets, this is Pluto's orbit right over here. This whole diagram is only sitting in right over there. You can barely see it. This whole diagram is just that dot in this. Then you can see the Oort cloud all around it."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This whole diagram is only sitting in right over there. You can barely see it. This whole diagram is just that dot in this. Then you can see the Oort cloud all around it. It's more of a spherical cloud. We think it exists. Obviously, it's hard to observe things at that distance."}, {"video_title": "Scale of solar system Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then you can see the Oort cloud all around it. It's more of a spherical cloud. We think it exists. Obviously, it's hard to observe things at that distance. Hopefully, that gives you a beginning sense of the scale of the solar system. What's really going to blow your mind, if this hasn't blown your mind already, is that this whole thing is going to start looking like a speck. When you even just look at the local area around our galaxy, much less the galaxy as a whole, much less the universe as a whole."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the left we have a 2,5-hexane dione, and if we add sodium hydroxide and we heat things up, we would form this compound over here on the right. So we would actually form a ring. Let's go ahead and number our dione over here. It's symmetrical, so we could start from either side. So I could say this is carbon one, carbon two, three, four, five, and six. We have 2,5-hexane dione here. And if we add sodium hydroxide, we're going to deprotonate at the alpha carbon."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's symmetrical, so we could start from either side. So I could say this is carbon one, carbon two, three, four, five, and six. We have 2,5-hexane dione here. And if we add sodium hydroxide, we're going to deprotonate at the alpha carbon. So I need to find my alpha carbons here. So that's the carbon next to the carbonyl. So carbon one could be an alpha carbon."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if we add sodium hydroxide, we're going to deprotonate at the alpha carbon. So I need to find my alpha carbons here. So that's the carbon next to the carbonyl. So carbon one could be an alpha carbon. Carbon three could be an alpha carbon. And then over here, these would also be alpha carbons to this carbonyl. But since it's symmetrical, we don't need to worry about those."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So carbon one could be an alpha carbon. Carbon three could be an alpha carbon. And then over here, these would also be alpha carbons to this carbonyl. But since it's symmetrical, we don't need to worry about those. So we just need to focus in on the possible alpha carbons at one and three. Let's first do this thinking about deprotonating at alpha carbon one here, at alpha carbon, which is carbon one. So I put my alpha proton on here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But since it's symmetrical, we don't need to worry about those. So we just need to focus in on the possible alpha carbons at one and three. Let's first do this thinking about deprotonating at alpha carbon one here, at alpha carbon, which is carbon one. So I put my alpha proton on here. And I think about hydroxide taking this alpha proton, leaving these electrons behind on this carbon. And so here I have the dione drawn in a different conformation. And let's go ahead and number this one."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I put my alpha proton on here. And I think about hydroxide taking this alpha proton, leaving these electrons behind on this carbon. And so here I have the dione drawn in a different conformation. And let's go ahead and number this one. So this would be carbon one, two, three, four, five, and six. This is just going to help us when we're thinking about forming our product. And so if we're deprotonating at carbon one, so these electrons in here would end up on carbon one to form a carbanion."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and number this one. So this would be carbon one, two, three, four, five, and six. This is just going to help us when we're thinking about forming our product. And so if we're deprotonating at carbon one, so these electrons in here would end up on carbon one to form a carbanion. So I'm showing those electrons in magenta on carbon one. So that's a negative one formal charge on this carbon. And for this mechanism, I'm going to show the carbanion functioning as a nucleophile."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so if we're deprotonating at carbon one, so these electrons in here would end up on carbon one to form a carbanion. So I'm showing those electrons in magenta on carbon one. So that's a negative one formal charge on this carbon. And for this mechanism, I'm going to show the carbanion functioning as a nucleophile. So it's going to attack this carbonyl over here. We know this carbonyl has a partial negative oxygen, a partial positive carbon. So this carbon right here is electrophilic."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And for this mechanism, I'm going to show the carbanion functioning as a nucleophile. So it's going to attack this carbonyl over here. We know this carbonyl has a partial negative oxygen, a partial positive carbon. So this carbon right here is electrophilic. So our nucleophile is going to attack our electrophile. So these electrons in here are going to attack here, pushing these electrons off onto our oxygen. So let's go ahead and show the result of that."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here is electrophilic. So our nucleophile is going to attack our electrophile. So these electrons in here are going to attack here, pushing these electrons off onto our oxygen. So let's go ahead and show the result of that. We would actually form our ring here. So I'm going to draw our five-membered ring. We would have our carbonyl right here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the result of that. We would actually form our ring here. So I'm going to draw our five-membered ring. We would have our carbonyl right here. And we would form an alkoxide. And then we could think about protonating to form our aldol. So I'm just going to go ahead and draw the aldol here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We would have our carbonyl right here. And we would form an alkoxide. And then we could think about protonating to form our aldol. So I'm just going to go ahead and draw the aldol here. And then we would have a methyl group right here. So let's follow some of those electrons. So the electrons in magenta moved in here to form our bond."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just going to go ahead and draw the aldol here. And then we would have a methyl group right here. So let's follow some of those electrons. So the electrons in magenta moved in here to form our bond. And then let's number our carbons, too. So this would be carbon 1, carbon 2, 3, 4, 5, and 6, in terms of the numbering system that we used over here on the left. So it just helps us to follow those carbons along here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta moved in here to form our bond. And then let's number our carbons, too. So this would be carbon 1, carbon 2, 3, 4, 5, and 6, in terms of the numbering system that we used over here on the left. So it just helps us to follow those carbons along here. And so this would be our aldol. And then we know that this is our alpha carbon right here. So we could deprotonate that."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it just helps us to follow those carbons along here. And so this would be our aldol. And then we know that this is our alpha carbon right here. So we could deprotonate that. So I'm going to go ahead and draw an alpha proton right here. And so we could think about hydroxide coming along, functioning as a base, and taking that proton. So the hydroxide takes this proton."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we could deprotonate that. So I'm going to go ahead and draw an alpha proton right here. And so we could think about hydroxide coming along, functioning as a base, and taking that proton. So the hydroxide takes this proton. These electrons move in here to form your double bond. And you have hydroxide as your leaving group. So when we draw the product, we have our ring."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the hydroxide takes this proton. These electrons move in here to form your double bond. And you have hydroxide as your leaving group. So when we draw the product, we have our ring. We have our carbonyl right here. We would have a double bond now. And then we would have our methyl group like that."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So when we draw the product, we have our ring. We have our carbonyl right here. We would have a double bond now. And then we would have our methyl group like that. And so that's obviously our product. Let's follow those electrons. We already had a carbon-carbon bond formed in here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have our methyl group like that. And so that's obviously our product. Let's follow those electrons. We already had a carbon-carbon bond formed in here. But if we take the proton, these electrons in blue can move in here to form our double bond. And then we have our product. And so this would be the major product for this reaction."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We already had a carbon-carbon bond formed in here. But if we take the proton, these electrons in blue can move in here to form our double bond. And then we have our product. And so this would be the major product for this reaction. So that's showing deprotonation at the alpha carbon, which we labeled 1 right here. Let's show what would happen if we deprotonated the alpha carbon that we numbered 3. So we'll do that in the next area we have down here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this would be the major product for this reaction. So that's showing deprotonation at the alpha carbon, which we labeled 1 right here. Let's show what would happen if we deprotonated the alpha carbon that we numbered 3. So we'll do that in the next area we have down here. So the same confirmation we're starting with before. So 1, 2, 3, 4, 5, 6. If we deprotonate at carbon 3 this time, then that's where we'd form our carbanion."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we'll do that in the next area we have down here. So the same confirmation we're starting with before. So 1, 2, 3, 4, 5, 6. If we deprotonate at carbon 3 this time, then that's where we'd form our carbanion. So let me go ahead and show the lone pair of electrons on this carbon 3 right here. That gives that carbon a negative 1 formal charge. And so we could think about this being our nucleophile and attacking right here at this carbon."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we deprotonate at carbon 3 this time, then that's where we'd form our carbanion. So let me go ahead and show the lone pair of electrons on this carbon 3 right here. That gives that carbon a negative 1 formal charge. And so we could think about this being our nucleophile and attacking right here at this carbon. So that would push these electrons in here off onto the oxygen. So let's go ahead and draw the results of that. This time we would actually form a three-membered ring."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we could think about this being our nucleophile and attacking right here at this carbon. So that would push these electrons in here off onto the oxygen. So let's go ahead and draw the results of that. This time we would actually form a three-membered ring. So let's go ahead and draw everything in. And then let's follow everything around here. So we have a methyl group right here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This time we would actually form a three-membered ring. So let's go ahead and draw everything in. And then let's follow everything around here. So we have a methyl group right here. And then we would have an O. And that would be our alkoxide, which we could protonate to form our aldol as our intermediate here. So let's once again number these carbons, even though this isn't how you would actually number it."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have a methyl group right here. And then we would have an O. And that would be our alkoxide, which we could protonate to form our aldol as our intermediate here. So let's once again number these carbons, even though this isn't how you would actually number it. It just helps us to follow everything along here. So let me go ahead and number in red. So 1, 2, 3, 4, 5."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's once again number these carbons, even though this isn't how you would actually number it. It just helps us to follow everything along here. So let me go ahead and number in red. So 1, 2, 3, 4, 5. And this is where carbon 6, again, using these numbering system over here on the left just to make it a little bit easier for us. So the electrons in magenta are the ones that form this bond between carbon 3 and carbon 5 like that. And once again, we know that carbon 3 is an alpha carbon."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5. And this is where carbon 6, again, using these numbering system over here on the left just to make it a little bit easier for us. So the electrons in magenta are the ones that form this bond between carbon 3 and carbon 5 like that. And once again, we know that carbon 3 is an alpha carbon. So there is an alpha proton on here. And so we could think about hydroxide coming along and functioning as a base. So hydroxide is going to take this proton, leave these electrons behind."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we know that carbon 3 is an alpha carbon. So there is an alpha proton on here. And so we could think about hydroxide coming along and functioning as a base. So hydroxide is going to take this proton, leave these electrons behind. And then we have hydroxide as a leaving group. And so another possible product would be this three-membered ring right here. And then we would have a methyl group."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So hydroxide is going to take this proton, leave these electrons behind. And then we have hydroxide as a leaving group. And so another possible product would be this three-membered ring right here. And then we would have a methyl group. And then we would also have our ketone over here like that. And so we had already formed a carbon-carbon bond with the electrons in magenta. And then if we take this proton, these electrons in blue can move in here to form our double bond."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have a methyl group. And then we would also have our ketone over here like that. And so we had already formed a carbon-carbon bond with the electrons in magenta. And then if we take this proton, these electrons in blue can move in here to form our double bond. And so it turns out that this small ring here is possible to form this. But it's not isolated in large yield in this reaction because there's too much angle strain for this three-membered ring. And so this isn't really isolated as your product."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then if we take this proton, these electrons in blue can move in here to form our double bond. And so it turns out that this small ring here is possible to form this. But it's not isolated in large yield in this reaction because there's too much angle strain for this three-membered ring. And so this isn't really isolated as your product. Your product is this one. There's less angle strain to form this five-membered ring. But it is possible to get a three-membered ring in this."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this isn't really isolated as your product. Your product is this one. There's less angle strain to form this five-membered ring. But it is possible to get a three-membered ring in this. But the five-membered ring is going to be your major product here. So let's do one more intramolecular aldol condensation. So very similar to the last one, actually."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But it is possible to get a three-membered ring in this. But the five-membered ring is going to be your major product here. So let's do one more intramolecular aldol condensation. So very similar to the last one, actually. So it looks a little bit more complicated. But we can analyze it the same way. So if we're looking for our alpha carbons, we know those are the ones next to our carbonyls."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So very similar to the last one, actually. So it looks a little bit more complicated. But we can analyze it the same way. So if we're looking for our alpha carbons, we know those are the ones next to our carbonyls. Let's go ahead and number this compound. This would be carbon 1, 2, 3, 4, 5, and 6. And so for our alpha carbons, we know carbon 1 would be an alpha carbon."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we're looking for our alpha carbons, we know those are the ones next to our carbonyls. Let's go ahead and number this compound. This would be carbon 1, 2, 3, 4, 5, and 6. And so for our alpha carbons, we know carbon 1 would be an alpha carbon. And we know carbon 3 would be an alpha carbon, carbon 4, and then carbon 6. So those are our possibilities. So carbons 3 and 4, if you think about these being alpha carbons, these would give us a ring with a little bit too much angle strain."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so for our alpha carbons, we know carbon 1 would be an alpha carbon. And we know carbon 3 would be an alpha carbon, carbon 4, and then carbon 6. So those are our possibilities. So carbons 3 and 4, if you think about these being alpha carbons, these would give us a ring with a little bit too much angle strain. So we can kind of rule those out. And so now it's down to thinking about carbon 1 or carbon 6. And for that, we need to look at the base."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So carbons 3 and 4, if you think about these being alpha carbons, these would give us a ring with a little bit too much angle strain. So we can kind of rule those out. And so now it's down to thinking about carbon 1 or carbon 6. And for that, we need to look at the base. So we're using potassium hydroxide here. So we're using a non-sterically hindered base. And it isn't very strong."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And for that, we need to look at the base. So we're using potassium hydroxide here. So we're using a non-sterically hindered base. And it isn't very strong. And so it's going to favor the formation of the thermodynamic enolate, which is the more stable one because it's more substituted. So we talked about this in an earlier video. And so if you look at deprotonation at alpha carbon, which we have labeled 1 here, that would give you the kinetic enolate, which is formed the fastest but is not the most stable."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And it isn't very strong. And so it's going to favor the formation of the thermodynamic enolate, which is the more stable one because it's more substituted. So we talked about this in an earlier video. And so if you look at deprotonation at alpha carbon, which we have labeled 1 here, that would give you the kinetic enolate, which is formed the fastest but is not the most stable. If you deprotonate at the alpha carbon that's 6, that would give you the thermodynamic enolate. So it's more substituted than the kinetic enolate, so it's more stable. So if you think about the fact that on this alpha carbon, there are two alpha protons right here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so if you look at deprotonation at alpha carbon, which we have labeled 1 here, that would give you the kinetic enolate, which is formed the fastest but is not the most stable. If you deprotonate at the alpha carbon that's 6, that would give you the thermodynamic enolate. So it's more substituted than the kinetic enolate, so it's more stable. So if you think about the fact that on this alpha carbon, there are two alpha protons right here. And if your base takes one of these alpha protons, then these electrons would move into here, push these electrons off onto your oxygen. So let's do the same thing down here where we have a different conformation, so very similar to how we did the last one. So one of these acidic protons right here on our alpha carbon, I'll go ahead and draw it in right here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if you think about the fact that on this alpha carbon, there are two alpha protons right here. And if your base takes one of these alpha protons, then these electrons would move into here, push these electrons off onto your oxygen. So let's do the same thing down here where we have a different conformation, so very similar to how we did the last one. So one of these acidic protons right here on our alpha carbon, I'll go ahead and draw it in right here. So we have our hydroxide come along and function as a base. So hydroxide anion, so negative 1 charge is going to take this proton, leaving these electrons into here, pushing these electrons off onto the oxygen. So I'm going to draw the oxyanion this time."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So one of these acidic protons right here on our alpha carbon, I'll go ahead and draw it in right here. So we have our hydroxide come along and function as a base. So hydroxide anion, so negative 1 charge is going to take this proton, leaving these electrons into here, pushing these electrons off onto the oxygen. So I'm going to draw the oxyanion this time. So this is going to be in equilibrium. So let's go ahead and show. First, I'll show what we would get, and then we'll follow some electrons around."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to draw the oxyanion this time. So this is going to be in equilibrium. So let's go ahead and show. First, I'll show what we would get, and then we'll follow some electrons around. So I have my carbonyl here. And then we're going to form our oxyanion. So I'm going to go ahead and draw negative 1 formal charge on this oxygen."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "First, I'll show what we would get, and then we'll follow some electrons around. So I have my carbonyl here. And then we're going to form our oxyanion. So I'm going to go ahead and draw negative 1 formal charge on this oxygen. There's now a double bond right here. And then we'll go ahead and draw in the rest of our carbons. So this is the thermodynamic enolate."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and draw negative 1 formal charge on this oxygen. There's now a double bond right here. And then we'll go ahead and draw in the rest of our carbons. So this is the thermodynamic enolate. So if you look at this double bond, this is going to be the most stable enolate that forms because of the substitution of this double bond here. So following those electrons, let's go ahead and make those electrons in here blue. So these electrons in here would move in here like that."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is the thermodynamic enolate. So if you look at this double bond, this is going to be the most stable enolate that forms because of the substitution of this double bond here. So following those electrons, let's go ahead and make those electrons in here blue. So these electrons in here would move in here like that. And then you could think about your electrons, your pi electrons in here, moving off onto your oxygen to form your oxyanion like that. And so once again, this is the thermodynamic enolate because this reaction is under thermodynamic control because of our choice of base. Next, we're going to show our oxyanion functioning as a nucleophile."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here would move in here like that. And then you could think about your electrons, your pi electrons in here, moving off onto your oxygen to form your oxyanion like that. And so once again, this is the thermodynamic enolate because this reaction is under thermodynamic control because of our choice of base. Next, we're going to show our oxyanion functioning as a nucleophile. So once again, our carbonyl is polarized, partial negative, partial positive. So we could think about these electrons in red moving into here. And then we could think about these electrons, pretend these are our pi electrons in here, attacking at our carbonyl, pushing these electrons off onto our oxygen."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next, we're going to show our oxyanion functioning as a nucleophile. So once again, our carbonyl is polarized, partial negative, partial positive. So we could think about these electrons in red moving into here. And then we could think about these electrons, pretend these are our pi electrons in here, attacking at our carbonyl, pushing these electrons off onto our oxygen. So let's go ahead and draw the result of our nucleophilic attack right here. So we would form our ring. It'd be a five-membered ring."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we could think about these electrons, pretend these are our pi electrons in here, attacking at our carbonyl, pushing these electrons off onto our oxygen. So let's go ahead and draw the result of our nucleophilic attack right here. So we would form our ring. It'd be a five-membered ring. So let's go ahead and draw our five-membered ring here. We would form a carbonyl. And then, let's see, at this carbon, we would have this group coming off here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It'd be a five-membered ring. So let's go ahead and draw our five-membered ring here. We would form a carbonyl. And then, let's see, at this carbon, we would have this group coming off here. And then we would form an alkoxide, which you could protonate to form our aldol. And then we have a methyl group right here. So let's follow those electrons."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then, let's see, at this carbon, we would have this group coming off here. And then we would form an alkoxide, which you could protonate to form our aldol. And then we have a methyl group right here. So let's follow those electrons. So a lot happened here. So if the oxyanion acts as your nucleophile, so these electrons in here, which I'm saying are your pi electrons, are going to form this bond right here. So very similar to how we did it in the previous video, except we're using an oxyanion this time as our nucleophile."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons. So a lot happened here. So if the oxyanion acts as your nucleophile, so these electrons in here, which I'm saying are your pi electrons, are going to form this bond right here. So very similar to how we did it in the previous video, except we're using an oxyanion this time as our nucleophile. These electrons in here move in to reform your carbonyl. And if we're thinking about carbons, so this carbon right here is this carbon. And let's identify some other carbons."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So very similar to how we did it in the previous video, except we're using an oxyanion this time as our nucleophile. These electrons in here move in to reform your carbonyl. And if we're thinking about carbons, so this carbon right here is this carbon. And let's identify some other carbons. So this carbon right here is this carbon. And then this carbon is this carbon, just so it's not getting a little bit confusing in terms of what we have. So if we go back to our alpha carbon here, so there's an acidic proton on that alpha carbon."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's identify some other carbons. So this carbon right here is this carbon. And then this carbon is this carbon, just so it's not getting a little bit confusing in terms of what we have. So if we go back to our alpha carbon here, so there's an acidic proton on that alpha carbon. So I could go ahead and draw that in. And then our base could come along. So hydroxide could come along and take that acidic proton."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we go back to our alpha carbon here, so there's an acidic proton on that alpha carbon. So I could go ahead and draw that in. And then our base could come along. So hydroxide could come along and take that acidic proton. So let's go ahead and show that. So negative 1 formal charge takes this proton. These electrons move in here, and you lose hydroxide."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So hydroxide could come along and take that acidic proton. So let's go ahead and show that. So negative 1 formal charge takes this proton. These electrons move in here, and you lose hydroxide. And then you have your products. Let's go ahead and draw in our products. So we have a five-membered ring."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These electrons move in here, and you lose hydroxide. And then you have your products. Let's go ahead and draw in our products. So we have a five-membered ring. We have a ketone here. We have a double bond here. And then we retain the stereochemistry of this group that's coming off."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have a five-membered ring. We have a ketone here. We have a double bond here. And then we retain the stereochemistry of this group that's coming off. And then we have a methyl group. So we had already formed a carbon-carbon bond in here. And then let's go ahead and make these electrons in here."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we retain the stereochemistry of this group that's coming off. And then we have a methyl group. So we had already formed a carbon-carbon bond in here. And then let's go ahead and make these electrons in here. Let's make them magenta. So these electrons in here are going to move in to form your pi bond to give you your product. So the product is called cis-jasmine."}, {"video_title": "Intramolecular aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then let's go ahead and make these electrons in here. Let's make them magenta. So these electrons in here are going to move in to form your pi bond to give you your product. So the product is called cis-jasmine. And it's found in jasmine flowers. And so it gives it its scent. So this is a pretty cool reaction, intramolecular aldol condensation, which gives you a product that's used in the perfume industry and a pretty good yield."}, {"video_title": "Ka and acid strength Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so here's our equilibrium expression, and the ionization constant Ka for a weak acid, we already talked about the fact that it's going to be less than one. So here we have three weak acids. So hydrofluoric acid, acetic acid, and methanol. And over here are the Ka values. So you can see that hydrofluoric acid has the largest Ka value. So even though they're all considered to be weak acids, 3.5 times 10 to the negative four is larger than 1.8 times 10 to the negative five. So hydrofluoric acid is stronger than acetic acid, and acetic acid is stronger than methanol."}, {"video_title": "Ka and acid strength Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And over here are the Ka values. So you can see that hydrofluoric acid has the largest Ka value. So even though they're all considered to be weak acids, 3.5 times 10 to the negative four is larger than 1.8 times 10 to the negative five. So hydrofluoric acid is stronger than acetic acid, and acetic acid is stronger than methanol. But again, they're all considered to be weak acids relative to the stronger ones. So let's talk about pKa. So the pKa is defined as the negative log of the Ka."}, {"video_title": "Ka and acid strength Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So hydrofluoric acid is stronger than acetic acid, and acetic acid is stronger than methanol. But again, they're all considered to be weak acids relative to the stronger ones. So let's talk about pKa. So the pKa is defined as the negative log of the Ka. So if we wanted to find the pKa for methanol, all we have to do is take the Ka and take the negative log of it. So the pKa is equal to the negative log, negative log of 2.9 times 10 to the negative 16. So let's get out the calculator and let's do that."}, {"video_title": "Ka and acid strength Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the pKa is defined as the negative log of the Ka. So if we wanted to find the pKa for methanol, all we have to do is take the Ka and take the negative log of it. So the pKa is equal to the negative log, negative log of 2.9 times 10 to the negative 16. So let's get out the calculator and let's do that. Negative log of 2.9 times 10 to the negative 16. And this gives us 15.54. Alright, when we round that, so the pKa of methanol, the pKa of methanol is equal to 15.54."}, {"video_title": "Ka and acid strength Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's get out the calculator and let's do that. Negative log of 2.9 times 10 to the negative 16. And this gives us 15.54. Alright, when we round that, so the pKa of methanol, the pKa of methanol is equal to 15.54. So we could write in a pKa column right here, and for methanol it's 15.54. If you did the same calculation for acetic acid, you would get 4.74. And once again, if you did this for hydrofluoric acid, you would get 3.46."}, {"video_title": "Ka and acid strength Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Alright, when we round that, so the pKa of methanol, the pKa of methanol is equal to 15.54. So we could write in a pKa column right here, and for methanol it's 15.54. If you did the same calculation for acetic acid, you would get 4.74. And once again, if you did this for hydrofluoric acid, you would get 3.46. So as we go up on our table here, we're increasing in acid strength. So out of our three weak acids, hydrofluoric acid is the strongest, so it has the largest value for Ka, but notice it has the smallest value for the pKa. So the lower the value for pKa, the more acidic your acid."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the answer there comes from the same technique that we saw Mohorovichik use in 1909 to essentially see how the behavior, or when you measure the seismic waves, or whether you can even measure the seismic waves, the different distances from an earthquake. So if there's an earthquake right here, and we're calling that 0 degrees, let's remember a couple of things here. Let's remember that P waves can travel through anything. They can travel through solid or liquid, or air for that matter. So they can travel through anything. But S waves can only, S for secondary, these are the transverse waves, these can only travel through solids. So it turns out that if an earthquake happens at 0 degrees, and you have seismograph stations all over the world, and these are extremely sensitive in order to be able to measure earthquakes that are happening thousands of kilometers away, it turns out that there's something called an S shadow, an S wave shadow."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They can travel through solid or liquid, or air for that matter. So they can travel through anything. But S waves can only, S for secondary, these are the transverse waves, these can only travel through solids. So it turns out that if an earthquake happens at 0 degrees, and you have seismograph stations all over the world, and these are extremely sensitive in order to be able to measure earthquakes that are happening thousands of kilometers away, it turns out that there's something called an S shadow, an S wave shadow. You can measure, if these are S waves, you can measure them here. You can measure them here. They can go all the way over here."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it turns out that if an earthquake happens at 0 degrees, and you have seismograph stations all over the world, and these are extremely sensitive in order to be able to measure earthquakes that are happening thousands of kilometers away, it turns out that there's something called an S shadow, an S wave shadow. You can measure, if these are S waves, you can measure them here. You can measure them here. They can go all the way over here. They can go over here. They can go over there. You can measure them over here."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They can go all the way over here. They can go over here. They can go over there. You can measure them over here. So you can measure them at all of these points, but then all of a sudden, at 105 degrees, and so we're measuring 0 degrees here and we're going outwards like that, all of a sudden at 105 degrees and further, you stop measuring S waves. They don't get, for some reason, you would think that the S waves would get over here. Maybe they would be a little bit weaker, but they would be able to get all the way over here."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can measure them over here. So you can measure them at all of these points, but then all of a sudden, at 105 degrees, and so we're measuring 0 degrees here and we're going outwards like that, all of a sudden at 105 degrees and further, you stop measuring S waves. They don't get, for some reason, you would think that the S waves would get over here. Maybe they would be a little bit weaker, but they would be able to get all the way over here. But they just abruptly stop. No more S waves. So in this whole area right over here, you get no S waves."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe they would be a little bit weaker, but they would be able to get all the way over here. But they just abruptly stop. No more S waves. So in this whole area right over here, you get no S waves. And obviously, I could flip this picture over and you'd see a symmetric thing on the other side of the globe that all of this area over here, you also would not see. You would also not see S waves. You'd only see them from 105 degrees in this direction and 105 degrees in that direction."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in this whole area right over here, you get no S waves. And obviously, I could flip this picture over and you'd see a symmetric thing on the other side of the globe that all of this area over here, you also would not see. You would also not see S waves. You'd only see them from 105 degrees in this direction and 105 degrees in that direction. And the only reasonable explanation that we can give is that there must be some material that an S wave cannot travel through, that it would have to travel through to get to these points beyond 105 degrees. And we know that S waves only travel in solids. So the assumption there is that at some point beyond 105 degrees, it's hitting liquid."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You'd only see them from 105 degrees in this direction and 105 degrees in that direction. And the only reasonable explanation that we can give is that there must be some material that an S wave cannot travel through, that it would have to travel through to get to these points beyond 105 degrees. And we know that S waves only travel in solids. So the assumption there is that at some point beyond 105 degrees, it's hitting liquid. So that's what tells us that this right here is probably a liquid. So it's hitting some layer that is liquid. So that tells us that there's a core, and at least the outer part of that core is liquid, enough to stop S waves."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the assumption there is that at some point beyond 105 degrees, it's hitting liquid. So that's what tells us that this right here is probably a liquid. So it's hitting some layer that is liquid. So that tells us that there's a core, and at least the outer part of that core is liquid, enough to stop S waves. So the S waves, because it only travels in solids, it leads to this S wave shadow. And this tells us that we have a core. And that core, at least the outer part, is liquid."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that tells us that there's a core, and at least the outer part of that core is liquid, enough to stop S waves. So the S waves, because it only travels in solids, it leads to this S wave shadow. And this tells us that we have a core. And that core, at least the outer part, is liquid. We don't know yet whether the inner part is liquid or solid. Now the next point of evidence is how do we know that there's an inner core? And we can use P waves for that."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that core, at least the outer part, is liquid. We don't know yet whether the inner part is liquid or solid. Now the next point of evidence is how do we know that there's an inner core? And we can use P waves for that. The P wave can travel through anything. But remember, as you get denser material, in general, for the same type of material, if you get denser material, it's going to move faster. So it's going to refract outwards like we've seen over here."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we can use P waves for that. The P wave can travel through anything. But remember, as you get denser material, in general, for the same type of material, if you get denser material, it's going to move faster. So it's going to refract outwards like we've seen over here. But if it goes into a liquid, in general, sound waves, or I should say P waves, seismic waves, move slower in liquids. And so the refraction patterns we get when we do measure from seismograph stations around the world is that it looks like the P waves are doing what you would expect in the mantle. But then they're getting refracted as if they're going into a slower medium as they go through the outer core."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's going to refract outwards like we've seen over here. But if it goes into a liquid, in general, sound waves, or I should say P waves, seismic waves, move slower in liquids. And so the refraction patterns we get when we do measure from seismograph stations around the world is that it looks like the P waves are doing what you would expect in the mantle. But then they're getting refracted as if they're going into a slower medium as they go through the outer core. And we see that right over here. And then they get refracted again to get to some point on the other side. Now that is just what you would expect if it was all liquid."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But then they're getting refracted as if they're going into a slower medium as they go through the outer core. And we see that right over here. And then they get refracted again to get to some point on the other side. Now that is just what you would expect if it was all liquid. But if you go to stations that are even further out, it looks like, if you just look at the refraction patterns, and you can now model this with fancy computers and get all the data points, but you could say, well, the only way that reality can fit the data that we get based on when things reach here is if the P waves are being first refracted through the outer core. But then they're refracted in a way that they're going through denser material, significantly denser material in the inner core. And then they're just continuing to refract the way you would expect."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now that is just what you would expect if it was all liquid. But if you go to stations that are even further out, it looks like, if you just look at the refraction patterns, and you can now model this with fancy computers and get all the data points, but you could say, well, the only way that reality can fit the data that we get based on when things reach here is if the P waves are being first refracted through the outer core. But then they're refracted in a way that they're going through denser material, significantly denser material in the inner core. And then they're just continuing to refract the way you would expect. So it's really the refraction pattern of the P waves. And frankly, the fact that there's this what you call a P wave shadow. The P wave shadow by itself, all that tells you is that kind of roughly crazy things are happening someplace in the core."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then they're just continuing to refract the way you would expect. So it's really the refraction pattern of the P waves. And frankly, the fact that there's this what you call a P wave shadow. The P wave shadow by itself, all that tells you is that kind of roughly crazy things are happening someplace in the core. But the real way to know that we have an inner core that's solid, as opposed to the whole thing being liquid, is that the P waves is the pattern of how the P waves reach essentially the other side of the globe. And then you can kind of, based on modeling how waves would travel through different densities and different types of mediums, you could say, well, there's got to be an inner core right over here. And obviously, it's a lot more math than I'm going into."}, {"video_title": "How we know about the earth's core Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The P wave shadow by itself, all that tells you is that kind of roughly crazy things are happening someplace in the core. But the real way to know that we have an inner core that's solid, as opposed to the whole thing being liquid, is that the P waves is the pattern of how the P waves reach essentially the other side of the globe. And then you can kind of, based on modeling how waves would travel through different densities and different types of mediums, you could say, well, there's got to be an inner core right over here. And obviously, it's a lot more math than I'm going into. But if you do the math based on the shadow and you know the speed of the material and all of that type of thing, then you can figure out the depth at which these transitions occur. We know that we have a transition from mantle to outer core here, and then a transition from outer core to core there. So hopefully that satiates your questions about how do we know what the composition of the earth is without ever having to dig down there, because we've never even gotten below our crust."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the end result is to reduce the benzene ring to form 1,4-cyclohexadiene. Let's look at the mechanism for the Birch reduction. So we know that sodium is in group one of the periodic table. And so it has one valence electron, which I will go ahead and color magenta there. And so the start of the mechanism is for sodium to donate its one valence electron to the benzene ring. And so we can show the movement of that one electron with a fishhook arrow or a half-headed arrow here where we show that one electron moving over here to this carbon, so this carbon right here. Now, we're also going to get some movement of electrons in our benzene ring."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it has one valence electron, which I will go ahead and color magenta there. And so the start of the mechanism is for sodium to donate its one valence electron to the benzene ring. And so we can show the movement of that one electron with a fishhook arrow or a half-headed arrow here where we show that one electron moving over here to this carbon, so this carbon right here. Now, we're also going to get some movement of electrons in our benzene ring. So when I think about these electrons in here, so these pi electrons in red, we know that there are two electrons there. So let me go ahead and show those two electrons like that. So those two electrons are also going to move."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, we're also going to get some movement of electrons in our benzene ring. So when I think about these electrons in here, so these pi electrons in red, we know that there are two electrons there. So let me go ahead and show those two electrons like that. So those two electrons are also going to move. So this electron over here is going to come off on this carbon as well. And then this other electron in red is going to move over to here. So let's go ahead and show the start of the movement of those electrons."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So those two electrons are also going to move. So this electron over here is going to come off on this carbon as well. And then this other electron in red is going to move over to here. So let's go ahead and show the start of the movement of those electrons. We're going to come back and move some more. But I just want to do this really slowly here so we can follow along. So we had our hydrogens attached to our ring."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the start of the movement of those electrons. We're going to come back and move some more. But I just want to do this really slowly here so we can follow along. So we had our hydrogens attached to our ring. So let me go ahead and sketch those in really fast here. So these pi electrons are going to stay put for our mechanism. And let me show the electron in magenta, the one that sodium donated."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we had our hydrogens attached to our ring. So let me go ahead and sketch those in really fast here. So these pi electrons are going to stay put for our mechanism. And let me show the electron in magenta, the one that sodium donated. So it ends up being on this carbon right here. And then one of the electrons in red also moved onto that carbon. So let me show that electron in red right there."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let me show the electron in magenta, the one that sodium donated. So it ends up being on this carbon right here. And then one of the electrons in red also moved onto that carbon. So let me show that electron in red right there. So that carbon gets a negative 1 formal charge. So we form an anion here. One of those red electrons is going to move over to this position right here."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me show that electron in red right there. So that carbon gets a negative 1 formal charge. So we form an anion here. One of those red electrons is going to move over to this position right here. And then we are also going to show these electrons moving around. So the electrons in green here, let me just go ahead and highlight those. So these electrons, so there's one and there's two."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "One of those red electrons is going to move over to this position right here. And then we are also going to show these electrons moving around. So the electrons in green here, let me just go ahead and highlight those. So these electrons, so there's one and there's two. So one of them is going to move to the same position that the red one did in there. And the other one is going to come off onto this carbon. So let me see if I can show that."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons, so there's one and there's two. So one of them is going to move to the same position that the red one did in there. And the other one is going to come off onto this carbon. So let me see if I can show that. So one of them moved in here like that. And the other one moved off onto this carbon like that. So the one in red and the one in green are, of course, now a pi bond."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me see if I can show that. So one of them moved in here like that. And the other one moved off onto this carbon like that. So the one in red and the one in green are, of course, now a pi bond. You could think about it that way. And so we've now generated what's called a radical anion here. So this is a radical anion."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the one in red and the one in green are, of course, now a pi bond. You could think about it that way. And so we've now generated what's called a radical anion here. So this is a radical anion. So it's a radical because you have that one unpaired electron in green. And it's an anion since you have a negative 1 formal charge over here on this top carbon. And then I forgot to put in this hydrogen."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is a radical anion. So it's a radical because you have that one unpaired electron in green. And it's an anion since you have a negative 1 formal charge over here on this top carbon. And then I forgot to put in this hydrogen. So let me go ahead and add that one in there like that. Second step of our mechanism, our alcohol comes along. So we have our generic alcohol, which is going to function as an acid because the negative 1 formal charge, the anion here, the carbon is going to function as a base."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then I forgot to put in this hydrogen. So let me go ahead and add that one in there like that. Second step of our mechanism, our alcohol comes along. So we have our generic alcohol, which is going to function as an acid because the negative 1 formal charge, the anion here, the carbon is going to function as a base. And so these electrons here are going to pick up a proton from the alcohol. So these electrons would kick off onto the oxygen here. And so let's go ahead and draw the result of that acid-base reaction."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our generic alcohol, which is going to function as an acid because the negative 1 formal charge, the anion here, the carbon is going to function as a base. And so these electrons here are going to pick up a proton from the alcohol. So these electrons would kick off onto the oxygen here. And so let's go ahead and draw the result of that acid-base reaction. So we have these pi electrons in here. And we now have two hydrogens on that top carbon. We have one hydrogen on each of our other carbons here."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and draw the result of that acid-base reaction. So we have these pi electrons in here. And we now have two hydrogens on that top carbon. We have one hydrogen on each of our other carbons here. And we still have a radical. So let me go ahead and show this electron is still on that carbon. So now we have a radical instead of a radical anion."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have one hydrogen on each of our other carbons here. And we still have a radical. So let me go ahead and show this electron is still on that carbon. So now we have a radical instead of a radical anion. And I should point out that for our radical anion and for our radical, the electron density can be delocalized throughout the ring. But here we're just trying to show just moving around some electrons. And so let me go ahead and highlight these two electrons here."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So now we have a radical instead of a radical anion. And I should point out that for our radical anion and for our radical, the electron density can be delocalized throughout the ring. But here we're just trying to show just moving around some electrons. And so let me go ahead and highlight these two electrons here. So the electron in red and the electron in magenta are forming a bond with that proton right here on our ring. So next step in our mechanism, we get some more sodium. So some more sodium comes along here."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so let me go ahead and highlight these two electrons here. So the electron in red and the electron in magenta are forming a bond with that proton right here on our ring. So next step in our mechanism, we get some more sodium. So some more sodium comes along here. Let me go ahead and show that. And of course, once again, sodium has one valence electron. So here is sodium's one valence electron."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So some more sodium comes along here. Let me go ahead and show that. And of course, once again, sodium has one valence electron. So here is sodium's one valence electron. The sodium can donate that valence electron to our benzene ring. And so it's going to donate it over here to this carbon, the carbon that had the green electron on it already. And so let's go ahead and show the results of that."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So here is sodium's one valence electron. The sodium can donate that valence electron to our benzene ring. And so it's going to donate it over here to this carbon, the carbon that had the green electron on it already. And so let's go ahead and show the results of that. We would have our ring. We would have our pi electrons. We had two hydrogens bonded to the top carbon."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and show the results of that. We would have our ring. We would have our pi electrons. We had two hydrogens bonded to the top carbon. We had these hydrogens around my ring like that. The bottom carbon still has a hydrogen bonded to it. And we started with a green electron on that carbon."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We had two hydrogens bonded to the top carbon. We had these hydrogens around my ring like that. The bottom carbon still has a hydrogen bonded to it. And we started with a green electron on that carbon. And now we're going to add a magenta electron, giving that carbon a negative 1 formal charge. So we form an anion again. So now that we have an anion, the last step of the mechanism is another acid-base reaction."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we started with a green electron on that carbon. And now we're going to add a magenta electron, giving that carbon a negative 1 formal charge. So we form an anion again. So now that we have an anion, the last step of the mechanism is another acid-base reaction. So our alcohol comes along. And the carbanion is going to function as a base and pick up a proton from our alcohol. So the same step that we saw before, pretty much."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So now that we have an anion, the last step of the mechanism is another acid-base reaction. So our alcohol comes along. And the carbanion is going to function as a base and pick up a proton from our alcohol. So the same step that we saw before, pretty much. And we go ahead and draw our final product. So we have those pi electrons. We had these hydrogens on our top carbon."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the same step that we saw before, pretty much. And we go ahead and draw our final product. So we have those pi electrons. We had these hydrogens on our top carbon. These carbons all get hydrogens. And then finally, we have added on a proton to this bottom carbon here. So let me go ahead and highlight those."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We had these hydrogens on our top carbon. These carbons all get hydrogens. And then finally, we have added on a proton to this bottom carbon here. So let me go ahead and highlight those. So these electrons here, those two electrons, pick up a proton. So we protonate our ring and we finish. So this is our 1,4-cyclohexadiene product here."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight those. So these electrons here, those two electrons, pick up a proton. So we protonate our ring and we finish. So this is our 1,4-cyclohexadiene product here. Now, a simple way of thinking about this mechanism for the Birch reduction is to break it into these four steps and to make those steps very simple. So in the first step, sodium is donating an electron. So you could think electron for step one."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is our 1,4-cyclohexadiene product here. Now, a simple way of thinking about this mechanism for the Birch reduction is to break it into these four steps and to make those steps very simple. So in the first step, sodium is donating an electron. So you could think electron for step one. Second step, we know that the anion is picking up a proton from our alcohol. So you could think proton for the second step. For the third step, once again, sodium is going to donate an electron."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could think electron for step one. Second step, we know that the anion is picking up a proton from our alcohol. So you could think proton for the second step. For the third step, once again, sodium is going to donate an electron. So you could think electron. And then finally, once again, the anion is picking up a proton. So you could think proton."}, {"video_title": "Birch reduction I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "For the third step, once again, sodium is going to donate an electron. So you could think electron. And then finally, once again, the anion is picking up a proton. So you could think proton. So you could think electron, proton, electron, proton is a simple way of thinking about the steps for a Birch reduction. In the next video, we're going to look at what happens with the Birch reduction when you get a substituted benzene ring. And I'll show you mechanisms for the two possibilities that you might see on an exam."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Bond line structures contain the same information as a Lewis dot structure, but it's obviously much easier, much faster to draw the bond line structure on the right than the full Lewis dot structure on the left. What about three-dimensional bond line structures? So how could you represent this molecule in three dimensions using a flat sheet of paper? Well, on the left here is a picture where I made a model of this molecule, and this is gonna help us draw this molecule in three dimensions. So we have a flat sheet of paper. How could we represent this picture on our flat sheet of paper? Let's start with the carbon in the center."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, on the left here is a picture where I made a model of this molecule, and this is gonna help us draw this molecule in three dimensions. So we have a flat sheet of paper. How could we represent this picture on our flat sheet of paper? Let's start with the carbon in the center. So that's our carbon in magenta. So that's this one on our Lewis dot structure, this one on our bond line structure. Well, the carbon in magenta is sp3 hybridized."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with the carbon in the center. So that's our carbon in magenta. So that's this one on our Lewis dot structure, this one on our bond line structure. Well, the carbon in magenta is sp3 hybridized. This is sp3 hybridized, so we would expect tetrahedral geometry around that carbon. And if you look at that carbon on the picture here, you can see that this bond and this bond are in the same plane. So if you had a flat sheet of paper, you could say those bonds are in the same plane."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, the carbon in magenta is sp3 hybridized. This is sp3 hybridized, so we would expect tetrahedral geometry around that carbon. And if you look at that carbon on the picture here, you can see that this bond and this bond are in the same plane. So if you had a flat sheet of paper, you could say those bonds are in the same plane. So a line represents a bond in the plane of the paper. Let me go ahead and draw that. So this is the carbon in magenta, and then we have these two bonds here, and those bonds are in the plane of the paper."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if you had a flat sheet of paper, you could say those bonds are in the same plane. So a line represents a bond in the plane of the paper. Let me go ahead and draw that. So this is the carbon in magenta, and then we have these two bonds here, and those bonds are in the plane of the paper. Next, let's look at what else is connected to the carbon in magenta. Well, obviously there is an OH, so let me go ahead and circle that. So there's an OH, we can see there's an OH here."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this is the carbon in magenta, and then we have these two bonds here, and those bonds are in the plane of the paper. Next, let's look at what else is connected to the carbon in magenta. Well, obviously there is an OH, so let me go ahead and circle that. So there's an OH, we can see there's an OH here. And then the OH in our picture is coming out at us in space. So hopefully you can visualize that this bond in here is coming towards you in space, which is why this oxygen, this red oxygen atom, looks so big. So this is coming towards you."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So there's an OH, we can see there's an OH here. And then the OH in our picture is coming out at us in space. So hopefully you can visualize that this bond in here is coming towards you in space, which is why this oxygen, this red oxygen atom, looks so big. So this is coming towards you. We would represent that with a wedge. So let me go ahead and draw a wedge in here. And a wedge means that the bond is in front of your paper."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this is coming towards you. We would represent that with a wedge. So let me go ahead and draw a wedge in here. And a wedge means that the bond is in front of your paper. So this means the OH is coming out at you in space. So let me draw in the OH like that. Now let's look at what else is connected to that carbon in magenta."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And a wedge means that the bond is in front of your paper. So this means the OH is coming out at you in space. So let me draw in the OH like that. Now let's look at what else is connected to that carbon in magenta. We know there's a hydrogen, right? We didn't draw it over here, but we know there's a hydrogen connected to that carbon. And we can see that this hydrogen, this hydrogen right here, let me go ahead and switch colors, this hydrogen is going away from us in space."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now let's look at what else is connected to that carbon in magenta. We know there's a hydrogen, right? We didn't draw it over here, but we know there's a hydrogen connected to that carbon. And we can see that this hydrogen, this hydrogen right here, let me go ahead and switch colors, this hydrogen is going away from us in space. So this bond is going away from us in space or into the paper, or the bond is behind the paper. And we represent that with a dash. So I'm gonna draw a dash here showing that this hydrogen is going away from us."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we can see that this hydrogen, this hydrogen right here, let me go ahead and switch colors, this hydrogen is going away from us in space. So this bond is going away from us in space or into the paper, or the bond is behind the paper. And we represent that with a dash. So I'm gonna draw a dash here showing that this hydrogen is going away from us. So we're imagining our flat sheet of paper and the OH coming out at us and that hydrogen going away from us. All right, next let's look at the carbon on the left here. So this carbon in blue."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna draw a dash here showing that this hydrogen is going away from us. So we're imagining our flat sheet of paper and the OH coming out at us and that hydrogen going away from us. All right, next let's look at the carbon on the left here. So this carbon in blue. So that's this carbon. And I'll say that's this carbon over here on the left. So we know that this carbon, we can see that this bond and this bond are in the same plane."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon in blue. So that's this carbon. And I'll say that's this carbon over here on the left. So we know that this carbon, we can see that this bond and this bond are in the same plane. So let's go ahead and draw in the carbon. So the carbon that I just put in is the carbon in blue. And this hydrogen over here on the left, this bond is in the same plane."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we know that this carbon, we can see that this bond and this bond are in the same plane. So let's go ahead and draw in the carbon. So the carbon that I just put in is the carbon in blue. And this hydrogen over here on the left, this bond is in the same plane. So I'm gonna draw a line representing the bond is in the plane of the paper. And so we have a hydrogen right here. What about the other two hydrogens?"}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And this hydrogen over here on the left, this bond is in the same plane. So I'm gonna draw a line representing the bond is in the plane of the paper. And so we have a hydrogen right here. What about the other two hydrogens? Well, let me highlight those. So this hydrogen, hopefully you can see that this is coming out at us in space. So we represent that with a wedge."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "What about the other two hydrogens? Well, let me highlight those. So this hydrogen, hopefully you can see that this is coming out at us in space. So we represent that with a wedge. So we draw a wedge right here and then we draw in the hydrogen. So the bond is in front of the paper. The bond is coming towards us in space."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we represent that with a wedge. So we draw a wedge right here and then we draw in the hydrogen. So the bond is in front of the paper. The bond is coming towards us in space. And then there's another hydrogen bonded to the carbon in blue. And my thumb here is blocking it a little bit, so hopefully you can see that's going away from us in space. So this hydrogen is going away from us in space."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The bond is coming towards us in space. And then there's another hydrogen bonded to the carbon in blue. And my thumb here is blocking it a little bit, so hopefully you can see that's going away from us in space. So this hydrogen is going away from us in space. So the carbon in blue is also sp3 hybridized. This carbon is sp3 hybridized. We would expect tetrahedral geometry around that carbon."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen is going away from us in space. So the carbon in blue is also sp3 hybridized. This carbon is sp3 hybridized. We would expect tetrahedral geometry around that carbon. And then finally, let's look at the last carbon, so this carbon right here in red. So that's this one right here and this one right here. This carbon is also sp3 hybridized."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We would expect tetrahedral geometry around that carbon. And then finally, let's look at the last carbon, so this carbon right here in red. So that's this one right here and this one right here. This carbon is also sp3 hybridized. So we expect tetrahedral geometry. And if we look at this carbon, let me use yellow again, this bond and this bond are in the same plane. So let's draw in the carbon in red and we can draw in the hydrogen right here in the same plane."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is also sp3 hybridized. So we expect tetrahedral geometry. And if we look at this carbon, let me use yellow again, this bond and this bond are in the same plane. So let's draw in the carbon in red and we can draw in the hydrogen right here in the same plane. And then let's visualize the other two hydrogens. So this hydrogen is coming out at us in space, so we represent that with a wedge like that. So we have a hydrogen coming out at us in space."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw in the carbon in red and we can draw in the hydrogen right here in the same plane. And then let's visualize the other two hydrogens. So this hydrogen is coming out at us in space, so we represent that with a wedge like that. So we have a hydrogen coming out at us in space. And this other hydrogen here is going away from us in space. It looks a little bit smaller, so that's a dash right here. So we've drawn everything out."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we have a hydrogen coming out at us in space. And this other hydrogen here is going away from us in space. It looks a little bit smaller, so that's a dash right here. So we've drawn everything out. And notice when you have a tetrahedral carbon, an sp3 hybridized carbon, you have this pattern. You have this pattern of two bonds in the plane of the paper and one wedge and one dash. So that's how we're representing our tetrahedral geometry around those carbons, around those sp3 hybridized carbons."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we've drawn everything out. And notice when you have a tetrahedral carbon, an sp3 hybridized carbon, you have this pattern. You have this pattern of two bonds in the plane of the paper and one wedge and one dash. So that's how we're representing our tetrahedral geometry around those carbons, around those sp3 hybridized carbons. Usually you don't see the hydrogens drawn in on one of these, so we could simplify our three-dimensional bond line structure even more. We could just say we have an OH coming out at us in space. So I draw a wedge right here."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that's how we're representing our tetrahedral geometry around those carbons, around those sp3 hybridized carbons. Usually you don't see the hydrogens drawn in on one of these, so we could simplify our three-dimensional bond line structure even more. We could just say we have an OH coming out at us in space. So I draw a wedge right here. And you could draw it like that, which is implied that there's a hydrogen going away from you. So if you draw an OH coming out at you, that implies that there is a hydrogen going away from you in space. So three-dimensional bond line structures are an important skill to be able to visualize."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I draw a wedge right here. And you could draw it like that, which is implied that there's a hydrogen going away from you. So if you draw an OH coming out at you, that implies that there is a hydrogen going away from you in space. So three-dimensional bond line structures are an important skill to be able to visualize. For this molecule, it's not so important that the OH is coming out at us. That's just how I drew it here, because that's what it looks like in the picture. But later in organic chemistry, it's very important to understand what's coming out at you in space, what's going away from you, what does the molecule look like in three dimensions?"}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So three-dimensional bond line structures are an important skill to be able to visualize. For this molecule, it's not so important that the OH is coming out at us. That's just how I drew it here, because that's what it looks like in the picture. But later in organic chemistry, it's very important to understand what's coming out at you in space, what's going away from you, what does the molecule look like in three dimensions? And bond line structures, three-dimensional bond line structures, allow us to visualize that. Model sets help, so you should definitely purchase a model set at this point in your study of organic chemistry, because it's going to help you a lot later in the course. On the left is the Lewis dot structure for acetone."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But later in organic chemistry, it's very important to understand what's coming out at you in space, what's going away from you, what does the molecule look like in three dimensions? And bond line structures, three-dimensional bond line structures, allow us to visualize that. Model sets help, so you should definitely purchase a model set at this point in your study of organic chemistry, because it's going to help you a lot later in the course. On the left is the Lewis dot structure for acetone. And we could turn that into a bond line structure really quickly. So here is the bond line structure for acetone. I could put in lone pairs of electrons on the oxygen, or I could leave them off."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "On the left is the Lewis dot structure for acetone. And we could turn that into a bond line structure really quickly. So here is the bond line structure for acetone. I could put in lone pairs of electrons on the oxygen, or I could leave them off. I'll just go ahead and put those lone pairs in there, like that. What would be a three-dimensional bond line structure for acetone? Well, on the left here is a model of the acetone molecule."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I could put in lone pairs of electrons on the oxygen, or I could leave them off. I'll just go ahead and put those lone pairs in there, like that. What would be a three-dimensional bond line structure for acetone? Well, on the left here is a model of the acetone molecule. And hopefully you can see that these atoms right here, these atoms are all in the same plane of the page. And so is oxygen, actually. So so is this oxygen here."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, on the left here is a model of the acetone molecule. And hopefully you can see that these atoms right here, these atoms are all in the same plane of the page. And so is oxygen, actually. So so is this oxygen here. And it's a little bit easier to see in the picture on the right. So in the picture on the right, so these are the atoms that we were just talking about. So these are all in the same plane."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So so is this oxygen here. And it's a little bit easier to see in the picture on the right. So in the picture on the right, so these are the atoms that we were just talking about. So these are all in the same plane. So if you can, hopefully you can visualize like a sheet of paper. So let me see if I can sketch in a sheet of paper right here like this. And so all of those atoms, all those atoms are in the same plane of that paper."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So these are all in the same plane. So if you can, hopefully you can visualize like a sheet of paper. So let me see if I can sketch in a sheet of paper right here like this. And so all of those atoms, all those atoms are in the same plane of that paper. So we could draw that in like this. We could have our carbons over here like that, our three carbons. And according to this picture, these two hydrogens here, so if we're trying to copy this picture, those two hydrogens are in the plane of the page."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so all of those atoms, all those atoms are in the same plane of that paper. So we could draw that in like this. We could have our carbons over here like that, our three carbons. And according to this picture, these two hydrogens here, so if we're trying to copy this picture, those two hydrogens are in the plane of the page. And so is this oxygen. So this oxygen right here is in the plane of the page. So we'll draw it in there like that."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And according to this picture, these two hydrogens here, so if we're trying to copy this picture, those two hydrogens are in the plane of the page. And so is this oxygen. So this oxygen right here is in the plane of the page. So we'll draw it in there like that. I'll go ahead and put in lone pairs of electrons on the oxygen. All right, let's focus in on the carbon in the center for right now. I'm not done with my three-dimensional bond line structure."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we'll draw it in there like that. I'll go ahead and put in lone pairs of electrons on the oxygen. All right, let's focus in on the carbon in the center for right now. I'm not done with my three-dimensional bond line structure. I just want to point something out right here. So this carbon right here in magenta, right, is this carbon, which is this one, just it's hiding back here. But we know that carbon is sp2 hybridized."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I'm not done with my three-dimensional bond line structure. I just want to point something out right here. So this carbon right here in magenta, right, is this carbon, which is this one, just it's hiding back here. But we know that carbon is sp2 hybridized. So we would expect trigonal planar geometry around that carbon. So trigonal planar, right, we would expect everything to be planar. So the atoms connected to the carbon in magenta, we would expect those to be in the same plane."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But we know that carbon is sp2 hybridized. So we would expect trigonal planar geometry around that carbon. So trigonal planar, right, we would expect everything to be planar. So the atoms connected to the carbon in magenta, we would expect those to be in the same plane. Now it's a little bit easier to see that on the right here. So these carbons and these carbons and this oxygen, right, those are all in the same plane. We have sp2 hybridization."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the atoms connected to the carbon in magenta, we would expect those to be in the same plane. Now it's a little bit easier to see that on the right here. So these carbons and these carbons and this oxygen, right, those are all in the same plane. We have sp2 hybridization. All right, let's now look at the other carbons. So let's do blue here. So carbon on the left right here."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have sp2 hybridization. All right, let's now look at the other carbons. So let's do blue here. So carbon on the left right here. I'm saying that's this one. This carbon is sp3 hybridized, right, sp3 hybridized, which means tetrahedral geometry. We would expect tetrahedral geometry around that carbon."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So carbon on the left right here. I'm saying that's this one. This carbon is sp3 hybridized, right, sp3 hybridized, which means tetrahedral geometry. We would expect tetrahedral geometry around that carbon. And so we have these two bonds in the plane of the page, right, and then this hydrogen is coming out at us in space, and then this other hydrogen back here is going away from us in space. So on our three-dimensional bond line structure, we could draw hydrogen coming out at us in space, right, so that's a wedge, and then hydrogen going away from us in space, so that would be a dash. And then the carbon on the right, which I'll make red, this is also sp3 hybridized, so we would expect tetrahedral geometry."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We would expect tetrahedral geometry around that carbon. And so we have these two bonds in the plane of the page, right, and then this hydrogen is coming out at us in space, and then this other hydrogen back here is going away from us in space. So on our three-dimensional bond line structure, we could draw hydrogen coming out at us in space, right, so that's a wedge, and then hydrogen going away from us in space, so that would be a dash. And then the carbon on the right, which I'll make red, this is also sp3 hybridized, so we would expect tetrahedral geometry. And we can see this bond is in the plane, this bond is in the plane of the page, and this hydrogen is coming out at us, and then this one's going away from us. So we could complete our three-dimensional bond line structure over here on the right by showing a hydrogen coming out at us and a hydrogen going away from us in space. All right, if we look at this picture on the right again, and we've already visualized our flat sheet of paper, imagine your eye is right here."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then the carbon on the right, which I'll make red, this is also sp3 hybridized, so we would expect tetrahedral geometry. And we can see this bond is in the plane, this bond is in the plane of the page, and this hydrogen is coming out at us, and then this one's going away from us. So we could complete our three-dimensional bond line structure over here on the right by showing a hydrogen coming out at us and a hydrogen going away from us in space. All right, if we look at this picture on the right again, and we've already visualized our flat sheet of paper, imagine your eye is right here. Imagine your eye is looking down on your flat sheet of paper. So your eye would see, let me use dark blue for this, these two hydrogens coming out at you in space, right, so that's these two hydrogens coming out at you. So hopefully this just helps you visualize it a little bit better."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "All right, if we look at this picture on the right again, and we've already visualized our flat sheet of paper, imagine your eye is right here. Imagine your eye is looking down on your flat sheet of paper. So your eye would see, let me use dark blue for this, these two hydrogens coming out at you in space, right, so that's these two hydrogens coming out at you. So hopefully this just helps you visualize it a little bit better. And then your eye would see these two hydrogens, these would be behind the plane of the paper, right, so that's going away from you in space. So the hydrogens in green would be going away from you, so we represent that with a dash. So hopefully that helps you visualize it a little bit better here."}, {"video_title": "Three-dimensional bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So hopefully this just helps you visualize it a little bit better. And then your eye would see these two hydrogens, these would be behind the plane of the paper, right, so that's going away from you in space. So the hydrogens in green would be going away from you, so we represent that with a dash. So hopefully that helps you visualize it a little bit better here. Now for acetone, you normally wouldn't draw out a three-dimensional bond line structure. There's not much of a point to drawing the structure on the right. I just did it to help you visualize things a little bit better and to contrast an sp2 hybridized carbon with an sp3 hybridized carbon and to think about what it looks like in three dimensions."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Here we have a representative carboxylic acid derivative with this Y substituent here bonded to the carbonyl. We know that carbonyls are reactive because this oxygen is withdrawing some electron density away from our carbonyl carbon, making it partially positive. And for carboxylic acid derivatives, our Y substituent is an electronegative atom too, and so that's going to withdraw even more electron density from our carbonyl carbon. So let's go ahead and write down the first effect, the inductive effect. So induction is an electron withdrawing effect. We're withdrawing electron density from our carbonyl carbon. That makes our carbonyl carbon more partially positive, so it's more electrophilic and better able to react with a nucleophile."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write down the first effect, the inductive effect. So induction is an electron withdrawing effect. We're withdrawing electron density from our carbonyl carbon. That makes our carbonyl carbon more partially positive, so it's more electrophilic and better able to react with a nucleophile. And so induction increases the reactivity of carboxylic acid derivatives. So this effect increases the reactivity. We have a competing effect of induction with resonance."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "That makes our carbonyl carbon more partially positive, so it's more electrophilic and better able to react with a nucleophile. And so induction increases the reactivity of carboxylic acid derivatives. So this effect increases the reactivity. We have a competing effect of induction with resonance. So let's think about resonance next. So our Y substituent with a lone pair of electrons can donate some electron density to our carbonyl carbon. When we draw our resonance structure, we can see that our top oxygen is going to have a negative one formal charge, and we would have a bond between, a pi bond between our carbon and our Y substituent, giving our Y a plus one formal charge."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We have a competing effect of induction with resonance. So let's think about resonance next. So our Y substituent with a lone pair of electrons can donate some electron density to our carbonyl carbon. When we draw our resonance structure, we can see that our top oxygen is going to have a negative one formal charge, and we would have a bond between, a pi bond between our carbon and our Y substituent, giving our Y a plus one formal charge. So once again, this concept of increasing the electron density from this lone pair of electrons to our carbonyl carbon, that increases the electron density. That's an electron donating effect. And if you're donating electron density, you're decreasing the partial positive charge, making it less electrophilic, and therefore making it less reactive with a nucleophile."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "When we draw our resonance structure, we can see that our top oxygen is going to have a negative one formal charge, and we would have a bond between, a pi bond between our carbon and our Y substituent, giving our Y a plus one formal charge. So once again, this concept of increasing the electron density from this lone pair of electrons to our carbonyl carbon, that increases the electron density. That's an electron donating effect. And if you're donating electron density, you're decreasing the partial positive charge, making it less electrophilic, and therefore making it less reactive with a nucleophile. So resonance will decrease the reactivity of a carboxylic acid derivative. And so we have these two competing effects of induction versus resonance. And whichever one is going to win, we can think about this balance for helping us to determine the reactivity of our carboxylic acid derivatives."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And if you're donating electron density, you're decreasing the partial positive charge, making it less electrophilic, and therefore making it less reactive with a nucleophile. So resonance will decrease the reactivity of a carboxylic acid derivative. And so we have these two competing effects of induction versus resonance. And whichever one is going to win, we can think about this balance for helping us to determine the reactivity of our carboxylic acid derivatives. So we start with an acyl or acid chloride with, once again, the inductive effect. We know the oxygen withdraws some electron density from our carbonyl carbon, and so does our chlorine. So we have a strong inductive effect."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And whichever one is going to win, we can think about this balance for helping us to determine the reactivity of our carboxylic acid derivatives. So we start with an acyl or acid chloride with, once again, the inductive effect. We know the oxygen withdraws some electron density from our carbonyl carbon, and so does our chlorine. So we have a strong inductive effect. If we think about the possibility of resonance, I would move these electrons into here and push those electrons off onto the oxygen. So when we draw in the possible resonance structure, once again, a negative one formal charge on the oxygen and a plus one formal charge on the chlorine. It has only two lone pairs of electrons around it now like that."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we have a strong inductive effect. If we think about the possibility of resonance, I would move these electrons into here and push those electrons off onto the oxygen. So when we draw in the possible resonance structure, once again, a negative one formal charge on the oxygen and a plus one formal charge on the chlorine. It has only two lone pairs of electrons around it now like that. So if we think about this resonance structure, we have a pi bond between carbon and chlorine. And if we draw the p orbital, carbon's in the second period, so we draw a p orbital for the second period. And if we think about chlorine, chlorine's in the third period, so it has a bigger p orbital."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It has only two lone pairs of electrons around it now like that. So if we think about this resonance structure, we have a pi bond between carbon and chlorine. And if we draw the p orbital, carbon's in the second period, so we draw a p orbital for the second period. And if we think about chlorine, chlorine's in the third period, so it has a bigger p orbital. And it turns out that when you mismatch these sizes, they can overlap as well. And so poor orbital overlap means that chlorine is not donating a lot of electron density to our carbonyl carbon here. So this is not a major contributor in the overall resonance hybrid."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And if we think about chlorine, chlorine's in the third period, so it has a bigger p orbital. And it turns out that when you mismatch these sizes, they can overlap as well. And so poor orbital overlap means that chlorine is not donating a lot of electron density to our carbonyl carbon here. So this is not a major contributor in the overall resonance hybrid. And so therefore, induction is going to dominate. So let me go ahead and write that here. I'll go ahead and use this color here."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this is not a major contributor in the overall resonance hybrid. And so therefore, induction is going to dominate. So let me go ahead and write that here. I'll go ahead and use this color here. So induction dominates. And if induction dominates, then we would expect acyl or acid chlorides to be extremely reactive, and indeed they are. They will react with water, sometimes violently at room temperature."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "I'll go ahead and use this color here. So induction dominates. And if induction dominates, then we would expect acyl or acid chlorides to be extremely reactive, and indeed they are. They will react with water, sometimes violently at room temperature. So let's look at our next carboxylic acid derivative, which is an acid anhydride. So once again, we think about induction first. So this oxygen's withdrawing some electron density from this carbon."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "They will react with water, sometimes violently at room temperature. So let's look at our next carboxylic acid derivative, which is an acid anhydride. So once again, we think about induction first. So this oxygen's withdrawing some electron density from this carbon. So is this oxygen. And if you think about this as your y substituent, you have this other oxygen here, which could contribute. So once again, we have a strong inductive effect."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen's withdrawing some electron density from this carbon. So is this oxygen. And if you think about this as your y substituent, you have this other oxygen here, which could contribute. So once again, we have a strong inductive effect. When we think about resonance, I could move this lone pair of electrons from oxygen into here and push those electrons off. And so when we think about overlapping our orbitals for oxygen and carbon, this is a better situation than before because carbon and oxygen are in the same period on the periodic table. So here we have carbon and oxygen."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we have a strong inductive effect. When we think about resonance, I could move this lone pair of electrons from oxygen into here and push those electrons off. And so when we think about overlapping our orbitals for oxygen and carbon, this is a better situation than before because carbon and oxygen are in the same period on the periodic table. So here we have carbon and oxygen. It's the same period, so similar size p orbitals. So better overlap. And so therefore, there's more of a contribution, there's more of an electron donating effect than in our previous example."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So here we have carbon and oxygen. It's the same period, so similar size p orbitals. So better overlap. And so therefore, there's more of a contribution, there's more of an electron donating effect than in our previous example. So you can think about a lone pair of electrons from the oxygen increasing electron density around this carbonyl carbon here, therefore decreasing the reactivity. However, the induction effect still dominates the resonance effect. One way to think about that is we have a competing resonance structure."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, there's more of a contribution, there's more of an electron donating effect than in our previous example. So you can think about a lone pair of electrons from the oxygen increasing electron density around this carbonyl carbon here, therefore decreasing the reactivity. However, the induction effect still dominates the resonance effect. One way to think about that is we have a competing resonance structure. So this lone pair of electrons can move over to here and those electrons come off onto this oxygen. So some of the electron density, not all of it is being donated to the carbonyl carbon on the left. Some of the electron density is going to the carbonyl carbon on the right here."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "One way to think about that is we have a competing resonance structure. So this lone pair of electrons can move over to here and those electrons come off onto this oxygen. So some of the electron density, not all of it is being donated to the carbonyl carbon on the left. Some of the electron density is going to the carbonyl carbon on the right here. And so resonance is not as big of an effect as induction. And so induction still dominates here. It's much stronger."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Some of the electron density is going to the carbonyl carbon on the right here. And so resonance is not as big of an effect as induction. And so induction still dominates here. It's much stronger. So let's go ahead and write that. So induction is much stronger than resonance. Induction is much stronger than resonance."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's much stronger. So let's go ahead and write that. So induction is much stronger than resonance. Induction is much stronger than resonance. So we would expect an acid anhydride to be pretty reactive because induction increases the reactivity. And that is again what we observe. Acid anhydrides are reactive, will react with water."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Induction is much stronger than resonance. So we would expect an acid anhydride to be pretty reactive because induction increases the reactivity. And that is again what we observe. Acid anhydrides are reactive, will react with water. Something like acetic anhydride will react with water at room temperature. So induction is the stronger effect again. Let's go to the next carboxylic acid derivative which is an ester."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Acid anhydrides are reactive, will react with water. Something like acetic anhydride will react with water at room temperature. So induction is the stronger effect again. Let's go to the next carboxylic acid derivative which is an ester. So once again we think about induction. So this oxygen withdraws some electron density so does this one. When we think about resonance, we move this lone pair into here and move those electrons off onto the oxygen."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's go to the next carboxylic acid derivative which is an ester. So once again we think about induction. So this oxygen withdraws some electron density so does this one. When we think about resonance, we move this lone pair into here and move those electrons off onto the oxygen. We don't have a competing resonance structure this time so the resonance effect is a little bit more important than before. And so the resonance structure is a little bit more important than before. And so there's a closer balance between induction and resonance."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "When we think about resonance, we move this lone pair into here and move those electrons off onto the oxygen. We don't have a competing resonance structure this time so the resonance effect is a little bit more important than before. And so the resonance structure is a little bit more important than before. And so there's a closer balance between induction and resonance. However, induction still wins. So induction is stronger but it's closer than the previous example. So induction is stronger."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so there's a closer balance between induction and resonance. However, induction still wins. So induction is stronger but it's closer than the previous example. So induction is stronger. Let's move now to our final carboxylic acid derivative which is our amide. So once again this oxygen withdraws some electron density from this carbon. Nitrogen is a little bit more electronegative than carbon so we can think about that possibility."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So induction is stronger. Let's move now to our final carboxylic acid derivative which is our amide. So once again this oxygen withdraws some electron density from this carbon. Nitrogen is a little bit more electronegative than carbon so we can think about that possibility. When we consider the resonance effect, move this lone pair of electrons into here, push those electrons off onto your oxygen, and we draw the resonance structure for our amide. Our top oxygen gets a negative one formal charge and we would have our nitrogen now, double bonded to this carbon, put in these hydrogens here, and then this would be a plus one formal charge on the nitrogen. And so in this case, this is a major contributor to the overall hybrid."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Nitrogen is a little bit more electronegative than carbon so we can think about that possibility. When we consider the resonance effect, move this lone pair of electrons into here, push those electrons off onto your oxygen, and we draw the resonance structure for our amide. Our top oxygen gets a negative one formal charge and we would have our nitrogen now, double bonded to this carbon, put in these hydrogens here, and then this would be a plus one formal charge on the nitrogen. And so in this case, this is a major contributor to the overall hybrid. So this resonance structure right here, let me go ahead and identify it, this is a major contributor to the overall hybrid. And we know this because the carbon-nitrogen bond has significant double bond character due to this resonance structure. And this is much more of an important resonance structure than say the one that I didn't draw but we could think about here for the ester."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so in this case, this is a major contributor to the overall hybrid. So this resonance structure right here, let me go ahead and identify it, this is a major contributor to the overall hybrid. And we know this because the carbon-nitrogen bond has significant double bond character due to this resonance structure. And this is much more of an important resonance structure than say the one that I didn't draw but we could think about here for the ester. And the reason why is because nitrogen is not as electronegative as oxygen. So nitrogen is more willing to donate its lone pair of electrons than this oxygen is. And therefore, this resonance structure is more of a contributor."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this is much more of an important resonance structure than say the one that I didn't draw but we could think about here for the ester. And the reason why is because nitrogen is not as electronegative as oxygen. So nitrogen is more willing to donate its lone pair of electrons than this oxygen is. And therefore, this resonance structure is more of a contributor. Another way to say that is the least electronegative element is the one that's most likely to form a plus one charge. And so we're donating a lot of electron density to our carbonyl carbon, therefore we're decreasing the reactivity. And since we have a major contributor to the overall hybrid here, it turns out that the resonance effect is more important than the inductive effect."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, this resonance structure is more of a contributor. Another way to say that is the least electronegative element is the one that's most likely to form a plus one charge. And so we're donating a lot of electron density to our carbonyl carbon, therefore we're decreasing the reactivity. And since we have a major contributor to the overall hybrid here, it turns out that the resonance effect is more important than the inductive effect. So I go ahead and write here this time, resonance wins. So resonance dominates induction. And if resonance dominates induction, then we would expect MIs to be relatively unreactive."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And since we have a major contributor to the overall hybrid here, it turns out that the resonance effect is more important than the inductive effect. So I go ahead and write here this time, resonance wins. So resonance dominates induction. And if resonance dominates induction, then we would expect MIs to be relatively unreactive. And that is of course what we observe. So we talked about induction and resonance for these four carboxylic acid derivatives. And we can see a clear trend now in terms of reactivity, right?"}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And if resonance dominates induction, then we would expect MIs to be relatively unreactive. And that is of course what we observe. So we talked about induction and resonance for these four carboxylic acid derivatives. And we can see a clear trend now in terms of reactivity, right? As you move up in this direction, you get more reactive. So acyl or acid chlorides are the most reactive because induction dominates. And amides are the least reactive because resonance dominates."}, {"video_title": "Reactivity of carboxylic acid derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we can see a clear trend now in terms of reactivity, right? As you move up in this direction, you get more reactive. So acyl or acid chlorides are the most reactive because induction dominates. And amides are the least reactive because resonance dominates. And it's important to understand this trend for reactivity, and especially if you think about biology, because in the human body, there are a lot of esters, and there are a lot of amides. And these are the two least reactive ones that we talked about. There are no acid chlorides or acid anhydrides, they would just be too reactive for the human body."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It was a time we had come out of the Renaissance. We had rediscovered science and reason. In the 1700s, we saw that come about with even more progress of society. As we exit the 1700s and enter into the 1800s, we start having the Industrial Revolution. People saw the steady march of human reason, of human progress. Because of this, a lot of people were saying, humanity will continue to improve and improve forever to a point that poverty will go away. We will turn into this perfect utopian civilization without wars, without strife of any kind."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As we exit the 1700s and enter into the 1800s, we start having the Industrial Revolution. People saw the steady march of human reason, of human progress. Because of this, a lot of people were saying, humanity will continue to improve and improve forever to a point that poverty will go away. We will turn into this perfect utopian civilization without wars, without strife of any kind. There was something to be said about that. You had significant improvements. In fact, you had even more dramatic improvements once the Industrial Revolution started."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We will turn into this perfect utopian civilization without wars, without strife of any kind. There was something to be said about that. You had significant improvements. In fact, you had even more dramatic improvements once the Industrial Revolution started. But not everyone in the late 1700s was as optimistic. One of the more famous not-so-optimistic people was Thomas Malthus. Thomas Malthus, right over here."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, you had even more dramatic improvements once the Industrial Revolution started. But not everyone in the late 1700s was as optimistic. One of the more famous not-so-optimistic people was Thomas Malthus. Thomas Malthus, right over here. And I will just quote him directly. This is from his essay on the principle of population. The power of population is so superior to the power of the earth to produce subsistence for man that premature death must in some shape or other visit the human race."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Thomas Malthus, right over here. And I will just quote him directly. This is from his essay on the principle of population. The power of population is so superior to the power of the earth to produce subsistence for man that premature death must in some shape or other visit the human race. This is uplifting. The vices of mankind are active and able ministers of deep population. They are the precursors in the great army of destruction and often finish the dreadful work themselves."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The power of population is so superior to the power of the earth to produce subsistence for man that premature death must in some shape or other visit the human race. This is uplifting. The vices of mankind are active and able ministers of deep population. They are the precursors in the great army of destruction and often finish the dreadful work themselves. But should they fail in this war of extermination, sickly seasons, epidemics, pestilence and plague advance in terrific array and sweep off their thousands and tens of thousands, should success still be incomplete, gigantic inevitable famine stalks in the rear and with one mighty blow levels the population with the food of the world. So not so uplifting of a little quote right over here. But this was his general sense."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They are the precursors in the great army of destruction and often finish the dreadful work themselves. But should they fail in this war of extermination, sickly seasons, epidemics, pestilence and plague advance in terrific array and sweep off their thousands and tens of thousands, should success still be incomplete, gigantic inevitable famine stalks in the rear and with one mighty blow levels the population with the food of the world. So not so uplifting of a little quote right over here. But this was his general sense. He lived in a time where people were being very optimistic that the march of progress would go on forever until we got to some utopian civilization. But from Thomas Malthus' point of view, he felt that if people could reproduce and increase the population, they will, that there's no way of stopping them. So from his point of view, the way he saw it, so let me on that axis, let's say that that is the population, and this axis right over here, let's say that that is time."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this was his general sense. He lived in a time where people were being very optimistic that the march of progress would go on forever until we got to some utopian civilization. But from Thomas Malthus' point of view, he felt that if people could reproduce and increase the population, they will, that there's no way of stopping them. So from his point of view, the way he saw it, so let me on that axis, let's say that that is the population, and this axis right over here, let's say that that is time. So by his thinking and everything that he'd seen in reality up to that point would back this up, that if people had enough food and time, they would reproduce and they would reproduce in numbers that would grow the population. So in his mind, the population would just keep on increasing until it can't support itself anymore, until the actual productivity of the land can't produce enough calories to feed all of those people. So in his mind, there would be some natural upper bound based on the actual amount of food that the earth could support."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So from his point of view, the way he saw it, so let me on that axis, let's say that that is the population, and this axis right over here, let's say that that is time. So by his thinking and everything that he'd seen in reality up to that point would back this up, that if people had enough food and time, they would reproduce and they would reproduce in numbers that would grow the population. So in his mind, the population would just keep on increasing until it can't support itself anymore, until the actual productivity of the land can't produce enough calories to feed all of those people. So in his mind, there would be some natural upper bound based on the actual amount of food that the earth could support. Let me do that in a different color. So in his mind, there was some upper bound. And once you get to that upper bound, then all of a sudden the vices of mankind will show up, and if those don't start killing people, then all of these other things will."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in his mind, there would be some natural upper bound based on the actual amount of food that the earth could support. Let me do that in a different color. So in his mind, there was some upper bound. And once you get to that upper bound, then all of a sudden the vices of mankind will show up, and if those don't start killing people, then all of these other things will. Epidemics, pestilence, plague, and then famine. People are actually starving to death. So in his mind, once you got to this level, maybe you had a couple of good crops, but then all of a sudden you have a bad crop."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And once you get to that upper bound, then all of a sudden the vices of mankind will show up, and if those don't start killing people, then all of these other things will. Epidemics, pestilence, plague, and then famine. People are actually starving to death. So in his mind, once you got to this level, maybe you had a couple of good crops, but then all of a sudden you have a bad crop. Or because you have a bad crop, people start fighting over resources and wars happen. Or maybe the population is so dense that a plague develops, and then you have a massive wave of depopulation. And so you would just oscillate around this limit."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in his mind, once you got to this level, maybe you had a couple of good crops, but then all of a sudden you have a bad crop. Or because you have a bad crop, people start fighting over resources and wars happen. Or maybe the population is so dense that a plague develops, and then you have a massive wave of depopulation. And so you would just oscillate around this limit. And this limit, some people refer to as a Malthusian limit, but it's just really the limit at which the population can sustain itself. And from Thomas Malthus' point of view, he did recognize that there were technological improvements, especially in things like agriculture, and that this line was moving up. He had seen it in his own lifetime that this line had moved up."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you would just oscillate around this limit. And this limit, some people refer to as a Malthusian limit, but it's just really the limit at which the population can sustain itself. And from Thomas Malthus' point of view, he did recognize that there were technological improvements, especially in things like agriculture, and that this line was moving up. He had seen it in his own lifetime that this line had moved up. But from his point of view, however far you move this line up, the population will always compensate for it and catch up to it and eventually get to this limit, and then the same kind of not-so-positive things that he talks about would actually happen. And some people now say, Oh, Thomas Malthus, he was so pessimistic. He was obviously wrong."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "He had seen it in his own lifetime that this line had moved up. But from his point of view, however far you move this line up, the population will always compensate for it and catch up to it and eventually get to this limit, and then the same kind of not-so-positive things that he talks about would actually happen. And some people now say, Oh, Thomas Malthus, he was so pessimistic. He was obviously wrong. Look at what's happened. We have so much food on this planet right now. We've gone through multiple agricultural revolutions."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "He was obviously wrong. Look at what's happened. We have so much food on this planet right now. We've gone through multiple agricultural revolutions. And they are right. In the last 200 years since Malthus, so since the early 1800s, we really have been able to outstrip population. So this line up here has been moving up much faster than even population."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We've gone through multiple agricultural revolutions. And they are right. In the last 200 years since Malthus, so since the early 1800s, we really have been able to outstrip population. So this line up here has been moving up much faster than even population. So right now we actually do have more calories per person on the planet than we've had at any time in history. But it's not saying that Thomas Malthus was wrong. It's just saying that maybe he was just a little bit pessimistic in when that limit will be reached."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this line up here has been moving up much faster than even population. So right now we actually do have more calories per person on the planet than we've had at any time in history. But it's not saying that Thomas Malthus was wrong. It's just saying that maybe he was just a little bit pessimistic in when that limit will be reached. Now the other dimension where you might say that he was maybe wrong was in this principle that a population will increase if it can increase. If there is food and if there is time, people will reproduce. And a good counterpoint to that is what we've now observed in modern developed nations."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's just saying that maybe he was just a little bit pessimistic in when that limit will be reached. Now the other dimension where you might say that he was maybe wrong was in this principle that a population will increase if it can increase. If there is food and if there is time, people will reproduce. And a good counterpoint to that is what we've now observed in modern developed nations. And so this right over here shows the population growth. I got this from the World Bank. But the population growth of some modern developed nations."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And a good counterpoint to that is what we've now observed in modern developed nations. And so this right over here shows the population growth. I got this from the World Bank. But the population growth of some modern developed nations. And you can see the United States is pretty low, but it's still positive. It's still over half a percent. But even that adds up when you compound it."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the population growth of some modern developed nations. And you can see the United States is pretty low, but it's still positive. It's still over half a percent. But even that adds up when you compound it. But if you look over here, Japan and Germany have less immigration than the United States, especially Japan, they are actually negative. So just this population left to its own devices, especially if you account for people not going across borders, just the population itself growing, they actually have negative growth. So there's some reason to believe that this is evidence that Thomas Malthus was wrong, or not completely right."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But even that adds up when you compound it. But if you look over here, Japan and Germany have less immigration than the United States, especially Japan, they are actually negative. So just this population left to its own devices, especially if you account for people not going across borders, just the population itself growing, they actually have negative growth. So there's some reason to believe that this is evidence that Thomas Malthus was wrong, or not completely right. He didn't put into account that maybe once a society becomes rich enough and educated enough that they might not just populate the world or have as many kids as they want. They might try to do other things with their time, whatever that might be. So I just wanted to expose you to this idea."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So there's some reason to believe that this is evidence that Thomas Malthus was wrong, or not completely right. He didn't put into account that maybe once a society becomes rich enough and educated enough that they might not just populate the world or have as many kids as they want. They might try to do other things with their time, whatever that might be. So I just wanted to expose you to this idea. Time will tell if Thomas Malthus, if we can always keep this line of food productivity growing faster than the population, and time will tell whether our populations can become, I guess you could say, developed enough so that they don't inexorably, I can never say that word, they don't always just keep growing. Maybe they do become a Japan or a Germany situation in the world population, especially if we have a high rate of literacy eventually does level off, so it never even has a chance of hitting up against that Malthusian limit. But I thought I would introduce you to the idea, and now you can go to parties and you can talk about things like Malthusian limits."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I just wanted to expose you to this idea. Time will tell if Thomas Malthus, if we can always keep this line of food productivity growing faster than the population, and time will tell whether our populations can become, I guess you could say, developed enough so that they don't inexorably, I can never say that word, they don't always just keep growing. Maybe they do become a Japan or a Germany situation in the world population, especially if we have a high rate of literacy eventually does level off, so it never even has a chance of hitting up against that Malthusian limit. But I thought I would introduce you to the idea, and now you can go to parties and you can talk about things like Malthusian limits. And if you want to know what country is maybe closest to the Malthusian limit right now, and we've talked about this before, but a good case example is something like Bangladesh. They are right now the most population-dense country in the world. They have 900 people per square kilometer, and just to give you a sense of perspective, that's 30 times more dense than the United States is."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I thought I would introduce you to the idea, and now you can go to parties and you can talk about things like Malthusian limits. And if you want to know what country is maybe closest to the Malthusian limit right now, and we've talked about this before, but a good case example is something like Bangladesh. They are right now the most population-dense country in the world. They have 900 people per square kilometer, and just to give you a sense of perspective, that's 30 times more dense than the United States is. So if you took every person in the United States and turned them into 30 people in the United States, that would give you a sense of how dense Bangladesh is. And it's probably due to a certain degree that it's very fertile land. It's the river delta of the Ganges that essentially makes up the entire country."}, {"video_title": "Thomas Malthus and population growth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They have 900 people per square kilometer, and just to give you a sense of perspective, that's 30 times more dense than the United States is. So if you took every person in the United States and turned them into 30 people in the United States, that would give you a sense of how dense Bangladesh is. And it's probably due to a certain degree that it's very fertile land. It's the river delta of the Ganges that essentially makes up the entire country. But they have in the past had famines. They've gotten a little bit beyond that, but still you do have major problems with flooding and resources. So hopefully they'll be able to stay ahead of the curve."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it's going to put a chlorine onto the ring. The question is, where will the chlorine go? To figure that out, you have to look at your substituents and see what kind of directors they are. So we'll start right here with this methyl group, which we know is an ortho, para director. So it's going to direct the chlorine to the ortho and para positions. And so this would be the ortho position. And this is also an ortho position."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start right here with this methyl group, which we know is an ortho, para director. So it's going to direct the chlorine to the ortho and para positions. And so this would be the ortho position. And this is also an ortho position. But by symmetry, these are the same thing, if you think about the molecule here. The para position is occupied by this nitro group right here. And so we can't add anything on para in this example."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this is also an ortho position. But by symmetry, these are the same thing, if you think about the molecule here. The para position is occupied by this nitro group right here. And so we can't add anything on para in this example. If we look at the nitro group, so right here, I think about what kind of a director that is. I know that it's a meta director, from the reasons we talked about in previous videos. So the position that's meta to this nitro group would, of course, end up being the exact same carbons that we marked before, which again, by symmetry, are the same ones here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we can't add anything on para in this example. If we look at the nitro group, so right here, I think about what kind of a director that is. I know that it's a meta director, from the reasons we talked about in previous videos. So the position that's meta to this nitro group would, of course, end up being the exact same carbons that we marked before, which again, by symmetry, are the same ones here. So this turns out to be a case where both of my substituents direct to the same spot. And so I can go ahead and draw my product. So I have my benzene ring."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the position that's meta to this nitro group would, of course, end up being the exact same carbons that we marked before, which again, by symmetry, are the same ones here. So this turns out to be a case where both of my substituents direct to the same spot. And so I can go ahead and draw my product. So I have my benzene ring. And I have my substituents on there, my methyl group, and also my nitro group. And I could have picked either one of these two carbons. So this one or this one."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I have my benzene ring. And I have my substituents on there, my methyl group, and also my nitro group. And I could have picked either one of these two carbons. So this one or this one. It doesn't matter which one you choose. Again, because of symmetry, you will get an identical product here. So we're going to put a chlorine on in one of those positions, ortho to the methyl group, and meta to the nitro group."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this one or this one. It doesn't matter which one you choose. Again, because of symmetry, you will get an identical product here. So we're going to put a chlorine on in one of those positions, ortho to the methyl group, and meta to the nitro group. And so that's the only product that we will get. Let's do another one here. Let's look at this reaction."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to put a chlorine on in one of those positions, ortho to the methyl group, and meta to the nitro group. And so that's the only product that we will get. Let's do another one here. Let's look at this reaction. So I know that this is a sulfonation reaction. So I'm going to put an SO3H group onto my benzene ring. And the question is, where will that group go?"}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at this reaction. So I know that this is a sulfonation reaction. So I'm going to put an SO3H group onto my benzene ring. And the question is, where will that group go? And so once again, I analyze my substituents. And first, I look at the OH group, which I know is an ortho para director. So this is an ortho para director here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the question is, where will that group go? And so once again, I analyze my substituents. And first, I look at the OH group, which I know is an ortho para director. So this is an ortho para director here. And I look at where that would be on my ring. Well, once again, this would be the ortho position. And again, by symmetry, the same ortho position over there."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is an ortho para director here. And I look at where that would be on my ring. Well, once again, this would be the ortho position. And again, by symmetry, the same ortho position over there. The para position is once again taken up by this methyl group here. When I think about the methyl group, so let me just use a different color for that one. The methyl group is also an ortho para director."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And again, by symmetry, the same ortho position over there. The para position is once again taken up by this methyl group here. When I think about the methyl group, so let me just use a different color for that one. The methyl group is also an ortho para director. So what's ortho and para to the methyl group? Well, this would be ortho. These two spots would be ortho."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The methyl group is also an ortho para director. So what's ortho and para to the methyl group? Well, this would be ortho. These two spots would be ortho. Again, identical because of symmetry. The para spot is taken up by this OH here. And so now we have a case where we have two ortho para directors directing two different spots."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "These two spots would be ortho. Again, identical because of symmetry. The para spot is taken up by this OH here. And so now we have a case where we have two ortho para directors directing two different spots. And so the way to figure out which one wins is to think about the activating strength of these two ortho para directors. So the strongest activator is going to be the directing group. And the strongest activator here is, of course, the OH group here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so now we have a case where we have two ortho para directors directing two different spots. And so the way to figure out which one wins is to think about the activating strength of these two ortho para directors. So the strongest activator is going to be the directing group. And the strongest activator here is, of course, the OH group here. So this is a strong activator. And the methyl group we saw is a weak activator from some of the earlier videos here. So the spot that's going to get the SO3H is this carbon, which again, by symmetry, would be this one right here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the strongest activator here is, of course, the OH group here. So this is a strong activator. And the methyl group we saw is a weak activator from some of the earlier videos here. So the spot that's going to get the SO3H is this carbon, which again, by symmetry, would be this one right here. So we can go ahead and draw the product of this sulfonation reaction. So let me go ahead and sketch in my benzene ring. Let me do another one here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the spot that's going to get the SO3H is this carbon, which again, by symmetry, would be this one right here. So we can go ahead and draw the product of this sulfonation reaction. So let me go ahead and sketch in my benzene ring. Let me do another one here. That one wasn't very good. And we have our ring. We have our OH."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me do another one here. That one wasn't very good. And we have our ring. We have our OH. We have our methyl group. And since the OH is the strongest activator, we're going to go ahead and put SO3H ortho to the OH. And that is our final product."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our OH. We have our methyl group. And since the OH is the strongest activator, we're going to go ahead and put SO3H ortho to the OH. And that is our final product. Let's do one more example. So let's see what happens with this reaction here. So once again, I can identify this as being a halogenation reaction, where I add chlorine onto my ring."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that is our final product. Let's do one more example. So let's see what happens with this reaction here. So once again, I can identify this as being a halogenation reaction, where I add chlorine onto my ring. And let's go ahead and analyze the substituents once again. So I have a methyl group, again, which I know is an ortho para director. So I'll go ahead and mark the spots that are ortho and para to that methyl group."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, I can identify this as being a halogenation reaction, where I add chlorine onto my ring. And let's go ahead and analyze the substituents once again. So I have a methyl group, again, which I know is an ortho para director. So I'll go ahead and mark the spots that are ortho and para to that methyl group. So this would be the ortho position. This would be the ortho position. And then this time, the para position is free, so we could possibly put the chlorine there."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and mark the spots that are ortho and para to that methyl group. So this would be the ortho position. This would be the ortho position. And then this time, the para position is free, so we could possibly put the chlorine there. If I look at what else is on my ring, so I have another substituent here, which is a chlorine. We know that halogens are also ortho para directors, because of the lone pair of electrons that are on the chlorine there. So an ortho para director for the chlorine."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then this time, the para position is free, so we could possibly put the chlorine there. If I look at what else is on my ring, so I have another substituent here, which is a chlorine. We know that halogens are also ortho para directors, because of the lone pair of electrons that are on the chlorine there. So an ortho para director for the chlorine. So what's ortho to this chlorine here? Well, this spot's ortho. So is this spot."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So an ortho para director for the chlorine. So what's ortho to this chlorine here? Well, this spot's ortho. So is this spot. And so is this spot. And so we have a couple of different possible products here. And let's go ahead and start drawing them here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So is this spot. And so is this spot. And so we have a couple of different possible products here. And let's go ahead and start drawing them here. So if I think about the product, if we add a chlorine onto this position, let me go ahead and draw that. So we would have our benzene ring here. We would have our methyl group."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and start drawing them here. So if I think about the product, if we add a chlorine onto this position, let me go ahead and draw that. So we would have our benzene ring here. We would have our methyl group. We would have our chlorine. And it's possible that a chlorine would add on to that position. Let's next look at this position."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We would have our methyl group. We would have our chlorine. And it's possible that a chlorine would add on to that position. Let's next look at this position. So this could be another possible product here. So let me go ahead and draw that one as well. So we have our benzene ring."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's next look at this position. So this could be another possible product here. So let me go ahead and draw that one as well. So we have our benzene ring. And we have our methyl group. And we had the chlorine that was originally on our ring. And then we're adding on a chlorine to this position."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our benzene ring. And we have our methyl group. And we had the chlorine that was originally on our ring. And then we're adding on a chlorine to this position. And finally, let's go ahead and think about what would happen if a chlorine tried to add to this final position right here on the right. Well, it turns out that there's just too much steric hindrance for a chlorine to add on to this position. So if you think about this methyl group being bulky and this chlorine being bulky, that makes it difficult for another chlorine to add to here."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then we're adding on a chlorine to this position. And finally, let's go ahead and think about what would happen if a chlorine tried to add to this final position right here on the right. Well, it turns out that there's just too much steric hindrance for a chlorine to add on to this position. So if you think about this methyl group being bulky and this chlorine being bulky, that makes it difficult for another chlorine to add to here. So this is not observed in large amounts. So this product, we're going to say, does not form here. And so we're left with two major products for this reaction."}, {"video_title": "Multiple substituents Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if you think about this methyl group being bulky and this chlorine being bulky, that makes it difficult for another chlorine to add to here. So this is not observed in large amounts. So this product, we're going to say, does not form here. And so we're left with two major products for this reaction. And that's how to think about directing effects in synthesis when you're talking about multiple substituents. First, think about where the substituents will direct. And you have to think about activating strength."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So you might think it might attack right here and you lose your OH and you would form an amide or an amide as your product. However, this is not what happens at room temperature. The ammonia's gonna function as a base instead and it's gonna take the acidic proton on your carboxylic acid, leaving these electrons behind on your oxygen. So you would actually form your carboxylate anion here. So three lone pairs of electrons on your oxygen, giving it a negative one formal charge. If you add a proton to ammonia, you would form ammonium, right? So NH4 plus and we have our ammonium salt here."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So you would actually form your carboxylate anion here. So three lone pairs of electrons on your oxygen, giving it a negative one formal charge. If you add a proton to ammonia, you would form ammonium, right? So NH4 plus and we have our ammonium salt here. If you heat up this salt, you actually can sometimes form your amide. However, this is definitely not the best way to make an amide. A much better way would be to use something called DCC."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So NH4 plus and we have our ammonium salt here. If you heat up this salt, you actually can sometimes form your amide. However, this is definitely not the best way to make an amide. A much better way would be to use something called DCC. And so let's get some more room down here so we can see what DCC looks like. That's an acronym for Dicyclohexyl Carbodiimide or diamides, right? So we have the D, the C, and the C here."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "A much better way would be to use something called DCC. And so let's get some more room down here so we can see what DCC looks like. That's an acronym for Dicyclohexyl Carbodiimide or diamides, right? So we have the D, the C, and the C here. And so if this R double prime group, right, is a cyclohexyl group, then we would have DCC. And so if you start with your carboxylic acid and add an amine, the use of DCC allows your amine to function as a nucleophile and eventually form your amide as your product. So let's look at the mechanism."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So we have the D, the C, and the C here. And so if this R double prime group, right, is a cyclohexyl group, then we would have DCC. And so if you start with your carboxylic acid and add an amine, the use of DCC allows your amine to function as a nucleophile and eventually form your amide as your product. So let's look at the mechanism. The first step is for DCC to function as a base. And so this lone pair of electrons on the nitrogen take this proton, leave these electrons behind on your oxygen. So let's get some more room here."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at the mechanism. The first step is for DCC to function as a base. And so this lone pair of electrons on the nitrogen take this proton, leave these electrons behind on your oxygen. So let's get some more room here. So we're going to go ahead and show taking a proton away from your carboxylic acid gives you your carboxylate anion. So let's go ahead and draw in our carboxylate anion. So three lone pairs of electrons on our oxygen, so negative one formal charge."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's get some more room here. So we're going to go ahead and show taking a proton away from your carboxylic acid gives you your carboxylate anion. So let's go ahead and draw in our carboxylate anion. So three lone pairs of electrons on our oxygen, so negative one formal charge. This nitrogen here is going to pick up that proton. And so that nitrogen now has a plus one formal charge. So let's show some electrons."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So three lone pairs of electrons on our oxygen, so negative one formal charge. This nitrogen here is going to pick up that proton. And so that nitrogen now has a plus one formal charge. So let's show some electrons. So the electrons in magenta here are going to take this proton, so forming this bond after it took this proton here. And we could think about these electrons moving off onto our oxygen to form our carboxylate anion. Let's go ahead and draw in the rest of this."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's show some electrons. So the electrons in magenta here are going to take this proton, so forming this bond after it took this proton here. And we could think about these electrons moving off onto our oxygen to form our carboxylate anion. Let's go ahead and draw in the rest of this. So we have a nitrogen double bonded to a carbon, which is double bonded to another nitrogen with a lone pair of electrons. And we have our R double prime group. So because we just protonated this nitrogen, it really wants electrons."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw in the rest of this. So we have a nitrogen double bonded to a carbon, which is double bonded to another nitrogen with a lone pair of electrons. And we have our R double prime group. So because we just protonated this nitrogen, it really wants electrons. It can withdraw some electron density away from this carbon. And so this carbon is going to lose some electron density and become a little bit positive. And so this carbon is electrophilic now."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So because we just protonated this nitrogen, it really wants electrons. It can withdraw some electron density away from this carbon. And so this carbon is going to lose some electron density and become a little bit positive. And so this carbon is electrophilic now. And our carboxylate anion is going to function as a nucleophile in our next step. So the nucleophile attacks our electrophile, pushing these electrons off onto our nitrogen. So let's go ahead and show that."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And so this carbon is electrophilic now. And our carboxylate anion is going to function as a nucleophile in our next step. So the nucleophile attacks our electrophile, pushing these electrons off onto our nitrogen. So let's go ahead and show that. So now we would have our carbon double bonded to this oxygen. And then this oxygen down here has now formed a bond to this carbon. So those electrons in blue have formed this bond now."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. So now we would have our carbon double bonded to this oxygen. And then this oxygen down here has now formed a bond to this carbon. So those electrons in blue have formed this bond now. And this carbon is bonded to this nitrogen, bonded to this and our R double prime group. And if we showed, let's make these electrons in here green. These electrons move off onto our nitrogen like that."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So those electrons in blue have formed this bond now. And this carbon is bonded to this nitrogen, bonded to this and our R double prime group. And if we showed, let's make these electrons in here green. These electrons move off onto our nitrogen like that. And so our carbon is also double bonded to this nitrogen with R double prime group. So in the next step, our amine is going to function as a nucleophile. So let's go ahead and draw in our amine."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "These electrons move off onto our nitrogen like that. And so our carbon is also double bonded to this nitrogen with R double prime group. So in the next step, our amine is going to function as a nucleophile. So let's go ahead and draw in our amine. So we have our nitrogen with two hydrogens and an R prime group. Lone pair of electrons on our nitrogen makes this amine able to function as a nucleophile. So I'm going to go ahead and put those electrons in magenta."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in our amine. So we have our nitrogen with two hydrogens and an R prime group. Lone pair of electrons on our nitrogen makes this amine able to function as a nucleophile. So I'm going to go ahead and put those electrons in magenta. So right here, this oxygen is partially negative. It's going to withdraw some electron density from this carbon, so partially positive. And so we have a nucleophile that's going to attack our electrophile."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and put those electrons in magenta. So right here, this oxygen is partially negative. It's going to withdraw some electron density from this carbon, so partially positive. And so we have a nucleophile that's going to attack our electrophile. So our amine is going to attack this carbon, push these electrons off onto our oxygen. So let's go ahead and show the result of this nucleophilic attack. So let's get some room down here."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And so we have a nucleophile that's going to attack our electrophile. So our amine is going to attack this carbon, push these electrons off onto our oxygen. So let's go ahead and show the result of this nucleophilic attack. So let's get some room down here. We would have our carbon bonded to this oxygen. So let's go ahead and draw in all of those electrons. So negative 1 formal charge on this oxygen."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's get some room down here. We would have our carbon bonded to this oxygen. So let's go ahead and draw in all of those electrons. So negative 1 formal charge on this oxygen. So if these electrons in here in green move off onto our oxygen, we get a negative 1 formal charge. This carbon is bonded to an R group. It's also bonded to this nitrogen."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So negative 1 formal charge on this oxygen. So if these electrons in here in green move off onto our oxygen, we get a negative 1 formal charge. This carbon is bonded to an R group. It's also bonded to this nitrogen. This nitrogen now has a plus 1 formal charge. So a plus 1 formal charge on our nitrogen after the electrons in magenta move in here to form this bond. So we still have our carbon bonded to this oxygen."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "It's also bonded to this nitrogen. This nitrogen now has a plus 1 formal charge. So a plus 1 formal charge on our nitrogen after the electrons in magenta move in here to form this bond. So we still have our carbon bonded to this oxygen. And draw in our lone pairs of electrons. And then we have this carbon. We have a nitrogen."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So we still have our carbon bonded to this oxygen. And draw in our lone pairs of electrons. And then we have this carbon. We have a nitrogen. We have our R double prime group. We have a hydrogen. Lone pair of electrons double bonded to this nitrogen over here with our R double prime group."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "We have a nitrogen. We have our R double prime group. We have a hydrogen. Lone pair of electrons double bonded to this nitrogen over here with our R double prime group. So a lot of stuff going on here. So the use of DCC gives you a good leaving group. So if we think about all this stuff over here, this is an excellent leaving group."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "Lone pair of electrons double bonded to this nitrogen over here with our R double prime group. So a lot of stuff going on here. So the use of DCC gives you a good leaving group. So if we think about all this stuff over here, this is an excellent leaving group. So if we reform our carbonyl, let's go ahead and show that. So if these electrons in here move in to reform our carbonyl, these electrons could come off onto our oxygen. And we could even show them moving over to here to save some time."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about all this stuff over here, this is an excellent leaving group. So if we reform our carbonyl, let's go ahead and show that. So if these electrons in here move in to reform our carbonyl, these electrons could come off onto our oxygen. And we could even show them moving over to here to save some time. And if there's a proton out here, these electrons could pick up that proton. And we have an excellent leaving group. So this actually forms dicyclohexylurea over here on the right."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And we could even show them moving over to here to save some time. And if there's a proton out here, these electrons could pick up that proton. And we have an excellent leaving group. So this actually forms dicyclohexylurea over here on the right. So if I circle all of this stuff, we're going to get dicyclohexylurea. And let's go ahead and show what would happen. So if we reform our carbonyl, let's use those electrons in here in red."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So this actually forms dicyclohexylurea over here on the right. So if I circle all of this stuff, we're going to get dicyclohexylurea. And let's go ahead and show what would happen. So if we reform our carbonyl, let's use those electrons in here in red. So if those electrons in red move in, now we would have our R group. Our carbon is now double bonded to this oxygen with only two lone pairs of electrons. So the electrons in red move in to reform our carbonyl."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So if we reform our carbonyl, let's use those electrons in here in red. So if those electrons in red move in, now we would have our R group. Our carbon is now double bonded to this oxygen with only two lone pairs of electrons. So the electrons in red move in to reform our carbonyl. And then this carbon is still bonded to this nitrogen. So let's go ahead and draw this nitrogen in here. And if we think about a base taking this proton, let me go ahead and change colors."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in red move in to reform our carbonyl. And then this carbon is still bonded to this nitrogen. So let's go ahead and draw this nitrogen in here. And if we think about a base taking this proton, let me go ahead and change colors. A base taking this proton, leaving these electrons behind. So these electrons are going to be left behind on this nitrogen here. So I'll draw them in here in blue."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And if we think about a base taking this proton, let me go ahead and change colors. A base taking this proton, leaving these electrons behind. So these electrons are going to be left behind on this nitrogen here. So I'll draw them in here in blue. And so now we have only one hydrogen on this nitrogen. And then we have our prime group like that. So plus di-cyclohexylurea as our other product."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw them in here in blue. And so now we have only one hydrogen on this nitrogen. And then we have our prime group like that. So plus di-cyclohexylurea as our other product. So we formed our amide. So once again, DCC allows the amine to function as a nucleophile. So let's see some uses for DCC."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So plus di-cyclohexylurea as our other product. So we formed our amide. So once again, DCC allows the amine to function as a nucleophile. So let's see some uses for DCC. One of the most famous uses for DCC is to react amino acids together to form peptides. So if we have an amino acid over here on the left, so with an R group, we'll call it R1. And we have an amino acid over here on the right."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's see some uses for DCC. One of the most famous uses for DCC is to react amino acids together to form peptides. So if we have an amino acid over here on the left, so with an R group, we'll call it R1. And we have an amino acid over here on the right. With a different R group, we have R2. Not going to worry about stereochemistry here. We could join these amino acids using DCC."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And we have an amino acid over here on the right. With a different R group, we have R2. Not going to worry about stereochemistry here. We could join these amino acids using DCC. So if we look for our carboxylic acid over here on the left and our amine over here on the right, we know that DCC could form an amide. Now for something like this, you have to be a little bit more careful because you have an amine over here on this side and a carboxylic acid over here on this side. And so you would have to add a protecting group."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "We could join these amino acids using DCC. So if we look for our carboxylic acid over here on the left and our amine over here on the right, we know that DCC could form an amide. Now for something like this, you have to be a little bit more careful because you have an amine over here on this side and a carboxylic acid over here on this side. And so you would have to add a protecting group. So we'll just go ahead and think about a protecting group being over here on this side. And we could change this to a protecting group over here, something like OR prime, so an ester instead of our carboxylic acid. And when we add DCC, we can think about DCC as being a dehydrating agent."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And so you would have to add a protecting group. So we'll just go ahead and think about a protecting group being over here on this side. And we could change this to a protecting group over here, something like OR prime, so an ester instead of our carboxylic acid. And when we add DCC, we can think about DCC as being a dehydrating agent. So we can think about losing water. So let's go ahead and show that. So we would lose the OH from our carboxylic acid and the H from our amine."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And when we add DCC, we can think about DCC as being a dehydrating agent. So we can think about losing water. So let's go ahead and show that. So we would lose the OH from our carboxylic acid and the H from our amine. So we can see that's H2O. So if we think about minus H2O, we can stick those two amino acids together. So let's go ahead and do that."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So we would lose the OH from our carboxylic acid and the H from our amine. So we can see that's H2O. So if we think about minus H2O, we can stick those two amino acids together. So let's go ahead and do that. So we would have our carbonyl bonded to our nitrogen, bonded to this hydrogen here. And then we would have our R group. And then we have our carboxylic acid."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and do that. So we would have our carbonyl bonded to our nitrogen, bonded to this hydrogen here. And then we would have our R group. And then we have our carboxylic acid. And I'll go ahead and put an OR prime here. So our carboxylic acid was protected over here and the right to form an ester instead. Over here, we would have our R1."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our carboxylic acid. And I'll go ahead and put an OR prime here. So our carboxylic acid was protected over here and the right to form an ester instead. Over here, we would have our R1. We would have our nitrogen with the hydrogen. And depending on what protecting group you're using, I'm just going to go ahead and put that one in here. So we would form a dipeptide."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "Over here, we would have our R1. We would have our nitrogen with the hydrogen. And depending on what protecting group you're using, I'm just going to go ahead and put that one in here. So we would form a dipeptide. So right here, you could see our amide or our amide. And then right here is our peptide bond that formed. So this was developed in the 1950s at MIT."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So we would form a dipeptide. So right here, you could see our amide or our amide. And then right here is our peptide bond that formed. So this was developed in the 1950s at MIT. And it was published by Dr. Sheehan's group somewhere around 1955. And he actually calls this up here what I've done. We take the OH and the H and just kind of think about removing them to lose water as lasso chemistry."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So this was developed in the 1950s at MIT. And it was published by Dr. Sheehan's group somewhere around 1955. And he actually calls this up here what I've done. We take the OH and the H and just kind of think about removing them to lose water as lasso chemistry. So it's certainly not the best way to think about exactly what's happening. But it's a good way of thinking about sticking these two individual amino acids together to form your dipeptide. And so Dr. Sheehan's group was using DCC in the 50s."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "We take the OH and the H and just kind of think about removing them to lose water as lasso chemistry. So it's certainly not the best way to think about exactly what's happening. But it's a good way of thinking about sticking these two individual amino acids together to form your dipeptide. And so Dr. Sheehan's group was using DCC in the 50s. And eventually, of course, it became the standard for forming peptides. And of course, there are different versions of this now. But this was a breakthrough in the 1950s."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And so Dr. Sheehan's group was using DCC in the 50s. And eventually, of course, it became the standard for forming peptides. And of course, there are different versions of this now. But this was a breakthrough in the 1950s. And Dr. Sheehan was also involved in the total synthesis of penicillin in the 50s. And when he used DCC as a coupling agent to form his peptides in his lab, he thought that he could also use it to form penicillin. So let's take a look at the structure of penicillin."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "But this was a breakthrough in the 1950s. And Dr. Sheehan was also involved in the total synthesis of penicillin in the 50s. And when he used DCC as a coupling agent to form his peptides in his lab, he thought that he could also use it to form penicillin. So let's take a look at the structure of penicillin. So over here on the right is penicillin salt. And this is actually from Dr. Sheehan's total synthesis of penicillin. So he was the first to do so in 1957."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a look at the structure of penicillin. So over here on the right is penicillin salt. And this is actually from Dr. Sheehan's total synthesis of penicillin. So he was the first to do so in 1957. So if we look over here, we can see an amine. And we can see a carboxylic acid. And so if we do our lasso chemistry, so if we take this OH and this H, we think about losing water."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So he was the first to do so in 1957. So if we look over here, we can see an amine. And we can see a carboxylic acid. And so if we do our lasso chemistry, so if we take this OH and this H, we think about losing water. And you can see here we have an amide in our ring. So an amide in a ring is called a lactam. So this is very famous."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And so if we do our lasso chemistry, so if we take this OH and this H, we think about losing water. And you can see here we have an amide in our ring. So an amide in a ring is called a lactam. So this is very famous. This is a beta lactam. So penicillin is in the beta lactam antibiotic. So it's called a beta lactam because the carbon next to your carbonyl here would be your alpha carbon."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So this is very famous. This is a beta lactam. So penicillin is in the beta lactam antibiotic. So it's called a beta lactam because the carbon next to your carbonyl here would be your alpha carbon. And then the carbon next to that would be your beta carbon. So here we have a four-membered ring to form our beta lactam. And so Dr. Sheehan used DCC in his synthesis to join this together."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So it's called a beta lactam because the carbon next to your carbonyl here would be your alpha carbon. And then the carbon next to that would be your beta carbon. So here we have a four-membered ring to form our beta lactam. And so Dr. Sheehan used DCC in his synthesis to join this together. And this beta lactam was extremely difficult for other chemists to make. And so during World War II, there were several labs that were working on this. So I think Dr. Sheehan wrote in his book, it's called The Enchanted Ring, because making this beta lactam ring was extremely difficult."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "And so Dr. Sheehan used DCC in his synthesis to join this together. And this beta lactam was extremely difficult for other chemists to make. And so during World War II, there were several labs that were working on this. So I think Dr. Sheehan wrote in his book, it's called The Enchanted Ring, because making this beta lactam ring was extremely difficult. This was the tricky part. How do you form a beta lactam ring? It was extremely difficult for most chemists to do so."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "So I think Dr. Sheehan wrote in his book, it's called The Enchanted Ring, because making this beta lactam ring was extremely difficult. This was the tricky part. How do you form a beta lactam ring? It was extremely difficult for most chemists to do so. But the use of DCC, which can be used in water. You can do this at room temperature. You can do this at relatively normal reaction conditions."}, {"video_title": "Preparation of amides using DCC Organic chemistry Khan Academy.mp3", "Sentence": "It was extremely difficult for most chemists to do so. But the use of DCC, which can be used in water. You can do this at room temperature. You can do this at relatively normal reaction conditions. And so that made him the first one to synthesize penicillin, the first total synthesis of penicillin. So according to him in his book, he said there was no competition at the time. And he thought that other chemists were simply tired of trying."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the general mechanism for a nucleophilic acyl substitution reaction. So here we have our carboxylic acid derivative, and we know that this carbon right here is our electrophilic portion of the molecule, right? It's partially positive, the oxygen's withdrawing some electron density, and we talked about the relative reactivities of carboxylic acid derivatives in the last video. And so that carbon is where our nucleophile is going to attack. So our nucleophile attacks here, these electrons kick off onto the oxygen, so let's go ahead and draw the result of that. We would have an R group over here, we would have a carbon, we would have on this left side here an oxygen with three lone pairs of electrons, giving it a negative one formal charge. So let's say that these electrons in here, in magenta, move off onto our oxygen, so a negative one formal charge."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so that carbon is where our nucleophile is going to attack. So our nucleophile attacks here, these electrons kick off onto the oxygen, so let's go ahead and draw the result of that. We would have an R group over here, we would have a carbon, we would have on this left side here an oxygen with three lone pairs of electrons, giving it a negative one formal charge. So let's say that these electrons in here, in magenta, move off onto our oxygen, so a negative one formal charge. Over here on the right we have our Y substituent, and now our nucleophile is bonded to our carbon, so let's show those electrons in blue here. So these electrons in blue, right, attacked our carbonyl carbon and formed this bond. To go from our tetrahedral intermediate here to our final product, we must reform our carbonyl, right?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that these electrons in here, in magenta, move off onto our oxygen, so a negative one formal charge. Over here on the right we have our Y substituent, and now our nucleophile is bonded to our carbon, so let's show those electrons in blue here. So these electrons in blue, right, attacked our carbonyl carbon and formed this bond. To go from our tetrahedral intermediate here to our final product, we must reform our carbonyl, right? So these electrons would have to move into here, and then that would push these electrons off onto our Y substituent, and we would form, if we started with a, right, this would be a Y negative now, so we have a negative one charge on our Y, right, when those two electrons move off on it, so let's show those electrons here in green. So these electrons in here move off onto the Y to give us a negative one formal charge, and this is our leaving group. This is our leaving group right here, and we can see the end result, right?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "To go from our tetrahedral intermediate here to our final product, we must reform our carbonyl, right? So these electrons would have to move into here, and then that would push these electrons off onto our Y substituent, and we would form, if we started with a, right, this would be a Y negative now, so we have a negative one charge on our Y, right, when those two electrons move off on it, so let's show those electrons here in green. So these electrons in here move off onto the Y to give us a negative one formal charge, and this is our leaving group. This is our leaving group right here, and we can see the end result, right? The end result is to substitute our nucleophile for our Y substituent, and this portion is called an acyl group. So we have nucleophilic acyl substitution, where our nucleophile substitutes for the Y group. And so there are several aspects to this mechanism that we need to talk about."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This is our leaving group right here, and we can see the end result, right? The end result is to substitute our nucleophile for our Y substituent, and this portion is called an acyl group. So we have nucleophilic acyl substitution, where our nucleophile substitutes for the Y group. And so there are several aspects to this mechanism that we need to talk about. We've already talked about the reactivity of carboxylic acid derivatives in the previous video, so what does the substituent do to the reactivity? How does it affect the partial positive charge on the carbonyl carbon? We saw that acyl chlorides are the most reactive, so we compared the inductive effects versus resonance effects for our carboxylic acid derivatives."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so there are several aspects to this mechanism that we need to talk about. We've already talked about the reactivity of carboxylic acid derivatives in the previous video, so what does the substituent do to the reactivity? How does it affect the partial positive charge on the carbonyl carbon? We saw that acyl chlorides are the most reactive, so we compared the inductive effects versus resonance effects for our carboxylic acid derivatives. So other things that could affect this mechanism, one thing could be steric hindrance. So thinking about this R group on our carboxylic acid derivative, if the R group was a methyl versus a tert-butyl, the tert-butyl group would have increased steric hindrance. It's much bigger, and so that could prevent the nucleophile from attacking."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We saw that acyl chlorides are the most reactive, so we compared the inductive effects versus resonance effects for our carboxylic acid derivatives. So other things that could affect this mechanism, one thing could be steric hindrance. So thinking about this R group on our carboxylic acid derivative, if the R group was a methyl versus a tert-butyl, the tert-butyl group would have increased steric hindrance. It's much bigger, and so that could prevent the nucleophile from attacking. So steric effects are something to think about. So the reactivity of your carboxylic acid derivative is something to think about. The strength of the nucleophile is another thing to think about."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's much bigger, and so that could prevent the nucleophile from attacking. So steric effects are something to think about. So the reactivity of your carboxylic acid derivative is something to think about. The strength of the nucleophile is another thing to think about. So you want a strong nucleophile to attack your carbonyl carbon. And then in this video, we're gonna focus mostly on the leaving group, so the stability of the leaving group. You want something that's stable with a negative one formal charge on it."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "The strength of the nucleophile is another thing to think about. So you want a strong nucleophile to attack your carbonyl carbon. And then in this video, we're gonna focus mostly on the leaving group, so the stability of the leaving group. You want something that's stable with a negative one formal charge on it. It's much more likely to leave if it's stable. And so let's look at an example where we try to identify the leaving group. So which one is the most stable and why?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "You want something that's stable with a negative one formal charge on it. It's much more likely to leave if it's stable. And so let's look at an example where we try to identify the leaving group. So which one is the most stable and why? So let's look at this reaction here. So we have an acyl chloride, so we have acetyl chloride here. And then we're gonna react that with sodium formate."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So which one is the most stable and why? So let's look at this reaction here. So we have an acyl chloride, so we have acetyl chloride here. And then we're gonna react that with sodium formate. So acetyl chloride, this is our reactive portion of the molecule, so oxygen's more electronegative. Chlorine is more electronegative, so that means that this carbon right here is partially positive. So that's our electrophile."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then we're gonna react that with sodium formate. So acetyl chloride, this is our reactive portion of the molecule, so oxygen's more electronegative. Chlorine is more electronegative, so that means that this carbon right here is partially positive. So that's our electrophile. Our nucleophile is sodium formate, right? Negative one formal charge on the oxygen, so this is going to be our nucleophile and attack our electrophile. And so these electrons are gonna come off onto the oxygen."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So that's our electrophile. Our nucleophile is sodium formate, right? Negative one formal charge on the oxygen, so this is going to be our nucleophile and attack our electrophile. And so these electrons are gonna come off onto the oxygen. So let's go ahead and draw our intermediate. We would have our oxygen here, now has three lone pairs of electrons, so negative one formal charge. So let's follow those electrons in here."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so these electrons are gonna come off onto the oxygen. So let's go ahead and draw our intermediate. We would have our oxygen here, now has three lone pairs of electrons, so negative one formal charge. So let's follow those electrons in here. So in magenta, these electrons move off onto our oxygen. And what else do we have bonded to this carbon? We have a CH3 over here on the left."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons in here. So in magenta, these electrons move off onto our oxygen. And what else do we have bonded to this carbon? We have a CH3 over here on the left. On the right, we have a chlorine. Let me go ahead and put in lone pairs of electrons on the chlorine. And then now, this oxygen is bonded to this carbon."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We have a CH3 over here on the left. On the right, we have a chlorine. Let me go ahead and put in lone pairs of electrons on the chlorine. And then now, this oxygen is bonded to this carbon. So let's go ahead and finish this off here and put in some lone pairs of electrons on our oxygen. So let's be consistent. Let's show these electrons in here in blue, right?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then now, this oxygen is bonded to this carbon. So let's go ahead and finish this off here and put in some lone pairs of electrons on our oxygen. So let's be consistent. Let's show these electrons in here in blue, right? Forming this bond between this carbon and this oxygen. So we have our tetrahedral intermediate. All right, we know the next step in the mechanism for nucleophilic ACL substitution, right, is to reform our carbonyl, all right?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's show these electrons in here in blue, right? Forming this bond between this carbon and this oxygen. So we have our tetrahedral intermediate. All right, we know the next step in the mechanism for nucleophilic ACL substitution, right, is to reform our carbonyl, all right? And when we reform our carbonyl, right, we can't have five bonds to carbon, so we have to lose something as a leaving group. And we have several possibilities here. So one possibility would be to have these electrons come off onto the chlorine to form the chloride anion as a leaving group."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "All right, we know the next step in the mechanism for nucleophilic ACL substitution, right, is to reform our carbonyl, all right? And when we reform our carbonyl, right, we can't have five bonds to carbon, so we have to lose something as a leaving group. And we have several possibilities here. So one possibility would be to have these electrons come off onto the chlorine to form the chloride anion as a leaving group. So let me go ahead and draw that over here as one of our possibilities. So the chloride anion could leave, right? So negative 1 formal charge on our chloride anion."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So one possibility would be to have these electrons come off onto the chlorine to form the chloride anion as a leaving group. So let me go ahead and draw that over here as one of our possibilities. So the chloride anion could leave, right? So negative 1 formal charge on our chloride anion. So that's one of our possibilities. Another possibility is when we reform our carbonyl, these electrons in blue come back off onto the oxygen, giving us our formate anion back, right? So that's another possibility."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So negative 1 formal charge on our chloride anion. So that's one of our possibilities. Another possibility is when we reform our carbonyl, these electrons in blue come back off onto the oxygen, giving us our formate anion back, right? So that's another possibility. Let me go ahead and sketch that in here. So another possibility would be the formate anion as a leaving group. All right, so let me go ahead and draw that in with our H here, all right?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So that's another possibility. Let me go ahead and sketch that in here. So another possibility would be the formate anion as a leaving group. All right, so let me go ahead and draw that in with our H here, all right? And then another possibility would be for these electrons to come off onto CH3, right? So we form a carb anion. So let me go ahead and draw on CH3, lone pair of electrons, negative 1 formal charge."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let me go ahead and draw that in with our H here, all right? And then another possibility would be for these electrons to come off onto CH3, right? So we form a carb anion. So let me go ahead and draw on CH3, lone pair of electrons, negative 1 formal charge. So those are our three possible leaving groups. And to think about which one of those is the best leaving group, a good way of doing it is thinking about the conjugate acid, all right? So the conjugate acid to the chloride anion would, of course, be HCl."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw on CH3, lone pair of electrons, negative 1 formal charge. So those are our three possible leaving groups. And to think about which one of those is the best leaving group, a good way of doing it is thinking about the conjugate acid, all right? So the conjugate acid to the chloride anion would, of course, be HCl. So let me go ahead and draw an HCl here, all right? And I could put in lone pairs of electrons. The conjugate acid to the formate anion would be formic acid."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the conjugate acid to the chloride anion would, of course, be HCl. So let me go ahead and draw an HCl here, all right? And I could put in lone pairs of electrons. The conjugate acid to the formate anion would be formic acid. So let me go ahead and draw in formic acid here. And the conjugate acid to our carb anion would, of course, be methane. So we have CH4."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "The conjugate acid to the formate anion would be formic acid. So let me go ahead and draw in formic acid here. And the conjugate acid to our carb anion would, of course, be methane. So we have CH4. When we think about the pKa values for these acids, the pKa of hydrochloric acid is approximately negative 7. The pKa of formic acid is approximately 5. And the pKa of methane, or alkanes in general, is somewhere around 50."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we have CH4. When we think about the pKa values for these acids, the pKa of hydrochloric acid is approximately negative 7. The pKa of formic acid is approximately 5. And the pKa of methane, or alkanes in general, is somewhere around 50. And so we have these different values for our pKa's. And remember what the pKa value tells you. The lower the pKa, the stronger the acid."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And the pKa of methane, or alkanes in general, is somewhere around 50. And so we have these different values for our pKa's. And remember what the pKa value tells you. The lower the pKa, the stronger the acid. So as you go this way, you have increasing acidity. So hydrochloric acid is the most acidic out of those three acids that we just talked about. And why is it the most acidic?"}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "The lower the pKa, the stronger the acid. So as you go this way, you have increasing acidity. So hydrochloric acid is the most acidic out of those three acids that we just talked about. And why is it the most acidic? It's the most acidic because it's the most willing to donate a proton. It's the most willing to donate a proton because its conjugate base, the chloride anion, is extremely stable on its own. So the stability of the chloride anion means that hydrochloric acid is most likely to donate a proton."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And why is it the most acidic? It's the most acidic because it's the most willing to donate a proton. It's the most willing to donate a proton because its conjugate base, the chloride anion, is extremely stable on its own. So the stability of the chloride anion means that hydrochloric acid is most likely to donate a proton. And so we have our answer. We know that the chloride anion is the most stable out of these three possible leaving groups. And we got that by thinking about the pKa of the conjugate acid."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the stability of the chloride anion means that hydrochloric acid is most likely to donate a proton. And so we have our answer. We know that the chloride anion is the most stable out of these three possible leaving groups. And we got that by thinking about the pKa of the conjugate acid. So going back to our mechanism, the chloride anion is going to leave. So these electrons in here in green are going to come off onto the chlorine. And we can go ahead and draw our final product now."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we got that by thinking about the pKa of the conjugate acid. So going back to our mechanism, the chloride anion is going to leave. So these electrons in here in green are going to come off onto the chlorine. And we can go ahead and draw our final product now. So when we reform our carbonyl, over here on the left, we had a CH3. And then over here on the right, we had an oxygen. And then we had over here."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we can go ahead and draw our final product now. So when we reform our carbonyl, over here on the left, we had a CH3. And then over here on the right, we had an oxygen. And then we had over here. And then this is a hydrogen. So this is our final product. And then let's go ahead and draw in our chloride anion as our leaving group."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then we had over here. And then this is a hydrogen. So this is our final product. And then let's go ahead and draw in our chloride anion as our leaving group. So negative 1 formal charge on our chloride anion. So electrons in green came off onto here as our leaving group. And we formed our final product, which is an acid anhydride."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then let's go ahead and draw in our chloride anion as our leaving group. So negative 1 formal charge on our chloride anion. So electrons in green came off onto here as our leaving group. And we formed our final product, which is an acid anhydride. So this is, if we were to name this acid anhydride, we would have acetic formic anhydride. So we formed an acid anhydride from an acyl chloride. And we thought about the stability of the leaving groups."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we formed our final product, which is an acid anhydride. So this is, if we were to name this acid anhydride, we would have acetic formic anhydride. So we formed an acid anhydride from an acyl chloride. And we thought about the stability of the leaving groups. So the chloride anion is the most stable leaving group. And so in general, we can look at the pKa values for the conjugate acid. So we could call these pKa and then, if you wanted to, H values."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we thought about the stability of the leaving groups. So the chloride anion is the most stable leaving group. And so in general, we can look at the pKa values for the conjugate acid. So we could call these pKa and then, if you wanted to, H values. And that just means the pKa of the conjugate acid. So thinking about your leaving groups, if you think about the pKa of the conjugate acid, the lower the pKaH, the better the leaving group. So as you go this way, you increase in terms of better leaving groups."}, {"video_title": "Nucleophilic acyl substitution Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we could call these pKa and then, if you wanted to, H values. And that just means the pKa of the conjugate acid. So thinking about your leaving groups, if you think about the pKa of the conjugate acid, the lower the pKaH, the better the leaving group. So as you go this way, you increase in terms of better leaving groups. So the chloride anion is a better leaving group than the formate anion, which is a better leaving group than our carb anion here as well. And so this is just a nice way of thinking about leaving groups in mechanisms. Think about the conjugate acid and think about the pKa."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So what do we have? This first molecule right here, I have a bunch of rings. This is a 1, 2, 3, 4, 5, 6 carbon ring. These are each four carbon rings. 1, 2, 3, 4. So the largest ring is essentially going to be our backbone. So it's going to be this six carbon ring right here."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "These are each four carbon rings. 1, 2, 3, 4. So the largest ring is essentially going to be our backbone. So it's going to be this six carbon ring right here. 1, 2, 3, 4, 5, 6 carbons. So that is a cyclohexane. All double bonds."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to be this six carbon ring right here. 1, 2, 3, 4, 5, 6 carbons. So that is a cyclohexane. All double bonds. That's why we get the ane. Six carbons, hex. It's in a cycle, cyclo."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "All double bonds. That's why we get the ane. Six carbons, hex. It's in a cycle, cyclo. And then we have two of these four carbon rings. So four carbons, we're dealing with the prefix bute. Bute for four."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "It's in a cycle, cyclo. And then we have two of these four carbon rings. So four carbons, we're dealing with the prefix bute. Bute for four. And we've got two of them. So both of these are butyl groups. Oops."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Bute for four. And we've got two of them. So both of these are butyl groups. Oops. Both of these are butyl groups. But they're in a cycle. So they're actually cyclobutyl groups."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Oops. Both of these are butyl groups. But they're in a cycle. So they're actually cyclobutyl groups. And we have two things attached to this ring right there. And if we only had one thing attached to it, you wouldn't have to number. But when you have two things, you start numbering at one of them."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So they're actually cyclobutyl groups. And we have two things attached to this ring right there. And if we only had one thing attached to it, you wouldn't have to number. But when you have two things, you start numbering at one of them. So let's say we start numbering here. And you go in the direction so that the next group has the lowest number. So in this case, we want to go in the counterclockwise direction."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "But when you have two things, you start numbering at one of them. So let's say we start numbering here. And you go in the direction so that the next group has the lowest number. So in this case, we want to go in the counterclockwise direction. So if we went this way, it'd be 1, 2, 3, 4, 5. This guy would be a 5. If we go in the counterclockwise, it would be 1, 2, 3."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So in this case, we want to go in the counterclockwise direction. So if we went this way, it'd be 1, 2, 3, 4, 5. This guy would be a 5. If we go in the counterclockwise, it would be 1, 2, 3. This guy will only be a 3. So this right here, we have two cyclobutyls. So it's dicyclobutyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "If we go in the counterclockwise, it would be 1, 2, 3. This guy will only be a 3. So this right here, we have two cyclobutyls. So it's dicyclobutyl. It is dicyclobutyl. We have two of them. And they are at the 1 and 3 position."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So it's dicyclobutyl. It is dicyclobutyl. We have two of them. And they are at the 1 and 3 position. So at the 1 and 3 position, I have two cyclobutyls on my cyclohexane, I guess you could call it, main ring. Let's try this one right here. So I have a 5-carbon ring."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And they are at the 1 and 3 position. So at the 1 and 3 position, I have two cyclobutyls on my cyclohexane, I guess you could call it, main ring. Let's try this one right here. So I have a 5-carbon ring. 1, 2, 3, 4, 5 right there. And then I have a 1, 2, 3, 4, 5, 6-carbon ring right there. So this is going to be the main ring."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So I have a 5-carbon ring. 1, 2, 3, 4, 5 right there. And then I have a 1, 2, 3, 4, 5, 6-carbon ring right there. So this is going to be the main ring. That is a cyclohexane. Has 6 carbons on them in a cycle, all single bonds. And attached to that, I have a cyclopentyl group, YL for the group."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be the main ring. That is a cyclohexane. Has 6 carbons on them in a cycle, all single bonds. And attached to that, I have a cyclopentyl group, YL for the group. So this is a cyclopentyl group on it. We don't have to number it, because it's only one group attached to the main ring. If there was another group, we would have to number it like we did up here."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And attached to that, I have a cyclopentyl group, YL for the group. So this is a cyclopentyl group on it. We don't have to number it, because it's only one group attached to the main ring. If there was another group, we would have to number it like we did up here. So this is cyclopentyl. That's this part right here. Cyclopentyl attached to cyclohexane."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "If there was another group, we would have to number it like we did up here. So this is cyclopentyl. That's this part right here. Cyclopentyl attached to cyclohexane. Cyclopentyl, cyclohexane. Now let's try this one over here. So the first thing we want to do, there's no cycles here, but we have to identify the longest chain."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Cyclopentyl attached to cyclohexane. Cyclopentyl, cyclohexane. Now let's try this one over here. So the first thing we want to do, there's no cycles here, but we have to identify the longest chain. To do that, let's just count it out. It could be 1, 2, 3, 4, 5, 6, 7. Is that the longest chain?"}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So the first thing we want to do, there's no cycles here, but we have to identify the longest chain. To do that, let's just count it out. It could be 1, 2, 3, 4, 5, 6, 7. Is that the longest chain? Maybe it's 1, 2, 3, 4, 5, 6, 7, 8. That looks like the longest chain. So let's make that the longest chain."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Is that the longest chain? Maybe it's 1, 2, 3, 4, 5, 6, 7, 8. That looks like the longest chain. So let's make that the longest chain. Let's make that the longest chain right over there. And we want to start numbering in the direction so we encounter the first attached groups first. So we do want to start numbering down here, because we have groups attached right on the 2-carbon."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let's make that the longest chain. Let's make that the longest chain right over there. And we want to start numbering in the direction so we encounter the first attached groups first. So we do want to start numbering down here, because we have groups attached right on the 2-carbon. If we started over here, we'd have to go pretty far until something's attached. So we go 1, 2, 3, 4, 5, 6, 7, 8. So we know we're dealing with an octane, all single bonds."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we do want to start numbering down here, because we have groups attached right on the 2-carbon. If we started over here, we'd have to go pretty far until something's attached. So we go 1, 2, 3, 4, 5, 6, 7, 8. So we know we're dealing with an octane, all single bonds. It's not a cyclooctane. It's not in a cycle. So we know we're dealing with an octane."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we know we're dealing with an octane, all single bonds. It's not a cyclooctane. It's not in a cycle. So we know we're dealing with an octane. And now we just have to add the groups to it. So what do we have here? This right here is just one carbon attached to the main chain."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we know we're dealing with an octane. And now we just have to add the groups to it. So what do we have here? This right here is just one carbon attached to the main chain. This is another carbon attached to the main chain. So both of these right here are methyl groups. Meth is for one carbon."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "This right here is just one carbon attached to the main chain. This is another carbon attached to the main chain. So both of these right here are methyl groups. Meth is for one carbon. Those are methyl groups. Now, if you look at all of them, these are the only methyl groups. These two up here aren't methyl groups."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Meth is for one carbon. Those are methyl groups. Now, if you look at all of them, these are the only methyl groups. These two up here aren't methyl groups. So we have two methyl groups on our entire chain. So it's going to be dimethyl. And both of the methyl groups are at our 2 position."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "These two up here aren't methyl groups. So we have two methyl groups on our entire chain. So it's going to be dimethyl. And both of the methyl groups are at our 2 position. So this is going to be 2, 2-dimethyl. This part right here is 2, 2. That right there is 2, 2-dimethyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And both of the methyl groups are at our 2 position. So this is going to be 2, 2-dimethyl. This part right here is 2, 2. That right there is 2, 2-dimethyl. We're going to decide what order to write it in after we figure out what these are called, because it has to be an alphabetical order. So this is 2, 2-dimethyl. The whole chain is an octane."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "That right there is 2, 2-dimethyl. We're going to decide what order to write it in after we figure out what these are called, because it has to be an alphabetical order. So this is 2, 2-dimethyl. The whole chain is an octane. What are these over here? Well, these, how many carbons do we have here? 1, 2, 3."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "The whole chain is an octane. What are these over here? Well, these, how many carbons do we have here? 1, 2, 3. They actually look the same. We have 1, 2, 3 here. We have 1, 2, 3 there."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3. They actually look the same. We have 1, 2, 3 here. We have 1, 2, 3 there. So these are both propyl groups. And if we deal with common names, this is kind of that Y shape. You could call it secpropyl, because this carbon right here that's attached to the main chain is attached to two other carbons."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "We have 1, 2, 3 there. So these are both propyl groups. And if we deal with common names, this is kind of that Y shape. You could call it secpropyl, because this carbon right here that's attached to the main chain is attached to two other carbons. But the more common one, because this part forms this Y shape, is isopropyl. It's isopropyl. And we have two isopropyl groups."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "You could call it secpropyl, because this carbon right here that's attached to the main chain is attached to two other carbons. But the more common one, because this part forms this Y shape, is isopropyl. It's isopropyl. And we have two isopropyl groups. These are both isopropyl. So we would want to call this, let me actually, so we have two isopropyl groups. So we would have diisopropyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And we have two isopropyl groups. These are both isopropyl. So we would want to call this, let me actually, so we have two isopropyl groups. So we would have diisopropyl. Let me write this. Diisopropyl. And they're occurring at the 4 and 5 positions."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we would have diisopropyl. Let me write this. Diisopropyl. And they're occurring at the 4 and 5 positions. Two isopropyls at 4 and 5. So this is 4 and 5 diisopropyl. That's that group and that group right there are counted for with this."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And they're occurring at the 4 and 5 positions. Two isopropyls at 4 and 5. So this is 4 and 5 diisopropyl. That's that group and that group right there are counted for with this. Now we have to just figure out the order that we write it in. You ignore the di or the tri out front, and you just look at them in alphabetical order. So we have an I for isopropyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "That's that group and that group right there are counted for with this. Now we have to just figure out the order that we write it in. You ignore the di or the tri out front, and you just look at them in alphabetical order. So we have an I for isopropyl. We have an M for methyl. Let's write the isopropyl first. And I've actually seen some people want to go for the P. But the main thing I ignore is just the di or the tri in front of the isopropyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we have an I for isopropyl. We have an M for methyl. Let's write the isopropyl first. And I've actually seen some people want to go for the P. But the main thing I ignore is just the di or the tri in front of the isopropyl. You shouldn't involve that. But everything after that, you do involve. So I'll write the isopropyl first."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And I've actually seen some people want to go for the P. But the main thing I ignore is just the di or the tri in front of the isopropyl. You shouldn't involve that. But everything after that, you do involve. So I'll write the isopropyl first. I comes before M. So this is going to be, if we were to write the whole thing, this is going to be 4, 5 diisopropyl 2, 2 dimethyl. Actually, this should be a comma here. 4, 5 diisopropyl 2, 2 dimethyl octane."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So I'll write the isopropyl first. I comes before M. So this is going to be, if we were to write the whole thing, this is going to be 4, 5 diisopropyl 2, 2 dimethyl. Actually, this should be a comma here. 4, 5 diisopropyl 2, 2 dimethyl octane. And we're done. But this was just the common name. You might remember that when we do with iso or sec or tert, butyl or propyl or whatever, that's the common name."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "4, 5 diisopropyl 2, 2 dimethyl octane. And we're done. But this was just the common name. You might remember that when we do with iso or sec or tert, butyl or propyl or whatever, that's the common name. If we want the systematic name, we could start at where we are attached to the main chain and view that as 1 and then make the longest chain with that as 1. And so you could say that we have a chain there. And this would be both of these cases."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "You might remember that when we do with iso or sec or tert, butyl or propyl or whatever, that's the common name. If we want the systematic name, we could start at where we are attached to the main chain and view that as 1 and then make the longest chain with that as 1. And so you could say that we have a chain there. And this would be both of these cases. So this is one, two carbons. Two carbons, we're dealing with an ethyl. And on the first carbon, you have a methyl attached to it."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And this would be both of these cases. So this is one, two carbons. Two carbons, we're dealing with an ethyl. And on the first carbon, you have a methyl attached to it. So you could also call each of these groups a 1-1-methyl ethyl instead of an isopropyl. So you could either say isopropyl for each of these groups, or you could call each of them a methyl ethyl if you do systematic naming. Now, we have two of these 1-methyl ethyl groups, just like we had two isopropyl groups."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "And on the first carbon, you have a methyl attached to it. So you could also call each of these groups a 1-1-methyl ethyl instead of an isopropyl. So you could either say isopropyl for each of these groups, or you could call each of them a methyl ethyl if you do systematic naming. Now, we have two of these 1-methyl ethyl groups, just like we had two isopropyl groups. If you're using common naming, you can say diisopropyl to say you have two of these groups. When you're using systematic naming, you don't say di-1-methyl ethyl, although that probably would get the point across. You use bis."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Now, we have two of these 1-methyl ethyl groups, just like we had two isopropyl groups. If you're using common naming, you can say diisopropyl to say you have two of these groups. When you're using systematic naming, you don't say di-1-methyl ethyl, although that probably would get the point across. You use bis. So this is, since we have two of them, instead of writing di, you write bis. 1-methyl ethyl. That means you have two of these things right there."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "You use bis. So this is, since we have two of them, instead of writing di, you write bis. 1-methyl ethyl. That means you have two of these things right there. And it's still in the 4 and the 5 position. And if you look at it in alphabetical order now, methyl ethyl comes after methyl. So the order will now change."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "That means you have two of these things right there. And it's still in the 4 and the 5 position. And if you look at it in alphabetical order now, methyl ethyl comes after methyl. So the order will now change. So now, if you want to write it with systematic naming, it would be written as 2, 2-dimethyl. That's these two guys. 2, 2-dimethyl ethyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So the order will now change. So now, if you want to write it with systematic naming, it would be written as 2, 2-dimethyl. That's these two guys. 2, 2-dimethyl ethyl. And then you would write this guy. So the order changed for the two groups, just based on how they're named. Bis."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "2, 2-dimethyl ethyl. And then you would write this guy. So the order changed for the two groups, just based on how they're named. Bis. And then over here, you have two 1-methyl ethyl groups. I know it's confusing, but when you just break it down, it actually makes a reasonable amount of sense. You have two of these methyl ethyl groups."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Bis. And then over here, you have two 1-methyl ethyl groups. I know it's confusing, but when you just break it down, it actually makes a reasonable amount of sense. You have two of these methyl ethyl groups. Oh, sorry. I forgot where they're located. So we have them at the 4 and 5 position."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "You have two of these methyl ethyl groups. Oh, sorry. I forgot where they're located. So we have them at the 4 and 5 position. We have two bis 1-methyl ethyl groups. I know it's a little daunting now, but it all makes sense when you break it down. So we have 10 methyl ethyl groups."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we have them at the 4 and 5 position. We have two bis 1-methyl ethyl groups. I know it's a little daunting now, but it all makes sense when you break it down. So we have 10 methyl ethyl groups. And then we can just add the octane at the end. And we're done. Now, this might seem more confusing, but when you break it down, it makes sense."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we have 10 methyl ethyl groups. And then we can just add the octane at the end. And we're done. Now, this might seem more confusing, but when you break it down, it makes sense. You have octane as the backbone. We have two methyls. They're both sitting on the 2 position."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "Now, this might seem more confusing, but when you break it down, it makes sense. You have octane as the backbone. We have two methyls. They're both sitting on the 2 position. So you have two methyls sitting on the 2 position. And then you have two 1-methyl ethyl groups sitting at the 4 and 5 position. So the 4 and 5 position, 1-methyl ethyl."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "They're both sitting on the 2 position. So you have two methyls sitting on the 2 position. And then you have two 1-methyl ethyl groups sitting at the 4 and 5 position. So the 4 and 5 position, 1-methyl ethyl. You have an ethyl, and in the 1 position, you have a methyl. So that's all it's saying. Or another way to think about it, just in case this doesn't confuse you enough, you could call it that."}, {"video_title": "Organic chemistry naming examples 2 Organic chemistry Khan Academy.mp3", "Sentence": "So the 4 and 5 position, 1-methyl ethyl. You have an ethyl, and in the 1 position, you have a methyl. So that's all it's saying. Or another way to think about it, just in case this doesn't confuse you enough, you could call it that. Or you could say 4,5-diisopropyl. These two things are the same thing. Common naming, systematic."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And so if we start with an aldehyde here, and we add Tollens' reagent, which was created by Bernhard Tollens, a German chemist, you need a source of silver ions, so silver nitrate works well. You need some hydroxide anions, so sodium hydroxide, and some ammonia. And the order in which you add these and the concentrations will depend on which procedure you're using. But eventually you're gonna form this this diamine silver cation here, and that's going to oxidize your aldehyde to a carboxylate anion. So let's go ahead and show the formation of a carboxylate anion here. So I'm gonna go ahead and put in my electrons. So now the carbon is bonded to an oxygen over here on the right, so no longer bonded to a hydrogen, and that's an oxidation here."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "But eventually you're gonna form this this diamine silver cation here, and that's going to oxidize your aldehyde to a carboxylate anion. So let's go ahead and show the formation of a carboxylate anion here. So I'm gonna go ahead and put in my electrons. So now the carbon is bonded to an oxygen over here on the right, so no longer bonded to a hydrogen, and that's an oxidation here. Let's go ahead and draw out some electrons, and let's assign some oxidation states so we know that this is indeed an oxidation reaction here. So let's put in the electrons in these bonds. So these electrons, we know that each bond consists of two electrons, so I'm putting in those electrons in here like that."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So now the carbon is bonded to an oxygen over here on the right, so no longer bonded to a hydrogen, and that's an oxidation here. Let's go ahead and draw out some electrons, and let's assign some oxidation states so we know that this is indeed an oxidation reaction here. So let's put in the electrons in these bonds. So these electrons, we know that each bond consists of two electrons, so I'm putting in those electrons in here like that. One way to assign oxidation states is to think about differences in electronegativity. Oxygen is more electronegative than carbon, so we're gonna give all four of those electrons to oxygen. Carbon and carbon have the exact same value, so one carbon gets one electron, one carbon gets the other electron."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons, we know that each bond consists of two electrons, so I'm putting in those electrons in here like that. One way to assign oxidation states is to think about differences in electronegativity. Oxygen is more electronegative than carbon, so we're gonna give all four of those electrons to oxygen. Carbon and carbon have the exact same value, so one carbon gets one electron, one carbon gets the other electron. And between carbon and hydrogen, carbon is more electronegative than hydrogen, so it takes those two electrons. Carbon has four valence electrons, and here we have it surrounded by three electrons, so four minus three gives us an oxidation state of plus one. Over here for our carboxylate anion, let's go ahead and do the same thing here."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "Carbon and carbon have the exact same value, so one carbon gets one electron, one carbon gets the other electron. And between carbon and hydrogen, carbon is more electronegative than hydrogen, so it takes those two electrons. Carbon has four valence electrons, and here we have it surrounded by three electrons, so four minus three gives us an oxidation state of plus one. Over here for our carboxylate anion, let's go ahead and do the same thing here. Let's go ahead and draw everything out, and let's go ahead and use the same color. So we have carbon double bonded to our oxygen, and put in our electrons here, and then we have carbon bonded to a carbon, and then carbon now bonded to an oxygen over here on the right. So let's put in our electrons."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "Over here for our carboxylate anion, let's go ahead and do the same thing here. Let's go ahead and draw everything out, and let's go ahead and use the same color. So we have carbon double bonded to our oxygen, and put in our electrons here, and then we have carbon bonded to a carbon, and then carbon now bonded to an oxygen over here on the right. So let's put in our electrons. So we have our electrons in here like that. And once again, think about differences in electronegativity. Oxygen beats carbon, so oxygen gets all four of those electrons."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in our electrons. So we have our electrons in here like that. And once again, think about differences in electronegativity. Oxygen beats carbon, so oxygen gets all four of those electrons. We have a tie between these two carbons. And then over here now, oxygen takes both of those electrons. So carbon has four valence electrons."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen beats carbon, so oxygen gets all four of those electrons. We have a tie between these two carbons. And then over here now, oxygen takes both of those electrons. So carbon has four valence electrons. Here we have it surrounded by one, so four minus three gives us an oxidation state of, four minus one I should say, gives us an oxidation state of plus three. And so you can see that our oxidation state has increased. We go from an oxidation state of plus one for the carbonyl carbon or aldehyde, to an oxidation state of plus three for this carbonyl carbon for our carboxylate anion."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So carbon has four valence electrons. Here we have it surrounded by one, so four minus three gives us an oxidation state of, four minus one I should say, gives us an oxidation state of plus three. And so you can see that our oxidation state has increased. We go from an oxidation state of plus one for the carbonyl carbon or aldehyde, to an oxidation state of plus three for this carbonyl carbon for our carboxylate anion. So we have oxidized that carbon. And if you're oxidized something, you are reducing something else. And that something else is going to be your silver cation here."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "We go from an oxidation state of plus one for the carbonyl carbon or aldehyde, to an oxidation state of plus three for this carbonyl carbon for our carboxylate anion. So we have oxidized that carbon. And if you're oxidized something, you are reducing something else. And that something else is going to be your silver cation here. So we have a silver cation, Ag plus. So an oxidation state of plus one. And since we oxidized our aldehyde, we're going to reduce our silver ion."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And that something else is going to be your silver cation here. So we have a silver cation, Ag plus. So an oxidation state of plus one. And since we oxidized our aldehyde, we're going to reduce our silver ion. So we're going to reduce it to solid silver with an oxidation state of zero. So it gains an electron. So the silver cation is reduced to solid silver."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And since we oxidized our aldehyde, we're going to reduce our silver ion. So we're going to reduce it to solid silver with an oxidation state of zero. So it gains an electron. So the silver cation is reduced to solid silver. And this forms a silver mirror on your glassware if you do everything properly. So the formation of a silver mirror indicates the presence of an aldehyde. And so this is a rather cool way, a rather cool diagnostic test for the presence of an aldehyde."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So the silver cation is reduced to solid silver. And this forms a silver mirror on your glassware if you do everything properly. So the formation of a silver mirror indicates the presence of an aldehyde. And so this is a rather cool way, a rather cool diagnostic test for the presence of an aldehyde. And this will help distinguish an aldehyde from a ketone. Because in general, only aldehydes are going to react with the Tollens reagent. And you get a really cool silver mirror out of it."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And so this is a rather cool way, a rather cool diagnostic test for the presence of an aldehyde. And this will help distinguish an aldehyde from a ketone. Because in general, only aldehydes are going to react with the Tollens reagent. And you get a really cool silver mirror out of it. If you want to form the carboxylic acid, you would need to then protonate your carboxylate anion. And so that would give you your carboxylic acid here. And so oxidation of aldehyde and a reduction of your silver to form a silver mirror."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And you get a really cool silver mirror out of it. If you want to form the carboxylic acid, you would need to then protonate your carboxylate anion. And so that would give you your carboxylic acid here. And so oxidation of aldehyde and a reduction of your silver to form a silver mirror. Let's look at a reaction here. We're starting off with this compound. And the first reagent we're going to use, sodium dichromate, sulfuric acid, and water."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And so oxidation of aldehyde and a reduction of your silver to form a silver mirror. Let's look at a reaction here. We're starting off with this compound. And the first reagent we're going to use, sodium dichromate, sulfuric acid, and water. We know that's going to oxidize different functional groups. And it's going to oxidize both our aldehyde and our alcohol. So let's go ahead and draw the final product here."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And the first reagent we're going to use, sodium dichromate, sulfuric acid, and water. We know that's going to oxidize different functional groups. And it's going to oxidize both our aldehyde and our alcohol. So let's go ahead and draw the final product here. So if we oxidize our aldehyde, we're going to form a carboxylic acid. And if we oxidize our alcohol, we're going to form a ketone. So you oxidize a secondary alcohol and you're going to form a ketone here."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the final product here. So if we oxidize our aldehyde, we're going to form a carboxylic acid. And if we oxidize our alcohol, we're going to form a ketone. So you oxidize a secondary alcohol and you're going to form a ketone here. And if you do this reaction using Tollens' reagent and then go ahead and protonate, you're going to oxidize only the aldehyde. So the aldehyde gets oxidized to a carboxylic acid and the OH remains untouched. So once again, in the workup when you protonate, you'd form your carboxylic acid."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So you oxidize a secondary alcohol and you're going to form a ketone here. And if you do this reaction using Tollens' reagent and then go ahead and protonate, you're going to oxidize only the aldehyde. So the aldehyde gets oxidized to a carboxylic acid and the OH remains untouched. So once again, in the workup when you protonate, you'd form your carboxylic acid. So the Tollens' reagent is selective for your aldehyde. So it leaves the alcohol untouched. And this is because Tollens' reagent is a mild oxidizing agent."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So once again, in the workup when you protonate, you'd form your carboxylic acid. So the Tollens' reagent is selective for your aldehyde. So it leaves the alcohol untouched. And this is because Tollens' reagent is a mild oxidizing agent. And it's pretty easy to oxidize aldehydes. Much harder to oxidize something like a ketone. All right, let's look at another example where we're using Tollens' reagent."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And this is because Tollens' reagent is a mild oxidizing agent. And it's pretty easy to oxidize aldehydes. Much harder to oxidize something like a ketone. All right, let's look at another example where we're using Tollens' reagent. And this is a pretty cool experiment that's done in most undergraduate organic labs. And this is using glucose as your aldehyde. So right here, you can see the open chain form of glucose."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at another example where we're using Tollens' reagent. And this is a pretty cool experiment that's done in most undergraduate organic labs. And this is using glucose as your aldehyde. So right here, you can see the open chain form of glucose. And you can see the aldehyde functional group right here. And this is worth looking at because we've talked about some of this chemistry before. So a lone pair of electrons on this oxygen can attack this carbonyl, push these electrons off onto this oxygen."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So right here, you can see the open chain form of glucose. And you can see the aldehyde functional group right here. And this is worth looking at because we've talked about some of this chemistry before. So a lone pair of electrons on this oxygen can attack this carbonyl, push these electrons off onto this oxygen. And once you deprotonate and protonate, you're going to form a cyclic hemiacetal. So over here on the left, let's say this is the cyclic hemiacetal that you form. So let's analyze those carbons there."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on this oxygen can attack this carbonyl, push these electrons off onto this oxygen. And once you deprotonate and protonate, you're going to form a cyclic hemiacetal. So over here on the left, let's say this is the cyclic hemiacetal that you form. So let's analyze those carbons there. So this carbonyl carbon right here becomes the anomeric carbon for our cyclic hemiacetal. So this is one of the cyclic forms of glucose. And you can see this is the one with the OH down relative to the plane of the ring."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So let's analyze those carbons there. So this carbonyl carbon right here becomes the anomeric carbon for our cyclic hemiacetal. So this is one of the cyclic forms of glucose. And you can see this is the one with the OH down relative to the plane of the ring. And if you're thinking about which anomer this is, you can compare this OH to this CH2OH group, which is up relative to a flat plane of the ring. And so since they're on opposite sides, this would be the trans. They're trans to each other."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And you can see this is the one with the OH down relative to the plane of the ring. And if you're thinking about which anomer this is, you can compare this OH to this CH2OH group, which is up relative to a flat plane of the ring. And so since they're on opposite sides, this would be the trans. They're trans to each other. This is the alpha anomer. So this is the alpha-alpha glucose form in the cyclic hemiacetal form. So the other possibility, of course, would be to add that OH up relative to the plane of the ring."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "They're trans to each other. This is the alpha anomer. So this is the alpha-alpha glucose form in the cyclic hemiacetal form. So the other possibility, of course, would be to add that OH up relative to the plane of the ring. So since we have a planar aldehyde here, we could have added the OH up. And then they'd be on the same side as the CH2OH. So they're cis to each other."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So the other possibility, of course, would be to add that OH up relative to the plane of the ring. So since we have a planar aldehyde here, we could have added the OH up. And then they'd be on the same side as the CH2OH. So they're cis to each other. So this would be the beta anomer. So that would be beta glucose, the cyclic form of it. And the cyclic hemiacetal is actually favored."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So they're cis to each other. So this would be the beta anomer. So that would be beta glucose, the cyclic form of it. And the cyclic hemiacetal is actually favored. So we talked about that in the video on hemiacetal formation. And so glucose spends most of its time in these cyclic forms, in the beta and in the alpha form. But it's in equilibrium with the open chain form, with the open chain form containing the RR aldehyde functional group."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And the cyclic hemiacetal is actually favored. So we talked about that in the video on hemiacetal formation. And so glucose spends most of its time in these cyclic forms, in the beta and in the alpha form. But it's in equilibrium with the open chain form, with the open chain form containing the RR aldehyde functional group. And this is what we need to react with Tollens' reagent. So when we oxidize the aldehydes, we're going to form a carboxylate anion. And when we oxidize the aldehyde, we're going to reduce the silver."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "But it's in equilibrium with the open chain form, with the open chain form containing the RR aldehyde functional group. And this is what we need to react with Tollens' reagent. So when we oxidize the aldehydes, we're going to form a carboxylate anion. And when we oxidize the aldehyde, we're going to reduce the silver. So the silver ions go from Ag plus to Ag and forming our silver mirror. And the reason that glucose is used is because this is a highly water-soluble. So it just makes this reaction a lot easier because of all these OH groups on the glucose molecule."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And when we oxidize the aldehyde, we're going to reduce the silver. So the silver ions go from Ag plus to Ag and forming our silver mirror. And the reason that glucose is used is because this is a highly water-soluble. So it just makes this reaction a lot easier because of all these OH groups on the glucose molecule. And so you can form your silver mirror this way. And you can make some really cool silver mirrors using glucose. And let me show you pictures of a couple of these things."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "So it just makes this reaction a lot easier because of all these OH groups on the glucose molecule. And so you can form your silver mirror this way. And you can make some really cool silver mirrors using glucose. And let me show you pictures of a couple of these things. And so over here on the left, my students actually made me an ornament. So you can do this reaction using a glass ornament here. And then you can put your ornament on a chemist's tree."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And let me show you pictures of a couple of these things. And so over here on the left, my students actually made me an ornament. So you can do this reaction using a glass ornament here. And then you can put your ornament on a chemist's tree. So let me go ahead and write out here, chemist's tree, if it's the holiday season. And here's actually my wife holding the ornament that my students made for us. And then you can actually make flat mirrors out of this too."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And then you can put your ornament on a chemist's tree. So let me go ahead and write out here, chemist's tree, if it's the holiday season. And here's actually my wife holding the ornament that my students made for us. And then you can actually make flat mirrors out of this too. And so over here on the right, here I am reflected in the flat silver mirror that I made. And I did this by starting off with a microscope slide, which is glass. And then I put that at the bottom of a 600 mL beaker."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And then you can actually make flat mirrors out of this too. And so over here on the right, here I am reflected in the flat silver mirror that I made. And I did this by starting off with a microscope slide, which is glass. And then I put that at the bottom of a 600 mL beaker. And then I silvered the top portion of it. And then I was able to make sure the other side was just glass. And then if you put some varnish on there, you'll have a really nice silver mirror."}, {"video_title": "Oxidation of aldehydes using Tollens' reagent Organic chemistry Khan Academy.mp3", "Sentence": "And then I put that at the bottom of a 600 mL beaker. And then I silvered the top portion of it. And then I was able to make sure the other side was just glass. And then if you put some varnish on there, you'll have a really nice silver mirror. And so making silver mirrors using chemistry is something that was perfected by the German chemist Justus von Liebig in the 1830s. And so even though Bernhard Tollens is the creator of the Tollens reagent, Liebig is the one who perfected the formation of these silver mirrors. And it's a lot of fun to make one."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when we think of farming, we imagine a farmer planting seeds and later harvesting the crops, or maybe having cattle that they can allow to graze and then using that cattle for either meat or milk or wool. But there's actually a different type of farming that predates this association with kind of, I guess what we could call the traditional form of farming. And it predates it by several tens of thousands of years. And we believe that it started with the original inhabitants of Australia. And what they did is, and this is why we call it farming, because if you think about farming in the most general sense, it's really humans using technology to manipulate their environment so it becomes more suitable for humans, so it becomes more suitable for things that humans might want to eat or get milk from or whatever. And this type of farming is called fire stick farming. Fire stick farming."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we believe that it started with the original inhabitants of Australia. And what they did is, and this is why we call it farming, because if you think about farming in the most general sense, it's really humans using technology to manipulate their environment so it becomes more suitable for humans, so it becomes more suitable for things that humans might want to eat or get milk from or whatever. And this type of farming is called fire stick farming. Fire stick farming. And I think you can already imagine what it might involve. It involves using fire, which is really a form of technology, or it can be a form of technology, using fire to make the environment more suitable for human activity. And so what the original Australians did, the indigenous Australians, or sometimes referred to as the Aboriginal Australians, Aboriginal Australians, and if you're wondering where the word Aboriginal comes from, you might recognize some parts of it."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Fire stick farming. And I think you can already imagine what it might involve. It involves using fire, which is really a form of technology, or it can be a form of technology, using fire to make the environment more suitable for human activity. And so what the original Australians did, the indigenous Australians, or sometimes referred to as the Aboriginal Australians, Aboriginal Australians, and if you're wondering where the word Aboriginal comes from, you might recognize some parts of it. Original, you know what that means, the first things, the things that were there from the beginning. And then you have ab, which is Latin for from. So this is literally from the beginning."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so what the original Australians did, the indigenous Australians, or sometimes referred to as the Aboriginal Australians, Aboriginal Australians, and if you're wondering where the word Aboriginal comes from, you might recognize some parts of it. Original, you know what that means, the first things, the things that were there from the beginning. And then you have ab, which is Latin for from. So this is literally from the beginning. So when you say Aboriginal Australians, you're really kind of saying the Australians that were there from the beginning. And so what they would do is, is that we believe if you go back 50 or 60,000 years, before the first Aboriginal Australians settled Australia, Australia had much more forest. It still has forest."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is literally from the beginning. So when you say Aboriginal Australians, you're really kind of saying the Australians that were there from the beginning. And so what they would do is, is that we believe if you go back 50 or 60,000 years, before the first Aboriginal Australians settled Australia, Australia had much more forest. It still has forest. This is a modern picture, obviously, of an Australian forest. But what they did is that they set up controlled burns. And what these controlled burns did is that they cleared away a lot of the forest, they cleared away a lot of the brush that's at the bottom of, you know, that's over here, and it made it much more suitable for grassland to develop."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It still has forest. This is a modern picture, obviously, of an Australian forest. But what they did is that they set up controlled burns. And what these controlled burns did is that they cleared away a lot of the forest, they cleared away a lot of the brush that's at the bottom of, you know, that's over here, and it made it much more suitable for grassland to develop. And the reason why they liked grassland, so let's make a little cycle here of what they did. So they have controlled burns, controlled fires, those controlled fires helped promote grassland. And then once you have grassland, that made the environment more suitable for animals that the original human settlers could essentially live off of, that they could hunt, that they could potentially eat their meat."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what these controlled burns did is that they cleared away a lot of the forest, they cleared away a lot of the brush that's at the bottom of, you know, that's over here, and it made it much more suitable for grassland to develop. And the reason why they liked grassland, so let's make a little cycle here of what they did. So they have controlled burns, controlled fires, those controlled fires helped promote grassland. And then once you have grassland, that made the environment more suitable for animals that the original human settlers could essentially live off of, that they could hunt, that they could potentially eat their meat. And so, for example, things like kangaroos, and these supported the human population, which obviously would then do the controlled burns. And you see here, so we could have started off with something like this. Someone provides a controlled burn."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then once you have grassland, that made the environment more suitable for animals that the original human settlers could essentially live off of, that they could hunt, that they could potentially eat their meat. And so, for example, things like kangaroos, and these supported the human population, which obviously would then do the controlled burns. And you see here, so we could have started off with something like this. Someone provides a controlled burn. And they were actually pretty scientific about how they did it. They wouldn't just go at the end of summer when everything was hot and ready to just blow up and then start a fire that they couldn't control. They would often do these in seasons knowing that it had a certain level of moisture in the air, it wasn't too hot."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Someone provides a controlled burn. And they were actually pretty scientific about how they did it. They wouldn't just go at the end of summer when everything was hot and ready to just blow up and then start a fire that they couldn't control. They would often do these in seasons knowing that it had a certain level of moisture in the air, it wasn't too hot. And to a large degree, by doing these controlled burns, not only did it provide an environment, kind of do this fire stick farming, not only did it provide an environment that was suitable for things like kangaroos, some type of things that humans could eat, but it also prevented major fires. And you still see forest strangers doing this type of thing. And there's some reason to believe that what the original Australians did, on some level, was more nuanced and more fine-tuned than even what we do in a modern sense in controlled burns."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They would often do these in seasons knowing that it had a certain level of moisture in the air, it wasn't too hot. And to a large degree, by doing these controlled burns, not only did it provide an environment, kind of do this fire stick farming, not only did it provide an environment that was suitable for things like kangaroos, some type of things that humans could eat, but it also prevented major fires. And you still see forest strangers doing this type of thing. And there's some reason to believe that what the original Australians did, on some level, was more nuanced and more fine-tuned than even what we do in a modern sense in controlled burns. So these controlled fires also prevented major uncontrollable fires. Because what happens is if you don't have these controlled fires, then you have brush building up year after year after year. You have stuff building up."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there's some reason to believe that what the original Australians did, on some level, was more nuanced and more fine-tuned than even what we do in a modern sense in controlled burns. So these controlled fires also prevented major uncontrollable fires. Because what happens is if you don't have these controlled fires, then you have brush building up year after year after year. You have stuff building up. And then when the fires do occur, they're not going to occur, or they're less likely, the uncontrolled fires are less likely to be started during the winter when the air is cool or when there might be some moisture. They're more likely to occur in the dry season. So you have all this stuff build up, and then when the fire does happen, it happens in the driest season."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have stuff building up. And then when the fires do occur, they're not going to occur, or they're less likely, the uncontrolled fires are less likely to be started during the winter when the air is cool or when there might be some moisture. They're more likely to occur in the dry season. So you have all this stuff build up, and then when the fire does happen, it happens in the driest season. And then when it happens with all of the stuff build up in the dry season, it just becomes uncontrollable. One of the byproducts, or actually there's several byproducts, of this fire stick farming, we believe, is a lot of the grassland in Australia now might have been more forested before. And even when the first European settlers came in the late 1700s, they were kind of surprised when they went into what is now Sydney Harbor, and they said, wow, look at all the grassland here."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have all this stuff build up, and then when the fire does happen, it happens in the driest season. And then when it happens with all of the stuff build up in the dry season, it just becomes uncontrollable. One of the byproducts, or actually there's several byproducts, of this fire stick farming, we believe, is a lot of the grassland in Australia now might have been more forested before. And even when the first European settlers came in the late 1700s, they were kind of surprised when they went into what is now Sydney Harbor, and they said, wow, look at all the grassland here. It almost looks like park space. And then they would let their sheep graze there, and they were surprised, because they had driven out the original inhabitants, and then they were surprised when forests just started to grow up in that grassland. And it was because the original Australians were actually controlling that forest growth to make it more inhabitable for things like kangaroos."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even when the first European settlers came in the late 1700s, they were kind of surprised when they went into what is now Sydney Harbor, and they said, wow, look at all the grassland here. It almost looks like park space. And then they would let their sheep graze there, and they were surprised, because they had driven out the original inhabitants, and then they were surprised when forests just started to grow up in that grassland. And it was because the original Australians were actually controlling that forest growth to make it more inhabitable for things like kangaroos. And then when the English settlers came, they started to have their sheep graze in those grasslands. And it also is responsible for the disappearance, we think, of many major, I guess for lack of a better word, megafauna. So really large animals that inhabited Australia for really millions of years until humans showed up."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it was because the original Australians were actually controlling that forest growth to make it more inhabitable for things like kangaroos. And then when the English settlers came, they started to have their sheep graze in those grasslands. And it also is responsible for the disappearance, we think, of many major, I guess for lack of a better word, megafauna. So really large animals that inhabited Australia for really millions of years until humans showed up. And this is one of them. It's just neat to look at them. This is called Driptodon optimum, or another way to think of it, the giant wombat."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So really large animals that inhabited Australia for really millions of years until humans showed up. And this is one of them. It's just neat to look at them. This is called Driptodon optimum, or another way to think of it, the giant wombat. And there's fossils of the giant wombat around 40,000, 50,000 years ago. But they disappeared with humans showing up. And there's multiple ways that you could think about why they disappeared."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is called Driptodon optimum, or another way to think of it, the giant wombat. And there's fossils of the giant wombat around 40,000, 50,000 years ago. But they disappeared with humans showing up. And there's multiple ways that you could think about why they disappeared. They might have, and this is probably the case, they might have been more dependent on the forest habitat. Or this was a more favorable habitat for them than the grasslands, maybe because they ate leaves that were high up. Or another thing is once the forest habitat goes away, they were actually also easier to hunt down."}, {"video_title": "Firestick farming Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there's multiple ways that you could think about why they disappeared. They might have, and this is probably the case, they might have been more dependent on the forest habitat. Or this was a more favorable habitat for them than the grasslands, maybe because they ate leaves that were high up. Or another thing is once the forest habitat goes away, they were actually also easier to hunt down. Or either way you think about it, they might have just been hunted by humans. But we do see that with humans coming to the Australian continent, you start to see the disappearance. And this isn't the only one, but there was several major species of megafauna, of super large animals that disappeared at that time period."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And what happens in electrophilic aromatic substitution, we're going to substitute the electrophile for a proton on our benzene ring. And so over here we can see the electrophile is now in place of that proton. So that's where the electrophilic part comes in and that's where the substitution part comes in. We're using an electrophile for a proton. The aromatic comes in because you're going to reform an aromatic ring in your mechanism. Electrophilic aromatic substitution requires a catalyst and the point of a catalyst is to generate your electrophile. So down here you can see that the catalyst is going to react to produce the positively charged electrophile."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "We're using an electrophile for a proton. The aromatic comes in because you're going to reform an aromatic ring in your mechanism. Electrophilic aromatic substitution requires a catalyst and the point of a catalyst is to generate your electrophile. So down here you can see that the catalyst is going to react to produce the positively charged electrophile. So remember electrophile means loving electrons. So something that's positively charged, it's going to love electrons. We also form this catalyst complex over here which is going to factor into our mechanism."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So down here you can see that the catalyst is going to react to produce the positively charged electrophile. So remember electrophile means loving electrons. So something that's positively charged, it's going to love electrons. We also form this catalyst complex over here which is going to factor into our mechanism. So now that we've formed our electrophile, let's look in more detail as to what happens in electrophilic aromatic substitution. So we start with our benzene ring and I'm showing one of the hydrogens on the benzene ring. It could be any of the six since they are all equivalent."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "We also form this catalyst complex over here which is going to factor into our mechanism. So now that we've formed our electrophile, let's look in more detail as to what happens in electrophilic aromatic substitution. So we start with our benzene ring and I'm showing one of the hydrogens on the benzene ring. It could be any of the six since they are all equivalent. And now we've formed our electrophile from our catalyst. So the pi electrons in the benzene ring can be attracted to the positively charged electrophile because negative charges are attracted to positive charges. And so pi electrons in your benzene ring are going to function as a nucleophile and those electrons are going to attack the electrophile."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "It could be any of the six since they are all equivalent. And now we've formed our electrophile from our catalyst. So the pi electrons in the benzene ring can be attracted to the positively charged electrophile because negative charges are attracted to positive charges. And so pi electrons in your benzene ring are going to function as a nucleophile and those electrons are going to attack the electrophile. So this is a nucleophile-electrophile attack where those pi electrons are going to bond to that electrophile there. So those pi electrons are going to form a covalent bond with your electrophile. So let's go ahead and show that."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And so pi electrons in your benzene ring are going to function as a nucleophile and those electrons are going to attack the electrophile. So this is a nucleophile-electrophile attack where those pi electrons are going to bond to that electrophile there. So those pi electrons are going to form a covalent bond with your electrophile. So let's go ahead and show that. So these pi electrons didn't do anything. The hydrogen stays there. Now I could show the electrophile adding to either of the two carbons on the side of the double bond."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. So these pi electrons didn't do anything. The hydrogen stays there. Now I could show the electrophile adding to either of the two carbons on the side of the double bond. So it could be that carbon or it could be this carbon. Since I've drawn this hydrogen up here at the top, I'm going to go ahead and say the electrophile adds to the top carbon there. So there's my electrophile."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Now I could show the electrophile adding to either of the two carbons on the side of the double bond. So it could be that carbon or it could be this carbon. Since I've drawn this hydrogen up here at the top, I'm going to go ahead and say the electrophile adds to the top carbon there. So there's my electrophile. Let me go ahead and highlight the electrons that are forming that covalent bond. So these pi electrons here are the ones that are functioning as a nucleophile and those pi electrons are going to form this bond right here. Now in forming that bond, we're taking a bond away from this bottom carbon here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So there's my electrophile. Let me go ahead and highlight the electrons that are forming that covalent bond. So these pi electrons here are the ones that are functioning as a nucleophile and those pi electrons are going to form this bond right here. Now in forming that bond, we're taking a bond away from this bottom carbon here. And so that bottom carbon is going to be left with a positive 1 formal charge. Therefore we can draw a resonance structure for this cation. So let's go ahead and show a possible resonance structure here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Now in forming that bond, we're taking a bond away from this bottom carbon here. And so that bottom carbon is going to be left with a positive 1 formal charge. Therefore we can draw a resonance structure for this cation. So let's go ahead and show a possible resonance structure here. So these pi electrons could move over to here. And let's go ahead and draw what would result if that happens. So now we have these pi electrons up here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show a possible resonance structure here. So these pi electrons could move over to here. And let's go ahead and draw what would result if that happens. So now we have these pi electrons up here. We have our hydrogen. We have our electrophile. And the electrons moved over to this position."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So now we have these pi electrons up here. We have our hydrogen. We have our electrophile. And the electrons moved over to this position. So let me go ahead and highlight those in magenta. So I'm saying that these pi electrons right here moved over to here. And when those electrons move over to there, we're taking a bond away from this carbon this time."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons moved over to this position. So let me go ahead and highlight those in magenta. So I'm saying that these pi electrons right here moved over to here. And when those electrons move over to there, we're taking a bond away from this carbon this time. So that is the carbon that's going to get a plus 1 formal charge like that. So we can draw another resonance structure. So let's go ahead and do that."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And when those electrons move over to there, we're taking a bond away from this carbon this time. So that is the carbon that's going to get a plus 1 formal charge like that. So we can draw another resonance structure. So let's go ahead and do that. So we could take these pi electrons and move them into here. So let's go ahead and show what that would look like. So if those pi electrons moved into there, we would now have, again, our hydrogen, our electrophile, these pi electrons, and then these pi electrons right here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and do that. So we could take these pi electrons and move them into here. So let's go ahead and show what that would look like. So if those pi electrons moved into there, we would now have, again, our hydrogen, our electrophile, these pi electrons, and then these pi electrons right here. So once again, let me go ahead and highlight those. This time I'll use blue. So these pi electrons are going to move over to here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So if those pi electrons moved into there, we would now have, again, our hydrogen, our electrophile, these pi electrons, and then these pi electrons right here. So once again, let me go ahead and highlight those. This time I'll use blue. So these pi electrons are going to move over to here. And once again, we're taking a bond away from a carbon. This time it's this top carbon up here. So that's the carbon that's going to get the plus 1 formal charge like that."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons are going to move over to here. And once again, we're taking a bond away from a carbon. This time it's this top carbon up here. So that's the carbon that's going to get the plus 1 formal charge like that. So these are all resonance structures. And remember, the actual cation would be a hybrid of these resonance structures. And we call that hybrid a sigma complex."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So that's the carbon that's going to get the plus 1 formal charge like that. So these are all resonance structures. And remember, the actual cation would be a hybrid of these resonance structures. And we call that hybrid a sigma complex. So you have a positive 1 formal charge delocalized over three carbons in your sigma complex. So the next step in the mechanism, I'm just going to redraw the first resonance structure that we did here. So I'm going to go ahead and redraw that down here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And we call that hybrid a sigma complex. So you have a positive 1 formal charge delocalized over three carbons in your sigma complex. So the next step in the mechanism, I'm just going to redraw the first resonance structure that we did here. So I'm going to go ahead and redraw that down here. So let's go ahead and show the first resonance structure. So in our first resonance structure, we had our hydrogen here, our electrophile already bonded to our ring. And we had a positive 1 formal charge on this carbon right here."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and redraw that down here. So let's go ahead and show the first resonance structure. So in our first resonance structure, we had our hydrogen here, our electrophile already bonded to our ring. And we had a positive 1 formal charge on this carbon right here. Well remember, the catalyst had formed a complex. And I represent it like this. So something bonded to your catalyst like that."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And we had a positive 1 formal charge on this carbon right here. Well remember, the catalyst had formed a complex. And I represent it like this. So something bonded to your catalyst like that. So let's just go up here and refresh our memory. So right up here, when we generated our electrophile, we also generated this catalyst complex up here. So Y bonded to a catalyst."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So something bonded to your catalyst like that. So let's just go up here and refresh our memory. So right up here, when we generated our electrophile, we also generated this catalyst complex up here. So Y bonded to a catalyst. So I have Y bonded to a catalyst down here. And you could think about this as functioning as a base, or it's going to accept a proton. So I could show these electrons in here taking this proton."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So Y bonded to a catalyst. So I have Y bonded to a catalyst down here. And you could think about this as functioning as a base, or it's going to accept a proton. So I could show these electrons in here taking this proton. And if it takes that proton, that leaves these electrons behind. And those electrons are going to move in here to reform your benzene ring and take away that positive 1 formal charge. Let's go ahead and show that."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So I could show these electrons in here taking this proton. And if it takes that proton, that leaves these electrons behind. And those electrons are going to move in here to reform your benzene ring and take away that positive 1 formal charge. Let's go ahead and show that. So we now have our benzene ring back. And our electrophile is now bonded to our ring. And the proton has left."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and show that. So we now have our benzene ring back. And our electrophile is now bonded to our ring. And the proton has left. So the electrophile has completely substituted for that proton. Let's follow those electrons again. So the electrons in magenta in here, so those are the ones that are going to move in here to reform your aromatic ring."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "And the proton has left. So the electrophile has completely substituted for that proton. Let's follow those electrons again. So the electrons in magenta in here, so those are the ones that are going to move in here to reform your aromatic ring. So deprotonation of the sigma complex restores the aromatic ring. And so we have a stable product here. So the other product you could think about, this Y here is now going to be bonded to that proton."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta in here, so those are the ones that are going to move in here to reform your aromatic ring. So deprotonation of the sigma complex restores the aromatic ring. And so we have a stable product here. So the other product you could think about, this Y here is now going to be bonded to that proton. So you could have the Y here bonded to that proton. And you could highlight those electrons. You could say that these electrons right here are now these electrons."}, {"video_title": "Electrophilic aromatic substitution mechanism Organic chemistry Khan Academy.mp3", "Sentence": "So the other product you could think about, this Y here is now going to be bonded to that proton. So you could have the Y here bonded to that proton. And you could highlight those electrons. You could say that these electrons right here are now these electrons. And taking those electrons away from the catalyst would, of course, regenerate your catalyst. And so it's free to then catalyze another reaction. And so this is the general mechanism for electrophilic aromatic substitution, which the reactions that we're going to see are pretty much going to follow this general mechanism."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "That proton has a pKa value of approximately 4.8. So if acetic acid donates that proton, these electrons in red here are left behind on the oxygen, which gives the oxygen a negative one formal charge. So on the right, this would be the conjugate base to acetic acid. If we compare acetic acid to our next compound, this is chloroacetic acid. Notice we now have a chlorine attached to this carbon. Now, I didn't draw in lone pairs of electrons for chlorine just to make it easier to see. So the acidic proton on chloroacetic acid is this proton, so then these electrons in red are left behind on the oxygen giving the oxygen a negative one formal charge."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "If we compare acetic acid to our next compound, this is chloroacetic acid. Notice we now have a chlorine attached to this carbon. Now, I didn't draw in lone pairs of electrons for chlorine just to make it easier to see. So the acidic proton on chloroacetic acid is this proton, so then these electrons in red are left behind on the oxygen giving the oxygen a negative one formal charge. If we look at the pKa value, the approximate pKa value for this proton, it's about 2.9. So think about the difference in acidity between chloroacetic acid and acetic acid. You're going from 4.8 to 2.9."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So the acidic proton on chloroacetic acid is this proton, so then these electrons in red are left behind on the oxygen giving the oxygen a negative one formal charge. If we look at the pKa value, the approximate pKa value for this proton, it's about 2.9. So think about the difference in acidity between chloroacetic acid and acetic acid. You're going from 4.8 to 2.9. The lower the pKa, the more acidic the compound. And that's approximately 100 times more acidic, right? This is two pH units."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "You're going from 4.8 to 2.9. The lower the pKa, the more acidic the compound. And that's approximately 100 times more acidic, right? This is two pH units. From 4.8 to 2.9 is pretty close to two pH units. So that would be 10 to the second power, or 100 times more acidic. And so chloroacetic acid is much more acidic than acetic acid."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "This is two pH units. From 4.8 to 2.9 is pretty close to two pH units. So that would be 10 to the second power, or 100 times more acidic. And so chloroacetic acid is much more acidic than acetic acid. And if we look at the conjugate bases, we can understand why. So this conjugate base must be more stable than this conjugate base. And we can explain this in terms of induction."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "And so chloroacetic acid is much more acidic than acetic acid. And if we look at the conjugate bases, we can understand why. So this conjugate base must be more stable than this conjugate base. And we can explain this in terms of induction. So if we look at the difference, right, we know we have chlorine here, and chlorine is an electronegative element. It's much more electronegative than carbon. So chlorine's going to withdraw some electron density this way."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "And we can explain this in terms of induction. So if we look at the difference, right, we know we have chlorine here, and chlorine is an electronegative element. It's much more electronegative than carbon. So chlorine's going to withdraw some electron density this way. And if you withdraw electron density, right, you delocalize this negative charge. You spread out this negative charge. And that stabilizes the conjugate base."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So chlorine's going to withdraw some electron density this way. And if you withdraw electron density, right, you delocalize this negative charge. You spread out this negative charge. And that stabilizes the conjugate base. And since this conjugate base is more stable than this conjugate base, chloroacetic acid is more likely to donate its proton than acetic acid. And we can see what happens as we increase the number of chlorines. So down here is trichloroacetic acid."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "And that stabilizes the conjugate base. And since this conjugate base is more stable than this conjugate base, chloroacetic acid is more likely to donate its proton than acetic acid. And we can see what happens as we increase the number of chlorines. So down here is trichloroacetic acid. We have three chlorines. And the pKa has lowered even more, right, because now we have all of these chlorines here withdrawing some electron density. So all of these electron withdrawing groups, if you will, are withdrawing electron density, and that's stabilizing this conjugate base."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So down here is trichloroacetic acid. We have three chlorines. And the pKa has lowered even more, right, because now we have all of these chlorines here withdrawing some electron density. So all of these electron withdrawing groups, if you will, are withdrawing electron density, and that's stabilizing this conjugate base. That's spreading out this negative charge. So this is the most stable conjugate base out of these three. Therefore, this is the most acidic compound out of these three."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So all of these electron withdrawing groups, if you will, are withdrawing electron density, and that's stabilizing this conjugate base. That's spreading out this negative charge. So this is the most stable conjugate base out of these three. Therefore, this is the most acidic compound out of these three. The inductive effect falls off over distance. So if we look at this acid here, this is called butanoic acid. So this is carbon one."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, this is the most acidic compound out of these three. The inductive effect falls off over distance. So if we look at this acid here, this is called butanoic acid. So this is carbon one. This is carbon two. This is carbon three. And this is carbon four."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So this is carbon one. This is carbon two. This is carbon three. And this is carbon four. So this proton has a pKa value of approximately 4.8. If we compare butanoic acid to chlorobutanoic acid, so this would be with a chlorine on carbon two, so two chlorobutanoic acids, the pKa value has dropped to 2.8. So again, that's because of the inductive effect, right?"}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "And this is carbon four. So this proton has a pKa value of approximately 4.8. If we compare butanoic acid to chlorobutanoic acid, so this would be with a chlorine on carbon two, so two chlorobutanoic acids, the pKa value has dropped to 2.8. So again, that's because of the inductive effect, right? We have an electronegative atom withdrawing electron density, stabilizing our conjugate base, therefore lowering the pKa value for this proton. Now, if we move the chlorine to the third position, right, this is carbon one, this is carbon two, this is carbon three. So now the chlorine's on the third position."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So again, that's because of the inductive effect, right? We have an electronegative atom withdrawing electron density, stabilizing our conjugate base, therefore lowering the pKa value for this proton. Now, if we move the chlorine to the third position, right, this is carbon one, this is carbon two, this is carbon three. So now the chlorine's on the third position. So this is three chlorobutanoic acid. The pKa value is still lower than 4.8, right? 4.8 was the original pKa value for this proton."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "So now the chlorine's on the third position. So this is three chlorobutanoic acid. The pKa value is still lower than 4.8, right? 4.8 was the original pKa value for this proton. Now it's 4.1, but notice, it's not as low as it was in the previous example. So with the chlorine on carbon two, the pKa value was 2.8. When the chlorine's on carbon three, the pKa value is 4.1."}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "4.8 was the original pKa value for this proton. Now it's 4.1, but notice, it's not as low as it was in the previous example. So with the chlorine on carbon two, the pKa value was 2.8. When the chlorine's on carbon three, the pKa value is 4.1. So still more acidic than the original butanoic acid, but you can see the electronegative atom is further away from the negative charge and that decreases, that decreases the effect, right? The chlorine is further away from the negative charge than in this example. And finally, you can see the effect even more if you move the chlorine to the fourth position, right?"}, {"video_title": "Stabilization of a conjugate base induction Organic chemistry Khan Academy.mp3", "Sentence": "When the chlorine's on carbon three, the pKa value is 4.1. So still more acidic than the original butanoic acid, but you can see the electronegative atom is further away from the negative charge and that decreases, that decreases the effect, right? The chlorine is further away from the negative charge than in this example. And finally, you can see the effect even more if you move the chlorine to the fourth position, right? This is carbon one, two, three, and four. If you move the chlorine to the fourth position, now your pKa value is almost back up to 4.8, so it's about 4.5. So this chlorine is further away from this negative charge and that decreases the inductive effect."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "For this reaction, we have a tertiary alkyl halide, and we know that a tertiary alkyl halide will form a tertiary carbocation, which is a stable carbocation, and therefore an SN1 reaction is possible. An SN2 reaction is not possible because this tertiary alkyl halide has too much steric hindrance to undergo an SN2 mechanism. An E1 mechanism is also possible, because an E1 mechanism requires a stable carbocation, and an E2 reaction is also possible. So we have three choices for a tertiary substrate. Next, you want to look at your reagent. Here we have sodium chloride, and we know that the chloride ion is a very weak base, and it's gonna function only as a nucleophile. So if we have a nucleophile here, we're gonna go with substitution reactions."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So we have three choices for a tertiary substrate. Next, you want to look at your reagent. Here we have sodium chloride, and we know that the chloride ion is a very weak base, and it's gonna function only as a nucleophile. So if we have a nucleophile here, we're gonna go with substitution reactions. So an SN1 reaction must be the case for this example. And we know in an SN1 mechanism, first step should be loss of our leaving group for this tertiary alkyl halide. So we form the iodide anion, and we take a bond away from this carbon in red."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So if we have a nucleophile here, we're gonna go with substitution reactions. So an SN1 reaction must be the case for this example. And we know in an SN1 mechanism, first step should be loss of our leaving group for this tertiary alkyl halide. So we form the iodide anion, and we take a bond away from this carbon in red. So we form a tertiary carbocation. So let me draw this in. We have our six-membered ring."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So we form the iodide anion, and we take a bond away from this carbon in red. So we form a tertiary carbocation. So let me draw this in. We have our six-membered ring. We have our methyl group, and the carbon in red is this one. It gets a plus one formal charge, so a positive charge on this carbon. Once we've formed our carbocation, in the next step of our mechanism, our nucleophile attacks, and our nucleophile is the chloride ion."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "We have our six-membered ring. We have our methyl group, and the carbon in red is this one. It gets a plus one formal charge, so a positive charge on this carbon. Once we've formed our carbocation, in the next step of our mechanism, our nucleophile attacks, and our nucleophile is the chloride ion. So a lone pair of electrons on the chlorine are gonna form a bond with that carbon in red, and we end up with our product. So let me draw in our ring. We have a methyl group, and we have a bond to a chlorine now."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Once we've formed our carbocation, in the next step of our mechanism, our nucleophile attacks, and our nucleophile is the chloride ion. So a lone pair of electrons on the chlorine are gonna form a bond with that carbon in red, and we end up with our product. So let me draw in our ring. We have a methyl group, and we have a bond to a chlorine now. So let me highlight a lone pair of electrons in magenta, form this bond between the chlorine and the carbon in red. We don't have any stereochemistry to worry about here because we don't have any chiral centers. So this was an SN1 mechanism."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "We have a methyl group, and we have a bond to a chlorine now. So let me highlight a lone pair of electrons in magenta, form this bond between the chlorine and the carbon in red. We don't have any stereochemistry to worry about here because we don't have any chiral centers. So this was an SN1 mechanism. First, we look at our substrate, and we can see that this is a tertiary alkyl halide. The carbon that is directly bonded to our halogen is bonded to three other carbons. So a tertiary alkyl halide is too sterically hindered to undergo an SN2 reaction, so immediately we know that that is out."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So this was an SN1 mechanism. First, we look at our substrate, and we can see that this is a tertiary alkyl halide. The carbon that is directly bonded to our halogen is bonded to three other carbons. So a tertiary alkyl halide is too sterically hindered to undergo an SN2 reaction, so immediately we know that that is out. And when we look at our reagent, this is potassium tert-butoxide. So let me draw out the structure for that. We would have a negative one formal charge on this oxygen, and this potassium would be a plus one charge."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So a tertiary alkyl halide is too sterically hindered to undergo an SN2 reaction, so immediately we know that that is out. And when we look at our reagent, this is potassium tert-butoxide. So let me draw out the structure for that. We would have a negative one formal charge on this oxygen, and this potassium would be a plus one charge. So potassium tert-butoxide is too sterically hindered to function as a nucleophile. So instantly we know that SN1 is out, and potassium tert-butoxide is a strong base, so we know that E1 is out. And that leaves an E2 mechanism for this reaction."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "We would have a negative one formal charge on this oxygen, and this potassium would be a plus one charge. So potassium tert-butoxide is too sterically hindered to function as a nucleophile. So instantly we know that SN1 is out, and potassium tert-butoxide is a strong base, so we know that E1 is out. And that leaves an E2 mechanism for this reaction. So let me draw the alkyl halide over here again. So let me put these in here. And we know that the carbon that's bonded to our halogen is our alpha carbon, and the beta carbons are bonded to that."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And that leaves an E2 mechanism for this reaction. So let me draw the alkyl halide over here again. So let me put these in here. And we know that the carbon that's bonded to our halogen is our alpha carbon, and the beta carbons are bonded to that. So this would be a beta carbon, this would be a beta carbon, and this would be a beta carbon. Let's say our sterically hindered strong base comes along and takes a proton from this beta carbon. So our base takes this proton, these electrons move into here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And we know that the carbon that's bonded to our halogen is our alpha carbon, and the beta carbons are bonded to that. So this would be a beta carbon, this would be a beta carbon, and this would be a beta carbon. Let's say our sterically hindered strong base comes along and takes a proton from this beta carbon. So our base takes this proton, these electrons move into here. At the same time, these electrons come off onto the bromine to form the bromide ion. And we will get one product for this reaction. So let me draw it in here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So our base takes this proton, these electrons move into here. At the same time, these electrons come off onto the bromine to form the bromide ion. And we will get one product for this reaction. So let me draw it in here. We would now have a double bond. And let me show those electrons. So the electrons in magenta moved into here to form our double bond."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So let me draw it in here. We would now have a double bond. And let me show those electrons. So the electrons in magenta moved into here to form our double bond. Now let's say we took a proton from a different beta carbon. So let me redraw, let me just redraw this. So I'll draw the alkyl halide in here like this."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So the electrons in magenta moved into here to form our double bond. Now let's say we took a proton from a different beta carbon. So let me redraw, let me just redraw this. So I'll draw the alkyl halide in here like this. And again, we know that this is the alpha carbon. And let's take a proton from this beta carbon now. So I'll draw in a proton, and just think about our base coming along."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So I'll draw the alkyl halide in here like this. And again, we know that this is the alpha carbon. And let's take a proton from this beta carbon now. So I'll draw in a proton, and just think about our base coming along. So our base coming along with a negative charge, taking this proton, these electrons move into here. At the same time, these electrons come off onto the bromine. So that's our E2 mechanism."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So I'll draw in a proton, and just think about our base coming along. So our base coming along with a negative charge, taking this proton, these electrons move into here. At the same time, these electrons come off onto the bromine. So that's our E2 mechanism. So I'll draw the resulting alkene. So there's a double bond in here this time. The electrons in magenta moved into here to form our double bond."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So that's our E2 mechanism. So I'll draw the resulting alkene. So there's a double bond in here this time. The electrons in magenta moved into here to form our double bond. When we look at our products, we know that the one on the right is actually more stable because the one on the right is a tri-substituted alkene, whereas the one on the left is only a di-substituted alkene. But when you're dealing with a sterically hindered base, like potassium tert-butoxide, it turns out that the di-substituted, the less substituted alkene is the major product. And that's just because our base is so sterically hindered."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "The electrons in magenta moved into here to form our double bond. When we look at our products, we know that the one on the right is actually more stable because the one on the right is a tri-substituted alkene, whereas the one on the left is only a di-substituted alkene. But when you're dealing with a sterically hindered base, like potassium tert-butoxide, it turns out that the di-substituted, the less substituted alkene is the major product. And that's just because our base is so sterically hindered. Also notice, if our base came along and took a proton from this beta carbon, we would get the same product as right here. Here's another tertiary substrate. So we know an SN2 mechanism is out."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And that's just because our base is so sterically hindered. Also notice, if our base came along and took a proton from this beta carbon, we would get the same product as right here. Here's another tertiary substrate. So we know an SN2 mechanism is out. And when we look at our reagent, which is sodium methoxide, sodium has a plus one formal charge, and the oxygen has a negative one formal charge. The methoxide ion, we saw in an earlier video, is a strong nucleophile and a strong base. And whenever you think strong base, think E2 reaction."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So we know an SN2 mechanism is out. And when we look at our reagent, which is sodium methoxide, sodium has a plus one formal charge, and the oxygen has a negative one formal charge. The methoxide ion, we saw in an earlier video, is a strong nucleophile and a strong base. And whenever you think strong base, think E2 reaction. So the E2 reaction is going to dominate here. And an E2 reaction means the methoxide ion is gonna function as a base and take a proton from our alkyl halide. So next we analyze our alkyl halide."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And whenever you think strong base, think E2 reaction. So the E2 reaction is going to dominate here. And an E2 reaction means the methoxide ion is gonna function as a base and take a proton from our alkyl halide. So next we analyze our alkyl halide. The carbon bonded to our halogen is our alpha carbon. And the carbons directly bonded to the alpha carbon are the beta carbons. Here the beta carbons are identical, so I'm gonna say that's our beta carbon."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So next we analyze our alkyl halide. The carbon bonded to our halogen is our alpha carbon. And the carbons directly bonded to the alpha carbon are the beta carbons. Here the beta carbons are identical, so I'm gonna say that's our beta carbon. And let's draw in one beta hydrogen. The methoxide ion comes along, so I'll draw that in. So we have a negative one formal charge on our oxygen."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Here the beta carbons are identical, so I'm gonna say that's our beta carbon. And let's draw in one beta hydrogen. The methoxide ion comes along, so I'll draw that in. So we have a negative one formal charge on our oxygen. This is going to take this proton. At the same time, these electrons move in here to form our double bond, and these electrons come off to form our iodide ion as our leaving group. So for our final product, we're gonna have a double bond here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So we have a negative one formal charge on our oxygen. This is going to take this proton. At the same time, these electrons move in here to form our double bond, and these electrons come off to form our iodide ion as our leaving group. So for our final product, we're gonna have a double bond here. And let's change colors again. The electrons in magenta moved into here to form our double bond. Since our beta carbons are all the same, this is the only product for this reaction."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So for our final product, we're gonna have a double bond here. And let's change colors again. The electrons in magenta moved into here to form our double bond. Since our beta carbons are all the same, this is the only product for this reaction. Here's the same tertiary alkyl halide we saw in the previous problem. So an SN2 reaction is out. And when we analyze our reagent, we know that water is a weak nucleophile and a weak base."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Since our beta carbons are all the same, this is the only product for this reaction. Here's the same tertiary alkyl halide we saw in the previous problem. So an SN2 reaction is out. And when we analyze our reagent, we know that water is a weak nucleophile and a weak base. And since water is a weak base, the E2 reaction is out. That leaves the E1 reaction and the SN1 reaction, which both proceed via a carbocation. So let's draw the carbocation that we would form."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And when we analyze our reagent, we know that water is a weak nucleophile and a weak base. And since water is a weak base, the E2 reaction is out. That leaves the E1 reaction and the SN1 reaction, which both proceed via a carbocation. So let's draw the carbocation that we would form. These electrons come off to form the iodide ion, and we take a bond away from the carbon in red to form our tertiary carbocation. So our tertiary carbocation has a plus one formal charge on our central carbon, so this carbon in red. If water acts as a weak base, water can take a proton from one of the carbons next door to the carbon in red."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So let's draw the carbocation that we would form. These electrons come off to form the iodide ion, and we take a bond away from the carbon in red to form our tertiary carbocation. So our tertiary carbocation has a plus one formal charge on our central carbon, so this carbon in red. If water acts as a weak base, water can take a proton from one of the carbons next door to the carbon in red. So I'm drawing a hydrogen on this carbon. And if water takes this proton, these electrons would move into here to form a double bond. So that's one of the possible products for this reaction."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "If water acts as a weak base, water can take a proton from one of the carbons next door to the carbon in red. So I'm drawing a hydrogen on this carbon. And if water takes this proton, these electrons would move into here to form a double bond. So that's one of the possible products for this reaction. So we form an alkene, and our electrons in magenta moved into here to form our double bond. So that's when water acts as a weak base, and that would be an E1 mechanism. But we also have the possibility of an SN1 mechanism."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So that's one of the possible products for this reaction. So we form an alkene, and our electrons in magenta moved into here to form our double bond. So that's when water acts as a weak base, and that would be an E1 mechanism. But we also have the possibility of an SN1 mechanism. So let's draw our carbocation again. So here's our carbocation. Our carbon in red is this one with a plus one formal charge."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "But we also have the possibility of an SN1 mechanism. So let's draw our carbocation again. So here's our carbocation. Our carbon in red is this one with a plus one formal charge. And this time, we're gonna show water acting as a nucleophile. So let's draw in our water molecule with two lone pairs of electrons on the oxygen. So the nucleophile attacks our electrophile, and we form a bond between the oxygen and the carbon in red."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Our carbon in red is this one with a plus one formal charge. And this time, we're gonna show water acting as a nucleophile. So let's draw in our water molecule with two lone pairs of electrons on the oxygen. So the nucleophile attacks our electrophile, and we form a bond between the oxygen and the carbon in red. So let's draw that in here. So now we have a bond between our oxygen and our carbon in red, which is this one. So a lone pair of electrons on our oxygen formed that bond in here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So the nucleophile attacks our electrophile, and we form a bond between the oxygen and the carbon in red. So let's draw that in here. So now we have a bond between our oxygen and our carbon in red, which is this one. So a lone pair of electrons on our oxygen formed that bond in here. And our oxygen is still bonded to two other hydrogens, and still has one lone pair of electrons on the oxygen, which gives the oxygen a plus one formal charge. And we would also need, in our next step, to have an acid-base reaction. Another molecule of water could come along and take one of these protons."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So a lone pair of electrons on our oxygen formed that bond in here. And our oxygen is still bonded to two other hydrogens, and still has one lone pair of electrons on the oxygen, which gives the oxygen a plus one formal charge. And we would also need, in our next step, to have an acid-base reaction. Another molecule of water could come along and take one of these protons. So in the interest of time and space, I'm just gonna write minus H plus here. We lose one of our protons, and we form our product, which would be this tertiary alcohol. So we have two possibilities for this reaction."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Another molecule of water could come along and take one of these protons. So in the interest of time and space, I'm just gonna write minus H plus here. We lose one of our protons, and we form our product, which would be this tertiary alcohol. So we have two possibilities for this reaction. We could have either an E1 reaction, which would give us this alkene, or an SN1 reaction, which would give us this alcohol. For our last example, let's look at this tertiary alcohol. We know SN2 is out, and we have a strong acid present, not a strong base, so an E2 reaction is out."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So we have two possibilities for this reaction. We could have either an E1 reaction, which would give us this alkene, or an SN1 reaction, which would give us this alcohol. For our last example, let's look at this tertiary alcohol. We know SN2 is out, and we have a strong acid present, not a strong base, so an E2 reaction is out. And whenever you see a tertiary alcohol with something like sulfuric acid or phosphoric acid in heat, think E1 reaction. So we need to form a carbocation here. But before we do that, let's analyze the structure of our alcohol."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "We know SN2 is out, and we have a strong acid present, not a strong base, so an E2 reaction is out. And whenever you see a tertiary alcohol with something like sulfuric acid or phosphoric acid in heat, think E1 reaction. So we need to form a carbocation here. But before we do that, let's analyze the structure of our alcohol. The carbon bonded to the OH is our alpha carbon, and the carbons directly bonded to that are our beta carbons. I'll call this one beta one, this one beta two, and this one beta three. We can't have loss of a leaving group right away because the hydroxide ion is a poor leaving group."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "But before we do that, let's analyze the structure of our alcohol. The carbon bonded to the OH is our alpha carbon, and the carbons directly bonded to that are our beta carbons. I'll call this one beta one, this one beta two, and this one beta three. We can't have loss of a leaving group right away because the hydroxide ion is a poor leaving group. But our sulfuric acid is a source of protons, so our first step is to protonate the alcohol. So the lone pair of electrons on the oxygen pick up this proton, and let's draw what we would form here. So the oxygen would now be bonded to two hydrogens."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "We can't have loss of a leaving group right away because the hydroxide ion is a poor leaving group. But our sulfuric acid is a source of protons, so our first step is to protonate the alcohol. So the lone pair of electrons on the oxygen pick up this proton, and let's draw what we would form here. So the oxygen would now be bonded to two hydrogens. Still has a lone pair of electrons, and a plus one formal charge on the oxygen. So these electrons here in magenta pick up this proton to form this bond. And now we are ready for loss of a leaving group."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So the oxygen would now be bonded to two hydrogens. Still has a lone pair of electrons, and a plus one formal charge on the oxygen. So these electrons here in magenta pick up this proton to form this bond. And now we are ready for loss of a leaving group. When these electrons come off onto the oxygen, they would make water, which we know is a stable molecule. It's a good leaving group. And now we're taking a bond away from this carbon in red."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And now we are ready for loss of a leaving group. When these electrons come off onto the oxygen, they would make water, which we know is a stable molecule. It's a good leaving group. And now we're taking a bond away from this carbon in red. So now we're gonna form a carbocation. So let's draw in our carbocation. So actually, let me make this a better methyl group here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "And now we're taking a bond away from this carbon in red. So now we're gonna form a carbocation. So let's draw in our carbocation. So actually, let me make this a better methyl group here. And our carbon in red is this one. So that gets a plus one formal charge. In our E1 mechanism, we know we're gonna have a weak base come along and take a proton from one of the beta carbons."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So actually, let me make this a better methyl group here. And our carbon in red is this one. So that gets a plus one formal charge. In our E1 mechanism, we know we're gonna have a weak base come along and take a proton from one of the beta carbons. So I'll go with beta three for this carbocation here. So I'll draw in a hydrogen coming off of this carbon. Let me mark it in blue here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "In our E1 mechanism, we know we're gonna have a weak base come along and take a proton from one of the beta carbons. So I'll go with beta three for this carbocation here. So I'll draw in a hydrogen coming off of this carbon. Let me mark it in blue here. This carbon in blue was this carbon on our starting alcohol. So a weak base would have to be water. We just formed water in our previous step."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Let me mark it in blue here. This carbon in blue was this carbon on our starting alcohol. So a weak base would have to be water. We just formed water in our previous step. So water comes along and takes that proton. So let's draw in our lone pairs of electrons. So we're gonna take this proton, and these electrons are gonna move in here to form our double bond."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "We just formed water in our previous step. So water comes along and takes that proton. So let's draw in our lone pairs of electrons. So we're gonna take this proton, and these electrons are gonna move in here to form our double bond. And that would, of course, get rid of the formal charge on the carbon in red and give us our product, which is an alkene. So let's draw that in here. So our electrons in magenta moved into here to form our double bond."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So we're gonna take this proton, and these electrons are gonna move in here to form our double bond. And that would, of course, get rid of the formal charge on the carbon in red and give us our product, which is an alkene. So let's draw that in here. So our electrons in magenta moved into here to form our double bond. And that's a tri-substituted alkene. What about if we took a proton from one of the other beta carbons? So going back over here to our picture of the tertiary alcohol, beta one and beta two would actually both give us the same product."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So our electrons in magenta moved into here to form our double bond. And that's a tri-substituted alkene. What about if we took a proton from one of the other beta carbons? So going back over here to our picture of the tertiary alcohol, beta one and beta two would actually both give us the same product. So we'll just take one of them. And first I'd have to redraw my tertiary carbocation. So here it is, plus one formal charge on the carbon in red."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So going back over here to our picture of the tertiary alcohol, beta one and beta two would actually both give us the same product. So we'll just take one of them. And first I'd have to redraw my tertiary carbocation. So here it is, plus one formal charge on the carbon in red. And I'll draw in a hydrogen on this carbon. So water is going to function as a base. Let me draw in our water molecule here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "So here it is, plus one formal charge on the carbon in red. And I'll draw in a hydrogen on this carbon. So water is going to function as a base. Let me draw in our water molecule here. And it's gonna take this proton and these electrons are gonna move in to form our double bond. So when we draw in our product, we're gonna have a double bond here. And our electrons in magenta would move in to here."}, {"video_title": "Elimination vs substitution tertiary substrate.mp3", "Sentence": "Let me draw in our water molecule here. And it's gonna take this proton and these electrons are gonna move in to form our double bond. So when we draw in our product, we're gonna have a double bond here. And our electrons in magenta would move in to here. This is only a di-substituted alkene, so this would be the minor product. And the major product would be the tri-substituted alkene, which is the more stable one. So this is the major product for our E1 reaction."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "At the center here, at the intersection of these lines, we have a carbon, and this carbon is a chirality center. There are four different groups attached to this carbon. There's a hydrogen, there's an OH, there's an aldehyde, and there's a CH2OH. So in this picture, you can see I've drawn in the carbon here, and on the right is a picture of the actual molecule. So this is our carbon, this is our chirality center. We're actually staring straight down at it. A horizontal line means a bond that's coming out of the page."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in this picture, you can see I've drawn in the carbon here, and on the right is a picture of the actual molecule. So this is our carbon, this is our chirality center. We're actually staring straight down at it. A horizontal line means a bond that's coming out of the page. So this line right here indicates a bond that's coming out of the page, so we would represent that with a wedge. So this hydrogen is coming out at us in space, and hopefully the picture, it's a little bit easier to see this hydrogen is actually going up. It's coming out at us in space."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "A horizontal line means a bond that's coming out of the page. So this line right here indicates a bond that's coming out of the page, so we would represent that with a wedge. So this hydrogen is coming out at us in space, and hopefully the picture, it's a little bit easier to see this hydrogen is actually going up. It's coming out at us in space. Same thing with this horizontal line right here to the OH. That means a bond that's coming out of the page. So we represent that with a wedge."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's coming out at us in space. Same thing with this horizontal line right here to the OH. That means a bond that's coming out of the page. So we represent that with a wedge. The OH here is coming out at us in space. A vertical line means a bond going away from us in space. So this line right here means a bond to an aldehyde that's going away from us, so that is represented by a dash, and in the picture, this bond here is going away from us."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we represent that with a wedge. The OH here is coming out at us in space. A vertical line means a bond going away from us in space. So this line right here means a bond to an aldehyde that's going away from us, so that is represented by a dash, and in the picture, this bond here is going away from us. It's going into the page. Same thing with this vertical line here to the CH2OH. That means going away from us in space."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this line right here means a bond to an aldehyde that's going away from us, so that is represented by a dash, and in the picture, this bond here is going away from us. It's going into the page. Same thing with this vertical line here to the CH2OH. That means going away from us in space. We draw a dash here. The bond is going into the page. So this line right here is showing a bond going away from us in space."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That means going away from us in space. We draw a dash here. The bond is going into the page. So this line right here is showing a bond going away from us in space. If we wanted to assign a configuration to our chirality center, there are several methods that you can use. I'll show you the two that I like to use. The first way that I like to do it is to think about the priority of these four groups."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this line right here is showing a bond going away from us in space. If we wanted to assign a configuration to our chirality center, there are several methods that you can use. I'll show you the two that I like to use. The first way that I like to do it is to think about the priority of these four groups. We know from earlier videos that hydrogen is gonna have the lowest priority, and we want the lowest priority group pointing away from us in space. So the only way to do that would be to put our eye right here and to stare at our chiral center this way. From that perspective, the hydrogen is going away from us in space."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The first way that I like to do it is to think about the priority of these four groups. We know from earlier videos that hydrogen is gonna have the lowest priority, and we want the lowest priority group pointing away from us in space. So the only way to do that would be to put our eye right here and to stare at our chiral center this way. From that perspective, the hydrogen is going away from us in space. So let me go to a video where it's much easier to visualize what's going on here. So here we are staring down at our chiral center, and you can see there's an OH coming out at us. There's a hydrogen coming out at us."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "From that perspective, the hydrogen is going away from us in space. So let me go to a video where it's much easier to visualize what's going on here. So here we are staring down at our chiral center, and you can see there's an OH coming out at us. There's a hydrogen coming out at us. There's an aldehyde going away from us in space, and a CH2OH going away from us. If we stare at our chiral center from this direction, let me go ahead and rotate the molecule, now we can see that there's an OH coming out at us in space, and a hydrogen going away from us, and then we have our aldehyde going down and to the right, and our CH2OH is going down and to the left. So let's draw what we saw in the video."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's a hydrogen coming out at us. There's an aldehyde going away from us in space, and a CH2OH going away from us. If we stare at our chiral center from this direction, let me go ahead and rotate the molecule, now we can see that there's an OH coming out at us in space, and a hydrogen going away from us, and then we have our aldehyde going down and to the right, and our CH2OH is going down and to the left. So let's draw what we saw in the video. So here's our picture, and we'll start with our chiral center. So right here, so let's draw in our carbon, and then our OH is coming out at us in space, so we put that on a wedge. So let me put an OH here."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw what we saw in the video. So here's our picture, and we'll start with our chiral center. So right here, so let's draw in our carbon, and then our OH is coming out at us in space, so we put that on a wedge. So let me put an OH here. The hydrogen is now going away from us in space, so now we would represent that with a dash. The aldehyde is going down and to the right with a bond that's in the plane of how we're viewing it anyway, and let's put the carbon double bond to an oxygen here, and then we have our CH2OH going down and to the left. So let's put in our, I'll go ahead and draw in the carbon with two hydrogens, and then our OH down here."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me put an OH here. The hydrogen is now going away from us in space, so now we would represent that with a dash. The aldehyde is going down and to the right with a bond that's in the plane of how we're viewing it anyway, and let's put the carbon double bond to an oxygen here, and then we have our CH2OH going down and to the left. So let's put in our, I'll go ahead and draw in the carbon with two hydrogens, and then our OH down here. So let's assign priority to our four groups. So here is our chiral center. We look at the atoms directly bonded to our chiral center, and that would be a hydrogen, an oxygen, a carbon, and a carbon."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in our, I'll go ahead and draw in the carbon with two hydrogens, and then our OH down here. So let's assign priority to our four groups. So here is our chiral center. We look at the atoms directly bonded to our chiral center, and that would be a hydrogen, an oxygen, a carbon, and a carbon. We know that oxygen has the highest atomic number out of those atoms, so the OH group gets a number one. Hydrogen has the lowest atomic number, so hydrogen gets the lowest priority, and we say that's group four. But we have a tie between our two carbons because carbon has the same atomic number."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We look at the atoms directly bonded to our chiral center, and that would be a hydrogen, an oxygen, a carbon, and a carbon. We know that oxygen has the highest atomic number out of those atoms, so the OH group gets a number one. Hydrogen has the lowest atomic number, so hydrogen gets the lowest priority, and we say that's group four. But we have a tie between our two carbons because carbon has the same atomic number. So to break the tie, we need to look at what those carbons are bonded to. The carbon on the left is directly bonded to an oxygen and two hydrogens. So we write down here oxygen, hydrogen, hydrogen, so in order of decreasing atomic number."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But we have a tie between our two carbons because carbon has the same atomic number. So to break the tie, we need to look at what those carbons are bonded to. The carbon on the left is directly bonded to an oxygen and two hydrogens. So we write down here oxygen, hydrogen, hydrogen, so in order of decreasing atomic number. This carbon on the right is double bonded to this oxygen, and we saw in an earlier video how to handle that. We treat that like a carbon bonded to two different oxygens, even though it's not really, it has a double bond to one, but this helps us when we are assigning priority, and this carbon is also bonded to a hydrogen, so this one. So that would be oxygen, oxygen, hydrogen."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we write down here oxygen, hydrogen, hydrogen, so in order of decreasing atomic number. This carbon on the right is double bonded to this oxygen, and we saw in an earlier video how to handle that. We treat that like a carbon bonded to two different oxygens, even though it's not really, it has a double bond to one, but this helps us when we are assigning priority, and this carbon is also bonded to a hydrogen, so this one. So that would be oxygen, oxygen, hydrogen. So we write oxygen, oxygen, hydrogen. Next we compare and look for the first point of difference. So this is an oxygen versus an oxygen, so that's a tie."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that would be oxygen, oxygen, hydrogen. So we write oxygen, oxygen, hydrogen. Next we compare and look for the first point of difference. So this is an oxygen versus an oxygen, so that's a tie. We go to the next atom, and we have an oxygen versus a hydrogen. Obviously oxygen wins, so this group wins. The aldehyde is higher priority than the CH2OH."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is an oxygen versus an oxygen, so that's a tie. We go to the next atom, and we have an oxygen versus a hydrogen. Obviously oxygen wins, so this group wins. The aldehyde is higher priority than the CH2OH. So the aldehyde must get a number two, and the CH2OH should get a number three. So for assigning R or S, we know that the hydrogen is going away from us in space, so we don't have to worry about that. We're done with step one and step two from the earlier videos."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The aldehyde is higher priority than the CH2OH. So the aldehyde must get a number two, and the CH2OH should get a number three. So for assigning R or S, we know that the hydrogen is going away from us in space, so we don't have to worry about that. We're done with step one and step two from the earlier videos. Next we go around in a circle from one to two to three. So we're going around from one to two to three, so we're going around this way, and that is clockwise, and we know that clockwise is R. So the configuration at our chirality center is R. I wanted to take a minute to show how to go from this drawing to this picture. So if the hydrogen is on this side, we want to put our eye on this side, so the hydrogen is going away from us, and stare down at our chiral center, so this carbon."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We're done with step one and step two from the earlier videos. Next we go around in a circle from one to two to three. So we're going around from one to two to three, so we're going around this way, and that is clockwise, and we know that clockwise is R. So the configuration at our chirality center is R. I wanted to take a minute to show how to go from this drawing to this picture. So if the hydrogen is on this side, we want to put our eye on this side, so the hydrogen is going away from us, and stare down at our chiral center, so this carbon. I like to imagine this carbon as being in the plane of the page. So here is our chirality center. Imagine a flat sheet of paper right here, and that sheet of paper is passing through your chiral center."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if the hydrogen is on this side, we want to put our eye on this side, so the hydrogen is going away from us, and stare down at our chiral center, so this carbon. I like to imagine this carbon as being in the plane of the page. So here is our chirality center. Imagine a flat sheet of paper right here, and that sheet of paper is passing through your chiral center. The OH we know is up in space. There's a wedge here, so when we're looking at it from this perspective, the OH should be up relative to that flat sheet of paper, and we can see it is. This is going up."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Imagine a flat sheet of paper right here, and that sheet of paper is passing through your chiral center. The OH we know is up in space. There's a wedge here, so when we're looking at it from this perspective, the OH should be up relative to that flat sheet of paper, and we can see it is. This is going up. The aldehyde here would be going down because this is a dash, and this would be your right side if your eye is right here. So the aldehyde is going down relative to the sheet of paper and it's to the right. So here's our aldehyde going down and to the right, and then this would be your left side."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is going up. The aldehyde here would be going down because this is a dash, and this would be your right side if your eye is right here. So the aldehyde is going down relative to the sheet of paper and it's to the right. So here's our aldehyde going down and to the right, and then this would be your left side. The CH2OH is also going down, but it'd be going down and to the left. So here we can see the CH2OH going down and to the left. And once you have this picture, it's easy to assign a configuration to your chiral center."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's our aldehyde going down and to the right, and then this would be your left side. The CH2OH is also going down, but it'd be going down and to the left. So here we can see the CH2OH going down and to the left. And once you have this picture, it's easy to assign a configuration to your chiral center. So that was the first method. The second method is, in my opinion, even easier, and this is the way that I usually use. We already know that the OH group gets the highest priority, so that's a number one."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And once you have this picture, it's easy to assign a configuration to your chiral center. So that was the first method. The second method is, in my opinion, even easier, and this is the way that I usually use. We already know that the OH group gets the highest priority, so that's a number one. The aldehyde got a number two. The CH2OH got a number three. And the hydrogen got a number four."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We already know that the OH group gets the highest priority, so that's a number one. The aldehyde got a number two. The CH2OH got a number three. And the hydrogen got a number four. So the trick I showed you in earlier videos is to ignore, ignore the fact that the hydrogen is actually coming out at you in space. We know that because this horizontal line here in a Fischer projection means a wedge. So just ignore the hydrogen."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the hydrogen got a number four. So the trick I showed you in earlier videos is to ignore, ignore the fact that the hydrogen is actually coming out at you in space. We know that because this horizontal line here in a Fischer projection means a wedge. So just ignore the hydrogen. Look at one, two, and three. And one to two to three is going around in this direction, which we know is counterclockwise. So it looks like it's S. So it looks like it's S for this chiral center."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So just ignore the hydrogen. Look at one, two, and three. And one to two to three is going around in this direction, which we know is counterclockwise. So it looks like it's S. So it looks like it's S for this chiral center. However, since the hydrogen is actually coming out at us in space, we saw in an earlier video, the trick is just to take the opposite of how it looks. So if it looks S, it's actually R. And this trick should always work when you're working with Fischer projections. So there are many ways to do this."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it looks like it's S. So it looks like it's S for this chiral center. However, since the hydrogen is actually coming out at us in space, we saw in an earlier video, the trick is just to take the opposite of how it looks. So if it looks S, it's actually R. And this trick should always work when you're working with Fischer projections. So there are many ways to do this. In my opinion, you should get a model set and figure out a method that works the best for you. Finally, let's draw the enantiomer of this compound. So the mirror method works the best when you're working with Fischer projections."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there are many ways to do this. In my opinion, you should get a model set and figure out a method that works the best for you. Finally, let's draw the enantiomer of this compound. So the mirror method works the best when you're working with Fischer projections. So on the left is a model of our compound. On the right is its mirror image. We can see that this OH is reflected in our mirror."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the mirror method works the best when you're working with Fischer projections. So on the left is a model of our compound. On the right is its mirror image. We can see that this OH is reflected in our mirror. So let's go down here. Let's draw a line to represent our mirror. And let's reflect this OH in our mirror."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can see that this OH is reflected in our mirror. So let's go down here. Let's draw a line to represent our mirror. And let's reflect this OH in our mirror. Then we need to draw a horizontal line right here, which represents these two bonds. And we have a hydrogen on the right side, so we draw in our hydrogen. Next, we have a vertical line like that, so let me put in the vertical line."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's reflect this OH in our mirror. Then we need to draw a horizontal line right here, which represents these two bonds. And we have a hydrogen on the right side, so we draw in our hydrogen. Next, we have a vertical line like that, so let me put in the vertical line. And then we have an aldehyde at the top, so I'll draw in our aldehyde. And finally, a CH2OH at the bottom here, so a CH2OH. Our starting compound had only one chiral center, so this one right here, and here's the chiral center in the enantiomer."}, {"video_title": "Fischer projection introduction Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next, we have a vertical line like that, so let me put in the vertical line. And then we have an aldehyde at the top, so I'll draw in our aldehyde. And finally, a CH2OH at the bottom here, so a CH2OH. Our starting compound had only one chiral center, so this one right here, and here's the chiral center in the enantiomer. We don't have any more chiral centers in our compounds, so you don't have to worry much about the aldehyde or the CH2OH when you're talking about reflecting them in the mirror. Make sure to get this switched. So if this OH is on the right, then it'd be on the left for the enantiomer."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with bromination. So here's a benzene ring. And to it, we're going to add some bromine. And our catalyst will be aluminum bromide. And you could have used FeBr3 instead of AlBr3. That's fine. And the end result is substitution of a bromine atom for an aromatic proton on your ring."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And our catalyst will be aluminum bromide. And you could have used FeBr3 instead of AlBr3. That's fine. And the end result is substitution of a bromine atom for an aromatic proton on your ring. Let's look at the mechanism for this electrophilic aromatic substitution reaction. And if I look at the aluminum bromide catalyst and I can see there are six electrons around the aluminum atom. So here's two and four and then a total of six."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the end result is substitution of a bromine atom for an aromatic proton on your ring. Let's look at the mechanism for this electrophilic aromatic substitution reaction. And if I look at the aluminum bromide catalyst and I can see there are six electrons around the aluminum atom. So here's two and four and then a total of six. And so because of aluminum's position on the periodic table, it can actually accept two more electrons. So the aluminum bromide is going to function as an electron pair acceptor, which is the definition for a Lewis acid. The bromine is going to function as a Lewis base."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So here's two and four and then a total of six. And so because of aluminum's position on the periodic table, it can actually accept two more electrons. So the aluminum bromide is going to function as an electron pair acceptor, which is the definition for a Lewis acid. The bromine is going to function as a Lewis base. It's going to be an electron pair donor. So we could think about this lone pair of electrons in here being donated to the aluminum and a bond forming between that bromine and that aluminum. So let's go ahead and draw the result of that Lewis acid base reaction."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The bromine is going to function as a Lewis base. It's going to be an electron pair donor. So we could think about this lone pair of electrons in here being donated to the aluminum and a bond forming between that bromine and that aluminum. So let's go ahead and draw the result of that Lewis acid base reaction. And so now this bromine is bonded to this aluminum. This aluminum is still bonded to these other bromines here. And I'm not going to draw in those lone pairs electrons around those bromines just to save some time."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the result of that Lewis acid base reaction. And so now this bromine is bonded to this aluminum. This aluminum is still bonded to these other bromines here. And I'm not going to draw in those lone pairs electrons around those bromines just to save some time. Let's follow those electrons. So the electrons in magenta, those are the ones that were donated to the aluminum and forming this bond between the bromine and the aluminum. That would give the aluminum a negative 1 formal charge."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I'm not going to draw in those lone pairs electrons around those bromines just to save some time. Let's follow those electrons. So the electrons in magenta, those are the ones that were donated to the aluminum and forming this bond between the bromine and the aluminum. That would give the aluminum a negative 1 formal charge. And this bromine would get a plus 1 formal charge like that. Now technically, this is the complex that's going to react with our benzene ring in our mechanism. But it's kind of hard to see the electrophile in this complex."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That would give the aluminum a negative 1 formal charge. And this bromine would get a plus 1 formal charge like that. Now technically, this is the complex that's going to react with our benzene ring in our mechanism. But it's kind of hard to see the electrophile in this complex. So let me just go ahead and show you what you can think about the electrophile being. And then we'll come back to this complex in the mechanism with benzene. So if these electrons in here moved off onto the bromine on the right, the bromine on the left will have lost a bond."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But it's kind of hard to see the electrophile in this complex. So let me just go ahead and show you what you can think about the electrophile being. And then we'll come back to this complex in the mechanism with benzene. So if these electrons in here moved off onto the bromine on the right, the bromine on the left will have lost a bond. So it would now only be surrounded by three lone pairs of electrons, giving it a plus 1 formal charge. And it simplifies things to think about this as being the electrophile in your mechanism for electrophilic aromatic substitution, even though technically it's going to be this top complex here that's going to react with our benzene ring. So if we form Br plus over here on the right, we would have this bromine now still bonded to this aluminum."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if these electrons in here moved off onto the bromine on the right, the bromine on the left will have lost a bond. So it would now only be surrounded by three lone pairs of electrons, giving it a plus 1 formal charge. And it simplifies things to think about this as being the electrophile in your mechanism for electrophilic aromatic substitution, even though technically it's going to be this top complex here that's going to react with our benzene ring. So if we form Br plus over here on the right, we would have this bromine now still bonded to this aluminum. And it would have three lone pairs of electrons around it now. So let me go ahead and highlight these electrons in here in red. I'm saying they're going to kick off onto that bromine there like that."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if we form Br plus over here on the right, we would have this bromine now still bonded to this aluminum. And it would have three lone pairs of electrons around it now. So let me go ahead and highlight these electrons in here in red. I'm saying they're going to kick off onto that bromine there like that. And the aluminum, of course, is still bonded to these three other bromines. And it still has a negative 1 formal charge like that. So we've generated our electrophile."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'm saying they're going to kick off onto that bromine there like that. And the aluminum, of course, is still bonded to these three other bromines. And it still has a negative 1 formal charge like that. So we've generated our electrophile. Or this is one way of simplifying it, to think about the Br plus as being our electrophile. And so that could react with our benzene ring. So we come back to our benzene ring here."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we've generated our electrophile. Or this is one way of simplifying it, to think about the Br plus as being our electrophile. And so that could react with our benzene ring. So we come back to our benzene ring here. And we think about Br plus as being the electrophile. And so remember, electrophile means loving electrons. And so it is attracted to electrons."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we come back to our benzene ring here. And we think about Br plus as being the electrophile. And so remember, electrophile means loving electrons. And so it is attracted to electrons. And of course, electrons are negatively charged. And they're attracted to positive things. So you could think about these pi electrons in your benzene ring as functioning as a nucleophile."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it is attracted to electrons. And of course, electrons are negatively charged. And they're attracted to positive things. So you could think about these pi electrons in your benzene ring as functioning as a nucleophile. And so we get a nucleophile attacking an electrophile here. And so let's go ahead and show the result of that nucleophilic attack. So we have our ring."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about these pi electrons in your benzene ring as functioning as a nucleophile. And so we get a nucleophile attacking an electrophile here. And so let's go ahead and show the result of that nucleophilic attack. So we have our ring. We have our pi electrons in our ring. And you could show the bromine adding to either one of these carbons. It doesn't really matter, since they are equivalent."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our ring. We have our pi electrons in our ring. And you could show the bromine adding to either one of these carbons. It doesn't really matter, since they are equivalent. I'm going to show the bromine adding to the top carbon there. So the top carbon already has a hydrogen on it. And we're going to say that these electrons in here add on to the bromine."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't really matter, since they are equivalent. I'm going to show the bromine adding to the top carbon there. So the top carbon already has a hydrogen on it. And we're going to say that these electrons in here add on to the bromine. So let me go ahead and highlight the electrons that we're talking about. So these pi electrons in here function as a nucleophile, form a bond with that bromine like that. We took a bond away from this carbon down here."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to say that these electrons in here add on to the bromine. So let me go ahead and highlight the electrons that we're talking about. So these pi electrons in here function as a nucleophile, form a bond with that bromine like that. We took a bond away from this carbon down here. So we're actually going to get a plus 1 formal charge at that carbon. So we make a carbocation. Now remember, technically, it's actually this complex up here that's reacting with benzene."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We took a bond away from this carbon down here. So we're actually going to get a plus 1 formal charge at that carbon. So we make a carbocation. Now remember, technically, it's actually this complex up here that's reacting with benzene. So we could think about a better mechanism or a more accurate mechanism as being these electrons in magenta going up to here, attacking that bromine, and then the electrons in red here, once again, kicking off onto this bromine, forming this complex over here. And so that's the more accurate way of thinking about it. For me, it just simplifies things to think about Br plus as being an electrophile."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now remember, technically, it's actually this complex up here that's reacting with benzene. So we could think about a better mechanism or a more accurate mechanism as being these electrons in magenta going up to here, attacking that bromine, and then the electrons in red here, once again, kicking off onto this bromine, forming this complex over here. And so that's the more accurate way of thinking about it. For me, it just simplifies things to think about Br plus as being an electrophile. So now that we formed a carbocation here, we can actually draw some resonance structures. And so if I took these pi electrons and moved them into here, let's go ahead and draw the resulting resonance structure for that. So I would have my ring."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "For me, it just simplifies things to think about Br plus as being an electrophile. So now that we formed a carbocation here, we can actually draw some resonance structures. And so if I took these pi electrons and moved them into here, let's go ahead and draw the resulting resonance structure for that. So I would have my ring. I would have these pi electrons. I would have a hydrogen and a bromine still attached to my ring. And I would move those pi electrons over to here."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I would have my ring. I would have these pi electrons. I would have a hydrogen and a bromine still attached to my ring. And I would move those pi electrons over to here. So let me go ahead and highlight those. So these pi electrons moved over to here, took a bond away from this carbon this time. And so that's the one that's going to get a plus 1 formal charge like that."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I would move those pi electrons over to here. So let me go ahead and highlight those. So these pi electrons moved over to here, took a bond away from this carbon this time. And so that's the one that's going to get a plus 1 formal charge like that. We could draw another resonance structure. So these pi electrons up here could move down there. And let's go ahead and show the result of that."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's the one that's going to get a plus 1 formal charge like that. We could draw another resonance structure. So these pi electrons up here could move down there. And let's go ahead and show the result of that. So once again, we have our ring. We have a hydrogen, a bromine. These pi electrons are still there."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and show the result of that. So once again, we have our ring. We have a hydrogen, a bromine. These pi electrons are still there. And we get some pi electrons moving over to here. So let me highlight those now. So these pi electrons right here move over to here, took a bond away from this top carbon."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "These pi electrons are still there. And we get some pi electrons moving over to here. So let me highlight those now. So these pi electrons right here move over to here, took a bond away from this top carbon. And so that top carbon is going to get a plus 1 formal charge. And so we have three resonance structures that we could draw for this carbocation. And remember, the actual ion is a hybrid of our three resonance structures."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons right here move over to here, took a bond away from this top carbon. And so that top carbon is going to get a plus 1 formal charge. And so we have three resonance structures that we could draw for this carbocation. And remember, the actual ion is a hybrid of our three resonance structures. And we could think about that hybrid as being a sigma complex. So in electrophilic aromatic substitution, the last step of the mechanism is deprotonation of your sigma complex to reform your aromatic ring. And so we could think about this complex right here."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And remember, the actual ion is a hybrid of our three resonance structures. And we could think about that hybrid as being a sigma complex. So in electrophilic aromatic substitution, the last step of the mechanism is deprotonation of your sigma complex to reform your aromatic ring. And so we could think about this complex right here. It's going to function as a base. And I'm going to say that these electrons in here could take this proton. And then these electrons would move into here to reform our aromatic ring."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we could think about this complex right here. It's going to function as a base. And I'm going to say that these electrons in here could take this proton. And then these electrons would move into here to reform our aromatic ring. So let me go ahead and draw the product, which is bromobenzene. And since we have a lot of arrows going on there, let me go ahead and highlight some electrons so we can follow them. So let me go ahead and make these electrons in here blue."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons would move into here to reform our aromatic ring. So let me go ahead and draw the product, which is bromobenzene. And since we have a lot of arrows going on there, let me go ahead and highlight some electrons so we can follow them. So let me go ahead and make these electrons in here blue. So these electrons are going to move in here to reform our aromatic ring. And these electrons in here, you could think about this complex as functioning as a base. And so those electrons pick up that proton."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and make these electrons in here blue. So these electrons are going to move in here to reform our aromatic ring. And these electrons in here, you could think about this complex as functioning as a base. And so those electrons pick up that proton. And so you'd form HBr as another product. So we'd have HBr here. And I'll highlight those electrons in green like that."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so those electrons pick up that proton. And so you'd form HBr as another product. So we'd have HBr here. And I'll highlight those electrons in green like that. And then, of course, we would also reform our catalyst. So AlBl3, our catalyst, has been reformed in this reaction. And so that's the mechanism for the bromination of benzene."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I'll highlight those electrons in green like that. And then, of course, we would also reform our catalyst. So AlBl3, our catalyst, has been reformed in this reaction. And so that's the mechanism for the bromination of benzene. If you wanted to think about adding other halogens onto your benzene ring, let's go ahead and look at chlorination here. So if we started with a benzene ring and we wanted to put a chlorine on our benzene ring, we would add some Cl2. And our catalyst would be, you could use AlCl3, aluminum chloride, or you could use FeCl3."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's the mechanism for the bromination of benzene. If you wanted to think about adding other halogens onto your benzene ring, let's go ahead and look at chlorination here. So if we started with a benzene ring and we wanted to put a chlorine on our benzene ring, we would add some Cl2. And our catalyst would be, you could use AlCl3, aluminum chloride, or you could use FeCl3. It doesn't really matter which catalyst that you use here. The end result, of course, would be to substitute in a chlorine for one of the protons, one of the aromatic protons on your ring here. And so our product would be, let me see if I can draw this in here."}, {"video_title": "Halogenation Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And our catalyst would be, you could use AlCl3, aluminum chloride, or you could use FeCl3. It doesn't really matter which catalyst that you use here. The end result, of course, would be to substitute in a chlorine for one of the protons, one of the aromatic protons on your ring here. And so our product would be, let me see if I can draw this in here. So we would have our benzene ring. And we would have a chlorine on our benzene ring to form chlorobenzene. And if it helps you to think about HCl as being another product formed in this reaction, it is."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "If we turn it to the side, we can see this is not a planar molecule. This is called the chair conformation of cyclohexane. And if we stare down these two carbons, we'll be able to see the chair conformation from a Newman projection viewpoint. So now you can see that we have staggered hydrogens. Here we have a picture of the chair conformation from the video, and the reason why we call it a chair conformation is if you tilt it on its side a little bit, it looks a little bit like a chair. So it has these three parts to it, essentially. You have a place for your back and your head, so let's say that's this area right here."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So now you can see that we have staggered hydrogens. Here we have a picture of the chair conformation from the video, and the reason why we call it a chair conformation is if you tilt it on its side a little bit, it looks a little bit like a chair. So it has these three parts to it, essentially. You have a place for your back and your head, so let's say that's this area right here. There is a seat right here, and then finally there's a footrest. So because it looks a little bit like a chair, that's why we call this the chair conformation. Here's how to draw the chair conformation, and let's compare this drawing to the picture."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "You have a place for your back and your head, so let's say that's this area right here. There is a seat right here, and then finally there's a footrest. So because it looks a little bit like a chair, that's why we call this the chair conformation. Here's how to draw the chair conformation, and let's compare this drawing to the picture. So on the picture, we'll start with this carbon right here, which is actually carbon one, and that's this carbon here. And this carbon has two hydrogens, so there's the two hydrogens. Next we have this bond in here, which is represented by this bond, which takes us to carbon two, and carbon two also has two hydrogens."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Here's how to draw the chair conformation, and let's compare this drawing to the picture. So on the picture, we'll start with this carbon right here, which is actually carbon one, and that's this carbon here. And this carbon has two hydrogens, so there's the two hydrogens. Next we have this bond in here, which is represented by this bond, which takes us to carbon two, and carbon two also has two hydrogens. Of course, every carbon in cyclohexane has two hydrogens on it. Next we go to this bond right here, which is represented right here on our chair conformation, and we draw in our two hydrogens. Our next bond goes down a little bit in this direction, so that's this bond, and then we get to this carbon, and we put in our two hydrogens."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next we have this bond in here, which is represented by this bond, which takes us to carbon two, and carbon two also has two hydrogens. Of course, every carbon in cyclohexane has two hydrogens on it. Next we go to this bond right here, which is represented right here on our chair conformation, and we draw in our two hydrogens. Our next bond goes down a little bit in this direction, so that's this bond, and then we get to this carbon, and we put in our two hydrogens. Our next bond goes back like this, and so here's this bond, and then we'll put in our two hydrogens on this carbon. And then this bond is back here, if you can see it on the picture, and that's this bond that goes behind this front hydrogen here. And then this carbon has two hydrogens, so we'll draw those in."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Our next bond goes down a little bit in this direction, so that's this bond, and then we get to this carbon, and we put in our two hydrogens. Our next bond goes back like this, and so here's this bond, and then we'll put in our two hydrogens on this carbon. And then this bond is back here, if you can see it on the picture, and that's this bond that goes behind this front hydrogen here. And then this carbon has two hydrogens, so we'll draw those in. And then finally, this last bond here goes up a little bit in space, so this one goes up like that, and we're back to carbon one. So now let's look at the Newman projection. So this is what we saw in the video."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then this carbon has two hydrogens, so we'll draw those in. And then finally, this last bond here goes up a little bit in space, so this one goes up like that, and we're back to carbon one. So now let's look at the Newman projection. So this is what we saw in the video. Let's go ahead and draw that, and we'll start with this carbon right here. So we will represent that with a point, and we can see there's a hydrogen going straight up, so let's draw in our hydrogen going straight up. We have a hydrogen going down and to the left, so a hydrogen going down and to the left, and then we have down and to the right, we have a CH two."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this is what we saw in the video. Let's go ahead and draw that, and we'll start with this carbon right here. So we will represent that with a point, and we can see there's a hydrogen going straight up, so let's draw in our hydrogen going straight up. We have a hydrogen going down and to the left, so a hydrogen going down and to the left, and then we have down and to the right, we have a CH two. So I won't draw in CH two, we just know that's there. For the back carbon, you can actually see a little bit of the back carbon in the picture, just because of the perspective, but the front carbon should be completely eclipsing the back carbon, so we represent that with our circle. And then what's coming off of the back carbon?"}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We have a hydrogen going down and to the left, so a hydrogen going down and to the left, and then we have down and to the right, we have a CH two. So I won't draw in CH two, we just know that's there. For the back carbon, you can actually see a little bit of the back carbon in the picture, just because of the perspective, but the front carbon should be completely eclipsing the back carbon, so we represent that with our circle. And then what's coming off of the back carbon? Well, there's this hydrogen, so we'll draw in that hydrogen like that, so going up and to the left. There's a hydrogen going down, so we'll draw that hydrogen in here. And then we have, going to the right and up, we have a CH two."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then what's coming off of the back carbon? Well, there's this hydrogen, so we'll draw in that hydrogen like that, so going up and to the left. There's a hydrogen going down, so we'll draw that hydrogen in here. And then we have, going to the right and up, we have a CH two. So going to the right and up is a CH two group. Move over to this carbon over here on the right, so we represent that with a point, so we'll draw that in. Hydrogen going straight up on that carbon, so we'll draw in that hydrogen."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we have, going to the right and up, we have a CH two. So going to the right and up is a CH two group. Move over to this carbon over here on the right, so we represent that with a point, so we'll draw that in. Hydrogen going straight up on that carbon, so we'll draw in that hydrogen. We have a hydrogen going down and to the right, so we draw in that one. And then we have this bond connecting to that CH two group, so we can draw that in here like that. And then we have our back carbon, which we can't see in the picture, but we know there's a back carbon here, and bonding to the back carbon is a hydrogen going up and to the right, so we draw that one in, a hydrogen going straight down, so we draw that one in."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Hydrogen going straight up on that carbon, so we'll draw in that hydrogen. We have a hydrogen going down and to the right, so we draw in that one. And then we have this bond connecting to that CH two group, so we can draw that in here like that. And then we have our back carbon, which we can't see in the picture, but we know there's a back carbon here, and bonding to the back carbon is a hydrogen going up and to the right, so we draw that one in, a hydrogen going straight down, so we draw that one in. And then finally we have, let me go ahead and correct that one a little bit, so we have a hydrogen going straight down. And then finally we have a bond to the back carbon going to this CH two group, so we can draw that in here like this. So now we have a Newman projection for the chair conformation."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our back carbon, which we can't see in the picture, but we know there's a back carbon here, and bonding to the back carbon is a hydrogen going up and to the right, so we draw that one in, a hydrogen going straight down, so we draw that one in. And then finally we have, let me go ahead and correct that one a little bit, so we have a hydrogen going straight down. And then finally we have a bond to the back carbon going to this CH two group, so we can draw that in here like this. So now we have a Newman projection for the chair conformation. And the nice thing about the Newman projection is it shows you your hydrogens are all staggered here, so we have staggered hydrogens, so we don't have any torsional strain to worry about. And with a chair conformation, the bond angles are pretty close to the ideal bond angle of 109.5 degrees, so the carbon-carbon-carbon bond angles are approximately 111 degrees, so we don't have to worry about any angle strain. And we don't have to worry about any torsional strain."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So now we have a Newman projection for the chair conformation. And the nice thing about the Newman projection is it shows you your hydrogens are all staggered here, so we have staggered hydrogens, so we don't have any torsional strain to worry about. And with a chair conformation, the bond angles are pretty close to the ideal bond angle of 109.5 degrees, so the carbon-carbon-carbon bond angles are approximately 111 degrees, so we don't have to worry about any angle strain. And we don't have to worry about any torsional strain. So the chair conformation is the most stable conformation for cyclohexane. Next we'll take a look at the boat conformation. Here we have the boat conformation of cyclohexane."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And we don't have to worry about any torsional strain. So the chair conformation is the most stable conformation for cyclohexane. Next we'll take a look at the boat conformation. Here we have the boat conformation of cyclohexane. If you look at the carbons, it looks a little bit like a boat. There are a few things that destabilize the boat conformation, and one of them is these top hydrogens here. So they're close enough to get in the way of each other, and that's called flagpole interaction, and that increases the strain."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Here we have the boat conformation of cyclohexane. If you look at the carbons, it looks a little bit like a boat. There are a few things that destabilize the boat conformation, and one of them is these top hydrogens here. So they're close enough to get in the way of each other, and that's called flagpole interaction, and that increases the strain. If we look down at these two carbons, the one in the back and the one in front here, we'll see the boat conformation from a Newman projection viewpoint. So now we're looking at it from a Newman projection perspective. Now we can see there's another source of strain."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So they're close enough to get in the way of each other, and that's called flagpole interaction, and that increases the strain. If we look down at these two carbons, the one in the back and the one in front here, we'll see the boat conformation from a Newman projection viewpoint. So now we're looking at it from a Newman projection perspective. Now we can see there's another source of strain. There's torsional strain. For example, we have these front hydrogens here eclipsing these other hydrogens, and there's lots of examples of torsional strain destabilizing the boat conformation. So here's the boat conformation from the video, and let's analyze that and compare it to the drawing up here."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now we can see there's another source of strain. There's torsional strain. For example, we have these front hydrogens here eclipsing these other hydrogens, and there's lots of examples of torsional strain destabilizing the boat conformation. So here's the boat conformation from the video, and let's analyze that and compare it to the drawing up here. So this bond right here is this bond. I made it much longer on the drawing just to make it easier to see the hydrogens. So this carbon right here would be this one, and we'll draw in those two hydrogens, and then we have a bond going up to this carbon."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So here's the boat conformation from the video, and let's analyze that and compare it to the drawing up here. So this bond right here is this bond. I made it much longer on the drawing just to make it easier to see the hydrogens. So this carbon right here would be this one, and we'll draw in those two hydrogens, and then we have a bond going up to this carbon. So that's our bond going up right here, and then we have two hydrogens bonded to that carbon, and then this bond goes down and back to our carbon in the back here, this one, and then there are two hydrogens on this carbon, so here and here. Kind of hard to see those hydrogens on this carbon in the picture, but there's one, and then the other one, there's the tip of the other one sticking out. And then we have this bond back here, which is this one, and we'll draw in our hydrogens here as well, and then we go back up to this carbon."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here would be this one, and we'll draw in those two hydrogens, and then we have a bond going up to this carbon. So that's our bond going up right here, and then we have two hydrogens bonded to that carbon, and then this bond goes down and back to our carbon in the back here, this one, and then there are two hydrogens on this carbon, so here and here. Kind of hard to see those hydrogens on this carbon in the picture, but there's one, and then the other one, there's the tip of the other one sticking out. And then we have this bond back here, which is this one, and we'll draw in our hydrogens here as well, and then we go back up to this carbon. So we're going up to this carbon, we'll draw in our two hydrogens on that carbon, and then we go back down to here, and that's this carbon on the drawing, we'll put in our two hydrogens. So we can see the flagpole interaction, so when this hydrogen gets too close to this hydrogen, when they get in the way of each other, that increases our strain. And we also have this little boat shape here, so hopefully you can see, let me change colors here, so you can see how this looks kind of like a boat, if you just look at your carbons."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we have this bond back here, which is this one, and we'll draw in our hydrogens here as well, and then we go back up to this carbon. So we're going up to this carbon, we'll draw in our two hydrogens on that carbon, and then we go back down to here, and that's this carbon on the drawing, we'll put in our two hydrogens. So we can see the flagpole interaction, so when this hydrogen gets too close to this hydrogen, when they get in the way of each other, that increases our strain. And we also have this little boat shape here, so hopefully you can see, let me change colors here, so you can see how this looks kind of like a boat, if you just look at your carbons. So we have a boat like that, and because this is in a boat conformation, there's not really much in the way of angle strain, but we know there's some torsional strain, and the best way to see that is with our Newman projection. So next, look over here for our Newman projection, and let's draw what we see. We'll start with this carbon, so that's represented by a point, and then we have a hydrogen going up and to the left, that's this one right here, and then we have a hydrogen going down, so a hydrogen going down like that, and then up and to the right, we have a bond to a CH2, so up and to the right is our bond to a CH2, let me make that a little bit longer."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And we also have this little boat shape here, so hopefully you can see, let me change colors here, so you can see how this looks kind of like a boat, if you just look at your carbons. So we have a boat like that, and because this is in a boat conformation, there's not really much in the way of angle strain, but we know there's some torsional strain, and the best way to see that is with our Newman projection. So next, look over here for our Newman projection, and let's draw what we see. We'll start with this carbon, so that's represented by a point, and then we have a hydrogen going up and to the left, that's this one right here, and then we have a hydrogen going down, so a hydrogen going down like that, and then up and to the right, we have a bond to a CH2, so up and to the right is our bond to a CH2, let me make that a little bit longer. And then there's a back carbon, so I draw in a circle to represent the back carbon, and you can see the hydrogens in the back carbon are eclipsed by the hydrogens in the front carbon, so this hydrogen right here is pretty much eclipsed. I'll draw it a little bit out so we can still see it, and then this one is a little bit down, so we'll draw that one like this, and then the bond in the back, it's very hard to see, but we know there's a bond back there going to another CH2 group, so I'll draw that back here like this. Next, we look at this carbon, so we represent that with a point, and we're not really looking straight down this axis like we were on this situation, so I'll just treat the right the same way that we treated the left side."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We'll start with this carbon, so that's represented by a point, and then we have a hydrogen going up and to the left, that's this one right here, and then we have a hydrogen going down, so a hydrogen going down like that, and then up and to the right, we have a bond to a CH2, so up and to the right is our bond to a CH2, let me make that a little bit longer. And then there's a back carbon, so I draw in a circle to represent the back carbon, and you can see the hydrogens in the back carbon are eclipsed by the hydrogens in the front carbon, so this hydrogen right here is pretty much eclipsed. I'll draw it a little bit out so we can still see it, and then this one is a little bit down, so we'll draw that one like this, and then the bond in the back, it's very hard to see, but we know there's a bond back there going to another CH2 group, so I'll draw that back here like this. Next, we look at this carbon, so we represent that with a point, and we're not really looking straight down this axis like we were on this situation, so I'll just treat the right the same way that we treated the left side. So we'd have a hydrogen going up and to the right, we would have a hydrogen going down like this, and then we would have this bond going to this CH2 group, so we have to show this bond meeting up with this one right here, so there's a CH2 right here. And then the back carbon would be a circle, so we have a hydrogen going up and to the right, we have a hydrogen going down, and then finally we would have a bond from that back carbon going to the other CH2, so going to this other CH2 here. So this allows us to see all of the eclipsed hydrogens and all of the torsional strain."}, {"video_title": "Conformations of cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next, we look at this carbon, so we represent that with a point, and we're not really looking straight down this axis like we were on this situation, so I'll just treat the right the same way that we treated the left side. So we'd have a hydrogen going up and to the right, we would have a hydrogen going down like this, and then we would have this bond going to this CH2 group, so we have to show this bond meeting up with this one right here, so there's a CH2 right here. And then the back carbon would be a circle, so we have a hydrogen going up and to the right, we have a hydrogen going down, and then finally we would have a bond from that back carbon going to the other CH2, so going to this other CH2 here. So this allows us to see all of the eclipsed hydrogens and all of the torsional strain. So the bow conformation is much higher in energy compared to the chair conformation. The chair conformation is the lowest in energy, and there are actually other conformations of cyclohexane, so the boat conformation can actually twist a little bit to give you twist boat. There's also like half chair, but we're not really too concerned with those other conformations, and actually we're gonna focus on the chair conformation in future videos because cyclohexane spends most of its time in the chair conformation."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that ethane donates a proton, and let's make it this proton right here. So the electrons in this bond, the electrons in magenta, are left behind on that carbon to form the conjugate base. So here are the electrons in magenta, and this carbon is sp3 hybridized, which means that the electrons in magenta occupy an sp3 hybrid orbital. So that's meant to represent an sp3 hybrid orbital. We know from the videos on hybridization that an sp3 hybridized orbital has 25% s character and 75% p character. So I'm just gonna write down here 25% s character. Let's move on to ethene or ethylene."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So that's meant to represent an sp3 hybrid orbital. We know from the videos on hybridization that an sp3 hybridized orbital has 25% s character and 75% p character. So I'm just gonna write down here 25% s character. Let's move on to ethene or ethylene. This carbon is sp2 hybridized. So we know that this carbon in ethene is sp2 hybridized. If ethene donates this proton, the electrons in magenta are left behind."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Let's move on to ethene or ethylene. This carbon is sp2 hybridized. So we know that this carbon in ethene is sp2 hybridized. If ethene donates this proton, the electrons in magenta are left behind. So here are the electrons in magenta. This is the conjugate base to ethene. And this carbon is sp2 hybridized."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "If ethene donates this proton, the electrons in magenta are left behind. So here are the electrons in magenta. This is the conjugate base to ethene. And this carbon is sp2 hybridized. So the lone pair of electrons, the electrons in magenta, occupy an sp2 hybrid orbital. So that's supposed to represent an sp2 hybrid orbital. An sp2 hybridized orbital has approximately 33% s character."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon is sp2 hybridized. So the lone pair of electrons, the electrons in magenta, occupy an sp2 hybrid orbital. So that's supposed to represent an sp2 hybrid orbital. An sp2 hybridized orbital has approximately 33% s character. So I'm gonna write down here 33% s character. Finally, we have acetylene. This carbon in acetylene is sp hybridized."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "An sp2 hybridized orbital has approximately 33% s character. So I'm gonna write down here 33% s character. Finally, we have acetylene. This carbon in acetylene is sp hybridized. So if acetylene donates a proton, if acetylene donates this proton, then these electrons are left behind. So the electrons in magenta are these electrons. And this carbon is sp hybridized."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "This carbon in acetylene is sp hybridized. So if acetylene donates a proton, if acetylene donates this proton, then these electrons are left behind. So the electrons in magenta are these electrons. And this carbon is sp hybridized. So the electrons in magenta occupy an sp hybrid orbital. An sp hybrid orbital is 50% s character. So this is 50% s character."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon is sp hybridized. So the electrons in magenta occupy an sp hybrid orbital. An sp hybrid orbital is 50% s character. So this is 50% s character. Now let's look at pKa values. So the pKa for this proton on ethane is approximately 50. The pKa value for this proton on ethane is approximately 44."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So this is 50% s character. Now let's look at pKa values. So the pKa for this proton on ethane is approximately 50. The pKa value for this proton on ethane is approximately 44. And the pKa value for this proton on acetylene is about 25. We know the lower the pKa value, the stronger the acid. So as we move to the right, we see a decrease in pKa values."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "The pKa value for this proton on ethane is approximately 44. And the pKa value for this proton on acetylene is about 25. We know the lower the pKa value, the stronger the acid. So as we move to the right, we see a decrease in pKa values. And therefore, that's an increase in the acidity. So we're talking about increase in the acid strength. So acetylene is the strongest acid out of these three."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So as we move to the right, we see a decrease in pKa values. And therefore, that's an increase in the acidity. So we're talking about increase in the acid strength. So acetylene is the strongest acid out of these three. If acetylene is the strongest acid, that must mean it has the most stable conjugate base. So this conjugate base here to acetylene must be the most stable out of these three. So as we move to the right, as we move to the right, we are increasing in stability."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So acetylene is the strongest acid out of these three. If acetylene is the strongest acid, that must mean it has the most stable conjugate base. So this conjugate base here to acetylene must be the most stable out of these three. So as we move to the right, as we move to the right, we are increasing in stability. So increasing in the stability of the conjugate base. So how do we explain the increased stability of the conjugate base in terms of hybridization? Well, let's look at the hybrid orbitals that we were talking about here."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So as we move to the right, as we move to the right, we are increasing in stability. So increasing in the stability of the conjugate base. So how do we explain the increased stability of the conjugate base in terms of hybridization? Well, let's look at the hybrid orbitals that we were talking about here. For the first conjugate base, our lone pair of electrons occupied an sp3 hybridized orbital, and that was 25% s character. And as we went to the right for our conjugate bases, we increased in s character to 33% to 50%. So as we move to the right, we increase in stability."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "Well, let's look at the hybrid orbitals that we were talking about here. For the first conjugate base, our lone pair of electrons occupied an sp3 hybridized orbital, and that was 25% s character. And as we went to the right for our conjugate bases, we increased in s character to 33% to 50%. So as we move to the right, we increase in stability. We also increase in s character. So increasing in s character increases the stability of the conjugate base. And we can explain that by thinking about s and p orbitals."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So as we move to the right, we increase in stability. We also increase in s character. So increasing in s character increases the stability of the conjugate base. And we can explain that by thinking about s and p orbitals. On average, an s orbital has electron density closer to the nucleus than a p orbital. So as you increase in s character, you're increasing in electron density closest to the nucleus. So let me go ahead and point out what I mean here."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And we can explain that by thinking about s and p orbitals. On average, an s orbital has electron density closer to the nucleus than a p orbital. So as you increase in s character, you're increasing in electron density closest to the nucleus. So let me go ahead and point out what I mean here. So let's look at this lone pair of electrons in the conjugate base to ethane. We think about the distance of those electrons to the nucleus. An sp3 hybridized orbital has the smallest amount of s character."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and point out what I mean here. So let's look at this lone pair of electrons in the conjugate base to ethane. We think about the distance of those electrons to the nucleus. An sp3 hybridized orbital has the smallest amount of s character. Therefore, those electrons are, on average, further away from the nucleus. That's less stable, that's higher in energy. As we move to the right, we can see that that distance decreases."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "An sp3 hybridized orbital has the smallest amount of s character. Therefore, those electrons are, on average, further away from the nucleus. That's less stable, that's higher in energy. As we move to the right, we can see that that distance decreases. So the distance decreases. And finally, for an sp hybridized orbital, that's the shortest distance between that lone pair of electrons and the positively charged nucleus. If you decrease the distance between the positively charged nucleus and the electrons, that means you increase the force of attraction."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "As we move to the right, we can see that that distance decreases. So the distance decreases. And finally, for an sp hybridized orbital, that's the shortest distance between that lone pair of electrons and the positively charged nucleus. If you decrease the distance between the positively charged nucleus and the electrons, that means you increase the force of attraction. So this conjugate base is the most stable because there's a greater attraction to the nucleus for those electrons. So the nucleus is better able to hold onto those electrons, there's a greater force. And that means increased stability or lower energy."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "If you decrease the distance between the positively charged nucleus and the electrons, that means you increase the force of attraction. So this conjugate base is the most stable because there's a greater attraction to the nucleus for those electrons. So the nucleus is better able to hold onto those electrons, there's a greater force. And that means increased stability or lower energy. So this is the most stable conjugate base. If that's the most stable conjugate base, then acetylene is the most acidic compound out of those three. So this also has an effect on electronegativity."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "And that means increased stability or lower energy. So this is the most stable conjugate base. If that's the most stable conjugate base, then acetylene is the most acidic compound out of those three. So this also has an effect on electronegativity. If an sp hybridized carbon is better able to attract electrons, well think about our definition for electronegativity. It's the power of an atom to attract electrons to itself. So since the electrons are closer to the positively charged nucleus than an sp hybridized carbon, so that must mean that sp hybridized carbons are more electronegative."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So this also has an effect on electronegativity. If an sp hybridized carbon is better able to attract electrons, well think about our definition for electronegativity. It's the power of an atom to attract electrons to itself. So since the electrons are closer to the positively charged nucleus than an sp hybridized carbon, so that must mean that sp hybridized carbons are more electronegative. So an sp hybridized carbon is more electronegative than an sp2 hybridized carbon. And an sp2 hybridized carbon is more electronegative than an sp3 hybridized carbon. So it has to do with the amount of s character."}, {"video_title": "Stabilization of a conjugate base hybridization Organic chemistry Khan Academy.mp3", "Sentence": "So since the electrons are closer to the positively charged nucleus than an sp hybridized carbon, so that must mean that sp hybridized carbons are more electronegative. So an sp hybridized carbon is more electronegative than an sp2 hybridized carbon. And an sp2 hybridized carbon is more electronegative than an sp3 hybridized carbon. So it has to do with the amount of s character. And that might seem weird because so far we've said that carbon has a certain value for the electronegativity and we've always assumed that it's the same. But now we can see that it's different, right? An sp hybridized carbon is actually the most electronegative."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "And you're also going to make water in this process. It's important to note that the oxygen and the R prime group come from your alcohol. And so we'll see that in the mechanism. Also, this reaction is at equilibrium. So if you want to make more of your ester product, you have to shift the equilibrium to the right. And so there are a couple ways to do that. One thing you could do would be to decrease the concentration of water, which would shift your equilibrium to the right to make more of your ester."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "Also, this reaction is at equilibrium. So if you want to make more of your ester product, you have to shift the equilibrium to the right. And so there are a couple ways to do that. One thing you could do would be to decrease the concentration of water, which would shift your equilibrium to the right to make more of your ester. You could also do something like increase the concentration of the alcohol, and that would shift the equilibrium to the right as well. Let's take a look at the mechanism to form esters. So we start with our carboxylic acid, and we're going to protonate the oxygen."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "One thing you could do would be to decrease the concentration of water, which would shift your equilibrium to the right to make more of your ester. You could also do something like increase the concentration of the alcohol, and that would shift the equilibrium to the right as well. Let's take a look at the mechanism to form esters. So we start with our carboxylic acid, and we're going to protonate the oxygen. So a lone pair of electrons in the oxygen can pick up a proton, leave these electrons behind. And so we can show that we have now protonated the oxygen right here, so it gets a plus one formal charge. Protonation of your carbonyl activates your carbonyl."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we start with our carboxylic acid, and we're going to protonate the oxygen. So a lone pair of electrons in the oxygen can pick up a proton, leave these electrons behind. And so we can show that we have now protonated the oxygen right here, so it gets a plus one formal charge. Protonation of your carbonyl activates your carbonyl. It makes your carbon more electrophilic. So this carbon is now more electrophilic. So we saw that in some of the previous videos."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "Protonation of your carbonyl activates your carbonyl. It makes your carbon more electrophilic. So this carbon is now more electrophilic. So we saw that in some of the previous videos. And so if the carbon's more electrophilic, our next step, it makes sense, is going to be a nucleophilic attack. So a molecule of our alcohol comes along. And you can think about a lone pair of electrons on the oxygen attacking this carbon right here, pushing these electrons off onto your oxygen."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we saw that in some of the previous videos. And so if the carbon's more electrophilic, our next step, it makes sense, is going to be a nucleophilic attack. So a molecule of our alcohol comes along. And you can think about a lone pair of electrons on the oxygen attacking this carbon right here, pushing these electrons off onto your oxygen. So let's go ahead and show the result of our nucleophilic attack. So we now have an oxygen up here. It had one lone pair on the left."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "And you can think about a lone pair of electrons on the oxygen attacking this carbon right here, pushing these electrons off onto your oxygen. So let's go ahead and show the result of our nucleophilic attack. So we now have an oxygen up here. It had one lone pair on the left. Now it has two lone pairs, bonded to a hydrogen. And then over here on the right is the bond that we just formed, the bond between carbon and oxygen. So let's show those electrons in magenta."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "It had one lone pair on the left. Now it has two lone pairs, bonded to a hydrogen. And then over here on the right is the bond that we just formed, the bond between carbon and oxygen. So let's show those electrons in magenta. So this lone pair on the oxygen forms the bond between carbon and oxygen. So there it is. Also on this oxygen, there's still a hydrogen."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let's show those electrons in magenta. So this lone pair on the oxygen forms the bond between carbon and oxygen. So there it is. Also on this oxygen, there's still a hydrogen. There's still an R prime group, and one lone pair of electrons, giving this oxygen a plus one formal charge. And there's still an OH bonded to our carbon like that. The next step is to get rid of this plus one formal charge on our oxygen."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "Also on this oxygen, there's still a hydrogen. There's still an R prime group, and one lone pair of electrons, giving this oxygen a plus one formal charge. And there's still an OH bonded to our carbon like that. The next step is to get rid of this plus one formal charge on our oxygen. So another molecule of alcohol comes along, and this time acts as a base. So in the previous step, alcohol acted as a nucleophile. In this step, it's going to act as a base."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "The next step is to get rid of this plus one formal charge on our oxygen. So another molecule of alcohol comes along, and this time acts as a base. So in the previous step, alcohol acted as a nucleophile. In this step, it's going to act as a base. It's going to take this proton and leave these electrons behind on the oxygen. So let's go ahead and get some more room down here. So we're going to deprotonate."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "In this step, it's going to act as a base. It's going to take this proton and leave these electrons behind on the oxygen. So let's go ahead and get some more room down here. So we're going to deprotonate. So let's go ahead and show what we would make. So we have our carbon bonded to this top oxygen, two lone pairs of electrons on it, a hydrogen, an R group off to the left. I'm gonna draw this oxygen down here with lone pairs of electrons so we can show the next step."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to deprotonate. So let's go ahead and show what we would make. So we have our carbon bonded to this top oxygen, two lone pairs of electrons on it, a hydrogen, an R group off to the left. I'm gonna draw this oxygen down here with lone pairs of electrons so we can show the next step. And then this oxygen right here now has two lone pairs of electrons on it, and it's still bonded to our R prime group. So let's show those electrons. The electrons in blue here move off onto the oxygen, so that's the deprotonation step."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "I'm gonna draw this oxygen down here with lone pairs of electrons so we can show the next step. And then this oxygen right here now has two lone pairs of electrons on it, and it's still bonded to our R prime group. So let's show those electrons. The electrons in blue here move off onto the oxygen, so that's the deprotonation step. Next step is to protonate the OH at the bottom. So let's get some more, even more room to show that. So we're going to show this OH down at the bottom here being protonated."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in blue here move off onto the oxygen, so that's the deprotonation step. Next step is to protonate the OH at the bottom. So let's get some more, even more room to show that. So we're going to show this OH down at the bottom here being protonated. So I'm gonna draw in a source of protons right down here. So our protonated alcohol plus one formal charge on our oxygen, R prime right here. So lone pair of electrons on this oxygen can pick up a proton, leave these electrons behind."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to show this OH down at the bottom here being protonated. So I'm gonna draw in a source of protons right down here. So our protonated alcohol plus one formal charge on our oxygen, R prime right here. So lone pair of electrons on this oxygen can pick up a proton, leave these electrons behind. So we have a protonation step. And the reason why this protonation is favored is because this is going to create an excellent leaving group. So if you look closely, you're gonna see water there as our leaving group."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So lone pair of electrons on this oxygen can pick up a proton, leave these electrons behind. So we have a protonation step. And the reason why this protonation is favored is because this is going to create an excellent leaving group. So if you look closely, you're gonna see water there as our leaving group. So let's go ahead and show that. So now this oxygen down here has been protonated, so it has a plus one formal charge on this oxygen. So let's show those electrons."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So if you look closely, you're gonna see water there as our leaving group. So let's go ahead and show that. So now this oxygen down here has been protonated, so it has a plus one formal charge on this oxygen. So let's show those electrons. So these electrons right here on this oxygen pick up this proton, so forming this bond to that proton. So let's go ahead and draw in the rest of what we have. We have our oxygen with two lone pairs of electrons and our R prime group like that."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let's show those electrons. So these electrons right here on this oxygen pick up this proton, so forming this bond to that proton. So let's go ahead and draw in the rest of what we have. We have our oxygen with two lone pairs of electrons and our R prime group like that. So in the next step, we just formed water as our leaving group in here, if you can see it. If these electrons in here were to come off on the oxygen, you can see that's water. So when these electrons on this top oxygen move in here to reform our double bond, that's when these electrons in here in blue are going to come off onto our oxygen and water is an excellent leaving group."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "We have our oxygen with two lone pairs of electrons and our R prime group like that. So in the next step, we just formed water as our leaving group in here, if you can see it. If these electrons in here were to come off on the oxygen, you can see that's water. So when these electrons on this top oxygen move in here to reform our double bond, that's when these electrons in here in blue are going to come off onto our oxygen and water is an excellent leaving group. So let's draw what we have now. So once again, everything's at equilibrium. So we're going to reform our double bond and this top oxygen now has a plus one formal charge."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So when these electrons on this top oxygen move in here to reform our double bond, that's when these electrons in here in blue are going to come off onto our oxygen and water is an excellent leaving group. So let's draw what we have now. So once again, everything's at equilibrium. So we're going to reform our double bond and this top oxygen now has a plus one formal charge. So let's show those electrons. Let's make them red here. So these electrons in red are going to move in here to reform our double bond."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to reform our double bond and this top oxygen now has a plus one formal charge. So let's show those electrons. Let's make them red here. So these electrons in red are going to move in here to reform our double bond. This carbon is still bonded to an R group on the left side. This carbon is bonded to an oxygen and our R prime group over here, the oxygen has two lone pairs of electrons and we just lost water. So let me go ahead and draw water down here."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in red are going to move in here to reform our double bond. This carbon is still bonded to an R group on the left side. This carbon is bonded to an oxygen and our R prime group over here, the oxygen has two lone pairs of electrons and we just lost water. So let me go ahead and draw water down here. So loss of H2O at this step. We're almost to our final product because all we have to do is deprotonate right here and we'll form our ester. So we could show another molecule of alcohol coming along."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw water down here. So loss of H2O at this step. We're almost to our final product because all we have to do is deprotonate right here and we'll form our ester. So we could show another molecule of alcohol coming along. So R prime, two lone pairs of electrons, taking this proton right here, leaving these electrons behind on our oxygen. So let's go ahead and draw the final structure of our ester. So we would have our C double bond O with two lone pairs of electrons."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we could show another molecule of alcohol coming along. So R prime, two lone pairs of electrons, taking this proton right here, leaving these electrons behind on our oxygen. So let's go ahead and draw the final structure of our ester. So we would have our C double bond O with two lone pairs of electrons. So let's show these electrons in here, move off onto our oxygen like that. And then our carbon is still bonded to this oxygen and we have our R prime group. And so our end result is to form our ester and water."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our C double bond O with two lone pairs of electrons. So let's show these electrons in here, move off onto our oxygen like that. And then our carbon is still bonded to this oxygen and we have our R prime group. And so our end result is to form our ester and water. Okay, so that's a little bit of a long mechanism. Let's take a look at some reactions to form esters using the Fischer esterification reaction. So let's start with this molecule over here on the left."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "And so our end result is to form our ester and water. Okay, so that's a little bit of a long mechanism. Let's take a look at some reactions to form esters using the Fischer esterification reaction. So let's start with this molecule over here on the left. So this is salicylic acid. And if we add methanol and we use sulfuric acid as our source of protons, we're going to form an ester. And this is one of those famous labs that's always done in undergraduate organic chemistry."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with this molecule over here on the left. So this is salicylic acid. And if we add methanol and we use sulfuric acid as our source of protons, we're going to form an ester. And this is one of those famous labs that's always done in undergraduate organic chemistry. So if we think about the mechanism, remember that this oxygen on our alcohol, and in this case this methyl, are going to add. So we're going to lose this OH on our carboxylic acid and we're going to put this oxygen and this methyl group on in place. So let's go ahead and draw the product."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "And this is one of those famous labs that's always done in undergraduate organic chemistry. So if we think about the mechanism, remember that this oxygen on our alcohol, and in this case this methyl, are going to add. So we're going to lose this OH on our carboxylic acid and we're going to put this oxygen and this methyl group on in place. So let's go ahead and draw the product. So we would have our benzene ring right here. And we would have our carbon double bonded to our oxygen. And we would have the oxygen from the alcohol, from methanol, and then our methyl group like that."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the product. So we would have our benzene ring right here. And we would have our carbon double bonded to our oxygen. And we would have the oxygen from the alcohol, from methanol, and then our methyl group like that. And then we still have our OH right here. The reason why this is one of those classic undergraduate labs is this is wintergreen. So this is an incredible smell."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "And we would have the oxygen from the alcohol, from methanol, and then our methyl group like that. And then we still have our OH right here. The reason why this is one of those classic undergraduate labs is this is wintergreen. So this is an incredible smell. It's always a lot of fun to do this in an undergraduate lab because the lab smells great when you're done. So the synthesis of wintergreen. Let's look at another Fischer esterification."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So this is an incredible smell. It's always a lot of fun to do this in an undergraduate lab because the lab smells great when you're done. So the synthesis of wintergreen. Let's look at another Fischer esterification. This one is a little bit different. This one is an intramolecular Fischer esterification. So if we look at our starting molecule on the left, this time we have our carboxylic acid and our alcohol in the exact same molecule."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at another Fischer esterification. This one is a little bit different. This one is an intramolecular Fischer esterification. So if we look at our starting molecule on the left, this time we have our carboxylic acid and our alcohol in the exact same molecule. And we have all these single bonds in here, which we know we can have some free rotation. So if we draw the molecule in a different conformation, so let's go ahead and do that. So we have our carboxylic acid up here."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So if we look at our starting molecule on the left, this time we have our carboxylic acid and our alcohol in the exact same molecule. And we have all these single bonds in here, which we know we can have some free rotation. So if we draw the molecule in a different conformation, so let's go ahead and do that. So we have our carboxylic acid up here. And let's count how many carbons we have. So let me use red for that. So we have one."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we have our carboxylic acid up here. And let's count how many carbons we have. So let me use red for that. So we have one. We have carbon 1, 2, 3, 4, and 5. So we have five carbons. So let's go ahead and draw them in."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we have one. We have carbon 1, 2, 3, 4, and 5. So we have five carbons. So let's go ahead and draw them in. So there's carbon 1, 2, 3, 4, 5. And then we have our OH. So let me go ahead and number those carbons."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw them in. So there's carbon 1, 2, 3, 4, 5. And then we have our OH. So let me go ahead and number those carbons. So this is carbon 1, carbon 2, carbon 3, carbon 4, and carbon 5. And so in a different conformation, we can think about this oxygen attacking this carbonyl and the mechanism. So we know that we're going to lose this OH."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and number those carbons. So this is carbon 1, carbon 2, carbon 3, carbon 4, and carbon 5. And so in a different conformation, we can think about this oxygen attacking this carbonyl and the mechanism. So we know that we're going to lose this OH. We know we're going to lose this hydrogen. And so we can stick those together and think about our final product. So we're going to form an ester."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we know that we're going to lose this OH. We know we're going to lose this hydrogen. And so we can stick those together and think about our final product. So we're going to form an ester. But it's a different ester than what we've seen before. So we have our carbonyl right here. And then we have this oxygen."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to form an ester. But it's a different ester than what we've seen before. So we have our carbonyl right here. And then we have this oxygen. So this oxygen is a member of our ring now. And that came from this oxygen. So we can see our five carbons, 1, 2, 3, 4, 5."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "And then we have this oxygen. So this oxygen is a member of our ring now. And that came from this oxygen. So we can see our five carbons, 1, 2, 3, 4, 5. We can think about losing water here. And so our product is called a lactone. So this is a lactone here."}, {"video_title": "Preparation of esters via Fischer esterification Organic chemistry Khan Academy.mp3", "Sentence": "So we can see our five carbons, 1, 2, 3, 4, 5. We can think about losing water here. And so our product is called a lactone. So this is a lactone here. So it's an ester that's in a ring. Here we have a six-membered ring where oxygen is one of the members of the ring. So a pretty cool intramolecular Fischer esterification reaction to form a lactone."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "We start off with benzene. And to benzene, we add an acyl chloride. And so this right here, you can think about as an acyl group. We're also going to use aluminum chloride once again as our catalyst. And you can see the acyl group has substituted in for one of the aromatic protons. And so that's our electrophilic aromatic substitution reaction. The mechanism for an acylation is similar to an alkylation, although there is an important difference."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "We're also going to use aluminum chloride once again as our catalyst. And you can see the acyl group has substituted in for one of the aromatic protons. And so that's our electrophilic aromatic substitution reaction. The mechanism for an acylation is similar to an alkylation, although there is an important difference. But they start off the same in that aluminum chloride is going to function as a Lewis acid and accept a pair of electrons. And so this lone pair of electrons on this chlorine, you could think about that chlorine donating that pair of electrons and the aluminum accepting that electron pair. And so if I go ahead and draw the result of that Lewis acid base reaction, we have our carbonyl."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "The mechanism for an acylation is similar to an alkylation, although there is an important difference. But they start off the same in that aluminum chloride is going to function as a Lewis acid and accept a pair of electrons. And so this lone pair of electrons on this chlorine, you could think about that chlorine donating that pair of electrons and the aluminum accepting that electron pair. And so if I go ahead and draw the result of that Lewis acid base reaction, we have our carbonyl. We have our chlorine attached to our carbonyl carbon. The chlorine has two lone pairs of electrons. It's now formed a bond with the aluminum."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And so if I go ahead and draw the result of that Lewis acid base reaction, we have our carbonyl. We have our chlorine attached to our carbonyl carbon. The chlorine has two lone pairs of electrons. It's now formed a bond with the aluminum. And the aluminum is bonded to three other chlorines. I'm not going to draw the lone pairs of electrons on those other chlorines just to save time. But the aluminum gets a negative 1 formal charge."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "It's now formed a bond with the aluminum. And the aluminum is bonded to three other chlorines. I'm not going to draw the lone pairs of electrons on those other chlorines just to save time. But the aluminum gets a negative 1 formal charge. And this chlorine now has a positive 1 formal charge. So to highlight our electrons, these electrons right here in magenta are forming a bond between the chlorine and the aluminum like that. So in order to find our electrophile, you could think about these electrons in here kicking off onto the chlorine."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "But the aluminum gets a negative 1 formal charge. And this chlorine now has a positive 1 formal charge. So to highlight our electrons, these electrons right here in magenta are forming a bond between the chlorine and the aluminum like that. So in order to find our electrophile, you could think about these electrons in here kicking off onto the chlorine. And so I'm taking a bond away from that carbonyl carbon. And so if I take a bond away from that carbonyl carbon, that carbon is now positively charged. That carbon is still double bonded to an oxygen."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So in order to find our electrophile, you could think about these electrons in here kicking off onto the chlorine. And so I'm taking a bond away from that carbonyl carbon. And so if I take a bond away from that carbonyl carbon, that carbon is now positively charged. That carbon is still double bonded to an oxygen. And that carbon is bonded to an R group. And so we've created an acyl cation. And we can think about this acyl cation as being the electrophile in our mechanism for electrophilic aromatic substitution."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "That carbon is still double bonded to an oxygen. And that carbon is bonded to an R group. And so we've created an acyl cation. And we can think about this acyl cation as being the electrophile in our mechanism for electrophilic aromatic substitution. This cation is resonance stabilized. I could take a lone pair of electrons here on this oxygen and move them into here. So I could draw a resonance structure where now the carbon would be triple bonded to this top oxygen here."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And we can think about this acyl cation as being the electrophile in our mechanism for electrophilic aromatic substitution. This cation is resonance stabilized. I could take a lone pair of electrons here on this oxygen and move them into here. So I could draw a resonance structure where now the carbon would be triple bonded to this top oxygen here. This top oxygen would still have a lone pair of electrons and have a plus 1 formal charge like that. And this carbon is still bonded to an R group. So I'm saying that this lone pair of electrons here on this oxygen can move into here like that."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So I could draw a resonance structure where now the carbon would be triple bonded to this top oxygen here. This top oxygen would still have a lone pair of electrons and have a plus 1 formal charge like that. And this carbon is still bonded to an R group. So I'm saying that this lone pair of electrons here on this oxygen can move into here like that. And this resonance stabilization of the acyl cation is one difference between an acylation and an alkylation. Because our cation in an acylation is resonance stabilized, there is no rearrangement. And that's different from what we saw in our alkylation reaction."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So I'm saying that this lone pair of electrons here on this oxygen can move into here like that. And this resonance stabilization of the acyl cation is one difference between an acylation and an alkylation. Because our cation in an acylation is resonance stabilized, there is no rearrangement. And that's different from what we saw in our alkylation reaction. We formed a carbocation that was capable of rearranging to form a more stable carbocation in the previous video on alkylation. And so that made it a little bit difficult to control the types of products that we got. And so with the acylation, there is no rearrangement."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And that's different from what we saw in our alkylation reaction. We formed a carbocation that was capable of rearranging to form a more stable carbocation in the previous video on alkylation. And so that made it a little bit difficult to control the types of products that we got. And so with the acylation, there is no rearrangement. And again, it's due to this resonance stabilization of our acyl cation here. So we would also form this complex over here where the aluminum is bonded to four chlorines. So we could think about this chlorine now as having three lone pairs of electrons around it."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And so with the acylation, there is no rearrangement. And again, it's due to this resonance stabilization of our acyl cation here. So we would also form this complex over here where the aluminum is bonded to four chlorines. So we could think about this chlorine now as having three lone pairs of electrons around it. So I'm going to highlight those electrons in red. So these electrons in here kick off onto the chlorine like that. And once again, we still have a negative 1 formal charge on this aluminum like that."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So we could think about this chlorine now as having three lone pairs of electrons around it. So I'm going to highlight those electrons in red. So these electrons in here kick off onto the chlorine like that. And once again, we still have a negative 1 formal charge on this aluminum like that. So the catalyst has generated our electrophile. And now we can show our electrophile, our acyl cation, reacting with a benzene ring. And for that mechanism, you could show either one of these."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And once again, we still have a negative 1 formal charge on this aluminum like that. So the catalyst has generated our electrophile. And now we can show our electrophile, our acyl cation, reacting with a benzene ring. And for that mechanism, you could show either one of these. You could show either one of these resonance forms reacting with your benzene ring. I'm just going to take the one on the right and the one with the positive charge on the carbon. So we have our benzene ring."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And for that mechanism, you could show either one of these. You could show either one of these resonance forms reacting with your benzene ring. I'm just going to take the one on the right and the one with the positive charge on the carbon. So we have our benzene ring. And we have one of the protons on our benzene ring like that. And I'm choosing the resonance structure on the right. So I'm going to have a plus 1 formal charge on my carbon."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So we have our benzene ring. And we have one of the protons on our benzene ring like that. And I'm choosing the resonance structure on the right. So I'm going to have a plus 1 formal charge on my carbon. And I'm going to have an R group attached to that carbon as well. So we have a nucleophile electrophile situation. So once again, negative charges attract positive charges."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So I'm going to have a plus 1 formal charge on my carbon. And I'm going to have an R group attached to that carbon as well. So we have a nucleophile electrophile situation. So once again, negative charges attract positive charges. These pi electrons are going to function as a nucleophile, attack our electrophile. And so we can add our electrophile onto our benzene ring. And so once again, I'm going to show our electrophile adding to the top carbon here."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So once again, negative charges attract positive charges. These pi electrons are going to function as a nucleophile, attack our electrophile. And so we can add our electrophile onto our benzene ring. And so once again, I'm going to show our electrophile adding to the top carbon here. So the top carbon has a hydrogen. And now it's going to form a bond to our carbonyl carbon like that. So I put in my lone pairs of electrons."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And so once again, I'm going to show our electrophile adding to the top carbon here. So the top carbon has a hydrogen. And now it's going to form a bond to our carbonyl carbon like that. So I put in my lone pairs of electrons. And there's an R group attached to that carbon as well. So follow our electrons in magenta, functioning as a nucleophile, forming a bond between this carbon and this carbonyl carbon like that. We took a bond away from this carbon down here."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So I put in my lone pairs of electrons. And there's an R group attached to that carbon as well. So follow our electrons in magenta, functioning as a nucleophile, forming a bond between this carbon and this carbonyl carbon like that. We took a bond away from this carbon down here. So that's a plus 1 formal charge. And so of course, this is one possible resonance structure. And we could draw a few more possible resonance structures."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "We took a bond away from this carbon down here. So that's a plus 1 formal charge. And so of course, this is one possible resonance structure. And we could draw a few more possible resonance structures. I'm not going to do that for time reasons. I've done it in the earlier videos. So please watch the earlier videos if you're confused about resonance structures."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And we could draw a few more possible resonance structures. I'm not going to do that for time reasons. I've done it in the earlier videos. So please watch the earlier videos if you're confused about resonance structures. I'm going to use this resonance structure to represent our sigma complex. And of course, to finish our reaction, we need to deprotonate our sigma complex and regenerate our aromatic ring. So these electrons in here are going to pick up this proton, which would cause these electrons to move in to reform our aromatic ring and to take away the plus 1 formal charge."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So please watch the earlier videos if you're confused about resonance structures. I'm going to use this resonance structure to represent our sigma complex. And of course, to finish our reaction, we need to deprotonate our sigma complex and regenerate our aromatic ring. So these electrons in here are going to pick up this proton, which would cause these electrons to move in to reform our aromatic ring and to take away the plus 1 formal charge. And so when we do that, we form our benzene ring with our acyl group now attached to our benzene ring. So it's substituted for that proton there. So let's follow those electrons as well."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So these electrons in here are going to pick up this proton, which would cause these electrons to move in to reform our aromatic ring and to take away the plus 1 formal charge. And so when we do that, we form our benzene ring with our acyl group now attached to our benzene ring. So it's substituted for that proton there. So let's follow those electrons as well. So I'm going to make those electrons in here green. So these electrons in here are going to move into here. And then also, we could think about what else is formed."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So let's follow those electrons as well. So I'm going to make those electrons in here green. So these electrons in here are going to move into here. And then also, we could think about what else is formed. So these electrons up here in red are going to bond to that proton. And so we would also have HCl. So let me go ahead and highlight those electrons in here in red."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And then also, we could think about what else is formed. So these electrons up here in red are going to bond to that proton. And so we would also have HCl. So let me go ahead and highlight those electrons in here in red. And then of course, we would also regenerate our catalyst. So we would make AlCl3. So we've formed our product."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So let me go ahead and highlight those electrons in here in red. And then of course, we would also regenerate our catalyst. So we would make AlCl3. So we've formed our product. We've installed an acyl group on our benzene ring. And so that is Friedel-Crafts acylation. Let's look at a situation where a Friedel-Crafts acylation might be used instead of a Friedel-Crafts alkylation."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So we've formed our product. We've installed an acyl group on our benzene ring. And so that is Friedel-Crafts acylation. Let's look at a situation where a Friedel-Crafts acylation might be used instead of a Friedel-Crafts alkylation. So let's say that our goal was to go from benzene to butylbenzene. So let me go ahead and draw butylbenzene out here. So four carbon alkyl group coming off of our benzene ring."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "Let's look at a situation where a Friedel-Crafts acylation might be used instead of a Friedel-Crafts alkylation. So let's say that our goal was to go from benzene to butylbenzene. So let me go ahead and draw butylbenzene out here. So four carbon alkyl group coming off of our benzene ring. So we saw in the last video that a Friedel-Crafts alkylation would make butylbenzene as a minor product because of the rearrangement of the carbocation. And so this would be formed as a minor product. If you wanted to form it in a higher yield, you could use a Friedel-Crafts acylation."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So four carbon alkyl group coming off of our benzene ring. So we saw in the last video that a Friedel-Crafts alkylation would make butylbenzene as a minor product because of the rearrangement of the carbocation. And so this would be formed as a minor product. If you wanted to form it in a higher yield, you could use a Friedel-Crafts acylation. And so if I wanted to get to butylbenzene using an acylation, I would need to install an acyl group on my benzene ring that has the same number of carbons. So an acyl group that has four carbons on it. And so let me go ahead and draw that."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "If you wanted to form it in a higher yield, you could use a Friedel-Crafts acylation. And so if I wanted to get to butylbenzene using an acylation, I would need to install an acyl group on my benzene ring that has the same number of carbons. So an acyl group that has four carbons on it. And so let me go ahead and draw that. So we would show an acyl chloride that has a total of four carbons. So there's my acyl chloride. And I can highlight the four carbons."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And so let me go ahead and draw that. So we would show an acyl chloride that has a total of four carbons. So there's my acyl chloride. And I can highlight the four carbons. So this carbon, two, three, and four. So to that acyl chloride, we would also need to add our catalyst. So we need some aluminum chloride as well."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And I can highlight the four carbons. So this carbon, two, three, and four. So to that acyl chloride, we would also need to add our catalyst. So we need some aluminum chloride as well. So AlCl3 to catalyze this Friedel-Crafts acylation. And we're going to put this acyl group onto our benzene ring. So here's our acyl group."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So we need some aluminum chloride as well. So AlCl3 to catalyze this Friedel-Crafts acylation. And we're going to put this acyl group onto our benzene ring. So here's our acyl group. And you could think about just putting that onto your ring. So when I draw my ring, I know that the carbon on my ring is going to be directly attached to this carbonyl carbon. And there's a total of four carbons here."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "So here's our acyl group. And you could think about just putting that onto your ring. So when I draw my ring, I know that the carbon on my ring is going to be directly attached to this carbonyl carbon. And there's a total of four carbons here. So one, two, three, and then four. And so that's my Friedel-Crafts acylation. And so now I need to go from this compound to my target compound up here, my butylbenzene."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And there's a total of four carbons here. So one, two, three, and then four. And so that's my Friedel-Crafts acylation. And so now I need to go from this compound to my target compound up here, my butylbenzene. And so somehow I need to get rid of that carbonyl. And a reaction that's been historically used to do this is the Clemenson reduction, which involves a zinc amalgam with mercury and also a source of protons, so HCl. And the zinc amalgam is going to reduce that carbonyl to our alkyl group."}, {"video_title": "Friedel-Crafts acylation Aromatic Compounds Organic chemistry Khan Academy (2).mp3", "Sentence": "And so now I need to go from this compound to my target compound up here, my butylbenzene. And so somehow I need to get rid of that carbonyl. And a reaction that's been historically used to do this is the Clemenson reduction, which involves a zinc amalgam with mercury and also a source of protons, so HCl. And the zinc amalgam is going to reduce that carbonyl to our alkyl group. So it actually will form the desired alkyl group and gets rid of that carbonyl. And so the Clemenson reduction is a very useful reaction in synthesis. And so this is a way to make our butylbenzene molecule in high yield using an acylation, which is a little bit more reliable than our alkylation."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's focus in on this carbon right here on the formaldehyde molecule. Let's find the hybridization state of this carbon. So I'm gonna draw an arrow to this. And to find the hybridization state, one way to do it is to think about the steric number. Where the steric number is the number of sigma bonds plus the number of lone pairs of electrons. So to that carbon, let's count up some sigma bonds here. So we have a sigma bond to this hydrogen, a sigma bond to this hydrogen, and in our double bond here, one of those is a sigma bond and one of those is a pi bond."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And to find the hybridization state, one way to do it is to think about the steric number. Where the steric number is the number of sigma bonds plus the number of lone pairs of electrons. So to that carbon, let's count up some sigma bonds here. So we have a sigma bond to this hydrogen, a sigma bond to this hydrogen, and in our double bond here, one of those is a sigma bond and one of those is a pi bond. So we have a total of three sigma bonds. So three sigma bonds and zero lone pairs of electrons gives us a steric number of three, which we know means it must have three hybrid orbitals. And so this carbon is sp2 hybridized."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have a sigma bond to this hydrogen, a sigma bond to this hydrogen, and in our double bond here, one of those is a sigma bond and one of those is a pi bond. So we have a total of three sigma bonds. So three sigma bonds and zero lone pairs of electrons gives us a steric number of three, which we know means it must have three hybrid orbitals. And so this carbon is sp2 hybridized. So if I'm gonna go ahead and draw that carbon over here, so that carbon is sp2 hybridized, which means that it has three sp2 hybrid orbitals. I'm gonna go ahead and put those three sp2 hybrid orbitals in here like that. All right, and we know that carbon has unhybridized p orbitals."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so this carbon is sp2 hybridized. So if I'm gonna go ahead and draw that carbon over here, so that carbon is sp2 hybridized, which means that it has three sp2 hybrid orbitals. I'm gonna go ahead and put those three sp2 hybrid orbitals in here like that. All right, and we know that carbon has unhybridized p orbitals. I'm gonna go ahead and draw in that unhybridized p orbital right here. Next, let's think about those hydrogens. All right, so these hydrogens, let me go ahead and put them in red here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, and we know that carbon has unhybridized p orbitals. I'm gonna go ahead and draw in that unhybridized p orbital right here. Next, let's think about those hydrogens. All right, so these hydrogens, let me go ahead and put them in red here. So these hydrogens right here bond into that carbonyl carbon. All right, those have an electron and an s orbital, which we know is spherically shaped. So I can put an s orbital in here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so these hydrogens, let me go ahead and put them in red here. So these hydrogens right here bond into that carbonyl carbon. All right, those have an electron and an s orbital, which we know is spherically shaped. So I can put an s orbital in here. And the overlap, of course, would be a sigma bond. Right, so I have those sigma bonds right there. Next, let's look at the hybridization of the carbonyl oxygen here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I can put an s orbital in here. And the overlap, of course, would be a sigma bond. Right, so I have those sigma bonds right there. Next, let's look at the hybridization of the carbonyl oxygen here. So same idea, number of sigma bonds plus number of lone pairs of electrons. So there's one sigma bond between the oxygen and the carbon, and then we have two lone pairs of electrons. So one sigma bond and two lone pairs of electrons gives us a steric number of three, which means that oxygen must be sp2 hybridized as well."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at the hybridization of the carbonyl oxygen here. So same idea, number of sigma bonds plus number of lone pairs of electrons. So there's one sigma bond between the oxygen and the carbon, and then we have two lone pairs of electrons. So one sigma bond and two lone pairs of electrons gives us a steric number of three, which means that oxygen must be sp2 hybridized as well. So the oxygen has three sp2 hybrid orbitals. So let me go ahead and draw those. So put in the oxygen right here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So one sigma bond and two lone pairs of electrons gives us a steric number of three, which means that oxygen must be sp2 hybridized as well. So the oxygen has three sp2 hybrid orbitals. So let me go ahead and draw those. So put in the oxygen right here. The oxygen has three sp2 hybrid orbitals. So I'm gonna go ahead and draw in those sp2 hybrid orbitals. So there's one, and then I have these two over here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So put in the oxygen right here. The oxygen has three sp2 hybrid orbitals. So I'm gonna go ahead and draw in those sp2 hybrid orbitals. So there's one, and then I have these two over here. And so the lone pairs of electrons on the oxygen, right, one lone pair is gonna go into this sp2 hybrid orbital. The other lone pair is gonna go into this sp2 hybrid orbital. And then we have an overlap right here for this carbon."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So there's one, and then I have these two over here. And so the lone pairs of electrons on the oxygen, right, one lone pair is gonna go into this sp2 hybrid orbital. The other lone pair is gonna go into this sp2 hybrid orbital. And then we have an overlap right here for this carbon. So that's, of course, the sigma bond between the carbon and the oxygen. If the oxygen is sp2 hybridized, it must also have an unhybridized p orbital. So I'm gonna draw in the unhybridized p orbital on the oxygen here like that."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we have an overlap right here for this carbon. So that's, of course, the sigma bond between the carbon and the oxygen. If the oxygen is sp2 hybridized, it must also have an unhybridized p orbital. So I'm gonna draw in the unhybridized p orbital on the oxygen here like that. And then we can see that the pi bond, right, comes from the overlap, the side-by-side overlap of our p orbitals. And so let me go ahead and highlight the pi bond over here. So in that double bond again on the dot structure for formaldehyde, one of those is a sigma bond and one of those is a pi bond."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna draw in the unhybridized p orbital on the oxygen here like that. And then we can see that the pi bond, right, comes from the overlap, the side-by-side overlap of our p orbitals. And so let me go ahead and highlight the pi bond over here. So in that double bond again on the dot structure for formaldehyde, one of those is a sigma bond and one of those is a pi bond. And over here we can see it on the right. So this represents the bonding of the carbonyl. And that's gonna be important when we think about things like molecular geometry, right?"}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So in that double bond again on the dot structure for formaldehyde, one of those is a sigma bond and one of those is a pi bond. And over here we can see it on the right. So this represents the bonding of the carbonyl. And that's gonna be important when we think about things like molecular geometry, right? So if the carbon is sp2 hybridized, right, then we know that these atoms lie on the same plane here. They have bond angles close to 120 degrees. And so we'll talk more about that in a few minutes."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And that's gonna be important when we think about things like molecular geometry, right? So if the carbon is sp2 hybridized, right, then we know that these atoms lie on the same plane here. They have bond angles close to 120 degrees. And so we'll talk more about that in a few minutes. Next let's think about the polarization of that carbonyl. Alright, so once again we look at the, let's go down to this generic aldehyde here. And then we have this carbonyl carbon attached to this oxygen."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we'll talk more about that in a few minutes. Next let's think about the polarization of that carbonyl. Alright, so once again we look at the, let's go down to this generic aldehyde here. And then we have this carbonyl carbon attached to this oxygen. Oxygen is more electronegative than carbon, so it's going to withdraw some electron density. Remember we showed the polarization with this arrow here. Alright, the arrow points in the direction of the electrons, right?"}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we have this carbonyl carbon attached to this oxygen. Oxygen is more electronegative than carbon, so it's going to withdraw some electron density. Remember we showed the polarization with this arrow here. Alright, the arrow points in the direction of the electrons, right? So the electron's gonna be pulled closer to the oxygen. And so the oxygen is going to get a little bit partially negative. So we draw a partial negative sign here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Alright, the arrow points in the direction of the electrons, right? So the electron's gonna be pulled closer to the oxygen. And so the oxygen is going to get a little bit partially negative. So we draw a partial negative sign here. The oxygen's withdrawing some electron density from my carbonyl carbon right here, so my carbonyl carbon is going to be partially positive like that. And for an aldehyde, right, we know that alkyl groups are electron donating, right? So this R group right here on the left, you can think about that as being a little bit electron donating, which means this R group is going to donate some electron density."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we draw a partial negative sign here. The oxygen's withdrawing some electron density from my carbonyl carbon right here, so my carbonyl carbon is going to be partially positive like that. And for an aldehyde, right, we know that alkyl groups are electron donating, right? So this R group right here on the left, you can think about that as being a little bit electron donating, which means this R group is going to donate some electron density. So I'll draw an arrow showing that some electron density is being donated by the R group. So remember when we did carbocations and we put alkyl groups on our carbocations, the more alkyl groups we had, the more the carbocation was stabilized. So here we have one R group donating a little bit of electron density, attempting to stabilize that partially positive charge on the carbonyl carbon."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this R group right here on the left, you can think about that as being a little bit electron donating, which means this R group is going to donate some electron density. So I'll draw an arrow showing that some electron density is being donated by the R group. So remember when we did carbocations and we put alkyl groups on our carbocations, the more alkyl groups we had, the more the carbocation was stabilized. So here we have one R group donating a little bit of electron density, attempting to stabilize that partially positive charge on the carbonyl carbon. Let's go over here to the ketone, and we have a similar situation, right? We have once again the oxygen withdrawing some electron density from our carbonyl carbon, so we have a partial negative on our oxygen and our carbonyl carbon gets a partial positive. But this time we have two R groups."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So here we have one R group donating a little bit of electron density, attempting to stabilize that partially positive charge on the carbonyl carbon. Let's go over here to the ketone, and we have a similar situation, right? We have once again the oxygen withdrawing some electron density from our carbonyl carbon, so we have a partial negative on our oxygen and our carbonyl carbon gets a partial positive. But this time we have two R groups. So this R group on the left can donate some electron density, this R group on the right can donate some electron density. And once again, think about the carbocations, right? The more alkyl groups we had, the more our full positive charge in our carbocation was stabilized."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "But this time we have two R groups. So this R group on the left can donate some electron density, this R group on the right can donate some electron density. And once again, think about the carbocations, right? The more alkyl groups we had, the more our full positive charge in our carbocation was stabilized. Similar idea here, right? The more R groups you have, the more you stabilize the partial positive charge on your carbonyl. And so because of that, ketones are a little bit more stable than aldehydes, just thinking about the polarization."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The more alkyl groups we had, the more our full positive charge in our carbocation was stabilized. Similar idea here, right? The more R groups you have, the more you stabilize the partial positive charge on your carbonyl. And so because of that, ketones are a little bit more stable than aldehydes, just thinking about the polarization. So there's more of a polarization in an aldehyde carbonyl than in a ketone. All right, so now let's put these ideas together. Let's think about a nucleophilic addition reaction to a carbonyl."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so because of that, ketones are a little bit more stable than aldehydes, just thinking about the polarization. So there's more of a polarization in an aldehyde carbonyl than in a ketone. All right, so now let's put these ideas together. Let's think about a nucleophilic addition reaction to a carbonyl. And so I'm gonna go ahead and draw, I'm gonna draw a ketone down here, and I know that the geometry around the carbon and ketone is trigonal planar, right? Because we talked about the bonding already. So I'm gonna go ahead and draw this in."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about a nucleophilic addition reaction to a carbonyl. And so I'm gonna go ahead and draw, I'm gonna draw a ketone down here, and I know that the geometry around the carbon and ketone is trigonal planar, right? Because we talked about the bonding already. So I'm gonna go ahead and draw this in. So we're gonna have, let's make an R prime group coming out at us in space, and an R group going away from us in space. And we know that these all are on the same plane, like that. So I'm drawing in my plane here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna go ahead and draw this in. So we're gonna have, let's make an R prime group coming out at us in space, and an R group going away from us in space. And we know that these all are on the same plane, like that. So I'm drawing in my plane here. All right, we also know that there is a polarization of the carbonyl, all right? So the oxygen is more negative, and the carbon is a little bit positive, like that. So the carbon is partially positive."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'm drawing in my plane here. All right, we also know that there is a polarization of the carbonyl, all right? So the oxygen is more negative, and the carbon is a little bit positive, like that. So the carbon is partially positive. That means that once electrons, it's electrophilic, and that's extremely important when you're talking about reactions, right? Nucleophilic additions to carbonyls. So a nucleophile is gonna come along."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So the carbon is partially positive. That means that once electrons, it's electrophilic, and that's extremely important when you're talking about reactions, right? Nucleophilic additions to carbonyls. So a nucleophile is gonna come along. So let's go ahead and draw in a nucleophile here. So let's go ahead and make it a negatively charged nucleophile, so like that. So the nucleophile is going to be attracted to positive things, right?"}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So a nucleophile is gonna come along. So let's go ahead and draw in a nucleophile here. So let's go ahead and make it a negatively charged nucleophile, so like that. So the nucleophile is going to be attracted to positive things, right? The opposite charges attract. And so the nucleophile is going to attack the partially positive carbonyl carbon, like that. And when it does so, it's gonna form a bond, and therefore kick off these pi electrons here off onto this oxygen."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So the nucleophile is going to be attracted to positive things, right? The opposite charges attract. And so the nucleophile is going to attack the partially positive carbonyl carbon, like that. And when it does so, it's gonna form a bond, and therefore kick off these pi electrons here off onto this oxygen. So let's go ahead and draw the result of our nucleophilic attack here. So we're going to show a bond form between the carbon and the nucleophile, all right? So let me go ahead and highlight those electrons here."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And when it does so, it's gonna form a bond, and therefore kick off these pi electrons here off onto this oxygen. So let's go ahead and draw the result of our nucleophilic attack here. So we're going to show a bond form between the carbon and the nucleophile, all right? So let me go ahead and highlight those electrons here. So these electrons right here on the nucleophile have now formed a bond between the nucleophile and the carbon. And, let's see, we still have our oxygen. Our oxygen used to have two lone pairs of electrons, but it picked up another lone pair of electrons, so a negative one formal charge now."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight those electrons here. So these electrons right here on the nucleophile have now formed a bond between the nucleophile and the carbon. And, let's see, we still have our oxygen. Our oxygen used to have two lone pairs of electrons, but it picked up another lone pair of electrons, so a negative one formal charge now. And then we have our R groups. So let's go ahead and draw our R groups over here. So we have R prime, and then we have R over here, like that."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Our oxygen used to have two lone pairs of electrons, but it picked up another lone pair of electrons, so a negative one formal charge now. And then we have our R groups. So let's go ahead and draw our R groups over here. So we have R prime, and then we have R over here, like that. And so this would be an intermediate. And if we think about the geometry of this carbon, let me go ahead and label it here. So the geometry of this carbon now, if I calculate the steric number, I have four sigma bonds."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have R prime, and then we have R over here, like that. And so this would be an intermediate. And if we think about the geometry of this carbon, let me go ahead and label it here. So the geometry of this carbon now, if I calculate the steric number, I have four sigma bonds. And so four sigma bonds means a steric number for four hybrid orbitals. So this carbon must be sp3 hybridized now, so an sp3 hybridized carbon. And so therefore, your bond angle is closer to 109 degrees."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So the geometry of this carbon now, if I calculate the steric number, I have four sigma bonds. And so four sigma bonds means a steric number for four hybrid orbitals. So this carbon must be sp3 hybridized now, so an sp3 hybridized carbon. And so therefore, your bond angle is closer to 109 degrees. So in particular, we're going to think about these R groups here, and so this angle right in here is somewhere around 109 degrees. So I'm going to write that approximately 109 degrees. And so the bond angles change."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, your bond angle is closer to 109 degrees. So in particular, we're going to think about these R groups here, and so this angle right in here is somewhere around 109 degrees. So I'm going to write that approximately 109 degrees. And so the bond angles change. So go back over here to this situation on the left. This carbon right here was sp2 hybridized, and therefore, everything was planar. And so these bond angles were approximately equal at approximately 120 degrees."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so the bond angles change. So go back over here to this situation on the left. This carbon right here was sp2 hybridized, and therefore, everything was planar. And so these bond angles were approximately equal at approximately 120 degrees. And so we've gone from approximately 120 degrees to 109 degrees for our intermediate here. So over here on the right, our sp3 hybridized carbon has tetrahedral geometry. So we call this our tetrahedral intermediate."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so these bond angles were approximately equal at approximately 120 degrees. And so we've gone from approximately 120 degrees to 109 degrees for our intermediate here. So over here on the right, our sp3 hybridized carbon has tetrahedral geometry. So we call this our tetrahedral intermediate. So let me go ahead and write that. This is our tetrahedral intermediate, and it's an alkoxide anion. And so now we can compare aldehydes to ketones in terms of reactivity."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we call this our tetrahedral intermediate. So let me go ahead and write that. This is our tetrahedral intermediate, and it's an alkoxide anion. And so now we can compare aldehydes to ketones in terms of reactivity. And so the first factor is the polarization of the carbonyl. So we've already seen that aldehydes are more polarized than ketones, and so therefore, the carbonyl carbon is a little bit more positive. And that means the nucleophile can attack that positive charge a little bit more."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so now we can compare aldehydes to ketones in terms of reactivity. And so the first factor is the polarization of the carbonyl. So we've already seen that aldehydes are more polarized than ketones, and so therefore, the carbonyl carbon is a little bit more positive. And that means the nucleophile can attack that positive charge a little bit more. So that's one reason why aldehydes are more reactive than ketones, the polarization of the carbonyl. Another reason has to do with steric hindrance. So when you think about a ketone, and let's say we have some big bulky R groups here on this ketone."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And that means the nucleophile can attack that positive charge a little bit more. So that's one reason why aldehydes are more reactive than ketones, the polarization of the carbonyl. Another reason has to do with steric hindrance. So when you think about a ketone, and let's say we have some big bulky R groups here on this ketone. So those bulky R groups might prevent the nucleophile from attacking. So it turns out there's an optimum angle for the nucleophile to attack the carbonyl carbon. And if you have bulky R groups, they might prevent that."}, {"video_title": "Reactivity of aldehydes and ketones Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So when you think about a ketone, and let's say we have some big bulky R groups here on this ketone. So those bulky R groups might prevent the nucleophile from attacking. So it turns out there's an optimum angle for the nucleophile to attack the carbonyl carbon. And if you have bulky R groups, they might prevent that. They also would interfere with the formation of the tetrahedral intermediate, because if I had big bulky R groups, and I'm changing the bond angle from 120 degrees, approximately 120 degrees over here on the left, to 109 degrees, those bulky R groups have to get closer together in space. And they would, of course, repel. And so there's some steric hindrance upon formation of your tetrahedral intermediate as well."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So there's two types of curly arrows you will see. You will see a curly full arrow like this. So curly full arrow like this. And I made sure to draw it curly. You will always see them curly like this. And you'll see a curly half arrow that looks like this. So a curly half arrow or fishhook arrow."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "And I made sure to draw it curly. You will always see them curly like this. And you'll see a curly half arrow that looks like this. So a curly half arrow or fishhook arrow. So the convention is a full arrow or kind of a typical arrow that you're used to seeing. This is talking about the movement of pairs, of electron pairs. Movement of pairs is the convention."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So a curly half arrow or fishhook arrow. So the convention is a full arrow or kind of a typical arrow that you're used to seeing. This is talking about the movement of pairs, of electron pairs. Movement of pairs is the convention. And I'll show you in a second that I do a slight variation of that. And I do that because it helps me account for electrons. And it helps me at least visualize or conceptualize how things are happening a little bit better."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "Movement of pairs is the convention. And I'll show you in a second that I do a slight variation of that. And I do that because it helps me account for electrons. And it helps me at least visualize or conceptualize how things are happening a little bit better. But the general convention is that this is movement of pairs. And this is movement of electron by itself. So electron not part."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "And it helps me at least visualize or conceptualize how things are happening a little bit better. But the general convention is that this is movement of pairs. And this is movement of electron by itself. So electron not part. Electron by itself. Maybe I'll write it this way. By itself."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So electron not part. Electron by itself. Maybe I'll write it this way. By itself. And the full arrow is what you're going to see through most of organic chemistry. So this is the one that you're going to see most typically. The movement of pairs."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "By itself. And the full arrow is what you're going to see through most of organic chemistry. So this is the one that you're going to see most typically. The movement of pairs. The movement of electrons by itself. This is going to show up more in free radical reactions, which we do do. But this is later on."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "The movement of pairs. The movement of electrons by itself. This is going to show up more in free radical reactions, which we do do. But this is later on. And most of organic chemistry is going to be dealing with the movement of pairs. Now what I've drawn over here is a curly arrow showing the same thing happening. And I'm showing you the slight variation that I do."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "But this is later on. And most of organic chemistry is going to be dealing with the movement of pairs. Now what I've drawn over here is a curly arrow showing the same thing happening. And I'm showing you the slight variation that I do. And I do it because it helps me, once again, account for the electrons. And it helps me conceptualize what is going on. So the typical way that this type of mechanism will be shown is it'll say, OK, you have this electron pair on this oxygen."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "And I'm showing you the slight variation that I do. And I do it because it helps me, once again, account for the electrons. And it helps me conceptualize what is going on. So the typical way that this type of mechanism will be shown is it'll say, OK, you have this electron pair on this oxygen. And this electron pair, sometimes it'll be say, and you will learn about this reaction in not too long, is going to the carbon. Or I guess you could say it's attacking the carbon right over here. Now the reason why this I find a little bit less intuitive is that the whole pair is not going to the carbon."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So the typical way that this type of mechanism will be shown is it'll say, OK, you have this electron pair on this oxygen. And this electron pair, sometimes it'll be say, and you will learn about this reaction in not too long, is going to the carbon. Or I guess you could say it's attacking the carbon right over here. Now the reason why this I find a little bit less intuitive is that the whole pair is not going to the carbon. That the oxygen is still going to maintain half of this pair. And it's going to form a bond. Essentially, one end of this pair is going to end up at the carbon."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "Now the reason why this I find a little bit less intuitive is that the whole pair is not going to the carbon. That the oxygen is still going to maintain half of this pair. And it's going to form a bond. Essentially, one end of this pair is going to end up at the carbon. One end of this pair is going to end up at the oxygen. And they are going to form a bond. So the way I draw it, still drawing the full arrow."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "Essentially, one end of this pair is going to end up at the carbon. One end of this pair is going to end up at the oxygen. And they are going to form a bond. So the way I draw it, still drawing the full arrow. So here, I'm still talking about pairs. But I'm talking about the movement of an electron as part of a pair. So that's kind of the slight nonconventional thing that I do with the full arrow."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So the way I draw it, still drawing the full arrow. So here, I'm still talking about pairs. But I'm talking about the movement of an electron as part of a pair. So that's kind of the slight nonconventional thing that I do with the full arrow. So movement of electron as part of pair. And I'll oftentimes draw the back of the arrow from that electron. But it's important to recognize that that electron is not moving by itself."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So that's kind of the slight nonconventional thing that I do with the full arrow. So movement of electron as part of pair. And I'll oftentimes draw the back of the arrow from that electron. But it's important to recognize that that electron is not moving by itself. It's just ending up on one side of a bond. It is moving as part of a pair. And another way to think about it is this electron is going to be on the other side of the bond."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "But it's important to recognize that that electron is not moving by itself. It's just ending up on one side of a bond. It is moving as part of a pair. And another way to think about it is this electron is going to be on the other side of the bond. And I also want to be clear here again. When I talk about electrons at either side of bonds, I like to think about that because it helps me do it for accounting purposes. But we know that these covalent bonds, these one electron just doesn't sit on one side of the bond, and the other electron doesn't just sit on the other side of the bond."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "And another way to think about it is this electron is going to be on the other side of the bond. And I also want to be clear here again. When I talk about electrons at either side of bonds, I like to think about that because it helps me do it for accounting purposes. But we know that these covalent bonds, these one electron just doesn't sit on one side of the bond, and the other electron doesn't just sit on the other side of the bond. In fact, everything we do in organic chemistry is anywhere near as clean as the way we draw it. But I do this just to remind myself that there are two electrons here. And when you have a bond, there is some probability that one of the electrons is closer to the hydrogen."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "But we know that these covalent bonds, these one electron just doesn't sit on one side of the bond, and the other electron doesn't just sit on the other side of the bond. In fact, everything we do in organic chemistry is anywhere near as clean as the way we draw it. But I do this just to remind myself that there are two electrons here. And when you have a bond, there is some probability that one of the electrons is closer to the hydrogen. And there's some probability that that electron is closer to the carbon. And so you can kind of imagine that there are electrons on either sides of the bond. The actual reality is that there's a blur of them."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "And when you have a bond, there is some probability that one of the electrons is closer to the hydrogen. And there's some probability that that electron is closer to the carbon. And so you can kind of imagine that there are electrons on either sides of the bond. The actual reality is that there's a blur of them. And depending on which molecule is more electronegative, the probability blur is a little bit more weighted on one side or another. But of course, we like to clean things up with these formalisms right over here. So this is kind of the example when you have this attacking pair, why I like to think of the full arrow as the movement of an electron as part of a pair."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "The actual reality is that there's a blur of them. And depending on which molecule is more electronegative, the probability blur is a little bit more weighted on one side or another. But of course, we like to clean things up with these formalisms right over here. So this is kind of the example when you have this attacking pair, why I like to think of the full arrow as the movement of an electron as part of a pair. And then this breaking bond right over here is another example. So in the typical convention, you have this bond here. And remember, this covalent bond right over here is made up of two electrons."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So this is kind of the example when you have this attacking pair, why I like to think of the full arrow as the movement of an electron as part of a pair. And then this breaking bond right over here is another example. So in the typical convention, you have this bond here. And remember, this covalent bond right over here is made up of two electrons. And if they wanted to show this bond breaking and both of these electrons going to this bromine, the convention is to kind of go from the middle of the bond to the bromine. That I've never found that intuitive, because here, once again, bromine already essentially had part of the bond. It was already on one end of the bond."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "And remember, this covalent bond right over here is made up of two electrons. And if they wanted to show this bond breaking and both of these electrons going to this bromine, the convention is to kind of go from the middle of the bond to the bromine. That I've never found that intuitive, because here, once again, bromine already essentially had part of the bond. It was already on one end of the bond. So I like to visualize that it's getting the other electron, that it's now getting both electrons. One part of the bond was already closer to the bromine. Now it's getting the other part of the bond."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "It was already on one end of the bond. So I like to visualize that it's getting the other electron, that it's now getting both electrons. One part of the bond was already closer to the bromine. Now it's getting the other part of the bond. So we're going to use full arrows for these mechanisms, just as we would typically use full arrows. But I'll often conceptualize it as the movement of an electron as part of a pair, as opposed to the entire pair. But the full arrows are still used the way it would be conventionally used."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "Now it's getting the other part of the bond. So we're going to use full arrows for these mechanisms, just as we would typically use full arrows. But I'll often conceptualize it as the movement of an electron as part of a pair, as opposed to the entire pair. But the full arrows are still used the way it would be conventionally used. Later on, when we do free radical reactions, we're going to talk about an electron moving by itself. So notice, this electron right over here, it's moving, or it's doing something, and it's not part of a pair. It's not part of a pair, it's by itself, so we use the fishhook arrows."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "But the full arrows are still used the way it would be conventionally used. Later on, when we do free radical reactions, we're going to talk about an electron moving by itself. So notice, this electron right over here, it's moving, or it's doing something, and it's not part of a pair. It's not part of a pair, it's by itself, so we use the fishhook arrows. Right over here, we see a bond breaking. But instead of both electrons going to one of the atoms, or another one of the atoms, is right over here. So when both electrons went to one of the atoms, we use the full arrow."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "It's not part of a pair, it's by itself, so we use the fishhook arrows. Right over here, we see a bond breaking. But instead of both electrons going to one of the atoms, or another one of the atoms, is right over here. So when both electrons went to one of the atoms, we use the full arrow. This already, you can kind of say, had one, and now it's gaining another one, so we use a full arrow. But here, the bond is breaking, and each electron is going to a different atom. So once again, the electron is moving by itself."}, {"video_title": "Curly arrow conventions in organic chemistry.mp3", "Sentence": "So when both electrons went to one of the atoms, we use the full arrow. This already, you can kind of say, had one, and now it's gaining another one, so we use a full arrow. But here, the bond is breaking, and each electron is going to a different atom. So once again, the electron is moving by itself. Maybe I put this right, moving by itself, and here's the movement of the electron as part of a pair. So hopefully that clarifies it a little bit. And it's important to keep in mind, a lot of the notation I use is a departure from the traditional organic chemistry notation."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Achiral objects are objects that are superimposable on their mirror images. And in a minute I'm going to show you that a coffee cup is an example of an achiral object. Chiral objects are objects that are not superimposable on their mirror images. And the word chiral comes from the Greek word for hand. And so I'm going to show you how your hands are not superimposable on each other, but your left and your right hand are mirror images of each other. Let's take a look at a coffee cup reflected in a mirror. And so you can see on the left here is the actual coffee cup and in the mirror is the mirror image of the coffee cup."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the word chiral comes from the Greek word for hand. And so I'm going to show you how your hands are not superimposable on each other, but your left and your right hand are mirror images of each other. Let's take a look at a coffee cup reflected in a mirror. And so you can see on the left here is the actual coffee cup and in the mirror is the mirror image of the coffee cup. And I'm going to pull out the coffee cup to make some space and I'm going to put what I saw in the mirror, the mirror image, right next to it. So here I have another coffee cup. So that's its mirror image."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so you can see on the left here is the actual coffee cup and in the mirror is the mirror image of the coffee cup. And I'm going to pull out the coffee cup to make some space and I'm going to put what I saw in the mirror, the mirror image, right next to it. So here I have another coffee cup. So that's its mirror image. And I'm going to take the coffee cup on the right, I'm going to rotate it. And so as I rotate it, you can see that it is superimposable with the object on the left. So the mirror image is superimposable."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's its mirror image. And I'm going to take the coffee cup on the right, I'm going to rotate it. And so as I rotate it, you can see that it is superimposable with the object on the left. So the mirror image is superimposable. And that's the definition of an achiral object. So we say that a coffee cup is achiral. Now let's try the same thing with a molecule."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the mirror image is superimposable. And that's the definition of an achiral object. So we say that a coffee cup is achiral. Now let's try the same thing with a molecule. So this is difluoromethane. So the green are the fluorine atoms. And in the mirror, you can see the mirror image."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now let's try the same thing with a molecule. So this is difluoromethane. So the green are the fluorine atoms. And in the mirror, you can see the mirror image. So once again, I'm going to pull the molecule away and put what I saw in the mirror, the mirror image, right next to it. And I'm going to rotate the mirror image. So the one on the right, I'm going to rotate it to see if it's superimposable with the one on the left."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And in the mirror, you can see the mirror image. So once again, I'm going to pull the molecule away and put what I saw in the mirror, the mirror image, right next to it. And I'm going to rotate the mirror image. So the one on the right, I'm going to rotate it to see if it's superimposable with the one on the left. And so I take it and as you can see as I rotate, right there you can see that it is superimposable with the molecule on the left. And since the mirror image is superimposable, we say this is an achiral molecule. So these are actually two of the exact same molecules represented here."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the one on the right, I'm going to rotate it to see if it's superimposable with the one on the left. And so I take it and as you can see as I rotate, right there you can see that it is superimposable with the molecule on the left. And since the mirror image is superimposable, we say this is an achiral molecule. So these are actually two of the exact same molecules represented here. Now let's look at my hand. So my left hand and in the mirror you can see my right hand or what looks like my right hand. So I'm going to take my left and my right hand together."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these are actually two of the exact same molecules represented here. Now let's look at my hand. So my left hand and in the mirror you can see my right hand or what looks like my right hand. So I'm going to take my left and my right hand together. We've just shown that they're mirror images of each other. And then I'm going to try to rotate my right hand to see if it's superimposable with my left hand. And so you can see here I have my palms both up, but my thumbs are not pointing in the same direction."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to take my left and my right hand together. We've just shown that they're mirror images of each other. And then I'm going to try to rotate my right hand to see if it's superimposable with my left hand. And so you can see here I have my palms both up, but my thumbs are not pointing in the same direction. So they're not superimposable here. And so I'm going to try again, I'm going to try to rotate it, right? So now my thumbs are in the same position, but my palms are not in the same position."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so you can see here I have my palms both up, but my thumbs are not pointing in the same direction. So they're not superimposable here. And so I'm going to try again, I'm going to try to rotate it, right? So now my thumbs are in the same position, but my palms are not in the same position. And so no matter what I do, I can never superimpose my right hand on my left hand. So the mirror image is not superimposable upon the original object. So no matter what you do, you cannot accomplish this."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So now my thumbs are in the same position, but my palms are not in the same position. And so no matter what I do, I can never superimpose my right hand on my left hand. So the mirror image is not superimposable upon the original object. So no matter what you do, you cannot accomplish this. And since the mirror image is not superimposable, we say your hands are chiral. Now finally, let's take a look at a molecule. So the white is hydrogen, green is fluorine, red is bromine, and yellow is chlorine."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So no matter what you do, you cannot accomplish this. And since the mirror image is not superimposable, we say your hands are chiral. Now finally, let's take a look at a molecule. So the white is hydrogen, green is fluorine, red is bromine, and yellow is chlorine. And so I have the molecules on the left, and in the mirror is the mirror image. So I'm going to pull out that molecule, and once again leave some space, and put what I saw in the mirror right next to it. So there's the mirror image."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the white is hydrogen, green is fluorine, red is bromine, and yellow is chlorine. And so I have the molecules on the left, and in the mirror is the mirror image. So I'm going to pull out that molecule, and once again leave some space, and put what I saw in the mirror right next to it. So there's the mirror image. So I have these two mirror images of each other. I take the one on the right, and I try to rotate it to see if I can superimpose it with the one on the left. And so here I just rotated it a little bit there."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there's the mirror image. So I have these two mirror images of each other. I take the one on the right, and I try to rotate it to see if I can superimpose it with the one on the left. And so here I just rotated it a little bit there. And you can see that the red, the bromine atoms, are in the same position. However, the chlorine and the fluorine, the yellow and the green, are not in the same position. So I cannot superimpose it here."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so here I just rotated it a little bit there. And you can see that the red, the bromine atoms, are in the same position. However, the chlorine and the fluorine, the yellow and the green, are not in the same position. So I cannot superimpose it here. I try again, I rotate it some more, and you can see the yellow is in the same position, the chlorine, and also the hydrogens. But the red and the green are not. And so no matter how I rotate the mirror image, the one on the right, I can never get it to look like the one on the left."}, {"video_title": "Chiral vs achiral Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I cannot superimpose it here. I try again, I rotate it some more, and you can see the yellow is in the same position, the chlorine, and also the hydrogens. But the red and the green are not. And so no matter how I rotate the mirror image, the one on the right, I can never get it to look like the one on the left. And since the mirror image is not superimposable, we say that this is a chiral molecule. And this carbon here, this is a very important carbon with four different substituents attached to it, four different groups. So we call this a chiral center or a chirality center."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So that's CH4. If I want to draw a dot structure for methane, I would start with carbon and its four valence electrons. And then we put hydrogen around that. Each hydrogen has one valence electron. So we go ahead and draw in our hydrogens with one valence electron. And that gives us the Lewis dot structure. Usually, you see it drawn like this, with carbon with its four bonds to hydrogen around it like that."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Each hydrogen has one valence electron. So we go ahead and draw in our hydrogens with one valence electron. And that gives us the Lewis dot structure. Usually, you see it drawn like this, with carbon with its four bonds to hydrogen around it like that. And in methane, all of these bonds are equivalent in terms of things like bond length and energy. And so the four valence electrons that carbon brought to the table over here, let me go ahead and highlight those four valence electrons. Those should be equivalent."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Usually, you see it drawn like this, with carbon with its four bonds to hydrogen around it like that. And in methane, all of these bonds are equivalent in terms of things like bond length and energy. And so the four valence electrons that carbon brought to the table over here, let me go ahead and highlight those four valence electrons. Those should be equivalent. And if we look at the electron configuration for carbon, let's go ahead and do that right now. It's 1s2. So I go ahead and put in two electrons in the 1s orbital."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Those should be equivalent. And if we look at the electron configuration for carbon, let's go ahead and do that right now. It's 1s2. So I go ahead and put in two electrons in the 1s orbital. 2s2, I go ahead and put in two electrons in the 2s orbital. And then 2p2. And so I'm assuming you already know your electron configuration."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I go ahead and put in two electrons in the 1s orbital. 2s2, I go ahead and put in two electrons in the 2s orbital. And then 2p2. And so I'm assuming you already know your electron configuration. So it would look something like that. If we look at those four valence electrons on our orbital notation here, that would be these four electrons here, the valence electrons in the outer shell. And if we look at this, this implies that carbon would only form two bonds, because I have these unpaired electrons right here."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so I'm assuming you already know your electron configuration. So it would look something like that. If we look at those four valence electrons on our orbital notation here, that would be these four electrons here, the valence electrons in the outer shell. And if we look at this, this implies that carbon would only form two bonds, because I have these unpaired electrons right here. And everything's of different energies. And so what we see from the dot structure and experimentally doesn't quite match up with the electron configuration here. And so to explain this difference, Linus Pauling came up with the idea of hybridization."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And if we look at this, this implies that carbon would only form two bonds, because I have these unpaired electrons right here. And everything's of different energies. And so what we see from the dot structure and experimentally doesn't quite match up with the electron configuration here. And so to explain this difference, Linus Pauling came up with the idea of hybridization. And so the first thing that he said was you could go ahead and take out one of these electrons in the 2s and promote it up to the p orbital here. So let me go ahead and show that. So we've moved one of the electrons up to the 2p orbital."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so to explain this difference, Linus Pauling came up with the idea of hybridization. And so the first thing that he said was you could go ahead and take out one of these electrons in the 2s and promote it up to the p orbital here. So let me go ahead and show that. So we've moved one of the electrons up to the 2p orbital. So we're in the excited state now. And now we have the opportunity for carbon to form four bonds. However, those electrons are not of equivalent energy."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we've moved one of the electrons up to the 2p orbital. So we're in the excited state now. And now we have the opportunity for carbon to form four bonds. However, those electrons are not of equivalent energy. And so Linus Pauling said, let's do something else here. Let's go ahead and promote the 2s orbital. So we're going to take this s orbital and we're going to promote it in energy."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "However, those electrons are not of equivalent energy. And so Linus Pauling said, let's do something else here. Let's go ahead and promote the 2s orbital. So we're going to take this s orbital and we're going to promote it in energy. And we're going to take these p orbitals and demote them in energy. So we're going to lower those p orbitals like that. So we have our p orbitals here."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we're going to take this s orbital and we're going to promote it in energy. And we're going to take these p orbitals and demote them in energy. So we're going to lower those p orbitals like that. So we have our p orbitals here. And these had one electron in each of them. But we're going to hybridize them. So this is no longer going to be an s orbital."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have our p orbitals here. And these had one electron in each of them. But we're going to hybridize them. So this is no longer going to be an s orbital. It's going to be an sp3 hybrid orbital. This is no longer going to be a p orbital. It's going to be an sp3 hybrid orbital."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So this is no longer going to be an s orbital. It's going to be an sp3 hybrid orbital. This is no longer going to be a p orbital. It's going to be an sp3 hybrid orbital. And same with these. So the idea is you're taking some of the s character and some of the p character and you're hybridizing them together into brand new orbitals. And since you're taking this from one s orbital and three p orbitals, we're doing this using one s orbital and three p orbitals, we call this sp3 hybridization."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "It's going to be an sp3 hybrid orbital. And same with these. So the idea is you're taking some of the s character and some of the p character and you're hybridizing them together into brand new orbitals. And since you're taking this from one s orbital and three p orbitals, we're doing this using one s orbital and three p orbitals, we call this sp3 hybridization. So this is sp3 hybridization. We create four new hybrid orbitals. And now we have what we're looking for."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And since you're taking this from one s orbital and three p orbitals, we're doing this using one s orbital and three p orbitals, we call this sp3 hybridization. So this is sp3 hybridization. We create four new hybrid orbitals. And now we have what we're looking for. Because now we have four unpaired electrons. So carbon can form four bonds now. And they're equal in energy."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And now we have what we're looking for. Because now we have four unpaired electrons. So carbon can form four bonds now. And they're equal in energy. So that's the idea of hybridization. Let's think about the character or the shape of this new hybrid orbital. Well, we know that an s orbital is shaped like a sphere."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And they're equal in energy. So that's the idea of hybridization. Let's think about the character or the shape of this new hybrid orbital. Well, we know that an s orbital is shaped like a sphere. So we're taking one of those s orbitals here. So one s orbital. And we know that a p orbital is shaped like a dumbbell."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Well, we know that an s orbital is shaped like a sphere. So we're taking one of those s orbitals here. So one s orbital. And we know that a p orbital is shaped like a dumbbell. So we're taking three of these p orbitals here. So one of these s orbitals and three of these p orbitals. And then we're going to hybridize them together."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And we know that a p orbital is shaped like a dumbbell. So we're taking three of these p orbitals here. So one of these s orbitals and three of these p orbitals. And then we're going to hybridize them together. And when that happens, it turns out the shape of the new hybrid orbital you get has this large frontal lobe here like this and then a smaller back lobe back here like this. So we're going to make four of these. So once again, we have four sp3 hybrid orbitals."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then we're going to hybridize them together. And when that happens, it turns out the shape of the new hybrid orbital you get has this large frontal lobe here like this and then a smaller back lobe back here like this. So we're going to make four of these. So once again, we have four sp3 hybrid orbitals. And each one of these hybrid orbitals is going to have an electron in it. So we can see that each one of these sp3 hybrid orbitals has one electron in there like that. And so the final hybrid orbitals here contain 25% s character."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So once again, we have four sp3 hybrid orbitals. And each one of these hybrid orbitals is going to have an electron in it. So we can see that each one of these sp3 hybrid orbitals has one electron in there like that. And so the final hybrid orbitals here contain 25% s character. Let me go ahead and write this down here. So 25% s character. And 75% p character in this new hybrid orbital."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so the final hybrid orbitals here contain 25% s character. Let me go ahead and write this down here. So 25% s character. And 75% p character in this new hybrid orbital. Once again, that's because we started out with one s orbital and three p orbitals for our hybridization. Let's go back to methane. And let's go ahead and draw in a picture."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And 75% p character in this new hybrid orbital. Once again, that's because we started out with one s orbital and three p orbitals for our hybridization. Let's go back to methane. And let's go ahead and draw in a picture. Because now we know that this carbon is sp3 hybridized. So let's go ahead and draw a picture of that hybridized carbon here. So we're going to go ahead and draw in our carbon."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and draw in a picture. Because now we know that this carbon is sp3 hybridized. So let's go ahead and draw a picture of that hybridized carbon here. So we're going to go ahead and draw in our carbon. And we know that it has four sp3 hybrid orbitals. And once again, when we draw the orbitals, we're going to ignore the smaller backlobe here so it doesn't confuse us. So we go ahead and draw in here's one of our orbitals for carbons."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we're going to go ahead and draw in our carbon. And we know that it has four sp3 hybrid orbitals. And once again, when we draw the orbitals, we're going to ignore the smaller backlobe here so it doesn't confuse us. So we go ahead and draw in here's one of our orbitals for carbons. That's an sp3 hybrid orbital. Here's another sp3 hybrid orbital. Here's another one."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we go ahead and draw in here's one of our orbitals for carbons. That's an sp3 hybrid orbital. Here's another sp3 hybrid orbital. Here's another one. And then finally, a fourth one. So let's go back up here to this picture. Because once again, we need to show that each of these hybrid orbitals has one valence electron in it."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Here's another one. And then finally, a fourth one. So let's go back up here to this picture. Because once again, we need to show that each of these hybrid orbitals has one valence electron in it. So I can go ahead and put in my one valence electron in each of my hybrid orbitals like that. So here's our valence electron. If we're talking about methane, so carbon is bonded to four hydrogens."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Because once again, we need to show that each of these hybrid orbitals has one valence electron in it. So I can go ahead and put in my one valence electron in each of my hybrid orbitals like that. So here's our valence electron. If we're talking about methane, so carbon is bonded to four hydrogens. Each hydrogen has an unhybridized s orbital. And each hydrogen has one electron in that. So I'm going to go ahead and sketch that in."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "If we're talking about methane, so carbon is bonded to four hydrogens. Each hydrogen has an unhybridized s orbital. And each hydrogen has one electron in that. So I'm going to go ahead and sketch that in. Let's go ahead and use blue here. So here's an unhybridized s orbital. I'm going to go ahead and draw these in."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and sketch that in. Let's go ahead and use blue here. So here's an unhybridized s orbital. I'm going to go ahead and draw these in. So an unhybridized s orbital. Each one of these unhybridized s orbitals for hydrogen has one valence electron. So I'm going to go ahead and put in those one valence electrons in here like that."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and draw these in. So an unhybridized s orbital. Each one of these unhybridized s orbitals for hydrogen has one valence electron. So I'm going to go ahead and put in those one valence electrons in here like that. So same for here. And then finally for here. And so this is just one picture of the methane molecule."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and put in those one valence electrons in here like that. So same for here. And then finally for here. And so this is just one picture of the methane molecule. So this is hydrogen. These are all the hydrogens right here like that. So let's think about this bond that we've formed right here."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so this is just one picture of the methane molecule. So this is hydrogen. These are all the hydrogens right here like that. So let's think about this bond that we've formed right here. So here we have an overlap of orbitals, an overlap of an sp3 hybrid orbital from carbon with an unhybridized s orbital from hydrogen here. And so this is a head-on overlap. So we're sharing electrons here in this head-on overlap."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let's think about this bond that we've formed right here. So here we have an overlap of orbitals, an overlap of an sp3 hybrid orbital from carbon with an unhybridized s orbital from hydrogen here. And so this is a head-on overlap. So we're sharing electrons here in this head-on overlap. And a head-on overlap in chemistry is called a sigma bond. So this is a sigma bond here, a head-on overlap. And this happens three more times in the methane molecule."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we're sharing electrons here in this head-on overlap. And a head-on overlap in chemistry is called a sigma bond. So this is a sigma bond here, a head-on overlap. And this happens three more times in the methane molecule. So here's a head-on overlap. Here's a head-on overlap. And here's a head-on overlap."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And this happens three more times in the methane molecule. So here's a head-on overlap. Here's a head-on overlap. And here's a head-on overlap. And so we have a total of four sigma bonds in the methane molecule. So a single bond here is represented. Instead of saying a single bond now, we're saying it also can be called a sigma bond, so this head-on overlap."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And here's a head-on overlap. And so we have a total of four sigma bonds in the methane molecule. So a single bond here is represented. Instead of saying a single bond now, we're saying it also can be called a sigma bond, so this head-on overlap. Let's look at the ethane molecule now. So for ethane, we'll go ahead and draw that in. So ethane would be C2H6."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Instead of saying a single bond now, we're saying it also can be called a sigma bond, so this head-on overlap. Let's look at the ethane molecule now. So for ethane, we'll go ahead and draw that in. So ethane would be C2H6. So we have two carbons. So let's go ahead and draw in those two carbons. And then six hydrogens."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So ethane would be C2H6. So we have two carbons. So let's go ahead and draw in those two carbons. And then six hydrogens. So we put in our six hydrogens around there like that. When we're thinking about hybridization, we've just seen with methane that a carbon atom with four single bonds will be sp3 hybridized. So I go back up to here."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then six hydrogens. So we put in our six hydrogens around there like that. When we're thinking about hybridization, we've just seen with methane that a carbon atom with four single bonds will be sp3 hybridized. So I go back up to here. This carbon right here. Four single bonds. It's sp3 hybridized."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I go back up to here. This carbon right here. Four single bonds. It's sp3 hybridized. We can use that same logic and apply it to ethane here. Each of the carbons in ethane has four single bonds. So each carbon in ethane is sp3 hybridized."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "It's sp3 hybridized. We can use that same logic and apply it to ethane here. Each of the carbons in ethane has four single bonds. So each carbon in ethane is sp3 hybridized. So we go ahead and put sp3 hybridized here. So let's go ahead and draw the picture with the orbitals. So let's get some more room."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So each carbon in ethane is sp3 hybridized. So we go ahead and put sp3 hybridized here. So let's go ahead and draw the picture with the orbitals. So let's get some more room. If each carbon is sp3 hybridized, that means each carbon is going to have four sp3 hybrid orbitals. So I can go ahead and sketch in one carbon. Once again, I'm ignoring the back lobe."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let's get some more room. If each carbon is sp3 hybridized, that means each carbon is going to have four sp3 hybrid orbitals. So I can go ahead and sketch in one carbon. Once again, I'm ignoring the back lobe. One carbon with four sp3 hybrid orbitals. And we know the other carbon is also sp3 hybridized. So I can sketch in four sp3 hybrid orbitals for this one too."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Once again, I'm ignoring the back lobe. One carbon with four sp3 hybrid orbitals. And we know the other carbon is also sp3 hybridized. So I can sketch in four sp3 hybrid orbitals for this one too. So here's my four sp3 hybrid orbitals for this carbon. In terms of electrons, let's go ahead and put in our electrons here. So let's see."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I can sketch in four sp3 hybrid orbitals for this one too. So here's my four sp3 hybrid orbitals for this carbon. In terms of electrons, let's go ahead and put in our electrons here. So let's see. There's one electron in this orbital, one electron in this orbital, one electron in this orbital, one electron from this carbon. And then for this other carbon, so there's one electron in this orbital, one electron in this orbital, one electron in this orbital, and one electron in this orbital. And then we can go ahead and put in our hydrogens."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let's see. There's one electron in this orbital, one electron in this orbital, one electron in this orbital, one electron from this carbon. And then for this other carbon, so there's one electron in this orbital, one electron in this orbital, one electron in this orbital, and one electron in this orbital. And then we can go ahead and put in our hydrogens. So we know each hydrogen has an unhybridized s orbital with one valence electron. So I can go ahead and do that and draw in the rest of our hydrogens. So that's four hydrogens, five hydrogens, and then six hydrogens like that."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then we can go ahead and put in our hydrogens. So we know each hydrogen has an unhybridized s orbital with one valence electron. So I can go ahead and do that and draw in the rest of our hydrogens. So that's four hydrogens, five hydrogens, and then six hydrogens like that. We just said that a sigma bond is a head-on overlap of orbitals. So here we have a head-on overlap of orbitals, the bond between the two carbons. And then, of course, all of these are two."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So that's four hydrogens, five hydrogens, and then six hydrogens like that. We just said that a sigma bond is a head-on overlap of orbitals. So here we have a head-on overlap of orbitals, the bond between the two carbons. And then, of course, all of these are two. So when we count all of those up, that's four, five, six, and seven. So there are seven sigma bonds in the ethane molecule. So seven sigma bonds here."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then, of course, all of these are two. So when we count all of those up, that's four, five, six, and seven. So there are seven sigma bonds in the ethane molecule. So seven sigma bonds here. And we can go up here to this dot structure and look at them again. So here's one, two, three, four, five, six, and then seven. Seven sigma bonds."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So seven sigma bonds here. And we can go up here to this dot structure and look at them again. So here's one, two, three, four, five, six, and then seven. Seven sigma bonds. Let's focus in on the bond between the two carbons now. So this sigma bond right in here. So that's a sigma bond."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Seven sigma bonds. Let's focus in on the bond between the two carbons now. So this sigma bond right in here. So that's a sigma bond. And there's free rotation about this sigma bond. So if you can imagine rotating around this bond. So these carbons can rotate in space."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So that's a sigma bond. And there's free rotation about this sigma bond. So if you can imagine rotating around this bond. So these carbons can rotate in space. And that's going to give different conformations. So you can have different conformations of the ethane molecule, which is in later videos. So you have free rotation about sigma bonds."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So these carbons can rotate in space. And that's going to give different conformations. So you can have different conformations of the ethane molecule, which is in later videos. So you have free rotation about sigma bonds. Let me go ahead and write that, because that's pretty important. So free rotation about sigma bonds. And then the last thing I wanted to point out about the ethane molecule here is the bond length."}, {"video_title": "sp\u00b3 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So you have free rotation about sigma bonds. Let me go ahead and write that, because that's pretty important. So free rotation about sigma bonds. And then the last thing I wanted to point out about the ethane molecule here is the bond length. So the length between this carbon and this carbon. So this bond length in here turns out to be approximately 1.54 angstroms. So you'll see slightly different values in different textbooks."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video, we tried to relate what 13.7 billion years really means by compressing it to 10 years, which is still a pretty long time, at least from our perspective. And we saw that these huge periods of time, everything that's happened since the United States Declaration of Independence, all gets compressed into just five seconds. And that's not much when you think about 10 years. I mean, 10 years, you could wait around a lot. That's a huge amount of time. So hopefully that put things in perspective a little bit. What I want to do in this video is kind of do the same thing, but relate the 13.7 billion years, if that were a distance or the length of a timeline, how big each of these periods would be."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I mean, 10 years, you could wait around a lot. That's a huge amount of time. So hopefully that put things in perspective a little bit. What I want to do in this video is kind of do the same thing, but relate the 13.7 billion years, if that were a distance or the length of a timeline, how big each of these periods would be. So just to start off, the timeline, I'm recording this in high definition right now. There should be 1,280 pixels wide, depending on where you're watching it. So maybe you're watching it on a high def TV."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What I want to do in this video is kind of do the same thing, but relate the 13.7 billion years, if that were a distance or the length of a timeline, how big each of these periods would be. So just to start off, the timeline, I'm recording this in high definition right now. There should be 1,280 pixels wide, depending on where you're watching it. So maybe you're watching it on a high def TV. If the timeline was the width of my screen right over here, so if that was 13.7 billion years from there, let's say this is the beginning of the universe and this is our present time. If that was 13.7 billion years, the amount of time that humans have been on this planet, modern humans, the people who look and think like us, the amount of time that we have been on this planet will not even be a pixel. This little dot I drew is multiple pixels."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So maybe you're watching it on a high def TV. If the timeline was the width of my screen right over here, so if that was 13.7 billion years from there, let's say this is the beginning of the universe and this is our present time. If that was 13.7 billion years, the amount of time that humans have been on this planet, modern humans, the people who look and think like us, the amount of time that we have been on this planet will not even be a pixel. This little dot I drew is multiple pixels. The amount of time we've been here would not even be a pixel. In fact, a pixel, and that dot I drew here is about 4 pixels, but a pixel, just one little pixel on my screen, would represent 11 million years. 11, let me scroll over, it would represent 11 million years."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This little dot I drew is multiple pixels. The amount of time we've been here would not even be a pixel. In fact, a pixel, and that dot I drew here is about 4 pixels, but a pixel, just one little pixel on my screen, would represent 11 million years. 11, let me scroll over, it would represent 11 million years. So if this was the timeline, the dinosaurs would have been extinct about 6 pixels from the end. When the dinosaurs would have gotten extinct, and the amount of time that modern humans are on the planet wouldn't even register here. It would be 1 20th of that pixel over there."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "11, let me scroll over, it would represent 11 million years. So if this was the timeline, the dinosaurs would have been extinct about 6 pixels from the end. When the dinosaurs would have gotten extinct, and the amount of time that modern humans are on the planet wouldn't even register here. It would be 1 20th of that pixel over there. In fact, if we just expanded that very last pixel, I'll do it in yellow, that very last tiny pixel that you can't see, if we were to expand that to the entire length of this screen again, so if we were to just expand just that very last pixel, so I'm saying everything on this yellow line could be contained in that very last dot right over there, then every pixel in this would still be 8,000 years. So the entire period of time that humanity has been on this planet on this scale, so we've broken out this little tiny pixel here all the way out here, 200,000 years, that's on the order of about 25 of these pixels. So 25 of these pixels would be something like that."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would be 1 20th of that pixel over there. In fact, if we just expanded that very last pixel, I'll do it in yellow, that very last tiny pixel that you can't see, if we were to expand that to the entire length of this screen again, so if we were to just expand just that very last pixel, so I'm saying everything on this yellow line could be contained in that very last dot right over there, then every pixel in this would still be 8,000 years. So the entire period of time that humanity has been on this planet on this scale, so we've broken out this little tiny pixel here all the way out here, 200,000 years, that's on the order of about 25 of these pixels. So 25 of these pixels would be something like that. Not even that. So 25 pixels on this screen, at least at my resolution, is about that distance. So this is the fraction of just that pixel."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So 25 of these pixels would be something like that. Not even that. So 25 pixels on this screen, at least at my resolution, is about that distance. So this is the fraction of just that pixel. That is the amount of time humans have been on the planet. If you want the time since Jesus, it would be 1 4th of a pixel on this yellow scale. On this yellow scale, it would be 1 4th of a pixel."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the fraction of just that pixel. That is the amount of time humans have been on the planet. If you want the time since Jesus, it would be 1 4th of a pixel on this yellow scale. On this yellow scale, it would be 1 4th of a pixel. And the amount of time since the Declaration of Independence would be even a more minuscule amount of time. So that's one way to think about it. But these timelines, to some degree, also don't give justice."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "On this yellow scale, it would be 1 4th of a pixel. And the amount of time since the Declaration of Independence would be even a more minuscule amount of time. So that's one way to think about it. But these timelines, to some degree, also don't give justice. So another way to think about it, what if we had a timeline that stretched from here, where I am, in the San Francisco Bay Area, if it stretched 1,300 kilometers to Seattle. So if this thing stretched 1,300 kilometers all the way to Seattle, then how? So this thing, the Big Bang, would be sitting in Seattle."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But these timelines, to some degree, also don't give justice. So another way to think about it, what if we had a timeline that stretched from here, where I am, in the San Francisco Bay Area, if it stretched 1,300 kilometers to Seattle. So if this thing stretched 1,300 kilometers all the way to Seattle, then how? So this thing, the Big Bang, would be sitting in Seattle. Well, the formation of the Earth and the solar system, this would be about 200 kilometers away, a little over 200 kilometers. And in the direction of Seattle, I don't know, that would get us in probably some part of northern California with redwoods and whatnot. But just to give an idea of 200 kilometers, that would get me from where I am near the coast of California to about the Nevada border."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this thing, the Big Bang, would be sitting in Seattle. Well, the formation of the Earth and the solar system, this would be about 200 kilometers away, a little over 200 kilometers. And in the direction of Seattle, I don't know, that would get us in probably some part of northern California with redwoods and whatnot. But just to give an idea of 200 kilometers, that would get me from where I am near the coast of California to about the Nevada border. So still a pretty good distance relative to the entire timeline. The last land dinosaurs, when the Earth got hit by an asteroid, this, so remember, our whole timeline now stretches all the way to Seattle. This event now would have only occurred about 3,000 meters away, or maybe I should just say 3 kilometers away, or roughly about 2 miles."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But just to give an idea of 200 kilometers, that would get me from where I am near the coast of California to about the Nevada border. So still a pretty good distance relative to the entire timeline. The last land dinosaurs, when the Earth got hit by an asteroid, this, so remember, our whole timeline now stretches all the way to Seattle. This event now would have only occurred about 3,000 meters away, or maybe I should just say 3 kilometers away, or roughly about 2 miles. So that still seems like a pretty good distance. But remember, our timeline stretches all the way to Seattle from the San Francisco area. So it's a pretty big timeline."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This event now would have only occurred about 3,000 meters away, or maybe I should just say 3 kilometers away, or roughly about 2 miles. So that still seems like a pretty good distance. But remember, our timeline stretches all the way to Seattle from the San Francisco area. So it's a pretty big timeline. So this is already a pretty small distance. But it gets even smaller. Australopithecus, this right over here, when they were roaming the Earth 3 million years ago, this gets reduced to a little bit more than a football field."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's a pretty big timeline. So this is already a pretty small distance. But it gets even smaller. Australopithecus, this right over here, when they were roaming the Earth 3 million years ago, this gets reduced to a little bit more than a football field. About 150 meters, not kilometers. Let me write that in a color where there's some contrast. 150 meters is where, so 150 meters in the direction of Seattle, if that's the timeline that we're talking about."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Australopithecus, this right over here, when they were roaming the Earth 3 million years ago, this gets reduced to a little bit more than a football field. About 150 meters, not kilometers. Let me write that in a color where there's some contrast. 150 meters is where, so 150 meters in the direction of Seattle, if that's the timeline that we're talking about. And then the first humans, even shorter. It's only going to be 10 meters of this timeline that stretches all the way to Seattle. The birth of Jesus 2,000 years ago would be 10 centimeters."}, {"video_title": "Cosmological time scale 2 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "150 meters is where, so 150 meters in the direction of Seattle, if that's the timeline that we're talking about. And then the first humans, even shorter. It's only going to be 10 meters of this timeline that stretches all the way to Seattle. The birth of Jesus 2,000 years ago would be 10 centimeters. 10 centimeters on a timeline that stretches from San Francisco to Seattle. And then finally, the Declaration of Independence, the American Declaration of Independence would be 1 centimeter on a timeline that stretches to Seattle. Anyway, I don't know if that put things in more or less perspective, but it was worth giving it a shot."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Or I didn't explain why we made the decision. We said, look, this hydrogen is going to be partially positive because this guy is so electronegative. And maybe when it's partially positive, it'll be attracted, maybe it'll just bump in just the right way into one of these carbons. It'll maybe swipe its electron. We, somewhat arbitrarily in the last video, decided that it would swipe this guy's electron. But you could just as easily imagine a world where it swipes an electron from this guy. So let's draw a mechanism for that and just think about which one is more likely to actually happen."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It'll maybe swipe its electron. We, somewhat arbitrarily in the last video, decided that it would swipe this guy's electron. But you could just as easily imagine a world where it swipes an electron from this guy. So let's draw a mechanism for that and just think about which one is more likely to actually happen. So what happens? So once again, this guy, let me draw all of his valence electrons. So this is the bromine."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw a mechanism for that and just think about which one is more likely to actually happen. So what happens? So once again, this guy, let me draw all of his valence electrons. So this is the bromine. 1, 2, 3, 4, 5, 6, 7 valence electrons. You have the hydrogen. I'll do it the same color."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is the bromine. 1, 2, 3, 4, 5, 6, 7 valence electrons. You have the hydrogen. I'll do it the same color. The hydrogen has its electron right there. This is partially positive and this is partially negative. The hydrogen might want to swipe one of these electrons away."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'll do it the same color. The hydrogen has its electron right there. This is partially positive and this is partially negative. The hydrogen might want to swipe one of these electrons away. Let's do it from this guy right here. Let's do it from this guy. So he has this electron right over here."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen might want to swipe one of these electrons away. Let's do it from this guy right here. Let's do it from this guy. So he has this electron right over here. So the other side of that bond. It goes to the hydrogen when the hydrogen goes near it or maybe it's attracted to it. And when it goes to the hydrogen, then the hydrogen lets go of the electron that the bromine wanted all along because it's so electronegative."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So he has this electron right over here. So the other side of that bond. It goes to the hydrogen when the hydrogen goes near it or maybe it's attracted to it. And when it goes to the hydrogen, then the hydrogen lets go of the electron that the bromine wanted all along because it's so electronegative. So then that electron goes to the bromine. So after we do that, what will it look like? What will be the next step in our reaction?"}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And when it goes to the hydrogen, then the hydrogen lets go of the electron that the bromine wanted all along because it's so electronegative. So then that electron goes to the bromine. So after we do that, what will it look like? What will be the next step in our reaction? And it will look fundamentally different than this right over here. So now what happens? So we have a carbon bonded to two hydrogens and only has a single bond to the other carbon, which is bonded to the original hydrogen right over there."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What will be the next step in our reaction? And it will look fundamentally different than this right over here. So now what happens? So we have a carbon bonded to two hydrogens and only has a single bond to the other carbon, which is bonded to the original hydrogen right over there. The rest, let me write my hydrogens a little bit, actually let me write this whole thing a little bit neater. So you have your carbon bonded to a hydrogen and another hydrogen, and now it only has a single bond to this carbon right here, which is bonded to a hydrogen, and then the rest of the chain. Let me just draw the rest of the chain right here."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have a carbon bonded to two hydrogens and only has a single bond to the other carbon, which is bonded to the original hydrogen right over there. The rest, let me write my hydrogens a little bit, actually let me write this whole thing a little bit neater. So you have your carbon bonded to a hydrogen and another hydrogen, and now it only has a single bond to this carbon right here, which is bonded to a hydrogen, and then the rest of the chain. Let me just draw the rest of the chain right here. And now this electron went to the hydrogen. The other electron that it was paired with is still with this carbon. So now this carbon is now bonded to that hydrogen over there."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me just draw the rest of the chain right here. And now this electron went to the hydrogen. The other electron that it was paired with is still with this carbon. So now this carbon is now bonded to that hydrogen over there. So this blue electron is now with the hydrogen. So let me draw, so the blue electron that was here has now gone over to the orange hydrogen. Let me draw it a little bit neater than that."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now this carbon is now bonded to that hydrogen over there. So this blue electron is now with the hydrogen. So let me draw, so the blue electron that was here has now gone over to the orange hydrogen. Let me draw it a little bit neater than that. Has now gone over to this hydrogen right over there. And then the hydrogen lost its electron to the bromine. So the bromine originally had seven valence electrons."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw it a little bit neater than that. Has now gone over to this hydrogen right over there. And then the hydrogen lost its electron to the bromine. So the bromine originally had seven valence electrons. 1, 2, 3, 4, 5, 6, 7. And then it nabbed an extra electron from the hydrogen. So now it will have a negative charge."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the bromine originally had seven valence electrons. 1, 2, 3, 4, 5, 6, 7. And then it nabbed an extra electron from the hydrogen. So now it will have a negative charge. It is a negative ion. It is bromide. It's a bromide anion, I guess you could call it."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now it will have a negative charge. It is a negative ion. It is bromide. It's a bromide anion, I guess you could call it. And since this guy lost an electron, he had four valence electrons, lost one to the hydrogen, he now has a positive charge. He's a carbocation. And then from here, so notice the difference."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's a bromide anion, I guess you could call it. And since this guy lost an electron, he had four valence electrons, lost one to the hydrogen, he now has a positive charge. He's a carbocation. And then from here, so notice the difference. Before, this guy lost the electron, and so the hydrogen bonded to this carbon. In this situation, this guy lost the electron, so the hydrogen bonded to the other carbon. And so you can imagine from here, something very similar happens as what happened in the first video, but now it happens to this carbon right over there."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then from here, so notice the difference. Before, this guy lost the electron, and so the hydrogen bonded to this carbon. In this situation, this guy lost the electron, so the hydrogen bonded to the other carbon. And so you can imagine from here, something very similar happens as what happened in the first video, but now it happens to this carbon right over there. So let's do that. This carbon is positive. The bromide ion is obviously negative, so maybe he'll want to swipe his electron away."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so you can imagine from here, something very similar happens as what happened in the first video, but now it happens to this carbon right over there. So let's do that. This carbon is positive. The bromide ion is obviously negative, so maybe he'll want to swipe his electron away. So this electron then goes to the carbocation, and then it will form a bond. This green one will go to the carbocation, and then this purple one still stays with the bromine, and so they'll have a bond. They're paired up, you can imagine it."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The bromide ion is obviously negative, so maybe he'll want to swipe his electron away. So this electron then goes to the carbocation, and then it will form a bond. This green one will go to the carbocation, and then this purple one still stays with the bromine, and so they'll have a bond. They're paired up, you can imagine it. So then we're left with, we have a carbon, we have our original hydrogens, we have this carbon, that hydrogen, the rest of the chain, CH2CH3, and then you have this hydrogen right here that it bonded to. That was kind of our first step. And now the bromine has bonded to this carbon right over here, the bromine has bonded over to that carbon right over there."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "They're paired up, you can imagine it. So then we're left with, we have a carbon, we have our original hydrogens, we have this carbon, that hydrogen, the rest of the chain, CH2CH3, and then you have this hydrogen right here that it bonded to. That was kind of our first step. And now the bromine has bonded to this carbon right over here, the bromine has bonded over to that carbon right over there. And we're done. This is another possible mechanism. This one we ended up with 2-bromopentane, right, because it's on the number 2 carbon."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now the bromine has bonded to this carbon right over here, the bromine has bonded over to that carbon right over there. And we're done. This is another possible mechanism. This one we ended up with 2-bromopentane, right, because it's on the number 2 carbon. Here we have 1-bromopentane, 1, 2, 3, 4, 5, still 5 carbons. It's just the bromine is attached to the 1 carbon here, attached to the 2 carbon here. So we now need to think about, because on a first cut, these both seem like reasonable mechanisms."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This one we ended up with 2-bromopentane, right, because it's on the number 2 carbon. Here we have 1-bromopentane, 1, 2, 3, 4, 5, still 5 carbons. It's just the bromine is attached to the 1 carbon here, attached to the 2 carbon here. So we now need to think about, because on a first cut, these both seem like reasonable mechanisms. But if you did it experimentally, you would see that this is the one that you would really observe. I actually haven't done this exact experiment, so I don't know the proportions, but you're going to observe this one disproportionately. The great majority of the products that you see are going to be this one, not that one."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we now need to think about, because on a first cut, these both seem like reasonable mechanisms. But if you did it experimentally, you would see that this is the one that you would really observe. I actually haven't done this exact experiment, so I don't know the proportions, but you're going to observe this one disproportionately. The great majority of the products that you see are going to be this one, not that one. And so the question is, well, they both seem like reasonable things to do up here. Why is this one so much more likely to happen than that one? It all comes from something called Markovnikov's rule."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The great majority of the products that you see are going to be this one, not that one. And so the question is, well, they both seem like reasonable things to do up here. Why is this one so much more likely to happen than that one? It all comes from something called Markovnikov's rule. And there's a couple of ways to think about it. And when Markovnikov thought it up, or he observed it, it seemed to work. They weren't 100% sure about why it worked, but we can even think a little bit about why it worked."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It all comes from something called Markovnikov's rule. And there's a couple of ways to think about it. And when Markovnikov thought it up, or he observed it, it seemed to work. They weren't 100% sure about why it worked, but we can even think a little bit about why it worked. So Markovnikov's rule, a couple ways you could think about it. You could think of it as the thing that already has more hydrogens is more likely to get more hydrogens. So that's what happened here."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "They weren't 100% sure about why it worked, but we can even think a little bit about why it worked. So Markovnikov's rule, a couple ways you could think about it. You could think of it as the thing that already has more hydrogens is more likely to get more hydrogens. So that's what happened here. This thing had more hydrogens on it than the right carbon right here. This right carbon had a hydrogen, but it had some other, I guess you'd call it an alkyl group attached to it. And so the thing that had more hydrogens ended up with the hydrogen, and then the thing that had more groups, this character right here, had more groups."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's what happened here. This thing had more hydrogens on it than the right carbon right here. This right carbon had a hydrogen, but it had some other, I guess you'd call it an alkyl group attached to it. And so the thing that had more hydrogens ended up with the hydrogen, and then the thing that had more groups, this character right here, had more groups. He had one group over here. This carbon over here had no groups. He ended up with the bromine."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so the thing that had more hydrogens ended up with the hydrogen, and then the thing that had more groups, this character right here, had more groups. He had one group over here. This carbon over here had no groups. He ended up with the bromine. So the thing that has more hydrogens ends up with more hydrogens. The thing that has more groups ends up with more groups. So I guess you kind of go more in the direction that you were going in."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "He ended up with the bromine. So the thing that has more hydrogens ends up with more hydrogens. The thing that has more groups ends up with more groups. So I guess you kind of go more in the direction that you were going in. But that still is just a rule. So why does that make sense? It all makes sense, or it starts to make sense, when you think about that in both mechanisms we had to have a carbocation."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I guess you kind of go more in the direction that you were going in. But that still is just a rule. So why does that make sense? It all makes sense, or it starts to make sense, when you think about that in both mechanisms we had to have a carbocation. We talked about in the last video. Carbocation. Carbocation."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It all makes sense, or it starts to make sense, when you think about that in both mechanisms we had to have a carbocation. We talked about in the last video. Carbocation. Carbocation. We had a carbocation right over there. This is the left carbon being a carbocation. This is the right carbon being a carbocation."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Carbocation. We had a carbocation right over there. This is the left carbon being a carbocation. This is the right carbon being a carbocation. And Markovnikov's rule all comes from which carbocation is more stable, which one has a lower energy level. It turns out that the carbocation that is bonded to more, I guess, electron-rich molecules or atoms is going to be more stable. You can imagine that it has more things that, look, it's positive, but it has more carbons around it."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is the right carbon being a carbocation. And Markovnikov's rule all comes from which carbocation is more stable, which one has a lower energy level. It turns out that the carbocation that is bonded to more, I guess, electron-rich molecules or atoms is going to be more stable. You can imagine that it has more things that, look, it's positive, but it has more carbons around it. So it can share some of those electrons. The electron clouds will help it out a little bit to be a little bit more stable. This one right here is only bonded to one other carbon."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You can imagine that it has more things that, look, it's positive, but it has more carbons around it. So it can share some of those electrons. The electron clouds will help it out a little bit to be a little bit more stable. This one right here is only bonded to one other carbon. So not as much sharing. This is bonded to two. So in general, when you're only bonded to one other carbon, you're called a primary carbon."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This one right here is only bonded to one other carbon. So not as much sharing. This is bonded to two. So in general, when you're only bonded to one other carbon, you're called a primary carbon. And if you're a carbocation, this is a primary carbocation right here. This guy is bonded to two carbons, so he would be called a secondary carbon. Since it's a carbocation, it's a secondary carbocation."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in general, when you're only bonded to one other carbon, you're called a primary carbon. And if you're a carbocation, this is a primary carbocation right here. This guy is bonded to two carbons, so he would be called a secondary carbon. Since it's a carbocation, it's a secondary carbocation. So this right here is secondary. So a secondary carbocation is more stable than a primary and actually a tertiary. If you had another carbon group here or something else that had a lot of electrons around it, that would be even more stable."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Since it's a carbocation, it's a secondary carbocation. So this right here is secondary. So a secondary carbocation is more stable than a primary and actually a tertiary. If you had another carbon group here or something else that had a lot of electrons around it, that would be even more stable. So bonded to three things, more stable than two things. And when I say two things, two things other than hydrogen. And then more stable than one."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If you had another carbon group here or something else that had a lot of electrons around it, that would be even more stable. So bonded to three things, more stable than two things. And when I say two things, two things other than hydrogen. And then more stable than one. So Markovnikov's rule all is a byproduct of the fact that this carbocation is more stable than this one over here, and that's because it's secondary versus primary. Because it's secondary, it can kind of borrow electrons from some of its friends. It has more neighbors to borrow electrons than this one."}, {"video_title": "Markovnikov's rule and carbocations Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then more stable than one. So Markovnikov's rule all is a byproduct of the fact that this carbocation is more stable than this one over here, and that's because it's secondary versus primary. Because it's secondary, it can kind of borrow electrons from some of its friends. It has more neighbors to borrow electrons than this one. And since this is more stable, this is more likely to happen. This is a more likely intermediate to have. This is a less likely outcome to have in general."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But what happens when you start with a substituent already on your benzene ring? So we'll look at this molecule over here on the left, which is methoxybenzene. So now we have a methoxy substituent on our benzene ring. And if we react methoxybenzene with concentrated nitric and sulfuric acids, you should recognize that as being a nitration reaction, which will install a nitro group onto your aromatic ring. The presence of that methoxy substituent is going to affect where that nitro group goes on your ring. So for example, one of the observed products is to put the nitro group on this carbon, which is the carbon right next to the carbon that has our methoxy substituent. We say that these two groups are ortho to each other."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if we react methoxybenzene with concentrated nitric and sulfuric acids, you should recognize that as being a nitration reaction, which will install a nitro group onto your aromatic ring. The presence of that methoxy substituent is going to affect where that nitro group goes on your ring. So for example, one of the observed products is to put the nitro group on this carbon, which is the carbon right next to the carbon that has our methoxy substituent. We say that these two groups are ortho to each other. So this would be the ortho product. So if I go back over here on this diagram, I can label this carbon as being the ortho position on my benzene ring. The other product that we observe is nitro group installed on this carbon, which is on the opposite side from the carbon containing our methoxy substituent."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We say that these two groups are ortho to each other. So this would be the ortho product. So if I go back over here on this diagram, I can label this carbon as being the ortho position on my benzene ring. The other product that we observe is nitro group installed on this carbon, which is on the opposite side from the carbon containing our methoxy substituent. We call this the para product. So let me go ahead and label the para position on our benzene ring over here. So this would be the para position."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The other product that we observe is nitro group installed on this carbon, which is on the opposite side from the carbon containing our methoxy substituent. We call this the para product. So let me go ahead and label the para position on our benzene ring over here. So this would be the para position. Between the ortho and the para products, it turns out that the para product is the observed major product in the reaction of the nitration of methoxybenzene. And the ortho product is the minor product. And this has to do with mostly thinking about steric hindrance and this methoxy group as having some steric hindrance, which obviously would prevent this nitro group from adding onto the ortho position easily."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this would be the para position. Between the ortho and the para products, it turns out that the para product is the observed major product in the reaction of the nitration of methoxybenzene. And the ortho product is the minor product. And this has to do with mostly thinking about steric hindrance and this methoxy group as having some steric hindrance, which obviously would prevent this nitro group from adding onto the ortho position easily. Obviously, the para position would have much less steric hindrance. And that's the reason that you usually see for the para product being the major product for this reaction. Now, this isn't always the case."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this has to do with mostly thinking about steric hindrance and this methoxy group as having some steric hindrance, which obviously would prevent this nitro group from adding onto the ortho position easily. Obviously, the para position would have much less steric hindrance. And that's the reason that you usually see for the para product being the major product for this reaction. Now, this isn't always the case. Sometimes your ortho product is more than your para. But for this reaction, the para product is the major product. And once again, steric hindrance is one factor to think about when you're doing these kinds of reactions here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, this isn't always the case. Sometimes your ortho product is more than your para. But for this reaction, the para product is the major product. And once again, steric hindrance is one factor to think about when you're doing these kinds of reactions here. So there's, of course, another position on your aromatic ring. So if we installed the nitro group on this position, we would call this the meta product. And the meta product is not observed in high yields for this reaction."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, steric hindrance is one factor to think about when you're doing these kinds of reactions here. So there's, of course, another position on your aromatic ring. So if we installed the nitro group on this position, we would call this the meta product. And the meta product is not observed in high yields for this reaction. So we say that the methoxy group is an ortho para director. And we could also label this as being the ortho position on this side. And we could say this is the meta position because of symmetry."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the meta product is not observed in high yields for this reaction. So we say that the methoxy group is an ortho para director. And we could also label this as being the ortho position on this side. And we could say this is the meta position because of symmetry. But the meta product is not seen in a large yield, in a high yield. And let's go ahead and look at why by drawing a bunch of resonance structures and thinking about the mechanism for electrophilic aromatic substitution. And so let's go ahead and start with an ortho attack."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we could say this is the meta position because of symmetry. But the meta product is not seen in a large yield, in a high yield. And let's go ahead and look at why by drawing a bunch of resonance structures and thinking about the mechanism for electrophilic aromatic substitution. And so let's go ahead and start with an ortho attack. And so we know that when you're doing a nitration, the sulfuric acid acts as a catalyst to generate the nitronium ion from nitric acid. And that functions as the electrophile in your mechanism here. So if we're going to do an ortho attack, we need to show the nitro group adding onto the ortho position."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and start with an ortho attack. And so we know that when you're doing a nitration, the sulfuric acid acts as a catalyst to generate the nitronium ion from nitric acid. And that functions as the electrophile in your mechanism here. So if we're going to do an ortho attack, we need to show the nitro group adding onto the ortho position. So we need to show the nitro group adding onto this carbon. And so if the nitro group is going to add onto this carbon, then these are the pi electrons that can function as a nucleophile in our mechanism. So we have a nucleophile-electrophile reaction for the first step of our mechanism."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if we're going to do an ortho attack, we need to show the nitro group adding onto the ortho position. So we need to show the nitro group adding onto this carbon. And so if the nitro group is going to add onto this carbon, then these are the pi electrons that can function as a nucleophile in our mechanism. So we have a nucleophile-electrophile reaction for the first step of our mechanism. So the nucleophile, these pi electrons are going to attack that positively charged nitrogen, which kicks these electrons off onto the oxygen. So if we draw the result of that nucleophilic attack, we still have our methoxy substituent up here. I'm showing the nitro group adding onto the ortho position."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a nucleophile-electrophile reaction for the first step of our mechanism. So the nucleophile, these pi electrons are going to attack that positively charged nitrogen, which kicks these electrons off onto the oxygen. So if we draw the result of that nucleophilic attack, we still have our methoxy substituent up here. I'm showing the nitro group adding onto the ortho position. And remember, there's still a hydrogen attached to that carbon. So I have pi electrons over here, pi electrons over here. And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'm showing the nitro group adding onto the ortho position. And remember, there's still a hydrogen attached to that carbon. So I have pi electrons over here, pi electrons over here. And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that. That takes away a bond from this carbon. So that carbon gets a plus 1 formal charge. We can show some resonance structures."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I'm saying that these pi electrons are the ones that formed a bond with this nitrogen like that. That takes away a bond from this carbon. So that carbon gets a plus 1 formal charge. We can show some resonance structures. So we can show some resonance stabilization of this cation here. So I could show these pi electrons moving over to here. And we could draw another resonance structure."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We can show some resonance structures. So we can show some resonance stabilization of this cation here. So I could show these pi electrons moving over to here. And we could draw another resonance structure. So let's go ahead and show the movement of those pi electrons over to this position. So let me go ahead and draw in the rest of the ion here. So we have a hydrogen here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we could draw another resonance structure. So let's go ahead and show the movement of those pi electrons over to this position. So let me go ahead and draw in the rest of the ion here. So we have a hydrogen here. We have an NO2 here. And we took these pi electrons right here, moved them over to this position, took a bond away from that carbon. So we get a positive 1 formal charge on this carbon."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a hydrogen here. We have an NO2 here. And we took these pi electrons right here, moved them over to this position, took a bond away from that carbon. So we get a positive 1 formal charge on this carbon. And that's another resonance structure. We can draw another one. We can show the movement of these pi electrons into here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we get a positive 1 formal charge on this carbon. And that's another resonance structure. We can draw another one. We can show the movement of these pi electrons into here. So let's go ahead and show that. We have our ring. We have our methoxy group."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We can show the movement of these pi electrons into here. So let's go ahead and show that. We have our ring. We have our methoxy group. We have, once again, the nitro group in the ortho position. And we have these pi electrons here. And now we showed the movement of those pi electrons over to here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our methoxy group. We have, once again, the nitro group in the ortho position. And we have these pi electrons here. And now we showed the movement of those pi electrons over to here. So let me go ahead and highlight those. These electrons in red move down to here, took a bond away from this carbon. So that carbon is the one that gets a plus 1 formal charge now."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now we showed the movement of those pi electrons over to here. So let me go ahead and highlight those. These electrons in red move down to here, took a bond away from this carbon. So that carbon is the one that gets a plus 1 formal charge now. Since the oxygen is right next to this carbon, the oxygen has a lone pair of electrons. And so that lone pair of electrons can give us yet another resonance structure. So these electrons could move into here to draw a fourth resonance structure."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon is the one that gets a plus 1 formal charge now. Since the oxygen is right next to this carbon, the oxygen has a lone pair of electrons. And so that lone pair of electrons can give us yet another resonance structure. So these electrons could move into here to draw a fourth resonance structure. So the presence of that methoxy substituent with the lone pair of electrons on that oxygen allows you to draw a fourth resonance structure. So this would give this oxygen a plus 1 formal charge. We have these pi electrons over here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons could move into here to draw a fourth resonance structure. So the presence of that methoxy substituent with the lone pair of electrons on that oxygen allows you to draw a fourth resonance structure. So this would give this oxygen a plus 1 formal charge. We have these pi electrons over here. We have our nitro group, once again, in the ortho position. And let me just go ahead and show the movement of those electrons. So these electrons, I'll make them green."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have these pi electrons over here. We have our nitro group, once again, in the ortho position. And let me just go ahead and show the movement of those electrons. So these electrons, I'll make them green. These electrons right here are going to move in to form our pi bond like that. And we have a total of four possible resonance structures. Remember, the ion is actually a hybrid of these four resonance structures, which we call our sigma complex."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons, I'll make them green. These electrons right here are going to move in to form our pi bond like that. And we have a total of four possible resonance structures. Remember, the ion is actually a hybrid of these four resonance structures, which we call our sigma complex. And so again, the presence of this methoxy substituent with a lone pair of electrons right next to our aromatic ring gives us an extra resonance structure. This one is our extra resonance structure. And so we have a total of four resonance structures for an ortho attack."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Remember, the ion is actually a hybrid of these four resonance structures, which we call our sigma complex. And so again, the presence of this methoxy substituent with a lone pair of electrons right next to our aromatic ring gives us an extra resonance structure. This one is our extra resonance structure. And so we have a total of four resonance structures for an ortho attack. And the more resonance structures you can draw, the more that positive charge is delocalized and the more stable your sigma complex is. The more stable your sigma complex is, the more likely it is to form in your mechanism for electrophilic aromatic substitution. And so because we can draw four resonance structures for an ortho attack, that is a favored carbocation."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we have a total of four resonance structures for an ortho attack. And the more resonance structures you can draw, the more that positive charge is delocalized and the more stable your sigma complex is. The more stable your sigma complex is, the more likely it is to form in your mechanism for electrophilic aromatic substitution. And so because we can draw four resonance structures for an ortho attack, that is a favored carbocation. That's a stable sigma complex. And that's going to form much more easily. And of course, the last step in your mechanism is to deprotonate your sigma complex to reform your aromatic ring."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so because we can draw four resonance structures for an ortho attack, that is a favored carbocation. That's a stable sigma complex. And that's going to form much more easily. And of course, the last step in your mechanism is to deprotonate your sigma complex to reform your aromatic ring. But I'm not going to show that step here. I just wanted you to see the four possible resonance structures for an ortho attack. That's different from a meta attack."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, the last step in your mechanism is to deprotonate your sigma complex to reform your aromatic ring. But I'm not going to show that step here. I just wanted you to see the four possible resonance structures for an ortho attack. That's different from a meta attack. So let's go ahead and look at what would happen if we added on our nitro group meta. So the meta position would be, of course, this one right here. So we would use these pi electrons, so nucleophilic attack, and kicks these electrons off onto our oxygen."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That's different from a meta attack. So let's go ahead and look at what would happen if we added on our nitro group meta. So the meta position would be, of course, this one right here. So we would use these pi electrons, so nucleophilic attack, and kicks these electrons off onto our oxygen. So let's go ahead and show the result of our nucleophilic attack and adding on our nitro group in the meta position. So if we're going to show the nitro group in the meta position like that, let's highlight these electrons here. So these pi electrons come off onto and form a bond, I should say, with that nitrogen, taking a bond away from that top carbon."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we would use these pi electrons, so nucleophilic attack, and kicks these electrons off onto our oxygen. So let's go ahead and show the result of our nucleophilic attack and adding on our nitro group in the meta position. So if we're going to show the nitro group in the meta position like that, let's highlight these electrons here. So these pi electrons come off onto and form a bond, I should say, with that nitrogen, taking a bond away from that top carbon. So that top carbon is going to get a plus 1 formal charge. And of course, we have our other pi electrons in our ring like that. So resonance structure, I could draw a resonance structure for this ion here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons come off onto and form a bond, I should say, with that nitrogen, taking a bond away from that top carbon. So that top carbon is going to get a plus 1 formal charge. And of course, we have our other pi electrons in our ring like that. So resonance structure, I could draw a resonance structure for this ion here. I could take these pi electrons, move them over to here. So we'll go ahead and show another resonance structure. So I have those pi electrons moving over to there."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So resonance structure, I could draw a resonance structure for this ion here. I could take these pi electrons, move them over to here. So we'll go ahead and show another resonance structure. So I have those pi electrons moving over to there. I had these pi electrons here. I still have my nitro group in the meta position to my original methoxy substituent like that. And now my positive 1 formal charge is on this carbon right here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I have those pi electrons moving over to there. I had these pi electrons here. I still have my nitro group in the meta position to my original methoxy substituent like that. And now my positive 1 formal charge is on this carbon right here. So to save time, I'm not going to color coordinate these resonance structures. I can draw one more. I could show these pi electrons moving over to here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now my positive 1 formal charge is on this carbon right here. So to save time, I'm not going to color coordinate these resonance structures. I can draw one more. I could show these pi electrons moving over to here. And let's go ahead and draw that resonance structure. So I have those pi electrons move over to here. I have these pi electrons."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I could show these pi electrons moving over to here. And let's go ahead and draw that resonance structure. So I have those pi electrons move over to here. I have these pi electrons. I have my methoxy substituent. I have, once again, my nitro group. And of course, now my positive charge moves down to this carbon down here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I have these pi electrons. I have my methoxy substituent. I have, once again, my nitro group. And of course, now my positive charge moves down to this carbon down here. And that's it. I can only draw three resonance structures. So once again, the actual sigma complex is a hybrid of these three resonance structures."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, now my positive charge moves down to this carbon down here. And that's it. I can only draw three resonance structures. So once again, the actual sigma complex is a hybrid of these three resonance structures. And since I can only draw three, only a total of three resonance structures for this situation, this sigma complex is not as stable as the one that we saw for an ortho attack. So the sigma complex for an ortho attack was more stable because we could draw four resonance structures. And here we can only draw three."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, the actual sigma complex is a hybrid of these three resonance structures. And since I can only draw three, only a total of three resonance structures for this situation, this sigma complex is not as stable as the one that we saw for an ortho attack. So the sigma complex for an ortho attack was more stable because we could draw four resonance structures. And here we can only draw three. Let's go ahead and show a para attack. So if I want to show my nitro group adding onto the para position, so here is the para position. So I'm going to use these pi electrons."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And here we can only draw three. Let's go ahead and show a para attack. So if I want to show my nitro group adding onto the para position, so here is the para position. So I'm going to use these pi electrons. So nucleophilic attack kicks these electrons off. So this would be our para attack, adding on our nitro group in the para position. So once again, I have my methoxy group."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to use these pi electrons. So nucleophilic attack kicks these electrons off. So this would be our para attack, adding on our nitro group in the para position. So once again, I have my methoxy group. So this time I'm going to show my nitro group in the para position. So let's go ahead and follow these pi electrons. So these pi electrons in here, those pi electrons formed a bond with that nitrogen."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, I have my methoxy group. So this time I'm going to show my nitro group in the para position. So let's go ahead and follow these pi electrons. So these pi electrons in here, those pi electrons formed a bond with that nitrogen. And I took a bond away from this carbon. So that is where my plus 1 formal charge is going to be. I still have these pi electrons in my ring like that."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons in here, those pi electrons formed a bond with that nitrogen. And I took a bond away from this carbon. So that is where my plus 1 formal charge is going to be. I still have these pi electrons in my ring like that. So a resonance structure for this one, I could show these electrons in here moving over to there. And let's see what we would get. We would still have, once again, this substituent."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I still have these pi electrons in my ring like that. So a resonance structure for this one, I could show these electrons in here moving over to there. And let's see what we would get. We would still have, once again, this substituent. Our pi electrons moved over to here. These pi electrons were here. I had my nitro group para."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We would still have, once again, this substituent. Our pi electrons moved over to here. These pi electrons were here. I had my nitro group para. And now I took a bond away from this carbon. So that is the carbon that gets the plus 1 formal charge. So I can draw another resonance structure."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I had my nitro group para. And now I took a bond away from this carbon. So that is the carbon that gets the plus 1 formal charge. So I can draw another resonance structure. I could show these electrons in here moving to there. So let's get a little more room here for more resonance structures. So I can show my ring."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I can draw another resonance structure. I could show these electrons in here moving to there. So let's get a little more room here for more resonance structures. So I can show my ring. And I can show these pi electrons have moved over to here. I still have my substituent. And of course, my nitro group is still in the para position."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I can show my ring. And I can show these pi electrons have moved over to here. I still have my substituent. And of course, my nitro group is still in the para position. I took a bond away from this carbon. So I get a plus 1 formal charge. And once again, the presence of that lone pair of electrons on that oxygen right next to my benzene ring allows me to draw another resonance structure."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, my nitro group is still in the para position. I took a bond away from this carbon. So I get a plus 1 formal charge. And once again, the presence of that lone pair of electrons on that oxygen right next to my benzene ring allows me to draw another resonance structure. So I could think about these electrons on my oxygen moving into here to form a pi bond and pushing these electrons down to here. And so I can draw a fourth resonance structure. Showing the oxygen now double bonded to my ring."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, the presence of that lone pair of electrons on that oxygen right next to my benzene ring allows me to draw another resonance structure. So I could think about these electrons on my oxygen moving into here to form a pi bond and pushing these electrons down to here. And so I can draw a fourth resonance structure. Showing the oxygen now double bonded to my ring. So there's still a lone pair of electrons on that oxygen, which would give it a plus 1 formal charge. I still have pi electrons here. I showed pi electrons moving over to here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Showing the oxygen now double bonded to my ring. So there's still a lone pair of electrons on that oxygen, which would give it a plus 1 formal charge. I still have pi electrons here. I showed pi electrons moving over to here. And once again, my nitro group is still in the para position. And so I have a total of four resonance structures if I show the nitro group adding on in a para fashion as well. So the para attack also has four resonance structures."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I showed pi electrons moving over to here. And once again, my nitro group is still in the para position. And so I have a total of four resonance structures if I show the nitro group adding on in a para fashion as well. So the para attack also has four resonance structures. And the ortho had four resonance structures. And so that helps to explain why the methoxy substituent functions as an ortho para director. If you show ortho or a para attack, you can draw a total of four resonance structures, which stabilizes the sigma complex more than a meta attack."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the para attack also has four resonance structures. And the ortho had four resonance structures. And so that helps to explain why the methoxy substituent functions as an ortho para director. If you show ortho or a para attack, you can draw a total of four resonance structures, which stabilizes the sigma complex more than a meta attack. And that's the reason for the regiochemistry that we see in the nitration of methoxybenzene. Now, I drew these resonance structures because I started with my benzene ring. I started with my electrons in this position right here."}, {"video_title": "Ortho-para directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If you show ortho or a para attack, you can draw a total of four resonance structures, which stabilizes the sigma complex more than a meta attack. And that's the reason for the regiochemistry that we see in the nitration of methoxybenzene. Now, I drew these resonance structures because I started with my benzene ring. I started with my electrons in this position right here. Now, if you started with a different benzene ring, so a resonance structure of that, then your resonance structures might look a little bit different from mine. And you'll see different versions in different textbooks. So make sure to, once again, always look at what your professor does in class or your textbook."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the Big Bang starts with all of the mass and space in the universe, an infinitely dense singularity. The singularity is just something that the math doesn't even apply to. We don't even know how to understand that. But immediately after the Big Bang, so this occurred 13.7 billion years ago, immediately after this little tiny, infinitely small singularity begins to expand. And so for the first 100,000 years, it's still pretty dense. So let me just show this right now. So then it starts to expand."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But immediately after the Big Bang, so this occurred 13.7 billion years ago, immediately after this little tiny, infinitely small singularity begins to expand. And so for the first 100,000 years, it's still pretty dense. So let me just show this right now. So then it starts to expand. So maybe it gets to this level right over here. And I do not know if the entire universe is infinite or finite, whether it's a four-dimensional sphere, or whether it goes infinitely in every direction, or whether it's just slightly curved here and there and maybe flat everywhere else. I won't go into all of that."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So then it starts to expand. So maybe it gets to this level right over here. And I do not know if the entire universe is infinite or finite, whether it's a four-dimensional sphere, or whether it goes infinitely in every direction, or whether it's just slightly curved here and there and maybe flat everywhere else. I won't go into all of that. But then it starts to expand a little bit from the singularity, but it's still extremely dense. So dense that atoms can't even form. So you just have the basic fundamental building blocks of atoms."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I won't go into all of that. But then it starts to expand a little bit from the singularity, but it's still extremely dense. So dense that atoms can't even form. So you just have the basic fundamental building blocks of atoms. They're just all flying around. Electrons and protons, they're just flying around in just this ultra-hot, white-hot, I could say, or maybe even white-hot plasma. So this is, I'll call it white-hot plasma."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you just have the basic fundamental building blocks of atoms. They're just all flying around. Electrons and protons, they're just flying around in just this ultra-hot, white-hot, I could say, or maybe even white-hot plasma. So this is, I'll call it white-hot plasma. And then if we fast forward a little bit more, and now this is a point that we think we understand well, but this number, I actually looked at some old physics books, and this number has changed in really the last 15, 20 years. So maybe it'll change more. But after 380,000 years from the beginning of the Big Bang, and obviously this is give or take a couple of years, the universe expands enough."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is, I'll call it white-hot plasma. And then if we fast forward a little bit more, and now this is a point that we think we understand well, but this number, I actually looked at some old physics books, and this number has changed in really the last 15, 20 years. So maybe it'll change more. But after 380,000 years from the beginning of the Big Bang, and obviously this is give or take a couple of years, the universe expands enough. The universe is now large enough, and obviously I'm not drawing things to scale, and sparse enough that it can cool down a little bit. You don't have as much bumping around. It's still a hot place, but now it cools down enough that electrons can be captured by protons, and you could actually have the first hydrogen atoms can begin to form."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But after 380,000 years from the beginning of the Big Bang, and obviously this is give or take a couple of years, the universe expands enough. The universe is now large enough, and obviously I'm not drawing things to scale, and sparse enough that it can cool down a little bit. You don't have as much bumping around. It's still a hot place, but now it cools down enough that electrons can be captured by protons, and you could actually have the first hydrogen atoms can begin to form. The first hydrogen atoms begin to form. They actually condense. And we estimate this temperature to be around 3,000 Kelvin."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's still a hot place, but now it cools down enough that electrons can be captured by protons, and you could actually have the first hydrogen atoms can begin to form. The first hydrogen atoms begin to form. They actually condense. And we estimate this temperature to be around 3,000 Kelvin. So we've cooled the 3,000 Kelvin, but this is still a temperature that you would not want to hang out in. It's still extremely, extremely hot. Now why is this moment important, the first atoms forming?"}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we estimate this temperature to be around 3,000 Kelvin. So we've cooled the 3,000 Kelvin, but this is still a temperature that you would not want to hang out in. It's still extremely, extremely hot. Now why is this moment important, the first atoms forming? So let's think about what's happening here. You have all of this bumping and interactions. And if because of a bump or some energy release, or because of the heat temperature, if a photon is released, it'll be immediately absorbed by something else."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now why is this moment important, the first atoms forming? So let's think about what's happening here. You have all of this bumping and interactions. And if because of a bump or some energy release, or because of the heat temperature, if a photon is released, it'll be immediately absorbed by something else. If something gets, if some energy gets released, it'll immediately be absorbed by something else, because the universe is so dense, especially with charged particles. Here, all of a sudden, it's not that dense. So over here, things that were being emitted could not travel long distances."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if because of a bump or some energy release, or because of the heat temperature, if a photon is released, it'll be immediately absorbed by something else. If something gets, if some energy gets released, it'll immediately be absorbed by something else, because the universe is so dense, especially with charged particles. Here, all of a sudden, it's not that dense. So over here, things that were being emitted could not travel long distances. They would immediately bump into something else. While you go over here, and the universe is starting to look like the universe we recognize. All of a sudden, if one of these really hot, and it's still nowhere near as hot as this universe over here, but if one of these hot atoms emits a photon, and they would because they are at 3,000 Kelvin, if they emit a photon, all of a sudden, there's actually space for that photon to travel."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So over here, things that were being emitted could not travel long distances. They would immediately bump into something else. While you go over here, and the universe is starting to look like the universe we recognize. All of a sudden, if one of these really hot, and it's still nowhere near as hot as this universe over here, but if one of these hot atoms emits a photon, and they would because they are at 3,000 Kelvin, if they emit a photon, all of a sudden, there's actually space for that photon to travel. So for the first time in the history of the universe, 380,000 years after the Big Bang, you now have photons. You now have electromagnetic radiation. You now have information that can travel over long, long distances."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "All of a sudden, if one of these really hot, and it's still nowhere near as hot as this universe over here, but if one of these hot atoms emits a photon, and they would because they are at 3,000 Kelvin, if they emit a photon, all of a sudden, there's actually space for that photon to travel. So for the first time in the history of the universe, 380,000 years after the Big Bang, you now have photons. You now have electromagnetic radiation. You now have information that can travel over long, long distances. So given that this happened, it's still roughly 13.7 billion years ago. 380,000 years is not a lot when you talk about 13.7. It still wouldn't even really change the dial, because we're talking in the hundreds of thousands."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You now have information that can travel over long, long distances. So given that this happened, it's still roughly 13.7 billion years ago. 380,000 years is not a lot when you talk about 13.7. It still wouldn't even really change the dial, because we're talking in the hundreds of thousands. 0.7 is 700 million years. So this is actually a very small number. So it's still approximately 13.7 billion."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It still wouldn't even really change the dial, because we're talking in the hundreds of thousands. 0.7 is 700 million years. So this is actually a very small number. So it's still approximately 13.7 billion. It's really 13.7 minus 380,000 years. But given that this was the first time that information could travel, that photons could travel through space without most of them having to bump into something, especially something that's probably charged. The other interesting thing is that these atoms that formed are now neutral."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's still approximately 13.7 billion. It's really 13.7 minus 380,000 years. But given that this was the first time that information could travel, that photons could travel through space without most of them having to bump into something, especially something that's probably charged. The other interesting thing is that these atoms that formed are now neutral. What could we expect to see today? Well, let's think about it. These photons were emitted 13.7 billion years ago."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The other interesting thing is that these atoms that formed are now neutral. What could we expect to see today? Well, let's think about it. These photons were emitted 13.7 billion years ago. And they were emitted from every point in the universe. So this is every point in the universe. The universe was a pretty uniform place at that time."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These photons were emitted 13.7 billion years ago. And they were emitted from every point in the universe. So this is every point in the universe. The universe was a pretty uniform place at that time. Very minor irregularities, but you could see, because it was this white hot thing, just began to condense. It hadn't formed a lot of the structures that we now associate with the universe. It was just kind of a fairly uniform spread of, at that time, reasonably hot hydrogen atoms."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The universe was a pretty uniform place at that time. Very minor irregularities, but you could see, because it was this white hot thing, just began to condense. It hadn't formed a lot of the structures that we now associate with the universe. It was just kind of a fairly uniform spread of, at that time, reasonably hot hydrogen atoms. So this is every point in the universe. So let's think about what's going on here. Let me draw another diagram."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It was just kind of a fairly uniform spread of, at that time, reasonably hot hydrogen atoms. So this is every point in the universe. So let's think about what's going on here. Let me draw another diagram. So we're talking about this point in the universe right over here. The universe is, even 380,000 years after the Big Bang, still much, much, much, much smaller than the universe today. But let's say that this is the point in the universe where we happen to be now."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me draw another diagram. So we're talking about this point in the universe right over here. The universe is, even 380,000 years after the Big Bang, still much, much, much, much smaller than the universe today. But let's say that this is the point in the universe where we happen to be now. At this point in time, there was no Earth, there was no solar system, there was no Milky Way. It was just a bunch of hot hydrogen atoms. Now, if we were at this point in the universe, there must have been points in the universe at that exact same time that were emitting this radiation."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's say that this is the point in the universe where we happen to be now. At this point in time, there was no Earth, there was no solar system, there was no Milky Way. It was just a bunch of hot hydrogen atoms. Now, if we were at this point in the universe, there must have been points in the universe at that exact same time that were emitting this radiation. And there were actually every point in the universe was emitting this radiation. The point in the universe where we are now is emitting this radiation. And some of that radiation, so the points that were closer to us, it was emitting that radiation, but it got to us much sooner."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, if we were at this point in the universe, there must have been points in the universe at that exact same time that were emitting this radiation. And there were actually every point in the universe was emitting this radiation. The point in the universe where we are now is emitting this radiation. And some of that radiation, so the points that were closer to us, it was emitting that radiation, but it got to us much sooner. It got to us billions of years ago. But there were some points that were far enough that that radiation must be getting to us right now. Or another way to think about it is that radiation has taken 13.7 billion years to reach us."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And some of that radiation, so the points that were closer to us, it was emitting that radiation, but it got to us much sooner. It got to us billions of years ago. But there were some points that were far enough that that radiation must be getting to us right now. Or another way to think about it is that radiation has taken 13.7 billion years to reach us. So if I were to draw the visible universe today, and you know from the video about the size, so it's not going to be a scale, it would have to be far, far larger than the circle I drew here. But let's say that this is the visible universe today. We should be receiving, and we're in the center of it because we can only look roughly the same distance in every direction."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or another way to think about it is that radiation has taken 13.7 billion years to reach us. So if I were to draw the visible universe today, and you know from the video about the size, so it's not going to be a scale, it would have to be far, far larger than the circle I drew here. But let's say that this is the visible universe today. We should be receiving, and we're in the center of it because we can only look roughly the same distance in every direction. We're not the center of the universe. I want to be clear. We're the center of the observable universe because we can only observe the same distance in all directions."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We should be receiving, and we're in the center of it because we can only look roughly the same distance in every direction. We're not the center of the universe. I want to be clear. We're the center of the observable universe because we can only observe the same distance in all directions. Now, we're receiving some light from 100,000 light years away, and then we're looking 100,000 years in the past. We should be receiving some light, maybe a million light years that was first emitted a million light years before. And that's like looking a million years in the past because the light we see was emitted a million years ago."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're the center of the observable universe because we can only observe the same distance in all directions. Now, we're receiving some light from 100,000 light years away, and then we're looking 100,000 years in the past. We should be receiving some light, maybe a million light years that was first emitted a million light years before. And that's like looking a million years in the past because the light we see was emitted a million years ago. I think that's a bit redundant. We could see light that was emitted that's just getting to us after traveling for a billion years. And so we're actually looking at those objects a billion years ago because that's when they emitted the light."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's like looking a million years in the past because the light we see was emitted a million years ago. I think that's a bit redundant. We could see light that was emitted that's just getting to us after traveling for a billion years. And so we're actually looking at those objects a billion years ago because that's when they emitted the light. So the same way, we can look at objects that emitted their light 13.7 billion years ago, right at the beginning, right at this stage over here, right after 380,000 years after the Big Bang. And so since that light is only just reaching us, we will see it as it was 13.7 billion years ago. So we should see this type of radiation."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we're actually looking at those objects a billion years ago because that's when they emitted the light. So the same way, we can look at objects that emitted their light 13.7 billion years ago, right at the beginning, right at this stage over here, right after 380,000 years after the Big Bang. And so since that light is only just reaching us, we will see it as it was 13.7 billion years ago. So we should see this type of radiation. Now, the other thing to remember, the universe was expanding. When this was emitted, the universe was expanding. The universe was expanding at a very, well, it's all relative, what's a fast rate and all of that, but it was expanding."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we should see this type of radiation. Now, the other thing to remember, the universe was expanding. When this was emitted, the universe was expanding. The universe was expanding at a very, well, it's all relative, what's a fast rate and all of that, but it was expanding. And we learned on the video in Redshift that when the source of the light is moving away from you, or the source of the electromagnetic radiation is moving away from you, the radiation itself gets redshifted. So even though this is at a relatively high frequency, you could almost imagine it was kind of red hot gas, it was at 3,000 Kelvin, because it was moving away from us, these things, and we learned in the video on the actual size of the observable universe. Even though these electromagnetic waves are taking 13.7 billion years to reach us, in that time, this point in space, the point in space that emitted those electromagnetic waves, are about 46 billion light years away, so that's our best estimate."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The universe was expanding at a very, well, it's all relative, what's a fast rate and all of that, but it was expanding. And we learned on the video in Redshift that when the source of the light is moving away from you, or the source of the electromagnetic radiation is moving away from you, the radiation itself gets redshifted. So even though this is at a relatively high frequency, you could almost imagine it was kind of red hot gas, it was at 3,000 Kelvin, because it was moving away from us, these things, and we learned in the video on the actual size of the observable universe. Even though these electromagnetic waves are taking 13.7 billion years to reach us, in that time, this point in space, the point in space that emitted those electromagnetic waves, are about 46 billion light years away, so that's our best estimate. So this is still stretching away. So theory, if you believe all of this, that this was about 3,000 Kelvin, and it gets redshifted, theory would have it that we should see not something analogous to electromagnetic waves being released from a 3,000 degree temperature atom, we should see something redshifted into the radio spectrum. So we should be observing radio waves."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Even though these electromagnetic waves are taking 13.7 billion years to reach us, in that time, this point in space, the point in space that emitted those electromagnetic waves, are about 46 billion light years away, so that's our best estimate. So this is still stretching away. So theory, if you believe all of this, that this was about 3,000 Kelvin, and it gets redshifted, theory would have it that we should see not something analogous to electromagnetic waves being released from a 3,000 degree temperature atom, we should see something redshifted into the radio spectrum. So we should be observing radio waves. And the reason why we're observing radio waves, and not something of a higher frequency, is because it got redshifted. It got redshifted down into a lower frequency. And remember, we should be seeing it from every point in the universe where the photons have been traveling for 13.7 billion years."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we should be observing radio waves. And the reason why we're observing radio waves, and not something of a higher frequency, is because it got redshifted. It got redshifted down into a lower frequency. And remember, we should be seeing it from every point in the universe where the photons have been traveling for 13.7 billion years. We should see it all around us. This is almost a necessity for us to really believe in the current Big Bang theory. And it turns out that we did observe this."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And remember, we should be seeing it from every point in the universe where the photons have been traveling for 13.7 billion years. We should see it all around us. This is almost a necessity for us to really believe in the current Big Bang theory. And it turns out that we did observe this. And I want to make this very unintuitive, because you look at any other point in the universe, it's non-uniform. Every other point in the universe, you have stars and galaxies. These aren't atoms anymore."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it turns out that we did observe this. And I want to make this very unintuitive, because you look at any other point in the universe, it's non-uniform. Every other point in the universe, you have stars and galaxies. These aren't atoms anymore. These are stars and galaxies and whatnot. And so there's some points in the universe where you see a lot of radiation. And there's other points in the universe where you see nothing, it's just black."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These aren't atoms anymore. These are stars and galaxies and whatnot. And so there's some points in the universe where you see a lot of radiation. And there's other points in the universe where you see nothing, it's just black. But if this is correct, if this really did happen, we should be able to observe uniform radio waves from every direction around us. And you go more than 360 degrees, we're going in three dimensions, any direction you point an antenna, a radio antenna, you should be receiving these radio waves that were at much higher frequency when they were emitted, they'd been redshifted then, but they were emitted 13.7 billion years ago. And it turns out in the late 1960s, they did find these radio waves from every direction."}, {"video_title": "Cosmic background radiation Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there's other points in the universe where you see nothing, it's just black. But if this is correct, if this really did happen, we should be able to observe uniform radio waves from every direction around us. And you go more than 360 degrees, we're going in three dimensions, any direction you point an antenna, a radio antenna, you should be receiving these radio waves that were at much higher frequency when they were emitted, they'd been redshifted then, but they were emitted 13.7 billion years ago. And it turns out in the late 1960s, they did find these radio waves from every direction. And these are called the cosmic microwave background radiation. And it's this in combination, so it's this data that we're getting, this observation, in combination with the fact that the further we look out to galaxies and clusters of galaxies, they all seem to be moving away from us. They're all redshifted."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the nomenclature and physical properties of carboxylic acids. We'll start with nomenclature. If we wanted to name this carboxylic acid, it's the simplest one possible. There's one carbon, so a one-carbon carboxylic acid. If we had a one-carbon alkane, we would call that methane. To name a carboxylic acid, you drop the E ending and add oic acid, and so this one would be methanoic acid. Let me go ahead and write this out."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "There's one carbon, so a one-carbon carboxylic acid. If we had a one-carbon alkane, we would call that methane. To name a carboxylic acid, you drop the E ending and add oic acid, and so this one would be methanoic acid. Let me go ahead and write this out. So methanoic acid, you can see we've dropped the E and added oic and then acid. Methanoic acid is the IUPAC name for this molecule. The common name for this is formic acid, which is what you'll hear more often."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and write this out. So methanoic acid, you can see we've dropped the E and added oic and then acid. Methanoic acid is the IUPAC name for this molecule. The common name for this is formic acid, which is what you'll hear more often. So formic acid. And the name comes from the Latin word for ant, because formic acid is found in ant venom. And there are lots of really cool carboxylic acids with interesting common names, such as this example here."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "The common name for this is formic acid, which is what you'll hear more often. So formic acid. And the name comes from the Latin word for ant, because formic acid is found in ant venom. And there are lots of really cool carboxylic acids with interesting common names, such as this example here. Let's look at this one. So a two-carbon carboxylic acid. So using IUPAC nomenclature, that would be ethanoic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And there are lots of really cool carboxylic acids with interesting common names, such as this example here. Let's look at this one. So a two-carbon carboxylic acid. So using IUPAC nomenclature, that would be ethanoic acid. So it would be ethane, drop the E, add oic acid. So ethanoic acid. The common name for this is acetic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So using IUPAC nomenclature, that would be ethanoic acid. So it would be ethane, drop the E, add oic acid. So ethanoic acid. The common name for this is acetic acid. And so once again, that's the one that you'll hear more often. So acetic acid. And this name comes from the Latin word for vinegar, because vinegar is just a dilute solution of acetic acid in water."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "The common name for this is acetic acid. And so once again, that's the one that you'll hear more often. So acetic acid. And this name comes from the Latin word for vinegar, because vinegar is just a dilute solution of acetic acid in water. All right, let's name this one down here. So you want to find the longest carbon chain that includes the carbon of your carboxylic acid. So that's going to make this carbon over here carbon number one."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And this name comes from the Latin word for vinegar, because vinegar is just a dilute solution of acetic acid in water. All right, let's name this one down here. So you want to find the longest carbon chain that includes the carbon of your carboxylic acid. So that's going to make this carbon over here carbon number one. So two, three, four, five. So a five-carbon carboxylic acid. That would be pentanoic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So that's going to make this carbon over here carbon number one. So two, three, four, five. So a five-carbon carboxylic acid. That would be pentanoic acid. So go ahead and write pentanoic acid here. So pentanoic acid. And then we have a bromine on carbon four."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "That would be pentanoic acid. So go ahead and write pentanoic acid here. So pentanoic acid. And then we have a bromine on carbon four. And so it'd be 4-bromo. And so the full IUPAC name would be 4-bromo pentanoic acid. All right, let's look at this one."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a bromine on carbon four. And so it'd be 4-bromo. And so the full IUPAC name would be 4-bromo pentanoic acid. All right, let's look at this one. So three carbons. So this would be carbon number one, two, and three. So a three-carbon carboxylic acid would be propanoic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at this one. So three carbons. So this would be carbon number one, two, and three. So a three-carbon carboxylic acid would be propanoic acid. But since we have a double bond present, we need to change the A to an E. So it'd be propenoic acid. So let me go ahead and write this out. So it'd be propenes."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So a three-carbon carboxylic acid would be propanoic acid. But since we have a double bond present, we need to change the A to an E. So it'd be propenoic acid. So let me go ahead and write this out. So it'd be propenes. We'd have an E here. Propenoic acid, like that. And then we could just go ahead and put in a two here to indicate the start of the double bond."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So it'd be propenes. We'd have an E here. Propenoic acid, like that. And then we could just go ahead and put in a two here to indicate the start of the double bond. So two propenoic acid. And we don't have to worry about the stereochemistry of the double bond, since that's a monosubstituted double bond. But for a longer carboxylic acid, you would have to think about the stereochemistry."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And then we could just go ahead and put in a two here to indicate the start of the double bond. So two propenoic acid. And we don't have to worry about the stereochemistry of the double bond, since that's a monosubstituted double bond. But for a longer carboxylic acid, you would have to think about the stereochemistry. All right, let's do some more examples. So let's look at this molecule over here on the bottom left. And we can see there's a benzene ring attached to a carboxylic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "But for a longer carboxylic acid, you would have to think about the stereochemistry. All right, let's do some more examples. So let's look at this molecule over here on the bottom left. And we can see there's a benzene ring attached to a carboxylic acid. And so we've seen this before. And we've called it benzoic acid. So let's go ahead and start that as our parent's name."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And we can see there's a benzene ring attached to a carboxylic acid. And so we've seen this before. And we've called it benzoic acid. So let's go ahead and start that as our parent's name. So we have benzoic acid right here. And then if we're going to name it as benzoic acid, that gives the carbon attached to the carboxylic acid, carbon number one. And then we want to give our substituent the lowest number possible."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and start that as our parent's name. So we have benzoic acid right here. And then if we're going to name it as benzoic acid, that gives the carbon attached to the carboxylic acid, carbon number one. And then we want to give our substituent the lowest number possible. So of course, this is going to be carbon number two. And we have a OH, carbon two. So it's 2-hydroxybenzoic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And then we want to give our substituent the lowest number possible. So of course, this is going to be carbon number two. And we have a OH, carbon two. So it's 2-hydroxybenzoic acid. So 2-hydroxybenzoic acid. Benzoic acid is actually a common name. But it's, again, used so frequently in organic chemistry it's been incorporated into IUPAC nomenclature."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So it's 2-hydroxybenzoic acid. So 2-hydroxybenzoic acid. Benzoic acid is actually a common name. But it's, again, used so frequently in organic chemistry it's been incorporated into IUPAC nomenclature. If we wanted to name this another way, we could say it's 2-hydroxy or ortho hydroxy. And then we see there's a benzene ring. So 2-hydroxybenzene."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "But it's, again, used so frequently in organic chemistry it's been incorporated into IUPAC nomenclature. If we wanted to name this another way, we could say it's 2-hydroxy or ortho hydroxy. And then we see there's a benzene ring. So 2-hydroxybenzene. And then we have our carboxylic acid. So 2-hydroxybenzene carboxylic acid is another IUPAC name for this molecule. You don't see people name it that way usually, just because it's so long."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So 2-hydroxybenzene. And then we have our carboxylic acid. So 2-hydroxybenzene carboxylic acid is another IUPAC name for this molecule. You don't see people name it that way usually, just because it's so long. And it's just much easier to say benzoic acid. The common name for this molecule is salicylic acid. Let me go ahead and write that."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "You don't see people name it that way usually, just because it's so long. And it's just much easier to say benzoic acid. The common name for this molecule is salicylic acid. Let me go ahead and write that. So salicylic acid is famous because it's a precursor to aspirin and wintergreen. So the name for salicylic acid comes from the Latin word for willow tree, because you can get this compound from the bark of the willow tree. And the Greeks knew about this."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and write that. So salicylic acid is famous because it's a precursor to aspirin and wintergreen. So the name for salicylic acid comes from the Latin word for willow tree, because you can get this compound from the bark of the willow tree. And the Greeks knew about this. And it would reduce fevers and decrease pain. And so that's, of course, why salicylic acid was transformed into aspirin. All right, let's look at this one over here on the right."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And the Greeks knew about this. And it would reduce fevers and decrease pain. And so that's, of course, why salicylic acid was transformed into aspirin. All right, let's look at this one over here on the right. And so instead of having a benzene ring, we have a cyclohexane ring. So this would be cyclohexane. And now we have our carboxylic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at this one over here on the right. And so instead of having a benzene ring, we have a cyclohexane ring. So this would be cyclohexane. And now we have our carboxylic acid. So over here, we could say cyclohexane carboxylic acid for this one. So cyclohexane, almost running out of room here, carboxylic acid, like that. So instead of saying benzene carboxylic acid, it's cyclohexane carboxylic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And now we have our carboxylic acid. So over here, we could say cyclohexane carboxylic acid for this one. So cyclohexane, almost running out of room here, carboxylic acid, like that. So instead of saying benzene carboxylic acid, it's cyclohexane carboxylic acid. So what do you do if you have two carboxylic acids in the same molecule? And so that's what we have here. We can see there are two carbons present."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So instead of saying benzene carboxylic acid, it's cyclohexane carboxylic acid. So what do you do if you have two carboxylic acids in the same molecule? And so that's what we have here. We can see there are two carbons present. So we can go ahead and start by writing ethane. So we have two carboxylic acids. So we're going to use di in here."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "We can see there are two carbons present. So we can go ahead and start by writing ethane. So we have two carboxylic acids. So we're going to use di in here. So ethane di, and now oic. So ethane di-oic acid would be the IUPAC name for this molecule. The common name for this molecule is oxalic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to use di in here. So ethane di, and now oic. So ethane di-oic acid would be the IUPAC name for this molecule. The common name for this molecule is oxalic acid. So oxalic acid, like that. All right, let's look at properties of carboxylic acid. So we'll start with boiling point."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "The common name for this molecule is oxalic acid. So oxalic acid, like that. All right, let's look at properties of carboxylic acid. So we'll start with boiling point. And so let's compare these two molecules in terms of their boiling points. So over here on the left, we have acetic acid, which has a boiling point of approximately 118 degrees Celsius. And over here on the right, we have ethanol."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start with boiling point. And so let's compare these two molecules in terms of their boiling points. So over here on the left, we have acetic acid, which has a boiling point of approximately 118 degrees Celsius. And over here on the right, we have ethanol. And the boiling point of ethanol is approximately 78 degrees Celsius. So acetic acid has a higher boiling point. And we can think about why by thinking about two molecules of acetic acid interacting and the intermolecular forces that are present."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And over here on the right, we have ethanol. And the boiling point of ethanol is approximately 78 degrees Celsius. So acetic acid has a higher boiling point. And we can think about why by thinking about two molecules of acetic acid interacting and the intermolecular forces that are present. So let's go ahead and draw another molecule of acetic acid. So I'm going to go ahead and draw it. So there's my carbonyl."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And we can think about why by thinking about two molecules of acetic acid interacting and the intermolecular forces that are present. So let's go ahead and draw another molecule of acetic acid. So I'm going to go ahead and draw it. So there's my carbonyl. And then we have an oxygen and then a hydrogen. And on this side, a methyl group. So there are opportunities for hydrogen bonding."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So there's my carbonyl. And then we have an oxygen and then a hydrogen. And on this side, a methyl group. So there are opportunities for hydrogen bonding. So there could be a hydrogen bond right here and a hydrogen bond right here. So remember, oxygen is more electronegative than hydrogen. So oxygen gets a partial negative."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So there are opportunities for hydrogen bonding. So there could be a hydrogen bond right here and a hydrogen bond right here. So remember, oxygen is more electronegative than hydrogen. So oxygen gets a partial negative. The hydrogen gets a partial positive. And then this oxygen over here is also partially negative. And so you have this opposite charge attracting, this partially negatively charged oxygen attracted to this partially positively charged hydrogen here."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen gets a partial negative. The hydrogen gets a partial positive. And then this oxygen over here is also partially negative. And so you have this opposite charge attracting, this partially negatively charged oxygen attracted to this partially positively charged hydrogen here. And this is your hydrogen bond, the strongest intermolecular force. The same thing down here. So we could think about two hydrogen bonds forming for two molecules of acetic acid."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And so you have this opposite charge attracting, this partially negatively charged oxygen attracted to this partially positively charged hydrogen here. And this is your hydrogen bond, the strongest intermolecular force. The same thing down here. So we could think about two hydrogen bonds forming for two molecules of acetic acid. And over here on the right, we have ethanol. So let's go ahead and draw in another molecule of ethanol. So here we have our second molecule of ethanol."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So we could think about two hydrogen bonds forming for two molecules of acetic acid. And over here on the right, we have ethanol. So let's go ahead and draw in another molecule of ethanol. So here we have our second molecule of ethanol. And we can see there could be a hydrogen bond right here. So once again, we have our partially negative oxygen, partially positive hydrogen, partial negative oxygen, like that. So we have one hydrogen bond."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So here we have our second molecule of ethanol. And we can see there could be a hydrogen bond right here. So once again, we have our partially negative oxygen, partially positive hydrogen, partial negative oxygen, like that. So we have one hydrogen bond. So there's more opportunities for hydrogen bonding in acetic acid than in ethanol. And so because there are more opportunities for hydrogen bonding, there's stronger forces holding these two molecules together. And so it takes more energy to pull them apart."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So we have one hydrogen bond. So there's more opportunities for hydrogen bonding in acetic acid than in ethanol. And so because there are more opportunities for hydrogen bonding, there's stronger forces holding these two molecules together. And so it takes more energy to pull them apart. And that's the reason why it has a higher boiling point. All right, let's think about solubility in water. And let's stick with thinking about acetic acid right here."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And so it takes more energy to pull them apart. And that's the reason why it has a higher boiling point. All right, let's think about solubility in water. And let's stick with thinking about acetic acid right here. So we know acetic acid is soluble in water because vinegar is acetic acid in water. And so if we don't have a lot of carbons, so acetic acid has only two carbons, this molecule is polar enough to dissolve in water. So we could show acetic acid interacting with water."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And let's stick with thinking about acetic acid right here. So we know acetic acid is soluble in water because vinegar is acetic acid in water. And so if we don't have a lot of carbons, so acetic acid has only two carbons, this molecule is polar enough to dissolve in water. So we could show acetic acid interacting with water. So let me go ahead and draw a water molecule right here. And I'll draw a water molecule over here. And so there's, of course, some hydrogen bonding that can go on."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "So we could show acetic acid interacting with water. So let me go ahead and draw a water molecule right here. And I'll draw a water molecule over here. And so there's, of course, some hydrogen bonding that can go on. So you can think about a hydrogen bond right here and a hydrogen bond right here. And so acetic acid is soluble in water. Water is a polar molecule."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "And so there's, of course, some hydrogen bonding that can go on. So you can think about a hydrogen bond right here and a hydrogen bond right here. And so acetic acid is soluble in water. Water is a polar molecule. And acetic acid is polar enough to dissolve in water. However, as you increase the number of carbons in your R group, so as you increase the number of carbons, you get more carbons and hydrogens. And so you get more of a nonpolar character."}, {"video_title": "Carboxylic acid nomenclature and properties Organic chemistry Khan Academy.mp3", "Sentence": "Water is a polar molecule. And acetic acid is polar enough to dissolve in water. However, as you increase the number of carbons in your R group, so as you increase the number of carbons, you get more carbons and hydrogens. And so you get more of a nonpolar character. And the more nonpolar you make this molecule, the more you decrease its solubility in water. So once you get past some around five or six carbons, you decrease the solubility dramatically. And so that's just a little bit into the physical properties and the nomenclature of carboxylic acids."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And I can see this carbon is attached to two other carbons. So this carbon would be secondary. Or I could think about hydrogen replacing that R group, and then it would be a primary carbon. So either primary or secondary. If I look at the carbon on the right side, this carbon right here, it's attached to three other carbons. So this carbon on the right would be tertiary. So if I react this epoxide with a weak nucleophile, so HNU is going to refer to my weak nucleophile, and I make it an acid catalyzed reaction, I can see the product over here on the right."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So either primary or secondary. If I look at the carbon on the right side, this carbon right here, it's attached to three other carbons. So this carbon on the right would be tertiary. So if I react this epoxide with a weak nucleophile, so HNU is going to refer to my weak nucleophile, and I make it an acid catalyzed reaction, I can see the product over here on the right. The nucleophile has added to the most substituted carbon. So here I can see my nucleophile. It added to my tertiary carbon over here on the right."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So if I react this epoxide with a weak nucleophile, so HNU is going to refer to my weak nucleophile, and I make it an acid catalyzed reaction, I can see the product over here on the right. The nucleophile has added to the most substituted carbon. So here I can see my nucleophile. It added to my tertiary carbon over here on the right. So that's the regiochemistry, a nucleophilic attack of the most substituted carbon. And I can see when the ring opens, the OH group is going to be anti to the nucleophile. Let's take a look at a reaction."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "It added to my tertiary carbon over here on the right. So that's the regiochemistry, a nucleophilic attack of the most substituted carbon. And I can see when the ring opens, the OH group is going to be anti to the nucleophile. Let's take a look at a reaction. And we can run through the entire mechanism for this ring opening. So if I start with my epoxide, so I'll go ahead and make it like this. I'm going to have my epoxide coming out at me in space."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "Let's take a look at a reaction. And we can run through the entire mechanism for this ring opening. So if I start with my epoxide, so I'll go ahead and make it like this. I'm going to have my epoxide coming out at me in space. And then at this top carbon here, I'm going to have a methyl group going away from me in space. So here's my methyl group. I'm going to react this epoxide with ethanol."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to have my epoxide coming out at me in space. And then at this top carbon here, I'm going to have a methyl group going away from me in space. So here's my methyl group. I'm going to react this epoxide with ethanol. So I'll go ahead and put ethanol in here. Ethanol is going to function as my weak nucleophile. And it's acid catalyzed, so I'll make sure and put my protons in there like that."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to react this epoxide with ethanol. So I'll go ahead and put ethanol in here. Ethanol is going to function as my weak nucleophile. And it's acid catalyzed, so I'll make sure and put my protons in there like that. OK, let's go ahead and redraw the epoxide so we can see it a little bit better. So we're going to be looking down on our epoxide like this. So therefore, the oxygen is going to be up relative to the plane of the ring."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And it's acid catalyzed, so I'll make sure and put my protons in there like that. OK, let's go ahead and redraw the epoxide so we can see it a little bit better. So we're going to be looking down on our epoxide like this. So therefore, the oxygen is going to be up relative to the plane of the ring. Lone pair of electrons on the oxygen like that. And then at this carbon, there's going to be a methyl group going down in space. The first step of the mechanism is an acid-base reaction."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So therefore, the oxygen is going to be up relative to the plane of the ring. Lone pair of electrons on the oxygen like that. And then at this carbon, there's going to be a methyl group going down in space. The first step of the mechanism is an acid-base reaction. This is an acid-catalyzed reaction. So there are H plus protons floating around. Lone pair of electrons on oxygen are going to pick up that proton."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "The first step of the mechanism is an acid-base reaction. This is an acid-catalyzed reaction. So there are H plus protons floating around. Lone pair of electrons on oxygen are going to pick up that proton. So I'm going to protonate my epoxide. So when I draw the product of that acid-base reaction, and I now have on my oxygen, I still have one lone pair of electrons. The other lone pair formed a bond with that proton."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "Lone pair of electrons on oxygen are going to pick up that proton. So I'm going to protonate my epoxide. So when I draw the product of that acid-base reaction, and I now have on my oxygen, I still have one lone pair of electrons. The other lone pair formed a bond with that proton. So now it is a protonated epoxide, which gives the oxygen a plus 1 formal charge. I still need to have my methyl group down relative to the ring like that. Now when I think about the next step, I need to think about the classification of the carbon atoms."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "The other lone pair formed a bond with that proton. So now it is a protonated epoxide, which gives the oxygen a plus 1 formal charge. I still need to have my methyl group down relative to the ring like that. Now when I think about the next step, I need to think about the classification of the carbon atoms. So if I look at this top carbon right here, I can see that this one is attached to three other carbons. So this is my tertiary carbon. The one down here is only secondary."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "Now when I think about the next step, I need to think about the classification of the carbon atoms. So if I look at this top carbon right here, I can see that this one is attached to three other carbons. So this is my tertiary carbon. The one down here is only secondary. And I know the nucleophile is going to attack the more highly substituted carbon, which would be this one down here like that. So this is the carbon that's going to get the attack. Now if I think about that oxygen being positively charged, oxygen doesn't like to be positively charged."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "The one down here is only secondary. And I know the nucleophile is going to attack the more highly substituted carbon, which would be this one down here like that. So this is the carbon that's going to get the attack. Now if I think about that oxygen being positively charged, oxygen doesn't like to be positively charged. Oxygen is very electronegative. So it's going to do its best to pull some of these electrons and the bond between that carbon and that oxygen a little bit closer to the oxygen there. So I'm going to withdraw some electron density from this carbon."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "Now if I think about that oxygen being positively charged, oxygen doesn't like to be positively charged. Oxygen is very electronegative. So it's going to do its best to pull some of these electrons and the bond between that carbon and that oxygen a little bit closer to the oxygen there. So I'm going to withdraw some electron density from this carbon. So this carbon is going to end up having a partial positive charge like that. And if I think about what kind of a carbocation would that be, it has a little bit of partial carbocation character. And I can think about the fact that that carbon is attached to those three other carbons."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to withdraw some electron density from this carbon. So this carbon is going to end up having a partial positive charge like that. And if I think about what kind of a carbocation would that be, it has a little bit of partial carbocation character. And I can think about the fact that that carbon is attached to those three other carbons. It's a little bit easier to see here. There's one, two, and three other carbons. And in earlier videos, we've seen that tertiary carbocations are the most stable, much more stable than secondary carbocations."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And I can think about the fact that that carbon is attached to those three other carbons. It's a little bit easier to see here. There's one, two, and three other carbons. And in earlier videos, we've seen that tertiary carbocations are the most stable, much more stable than secondary carbocations. And if I have a weak nucleophile attacking, it's going to attack the most substituted carbon. So it's going to attack the one that is the most stable carbocation. So again, we have more of a partial carbocation character here."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And in earlier videos, we've seen that tertiary carbocations are the most stable, much more stable than secondary carbocations. And if I have a weak nucleophile attacking, it's going to attack the most substituted carbon. So it's going to attack the one that is the most stable carbocation. So again, we have more of a partial carbocation character here. But when the weak nucleophile comes along, that lone pair of electrons is going to attack this carbon right here. And that is going to kick these electrons in here, the magenta ones, off onto your oxygen. So let's go ahead and draw the result of that nucleophilic attack."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So again, we have more of a partial carbocation character here. But when the weak nucleophile comes along, that lone pair of electrons is going to attack this carbon right here. And that is going to kick these electrons in here, the magenta ones, off onto your oxygen. So let's go ahead and draw the result of that nucleophilic attack. So I have my ring right here. And I've opened my epoxide. So now this oxygen is going to swing over to the left here."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the result of that nucleophilic attack. So I have my ring right here. And I've opened my epoxide. So now this oxygen is going to swing over to the left here. And it's bonded to a hydrogen. It had one lone pair of electrons. It just picked up one more lone pair of electrons, the magenta ones."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So now this oxygen is going to swing over to the left here. And it's bonded to a hydrogen. It had one lone pair of electrons. It just picked up one more lone pair of electrons, the magenta ones. And in the course of that nucleophilic attack, the methyl group that's down relative to the ring here gets pushed up. So the methyl group will end up relative to the ring. And I have a lone pair of electrons from the ethanol molecule that are now bonded to this carbon."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "It just picked up one more lone pair of electrons, the magenta ones. And in the course of that nucleophilic attack, the methyl group that's down relative to the ring here gets pushed up. So the methyl group will end up relative to the ring. And I have a lone pair of electrons from the ethanol molecule that are now bonded to this carbon. So I can go ahead and show those electrons. And there's an oxygen, a hydrogen, and then two carbons. And then there's still a lone pair of electrons left in that oxygen, which gives that oxygen a plus 1 formal charge."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And I have a lone pair of electrons from the ethanol molecule that are now bonded to this carbon. So I can go ahead and show those electrons. And there's an oxygen, a hydrogen, and then two carbons. And then there's still a lone pair of electrons left in that oxygen, which gives that oxygen a plus 1 formal charge. So let's just go ahead and highlight those electrons really fast. So let's see. I'll make them red here."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And then there's still a lone pair of electrons left in that oxygen, which gives that oxygen a plus 1 formal charge. So let's just go ahead and highlight those electrons really fast. So let's see. I'll make them red here. So these electrons right here, those are the ones that attacked the carbon. And those are the ones that formed this covalent bond, like that. So we're almost done."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "I'll make them red here. So these electrons right here, those are the ones that attacked the carbon. And those are the ones that formed this covalent bond, like that. So we're almost done. So the last step will be an acid-base reaction to get rid of that plus 1 formal charge on the oxygen. So another molecule of ethanol comes along. This one's going to act as a base."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So we're almost done. So the last step will be an acid-base reaction to get rid of that plus 1 formal charge on the oxygen. So another molecule of ethanol comes along. This one's going to act as a base. So I go ahead and draw another molecule of ethanol. And it has lone pairs of electrons on it. One of those lone pairs of electrons are going to pick up that proton, which kicks these two electrons off onto this oxygen."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "This one's going to act as a base. So I go ahead and draw another molecule of ethanol. And it has lone pairs of electrons on it. One of those lone pairs of electrons are going to pick up that proton, which kicks these two electrons off onto this oxygen. And we're going to end up with our product. So let's go ahead and draw the product right up here. And like usual, we're going to be looking down on our molecule this way."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "One of those lone pairs of electrons are going to pick up that proton, which kicks these two electrons off onto this oxygen. And we're going to end up with our product. So let's go ahead and draw the product right up here. And like usual, we're going to be looking down on our molecule this way. And so we're going to look down right here and draw what we see. So we're going to have our ring like that. And at the top carbon, which would correspond to this one right here, I can see that there's a methyl group coming out at me."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And like usual, we're going to be looking down on our molecule this way. And so we're going to look down right here and draw what we see. So we're going to have our ring like that. And at the top carbon, which would correspond to this one right here, I can see that there's a methyl group coming out at me. So I can go ahead and put my methyl group coming out at me like that. And then going away from me, I'm going to have my oxygen. So I'm, of course, talking about this oxygen down here like that."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And at the top carbon, which would correspond to this one right here, I can see that there's a methyl group coming out at me. So I can go ahead and put my methyl group coming out at me like that. And then going away from me, I'm going to have my oxygen. So I'm, of course, talking about this oxygen down here like that. So I can go ahead and draw that oxygen going away from me. And connected to that oxygen, of course, would be two carbons. So I can go ahead and do that."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So I'm, of course, talking about this oxygen down here like that. So I can go ahead and draw that oxygen going away from me. And connected to that oxygen, of course, would be two carbons. So I can go ahead and do that. And when I look at the carbon on the left over here, so now I'm talking about this carbon, I can see there's an OH group coming out at me in space. So I can go ahead and put that OH group coming out at me in space right there. So that's going to be our product."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and do that. And when I look at the carbon on the left over here, so now I'm talking about this carbon, I can see there's an OH group coming out at me in space. So I can go ahead and put that OH group coming out at me in space right there. So that's going to be our product. So again, I went ahead and showed, I can go ahead and put my two lone pairs of electrons on that oxygen, because I showed that proton being removed in the last step of that mechanism there. So let's look at the stereochemistry. And let's first start with this top carbon here."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So that's going to be our product. So again, I went ahead and showed, I can go ahead and put my two lone pairs of electrons on that oxygen, because I showed that proton being removed in the last step of that mechanism there. So let's look at the stereochemistry. And let's first start with this top carbon here. So I go back to my reactants, and I look at this top carbon. And I think about the fact that I have a wedge to an oxygen. So oxygen's coming out at me, and a carbon's going away from me in space."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "And let's first start with this top carbon here. So I go back to my reactants, and I look at this top carbon. And I think about the fact that I have a wedge to an oxygen. So oxygen's coming out at me, and a carbon's going away from me in space. When I look at my products, this would be the carbon that corresponds to that carbon. It's the opposite. Now I have a carbon coming out at me in space, and an oxygen going away from me."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen's coming out at me, and a carbon's going away from me in space. When I look at my products, this would be the carbon that corresponds to that carbon. It's the opposite. Now I have a carbon coming out at me in space, and an oxygen going away from me. So that's inversion of configuration. So at that top carbon there, so at this carbon right here, I'm going to get inversion of configuration. So inversion of absolute configuration here."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "Now I have a carbon coming out at me in space, and an oxygen going away from me. So that's inversion of configuration. So at that top carbon there, so at this carbon right here, I'm going to get inversion of configuration. So inversion of absolute configuration here. So let me go ahead and write that. Inversion of configuration. So there's some stereochemistry involved."}, {"video_title": "Ring opening reactions of epoxides Acid-catalyzed Organic chemistry Khan Academy.mp3", "Sentence": "So inversion of absolute configuration here. So let me go ahead and write that. Inversion of configuration. So there's some stereochemistry involved. And when I look at the other carbon, so when I look at this carbon down here, I can see that there's an oxygen coming out at me in space. And when I look at that carbon for my products, I see there's still an oxygen coming out at me in space. So it's the same absolute configuration at this carbon."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "On the left we have an alkyl halide, and we know that this bromine is a little bit more electronegative than this carbon, so the bromine withdraws some electron density away from that carbon, which makes this carbon a little bit positive, so we say partially positive. That's the electrophilic center, so this on the left is our electrophile. On the right, we know that this hydroxide ion, which we could get from something like sodium hydroxide, it's a negative one formal charge on the oxygen, which makes it a good nucleophile, so let me write down here, this is our nucleophile on the right, and on the left is our electrophile, which I'm also gonna refer to as a substrate in this video, so this alkyl halide is our substrate. We know from an earlier video that the nucleophile will attack the electrophile because opposite charges attract. This negative charge is attracted to this partially positive charge, so a lone pair of electrons on the oxygen will attack this partially positive carbon. At the same time, the two electrons in this bond come off onto the bromine, so let me draw the bromine over here. The bromine had three lone pairs of electrons on it, and it's gonna pick up another lone pair of electrons."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "We know from an earlier video that the nucleophile will attack the electrophile because opposite charges attract. This negative charge is attracted to this partially positive charge, so a lone pair of electrons on the oxygen will attack this partially positive carbon. At the same time, the two electrons in this bond come off onto the bromine, so let me draw the bromine over here. The bromine had three lone pairs of electrons on it, and it's gonna pick up another lone pair of electrons. Let me show those electrons in magenta. So this bond breaks, and these two electrons come off onto the bromine, which gives the bromine a negative one formal charge, so this is the bromide anion. And we're also forming a bond between the oxygen and this carbon, and this bond comes from this lone pair of electrons, which I've just marked in blue here."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "The bromine had three lone pairs of electrons on it, and it's gonna pick up another lone pair of electrons. Let me show those electrons in magenta. So this bond breaks, and these two electrons come off onto the bromine, which gives the bromine a negative one formal charge, so this is the bromide anion. And we're also forming a bond between the oxygen and this carbon, and this bond comes from this lone pair of electrons, which I've just marked in blue here. So those two electrons in blue form this bond, and we get our product, which is an alcohol. So the SN2 mechanism is a concerted mechanism because the nucleophile attacks the electrophile at the same time we get loss of a leaving group, so there's only one step in this mechanism. Let's say we did a series of experiments to determine the rate law for this reaction."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "And we're also forming a bond between the oxygen and this carbon, and this bond comes from this lone pair of electrons, which I've just marked in blue here. So those two electrons in blue form this bond, and we get our product, which is an alcohol. So the SN2 mechanism is a concerted mechanism because the nucleophile attacks the electrophile at the same time we get loss of a leaving group, so there's only one step in this mechanism. Let's say we did a series of experiments to determine the rate law for this reaction. So remember from general chemistry, rate laws are determined experimentally. So capital R is the rate of the reaction, and that's equal to the rate constant K times the concentration of our alkyl halide, and it's determined experimentally, this is to the first power, times the concentration of the hydroxide ion, also to the first power. So what does this mean?"}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "Let's say we did a series of experiments to determine the rate law for this reaction. So remember from general chemistry, rate laws are determined experimentally. So capital R is the rate of the reaction, and that's equal to the rate constant K times the concentration of our alkyl halide, and it's determined experimentally, this is to the first power, times the concentration of the hydroxide ion, also to the first power. So what does this mean? This means if we increased the concentration of our alkyl halide, so if we increase the concentration of our alkyl halide by a factor of two, what happens to the rate of the reaction? Well, the rate of the reaction is proportional to the concentration of the alkyl halide to the first power. So two to the first is equal to two, which means the overall rate of the reaction would increase by a factor of two."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "So what does this mean? This means if we increased the concentration of our alkyl halide, so if we increase the concentration of our alkyl halide by a factor of two, what happens to the rate of the reaction? Well, the rate of the reaction is proportional to the concentration of the alkyl halide to the first power. So two to the first is equal to two, which means the overall rate of the reaction would increase by a factor of two. So doubling the concentration of our alkyl halide while keeping this concentration, the hydroxide ion concentration, the same should double the rate of the reaction. And also, if we kept the concentration of alkyl halide the same, and we doubled the concentration of hydroxide, that would also increase the rate by a factor of two. And this experimentally determined rate law makes sense with our mechanism."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "So two to the first is equal to two, which means the overall rate of the reaction would increase by a factor of two. So doubling the concentration of our alkyl halide while keeping this concentration, the hydroxide ion concentration, the same should double the rate of the reaction. And also, if we kept the concentration of alkyl halide the same, and we doubled the concentration of hydroxide, that would also increase the rate by a factor of two. And this experimentally determined rate law makes sense with our mechanism. So if we increase the concentration of the nucleophile, or we increase the concentration of the electrophile, we increase the frequency of collisions between the two, which increases the overall rate of the reaction. So the fact that our rate law is proportional to the concentration of both the substrate and the nucleophile fits with our idea of a one-step mechanism. Finally, let's take a look at where this SN2 comes from."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "And this experimentally determined rate law makes sense with our mechanism. So if we increase the concentration of the nucleophile, or we increase the concentration of the electrophile, we increase the frequency of collisions between the two, which increases the overall rate of the reaction. So the fact that our rate law is proportional to the concentration of both the substrate and the nucleophile fits with our idea of a one-step mechanism. Finally, let's take a look at where this SN2 comes from. So we keep on saying an SN2 mechanism, an SN2 reaction. The S stands for substitution. So let me write in here, substitution, because our nucleophile is substituting for our leaving group."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "Finally, let's take a look at where this SN2 comes from. So we keep on saying an SN2 mechanism, an SN2 reaction. The S stands for substitution. So let me write in here, substitution, because our nucleophile is substituting for our leaving group. We can see in our final product here, the nucleophile has substituted for the leaving group. The N stands for nucleophilic, because of course, it is our nucleophile that is doing the substituting. And finally, the two here refers to the fact that this is bimolecular, which means that the rate depends on the concentration of two things, the substrate and the nucleophile."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "So let me write in here, substitution, because our nucleophile is substituting for our leaving group. We can see in our final product here, the nucleophile has substituted for the leaving group. The N stands for nucleophilic, because of course, it is our nucleophile that is doing the substituting. And finally, the two here refers to the fact that this is bimolecular, which means that the rate depends on the concentration of two things, the substrate and the nucleophile. So that's different from an SN1 mechanism, where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left, we have a methyl halide, followed by a primary alkyl halide."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "And finally, the two here refers to the fact that this is bimolecular, which means that the rate depends on the concentration of two things, the substrate and the nucleophile. So that's different from an SN1 mechanism, where the rate is dependent only on the concentration of one thing. The rate of the reaction also depends on the structure of the alkyl halide, on the structure of the substrate. On the left, we have a methyl halide, followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group, followed by a secondary alkyl halide. The carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. This carbon is bonded to three alkyl groups."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "On the left, we have a methyl halide, followed by a primary alkyl halide. The carbon bonded to our bromine is directly attached to one alkyl group, followed by a secondary alkyl halide. The carbon bonded to the bromine is bonded to two alkyl groups, followed by a tertiary alkyl halide. This carbon is bonded to three alkyl groups. Turns out that the methyl halide and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly, and tertiary alkyl halides react so, so slowly that we say they are unreactive toward an SN2 mechanism. And this makes sense when we think about the mechanism, because remember, the nucleophile has to attack the electrophile."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "This carbon is bonded to three alkyl groups. Turns out that the methyl halide and the primary alkyl halide react the fastest in an SN2 mechanism. Secondary alkyl halides react very slowly, and tertiary alkyl halides react so, so slowly that we say they are unreactive toward an SN2 mechanism. And this makes sense when we think about the mechanism, because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon to actually form a bond, and steric hindrance would prevent that from happening. So something like a tertiary alkyl halide has these big, bulky methyl groups which prevent the nucleophile for attacking. So let's look at a video so we can see this a little bit more clearly."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "And this makes sense when we think about the mechanism, because remember, the nucleophile has to attack the electrophile. The nucleophile needs to get close enough to the electrophilic carbon to actually form a bond, and steric hindrance would prevent that from happening. So something like a tertiary alkyl halide has these big, bulky methyl groups which prevent the nucleophile for attacking. So let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly bonded to a halogen, which I'm saying is yellow, and here's our nucleophile, which could be the hydroxide ion. The nucleophile approaches the electrophile for the side opposite of the leaving group, and you can see with the methyl halide, there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it."}, {"video_title": "Sn2 mechanism kinetics and substrate.mp3", "Sentence": "So let's look at a video so we can see this a little bit more clearly. Here's our methyl halide with our carbon directly bonded to a halogen, which I'm saying is yellow, and here's our nucleophile, which could be the hydroxide ion. The nucleophile approaches the electrophile for the side opposite of the leaving group, and you can see with the methyl halide, there's no steric hindrance. When we move to a primary alkyl halide, the carbon bonded to the halogen has only one alkyl group bonded to it. It's still easy for the nucleophile to approach. When we move to a secondary alkyl halide, so for a secondary, you can see that the carbon bonded to the halogen has two methyl groups attached to it now. It gets a little harder for the nucleophile to approach in the proper orientation, so these bulky methyl groups make it more difficult for the nucleophile to get close enough to that electrophilic carbon."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have seen that a compound or ion is aromatic if it contains a ring of continuously overlapping p orbitals, and also if it has 4n plus 2 pi electrons in the ring, where n is an integer. So for example, n could be 0, or 1, or 2, or so on. We can use these criteria to analyze the cyclopropenyl cation. So here's my cyclopropenyl cation right there. And if I see how many pi electrons it has, that would be 2 pi electrons in the ion. So let me go ahead and write 2 pi electrons here. When I look at the carbons in the ion, this carbon has a double bond to it, so it's sp2 hybridized."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So here's my cyclopropenyl cation right there. And if I see how many pi electrons it has, that would be 2 pi electrons in the ion. So let me go ahead and write 2 pi electrons here. When I look at the carbons in the ion, this carbon has a double bond to it, so it's sp2 hybridized. Same with this carbon. And then this top carbon here, this carbocation, is also sp2 hybridized. So we know that sp2 hybridized carbons have a free p orbital."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "When I look at the carbons in the ion, this carbon has a double bond to it, so it's sp2 hybridized. Same with this carbon. And then this top carbon here, this carbocation, is also sp2 hybridized. So we know that sp2 hybridized carbons have a free p orbital. So each carbon in this ion has a p orbital on it. So I can go ahead and show that on this diagram. So I'm sketching in the p orbital on each carbon in my ring."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we know that sp2 hybridized carbons have a free p orbital. So each carbon in this ion has a p orbital on it. So I can go ahead and show that on this diagram. So I'm sketching in the p orbital on each carbon in my ring. And so that allows me to see that my p orbitals could overlap side by side. So this ion fulfills the first criteria. It has a ring of continuously overlapping p orbitals."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'm sketching in the p orbital on each carbon in my ring. And so that allows me to see that my p orbitals could overlap side by side. So this ion fulfills the first criteria. It has a ring of continuously overlapping p orbitals. I know that p orbitals are atomic orbitals, and so I have a total of three atomic orbitals. And according to MO theory, those three atomic orbitals are going to combine to give me three molecular orbitals. And I can analyze the relative energy levels of those molecular orbitals using what's called a Frost circle."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It has a ring of continuously overlapping p orbitals. I know that p orbitals are atomic orbitals, and so I have a total of three atomic orbitals. And according to MO theory, those three atomic orbitals are going to combine to give me three molecular orbitals. And I can analyze the relative energy levels of those molecular orbitals using what's called a Frost circle. And so we've seen these Frost circles before. And I'm just going to go ahead and start by drawing a line here to divide my circle in half, which divides my bonding from my antibonding molecular orbitals. And we always start at the bottom of our Frost circle."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I can analyze the relative energy levels of those molecular orbitals using what's called a Frost circle. And so we've seen these Frost circles before. And I'm just going to go ahead and start by drawing a line here to divide my circle in half, which divides my bonding from my antibonding molecular orbitals. And we always start at the bottom of our Frost circle. And we inscribe a polygon to match the number of atoms in my ring. So in this case, I have a three-membered ring. And so I'm going to inscribe a triangle into my Frost circle."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we always start at the bottom of our Frost circle. And we inscribe a polygon to match the number of atoms in my ring. So in this case, I have a three-membered ring. And so I'm going to inscribe a triangle into my Frost circle. And so I'm going to attempt to draw a triangle in my Frost circle here. And not the best triangle, but the important thing is where your polygon intersects with your Frost circle represents the energy level of your molecular orbitals. So I can see I have a total of three molecular orbitals looking at my Frost circle."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I'm going to inscribe a triangle into my Frost circle. And so I'm going to attempt to draw a triangle in my Frost circle here. And not the best triangle, but the important thing is where your polygon intersects with your Frost circle represents the energy level of your molecular orbitals. So I can see I have a total of three molecular orbitals looking at my Frost circle. And when I go ahead and represent them over here, I know that I have one molecular orbital right here. And I have two molecular orbitals up here. Bonding molecular orbitals are always lower in energy."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I can see I have a total of three molecular orbitals looking at my Frost circle. And when I go ahead and represent them over here, I know that I have one molecular orbital right here. And I have two molecular orbitals up here. Bonding molecular orbitals are always lower in energy. So I know that this orbital down here is my bonding molecular orbital. And therefore, I have two antibonding molecular orbitals up here. I know I have two pi electrons to worry about."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Bonding molecular orbitals are always lower in energy. So I know that this orbital down here is my bonding molecular orbital. And therefore, I have two antibonding molecular orbitals up here. I know I have two pi electrons to worry about. So it's completely analogous to electron configurations. I'm going to fill my lowest energy orbital first. And so that would be my bonding molecular orbital."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I know I have two pi electrons to worry about. So it's completely analogous to electron configurations. I'm going to fill my lowest energy orbital first. And so that would be my bonding molecular orbital. And I only have two pi electrons. And so those two pi electrons are going to completely fill my bonding molecular orbital. Let's analyze this in terms of Huckel's rule."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that would be my bonding molecular orbital. And I only have two pi electrons. And so those two pi electrons are going to completely fill my bonding molecular orbital. Let's analyze this in terms of Huckel's rule. So I can see that this would be a 2 here to represent my two pi electrons. And then I have zero electrons in my antibonding molecular orbitals. So it would be like 4 times 0."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's analyze this in terms of Huckel's rule. So I can see that this would be a 2 here to represent my two pi electrons. And then I have zero electrons in my antibonding molecular orbitals. So it would be like 4 times 0. So 4 times 0 is, of course, 0 plus 2 gives me a total of 2, which is Huckel's rule. So I have two pi electrons in the ring. And this ion satisfies the second criteria as well."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it would be like 4 times 0. So 4 times 0 is, of course, 0 plus 2 gives me a total of 2, which is Huckel's rule. So I have two pi electrons in the ring. And this ion satisfies the second criteria as well. So the cyclopropenyl cation is aromatic. So if I go ahead and write this ion is aromatic over here. And so it's extra stable."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this ion satisfies the second criteria as well. So the cyclopropenyl cation is aromatic. So if I go ahead and write this ion is aromatic over here. And so it's extra stable. And this observation allows us to explain some of the properties associated with this molecule down here, which is cyclopropenone. Cyclopropenone has a huge amount of angle strain with this three-membered ring here. And it also has an increased dipole moment from what we might expect."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it's extra stable. And this observation allows us to explain some of the properties associated with this molecule down here, which is cyclopropenone. Cyclopropenone has a huge amount of angle strain with this three-membered ring here. And it also has an increased dipole moment from what we might expect. And we can explain both of those by looking at the resonance structure for the cyclopropenone molecule. And so if I think about drawing a resonance structure, I could take these pi electrons and move them off onto my oxygen here. And so the resonance structure would have my three-membered ring like that."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it also has an increased dipole moment from what we might expect. And we can explain both of those by looking at the resonance structure for the cyclopropenone molecule. And so if I think about drawing a resonance structure, I could take these pi electrons and move them off onto my oxygen here. And so the resonance structure would have my three-membered ring like that. And then it would have this oxygen with now three lone pairs of electrons around it, giving it a negative 1 formal charge. Took a bond away from my carbonyl carbon right here. So that carbon is going to get a plus 1 formal charge."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the resonance structure would have my three-membered ring like that. And then it would have this oxygen with now three lone pairs of electrons around it, giving it a negative 1 formal charge. Took a bond away from my carbonyl carbon right here. So that carbon is going to get a plus 1 formal charge. And so you can see that we've just formed the cyclopropenyl cation here, which we know is extra stable. This is aromatic. And so while this is an extremely minor resonance structure for most carbonyl compounds, for this one, this contributes a little bit more to the resonance hybrid because of the extra stability associated with the cyclopropenyl cation, the fact that it is aromatic."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon is going to get a plus 1 formal charge. And so you can see that we've just formed the cyclopropenyl cation here, which we know is extra stable. This is aromatic. And so while this is an extremely minor resonance structure for most carbonyl compounds, for this one, this contributes a little bit more to the resonance hybrid because of the extra stability associated with the cyclopropenyl cation, the fact that it is aromatic. And so that affords some extra stability to this molecule. And we could also draw it where we show that positive charge is being spread out throughout our three-membered ring like that. And then we have our oxygen over here with a negative charge."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so while this is an extremely minor resonance structure for most carbonyl compounds, for this one, this contributes a little bit more to the resonance hybrid because of the extra stability associated with the cyclopropenyl cation, the fact that it is aromatic. And so that affords some extra stability to this molecule. And we could also draw it where we show that positive charge is being spread out throughout our three-membered ring like that. And then we have our oxygen over here with a negative charge. And this picture allows us to see the increased dipole moment a little bit more. So because the part on the left is extra stable as a positive charge, it's aromatic, that means we have an increased dipole moment and also increasing the stability despite the significant angle strain associated with the cyclopropenone molecule. So you can use the concept of aromatic stability to analyze the structure of other molecules as well."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our oxygen over here with a negative charge. And this picture allows us to see the increased dipole moment a little bit more. So because the part on the left is extra stable as a positive charge, it's aromatic, that means we have an increased dipole moment and also increasing the stability despite the significant angle strain associated with the cyclopropenone molecule. So you can use the concept of aromatic stability to analyze the structure of other molecules as well. Let's do one more example. Let's look at this molecule, which is cyclobutadiene. So over here on the left, we have cyclobutadiene."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you can use the concept of aromatic stability to analyze the structure of other molecules as well. Let's do one more example. Let's look at this molecule, which is cyclobutadiene. So over here on the left, we have cyclobutadiene. If I look at the pi electrons, here's two pi electrons. And here's another two for a total of four pi electrons. So there are four pi electrons in my molecule."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the left, we have cyclobutadiene. If I look at the pi electrons, here's two pi electrons. And here's another two for a total of four pi electrons. So there are four pi electrons in my molecule. When I look at the carbons, each carbon has a double bond. So each carbon is sp2 hybridized. So each carbon has a free p orbital."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there are four pi electrons in my molecule. When I look at the carbons, each carbon has a double bond. So each carbon is sp2 hybridized. So each carbon has a free p orbital. And so I can sketch in my p orbitals over here on my diagram like that. Makes it a little bit easier to see that these p orbitals could overlap side by side. So the first criteria has been fulfilled."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So each carbon has a free p orbital. And so I can sketch in my p orbitals over here on my diagram like that. Makes it a little bit easier to see that these p orbitals could overlap side by side. So the first criteria has been fulfilled. I have a ring of continuously overlapping p orbitals. But when I look at my second criteria, our second criteria was 4n plus 2 pi electrons. And so I don't have that."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the first criteria has been fulfilled. I have a ring of continuously overlapping p orbitals. But when I look at my second criteria, our second criteria was 4n plus 2 pi electrons. And so I don't have that. I have a total of four pi electrons. And so that's really 4n, where n is equal to 1. So 4 times 1 is equal to 4."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I don't have that. I have a total of four pi electrons. And so that's really 4n, where n is equal to 1. So 4 times 1 is equal to 4. So I have 4n pi electrons. I don't have 4n plus 2. So I already know that this compound is not aromatic, just by looking at that."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So 4 times 1 is equal to 4. So I have 4n pi electrons. I don't have 4n plus 2. So I already know that this compound is not aromatic, just by looking at that. But let's go ahead and draw in our molecular orbitals and see where we put those four pi electrons in our molecular orbitals. So my four p orbitals here on this diagram means I have a total of four atomic orbitals for cyclobutadiene. Once again, MO theory tells me four atomic orbitals are going to give me four molecular orbitals."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I already know that this compound is not aromatic, just by looking at that. But let's go ahead and draw in our molecular orbitals and see where we put those four pi electrons in our molecular orbitals. So my four p orbitals here on this diagram means I have a total of four atomic orbitals for cyclobutadiene. Once again, MO theory tells me four atomic orbitals are going to give me four molecular orbitals. And I can draw in my frost circle right here. So once again, I'm going to draw a line through the center to separate my bonding from my antibonding molecular orbitals. You always start at the bottom of your frost circle."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Once again, MO theory tells me four atomic orbitals are going to give me four molecular orbitals. And I can draw in my frost circle right here. So once again, I'm going to draw a line through the center to separate my bonding from my antibonding molecular orbitals. You always start at the bottom of your frost circle. Four-membered ring. So we are going to draw a four-sided figure in our frost circle. So I'm going to attempt to draw this square in here."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You always start at the bottom of your frost circle. Four-membered ring. So we are going to draw a four-sided figure in our frost circle. So I'm going to attempt to draw this square in here. And once again, where our polygon intersects with our circle represents the energy level of our molecular orbitals. So I have a total of four molecular orbitals. And if I go over here, I have a molecular orbital below the line."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to attempt to draw this square in here. And once again, where our polygon intersects with our circle represents the energy level of our molecular orbitals. So I have a total of four molecular orbitals. And if I go over here, I have a molecular orbital below the line. So that's my bonding molecular orbital. I have a molecular orbital above the center line there. So that's my antibonding molecular orbital."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if I go over here, I have a molecular orbital below the line. So that's my bonding molecular orbital. I have a molecular orbital above the center line there. So that's my antibonding molecular orbital. And this time, I have two molecular orbitals that are right on the line, which represents non-bonding molecular orbitals. When I go ahead and put in my four pi electrons, once again, I fill lowest energy molecular orbital first. So that takes care of two pi electrons."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's my antibonding molecular orbital. And this time, I have two molecular orbitals that are right on the line, which represents non-bonding molecular orbitals. When I go ahead and put in my four pi electrons, once again, I fill lowest energy molecular orbital first. So that takes care of two pi electrons. And now I have two more. And so here I have the non-bonding molecular orbitals are on the same energy level. And so if you remember Hund's rule from electron configurations, you can't pair up these last two pi electrons because the orbitals are of equal energy."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that takes care of two pi electrons. And now I have two more. And so here I have the non-bonding molecular orbitals are on the same energy level. And so if you remember Hund's rule from electron configurations, you can't pair up these last two pi electrons because the orbitals are of equal energy. And so this is the picture that we get. And you can see that I have two unpaired electrons. And two unpaired electrons implies that this molecule is extremely reactive."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if you remember Hund's rule from electron configurations, you can't pair up these last two pi electrons because the orbitals are of equal energy. And so this is the picture that we get. And you can see that I have two unpaired electrons. And two unpaired electrons implies that this molecule is extremely reactive. And experimentally, it is. So cyclobutadiene will actually react with itself. So it's experimentally extremely reactive, which tells you that it's not extra stable."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And two unpaired electrons implies that this molecule is extremely reactive. And experimentally, it is. So cyclobutadiene will actually react with itself. So it's experimentally extremely reactive, which tells you that it's not extra stable. It's not aromatic. You don't have 4n plus 2 pi electrons. And so the second criteria is not fulfilled."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's experimentally extremely reactive, which tells you that it's not extra stable. It's not aromatic. You don't have 4n plus 2 pi electrons. And so the second criteria is not fulfilled. But this compound does satisfy the first criteria. And so the term for this compound is anti-aromatic. So it is anti-aromatic, which means it fulfills the first criteria, a ring of continuously overlapping p orbitals, but does not satisfy the second criteria."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the second criteria is not fulfilled. But this compound does satisfy the first criteria. And so the term for this compound is anti-aromatic. So it is anti-aromatic, which means it fulfills the first criteria, a ring of continuously overlapping p orbitals, but does not satisfy the second criteria. It does not have 4n plus 2 pi electrons. It has 4n pi electrons. And so we say that is anti-aromatic."}, {"video_title": "Aromatic stability III Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it is anti-aromatic, which means it fulfills the first criteria, a ring of continuously overlapping p orbitals, but does not satisfy the second criteria. It does not have 4n plus 2 pi electrons. It has 4n pi electrons. And so we say that is anti-aromatic. And there are actually very few examples of anti-aromatic compounds. But cyclobutadiene is considered to be anti-aromatic. In the next video, we're going to look at a few more examples of aromatic stability."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just so you know, the word nebula is kind of a general word for any interstellar cloud of gas or dust. So when we're talking about the Eagle Nebula, we're talking about a huge nebula. And actually, it's a nebula that spans. And just so you have a sense, this is just one of the pillars in the, this is called the Pillars of Creation. You've probably seen this image before. There's these three pillars here. And this is just a small part of the actual Eagle Nebula."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just so you have a sense, this is just one of the pillars in the, this is called the Pillars of Creation. You've probably seen this image before. There's these three pillars here. And this is just a small part of the actual Eagle Nebula. And just this pillar right over here, just so that you have a sense of how large it is. Just this pillar itself is 7 light years. It is 7 light years tall."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is just a small part of the actual Eagle Nebula. And just this pillar right over here, just so that you have a sense of how large it is. Just this pillar itself is 7 light years. It is 7 light years tall. So this is an enormous amount of distance. Remember, the distance from Earth to the nearest star was about 4 light years. It would take Voyager, if it was pointed in the right direction, at moving at 60,000 kilometers per hour, it would take Voyager 80,000 years to go 4 light years."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is 7 light years tall. So this is an enormous amount of distance. Remember, the distance from Earth to the nearest star was about 4 light years. It would take Voyager, if it was pointed in the right direction, at moving at 60,000 kilometers per hour, it would take Voyager 80,000 years to go 4 light years. Just this pillar is 7 light years. But I wanted to show you this because these type of nebulae, as the plural of nebula, are where stars can form. So this right here, you actually see, this is actually kind of the breeding ground for the birth of new stars."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would take Voyager, if it was pointed in the right direction, at moving at 60,000 kilometers per hour, it would take Voyager 80,000 years to go 4 light years. Just this pillar is 7 light years. But I wanted to show you this because these type of nebulae, as the plural of nebula, are where stars can form. So this right here, you actually see, this is actually kind of the breeding ground for the birth of new stars. This gas is condensing, just like we talked about a couple of videos ago, until it gets to that critical temperature, that critical density, where you can actually get fusion of hydrogen. So this is just a huge interstellar cloud of hydrogen gas. And over here, you can just see it's just this breeding ground for stars."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right here, you actually see, this is actually kind of the breeding ground for the birth of new stars. This gas is condensing, just like we talked about a couple of videos ago, until it gets to that critical temperature, that critical density, where you can actually get fusion of hydrogen. So this is just a huge interstellar cloud of hydrogen gas. And over here, you can just see it's just this breeding ground for stars. And we actually think that this structure doesn't even exist anymore. Because remember, this thing is very, very far away from us. In fact, it is, just so you have the number, this is 7,000 light years away."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And over here, you can just see it's just this breeding ground for stars. And we actually think that this structure doesn't even exist anymore. Because remember, this thing is very, very far away from us. In fact, it is, just so you have the number, this is 7,000 light years away. Which means that what we are seeing now, the photons that are reaching us right now, reaching our eyes, are reaching our telescopes right now, left this region of space 7,000 years ago. So we're seeing it as it was 7,000 years ago. So a lot of this gas, a lot of this hydrogen, may have already condensed into many, many, many more stars."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, it is, just so you have the number, this is 7,000 light years away. Which means that what we are seeing now, the photons that are reaching us right now, reaching our eyes, are reaching our telescopes right now, left this region of space 7,000 years ago. So we're seeing it as it was 7,000 years ago. So a lot of this gas, a lot of this hydrogen, may have already condensed into many, many, many more stars. So the structure might not be the way it looks right now. And actually, there was another supernova that happened that we think might have kind of blown away a lot of this dust. And we won't even be able to see what the effects of that supernova were for another 1,000 years."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So a lot of this gas, a lot of this hydrogen, may have already condensed into many, many, many more stars. So the structure might not be the way it looks right now. And actually, there was another supernova that happened that we think might have kind of blown away a lot of this dust. And we won't even be able to see what the effects of that supernova were for another 1,000 years. But anyway, this is just a pretty amazing photograph, in my opinion. Especially, and it's beautiful at any scale, but it's even more mind-blowing when you think that this is 7, this is a structure that is 7 light years tall. And this is really just part of the Eagle Nebula, one of the pillars of creation."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we won't even be able to see what the effects of that supernova were for another 1,000 years. But anyway, this is just a pretty amazing photograph, in my opinion. Especially, and it's beautiful at any scale, but it's even more mind-blowing when you think that this is 7, this is a structure that is 7 light years tall. And this is really just part of the Eagle Nebula, one of the pillars of creation. This right here is a star field. And this is as we're looking towards the center of our galaxy, the Milky Way. So this is the Sagittarius star field."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is really just part of the Eagle Nebula, one of the pillars of creation. This right here is a star field. And this is as we're looking towards the center of our galaxy, the Milky Way. So this is the Sagittarius star field. And the neat thing here is you just see such a diversity in stars. And this is also kind of mind-numbing, because every one of these stars are inside of our galaxy. This is looking towards the center of our galaxy."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the Sagittarius star field. And the neat thing here is you just see such a diversity in stars. And this is also kind of mind-numbing, because every one of these stars are inside of our galaxy. This is looking towards the center of our galaxy. We're not looking at, this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies. This is just stars here. But what's neat here is you see a huge variety."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is looking towards the center of our galaxy. We're not looking at, this isn't one of those where we're looking beyond our galaxy or looking at clusters of galaxies. This is just stars here. But what's neat here is you see a huge variety. You see some stars that are shining red right over here. And obviously, the apparent size, you cannot completely tell, because the stars are at different distances and at different intensities. But the redder stars, these are stars at their red giant phase, or they're probably at their red giant phase."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's neat here is you see a huge variety. You see some stars that are shining red right over here. And obviously, the apparent size, you cannot completely tell, because the stars are at different distances and at different intensities. But the redder stars, these are stars at their red giant phase, or they're probably at their red giant phase. I haven't done specific research on these stars, but that's what we suspect, that they're at the red giant phase. The ones that are kind of in the yellowish-white part of the spectrum, these are stars probably in their main sequence, probably not too different than our own sun. The ones that are in the yellowish-white, or closer to orange, yellowish-white part of the spectrum."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the redder stars, these are stars at their red giant phase, or they're probably at their red giant phase. I haven't done specific research on these stars, but that's what we suspect, that they're at the red giant phase. The ones that are kind of in the yellowish-white part of the spectrum, these are stars probably in their main sequence, probably not too different than our own sun. The ones that are in the yellowish-white, or closer to orange, yellowish-white part of the spectrum. And the ones that look a little bit more bluish, or a little bit more greenish, these are burning super fast. Let me see if I can find. This one looks a little bit bluish to me."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The ones that are in the yellowish-white, or closer to orange, yellowish-white part of the spectrum. And the ones that look a little bit more bluish, or a little bit more greenish, these are burning super fast. Let me see if I can find. This one looks a little bit bluish to me. Sorry, I had to pause the video for a cough. These are burning super, super fast. And so the supermassive stars, they burn kind of fast and furious and then just die out."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This one looks a little bit bluish to me. Sorry, I had to pause the video for a cough. These are burning super, super fast. And so the supermassive stars, they burn kind of fast and furious and then just die out. While the smaller stars, the ones with less mass, they burn slower over a much, much, much longer period of time. So the ones that are burning really fast are emitting a lot of energy at the smaller wavelength part of the light spectrum, and that's why they look bluer or greener. And these are going to be more massive stars, the ones that look whiter or bluer or greener."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the supermassive stars, they burn kind of fast and furious and then just die out. While the smaller stars, the ones with less mass, they burn slower over a much, much, much longer period of time. So the ones that are burning really fast are emitting a lot of energy at the smaller wavelength part of the light spectrum, and that's why they look bluer or greener. And these are going to be more massive stars, the ones that look whiter or bluer or greener. While the redder ones are less massive stars that are in kind of their supergiant phase, and so they are at this point cooler than kind of the main sequence stars. This right here is the Cat's Eye Nebula. And the word nebula, this is actually a planetary nebula, and I want to differentiate it."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these are going to be more massive stars, the ones that look whiter or bluer or greener. While the redder ones are less massive stars that are in kind of their supergiant phase, and so they are at this point cooler than kind of the main sequence stars. This right here is the Cat's Eye Nebula. And the word nebula, this is actually a planetary nebula, and I want to differentiate it. This right here is a planetary nebula. And it's called a nebula because it is kind of this gas that's kind of floating out in space, but it's at a completely different scale than the Eagle Nebula that we drew over here. So normally when people just talk about nebula, they're talking about something like the Eagle Nebula, this huge masses of interstellar gas."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the word nebula, this is actually a planetary nebula, and I want to differentiate it. This right here is a planetary nebula. And it's called a nebula because it is kind of this gas that's kind of floating out in space, but it's at a completely different scale than the Eagle Nebula that we drew over here. So normally when people just talk about nebula, they're talking about something like the Eagle Nebula, this huge masses of interstellar gas. When people talk about planetary nebula, this is still actually a huge, huge radius, but nowhere near 7 light years, but it's still a huge. But this is the byproduct of a star shedding off all of its outer material. So at the center of this, we see a fairly mature star here, and it's shed off kind of its outer layers, and it did that while it was in its red giant phase."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So normally when people just talk about nebula, they're talking about something like the Eagle Nebula, this huge masses of interstellar gas. When people talk about planetary nebula, this is still actually a huge, huge radius, but nowhere near 7 light years, but it's still a huge. But this is the byproduct of a star shedding off all of its outer material. So at the center of this, we see a fairly mature star here, and it's shed off kind of its outer layers, and it did that while it was in its red giant phase. So the core would keep flaring up, keep having these hot explosions, and every time you had one of those hot explosions, you had more and more of its outer layers getting pushed off into space, forming this planetary nebula. So as we see it right now, it's still not yet a white dwarf, it is still an active star, fusion is still occurring in this star, but it's well on its way to becoming a white dwarf. It wants all the fuel runs out."}, {"video_title": "Star field and nebula images Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So at the center of this, we see a fairly mature star here, and it's shed off kind of its outer layers, and it did that while it was in its red giant phase. So the core would keep flaring up, keep having these hot explosions, and every time you had one of those hot explosions, you had more and more of its outer layers getting pushed off into space, forming this planetary nebula. So as we see it right now, it's still not yet a white dwarf, it is still an active star, fusion is still occurring in this star, but it's well on its way to becoming a white dwarf. It wants all the fuel runs out. So it's past the red giant phase, it's thrown all of this material into space, and it's on its way to becoming a white dwarf. Anyway, hopefully you enjoyed that. I actually find all of these images to be pretty captivating, especially the star field one, because this is just inside of our galaxy."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "What I want to do in this video is to talk about nucleophilicity. Nucleophilicity. Nucleophilicity. And this is really just how good of a nucleophile something is. Or we could, I'll just make up a definition right now. The ability for an atom, let me say an atom, atom slash ion slash molecule to act as a nucleophile or to give away extra, I'll write that in quotes, extra electron and bond with a nucleus or with something else. I'll say with a nucleus."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And this is really just how good of a nucleophile something is. Or we could, I'll just make up a definition right now. The ability for an atom, let me say an atom, atom slash ion slash molecule to act as a nucleophile or to give away extra, I'll write that in quotes, extra electron and bond with a nucleus or with something else. I'll say with a nucleus. Let me say with, I'll just say and bond. I want to say with a nucleus because that's what nucleophilicity is saying. It loves nucleuses, especially positive ones, that it can give its extra electron to it."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "I'll say with a nucleus. Let me say with, I'll just say and bond. I want to say with a nucleus because that's what nucleophilicity is saying. It loves nucleuses, especially positive ones, that it can give its extra electron to it. Now as a first cut, if you want to identify a good nucleophile, it should have extra electrons to give away. And the best things that have extra electrons to give away are negative ions or anions. So if you were just at a very high level, something like the fluoride anion."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It loves nucleuses, especially positive ones, that it can give its extra electron to it. Now as a first cut, if you want to identify a good nucleophile, it should have extra electrons to give away. And the best things that have extra electrons to give away are negative ions or anions. So if you were just at a very high level, something like the fluoride anion. Normally fluorine has seven valence electrons, one, two, three, four, five, six, seven. But it's so electronegative it might be able to swipe off another electron from something else. And then it becomes the fluoride anion with a negative charge."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So if you were just at a very high level, something like the fluoride anion. Normally fluorine has seven valence electrons, one, two, three, four, five, six, seven. But it's so electronegative it might be able to swipe off another electron from something else. And then it becomes the fluoride anion with a negative charge. And you can do that for all of the halides. You can do that for chlorine, can become chloride. Bromine can be bromide."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And then it becomes the fluoride anion with a negative charge. And you can do that for all of the halides. You can do that for chlorine, can become chloride. Bromine can be bromide. Iodine can be iodide. Let me do iodine too. So iodine, once again, it's a halide."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Bromine can be bromide. Iodine can be iodide. Let me do iodine too. So iodine, once again, it's a halide. It has seven valence electrons. It has many, many more electrons than fluorine. But if you just look at its valence shell, it has seven electrons."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So iodine, once again, it's a halide. It has seven valence electrons. It has many, many more electrons than fluorine. But if you just look at its valence shell, it has seven electrons. And then it is also reasonably electronegative, not as electronegative as fluorine. Remember, the trend goes like this, from the bottom left to the top right. Fluorine is extremely electronegative."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "But if you just look at its valence shell, it has seven electrons. And then it is also reasonably electronegative, not as electronegative as fluorine. Remember, the trend goes like this, from the bottom left to the top right. Fluorine is extremely electronegative. But iodine is still pretty electronegative. It is a halogen. So it also might be able to swipe off an electron from someone else and become iodide."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Fluorine is extremely electronegative. But iodine is still pretty electronegative. It is a halogen. So it also might be able to swipe off an electron from someone else and become iodide. So in general, things with extra electrons, lone pairs of electrons, and especially a negative charge, are going to be pretty good nucleophiles. Another example that's not a halide is a hydroxide anion, OH. This is an example of something that is a molecule."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So it also might be able to swipe off an electron from someone else and become iodide. So in general, things with extra electrons, lone pairs of electrons, and especially a negative charge, are going to be pretty good nucleophiles. Another example that's not a halide is a hydroxide anion, OH. This is an example of something that is a molecule. So OH, traditionally water would look like this. So traditionally, this is just neutral water. And oxygen has two lone pairs like that."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "This is an example of something that is a molecule. So OH, traditionally water would look like this. So traditionally, this is just neutral water. And oxygen has two lone pairs like that. But oxygen is pretty electronegative. It is already kind of taking the electrons away from one of the hydrogens. And at some point, it might just take it all together."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And oxygen has two lone pairs like that. But oxygen is pretty electronegative. It is already kind of taking the electrons away from one of the hydrogens. And at some point, it might just take it all together. And then you have the hydroxide anion. So this will look like this. So you have your original two lone pairs just like that."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And at some point, it might just take it all together. And then you have the hydroxide anion. So this will look like this. So you have your original two lone pairs just like that. And then you have this pair that's going to be taken. It already had that electron. It takes that electron from the hydrogen."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So you have your original two lone pairs just like that. And then you have this pair that's going to be taken. It already had that electron. It takes that electron from the hydrogen. So now it has two more electrons. Let me color code it so you see what it took. It took this electron from the hydrogen."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It takes that electron from the hydrogen. So now it has two more electrons. Let me color code it so you see what it took. It took this electron from the hydrogen. And now this also has a negative charge. That's also a reasonably good nucleophile. And of course, you have your hydrogen now."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It took this electron from the hydrogen. And now this also has a negative charge. That's also a reasonably good nucleophile. And of course, you have your hydrogen now. It lost its one electron. It only has a proton in its nucleus. So when you ever see H+, this is really just a proton."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And of course, you have your hydrogen now. It lost its one electron. It only has a proton in its nucleus. So when you ever see H+, this is really just a proton. There's nothing else to that hydrogen. So that's hydroxide. So these are all reasonably good nucleophiles in that they have something to give away."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So when you ever see H+, this is really just a proton. There's nothing else to that hydrogen. So that's hydroxide. So these are all reasonably good nucleophiles in that they have something to give away. They have extra charge. Now, what I want to do is think about between these. How do you think about what's going to be a stronger or weaker nucleophile?"}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So these are all reasonably good nucleophiles in that they have something to give away. They have extra charge. Now, what I want to do is think about between these. How do you think about what's going to be a stronger or weaker nucleophile? And here, it becomes a little bit more nuanced. What we're going to do is differentiate what happens in a protic solvent versus what happens in an aprotic solvent. So let me write down."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "How do you think about what's going to be a stronger or weaker nucleophile? And here, it becomes a little bit more nuanced. What we're going to do is differentiate what happens in a protic solvent versus what happens in an aprotic solvent. So let me write down. Let me start with aprotic solvent. Protic solvent. I'll make two columns here."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So let me write down. Let me start with aprotic solvent. Protic solvent. I'll make two columns here. Protic solvent, and then we'll do aprotic solvent. Aprotic solvent. Solvent right over here."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "I'll make two columns here. Protic solvent, and then we'll do aprotic solvent. Aprotic solvent. Solvent right over here. And once again, these are fancy words, but they mean something pretty simple. Protic solvent is something that has hydrogens that can be taken away or it might have free protons flowing around. An example of a protic solvent is water, or really any alcohol."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Solvent right over here. And once again, these are fancy words, but they mean something pretty simple. Protic solvent is something that has hydrogens that can be taken away or it might have free protons flowing around. An example of a protic solvent is water, or really any alcohol. So water is the simplest example, or maybe the most common. The reason why we call it why you might have protons floating around is exactly this little reaction I showed right here. Maybe every now and then, hydroxide anion forms, or even more likely, maybe a water takes a hydrogen from one of the other waters."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "An example of a protic solvent is water, or really any alcohol. So water is the simplest example, or maybe the most common. The reason why we call it why you might have protons floating around is exactly this little reaction I showed right here. Maybe every now and then, hydroxide anion forms, or even more likely, maybe a water takes a hydrogen from one of the other waters. One of the water molecules takes a hydrogen from one of the other water molecules and becomes a hydronium, where it's not a proton necessarily, but it's an oxygen that has... Let's say if you start with water, and one of these electrons were to be given to some proton floating around, it would look like this. And it has a positive charge. And then this proton is very available."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Maybe every now and then, hydroxide anion forms, or even more likely, maybe a water takes a hydrogen from one of the other waters. One of the water molecules takes a hydrogen from one of the other water molecules and becomes a hydronium, where it's not a proton necessarily, but it's an oxygen that has... Let's say if you start with water, and one of these electrons were to be given to some proton floating around, it would look like this. And it has a positive charge. And then this proton is very available. You can almost imagine it's almost floating around because that oxygen really wants to take back that electron. So protic solvent is water. In water you might see a little bit of hydroxide, a little bit of proton, a little bit of hydronium."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And then this proton is very available. You can almost imagine it's almost floating around because that oxygen really wants to take back that electron. So protic solvent is water. In water you might see a little bit of hydroxide, a little bit of proton, a little bit of hydronium. You see all of it in there, but the bottom line is that there are protons that can react with other things. You have protons that can react with other things. Let me clear this away so that I have some real estate."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "In water you might see a little bit of hydroxide, a little bit of proton, a little bit of hydronium. You see all of it in there, but the bottom line is that there are protons that can react with other things. You have protons that can react with other things. Let me clear this away so that I have some real estate. Let me just write down water. So water is a protic solvent. Now, you might say, hey, wait, is that always the case?"}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Let me clear this away so that I have some real estate. Let me just write down water. So water is a protic solvent. Now, you might say, hey, wait, is that always the case? It seems like hydrogens are everywhere. Well, no, it's not always the case. Let me show you an aprotic solvent."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Now, you might say, hey, wait, is that always the case? It seems like hydrogens are everywhere. Well, no, it's not always the case. Let me show you an aprotic solvent. So diethyl ether looks like this. And just so you know the naming, it's an ether because it has an oxygen, and it's diethyl because it has two ethyl groups. That's one ethyl group right there, and that is the second."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Let me show you an aprotic solvent. So diethyl ether looks like this. And just so you know the naming, it's an ether because it has an oxygen, and it's diethyl because it has two ethyl groups. That's one ethyl group right there, and that is the second. So it's diethyl ether. Now, you might say, hey, this guy's got hydrogens flying around as well. Maybe those can get released."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "That's one ethyl group right there, and that is the second. So it's diethyl ether. Now, you might say, hey, this guy's got hydrogens flying around as well. Maybe those can get released. But no, these hydrogens are bonded to the carbon, and carbon is not anywhere near as electronegative as oxygen. So these are not going to be carbon is unlikely to steal these hydrogens, electrons, and these hydrogens to be loose. If they were bonded to the oxygen, that would have been a possibility."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Maybe those can get released. But no, these hydrogens are bonded to the carbon, and carbon is not anywhere near as electronegative as oxygen. So these are not going to be carbon is unlikely to steal these hydrogens, electrons, and these hydrogens to be loose. If they were bonded to the oxygen, that would have been a possibility. So in water, you have, you know, obviously you have HOH. In alcohols, you have some maybe carbon chain bonded to an oxygen, which is then bonded to a hydrogen. So in either of these cases, in either water or alcohol, you have these hydrogens where the electron might be taken by the oxygen because it's so electronegative, and then the hydrogen floats around."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "If they were bonded to the oxygen, that would have been a possibility. So in water, you have, you know, obviously you have HOH. In alcohols, you have some maybe carbon chain bonded to an oxygen, which is then bonded to a hydrogen. So in either of these cases, in either water or alcohol, you have these hydrogens where the electron might be taken by the oxygen because it's so electronegative, and then the hydrogen floats around. Anyway, that's a review of protic versus aprotic. Now, in a protic solvent, and this is actually a general rule of thumb, if a nucleophile is likely to react with its solvent, it will be bad at being a nucleophile. And think about it."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So in either of these cases, in either water or alcohol, you have these hydrogens where the electron might be taken by the oxygen because it's so electronegative, and then the hydrogen floats around. Anyway, that's a review of protic versus aprotic. Now, in a protic solvent, and this is actually a general rule of thumb, if a nucleophile is likely to react with its solvent, it will be bad at being a nucleophile. And think about it. If it's reacting with the solvent, it's not going to be able to do this. It's not going to be able to give its electrons away to what it needs to give it away, to maybe, you know, what we saw in, say, an SN2-type reaction. So in a protic solvent, what happens is, is the things that are really electronegative and really small, like a fluoride anion, so let me draw a fluoride anion."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And think about it. If it's reacting with the solvent, it's not going to be able to do this. It's not going to be able to give its electrons away to what it needs to give it away, to maybe, you know, what we saw in, say, an SN2-type reaction. So in a protic solvent, what happens is, is the things that are really electronegative and really small, like a fluoride anion, so let me draw a fluoride anion. A fluoride anion in a protic solvent, what's going to happen is it's going to be blocked by hydrogen bonds. So it's very negative, right? It has a negative charge."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So in a protic solvent, what happens is, is the things that are really electronegative and really small, like a fluoride anion, so let me draw a fluoride anion. A fluoride anion in a protic solvent, what's going to happen is it's going to be blocked by hydrogen bonds. So it's very negative, right? It has a negative charge. And it's also tightly packed. As you can see right here, its electrons are very close, tied in. It's a much smaller atom, or ion in this case, than if we looked at iodide."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It has a negative charge. And it's also tightly packed. As you can see right here, its electrons are very close, tied in. It's a much smaller atom, or ion in this case, than if we looked at iodide. Iodide has 53 electrons, many orbitals. Fluoride, or actually, iodide would have 54. It had the same as iodine, plus 1."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It's a much smaller atom, or ion in this case, than if we looked at iodide. Iodide has 53 electrons, many orbitals. Fluoride, or actually, iodide would have 54. It had the same as iodine, plus 1. Fluoride will have 10 electrons, 9 from fluorine, plus it gains another one. So it's a much smaller atom. And so when you have water hanging around it, let's say you have something like water, that has a negative charge."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It had the same as iodine, plus 1. Fluoride will have 10 electrons, 9 from fluorine, plus it gains another one. So it's a much smaller atom. And so when you have water hanging around it, let's say you have something like water, that has a negative charge. Water is polar. And actually, both of these are polar, so I should write down polar for both of these. This is a polar protic solvent."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And so when you have water hanging around it, let's say you have something like water, that has a negative charge. Water is polar. And actually, both of these are polar, so I should write down polar for both of these. This is a polar protic solvent. This is a polar aprotic solvent. In this case, water is still more electronegative than the carbon, so it still has a partial negative charge. These parts still have a partial negative."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "This is a polar protic solvent. This is a polar aprotic solvent. In this case, water is still more electronegative than the carbon, so it still has a partial negative charge. These parts still have a partial negative. Water still has partial negative. The hydrogen has a partial positive charge, so it is going to be attracted to the fluorine. And this is going to happen all around the fluorine."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "These parts still have a partial negative. Water still has partial negative. The hydrogen has a partial positive charge, so it is going to be attracted to the fluorine. And this is going to happen all around the fluorine. And if these waters are attracted to the fluorine and kind of forming a shell, a tight shell around it, it makes it hard for fluorine to react. So it's a worse nucleophile than, say, iodide or hydroxide in a polar protic solvent. Hydroxide has the same issue."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And this is going to happen all around the fluorine. And if these waters are attracted to the fluorine and kind of forming a shell, a tight shell around it, it makes it hard for fluorine to react. So it's a worse nucleophile than, say, iodide or hydroxide in a polar protic solvent. Hydroxide has the same issue. It's still forming hydrogen bonds, but if you wanted to compare them, iodide is a much bigger ion. It has all these electrons in here. And so it still will form hydrogen bonds with the water, because they're partially positive, but it's going to be less tightly packed."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Hydroxide has the same issue. It's still forming hydrogen bonds, but if you wanted to compare them, iodide is a much bigger ion. It has all these electrons in here. And so it still will form hydrogen bonds with the water, because they're partially positive, but it's going to be less tightly packed. And on top of that, iodide is more polarizable. Which means that its electron cloud is so big and the valence electrons are so far away from the nucleus that they can be influenced by things and then be more likely to react. So let's say this iodide is getting close to a carbon that has a partial positive charge."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "And so it still will form hydrogen bonds with the water, because they're partially positive, but it's going to be less tightly packed. And on top of that, iodide is more polarizable. Which means that its electron cloud is so big and the valence electrons are so far away from the nucleus that they can be influenced by things and then be more likely to react. So let's say this iodide is getting close to a carbon that has a partial positive charge. Let's say a carbon is attached to a bromine and then it's attached to three hydrogens. We've seen this will have a partial negative charge. It's more electronegative than the carbon, which will have a partial positive charge."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "So let's say this iodide is getting close to a carbon that has a partial positive charge. Let's say a carbon is attached to a bromine and then it's attached to three hydrogens. We've seen this will have a partial negative charge. It's more electronegative than the carbon, which will have a partial positive charge. And so when this guy, this big guy with the electrons really far away, gets close to this, the electrons are going to, more of the electron cloud is going to be attracted to the partial positive charge, so it'll get distorted a little bit. And so it is more likely to react in a polar protic solvent. Fluoride, on the other hand, very tightly packed, blocked by the hydrogen bonds, less likely to react."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It's more electronegative than the carbon, which will have a partial positive charge. And so when this guy, this big guy with the electrons really far away, gets close to this, the electrons are going to, more of the electron cloud is going to be attracted to the partial positive charge, so it'll get distorted a little bit. And so it is more likely to react in a polar protic solvent. Fluoride, on the other hand, very tightly packed, blocked by the hydrogen bonds, less likely to react. So if you were to look at the periodic table, if you were to look at, say, just the halogens, in a polar protic solvent, the halides, so this would be the ion version of the halogens, the halides, iodide will be the best nucleophile, fluoride will be the worst. So in a polar protic solvent, let me write this down, in polar protic solvent, a new color, polar protic solvent, then we have a situation where the iodide is the best nucleophile, iodide followed by bromide, followed by chloride, and then last of all is the fluoride. Now, the exact opposite is true in a aprotic solvent."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Fluoride, on the other hand, very tightly packed, blocked by the hydrogen bonds, less likely to react. So if you were to look at the periodic table, if you were to look at, say, just the halogens, in a polar protic solvent, the halides, so this would be the ion version of the halogens, the halides, iodide will be the best nucleophile, fluoride will be the worst. So in a polar protic solvent, let me write this down, in polar protic solvent, a new color, polar protic solvent, then we have a situation where the iodide is the best nucleophile, iodide followed by bromide, followed by chloride, and then last of all is the fluoride. Now, the exact opposite is true in a aprotic solvent. In an aprotic solvent, the fluoride, which is, you know, fluorine is far more electronegative, fluoride is more basic, it will be more stable if it is able to form a bond with something than iodide. Iodide is pretty stable. It's actually, if you look at hydrogen, hydrogen iodide is actually a highly acidic molecule, so iodide itself, the conjugate base of hydrogen iodide, is going to be a very bad base."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "Now, the exact opposite is true in a aprotic solvent. In an aprotic solvent, the fluoride, which is, you know, fluorine is far more electronegative, fluoride is more basic, it will be more stable if it is able to form a bond with something than iodide. Iodide is pretty stable. It's actually, if you look at hydrogen, hydrogen iodide is actually a highly acidic molecule, so iodide itself, the conjugate base of hydrogen iodide, is going to be a very bad base. So when you're dealing with an aprotic solvent, you go in the direction of basicity. And we're going to learn in the next video that actually basicity and nucleophilicity are related, but they aren't the same concept, and we're going to talk about that in a little bit. So if you're in an aprotic solvent, you're no longer as light, you're not reacting with the solvent as much, and in this situation, fluoride is actually the best nucleophile, followed by chloride, followed by bromide, followed by iodide."}, {"video_title": "Nucleophilicity (Nucleophile Strength).mp3", "Sentence": "It's actually, if you look at hydrogen, hydrogen iodide is actually a highly acidic molecule, so iodide itself, the conjugate base of hydrogen iodide, is going to be a very bad base. So when you're dealing with an aprotic solvent, you go in the direction of basicity. And we're going to learn in the next video that actually basicity and nucleophilicity are related, but they aren't the same concept, and we're going to talk about that in a little bit. So if you're in an aprotic solvent, you're no longer as light, you're not reacting with the solvent as much, and in this situation, fluoride is actually the best nucleophile, followed by chloride, followed by bromide, followed by iodide. So here you're going in the direction of basicity. This is the best, this is the worst. In an aprotic solvent, if it was in a protic solvent, this is flipped around."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with sodium borohydride. So if I wanted to draw the dot structure for this, it would have boron with four hydrogens, and that would give boron a negative one formal charge. And so then we'd also have our sodium cation here. You could think about the hydride reducing agents as hydride transfer agents. We're going to transfer a hydride to our aldehyde or our ketone here. And so to refresh your memory about what a hydride is, that would be hydrogen with two electrons, which would give it a negative one formal charge. And so let's go ahead and show the transfer of a hydride from sodium borohydride to our carbonyl."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "You could think about the hydride reducing agents as hydride transfer agents. We're going to transfer a hydride to our aldehyde or our ketone here. And so to refresh your memory about what a hydride is, that would be hydrogen with two electrons, which would give it a negative one formal charge. And so let's go ahead and show the transfer of a hydride from sodium borohydride to our carbonyl. So remember, our carbonyl is partially negative and partially positive. So this carbon right here wants electrons. And so it can get some electrons from right here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and show the transfer of a hydride from sodium borohydride to our carbonyl. So remember, our carbonyl is partially negative and partially positive. So this carbon right here wants electrons. And so it can get some electrons from right here. So these two electrons are going to attack this carbon, pushing these electrons off onto the oxygen. So let's go ahead and show what that's going to form. All right, so we're going to now have our carbon bonded to a hydrogen."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so it can get some electrons from right here. So these two electrons are going to attack this carbon, pushing these electrons off onto the oxygen. So let's go ahead and show what that's going to form. All right, so we're going to now have our carbon bonded to a hydrogen. And then over here on the left, we would have our oxygen, now with three lone pairs of electrons, so a negative one formal charge, and an R and an H. So let's follow those electrons. So the electrons in blue right up here are going to form this bond right here. And then this would be this hydrogen right here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we're going to now have our carbon bonded to a hydrogen. And then over here on the left, we would have our oxygen, now with three lone pairs of electrons, so a negative one formal charge, and an R and an H. So let's follow those electrons. So the electrons in blue right up here are going to form this bond right here. And then this would be this hydrogen right here. So we've transferred a hydride, so a hydrogen and these two electrons to our carbonyl. Now, the reason we need something like sodium borohydride is because you couldn't use something like sodium hydride by itself, because hydride by itself is not a great nucleophile. It's a good base."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And then this would be this hydrogen right here. So we've transferred a hydride, so a hydrogen and these two electrons to our carbonyl. Now, the reason we need something like sodium borohydride is because you couldn't use something like sodium hydride by itself, because hydride by itself is not a great nucleophile. It's a good base. So the orbital is too small to interact well with the carbonyl carbon here. And so that's why we need a hydride transfer agent, something like sodium borohydride. So once we've added on our hydride, we could then protonate it."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "It's a good base. So the orbital is too small to interact well with the carbonyl carbon here. And so that's why we need a hydride transfer agent, something like sodium borohydride. So once we've added on our hydride, we could then protonate it. And you'll see different ways to do this in textbooks. One way would be to, in a second step, just add protons and protonate your alkoxide to form your final product. And so that could be done in the workup."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So once we've added on our hydride, we could then protonate it. And you'll see different ways to do this in textbooks. One way would be to, in a second step, just add protons and protonate your alkoxide to form your final product. And so that could be done in the workup. Or sometimes you'll see textbooks just list sodium borohydride with either an alcohol, like methanol or ethanol, or water. And you could protonate your alkoxide that way, too. So just depending on how you work up this reaction."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so that could be done in the workup. Or sometimes you'll see textbooks just list sodium borohydride with either an alcohol, like methanol or ethanol, or water. And you could protonate your alkoxide that way, too. So just depending on how you work up this reaction. So the important thing is the formation of your alcohol and by transferring a hydride, so a hydrogen right here and then two electrons to the carbon there. And so you're reducing your carbonyl. So if I go back over here to the left, so let's look at this aldehyde here, or a ketone."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So just depending on how you work up this reaction. So the important thing is the formation of your alcohol and by transferring a hydride, so a hydrogen right here and then two electrons to the carbon there. And so you're reducing your carbonyl. So if I go back over here to the left, so let's look at this aldehyde here, or a ketone. So I have a carbon with two bonds to oxygen. And by adding a hydride, now the carbon has only one bond to oxygen. So if you assign some oxidation states, you will see that this carbon here in red has been reduced."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So if I go back over here to the left, so let's look at this aldehyde here, or a ketone. So I have a carbon with two bonds to oxygen. And by adding a hydride, now the carbon has only one bond to oxygen. So if you assign some oxidation states, you will see that this carbon here in red has been reduced. So the hydride reducing agents reduce the carbonyl to form an alcohol. Let's look at an example using sodium borohydride. So over here we have a ketone."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So if you assign some oxidation states, you will see that this carbon here in red has been reduced. So the hydride reducing agents reduce the carbonyl to form an alcohol. Let's look at an example using sodium borohydride. So over here we have a ketone. And we have sodium borohydride in methanol here. And so for our final products, we could go ahead and draw our ring is untouched. So we put that in here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So over here we have a ketone. And we have sodium borohydride in methanol here. And so for our final products, we could go ahead and draw our ring is untouched. So we put that in here. And then just think about adding a hydride. So a hydrogen and two electrons to our carbonyl carbon. So let's go ahead and draw this out here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So we put that in here. And then just think about adding a hydride. So a hydrogen and two electrons to our carbonyl carbon. So let's go ahead and draw this out here. So we're going to add hydrogen and two electrons to our carbonyl carbon. We're going to form an alkoxide. And then we're going to protonate the alkoxide in the workup to form an alcohol."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw this out here. So we're going to add hydrogen and two electrons to our carbonyl carbon. We're going to form an alkoxide. And then we're going to protonate the alkoxide in the workup to form an alcohol. So once again, let's go ahead and show the addition of that hydride. So this would be this hydrogen. And these two electrons add on to form your alcohol."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And then we're going to protonate the alkoxide in the workup to form an alcohol. So once again, let's go ahead and show the addition of that hydride. So this would be this hydrogen. And these two electrons add on to form your alcohol. And so here we're starting with a ketone. And we're ending with a secondary alcohol. So the carbon bonded to the OH is bonded to two other carbons."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And these two electrons add on to form your alcohol. And so here we're starting with a ketone. And we're ending with a secondary alcohol. So the carbon bonded to the OH is bonded to two other carbons. So that's formation of a secondary alcohol, reduction of a ketone to form a secondary alcohol. Another hydride reducing agent is lithium aluminum hydride. So let's look at this reaction here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So the carbon bonded to the OH is bonded to two other carbons. So that's formation of a secondary alcohol, reduction of a ketone to form a secondary alcohol. Another hydride reducing agent is lithium aluminum hydride. So let's look at this reaction here. So we have lithium aluminum hydride. Let's draw the structure for this one. So I would have aluminum, this time with four bonds to hydrogen like that, which would give aluminum a negative 1 formal charge."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this reaction here. So we have lithium aluminum hydride. Let's draw the structure for this one. So I would have aluminum, this time with four bonds to hydrogen like that, which would give aluminum a negative 1 formal charge. And so the lithium plus 1 ion would be there like that. And so once again, this is a hydride transfer agent. We're going to think about transferring a hydride from lithium aluminum hydride to the carbonyl."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So I would have aluminum, this time with four bonds to hydrogen like that, which would give aluminum a negative 1 formal charge. And so the lithium plus 1 ion would be there like that. And so once again, this is a hydride transfer agent. We're going to think about transferring a hydride from lithium aluminum hydride to the carbonyl. And so let's try to be consistent here with colors. So I'm going to say these two electrons and this hydrogen are going to come along. So we're going to attack here and push these electrons off onto the oxygen."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "We're going to think about transferring a hydride from lithium aluminum hydride to the carbonyl. And so let's try to be consistent here with colors. So I'm going to say these two electrons and this hydrogen are going to come along. So we're going to attack here and push these electrons off onto the oxygen. And this is definitely an oversimplified mechanism. But once again, this is the easiest to think about. So now we would have our alkoxide over here on the left."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to attack here and push these electrons off onto the oxygen. And this is definitely an oversimplified mechanism. But once again, this is the easiest to think about. So now we would have our alkoxide over here on the left. So let me go ahead and draw in that negative 1 formal charge on our oxygen. And then over here on the right, we've added a hydride. And we started with a hydrogen originally on our aldehyde."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So now we would have our alkoxide over here on the left. So let me go ahead and draw in that negative 1 formal charge on our oxygen. And then over here on the right, we've added a hydride. And we started with a hydrogen originally on our aldehyde. So let me go ahead and show those. So the hydride that we added, I'm going to say it's this right here and these electrons. And then we started with a hydrogen originally on our aldehyde."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And we started with a hydrogen originally on our aldehyde. So let me go ahead and show those. So the hydride that we added, I'm going to say it's this right here and these electrons. And then we started with a hydrogen originally on our aldehyde. So I'm going to say that's this hydrogen right here. And so in the next step, we need to protonate our alkoxide. And so in the second step, you could add water or dilute acid or something like that to protonate your alkoxide."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And then we started with a hydrogen originally on our aldehyde. So I'm going to say that's this hydrogen right here. And so in the next step, we need to protonate our alkoxide. And so in the second step, you could add water or dilute acid or something like that to protonate your alkoxide. So it grabs a proton from here, pushes these electrons off. And then you form your alcohol product. Let me go ahead and draw that in here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so in the second step, you could add water or dilute acid or something like that to protonate your alkoxide. So it grabs a proton from here, pushes these electrons off. And then you form your alcohol product. Let me go ahead and draw that in here. So we would form, let's see, this as our alcohol, so an OH. So let's go ahead and count our carbons here, just to make sure we did it right. So this is carbon 1, 2, 3, and 4 carbons to start with."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and draw that in here. So we would form, let's see, this as our alcohol, so an OH. So let's go ahead and count our carbons here, just to make sure we did it right. So this is carbon 1, 2, 3, and 4 carbons to start with. And then we have carbon 1, 2, 3, and 4 carbons to start with here. And so let's go ahead and show the hydride that we added. So we added this hydride onto here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So this is carbon 1, 2, 3, and 4 carbons to start with. And then we have carbon 1, 2, 3, and 4 carbons to start with here. And so let's go ahead and show the hydride that we added. So we added this hydride onto here. And then there was already a hydrogen on that carbon starting out from our aldehyde. And so that's our product. And so going from an aldehyde over here on the left, and if we reduce our aldehyde, then we're going to form a primary alcohol."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So we added this hydride onto here. And then there was already a hydrogen on that carbon starting out from our aldehyde. And so that's our product. And so going from an aldehyde over here on the left, and if we reduce our aldehyde, then we're going to form a primary alcohol. So let's go ahead and analyze this. This is a primary alcohol. The carbon that's bonded to our OH is bonded to one other carbon."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so going from an aldehyde over here on the left, and if we reduce our aldehyde, then we're going to form a primary alcohol. So let's go ahead and analyze this. This is a primary alcohol. The carbon that's bonded to our OH is bonded to one other carbon. So a reduction of an aldehyde using lithium aluminum hydride would give us a primary alcohol as our target. Now for lithium aluminum hydride, you definitely have to show these two different steps. Lithium aluminum hydride reacts violently with water."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "The carbon that's bonded to our OH is bonded to one other carbon. So a reduction of an aldehyde using lithium aluminum hydride would give us a primary alcohol as our target. Now for lithium aluminum hydride, you definitely have to show these two different steps. Lithium aluminum hydride reacts violently with water. And let's go ahead and show why. So if you have lithium aluminum hydride and water together, you get a violent and sometimes potentially dangerous reaction. So we would have our negatively charged aluminum here like that."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "Lithium aluminum hydride reacts violently with water. And let's go ahead and show why. So if you have lithium aluminum hydride and water together, you get a violent and sometimes potentially dangerous reaction. So we would have our negatively charged aluminum here like that. And if you mix these in at the same time, pretty much what's going to happen is you're going to transfer a hydride. You're going to get this and this. You're going to take this proton off of water."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our negatively charged aluminum here like that. And if you mix these in at the same time, pretty much what's going to happen is you're going to transfer a hydride. You're going to get this and this. You're going to take this proton off of water. And that's going to form hydrogen gas, so H2, which could be potentially dangerous. And so sodium borohydride is not quite as reactive as lithium aluminum hydride. And so you can use it in an alcohol."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "You're going to take this proton off of water. And that's going to form hydrogen gas, so H2, which could be potentially dangerous. And so sodium borohydride is not quite as reactive as lithium aluminum hydride. And so you can use it in an alcohol. But you can't use something a lot like lithium aluminum hydride. You have to make sure everything is completely dry when you're doing that. And so let's talk a little bit more about why lithium aluminum hydride is more reactive than sodium borohydride."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so you can use it in an alcohol. But you can't use something a lot like lithium aluminum hydride. You have to make sure everything is completely dry when you're doing that. And so let's talk a little bit more about why lithium aluminum hydride is more reactive than sodium borohydride. And let's look at this next picture here. And let's discuss the different reactivities. And so first, let's talk about sodium borohydride."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so let's talk a little bit more about why lithium aluminum hydride is more reactive than sodium borohydride. And let's look at this next picture here. And let's discuss the different reactivities. And so first, let's talk about sodium borohydride. So that would be the bond between boron and this hydrogen. And then let's think about lithium aluminum hydride. So that would be aluminum bonded to this hydrogen here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so first, let's talk about sodium borohydride. So that would be the bond between boron and this hydrogen. And then let's think about lithium aluminum hydride. So that would be aluminum bonded to this hydrogen here. So electronegativity differences. So boron has a value of approximately 2. And aluminum has a value of approximately 1.5."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So that would be aluminum bonded to this hydrogen here. So electronegativity differences. So boron has a value of approximately 2. And aluminum has a value of approximately 1.5. So if you think about the electrons that we've been discussing for our hydride transfer, so these electrons right in here, boron is more electronegative than aluminum. So it wants those electrons in blue a little bit more. And since aluminum doesn't want those electrons as much, its value is not as great."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And aluminum has a value of approximately 1.5. So if you think about the electrons that we've been discussing for our hydride transfer, so these electrons right in here, boron is more electronegative than aluminum. So it wants those electrons in blue a little bit more. And since aluminum doesn't want those electrons as much, its value is not as great. Therefore, it's easier to give them away. So it's easier for these electrons to be given away if it's bonded to aluminum. That's what makes it more reactive."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And since aluminum doesn't want those electrons as much, its value is not as great. Therefore, it's easier to give them away. So it's easier for these electrons to be given away if it's bonded to aluminum. That's what makes it more reactive. Aluminum is more willing to transfer the hydride either to water or to a carbonyl. And so therefore, lithium aluminum hydride is more reactive and it will reduce more functional groups than sodium borohydride will. And so let's look at what we have here for our starting material."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "That's what makes it more reactive. Aluminum is more willing to transfer the hydride either to water or to a carbonyl. And so therefore, lithium aluminum hydride is more reactive and it will reduce more functional groups than sodium borohydride will. And so let's look at what we have here for our starting material. So we have this compound. Let's say we first do a reduction with sodium borohydride. And sodium borohydride is going to react with our aldehyde, so right here on our molecule."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so let's look at what we have here for our starting material. So we have this compound. Let's say we first do a reduction with sodium borohydride. And sodium borohydride is going to react with our aldehyde, so right here on our molecule. But it's not strong enough to reduce our ester over here. And so it's only going to react with this portion of the molecules. Let's go ahead and draw the final product if sodium borohydride is added."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And sodium borohydride is going to react with our aldehyde, so right here on our molecule. But it's not strong enough to reduce our ester over here. And so it's only going to react with this portion of the molecules. Let's go ahead and draw the final product if sodium borohydride is added. So we're going to have our ring. It's going to reduce the aldehyde. And so if I think about the product, I could draw my OH here."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw the final product if sodium borohydride is added. So we're going to have our ring. It's going to reduce the aldehyde. And so if I think about the product, I could draw my OH here. And we started with one hydrogen on this carbon. And now we're going to have two. So let me go ahead and just go ahead and highlight those."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And so if I think about the product, I could draw my OH here. And we started with one hydrogen on this carbon. And now we're going to have two. So let me go ahead and just go ahead and highlight those. I'm saying one of these is the one we already started with. And then one of these is the one that we added. So let's say this one right here and these electrons."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and just go ahead and highlight those. I'm saying one of these is the one we already started with. And then one of these is the one that we added. So let's say this one right here and these electrons. This is our hydride transfer like that. And so we reduced our aldehyde to this alcohol. And the ester portion of the molecule remains untouched."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So let's say this one right here and these electrons. This is our hydride transfer like that. And so we reduced our aldehyde to this alcohol. And the ester portion of the molecule remains untouched. So let me go ahead and draw that. And so the sodium borohydride is selective for this aldehyde here. So not strong enough to reduce the ester."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And the ester portion of the molecule remains untouched. So let me go ahead and draw that. And so the sodium borohydride is selective for this aldehyde here. So not strong enough to reduce the ester. Lithium aluminum hydride, however, will reduce both of these functional groups. So let's go ahead and draw the product if that happens here. So let's go ahead and draw our ring."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So not strong enough to reduce the ester. Lithium aluminum hydride, however, will reduce both of these functional groups. So let's go ahead and draw the product if that happens here. So let's go ahead and draw our ring. And then we're going to reduce the aldehyde to, once again, our alcohol. So same way to think about it as before. So let's say this hydrogen was on there to start with."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw our ring. And then we're going to reduce the aldehyde to, once again, our alcohol. So same way to think about it as before. So let's say this hydrogen was on there to start with. And then we transferred a hydride. We transferred hydrogen with two electrons like that. And then we're also going to reduce our ester."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So let's say this hydrogen was on there to start with. And then we transferred a hydride. We transferred hydrogen with two electrons like that. And then we're also going to reduce our ester. So let me go ahead and draw the product there as well. So we're going to have an OH. And then we're going to have two hydrogens that were added on in this reduction, actually."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And then we're also going to reduce our ester. So let me go ahead and draw the product there as well. So we're going to have an OH. And then we're going to have two hydrogens that were added on in this reduction, actually. So let me go ahead and show those. So this hydrogen right here and these electrons, this hydrogen right here and these electrons. So in this case, it reduced it."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "And then we're going to have two hydrogens that were added on in this reduction, actually. So let me go ahead and show those. So this hydrogen right here and these electrons, this hydrogen right here and these electrons. So in this case, it reduced it. It added on two hydrides. And so we don't have time to talk about why in this video. But in a later video, we'll talk about the mechanism for reducing an ester using lithium aluminum hydride."}, {"video_title": "Formation of alcohols using hydride reducing agents Organic chemistry Khan Academy.mp3", "Sentence": "So in this case, it reduced it. It added on two hydrides. And so we don't have time to talk about why in this video. But in a later video, we'll talk about the mechanism for reducing an ester using lithium aluminum hydride. And then, of course, in the second step, we added a proton source here. So sodium borohydride, under normal conditions, will reduce aldehydes and ketones. Lithium aluminum hydride is much more reactive."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If I were to come over to your place and say, name this molecule, the first thing you would do is you would say, well, look, we have a double bond. So we're dealing with an alkene. We have four carbons in the longest chain. There's actually one chain here. So it's one, two, three, four carbons. So the prefix will be bute. And regardless of which end you count from, the double bond starts at the number two carbon."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "There's actually one chain here. So it's one, two, three, four carbons. So the prefix will be bute. And regardless of which end you count from, the double bond starts at the number two carbon. So this is bute 2-ene. I'd say, well, that's very good. And that is correct based on everything we've learned."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And regardless of which end you count from, the double bond starts at the number two carbon. So this is bute 2-ene. I'd say, well, that's very good. And that is correct based on everything we've learned. But what about this molecule? What about the molecule that looks like this? And you'd go through the same process."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that is correct based on everything we've learned. But what about this molecule? What about the molecule that looks like this? And you'd go through the same process. You'd say it has a double bond. It's an alkene. Our longest chain has four carbons."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And you'd go through the same process. You'd say it has a double bond. It's an alkene. Our longest chain has four carbons. One, two, three, four. Regardless of which end you start counting from, the double bond starts at the number two carbon. One, two, three, four."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Our longest chain has four carbons. One, two, three, four. Regardless of which end you start counting from, the double bond starts at the number two carbon. One, two, three, four. So once again, you'd say, hey, Sal, that's also bute 2-ene. And then I'd point out that these are each fundamentally different molecules. And the reason why they're fundamentally different molecules is because this double bond can't rotate."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four. So once again, you'd say, hey, Sal, that's also bute 2-ene. And then I'd point out that these are each fundamentally different molecules. And the reason why they're fundamentally different molecules is because this double bond can't rotate. It's got that pi bond there for that second bond that keeps this carbon-carbon bond rigid. So it's not like the molecule can switch configurations between this or it's not like it can switch conformations between this and that. It cannot rotate around this double bond."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why they're fundamentally different molecules is because this double bond can't rotate. It's got that pi bond there for that second bond that keeps this carbon-carbon bond rigid. So it's not like the molecule can switch configurations between this or it's not like it can switch conformations between this and that. It cannot rotate around this double bond. So these are different molecules that will behave different chemically. And because they behave different, we have to have different names for them. Now there's two different naming conventions."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It cannot rotate around this double bond. So these are different molecules that will behave different chemically. And because they behave different, we have to have different names for them. Now there's two different naming conventions. One is kind of, I guess, the simpler naming convention. And it works when we have one functional group on each of the carbons in the double bond. And then in the next few videos, we'll talk about the slightly more advanced naming scheme that will work when we have more than one functional group."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now there's two different naming conventions. One is kind of, I guess, the simpler naming convention. And it works when we have one functional group on each of the carbons in the double bond. And then in the next few videos, we'll talk about the slightly more advanced naming scheme that will work when we have more than one functional group. So the first way to name it is if you have your functional group, so let me circle the functional group. So in this top one, our functional groups are on opposite sides. They're apart."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then in the next few videos, we'll talk about the slightly more advanced naming scheme that will work when we have more than one functional group. So the first way to name it is if you have your functional group, so let me circle the functional group. So in this top one, our functional groups are on opposite sides. They're apart. They're on opposite sides of the carbon-carbon double bond. This is on the top. This is on the bottom."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "They're apart. They're on opposite sides of the carbon-carbon double bond. This is on the top. This is on the bottom. In this molecule right here, our functional groups are on the same side. So when our functional groups are on different sides, we could call it trans, literally for, I believe, Latin for apart or Latin for opposite. I'm not a Latin scholar, so forgive me."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is on the bottom. In this molecule right here, our functional groups are on the same side. So when our functional groups are on different sides, we could call it trans, literally for, I believe, Latin for apart or Latin for opposite. I'm not a Latin scholar, so forgive me. But the functional groups are apart, so we call it transbuteine. The other convention to use comes from the German for apart, and there we call it entgegen. And instead of writing out entgegen, we write e in parentheses."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'm not a Latin scholar, so forgive me. But the functional groups are apart, so we call it transbuteine. The other convention to use comes from the German for apart, and there we call it entgegen. And instead of writing out entgegen, we write e in parentheses. So we would call this e-bute-tween. These mean the same thing. But this is now kind of the standard convention."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And instead of writing out entgegen, we write e in parentheses. So we would call this e-bute-tween. These mean the same thing. But this is now kind of the standard convention. We'll see this notation where you use entgegen or e is actually more powerful. It can extend to when we have more than one functional group on each carbon. So let me just make it clear that the e stands for entgegen, which is German for apart."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But this is now kind of the standard convention. We'll see this notation where you use entgegen or e is actually more powerful. It can extend to when we have more than one functional group on each carbon. So let me just make it clear that the e stands for entgegen, which is German for apart. Or at least I believe it is. I can't speak German, so I'll just have to take people's words for it. Now, in this situation where our functional groups are on the same side, you could call this cisbuteine."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me just make it clear that the e stands for entgegen, which is German for apart. Or at least I believe it is. I can't speak German, so I'll just have to take people's words for it. Now, in this situation where our functional groups are on the same side, you could call this cisbuteine. Cis, I believe, comes from the Latin for together. Or you could call it z-bute-tween. And you could imagine this z comes from the German for together, and it stands for zusammen, which is German for together."}, {"video_title": "cis-trans and E-Z naming scheme for alkenes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, in this situation where our functional groups are on the same side, you could call this cisbuteine. Cis, I believe, comes from the Latin for together. Or you could call it z-bute-tween. And you could imagine this z comes from the German for together, and it stands for zusammen, which is German for together. And that's all you have to do. And in the next few videos, we'll do a bunch of examples here so we can really figure out whether to label something with either an e or a z. But I wanted to expose you to the cis and trans naming scheme, because that's sometimes used for simpler molecules where you only have one functional group on each side."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "In the previous video, we looked at aldol condensations with the same molecule. You can call those a simple aldol condensation. In this video, we're gonna look at mixed or crossed aldol condensation. So no longer are you starting with the same molecule. Here, we don't have two aldehydes that are the same. We have different aldehydes, right? We have benzaldehyde on the left and propanol on the right."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So no longer are you starting with the same molecule. Here, we don't have two aldehydes that are the same. We have different aldehydes, right? We have benzaldehyde on the left and propanol on the right. And so we need to figure out what sort of enolate anion that we're going to form. So when we add our sodium hydroxide as our base, what is going to be our enolate anion? To do that, we need to look for alpha carbons."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have benzaldehyde on the left and propanol on the right. And so we need to figure out what sort of enolate anion that we're going to form. So when we add our sodium hydroxide as our base, what is going to be our enolate anion? To do that, we need to look for alpha carbons. So we'll start with propanol. We know the alpha carbon is the one next to the carbonyl. And so right here, this would be an alpha carbon."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "To do that, we need to look for alpha carbons. So we'll start with propanol. We know the alpha carbon is the one next to the carbonyl. And so right here, this would be an alpha carbon. And there are two protons on that alpha carbon. So we have two alpha protons here. So I'm gonna go ahead and draw in those."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so right here, this would be an alpha carbon. And there are two protons on that alpha carbon. So we have two alpha protons here. So I'm gonna go ahead and draw in those. And so that's a possibility to form our enolate anion. This aldehyde hydrogen right here is not. We're not gonna form an enolate from that."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna go ahead and draw in those. And so that's a possibility to form our enolate anion. This aldehyde hydrogen right here is not. We're not gonna form an enolate from that. So we don't have to worry about the right side of our aldehyde. Let's look at benzaldehyde now. So if we think about this carbon right here, it is the carbon next to a carbonyl."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We're not gonna form an enolate from that. So we don't have to worry about the right side of our aldehyde. Let's look at benzaldehyde now. So if we think about this carbon right here, it is the carbon next to a carbonyl. However, this carbon already has four bonds to it. And so it doesn't have an alpha proton. So benzaldehyde does not have an alpha proton."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about this carbon right here, it is the carbon next to a carbonyl. However, this carbon already has four bonds to it. And so it doesn't have an alpha proton. So benzaldehyde does not have an alpha proton. So we don't need to worry about it forming an enolate anion. And so we know the enolate anion is going to form from propanol over here. And so let me just go ahead and redraw these in a way that makes it easier for me to see what the product is."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So benzaldehyde does not have an alpha proton. So we don't need to worry about it forming an enolate anion. And so we know the enolate anion is going to form from propanol over here. And so let me just go ahead and redraw these in a way that makes it easier for me to see what the product is. So I'm just gonna draw my ring here. And I'm gonna draw my carbonyl. And I like to leave off this hydrogen because it just gets in the way when I'm thinking about my mechanism."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so let me just go ahead and redraw these in a way that makes it easier for me to see what the product is. So I'm just gonna draw my ring here. And I'm gonna draw my carbonyl. And I like to leave off this hydrogen because it just gets in the way when I'm thinking about my mechanism. And for my propanol over here, I prefer to draw a different conformation of propanol to make it easier to see the product. So I like to put my carbonyl up here like this. And then I like to draw this carbon down like that."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I like to leave off this hydrogen because it just gets in the way when I'm thinking about my mechanism. And for my propanol over here, I prefer to draw a different conformation of propanol to make it easier to see the product. So I like to put my carbonyl up here like this. And then I like to draw this carbon down like that. And it's gonna make it easier to come up with the final result. All right, so if I think about deprotonation, all right, if I think about sodium hydroxide taking one of these alpha protons here, right, leaving electrons behind in my alpha carbon, I can think about the structure for the carbanion enolate anion, right? So there's a negative one formal charge on this carbon."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then I like to draw this carbon down like that. And it's gonna make it easier to come up with the final result. All right, so if I think about deprotonation, all right, if I think about sodium hydroxide taking one of these alpha protons here, right, leaving electrons behind in my alpha carbon, I can think about the structure for the carbanion enolate anion, right? So there's a negative one formal charge on this carbon. And that's my nucleophilic enolate anion. Once again, I prefer to draw the carbanion as opposed to the oxyanion. It just makes it easier for me when I'm trying to do a quick mechanism like on a test to figure out a product here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there's a negative one formal charge on this carbon. And that's my nucleophilic enolate anion. Once again, I prefer to draw the carbanion as opposed to the oxyanion. It just makes it easier for me when I'm trying to do a quick mechanism like on a test to figure out a product here. So now we have a nucleophile, right? So I'm gonna go ahead and put these electrons in magenta. So this is going to be our nucleophile."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It just makes it easier for me when I'm trying to do a quick mechanism like on a test to figure out a product here. So now we have a nucleophile, right? So I'm gonna go ahead and put these electrons in magenta. So this is going to be our nucleophile. And our nucleophile is going to attack the electrophilic portion of our benzaldehyde molecule. All right, so if we think about that, the oxygen is partial negative and this carbonyl carbon here is partially positive. So that's the electrophilic portion of the molecule."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be our nucleophile. And our nucleophile is going to attack the electrophilic portion of our benzaldehyde molecule. All right, so if we think about that, the oxygen is partial negative and this carbonyl carbon here is partially positive. So that's the electrophilic portion of the molecule. And so that lone pair of electrons here on this carbon is going to attack this carbon, right, pushing these electrons off onto the oxygen. And that would form an alkoxide intermediate. So let me just go ahead and sketch in here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's the electrophilic portion of the molecule. And so that lone pair of electrons here on this carbon is going to attack this carbon, right, pushing these electrons off onto the oxygen. And that would form an alkoxide intermediate. So let me just go ahead and sketch in here. All right, so we would form an oxygen right there with three lone pairs of electrons giving it a negative one formal charge. And I'll come back to that. The important carbon-carbon bond that's formed is right here and then we have, we can go ahead and draw in everything else."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me just go ahead and sketch in here. All right, so we would form an oxygen right there with three lone pairs of electrons giving it a negative one formal charge. And I'll come back to that. The important carbon-carbon bond that's formed is right here and then we have, we can go ahead and draw in everything else. So we have our aldehyde and then don't forget about this down here. So in magenta, right, these electrons in magenta, right, formed our new bond right here. And then we would have an alkoxide intermediate."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The important carbon-carbon bond that's formed is right here and then we have, we can go ahead and draw in everything else. So we have our aldehyde and then don't forget about this down here. So in magenta, right, these electrons in magenta, right, formed our new bond right here. And then we would have an alkoxide intermediate. So we can think about just protonating it. So I'm not concerned with an exact mechanism here. I'm more concerned about figuring out the product."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have an alkoxide intermediate. So we can think about just protonating it. So I'm not concerned with an exact mechanism here. I'm more concerned about figuring out the product. All right, so we go ahead and protonate our alkoxide to form our aldol intermediate. And so in the next step, right, we add heat, you're most likely going to form your conjugated product. So let's think about what would happen next, right?"}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm more concerned about figuring out the product. All right, so we go ahead and protonate our alkoxide to form our aldol intermediate. And so in the next step, right, we add heat, you're most likely going to form your conjugated product. So let's think about what would happen next, right? So we still have an alpha carbon. So this carbon right here next to our carbonyl is an alpha carbon. It still has an alpha proton on it, right?"}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about what would happen next, right? So we still have an alpha carbon. So this carbon right here next to our carbonyl is an alpha carbon. It still has an alpha proton on it, right? So we could show a proton right here and we could think about hydroxide once again acting as a base. So hydroxide acting as a base, coming along and taking this proton, right? And then we could even just go ahead and think about the final product."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It still has an alpha proton on it, right? So we could show a proton right here and we could think about hydroxide once again acting as a base. So hydroxide acting as a base, coming along and taking this proton, right? And then we could even just go ahead and think about the final product. We could think about these electrons moving into here and then these electrons moving out onto here. Again, I'm not so concerned with an exact detailed mechanism. I'm more worried about figuring out how to figure out our product here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we could even just go ahead and think about the final product. We could think about these electrons moving into here and then these electrons moving out onto here. Again, I'm not so concerned with an exact detailed mechanism. I'm more worried about figuring out how to figure out our product here. So when I draw our product, right, we have our ring and then we now have a double bond that forms here, right? We have this going down and then we have our aldehyde. So we form a conjugated product."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm more worried about figuring out how to figure out our product here. So when I draw our product, right, we have our ring and then we now have a double bond that forms here, right? We have this going down and then we have our aldehyde. So we form a conjugated product. We have an enal as our product here. So let's follow those electrons, right? So let's make these electrons in blue here, right?"}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we form a conjugated product. We have an enal as our product here. So let's follow those electrons, right? So let's make these electrons in blue here, right? So these electrons in blue moved in here to form our double bond. And we already formed this bond in magenta, right? The carbon-carbon bond forming part to form our aldol."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's make these electrons in blue here, right? So these electrons in blue moved in here to form our double bond. And we already formed this bond in magenta, right? The carbon-carbon bond forming part to form our aldol. And then we lost hydroxide. And so this would be our final product, our conjugated product. So again, the goal is just to figure out how to draw a product, right, from these reactants and thinking about where is the alpha carbon and thinking about what is the enolate that forms."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The carbon-carbon bond forming part to form our aldol. And then we lost hydroxide. And so this would be our final product, our conjugated product. So again, the goal is just to figure out how to draw a product, right, from these reactants and thinking about where is the alpha carbon and thinking about what is the enolate that forms. All right, let's do another one where I just kind of walk you through my thought process here. So let's look at this reaction. So say you had this on an exam and your task was to draw the product."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So again, the goal is just to figure out how to draw a product, right, from these reactants and thinking about where is the alpha carbon and thinking about what is the enolate that forms. All right, let's do another one where I just kind of walk you through my thought process here. So let's look at this reaction. So say you had this on an exam and your task was to draw the product. All right, so we have, once again, some different possible alpha carbons. So let's first focus on cyclohexanone, right? So we know that this could be an alpha carbon and we know that this could be an alpha carbon, all right?"}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So say you had this on an exam and your task was to draw the product. All right, so we have, once again, some different possible alpha carbons. So let's first focus on cyclohexanone, right? So we know that this could be an alpha carbon and we know that this could be an alpha carbon, all right? And then for this compound over here, we have these two carbonyls, right, in this molecule. So we know that this could be an alpha carbon and we know that this could be an alpha carbon. And so now we have to figure out which one of those alpha carbons is going to be deprotonated when we add sodium ethoxide, right, as our base right here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we know that this could be an alpha carbon and we know that this could be an alpha carbon, all right? And then for this compound over here, we have these two carbonyls, right, in this molecule. So we know that this could be an alpha carbon and we know that this could be an alpha carbon. And so now we have to figure out which one of those alpha carbons is going to be deprotonated when we add sodium ethoxide, right, as our base right here. And if you remember in one of the earlier videos, we talked about an alpha carbon between two carbonyls as having the most acidic protons. So this alpha carbon right here has two protons on it and it's easy to deprotonate, all right, because of the resonance stabilization that we can draw because of those carbonyls. And so these protons are the most acidic, right, much more acidic than the protons on our ketone here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so now we have to figure out which one of those alpha carbons is going to be deprotonated when we add sodium ethoxide, right, as our base right here. And if you remember in one of the earlier videos, we talked about an alpha carbon between two carbonyls as having the most acidic protons. So this alpha carbon right here has two protons on it and it's easy to deprotonate, all right, because of the resonance stabilization that we can draw because of those carbonyls. And so these protons are the most acidic, right, much more acidic than the protons on our ketone here. And so these are the ones that are going to be, one of these protons could be deprotonated when we add sodium ethoxide here. So sodium ethoxide is going to come along, right, take one of these protons here, leave these electrons behind. And so let's go ahead and redraw what we would form here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so these protons are the most acidic, right, much more acidic than the protons on our ketone here. And so these are the ones that are going to be, one of these protons could be deprotonated when we add sodium ethoxide here. So sodium ethoxide is going to come along, right, take one of these protons here, leave these electrons behind. And so let's go ahead and redraw what we would form here. So we're going to form our enolate anion. So let me go ahead and draw that. So we would now have a lone pair of electrons on our carbon, giving that carbon a negative one formal charge."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and redraw what we would form here. So we're going to form our enolate anion. So let me go ahead and draw that. So we would now have a lone pair of electrons on our carbon, giving that carbon a negative one formal charge. So let me follow those electrons. These electrons in magenta are now on our carbon, forming our carb anion. And I'm not going to take the time to draw the resonance structures for the oxyanions because I'm just concerned about figuring out the product here."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we would now have a lone pair of electrons on our carbon, giving that carbon a negative one formal charge. So let me follow those electrons. These electrons in magenta are now on our carbon, forming our carb anion. And I'm not going to take the time to draw the resonance structures for the oxyanions because I'm just concerned about figuring out the product here. So now we have a nucleophilic enolate anion and we know that's going to attack the carbonyl of our ketone, right. So going back over here to our ketone, right, the oxygen is partial negative. This carbon is partially positive."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I'm not going to take the time to draw the resonance structures for the oxyanions because I'm just concerned about figuring out the product here. So now we have a nucleophilic enolate anion and we know that's going to attack the carbonyl of our ketone, right. So going back over here to our ketone, right, the oxygen is partial negative. This carbon is partially positive. And so our nucleophile is going to attack our electrophile. So you can think about these electrons, right, attacking here, pushing those electrons off onto your oxygen. So if we were to draw the intermediate here, right, we would have our ring."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is partially positive. And so our nucleophile is going to attack our electrophile. So you can think about these electrons, right, attacking here, pushing those electrons off onto your oxygen. So if we were to draw the intermediate here, right, we would have our ring. And we're going to form an alkoxide. So I'm not going to draw all the lone pairs on that oxygen right now. I'm more concerned right now with showing the formation of this bond."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we were to draw the intermediate here, right, we would have our ring. And we're going to form an alkoxide. So I'm not going to draw all the lone pairs on that oxygen right now. I'm more concerned right now with showing the formation of this bond. So let me go ahead and draw in everything and then we'll follow some electrons. So I put in my carbonyls and then I have these guys over here like that. So the electrons in magenta, right, these electrons formed our carbon-carbon bond."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm more concerned right now with showing the formation of this bond. So let me go ahead and draw in everything and then we'll follow some electrons. So I put in my carbonyls and then I have these guys over here like that. So the electrons in magenta, right, these electrons formed our carbon-carbon bond. So they formed this bond right here. And then we would form an alkoxide. And then we would just go ahead and protonate that to form our aldol."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta, right, these electrons formed our carbon-carbon bond. So they formed this bond right here. And then we would form an alkoxide. And then we would just go ahead and protonate that to form our aldol. So once again, just to save time, not an exact mechanism, but thinking about our intermediate as being this aldol. All right, so because we have heat, we're probably going to, once again, form a conjugated product here and keep going for the complete aldol condensation. And so next we can think about this alpha carbon right here, still having an acidic proton on it, right?"}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we would just go ahead and protonate that to form our aldol. So once again, just to save time, not an exact mechanism, but thinking about our intermediate as being this aldol. All right, so because we have heat, we're probably going to, once again, form a conjugated product here and keep going for the complete aldol condensation. And so next we can think about this alpha carbon right here, still having an acidic proton on it, right? So there's still a proton attached to that alpha carbon. And so sodium methoxide can come along. So we have a methoxide anion."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so next we can think about this alpha carbon right here, still having an acidic proton on it, right? So there's still a proton attached to that alpha carbon. And so sodium methoxide can come along. So we have a methoxide anion. Let's go ahead and draw that in here, which could function as our base. So it could take that proton and think about these electrons moving into here to form our double bond. We could think about hydroxide as a leaving group."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have a methoxide anion. Let's go ahead and draw that in here, which could function as our base. So it could take that proton and think about these electrons moving into here to form our double bond. We could think about hydroxide as a leaving group. And then that just allows us to figure out our product. So we have our ring. And then we now have a double bond."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could think about hydroxide as a leaving group. And then that just allows us to figure out our product. So we have our ring. And then we now have a double bond. And then we have a carbonyl over here, and then a carbonyl over here, and then our oxygen and our ethyl like that. So let's think about those electrons. So the electrons in blue here move in to form our double bond."}, {"video_title": "Mixed (crossed) aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we now have a double bond. And then we have a carbonyl over here, and then a carbonyl over here, and then our oxygen and our ethyl like that. So let's think about those electrons. So the electrons in blue here move in to form our double bond. And we had already formed a carbon-carbon bond before. So let's say it's these electrons right here. And then we have a stable conjugated product."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Here we have one chair conformation of methylcyclohexane, and this is carbon one. You can see we have a methyl group that is axial up at carbon one. We also have a hydrogen, and I've made the hydrogen green so we can tell it apart from the other hydrogens. And this hydrogen is equatorial down at carbon one. Now this chair conformation is in equilibrium with another chair conformation, and we can get to the other chair conformation by what we call a ring flip. So if I lift this carbon up and I pull this other carbon down here, and then we rotate it a little bit, we can see the other chair conformation of methylcyclohexane, so here it is. Now when we look at carbon one, we can see we have a difference now."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And this hydrogen is equatorial down at carbon one. Now this chair conformation is in equilibrium with another chair conformation, and we can get to the other chair conformation by what we call a ring flip. So if I lift this carbon up and I pull this other carbon down here, and then we rotate it a little bit, we can see the other chair conformation of methylcyclohexane, so here it is. Now when we look at carbon one, we can see we have a difference now. At carbon one, the methyl group is now equatorial. Notice it's still up relative to the plane of the ring, but it's an equatorial methyl group. And the hydrogen is now axial, so it's still down, but it's an axial hydrogen."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now when we look at carbon one, we can see we have a difference now. At carbon one, the methyl group is now equatorial. Notice it's still up relative to the plane of the ring, but it's an equatorial methyl group. And the hydrogen is now axial, so it's still down, but it's an axial hydrogen. So that's what happens in a ring flip. We just saw the video of methylcyclohexane undergoing a ring flip, and now we have to draw our two chair conformations. So we saw how to do this in an earlier video."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And the hydrogen is now axial, so it's still down, but it's an axial hydrogen. So that's what happens in a ring flip. We just saw the video of methylcyclohexane undergoing a ring flip, and now we have to draw our two chair conformations. So we saw how to do this in an earlier video. If I'm trying to draw the chair conformation on the left, I start with two parallel lines that are offset from each other, so something like that. And then I draw a dotted line here that intersects with the top point on the top line, and a dotted line that intersects with the bottom point on the bottom line. And next, we draw a line from the top dotted line down to the bottom, and then from the bottom up to the top, and those are parallel."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we saw how to do this in an earlier video. If I'm trying to draw the chair conformation on the left, I start with two parallel lines that are offset from each other, so something like that. And then I draw a dotted line here that intersects with the top point on the top line, and a dotted line that intersects with the bottom point on the bottom line. And next, we draw a line from the top dotted line down to the bottom, and then from the bottom up to the top, and those are parallel. And then finally, we have another set of parallel lines to give us a total of three sets of parallel lines. We start at carbon one, and we start axial up. So there's axial up, and then at carbon two would be axial down."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And next, we draw a line from the top dotted line down to the bottom, and then from the bottom up to the top, and those are parallel. And then finally, we have another set of parallel lines to give us a total of three sets of parallel lines. We start at carbon one, and we start axial up. So there's axial up, and then at carbon two would be axial down. Carbon three, axial up. Carbon four, axial down. Five is up, and six is down."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So there's axial up, and then at carbon two would be axial down. Carbon three, axial up. Carbon four, axial down. Five is up, and six is down. Next, we put in equatorial. So at carbon one, we start equatorial down, so I draw that down. At two, it'd be equatorial up."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Five is up, and six is down. Next, we put in equatorial. So at carbon one, we start equatorial down, so I draw that down. At two, it'd be equatorial up. At three, it would be down. At four, it would be up. At five, it would be down, and six would be up."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "At two, it'd be equatorial up. At three, it would be down. At four, it would be up. At five, it would be down, and six would be up. All right, this chair conformation is in equilibrium with another chair conformation, so let's go ahead and draw the other chair, and then I'll go back and put in the different groups. So for the other chair, we approach it the same way. We draw two parallel lines that are offset from each other, so those are my two parallel lines, and then we draw our dotted line that intersects pretty close to this top point here on the top line, and then we draw another dotted line which is pretty close to this bottom point on the bottom line."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "At five, it would be down, and six would be up. All right, this chair conformation is in equilibrium with another chair conformation, so let's go ahead and draw the other chair, and then I'll go back and put in the different groups. So for the other chair, we approach it the same way. We draw two parallel lines that are offset from each other, so those are my two parallel lines, and then we draw our dotted line that intersects pretty close to this top point here on the top line, and then we draw another dotted line which is pretty close to this bottom point on the bottom line. Next, we put in another set of parallel bonds. So we put this one in, and then we draw, think about drawing a line parallel to that, so like that, and then finally, we can go ahead and connect our dots, so we go like this, and these lines should be parallel too. Now this is carbon one, and we do the opposite of what we did before."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We draw two parallel lines that are offset from each other, so those are my two parallel lines, and then we draw our dotted line that intersects pretty close to this top point here on the top line, and then we draw another dotted line which is pretty close to this bottom point on the bottom line. Next, we put in another set of parallel bonds. So we put this one in, and then we draw, think about drawing a line parallel to that, so like that, and then finally, we can go ahead and connect our dots, so we go like this, and these lines should be parallel too. Now this is carbon one, and we do the opposite of what we did before. We start axial down, so axial down at carbon one. This is carbon two. We go axial up."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now this is carbon one, and we do the opposite of what we did before. We start axial down, so axial down at carbon one. This is carbon two. We go axial up. At carbon three is axial down. Carbon four, axial up. Carbon five, axial down, and finally, carbon six would be axial up."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We go axial up. At carbon three is axial down. Carbon four, axial up. Carbon five, axial down, and finally, carbon six would be axial up. Go back to carbon one, and we're putting in the equatorial bonds now. Since we started axial down, then we would do equatorial up, so this would be equatorial up, and then equatorial down at carbon two. At carbon three would be equatorial up."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Carbon five, axial down, and finally, carbon six would be axial up. Go back to carbon one, and we're putting in the equatorial bonds now. Since we started axial down, then we would do equatorial up, so this would be equatorial up, and then equatorial down at carbon two. At carbon three would be equatorial up. At carbon four, equatorial down. Carbon five, equatorial up, and finally, carbon six, equatorial down. Now let's finally put in our groups."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "At carbon three would be equatorial up. At carbon four, equatorial down. Carbon five, equatorial up, and finally, carbon six, equatorial down. Now let's finally put in our groups. So we go back up here to the diagram, and this was carbon one. We had a methyl group axial up, so that methyl group is axial up, and go ahead and put in a CH three right here at carbon one, and then we had this green hydrogen here, which was equatorial down if you look at how it's drawn. So let me go ahead and put in that hydrogen."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now let's finally put in our groups. So we go back up here to the diagram, and this was carbon one. We had a methyl group axial up, so that methyl group is axial up, and go ahead and put in a CH three right here at carbon one, and then we had this green hydrogen here, which was equatorial down if you look at how it's drawn. So let me go ahead and put in that hydrogen. So this is our green hydrogen. Let me go ahead and circle it and make it green so we can tell the difference here, and I could put in the other hydrogens, but I'm just not going to on our chair conformations. It's really good practice to draw all of these bonds, and you could put in the hydrogens, but I'm not worried about it."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and put in that hydrogen. So this is our green hydrogen. Let me go ahead and circle it and make it green so we can tell the difference here, and I could put in the other hydrogens, but I'm just not going to on our chair conformations. It's really good practice to draw all of these bonds, and you could put in the hydrogens, but I'm not worried about it. Our goal is to show the ring flip. So at carbon one, let me go ahead and highlight it, this carbon one became this carbon right here when we did the ring flip, and notice what happened to our methyl group just to repeat what we saw in the video. We started axial up for our methyl group, and we still have this methyl group up, but notice it is now equatorial."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It's really good practice to draw all of these bonds, and you could put in the hydrogens, but I'm not worried about it. Our goal is to show the ring flip. So at carbon one, let me go ahead and highlight it, this carbon one became this carbon right here when we did the ring flip, and notice what happened to our methyl group just to repeat what we saw in the video. We started axial up for our methyl group, and we still have this methyl group up, but notice it is now equatorial. So we need to put in that methyl group equatorial at carbon one, so we go ahead and do that. Remember, this is carbon one, and then our hydrogen, the green hydrogen, is still down relative to the plane of the ring, but it's now axial. So this hydrogen right here is the green one."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We started axial up for our methyl group, and we still have this methyl group up, but notice it is now equatorial. So we need to put in that methyl group equatorial at carbon one, so we go ahead and do that. Remember, this is carbon one, and then our hydrogen, the green hydrogen, is still down relative to the plane of the ring, but it's now axial. So this hydrogen right here is the green one. Let me go ahead and highlight it in green. It is now down relative to our plane. So the point of a ring flip is any group that's axial becomes equatorial."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen right here is the green one. Let me go ahead and highlight it in green. It is now down relative to our plane. So the point of a ring flip is any group that's axial becomes equatorial. Any group that's equatorial becomes axial. So that changes, but what doesn't change is is it up or down relative to the plane of the ring? So let me go ahead and highlight this here."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So the point of a ring flip is any group that's axial becomes equatorial. Any group that's equatorial becomes axial. So that changes, but what doesn't change is is it up or down relative to the plane of the ring? So let me go ahead and highlight this here. The methyl group stays up, so here it's axial up, and over here it's still up relative to the plane, but it turns equatorial. And what was down, which was our hydrogens, let me go ahead and highlight that here, this was equatorial down, the hydrogen stays down relative to the plane of the ring, but it turns axial. So you need to practice drawing all this stuff out."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight this here. The methyl group stays up, so here it's axial up, and over here it's still up relative to the plane, but it turns equatorial. And what was down, which was our hydrogens, let me go ahead and highlight that here, this was equatorial down, the hydrogen stays down relative to the plane of the ring, but it turns axial. So you need to practice drawing all this stuff out. Now let's think about which conformation is more stable. So we have two chair conformations, and one is more stable than the other. And let's go back up to this picture here, and let's look at the interaction of this methyl group with these other axial hydrogens."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So you need to practice drawing all this stuff out. Now let's think about which conformation is more stable. So we have two chair conformations, and one is more stable than the other. And let's go back up to this picture here, and let's look at the interaction of this methyl group with these other axial hydrogens. So think about this hydrogen right here, and then this hydrogen right back here. This methyl group takes up a lot of space, and it's pretty close to these hydrogens in space. If you rotate this, there's some serious steric hindrance with putting the methyl group in the axial position."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And let's go back up to this picture here, and let's look at the interaction of this methyl group with these other axial hydrogens. So think about this hydrogen right here, and then this hydrogen right back here. This methyl group takes up a lot of space, and it's pretty close to these hydrogens in space. If you rotate this, there's some serious steric hindrance with putting the methyl group in the axial position. However, for this conformation on the right, with this methyl group out to the side in an equatorial position, now this methyl group doesn't really interfere with any of the axial hydrogens. So down here we have a hydrogen, right, which could interact a little bit with these other hydrogens but nowhere near as much as putting that methyl group in this position. So the goal with a relatively bulky substituent is to put it in the equatorial position, and that's gonna be the most stable conformation."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "If you rotate this, there's some serious steric hindrance with putting the methyl group in the axial position. However, for this conformation on the right, with this methyl group out to the side in an equatorial position, now this methyl group doesn't really interfere with any of the axial hydrogens. So down here we have a hydrogen, right, which could interact a little bit with these other hydrogens but nowhere near as much as putting that methyl group in this position. So the goal with a relatively bulky substituent is to put it in the equatorial position, and that's gonna be the most stable conformation. It turns out at equilibrium, at room temperature, this conformation, this chair conformation on the right is about 95% of all the chair conformations of cyclohexane, and the one on the left is only about 5%. So this is a substantial effect. The steric hindrance that destabilizes this conformation, you could call this a 1,3-diaxial interaction."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So the goal with a relatively bulky substituent is to put it in the equatorial position, and that's gonna be the most stable conformation. It turns out at equilibrium, at room temperature, this conformation, this chair conformation on the right is about 95% of all the chair conformations of cyclohexane, and the one on the left is only about 5%. So this is a substantial effect. The steric hindrance that destabilizes this conformation, you could call this a 1,3-diaxial interaction. So this is carbon one right here, so carbon one, we have something that's axial, and then this would be carbon two, and this would be carbon three. So carbon three, we have something else that's axial, and these two can interfere with each other. So that's why we call this a 1,3-diaxial, so two things that are axial interactions, so 1,3-diaxial interaction."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "The steric hindrance that destabilizes this conformation, you could call this a 1,3-diaxial interaction. So this is carbon one right here, so carbon one, we have something that's axial, and then this would be carbon two, and this would be carbon three. So carbon three, we have something else that's axial, and these two can interfere with each other. So that's why we call this a 1,3-diaxial, so two things that are axial interactions, so 1,3-diaxial interaction. It's the same idea for this hydrogen back here. And this steric hindrance destabilizes this conformation. The effect is even more pronounced when you have a larger group that's interfering with these axial hydrogens."}, {"video_title": "Monosubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So that's why we call this a 1,3-diaxial, so two things that are axial interactions, so 1,3-diaxial interaction. It's the same idea for this hydrogen back here. And this steric hindrance destabilizes this conformation. The effect is even more pronounced when you have a larger group that's interfering with these axial hydrogens. For example, if we changed our group from a methyl group to a tert-butyl group, let me go down here so we can see a tert-butyl group, and I only drew part of the bonds here. Here you can see our tert-butyl group at carbon one, and if you think about 1,3-diaxial interactions, I could draw in some hydrogens here. Since you have all of these methyl groups here, this is gonna be even more steric hindrance."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "A liquid boils when its molecules have enough energy to break free of the attractions that exist between those molecules. And those attractions between the molecules are called the intermolecular forces. Let's compare two molecules, pentane on the left and hexane on the right. These are both hydrocarbons, which means they contain only hydrogen and carbon. Pentane has five carbons, one, two, three, four, five. So five carbons for pentane, and pentane has a boiling point of 36 degrees Celsius. Hexane has six carbons, one, two, three, four, five, and six."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "These are both hydrocarbons, which means they contain only hydrogen and carbon. Pentane has five carbons, one, two, three, four, five. So five carbons for pentane, and pentane has a boiling point of 36 degrees Celsius. Hexane has six carbons, one, two, three, four, five, and six. So six carbons and a higher boiling point of 69 degrees C. Let's draw on another molecule of pentane right here. So there's five carbons. Let's think about the intermolecular forces that exist between those two molecules of pentane."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Hexane has six carbons, one, two, three, four, five, and six. So six carbons and a higher boiling point of 69 degrees C. Let's draw on another molecule of pentane right here. So there's five carbons. Let's think about the intermolecular forces that exist between those two molecules of pentane. Pentane is a non-polar molecule, and we know the only intermolecular force that exists between two non-polar molecules, that would of course be the London dispersion forces. So London dispersion forces exist between these two molecules of pentane. London dispersion forces are the weakest of our intermolecular forces."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about the intermolecular forces that exist between those two molecules of pentane. Pentane is a non-polar molecule, and we know the only intermolecular force that exists between two non-polar molecules, that would of course be the London dispersion forces. So London dispersion forces exist between these two molecules of pentane. London dispersion forces are the weakest of our intermolecular forces. They're attractions between molecules that only exist for a short period of time. So I could represent the London dispersion forces like this. So I'm showing the brief, the transient attractive forces between these two molecules of pentane."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "London dispersion forces are the weakest of our intermolecular forces. They're attractions between molecules that only exist for a short period of time. So I could represent the London dispersion forces like this. So I'm showing the brief, the transient attractive forces between these two molecules of pentane. If I draw on another molecule of hexane, so over here I'll draw on another one, hexane is a larger hydrocarbon with more surface area. And more surface area means we have more opportunity for London dispersion forces. So I can show even more attraction between these two molecules of hexane."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I'm showing the brief, the transient attractive forces between these two molecules of pentane. If I draw on another molecule of hexane, so over here I'll draw on another one, hexane is a larger hydrocarbon with more surface area. And more surface area means we have more opportunity for London dispersion forces. So I can show even more attraction between these two molecules of hexane. So the two molecules of hexane attract each other more than the two molecules of pentane. That increased attraction means it takes more energy for those molecules to pull apart from each other. More energy means an increased boiling point."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I can show even more attraction between these two molecules of hexane. So the two molecules of hexane attract each other more than the two molecules of pentane. That increased attraction means it takes more energy for those molecules to pull apart from each other. More energy means an increased boiling point. So hexane has a higher boiling point than pentane. So as you increase the number of carbons in your carbon chain, you get an increase in the boiling point of your compound. So this is an example comparing two molecules that have straight chains."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "More energy means an increased boiling point. So hexane has a higher boiling point than pentane. So as you increase the number of carbons in your carbon chain, you get an increase in the boiling point of your compound. So this is an example comparing two molecules that have straight chains. Let's compare a straight chain to a branched hydrocarbon. So on the left down here, once again we have pentane with a boiling point of 36 degrees C. Let's write down its molecular formula. We already know there are five carbons and if we count up our hydrogens, one, two, three, four, five, six, seven, eight, nine, 10, 11, and 12."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this is an example comparing two molecules that have straight chains. Let's compare a straight chain to a branched hydrocarbon. So on the left down here, once again we have pentane with a boiling point of 36 degrees C. Let's write down its molecular formula. We already know there are five carbons and if we count up our hydrogens, one, two, three, four, five, six, seven, eight, nine, 10, 11, and 12. So there are 12 hydrogens, so H12. C5H12 is the molecular formula for pentane. What about neopentane on the right?"}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We already know there are five carbons and if we count up our hydrogens, one, two, three, four, five, six, seven, eight, nine, 10, 11, and 12. So there are 12 hydrogens, so H12. C5H12 is the molecular formula for pentane. What about neopentane on the right? Well, there's one, two, three, four, five carbons, so five carbons, and one, two, three, four, five, six, seven, eight, nine, 10, 11, and 12 hydrogens. So C5H12. So these two compounds have the same molecular formula."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "What about neopentane on the right? Well, there's one, two, three, four, five carbons, so five carbons, and one, two, three, four, five, six, seven, eight, nine, 10, 11, and 12 hydrogens. So C5H12. So these two compounds have the same molecular formula. So the same molecular formula, C5H12. The difference is neopentane has some branching. So neopentane has branching, whereas pentane doesn't."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So these two compounds have the same molecular formula. So the same molecular formula, C5H12. The difference is neopentane has some branching. So neopentane has branching, whereas pentane doesn't. It's a straight chain. All right, let's think about the boiling points. Pentane's boiling point is 36 degrees C. Neopentane's drops down to 10 degrees C. Now let's try to figure out why."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So neopentane has branching, whereas pentane doesn't. It's a straight chain. All right, let's think about the boiling points. Pentane's boiling point is 36 degrees C. Neopentane's drops down to 10 degrees C. Now let's try to figure out why. If I draw in another molecule of pentane, we just talked about the fact that London dispersion forces exist between these two molecules of pentane. So let me draw in those transient attractive forces between those two molecules. Neopentane is also a hydrocarbon."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Pentane's boiling point is 36 degrees C. Neopentane's drops down to 10 degrees C. Now let's try to figure out why. If I draw in another molecule of pentane, we just talked about the fact that London dispersion forces exist between these two molecules of pentane. So let me draw in those transient attractive forces between those two molecules. Neopentane is also a hydrocarbon. It's nonpolar. So if I draw in another molecule of neopentane, and I think about the attractive forces between these two molecules of neopentane, it must once again be London dispersion forces. Because of this branching, the shape of neopentane in three dimensions resembles a sphere."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Neopentane is also a hydrocarbon. It's nonpolar. So if I draw in another molecule of neopentane, and I think about the attractive forces between these two molecules of neopentane, it must once again be London dispersion forces. Because of this branching, the shape of neopentane in three dimensions resembles a sphere. So it's just an approximation, but if you could imagine this molecule of neopentane on the left as being a sphere, so spherical, and just try to imagine this molecule of neopentane on the right as being roughly spherical. And if you think about the surface area for an attraction between these two molecules, it's a much smaller surface area than for the two molecules of pentane. We can kind of stack these two molecules of pentane on top of each other and get increased surface area and increased attractive forces."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Because of this branching, the shape of neopentane in three dimensions resembles a sphere. So it's just an approximation, but if you could imagine this molecule of neopentane on the left as being a sphere, so spherical, and just try to imagine this molecule of neopentane on the right as being roughly spherical. And if you think about the surface area for an attraction between these two molecules, it's a much smaller surface area than for the two molecules of pentane. We can kind of stack these two molecules of pentane on top of each other and get increased surface area and increased attractive forces. But these two neopentane molecules, because of their shape, because of this branching, we don't get as much surface area. And that means that there's decreased attractive forces between molecules of neopentane. And because there's decreased attractive forces, that lowers the boiling point."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We can kind of stack these two molecules of pentane on top of each other and get increased surface area and increased attractive forces. But these two neopentane molecules, because of their shape, because of this branching, we don't get as much surface area. And that means that there's decreased attractive forces between molecules of neopentane. And because there's decreased attractive forces, that lowers the boiling point. So the boiling point is down to 10 degrees C. I always think of room temperatures being pretty close to 25 degrees C. So most of the time you see it listed as being between 20 and 25. But if room temperature is pretty close to 25 degrees C, think about the state of matter of neopentane. We're already higher than the boiling point of neopentane."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And because there's decreased attractive forces, that lowers the boiling point. So the boiling point is down to 10 degrees C. I always think of room temperatures being pretty close to 25 degrees C. So most of the time you see it listed as being between 20 and 25. But if room temperature is pretty close to 25 degrees C, think about the state of matter of neopentane. We're already higher than the boiling point of neopentane. So at room temperature and room pressure, neopentane is a gas. The molecules have enough energy already to break free of each other. And so neopentane is a gas at room temperature and pressure."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We're already higher than the boiling point of neopentane. So at room temperature and room pressure, neopentane is a gas. The molecules have enough energy already to break free of each other. And so neopentane is a gas at room temperature and pressure. Whereas if you look at pentane, pentane has a boiling point of 36 degrees C, which is higher than room temperature. So we haven't reached the boiling point of pentane, which means at room temperature and pressure, pentane is still a liquid. So pentane is a liquid."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so neopentane is a gas at room temperature and pressure. Whereas if you look at pentane, pentane has a boiling point of 36 degrees C, which is higher than room temperature. So we haven't reached the boiling point of pentane, which means at room temperature and pressure, pentane is still a liquid. So pentane is a liquid. And let's think about the trend for branching here. So we have the same number of carbons, same number of carbons, same number of hydrogens, but we have different boiling points. Neopentane has more branching and a decreased boiling point."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So pentane is a liquid. And let's think about the trend for branching here. So we have the same number of carbons, same number of carbons, same number of hydrogens, but we have different boiling points. Neopentane has more branching and a decreased boiling point. So we can say for our trend here, as you increase the branching, so not talking about number of carbons here, we're just talking about branching. As you increase the branching, you decrease the boiling point because you decrease the surface area for the attractive forces. Let's compare three more molecules here to finish this off."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Neopentane has more branching and a decreased boiling point. So we can say for our trend here, as you increase the branching, so not talking about number of carbons here, we're just talking about branching. As you increase the branching, you decrease the boiling point because you decrease the surface area for the attractive forces. Let's compare three more molecules here to finish this off. Let's look at these three molecules. Let's see if we can explain these different boiling points. So once again, we've talked about hexane already with a boiling point of 69 degrees C. If we draw in another molecule of hexane, our only intermolecular force is of course the London dispersion forces."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's compare three more molecules here to finish this off. Let's look at these three molecules. Let's see if we can explain these different boiling points. So once again, we've talked about hexane already with a boiling point of 69 degrees C. If we draw in another molecule of hexane, our only intermolecular force is of course the London dispersion forces. So I'll just write London here. So London dispersion forces, which exist between these two nonpolar hexane molecules. Next let's look at 3-hexanone."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we've talked about hexane already with a boiling point of 69 degrees C. If we draw in another molecule of hexane, our only intermolecular force is of course the London dispersion forces. So I'll just write London here. So London dispersion forces, which exist between these two nonpolar hexane molecules. Next let's look at 3-hexanone. Hexane has six carbons and so does 3-hexanone. One, two, three, four, five, and six. So don't worry about the names of these molecules at this point if you're just getting started with organic chemistry."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next let's look at 3-hexanone. Hexane has six carbons and so does 3-hexanone. One, two, three, four, five, and six. So don't worry about the names of these molecules at this point if you're just getting started with organic chemistry. Just try to think about what intermolecular forces are present in this video. So 3-hexanone also has six carbons. And let me draw in another molecule of 3-hexanone."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So don't worry about the names of these molecules at this point if you're just getting started with organic chemistry. Just try to think about what intermolecular forces are present in this video. So 3-hexanone also has six carbons. And let me draw in another molecule of 3-hexanone. So there's our other molecule. Let's think about electronegativity. And we'll compare this oxygen to this carbon right here."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And let me draw in another molecule of 3-hexanone. So there's our other molecule. Let's think about electronegativity. And we'll compare this oxygen to this carbon right here. Oxygen is more electronegative than carbon, so oxygen withdraws some electron density and oxygen becomes partially negative. This carbon here, this carbon would therefore become partially positive. And so this is a dipole."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we'll compare this oxygen to this carbon right here. Oxygen is more electronegative than carbon, so oxygen withdraws some electron density and oxygen becomes partially negative. This carbon here, this carbon would therefore become partially positive. And so this is a dipole. So we have a dipole for this molecule and we have the same dipole for this molecule, 3-hexanone down here. Partially negative oxygen, partially positive carbon. And since opposites attract, the partially negative oxygen is attracted to the partially positive carbon on the other molecule of 3-hexanone."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so this is a dipole. So we have a dipole for this molecule and we have the same dipole for this molecule, 3-hexanone down here. Partially negative oxygen, partially positive carbon. And since opposites attract, the partially negative oxygen is attracted to the partially positive carbon on the other molecule of 3-hexanone. And so what intermolecular force is that? We have dipoles interacting with dipoles. So this would be a dipole-dipole interaction."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And since opposites attract, the partially negative oxygen is attracted to the partially positive carbon on the other molecule of 3-hexanone. And so what intermolecular force is that? We have dipoles interacting with dipoles. So this would be a dipole-dipole interaction. So let me write that down here. So we're talking about a dipole-dipole interaction. Obviously, London dispersion forces would also be present."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this would be a dipole-dipole interaction. So let me write that down here. So we're talking about a dipole-dipole interaction. Obviously, London dispersion forces would also be present. So if you think about this area over here, you could think about London dispersion forces. But dipole-dipole is a stronger intermolecular force compared to London dispersion forces. And therefore, the two molecules here of 3-hexanone are attracted to each other more than the two molecules of hexane."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Obviously, London dispersion forces would also be present. So if you think about this area over here, you could think about London dispersion forces. But dipole-dipole is a stronger intermolecular force compared to London dispersion forces. And therefore, the two molecules here of 3-hexanone are attracted to each other more than the two molecules of hexane. And so therefore, it would take more energy for these molecules to pull apart from each other. And that's why you see the higher temperature for the boiling point. 3-hexanone has a much higher boiling point than hexane."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, the two molecules here of 3-hexanone are attracted to each other more than the two molecules of hexane. And so therefore, it would take more energy for these molecules to pull apart from each other. And that's why you see the higher temperature for the boiling point. 3-hexanone has a much higher boiling point than hexane. And that's because dipole-dipole interactions are a stronger intermolecular force compared to London dispersion forces. And finally, we have 3-hexanol over here on the right, which also has six carbons. 1, 2, 3, 4, 5, 6."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "3-hexanone has a much higher boiling point than hexane. And that's because dipole-dipole interactions are a stronger intermolecular force compared to London dispersion forces. And finally, we have 3-hexanol over here on the right, which also has six carbons. 1, 2, 3, 4, 5, 6. So we're still dealing with six carbons. If I draw in another molecule of 3-hexanol, let me do that up here. So let me sketch in the six carbons."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6. So we're still dealing with six carbons. If I draw in another molecule of 3-hexanol, let me do that up here. So let me sketch in the six carbons. And then we have our oxygen here and then the hydrogen like that. We know that there's opportunity for hydrogen bonding. Oxygen is more electronegative than hydrogen."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me sketch in the six carbons. And then we have our oxygen here and then the hydrogen like that. We know that there's opportunity for hydrogen bonding. Oxygen is more electronegative than hydrogen. So the oxygen is partially negative and the hydrogen is partially positive. The same setup over here on this other molecule of 3-hexanol. So partially negative oxygen, partially positive hydrogen."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen is more electronegative than hydrogen. So the oxygen is partially negative and the hydrogen is partially positive. The same setup over here on this other molecule of 3-hexanol. So partially negative oxygen, partially positive hydrogen. And so hydrogen bonding is possible. Let me draw that in. So we have a hydrogen bond right here."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So partially negative oxygen, partially positive hydrogen. And so hydrogen bonding is possible. Let me draw that in. So we have a hydrogen bond right here. So there's opportunities for hydrogen bonding between two molecules of 3-hexanol. So let me use deep blue for that. So now we're talking about hydrogen bonding."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we have a hydrogen bond right here. So there's opportunities for hydrogen bonding between two molecules of 3-hexanol. So let me use deep blue for that. So now we're talking about hydrogen bonding. And we know that hydrogen bonding is really just a stronger type of dipole-dipole interaction. So hydrogen bonding is our strongest intermolecular force. And so we have an increased attractive force holding these two molecules of 3-hexanol together."}, {"video_title": "Boiling points of organic compounds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So now we're talking about hydrogen bonding. And we know that hydrogen bonding is really just a stronger type of dipole-dipole interaction. So hydrogen bonding is our strongest intermolecular force. And so we have an increased attractive force holding these two molecules of 3-hexanol together. And so therefore, it takes even more energy for these molecules to pull apart from each other. And that's reflected in the higher boiling point for 3-hexanol. 3-hexanol has a higher boiling point than 3-hexanone and also more than hexane."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If I were to draw a hand, let me just draw a hand really fast. So I'll draw a left hand. It looks something like that. That is a left hand. Now if I were to take its mirror image, so let's say that this is a mirror right there, and I want to take its mirror image. I'll draw the mirror image in green. So its mirror image would look something like this."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That is a left hand. Now if I were to take its mirror image, so let's say that this is a mirror right there, and I want to take its mirror image. I'll draw the mirror image in green. So its mirror image would look something like this. It looks something like this. Not exact, but you get the idea. The mirror image of a left hand looks a lot like a right hand."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So its mirror image would look something like this. It looks something like this. Not exact, but you get the idea. The mirror image of a left hand looks a lot like a right hand. Now, no matter how I try to shift or rotate this hand like this, I might try to maybe rotate it 180 degrees so that the thumb is on the other side, like this image right here. But no matter what I do, I will never be able to make this thing look like that thing. I can shift it and rotate it."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The mirror image of a left hand looks a lot like a right hand. Now, no matter how I try to shift or rotate this hand like this, I might try to maybe rotate it 180 degrees so that the thumb is on the other side, like this image right here. But no matter what I do, I will never be able to make this thing look like that thing. I can shift it and rotate it. It'll just never happen. I will never be able to superimpose the blue hand on top of this green hand. When I say superimpose, literally put it exactly on top of the green hand."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I can shift it and rotate it. It'll just never happen. I will never be able to superimpose the blue hand on top of this green hand. When I say superimpose, literally put it exactly on top of the green hand. So whenever something is not superimposable on its mirror image, let me write this down. So this is not superimposable on its mirror image. We call it chiral."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "When I say superimpose, literally put it exactly on top of the green hand. So whenever something is not superimposable on its mirror image, let me write this down. So this is not superimposable on its mirror image. We call it chiral. So this hand drawing right here is an example of a chiral object. Or I guess the hand is an example of a chiral object. This is not superimposable on its mirror image."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We call it chiral. So this hand drawing right here is an example of a chiral object. Or I guess the hand is an example of a chiral object. This is not superimposable on its mirror image. And it makes sense that it's called chiral because the word chiral comes from the Greek word for hand. Greek for hand. And this definition of not being able to be superimposable on its mirror image, this applies whether you're dealing with chemistry or mathematics or I guess just hands in general."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is not superimposable on its mirror image. And it makes sense that it's called chiral because the word chiral comes from the Greek word for hand. Greek for hand. And this definition of not being able to be superimposable on its mirror image, this applies whether you're dealing with chemistry or mathematics or I guess just hands in general. So if we extend this definition to chemistry, because that's what we're talking about, there's two concepts here. Chiral molecules. And then there are chiral centers or chiral atoms."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And this definition of not being able to be superimposable on its mirror image, this applies whether you're dealing with chemistry or mathematics or I guess just hands in general. So if we extend this definition to chemistry, because that's what we're talking about, there's two concepts here. Chiral molecules. And then there are chiral centers or chiral atoms. They tend to be carbon atoms, so sometimes they call them chiral carbons. So you have these chiral atoms. Now, chiral molecules are literally molecules that are not superimposable on their mirror image."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then there are chiral centers or chiral atoms. They tend to be carbon atoms, so sometimes they call them chiral carbons. So you have these chiral atoms. Now, chiral molecules are literally molecules that are not superimposable on their mirror image. I'm not going to write the whole thing. Not superimposable on mirror image. Now, for chiral atoms, this is essentially true, but when you look for chiral atoms within a molecule, the best way to kind of spot them is to recognize that these generally, or maybe I should say usually, are carbons, especially when we're dealing with organic chemistry, but they could be phosphorouses or sulfurs, but usually are carbons bonded to four different groups."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now, chiral molecules are literally molecules that are not superimposable on their mirror image. I'm not going to write the whole thing. Not superimposable on mirror image. Now, for chiral atoms, this is essentially true, but when you look for chiral atoms within a molecule, the best way to kind of spot them is to recognize that these generally, or maybe I should say usually, are carbons, especially when we're dealing with organic chemistry, but they could be phosphorouses or sulfurs, but usually are carbons bonded to four different groups. And I want to emphasize groups, not just four different atoms. But to kind of highlight a molecule that contains a chiral atom or a chiral carbon, we can just think of one. So let's say that I have a carbon right here, and I'm going to set this up so this is actually a chiral atom, that the carbon specific is a chiral atom, but it's part of a chiral molecule."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now, for chiral atoms, this is essentially true, but when you look for chiral atoms within a molecule, the best way to kind of spot them is to recognize that these generally, or maybe I should say usually, are carbons, especially when we're dealing with organic chemistry, but they could be phosphorouses or sulfurs, but usually are carbons bonded to four different groups. And I want to emphasize groups, not just four different atoms. But to kind of highlight a molecule that contains a chiral atom or a chiral carbon, we can just think of one. So let's say that I have a carbon right here, and I'm going to set this up so this is actually a chiral atom, that the carbon specific is a chiral atom, but it's part of a chiral molecule. And then we'll see examples that are one or both of these are true. So let's say it's bonded to a methyl group. And from that bond it kind of pops out of the page."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that I have a carbon right here, and I'm going to set this up so this is actually a chiral atom, that the carbon specific is a chiral atom, but it's part of a chiral molecule. And then we'll see examples that are one or both of these are true. So let's say it's bonded to a methyl group. And from that bond it kind of pops out of the page. Let's say there's a bromine over here. Let's say behind it there is a hydrogen, and then above it we have a fluorine. Now if I were to take the mirror image of this thing right here, we have your carbon in the center."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And from that bond it kind of pops out of the page. Let's say there's a bromine over here. Let's say behind it there is a hydrogen, and then above it we have a fluorine. Now if I were to take the mirror image of this thing right here, we have your carbon in the center. I want to do that same blue. You have the carbon in the center, and then you have the fluorine above the carbon. You have your bromine now going in this direction."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now if I were to take the mirror image of this thing right here, we have your carbon in the center. I want to do that same blue. You have the carbon in the center, and then you have the fluorine above the carbon. You have your bromine now going in this direction. You have this methyl group. It's still popping out of the page, but it's now going to the right instead of to the left. So CH3."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You have your bromine now going in this direction. You have this methyl group. It's still popping out of the page, but it's now going to the right instead of to the left. So CH3. And then you have the hydrogen still in the back. These are mirror images if you view this as kind of the mirror, and you can see on both sides of the mirror. Now why is this chiral?"}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So CH3. And then you have the hydrogen still in the back. These are mirror images if you view this as kind of the mirror, and you can see on both sides of the mirror. Now why is this chiral? Well, no matter how, and it's a little bit of a visualization challenge, but no matter how you try to rotate this thing right here, you will never make it exactly like this thing. You might try to rotate it. You might try to rotate it around like that and try to get the methyl group over here to get it over there."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now why is this chiral? Well, no matter how, and it's a little bit of a visualization challenge, but no matter how you try to rotate this thing right here, you will never make it exactly like this thing. You might try to rotate it. You might try to rotate it around like that and try to get the methyl group over here to get it over there. So let's try to do that. But if we try to get the methyl group over there, what's going to happen to the other groups? Well, then the hydrogen group is going, or the hydrogen, I should say the hydrogen atom, is going to move there, and the bromine is going to move there."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You might try to rotate it around like that and try to get the methyl group over here to get it over there. So let's try to do that. But if we try to get the methyl group over there, what's going to happen to the other groups? Well, then the hydrogen group is going, or the hydrogen, I should say the hydrogen atom, is going to move there, and the bromine is going to move there. But this has the bromine in. So this would be superimposable if this was a hydrogen and this was a bromine, but it's not. You can imagine the hydrogen and bromine are switched."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, then the hydrogen group is going, or the hydrogen, I should say the hydrogen atom, is going to move there, and the bromine is going to move there. But this has the bromine in. So this would be superimposable if this was a hydrogen and this was a bromine, but it's not. You can imagine the hydrogen and bromine are switched. And you could flip it and do whatever else you want or try to rotate it in any direction, but you're not going to be able to superimpose them. So this molecule right here is a chiral molecule, and this carbon is a chiral center. So this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can imagine the hydrogen and bromine are switched. And you could flip it and do whatever else you want or try to rotate it in any direction, but you're not going to be able to superimpose them. So this molecule right here is a chiral molecule, and this carbon is a chiral center. So this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center. Sometimes you'll hear something called a stereocenter. A stereocenter is a more general term for any point in a molecule that is asymmetric relative to the different groups that it is joined to. But all of these, especially when you're kind of in an introductory organic chemistry class, tends to be a carbon bonded to four different groups."}, {"video_title": "Introduction to chirality Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon is a chiral carbon, sometimes called an asymmetric carbon or a chiral center. Sometimes you'll hear something called a stereocenter. A stereocenter is a more general term for any point in a molecule that is asymmetric relative to the different groups that it is joined to. But all of these, especially when you're kind of in an introductory organic chemistry class, tends to be a carbon bonded to four different groups. And I want to stress that it's not four different atoms. You could have had a methyl group here and a propyl group here, and the carbon would still be bonded directly to a carbon in either case, but that would still be a chiral carbon. And this would still actually be a chiral molecule."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And if we look at this general reaction, this is either an aldehyde or a ketone over here. If we add lithium aluminum hydride in the first step and then a source of protons in the second step, which is water, we will form either a primary or a secondary alcohol depending on our starting material. So in that respect, lithium aluminum hydride will react in the same way as sodium borohydride. However, sodium borohydride will only reduce aldehydes or ketones. It won't reduce carboxylic acids or esters. And that's what lithium aluminum hydride does. So we can see that if this is an OH right here, that would be a carboxylic acid functional group."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "However, sodium borohydride will only reduce aldehydes or ketones. It won't reduce carboxylic acids or esters. And that's what lithium aluminum hydride does. So we can see that if this is an OH right here, that would be a carboxylic acid functional group. And if we take off the H and put an alkyl group in our prime group there, we would have an ester. So lithium aluminum hydride not only reduces aldehydes and ketones, it also reduces carboxylic acids and esters since it's more reactive, which is also why we have to separate these two. We can't have water in the same reaction vessel as our lithium aluminum hydride because it will react with that faster."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we can see that if this is an OH right here, that would be a carboxylic acid functional group. And if we take off the H and put an alkyl group in our prime group there, we would have an ester. So lithium aluminum hydride not only reduces aldehydes and ketones, it also reduces carboxylic acids and esters since it's more reactive, which is also why we have to separate these two. We can't have water in the same reaction vessel as our lithium aluminum hydride because it will react with that faster. And so once again, our product will depend on what our starting material is. So the mechanism for the reduction of aldehydes or ketones with lithium aluminum hydride is just like the one for sodium borohydride. So we'll move on to a mechanism for the reduction of an ester."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "We can't have water in the same reaction vessel as our lithium aluminum hydride because it will react with that faster. And so once again, our product will depend on what our starting material is. So the mechanism for the reduction of aldehydes or ketones with lithium aluminum hydride is just like the one for sodium borohydride. So we'll move on to a mechanism for the reduction of an ester. So let's go ahead and do that. So let's start with an ester down here. So we have our carbonyl like that."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we'll move on to a mechanism for the reduction of an ester. So let's go ahead and do that. So let's start with an ester down here. So we have our carbonyl like that. So we'll put in our lone pairs. And down here we have our R prime group like that. So there's our ester."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we have our carbonyl like that. So we'll put in our lone pairs. And down here we have our R prime group like that. So there's our ester. And we add lithium aluminum hydride in excess. So in terms of molar equivalents, let's go ahead and put lithium aluminum hydride down here, Li plus. And then we have Al bonded to four hydrogens like that."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So there's our ester. And we add lithium aluminum hydride in excess. So in terms of molar equivalents, let's go ahead and put lithium aluminum hydride down here, Li plus. And then we have Al bonded to four hydrogens like that. And that's going to give the aluminum a negative 1 formal charge like that. So the first step in the mechanism is just like the one we did in the previous video. We think about the carbonyl up here and the difference in electronegativity between this carbon and this oxygen, oxygen being more electronegative, pulling these electrons closer towards it in the double bond, giving it a partial negative charge."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have Al bonded to four hydrogens like that. And that's going to give the aluminum a negative 1 formal charge like that. So the first step in the mechanism is just like the one we did in the previous video. We think about the carbonyl up here and the difference in electronegativity between this carbon and this oxygen, oxygen being more electronegative, pulling these electrons closer towards it in the double bond, giving it a partial negative charge. Whereas the carbon down here is losing some electron density, becoming partially positive. So the carbonyl carbon is an electrophile. It wants electrons."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "We think about the carbonyl up here and the difference in electronegativity between this carbon and this oxygen, oxygen being more electronegative, pulling these electrons closer towards it in the double bond, giving it a partial negative charge. Whereas the carbon down here is losing some electron density, becoming partially positive. So the carbonyl carbon is an electrophile. It wants electrons. And of course, it's going to get electrons from these two electrons in here. So these two electrons are going to attack this carbon, so nucleophilic attack, and kick these electrons off onto our oxygen. So that's our first step, the nucleophilic attack portion."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "It wants electrons. And of course, it's going to get electrons from these two electrons in here. So these two electrons are going to attack this carbon, so nucleophilic attack, and kick these electrons off onto our oxygen. So that's our first step, the nucleophilic attack portion. So now we have R. And carbon used to have two bonds to carbon-oxygen. Now it has only one bond because those electrons moved off onto the oxygen, giving the oxygen a negative 1 formal charge like that. So we added on our hydrogen like that."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So that's our first step, the nucleophilic attack portion. So now we have R. And carbon used to have two bonds to carbon-oxygen. Now it has only one bond because those electrons moved off onto the oxygen, giving the oxygen a negative 1 formal charge like that. So we added on our hydrogen like that. And then we still have our O and our R prime group over here. So it will look like that. In the next step of this reaction, the carbonyl is going to reform."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we added on our hydrogen like that. And then we still have our O and our R prime group over here. So it will look like that. In the next step of this reaction, the carbonyl is going to reform. So the electrons in here are going to kick back into here to reform our carbonyl. That would mean five bonds to carbon, which we know never happens, so that these electrons are going to have to break and come off onto the oxygen. So let's go ahead and draw the result of that."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "In the next step of this reaction, the carbonyl is going to reform. So the electrons in here are going to kick back into here to reform our carbonyl. That would mean five bonds to carbon, which we know never happens, so that these electrons are going to have to break and come off onto the oxygen. So let's go ahead and draw the result of that. So now we have R bonded to carbon. Now we reformed our carbonyl. So now we have only two lone pairs of electrons on that oxygen now."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the result of that. So now we have R bonded to carbon. Now we reformed our carbonyl. So now we have only two lone pairs of electrons on that oxygen now. And then we still have a hydrogen right here. So we lost an oxygen with an R prime group. And that oxygen is going to have three lone pairs around it with a negative 1 formal charge."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So now we have only two lone pairs of electrons on that oxygen now. And then we still have a hydrogen right here. So we lost an oxygen with an R prime group. And that oxygen is going to have three lone pairs around it with a negative 1 formal charge. Oxygen being relatively electronegative, it can handle that formal charge fairly well and be relatively stable. So now we have an aldehyde. And we know that this reaction can occur again with an aldehyde."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And that oxygen is going to have three lone pairs around it with a negative 1 formal charge. Oxygen being relatively electronegative, it can handle that formal charge fairly well and be relatively stable. So now we have an aldehyde. And we know that this reaction can occur again with an aldehyde. So since you have extra lithium aluminum hydride floating around, what's going to happen is another reaction, so another reaction just like the one we just did. So we're going to have our lithium aluminum hydride again. So it's so reactive, you can't stop this reaction from occurring a second time."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And we know that this reaction can occur again with an aldehyde. So since you have extra lithium aluminum hydride floating around, what's going to happen is another reaction, so another reaction just like the one we just did. So we're going to have our lithium aluminum hydride again. So it's so reactive, you can't stop this reaction from occurring a second time. And same idea, exact same idea. Partial negative oxygen, partial positive carbon. The carbon wants electrons."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So it's so reactive, you can't stop this reaction from occurring a second time. And same idea, exact same idea. Partial negative oxygen, partial positive carbon. The carbon wants electrons. It's going to get these electrons right in here. So these electrons are going to attack this carbon, bring the hydrogen along with it. And then these electrons kick off onto the oxygen."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "The carbon wants electrons. It's going to get these electrons right in here. So these electrons are going to attack this carbon, bring the hydrogen along with it. And then these electrons kick off onto the oxygen. So let's go ahead and draw the product of that nucleophilic attack. Right now we have an R group here. And we have our carbon."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons kick off onto the oxygen. So let's go ahead and draw the product of that nucleophilic attack. Right now we have an R group here. And we have our carbon. Carbon bonded to an oxygen. Once again, used to have two lone pairs. Now it has three, giving a negative 1 formal charge."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And we have our carbon. Carbon bonded to an oxygen. Once again, used to have two lone pairs. Now it has three, giving a negative 1 formal charge. And this hydrogen over here on the right is going to stay here. And we added on another hydrogen like that. So in the second step of this reaction, we're going to add water."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "Now it has three, giving a negative 1 formal charge. And this hydrogen over here on the right is going to stay here. And we added on another hydrogen like that. So in the second step of this reaction, we're going to add water. So we go back up here and refresh our memory. In the second step of this reaction, you add water as a proton source. So let's go ahead and draw water floating around there."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So in the second step of this reaction, we're going to add water. So we go back up here and refresh our memory. In the second step of this reaction, you add water as a proton source. So let's go ahead and draw water floating around there. So here is our H2O molecule. And we're going to get an acid-base reaction. So we're going to get a lone pair of electrons attacking, grabbing that proton there, kicking these electrons off onto the oxygen."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw water floating around there. So here is our H2O molecule. And we're going to get an acid-base reaction. So we're going to get a lone pair of electrons attacking, grabbing that proton there, kicking these electrons off onto the oxygen. So we're going to protonate that alkoxide anion to form our alcohol, finally. So we're finally done. Now we have R. And then we have oxygen bonded to that hydrogen there."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to get a lone pair of electrons attacking, grabbing that proton there, kicking these electrons off onto the oxygen. So we're going to protonate that alkoxide anion to form our alcohol, finally. So we're finally done. Now we have R. And then we have oxygen bonded to that hydrogen there. And now only two lone pairs. The formal charge goes away. And you can see what we've done here."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "Now we have R. And then we have oxygen bonded to that hydrogen there. And now only two lone pairs. The formal charge goes away. And you can see what we've done here. We've actually added on two hydrogens to our original carbonyl carbon. So if we look at this is my original carbonyl carbon. And if I look at it over here, there's actually no hydrogen attached to it over here."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And you can see what we've done here. We've actually added on two hydrogens to our original carbonyl carbon. So if we look at this is my original carbonyl carbon. And if I look at it over here, there's actually no hydrogen attached to it over here. Both of those hydrogens came from our lithium aluminum hydride. So this one and this one. So the reaction happened twice."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And if I look at it over here, there's actually no hydrogen attached to it over here. Both of those hydrogens came from our lithium aluminum hydride. So this one and this one. So the reaction happened twice. So if you're doing this reaction with a carboxylic acid, it's a similar mechanism. We don't have time to go through it. But you're going to end up with the same product."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So the reaction happened twice. So if you're doing this reaction with a carboxylic acid, it's a similar mechanism. We don't have time to go through it. But you're going to end up with the same product. You're going to add on two hydrogens onto that original carbonyl carbon like that. So let's look at the chemoselectivity of this reaction. So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "But you're going to end up with the same product. You're going to add on two hydrogens onto that original carbonyl carbon like that. So let's look at the chemoselectivity of this reaction. So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use. So if I start out here with our reactant. So let's make it a benzene ring like that. And let's put stuff on the benzene ring."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So now that we've covered sodium borohydride and lithium aluminum hydride, let's see how you can choose which one of those reagents is the best to use. So if I start out here with our reactant. So let's make it a benzene ring like that. And let's put stuff on the benzene ring. So let's go ahead and put a double bond here. And then we'll make this an aldehyde functional group on one end. And then over here on this end, I'm going to put an ester like that."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And let's put stuff on the benzene ring. So let's go ahead and put a double bond here. And then we'll make this an aldehyde functional group on one end. And then over here on this end, I'm going to put an ester like that. And let's look at how we can selectively transform different parts of this molecule using different reagents. So let's say we were to do a reaction where we add on sodium borohydride. And then we'll add on a proton source in the second step."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And then over here on this end, I'm going to put an ester like that. And let's look at how we can selectively transform different parts of this molecule using different reagents. So let's say we were to do a reaction where we add on sodium borohydride. And then we'll add on a proton source in the second step. So we need to think about what's going to happen. Sodium borohydride is selective for aldehydes and ketones only. It will not reduce carboxylic acids or esters."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll add on a proton source in the second step. So we need to think about what's going to happen. Sodium borohydride is selective for aldehydes and ketones only. It will not reduce carboxylic acids or esters. So it's only going to react with the aldehyde at the top right portion of this molecule. So let's see if we can draw this in here. So it's going to react with the aldehyde in the top right portion."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "It will not reduce carboxylic acids or esters. So it's only going to react with the aldehyde at the top right portion of this molecule. So let's see if we can draw this in here. So it's going to react with the aldehyde in the top right portion. So we are still going to have our double bond here. And the aldehyde is going to go away to form a primary alcohol. So we're going to get a primary alcohol where the aldehyde used to be."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to react with the aldehyde in the top right portion. So we are still going to have our double bond here. And the aldehyde is going to go away to form a primary alcohol. So we're going to get a primary alcohol where the aldehyde used to be. Sodium borohydride has reduced that carbonyl. And in the second step, the protonation of the alkoxide forms your primary alcohol as your product. And the rest of the molecule is going to stay the same."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to get a primary alcohol where the aldehyde used to be. Sodium borohydride has reduced that carbonyl. And in the second step, the protonation of the alkoxide forms your primary alcohol as your product. And the rest of the molecule is going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And the rest of the molecule is going to stay the same. So this ester is going to remain untouched down here. So it's the chemoselectivity of that reaction. Let's say we start with the same starting material. And the first step, this time we add lithium aluminum hydride like that. And the second step, we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we start with the same starting material. And the first step, this time we add lithium aluminum hydride like that. And the second step, we add some water. Well, lithium aluminum hydride will reduce aldehydes and ketones. And it will also reduce esters. So lithium aluminum hydride in excess, so let's just assume this is added in excess here, it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "Well, lithium aluminum hydride will reduce aldehydes and ketones. And it will also reduce esters. So lithium aluminum hydride in excess, so let's just assume this is added in excess here, it's going to react with this aldehyde portion of the molecule. It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "It's also going to react with this ester portion of the molecule. So it's going to reduce both of those and form alcohol. So let's go ahead and try to draw the product here. So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride is going to form a primary alcohol as well like that. And then down here, what used to be our ester functional group, we're going to add two molar equivalents of hydrogen to that carbonyl. And we're going to end up breaking that bond between the carbonyl carbon and that oxygen."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So we have our benzene ring, which is untouched. And up here, we know that lithium aluminum hydride is going to form a primary alcohol as well like that. And then down here, what used to be our ester functional group, we're going to add two molar equivalents of hydrogen to that carbonyl. And we're going to end up breaking that bond between the carbonyl carbon and that oxygen. So all of this over here is going to go away as a leaving group in our mechanism. We're going to add on two hydrogens to that carbon. And then that's going to form our alcohol."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to end up breaking that bond between the carbonyl carbon and that oxygen. So all of this over here is going to go away as a leaving group in our mechanism. We're going to add on two hydrogens to that carbon. And then that's going to form our alcohol. So we're going to add two hydrogens onto that carbon, forming a primary alcohol down here as well, just like in the mechanism that we just discussed. So reduction of esters using lithium aluminum hydride. What about if we were to add hydrogen gas and palladium as our metal catalyst here?"}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And then that's going to form our alcohol. So we're going to add two hydrogens onto that carbon, forming a primary alcohol down here as well, just like in the mechanism that we just discussed. So reduction of esters using lithium aluminum hydride. What about if we were to add hydrogen gas and palladium as our metal catalyst here? Well, this is also a reduction reaction that we talked about earlier. Hydrogenation is an example of a reduction reaction. And it's going to be chemoselective."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "What about if we were to add hydrogen gas and palladium as our metal catalyst here? Well, this is also a reduction reaction that we talked about earlier. Hydrogenation is an example of a reduction reaction. And it's going to be chemoselective. If you just use the normal conditions for a hydrogenation reaction, the only thing the hydrogenation reaction is going to touch is this double bond. It's going to reduce this double bond. So let's go ahead and draw the product."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And it's going to be chemoselective. If you just use the normal conditions for a hydrogenation reaction, the only thing the hydrogenation reaction is going to touch is this double bond. It's going to reduce this double bond. So let's go ahead and draw the product. It's not going to touch the aldehyde. It's not going to touch the ester. And it's not going to touch the benzene ring."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the product. It's not going to touch the aldehyde. It's not going to touch the ester. And it's not going to touch the benzene ring. So let's go ahead and draw the product. The benzene ring is not hydrogenated under normal conditions. But we're going to add on two hydrogens across that double bond, and the aldehyde is untouched."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "And it's not going to touch the benzene ring. So let's go ahead and draw the product. The benzene ring is not hydrogenated under normal conditions. But we're going to add on two hydrogens across that double bond, and the aldehyde is untouched. And down here, the ester is going to be untouched as well. So that would be our product from a hydrogenation. So three different reductions, three different products."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "But we're going to add on two hydrogens across that double bond, and the aldehyde is untouched. And down here, the ester is going to be untouched as well. So that would be our product from a hydrogenation. So three different reductions, three different products. Now, hydrogen will reduce carbonyls under the right conditions. Usually, if you have increased pressure and increased temperatures and like that, you actually can reduce those carbonyls. But again, you can control those conditions."}, {"video_title": "Preparation of alcohols using LiAlH4 Organic chemistry Khan Academy.mp3", "Sentence": "So three different reductions, three different products. Now, hydrogen will reduce carbonyls under the right conditions. Usually, if you have increased pressure and increased temperatures and like that, you actually can reduce those carbonyls. But again, you can control those conditions. So you can control what part of the molecule is reduced. So that sums up the ways to reduce carbonyl compounds using sodium borohydride and lithium aluminum hydride. In the next video or two, we'll take a look at organometallics and particularly the Grignard reagent."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The number of detectable civilizations in the galaxy is equal to, and they'll have this. And this is not the number of stars in the galaxy. This is the average rate of star formation per year in the galaxy. So star, so let me write this down. Average rate of star formation, which seems kind of unintuitive, and frankly it is, but hopefully we'll reconcile to show you that this and what we're going to show with the traditional Drake equation are actually the same thing. So that's the average rate of star formation. So I don't know what it is."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So star, so let me write this down. Average rate of star formation, which seems kind of unintuitive, and frankly it is, but hopefully we'll reconcile to show you that this and what we're going to show with the traditional Drake equation are actually the same thing. So that's the average rate of star formation. So I don't know what it is. Maybe it's 10 stars a year or something on that order. And then the rest of it looks pretty similar. So times the fraction of stars that have planets."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I don't know what it is. Maybe it's 10 stars a year or something on that order. And then the rest of it looks pretty similar. So times the fraction of stars that have planets. So this product would give you the average stars with planet formations per year. You multiply that times the average number of planets capable of sustaining life for a star that has planets, for a solar system that has planets. So essentially if you multiply this, this is the average new planets per year in our galaxy capable of sustaining life."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So times the fraction of stars that have planets. So this product would give you the average stars with planet formations per year. You multiply that times the average number of planets capable of sustaining life for a star that has planets, for a solar system that has planets. So essentially if you multiply this, this is the average new planets per year in our galaxy capable of sustaining life. You multiply that times this, which is the same exact fraction, the fraction of those planets that are capable of sustaining life. Or actually, this is capable of sustaining life. Now we're saying that the fraction that actually do develop life."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So essentially if you multiply this, this is the average new planets per year in our galaxy capable of sustaining life. You multiply that times this, which is the same exact fraction, the fraction of those planets that are capable of sustaining life. Or actually, this is capable of sustaining life. Now we're saying that the fraction that actually do develop life. And then of the life, we care about the fraction that actually does become intelligent. And of the fraction that actually does become intelligent, we care about the fraction that eventually becomes detectable, that can actually communicate. And then in the traditional Drake equation, we multiply that times this L over here, times the detectable life of the civilization."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now we're saying that the fraction that actually do develop life. And then of the life, we care about the fraction that actually does become intelligent. And of the fraction that actually does become intelligent, we care about the fraction that eventually becomes detectable, that can actually communicate. And then in the traditional Drake equation, we multiply that times this L over here, times the detectable life of the civilization. So how long is that civilization detectable? Are they releasing radio waves or something like it that a civilization like ours can detect? Maybe there are other ways to communicate, and we're just not advanced enough."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then in the traditional Drake equation, we multiply that times this L over here, times the detectable life of the civilization. So how long is that civilization detectable? Are they releasing radio waves or something like it that a civilization like ours can detect? Maybe there are other ways to communicate, and we're just not advanced enough. Maybe in a few years we'll discover, in a few decades or a few hundreds of years, we'll discover that all of the other advanced civilizations are using a much more sophisticated way of communicating that doesn't involve electromagnetic waves. Who knows? But this is what we're thinking right now."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe there are other ways to communicate, and we're just not advanced enough. Maybe in a few years we'll discover, in a few decades or a few hundreds of years, we'll discover that all of the other advanced civilizations are using a much more sophisticated way of communicating that doesn't involve electromagnetic waves. Who knows? But this is what we're thinking right now. But anyway, the whole point here is to reconcile this thing, which is less intuitive, for me at least, than with this thing. Because I started up here with the total number of stars in the galaxy. The traditional Drake equation starts with the average rate of star formation."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is what we're thinking right now. But anyway, the whole point here is to reconcile this thing, which is less intuitive, for me at least, than with this thing. Because I started up here with the total number of stars in the galaxy. The traditional Drake equation starts with the average rate of star formation. So I was like, well, how does the average rate of star formation gel with the total number of stars or civilizations that are now detectable? What I want to do is diagram that out a little bit. And I'm going to make a few assumptions."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The traditional Drake equation starts with the average rate of star formation. So I was like, well, how does the average rate of star formation gel with the total number of stars or civilizations that are now detectable? What I want to do is diagram that out a little bit. And I'm going to make a few assumptions. I'm going to assume that this is kind of constant, that we're in a steady state. So this is constant, and we are in a steady state. The reality is that what would matter is the rate of star formation maybe 4, 5, 6 billion years ago."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm going to make a few assumptions. I'm going to assume that this is kind of constant, that we're in a steady state. So this is constant, and we are in a steady state. The reality is that what would matter is the rate of star formation maybe 4, 5, 6 billion years ago. I don't know how long it has to be ago. So that now it starts to become realistic for real intelligence and real detectable intelligence to exist. But let's just assume that this number is constant for most of the life of the galaxy."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The reality is that what would matter is the rate of star formation maybe 4, 5, 6 billion years ago. I don't know how long it has to be ago. So that now it starts to become realistic for real intelligence and real detectable intelligence to exist. But let's just assume that this number is constant for most of the life of the galaxy. Obviously, we're making all sorts of crazy assumptions here, so why not make another one? But what I want to show is that this is equivalent to the number of stars in the galaxy divided by the average life of a star or the average life of a solar system. And if n divided by this t sub s, if that's the same thing as r star, then essentially we have the same formulas."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's just assume that this number is constant for most of the life of the galaxy. Obviously, we're making all sorts of crazy assumptions here, so why not make another one? But what I want to show is that this is equivalent to the number of stars in the galaxy divided by the average life of a star or the average life of a solar system. And if n divided by this t sub s, if that's the same thing as r star, then essentially we have the same formulas. And to see that they're the same, imagine this, that this year, so this is this year. So this is 2000 and, well, let me say this year. This year, let's say that we have r star, let's say that this number is 10."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if n divided by this t sub s, if that's the same thing as r star, then essentially we have the same formulas. And to see that they're the same, imagine this, that this year, so this is this year. So this is 2000 and, well, let me say this year. This year, let's say that we have r star, let's say that this number is 10. We have 10 new stars in the galaxy. So this is, I'll just say it's 10. So r star is equal to 10."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This year, let's say that we have r star, let's say that this number is 10. We have 10 new stars in the galaxy. So this is, I'll just say it's 10. So r star is equal to 10. So this height over here is 10. That's what I'm depicting. So if I were to slide, I could show this is 10 units high or whatever."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So r star is equal to 10. So this height over here is 10. That's what I'm depicting. So if I were to slide, I could show this is 10 units high or whatever. And then last year, there was also 10, so on and so forth. Now let's go to whatever, let's say that this number right over here is 10 billion years. The average star life is 10 billion years."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if I were to slide, I could show this is 10 units high or whatever. And then last year, there was also 10, so on and so forth. Now let's go to whatever, let's say that this number right over here is 10 billion years. The average star life is 10 billion years. So let's go back 10 billion years into the past. So this, so the average life of a star is equal to 10 billion years. And we're assuming that this is constant."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The average star life is 10 billion years. So let's go back 10 billion years into the past. So this, so the average life of a star is equal to 10 billion years. And we're assuming that this is constant. So 10 billion years ago this year, there were also 10 new stars came about. And every year in between, every year in between, you had 10 stars come about. Now, how many total stars would there be in our galaxy?"}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we're assuming that this is constant. So 10 billion years ago this year, there were also 10 new stars came about. And every year in between, every year in between, you had 10 stars come about. Now, how many total stars would there be in our galaxy? Well, any star that came about, so we could go beyond that. We could go to stars that were born more than 10 billion years ago, more than this T sub s years ago. So you could have a star that was born 10 billion and 1 years ago, on average."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, how many total stars would there be in our galaxy? Well, any star that came about, so we could go beyond that. We could go to stars that were born more than 10 billion years ago, more than this T sub s years ago. So you could have a star that was born 10 billion and 1 years ago, on average. We're talking about on averages here. On average, that star will not exist anymore. So that star is not in existence."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you could have a star that was born 10 billion and 1 years ago, on average. We're talking about on averages here. On average, that star will not exist anymore. So that star is not in existence. The stars that are in existence, once again, on average, are the ones that were born 10 billion years ago, all the way to the ones that were born this year. So you have 10 billion years of star birth, the ones that are still around. Each year, there's 10 of those years."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that star is not in existence. The stars that are in existence, once again, on average, are the ones that were born 10 billion years ago, all the way to the ones that were born this year. So you have 10 billion years of star birth, the ones that are still around. Each year, there's 10 of those years. So the total number of stars should be equal to the number of stars that are born each year, assuming that that is constant, times the average lifespan of the stars. And if you look at it, and once again, this works because the stars that were born before this lifespan don't exist anymore. They've died out, on average."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Each year, there's 10 of those years. So the total number of stars should be equal to the number of stars that are born each year, assuming that that is constant, times the average lifespan of the stars. And if you look at it, and once again, this works because the stars that were born before this lifespan don't exist anymore. They've died out, on average. We care about this area right over here. 10 stars per year times 10 billion years. And now, if you manipulate this a little bit, you'll see that we'll get the result we want."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They've died out, on average. We care about this area right over here. 10 stars per year times 10 billion years. And now, if you manipulate this a little bit, you'll see that we'll get the result we want. Let's solve for r, so we can just divide both sides by this t. So you get n star. So the number of stars in our galaxy now, making a bunch of assumptions, divided by the average life of the stars, is equal to the average number of new stars per year. And we get our result."}, {"video_title": "Detectable civilizations in our galaxy 3 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now, if you manipulate this a little bit, you'll see that we'll get the result we want. Let's solve for r, so we can just divide both sides by this t. So you get n star. So the number of stars in our galaxy now, making a bunch of assumptions, divided by the average life of the stars, is equal to the average number of new stars per year. And we get our result. If you replace this with the total number of stars divided by t, you get the exact same result that we had before. You just change the order a bit. We can take this divided by t, put it under this n, take it out of here, and then you get the exact same thing."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And the difference is, on one level, kind of subtle, but it's actually a very big difference. And I'll show you why it's kind of confusing the first time you learn it. When we studied SN2 reactions, you have a nucleophile that has an extra electron right here. It has a negative charge. And maybe you have a methyl carbon. Let me draw it. Maybe you have a hydrogen coming out."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "It has a negative charge. And maybe you have a methyl carbon. Let me draw it. Maybe you have a hydrogen coming out. You have a hydrogen behind it. You have a hydrogen up top. And then you have a leaving group right over there."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Maybe you have a hydrogen coming out. You have a hydrogen behind it. You have a hydrogen up top. And then you have a leaving group right over there. And in an SN2 reaction, the nucleophile will give this electron to the carbon. The carbon has a partial positive charge. Let me draw that."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And then you have a leaving group right over there. And in an SN2 reaction, the nucleophile will give this electron to the carbon. The carbon has a partial positive charge. Let me draw that. The leaving group has a partial negative charge, because it tends to be, or it will be, more electronegative. And so this electron is given to this carbon. Right when the carbon gets that, or simultaneously with it, this electronegative leaving group is able to completely take this electron away from the carbon."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Let me draw that. The leaving group has a partial negative charge, because it tends to be, or it will be, more electronegative. And so this electron is given to this carbon. Right when the carbon gets that, or simultaneously with it, this electronegative leaving group is able to completely take this electron away from the carbon. And then after you are done, it looks like this. We have our methyl carbon. So the hydrogen in the back, hydrogen in the front, hydrogen on top."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Right when the carbon gets that, or simultaneously with it, this electronegative leaving group is able to completely take this electron away from the carbon. And then after you are done, it looks like this. We have our methyl carbon. So the hydrogen in the back, hydrogen in the front, hydrogen on top. The leaving group has left. It had this electron right there, but now it also took that magenta electron. So it now has a negative charge."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So the hydrogen in the back, hydrogen in the front, hydrogen on top. The leaving group has left. It had this electron right there, but now it also took that magenta electron. So it now has a negative charge. And the nucleophile has given this electron right over here. And so now it is bonded to the carbon. Now the whole reason I did this is because, OK, this is acting as a nucleophile."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So it now has a negative charge. And the nucleophile has given this electron right over here. And so now it is bonded to the carbon. Now the whole reason I did this is because, OK, this is acting as a nucleophile. It loves nucleuses. It's giving away its extra electron. But it is also acting as a Lewis base."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Now the whole reason I did this is because, OK, this is acting as a nucleophile. It loves nucleuses. It's giving away its extra electron. But it is also acting as a Lewis base. And as a bit of a refresher, a Lewis base, which is really the most general, or I guess it covers the most examples of what it means to be a base. A Lewis base means that you are an electron donor. And that's exactly what's happening here."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "But it is also acting as a Lewis base. And as a bit of a refresher, a Lewis base, which is really the most general, or I guess it covers the most examples of what it means to be a base. A Lewis base means that you are an electron donor. And that's exactly what's happening here. This nucleophile is donating an electron to the carbon. So it's acting as a Lewis base. So the first time you see that, you're like, well, why did chemists even go through the pain of defining something like a nucleophile?"}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And that's exactly what's happening here. This nucleophile is donating an electron to the carbon. So it's acting as a Lewis base. So the first time you see that, you're like, well, why did chemists even go through the pain of defining something like a nucleophile? Why don't they just call it a base? Why are there two different concepts of nucleophilicity and basicity? And the difference is that nucleophilicity is a kinetic concept, which means how good is it at reacting?"}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So the first time you see that, you're like, well, why did chemists even go through the pain of defining something like a nucleophile? Why don't they just call it a base? Why are there two different concepts of nucleophilicity and basicity? And the difference is that nucleophilicity is a kinetic concept, which means how good is it at reacting? How fast is it reacting? How little extra energy does it need to react? So if something has good nucleophilicity, it is good at reacting."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And the difference is that nucleophilicity is a kinetic concept, which means how good is it at reacting? How fast is it reacting? How little extra energy does it need to react? So if something has good nucleophilicity, it is good at reacting. It doesn't tell you anything about how stable or unstable the reactants before and after are. It just tells you they're good at reacting with each other. Basicity is a thermodynamic concept."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So if something has good nucleophilicity, it is good at reacting. It doesn't tell you anything about how stable or unstable the reactants before and after are. It just tells you they're good at reacting with each other. Basicity is a thermodynamic concept. It's telling you how stable are the reactants or the products. So it tells you how badly something would like to react. For example, we saw the situation of fluorine."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Basicity is a thermodynamic concept. It's telling you how stable are the reactants or the products. So it tells you how badly something would like to react. For example, we saw the situation of fluorine. We saw the situation where actually I say fluoride. So fluoride looks like this. 7 valence electrons for fluorine."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "For example, we saw the situation of fluorine. We saw the situation where actually I say fluoride. So fluoride looks like this. 7 valence electrons for fluorine. And then you have it swiped 1 extra electron away. You get fluoride. So fluoride is reasonably basic."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "7 valence electrons for fluorine. And then you have it swiped 1 extra electron away. You get fluoride. So fluoride is reasonably basic. It is more basic than iodide. So this is more basic than iodide. But in a protic solution, but less nucleophilic in protic solution."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So fluoride is reasonably basic. It is more basic than iodide. So this is more basic than iodide. But in a protic solution, but less nucleophilic in protic solution. And a protic solution, once again, has hydrogen protons around. And the reason why this is, is fluoride, it wants to bond with, let's say a carbon or something else, more badly, or maybe even a hydrogen proton. It wants to bond with it more badly than an iodide anion."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "But in a protic solution, but less nucleophilic in protic solution. And a protic solution, once again, has hydrogen protons around. And the reason why this is, is fluoride, it wants to bond with, let's say a carbon or something else, more badly, or maybe even a hydrogen proton. It wants to bond with it more badly than an iodide anion. And if it did, it actually will be a stronger bond than the iodide anion will form. The fluoride anion is actually less stable in this form than the iodide is. If it were to be able to get a proton or give its electron away, it will be happier."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "It wants to bond with it more badly than an iodide anion. And if it did, it actually will be a stronger bond than the iodide anion will form. The fluoride anion is actually less stable in this form than the iodide is. If it were to be able to get a proton or give its electron away, it will be happier. But it's less nucleophilic. It's less good at reacting in a protic solution. And the whole reason it's less nucleophilic is because there are other things that are keeping from reacting."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "If it were to be able to get a proton or give its electron away, it will be happier. But it's less nucleophilic. It's less good at reacting in a protic solution. And the whole reason it's less nucleophilic is because there are other things that are keeping from reacting. And we saw in the video on what makes a good nucleophile that in the case of fluoride, it's because it forms hydrogen. It's a very small atom. Actually, I should say a very small ion."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And the whole reason it's less nucleophilic is because there are other things that are keeping from reacting. And we saw in the video on what makes a good nucleophile that in the case of fluoride, it's because it forms hydrogen. It's a very small atom. Actually, I should say a very small ion. And so it's very closely held. The electron claw is very tight. And so what it allows is the hydrogens from the water to form a very tight shell around."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Actually, I should say a very small ion. And so it's very closely held. The electron claw is very tight. And so what it allows is the hydrogens from the water to form a very tight shell around. These all have partial positive charges. So they are attracted to the negative anion. And so they form a very tight shell, kind of protecting the fluoride anion, which makes it harder for it to react in a protic solution."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And so what it allows is the hydrogens from the water to form a very tight shell around. These all have partial positive charges. So they are attracted to the negative anion. And so they form a very tight shell, kind of protecting the fluoride anion, which makes it harder for it to react in a protic solution. So it doesn't react as well. If it was able to react, it actually will form a stronger bond than the iodide anion. So that's the big difference."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "And so they form a very tight shell, kind of protecting the fluoride anion, which makes it harder for it to react in a protic solution. So it doesn't react as well. If it was able to react, it actually will form a stronger bond than the iodide anion. So that's the big difference. And just so we see the difference in trends. So basicity, it does not matter what your actual solvent is. It is a thermodynamic property of the molecule or the atom or of the anion."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So that's the big difference. And just so we see the difference in trends. So basicity, it does not matter what your actual solvent is. It is a thermodynamic property of the molecule or the atom or of the anion. So if you looked at just pure basicity, the strongest base you see, and I'll just write hydroxide here, it's normally something like sodium hydroxide or potassium hydroxide, but when you dissolve it in, say, something like water, the sodium and the hydroxide separate, and it's really the hydroxide that is acting as a base, something that wants to donate electrons. So hydroxide is a much stronger base than fluoride, which is a stronger base than chloride, which is a stronger base than bromide, which is a stronger base than iodide. Now, if you were to look at nucleophilicity, just to see the difference, we saw that it actually matters what's the solvent, because the solvent will affect how good something is at reacting."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "It is a thermodynamic property of the molecule or the atom or of the anion. So if you looked at just pure basicity, the strongest base you see, and I'll just write hydroxide here, it's normally something like sodium hydroxide or potassium hydroxide, but when you dissolve it in, say, something like water, the sodium and the hydroxide separate, and it's really the hydroxide that is acting as a base, something that wants to donate electrons. So hydroxide is a much stronger base than fluoride, which is a stronger base than chloride, which is a stronger base than bromide, which is a stronger base than iodide. Now, if you were to look at nucleophilicity, just to see the difference, we saw that it actually matters what's the solvent, because the solvent will affect how good something is at reacting. So nucleophilicity, there's a difference between a protic solvent and an aprotic solvent. In a protic solvent, the thing that has the best nucleophilicity is actually iodide, because it's not hindered by these hydrogen bonds as much. It doesn't have a tight shell."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Now, if you were to look at nucleophilicity, just to see the difference, we saw that it actually matters what's the solvent, because the solvent will affect how good something is at reacting. So nucleophilicity, there's a difference between a protic solvent and an aprotic solvent. In a protic solvent, the thing that has the best nucleophilicity is actually iodide, because it's not hindered by these hydrogen bonds as much. It doesn't have a tight shell. It has this big molecular cloud, and some people think it also has kind of a softness. It has this polarizability where that cloud can be pulled towards the carbon and do what it needs to do. So in this case, iodide is a better nucleophile."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "It doesn't have a tight shell. It has this big molecular cloud, and some people think it also has kind of a softness. It has this polarizability where that cloud can be pulled towards the carbon and do what it needs to do. So in this case, iodide is a better nucleophile. Let me just say, then, hydroxide, which is a better nucleophile than fluoride. Now, in an aprotic solution, where all of a sudden the interactions with the solvent are not going to be as significant, then things change. In this situation, basicity matters."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So in this case, iodide is a better nucleophile. Let me just say, then, hydroxide, which is a better nucleophile than fluoride. Now, in an aprotic solution, where all of a sudden the interactions with the solvent are not going to be as significant, then things change. In this situation, basicity matters. So in an aprotic solution, basicity and nucleophilicity correlate. And I'll put an asterisk here, because there's also one other aspect of nucleophilicity that I haven't talked about yet, but I'll talk about it in a second. But in this type of a situation, hydroxide will be better at reacting than fluoride, which will be better at reacting than iodide."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "In this situation, basicity matters. So in an aprotic solution, basicity and nucleophilicity correlate. And I'll put an asterisk here, because there's also one other aspect of nucleophilicity that I haven't talked about yet, but I'll talk about it in a second. But in this type of a situation, hydroxide will be better at reacting than fluoride, which will be better at reacting than iodide. And the whole reason why in both situations, even where it can interact with the solvent, it's still a pretty good nucleophile. Because if you think about hydroxide, and I have to think about this a lot, it has an extra electron. If you think about it, you can imagine it's water that took away, let me draw it this way, you can imagine it's water where a proton left or where an electron was taken from a proton."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "But in this type of a situation, hydroxide will be better at reacting than fluoride, which will be better at reacting than iodide. And the whole reason why in both situations, even where it can interact with the solvent, it's still a pretty good nucleophile. Because if you think about hydroxide, and I have to think about this a lot, it has an extra electron. If you think about it, you can imagine it's water that took away, let me draw it this way, you can imagine it's water where a proton left or where an electron was taken from a proton. So normally you have two pairs, and now you have a third pair right here. This oxygen has 1, 2, 3, 4, 5, 6, 7 valence electrons, 1 more than neutral oxygen, so it has a negative charge. So it already has an extra electron that gives this negative charge."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "If you think about it, you can imagine it's water that took away, let me draw it this way, you can imagine it's water where a proton left or where an electron was taken from a proton. So normally you have two pairs, and now you have a third pair right here. This oxygen has 1, 2, 3, 4, 5, 6, 7 valence electrons, 1 more than neutral oxygen, so it has a negative charge. So it already has an extra electron that gives this negative charge. But oxygen is also more electronegative than hydrogen, so it's also able to get this guy involved a little bit anyway, so it's a very basic molecule. So even when it might be interfered a little bit by a protic environment, like water, it's still a better nucleophile than something like fluoride. If you take the solvent out of the picture, it's a super strong base, and so it's also going to be a very, very good nucleophile."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So it already has an extra electron that gives this negative charge. But oxygen is also more electronegative than hydrogen, so it's also able to get this guy involved a little bit anyway, so it's a very basic molecule. So even when it might be interfered a little bit by a protic environment, like water, it's still a better nucleophile than something like fluoride. If you take the solvent out of the picture, it's a super strong base, and so it's also going to be a very, very good nucleophile. Now the last aspect of nucleophilicity, remember nucleophilicity is how good something reacts. Now let's imagine we have something here. So we have two hydroxide molecules."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "If you take the solvent out of the picture, it's a super strong base, and so it's also going to be a very, very good nucleophile. Now the last aspect of nucleophilicity, remember nucleophilicity is how good something reacts. Now let's imagine we have something here. So we have two hydroxide molecules. And let's just say this one is just a straight up hydroxide. And let's say this one over here has all sorts of things off of it. Let's say it just has this big chain of stuff."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So we have two hydroxide molecules. And let's just say this one is just a straight up hydroxide. And let's say this one over here has all sorts of things off of it. Let's say it just has this big chain of stuff. I don't know which one. Now if you were to look at these two molecules, if you were to try to guess which one is going to be a better nucleophile, you should just remember, nucleophilicity is how good something reacts. How good is it getting in there and making a reaction happen?"}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "Let's say it just has this big chain of stuff. I don't know which one. Now if you were to look at these two molecules, if you were to try to guess which one is going to be a better nucleophile, you should just remember, nucleophilicity is how good something reacts. How good is it getting in there and making a reaction happen? Now this thing has this big molecule all around it. It might actually make it very hard, if you go back to this circumstance up here, it might make it very hard for it to get in there. We've talked about steric hindrance from the point of view of the carbon, but we haven't really talked about it from the point of view of the nucleophile."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "How good is it getting in there and making a reaction happen? Now this thing has this big molecule all around it. It might actually make it very hard, if you go back to this circumstance up here, it might make it very hard for it to get in there. We've talked about steric hindrance from the point of view of the carbon, but we haven't really talked about it from the point of view of the nucleophile. In this nucleophile right here, it might be hard for this extra electron right here to actually get to the target nucleus. It will be hindered. While in this situation, it will be much easier."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "We've talked about steric hindrance from the point of view of the carbon, but we haven't really talked about it from the point of view of the nucleophile. In this nucleophile right here, it might be hard for this extra electron right here to actually get to the target nucleus. It will be hindered. While in this situation, it will be much easier. Even though the group that's reacting, this oxygen that has a negative charge, this extra electron, is on some level fairly, fairly equivalent. But this one right here will be much more, it's a much smaller molecule, it'll be less hindered, easier to get in. So this will be a better nucleophile."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "While in this situation, it will be much easier. Even though the group that's reacting, this oxygen that has a negative charge, this extra electron, is on some level fairly, fairly equivalent. But this one right here will be much more, it's a much smaller molecule, it'll be less hindered, easier to get in. So this will be a better nucleophile. And that's why I didn't want to make the strong statement that in an aprotic solution, basicity and nucleophilicity are completely correlated, because nucleophilicity still has that other element of how hindered is it? Is it in an environment or is it part of a molecule that will keep it from reacting? Even though it might be a very strong base, if it actually forms a bond, it'll be very strong."}, {"video_title": "Nucleophilicity vs. Basicity.mp3", "Sentence": "So this will be a better nucleophile. And that's why I didn't want to make the strong statement that in an aprotic solution, basicity and nucleophilicity are completely correlated, because nucleophilicity still has that other element of how hindered is it? Is it in an environment or is it part of a molecule that will keep it from reacting? Even though it might be a very strong base, if it actually forms a bond, it'll be very strong. So the big thing to remember is that they're just two fundamentally different concepts, and that's why they're two different terms for them. Nucleophilicity, how good is it at reacting, saying nothing about how good the resulting bond is. Basicity is how good is the bond?"}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "When you get a spectroscopy problem on a test, you're often given the molecular formula and asked to determine the structure of the molecule. And sometimes the hydrogen deficiency index can be useful for figuring out the structure of your molecule. Let's start off by looking at hexane. So here we have hexane with a molecular formula of C6H14. We say that hexane is completely saturated with hydrogen, so it has the maximum number of hydrogen atoms possible for the number of carbons. So if we have six carbons, 14 is the maximum number of hydrogens that we can have. And remember, when you're talking about alkanes and you have N carbons, you get 2N plus two hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here we have hexane with a molecular formula of C6H14. We say that hexane is completely saturated with hydrogen, so it has the maximum number of hydrogen atoms possible for the number of carbons. So if we have six carbons, 14 is the maximum number of hydrogens that we can have. And remember, when you're talking about alkanes and you have N carbons, you get 2N plus two hydrogens. So here N is equal to six, so we have six carbons. So two times six plus two gives us 14 hydrogens. Let's compare hexane with 1-hexene."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And remember, when you're talking about alkanes and you have N carbons, you get 2N plus two hydrogens. So here N is equal to six, so we have six carbons. So two times six plus two gives us 14 hydrogens. Let's compare hexane with 1-hexene. So now we have a double bond present, and that changes the molecular formula, of course. So now the molecular formula is C6H12. We have the same number of carbons, six, but we have 12 hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's compare hexane with 1-hexene. So now we have a double bond present, and that changes the molecular formula, of course. So now the molecular formula is C6H12. We have the same number of carbons, six, but we have 12 hydrogens. And up here we had 14, so we're missing two hydrogens. And so we say that's one degree of unsaturation, or a hydrogen deficiency index equal to one. So an HDI equal to one."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have the same number of carbons, six, but we have 12 hydrogens. And up here we had 14, so we're missing two hydrogens. And so we say that's one degree of unsaturation, or a hydrogen deficiency index equal to one. So an HDI equal to one. So it's like we're missing one pair of hydrogens here. Let's look at cyclohexane. So here we have cyclohexane, and the molecular formula is C6H12."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So an HDI equal to one. So it's like we're missing one pair of hydrogens here. Let's look at cyclohexane. So here we have cyclohexane, and the molecular formula is C6H12. Once again, we have six carbons, but we have only 12 hydrogens. So we're missing two compared to hexane. We're missing two hydrogens, we're missing one pair of hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here we have cyclohexane, and the molecular formula is C6H12. Once again, we have six carbons, but we have only 12 hydrogens. So we're missing two compared to hexane. We're missing two hydrogens, we're missing one pair of hydrogens. So the hydrogen deficiency index, the HDI, is equal to one. So if you were given this molecular formula, C6H12, right here it's the same molecular formula, and you think about how many hydrogens you're missing from the maximum number, you're missing two hydrogens. That's a hydrogen deficiency index equal to one."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're missing two hydrogens, we're missing one pair of hydrogens. So the hydrogen deficiency index, the HDI, is equal to one. So if you were given this molecular formula, C6H12, right here it's the same molecular formula, and you think about how many hydrogens you're missing from the maximum number, you're missing two hydrogens. That's a hydrogen deficiency index equal to one. Here we had one double bond, and here we had one ring. So with an HDI equal to one, one possibility is you have one ring, another possibility is you have one double bond. All right, let's look at benzene."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's a hydrogen deficiency index equal to one. Here we had one double bond, and here we had one ring. So with an HDI equal to one, one possibility is you have one ring, another possibility is you have one double bond. All right, let's look at benzene. So here we have benzene, a molecular formula of C6H6. So with six carbons, the maximum number of hydrogens you can have is 14, and here we have six. So 14 minus six is equal to eight."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at benzene. So here we have benzene, a molecular formula of C6H6. So with six carbons, the maximum number of hydrogens you can have is 14, and here we have six. So 14 minus six is equal to eight. So it's like we're missing eight hydrogens. So that's four pairs of hydrogens. So the HDI is equal to four."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 14 minus six is equal to eight. So it's like we're missing eight hydrogens. So that's four pairs of hydrogens. So the HDI is equal to four. So the hydrogen deficiency index is equal to four. And I like to think about why the HDI is equal to four, because, right, if this is four, we have three double bonds here. One, two, three, and we have one ring."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the HDI is equal to four. So the hydrogen deficiency index is equal to four. And I like to think about why the HDI is equal to four, because, right, if this is four, we have three double bonds here. One, two, three, and we have one ring. So we have one ring here. So three plus one is equal to four, and we get an HDI equal to four. Up here we had one ring, and we had an HDI equal to one."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, and we have one ring. So we have one ring here. So three plus one is equal to four, and we get an HDI equal to four. Up here we had one ring, and we had an HDI equal to one. Up here we had one double bond, and an HDI equal to one. We could also calculate the HDI using a formula. So we could write that the hydrogen deficiency index is equal to 1 1 2 times what's in the parentheses, which is two times number of carbons plus two."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Up here we had one ring, and we had an HDI equal to one. Up here we had one double bond, and an HDI equal to one. We could also calculate the HDI using a formula. So we could write that the hydrogen deficiency index is equal to 1 1 2 times what's in the parentheses, which is two times number of carbons plus two. That's the whole two N plus two idea, minus the number of hydrogens. So let's go ahead and do it for this molecular formula with six carbons and six hydrogens. So let's plug in those numbers."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we could write that the hydrogen deficiency index is equal to 1 1 2 times what's in the parentheses, which is two times number of carbons plus two. That's the whole two N plus two idea, minus the number of hydrogens. So let's go ahead and do it for this molecular formula with six carbons and six hydrogens. So let's plug in those numbers. So we have 1 1 2. This would be two times number of carbons, that's two times six, plus two minus number of hydrogens, that's six. So let's do the math."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's plug in those numbers. So we have 1 1 2. This would be two times number of carbons, that's two times six, plus two minus number of hydrogens, that's six. So let's do the math. This would be 1 1 2. Two times six plus two is 14, minus six. 14 minus six is eight, times 1 1 2."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's do the math. This would be 1 1 2. Two times six plus two is 14, minus six. 14 minus six is eight, times 1 1 2. We get an HDI equal to four. So if you have a spectroscopy problem, and you calculate the HDI equal to four, think about the possibility of a benzene ring being present in the structure of your molecule. Let's compare benzene to another molecule."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "14 minus six is eight, times 1 1 2. We get an HDI equal to four. So if you have a spectroscopy problem, and you calculate the HDI equal to four, think about the possibility of a benzene ring being present in the structure of your molecule. Let's compare benzene to another molecule. We're adding in a nitrogen. So here's benzene, and we know there are six hydrogens. One, two, three, four, five, six."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's compare benzene to another molecule. We're adding in a nitrogen. So here's benzene, and we know there are six hydrogens. One, two, three, four, five, six. So six hydrogens. Let's add in a nitrogen. So we are adding in a nitrogen."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six. So six hydrogens. Let's add in a nitrogen. So we are adding in a nitrogen. So now we have aniline as our compound here. How many hydrogens? So this would be one, two, three, four, five, six, seven."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we are adding in a nitrogen. So now we have aniline as our compound here. How many hydrogens? So this would be one, two, three, four, five, six, seven. So we have seven hydrogens here. So we've added one nitrogen, and we get one more hydrogen. So we need to modify the formula for HDI."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this would be one, two, three, four, five, six, seven. So we have seven hydrogens here. So we've added one nitrogen, and we get one more hydrogen. So we need to modify the formula for HDI. So let's go ahead and do that. So we would write the hydrogen deficiency index is equal to 1 1 2 times what's in our parentheses, which would be two times the number of carbons, plus two. And now we're going to add the number of nitrogens that we have."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we need to modify the formula for HDI. So let's go ahead and do that. So we would write the hydrogen deficiency index is equal to 1 1 2 times what's in our parentheses, which would be two times the number of carbons, plus two. And now we're going to add the number of nitrogens that we have. So add the number of nitrogens, and then subtract the number of hydrogens. So let's go ahead and do this calculation. So 1 1 2 times, once again we have six carbons."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And now we're going to add the number of nitrogens that we have. So add the number of nitrogens, and then subtract the number of hydrogens. So let's go ahead and do this calculation. So 1 1 2 times, once again we have six carbons. One, two, three, four, five, and six. So two times six, plus two, plus the number of nitrogens. We have one nitrogen, so we add a one in here, minus the number of hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 1 1 2 times, once again we have six carbons. One, two, three, four, five, and six. So two times six, plus two, plus the number of nitrogens. We have one nitrogen, so we add a one in here, minus the number of hydrogens. We said there were seven, so minus seven. And let's do this. This would be 1 1 2 times, this would be 14 plus one is 15."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have one nitrogen, so we add a one in here, minus the number of hydrogens. We said there were seven, so minus seven. And let's do this. This would be 1 1 2 times, this would be 14 plus one is 15. So 15 minus seven is equal to eight, times 1 1 2 is equal to four. So an HDI of four, remember an HDI of four, think about a benzene ring being present in the structure of your molecule. All right, let's do another one."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This would be 1 1 2 times, this would be 14 plus one is 15. So 15 minus seven is equal to eight, times 1 1 2 is equal to four. So an HDI of four, remember an HDI of four, think about a benzene ring being present in the structure of your molecule. All right, let's do another one. Let's compare benzene to chlorobenzene. So over here on the left, we have benzene. Once again, six hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's do another one. Let's compare benzene to chlorobenzene. So over here on the left, we have benzene. Once again, six hydrogens. And here we have chlorobenzene. So we've added in a halogen. And how many hydrogens do we have in chlorobenzene?"}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Once again, six hydrogens. And here we have chlorobenzene. So we've added in a halogen. And how many hydrogens do we have in chlorobenzene? One, two, three, four, five. So only five hydrogens this time. So we've added a halogen, and we've lost a hydrogen."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And how many hydrogens do we have in chlorobenzene? One, two, three, four, five. So only five hydrogens this time. So we've added a halogen, and we've lost a hydrogen. We went from six to five. So we need to modify our formula again. So HDI is equal to 1 1 2 times, what's in the parentheses, two times the number of carbons, plus two, plus the number of nitrogens, so plus number of nitrogens, minus the number of hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we've added a halogen, and we've lost a hydrogen. We went from six to five. So we need to modify our formula again. So HDI is equal to 1 1 2 times, what's in the parentheses, two times the number of carbons, plus two, plus the number of nitrogens, so plus number of nitrogens, minus the number of hydrogens. And we would also need to subtract the number of halogens. So minus the number of halogens that you have. All right, let's do it for chlorobenzene."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So HDI is equal to 1 1 2 times, what's in the parentheses, two times the number of carbons, plus two, plus the number of nitrogens, so plus number of nitrogens, minus the number of hydrogens. And we would also need to subtract the number of halogens. So minus the number of halogens that you have. All right, let's do it for chlorobenzene. So 1 1 2 times, how many carbons do we have? Still six. One, two, three, four, five, and six."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's do it for chlorobenzene. So 1 1 2 times, how many carbons do we have? Still six. One, two, three, four, five, and six. So two times six, plus two, zero nitrogens. So zero nitrogens, how many hydrogens? Five for chlorobenzene, so minus five."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, and six. So two times six, plus two, zero nitrogens. So zero nitrogens, how many hydrogens? Five for chlorobenzene, so minus five. And we have one halogen. We have one halogen here, it's the chlorine, so minus one. So let's do that math."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Five for chlorobenzene, so minus five. And we have one halogen. We have one halogen here, it's the chlorine, so minus one. So let's do that math. So we have 1 1 2 times, two times six, plus two is 14. So we have 14 minus six over here. So 14 minus six is eight, times a half is equal to four."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's do that math. So we have 1 1 2 times, two times six, plus two is 14. So we have 14 minus six over here. So 14 minus six is eight, times a half is equal to four. So we get an HDI equal to four. And once again, we think benzene ring. So HDI is equal to four."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 14 minus six is eight, times a half is equal to four. So we get an HDI equal to four. And once again, we think benzene ring. So HDI is equal to four. We think about a benzene ring being present. All right, let's think about one more example. So if we look on the left, once again we have benzene."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So HDI is equal to four. We think about a benzene ring being present. All right, let's think about one more example. So if we look on the left, once again we have benzene. So here's benzene with the six hydrogens. And we've added in an oxygen. So for this molecule, we've added in one oxygen."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we look on the left, once again we have benzene. So here's benzene with the six hydrogens. And we've added in an oxygen. So for this molecule, we've added in one oxygen. What happened to the number of hydrogens? So we have five on the ring, one, two, three, four, five, and then this one, four, six. So even though we've added in a hydrogen, there's no change in the number, even though we've added in oxygen, I should say, there's no change in the number of hydrogens."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So for this molecule, we've added in one oxygen. What happened to the number of hydrogens? So we have five on the ring, one, two, three, four, five, and then this one, four, six. So even though we've added in a hydrogen, there's no change in the number, even though we've added in oxygen, I should say, there's no change in the number of hydrogens. And since there's no change in the number of hydrogens, we don't need to modify our formula here for the hydrogen deficiency index. So I'm gonna go right back up here and box this. So this is our final version for calculating the HDI."}, {"video_title": "Hydrogen deficiency index Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So even though we've added in a hydrogen, there's no change in the number, even though we've added in oxygen, I should say, there's no change in the number of hydrogens. And since there's no change in the number of hydrogens, we don't need to modify our formula here for the hydrogen deficiency index. So I'm gonna go right back up here and box this. So this is our final version for calculating the HDI. And you don't have to use this formula every time. Sometimes you can just think about how many hydrogens are missing. But sometimes using this formula allows you to calculate the HDI, which helps you to think about what's present in the structure of your molecule."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And to help us, we're gonna look at this SN1 versus SN2 summary. And the first thing that we're going to look at is the structure of our substrate. For example, for this reaction down here, we have a primary alkyl halide, a primary substrate. So we need to think about an SN2 reaction, which requires decreased steric hindrance, and that's what we have with a primary alkyl halide. From an earlier video, we know that an SN2 mechanism has our nucleophile attack at the same time that we get loss of a leaving group. And for this reaction, our nucleophile must be our solvent here, which is ethanol. So our nucleophile is going to attack, and the oxygen is gonna form a bond with this carbon, which I'll go ahead and highlight in red."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So we need to think about an SN2 reaction, which requires decreased steric hindrance, and that's what we have with a primary alkyl halide. From an earlier video, we know that an SN2 mechanism has our nucleophile attack at the same time that we get loss of a leaving group. And for this reaction, our nucleophile must be our solvent here, which is ethanol. So our nucleophile is going to attack, and the oxygen is gonna form a bond with this carbon, which I'll go ahead and highlight in red. So our nucleophile attacks the same time we get loss of a leaving group. These electrons come off to form the iodide anion, which is an excellent leaving group. So let's draw what we would form."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So our nucleophile is going to attack, and the oxygen is gonna form a bond with this carbon, which I'll go ahead and highlight in red. So our nucleophile attacks the same time we get loss of a leaving group. These electrons come off to form the iodide anion, which is an excellent leaving group. So let's draw what we would form. So let's sketch in our carbon chain here. And we know that a bond forms between the oxygen and the carbon in red. So the carbon in red is this carbon."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's draw what we would form. So let's sketch in our carbon chain here. And we know that a bond forms between the oxygen and the carbon in red. So the carbon in red is this carbon. And let's make these electrons magenta. So those electrons form a bond between the oxygen and the carbon in red. The oxygen is still attached to this ethyl group here, so let's draw in those two carbons."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So the carbon in red is this carbon. And let's make these electrons magenta. So those electrons form a bond between the oxygen and the carbon in red. The oxygen is still attached to this ethyl group here, so let's draw in those two carbons. And the oxygen is still bonded to this hydrogen, so let's put in this hydrogen. We still have a lone pair of electrons left on this oxygen, so I'll put in that lone pair right here. And that gives us a plus one formal charge on the oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "The oxygen is still attached to this ethyl group here, so let's draw in those two carbons. And the oxygen is still bonded to this hydrogen, so let's put in this hydrogen. We still have a lone pair of electrons left on this oxygen, so I'll put in that lone pair right here. And that gives us a plus one formal charge on the oxygen. Next, we need to make a neutral molecule for our product. So we need to have another molecule of ethanol come along. So let's draw that in here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And that gives us a plus one formal charge on the oxygen. Next, we need to make a neutral molecule for our product. So we need to have another molecule of ethanol come along. So let's draw that in here. So ethanol is our solvent. And this time, the ethanol molecule is going to function as a base. We need to take this proton here, and these electrons are left behind on the oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's draw that in here. So ethanol is our solvent. And this time, the ethanol molecule is going to function as a base. We need to take this proton here, and these electrons are left behind on the oxygen. So let's draw our final product. So we need to sketch in our carbon chain. We have our oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "We need to take this proton here, and these electrons are left behind on the oxygen. So let's draw our final product. So we need to sketch in our carbon chain. We have our oxygen. We have these two carbons. And now we have two lone pairs of electrons on this oxygen. So let's make these electrons blue."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "We have our oxygen. We have these two carbons. And now we have two lone pairs of electrons on this oxygen. So let's make these electrons blue. So our second step is an acid-base reaction, where we take a proton, and these are these electrons in blue here to form our final product, which is an ether. Notice we don't have to worry about any stereochemistry for our final product. We don't have any chiral centers to worry about."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's make these electrons blue. So our second step is an acid-base reaction, where we take a proton, and these are these electrons in blue here to form our final product, which is an ether. Notice we don't have to worry about any stereochemistry for our final product. We don't have any chiral centers to worry about. Let's look at another reaction. So for this reaction, we're starting with a secondary alkyl halide. If I look at my summary over here, with a secondary substrate, we could have either an SN2 mechanism or an SN1 mechanism."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "We don't have any chiral centers to worry about. Let's look at another reaction. So for this reaction, we're starting with a secondary alkyl halide. If I look at my summary over here, with a secondary substrate, we could have either an SN2 mechanism or an SN1 mechanism. So we need to look at a few more things. First, let's look at the nucleophile. This is Na plus and SH minus."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "If I look at my summary over here, with a secondary substrate, we could have either an SN2 mechanism or an SN1 mechanism. So we need to look at a few more things. First, let's look at the nucleophile. This is Na plus and SH minus. So let me draw in the SH minus here, which that is going to be our nucleophile. And that's a strong nucleophile. A negative charge on a sulfur would make a strong nucleophile."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "This is Na plus and SH minus. So let me draw in the SH minus here, which that is going to be our nucleophile. And that's a strong nucleophile. A negative charge on a sulfur would make a strong nucleophile. And for our solvent, we saw in an earlier video, DMSO is a polar aprotic solvent, which favors an SN2 reaction. So with a strong nucleophile and a polar aprotic solvent, we need to think about an SN2 mechanism. So we know our nucleophile attacks at the same time that we get loss of our leaving group."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "A negative charge on a sulfur would make a strong nucleophile. And for our solvent, we saw in an earlier video, DMSO is a polar aprotic solvent, which favors an SN2 reaction. So with a strong nucleophile and a polar aprotic solvent, we need to think about an SN2 mechanism. So we know our nucleophile attacks at the same time that we get loss of our leaving group. So our nucleophile is going to attack this carbon. So again, I'll make this carbon red. At the same time that we get loss of a leaving group, so these electrons are going to come off onto the bromine to form the bromide anion."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So we know our nucleophile attacks at the same time that we get loss of our leaving group. So our nucleophile is going to attack this carbon. So again, I'll make this carbon red. At the same time that we get loss of a leaving group, so these electrons are going to come off onto the bromine to form the bromide anion. For this reaction, we need to think about the stereochemistry of our SN2 reaction. Our nucleophile has to attack from the opposite side of our leaving group. So we get inversion of configuration."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "At the same time that we get loss of a leaving group, so these electrons are going to come off onto the bromine to form the bromide anion. For this reaction, we need to think about the stereochemistry of our SN2 reaction. Our nucleophile has to attack from the opposite side of our leaving group. So we get inversion of configuration. So if we have a chiral center, we have to worry about our stereochemistry for this reaction. So we had the bromine on a wedge. So drawing the final product here, we need to have the SH going away from us in space."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So we get inversion of configuration. So if we have a chiral center, we have to worry about our stereochemistry for this reaction. So we had the bromine on a wedge. So drawing the final product here, we need to have the SH going away from us in space. So we put that on a dash. Again, we saw more details about this in an earlier video. So we get inversion of configuration for this SN2 reaction."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So drawing the final product here, we need to have the SH going away from us in space. So we put that on a dash. Again, we saw more details about this in an earlier video. So we get inversion of configuration for this SN2 reaction. First, let's look at our alkyl halide. The carbon that's bonded to our bromine is bonded to two other carbons. So this is a secondary alkyl halide."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So we get inversion of configuration for this SN2 reaction. First, let's look at our alkyl halide. The carbon that's bonded to our bromine is bonded to two other carbons. So this is a secondary alkyl halide. And so we know we could have either SN1 or SN2. We need to look at the nucleophile and the solvent next to decide which mechanism it is. Our nucleophile will be formic acid, which is a weak nucleophile."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So this is a secondary alkyl halide. And so we know we could have either SN1 or SN2. We need to look at the nucleophile and the solvent next to decide which mechanism it is. Our nucleophile will be formic acid, which is a weak nucleophile. And water is a polar protic solvent. So we know that these two things favor an SN1 type mechanism. The polar protic solvent water can stabilize the carbocation that would result."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "Our nucleophile will be formic acid, which is a weak nucleophile. And water is a polar protic solvent. So we know that these two things favor an SN1 type mechanism. The polar protic solvent water can stabilize the carbocation that would result. So the first step should be loss of our leaving group to form our carbocations. So these electrons come off onto our bromine to form the bromide ion. And we're taking a bond away from this carbon in red."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "The polar protic solvent water can stabilize the carbocation that would result. So the first step should be loss of our leaving group to form our carbocations. So these electrons come off onto our bromine to form the bromide ion. And we're taking a bond away from this carbon in red. So the carbon in red is going to have a plus 1 formal charge. So let's draw our carbocations. Let me put in this ring here with our pi electrons."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And we're taking a bond away from this carbon in red. So the carbon in red is going to have a plus 1 formal charge. So let's draw our carbocations. Let me put in this ring here with our pi electrons. And then let me highlight our carbon in red. The carbon in red is this one. So that should get a plus 1 formal charge."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "Let me put in this ring here with our pi electrons. And then let me highlight our carbon in red. The carbon in red is this one. So that should get a plus 1 formal charge. And let me highlight the other carbon over here. So this carbon in magenta, I moved it up to here to make it easier to see the mechanism. So this is our secondary carbocation."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So that should get a plus 1 formal charge. And let me highlight the other carbon over here. So this carbon in magenta, I moved it up to here to make it easier to see the mechanism. So this is our secondary carbocation. But this is actually a benzylic carbocation, which makes it even more stable than we would normally expect. The pi electrons in the ring can actually provide us with other resonance structures to stabilize this positive charge. So I don't have the time or the space to show that here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So this is our secondary carbocation. But this is actually a benzylic carbocation, which makes it even more stable than we would normally expect. The pi electrons in the ring can actually provide us with other resonance structures to stabilize this positive charge. So I don't have the time or the space to show that here. But we have our secondary benzylic carbocation, which will be our electrophile. And our next step is to have our nucleophile attack our electrophile. And our nucleophile is formic acid here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So I don't have the time or the space to show that here. But we have our secondary benzylic carbocation, which will be our electrophile. And our next step is to have our nucleophile attack our electrophile. And our nucleophile is formic acid here. And we have two oxygens in formic acid. And one of them is more nucleophilic than the other. So it turns out that this carbonyl oxygen is more nucleophilic than this oxygen down here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And our nucleophile is formic acid here. And we have two oxygens in formic acid. And one of them is more nucleophilic than the other. So it turns out that this carbonyl oxygen is more nucleophilic than this oxygen down here. And we'll go into more detail in a couple of minutes. But for right now, let's show our nucleophile attacking our electrophile. So a lone pair of electrons on our oxygen are going to form a bond with this carbon in red."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So it turns out that this carbonyl oxygen is more nucleophilic than this oxygen down here. And we'll go into more detail in a couple of minutes. But for right now, let's show our nucleophile attacking our electrophile. So a lone pair of electrons on our oxygen are going to form a bond with this carbon in red. So let's draw the result of our nucleophilic attack. We have our benzene ring. So I'll draw that in here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So a lone pair of electrons on our oxygen are going to form a bond with this carbon in red. So let's draw the result of our nucleophilic attack. We have our benzene ring. So I'll draw that in here. And our carbon in red is this one. And let's highlight the electrons on our oxygen. So this lone pair of electrons in blue forms a bond with our carbon in red."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So I'll draw that in here. And our carbon in red is this one. And let's highlight the electrons on our oxygen. So this lone pair of electrons in blue forms a bond with our carbon in red. So now let's draw in our oxygen. Our oxygen still has a lone pair of electrons on it. And our oxygen is double bonded to a carbon, which is bonded to a hydrogen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So this lone pair of electrons in blue forms a bond with our carbon in red. So now let's draw in our oxygen. Our oxygen still has a lone pair of electrons on it. And our oxygen is double bonded to a carbon, which is bonded to a hydrogen. And on the right, we have our oxygen and a hydrogen. Now we have a plus 1 formal charge on our oxygen. So this oxygen has a plus 1 formal charge."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And our oxygen is double bonded to a carbon, which is bonded to a hydrogen. And on the right, we have our oxygen and a hydrogen. Now we have a plus 1 formal charge on our oxygen. So this oxygen has a plus 1 formal charge. And this product is resonance stabilized. So we can move in a lone pair of electrons here and push these electrons off onto our oxygen. And let's go ahead and show the result of that."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So this oxygen has a plus 1 formal charge. And this product is resonance stabilized. So we can move in a lone pair of electrons here and push these electrons off onto our oxygen. And let's go ahead and show the result of that. So we'll put in our resonance brackets here and our resonance arrow. So we have our benzene ring. So let's draw that in."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And let's go ahead and show the result of that. So we'll put in our resonance brackets here and our resonance arrow. So we have our benzene ring. So let's draw that in. And we have our carbons. We have a bond to this oxygen. And I'll draw in that lone pair of electrons on that top oxygen here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's draw that in. And we have our carbons. We have a bond to this oxygen. And I'll draw in that lone pair of electrons on that top oxygen here. Should only be one bond now to this carbon. I'll put in my hydrogen. And now actually, we have a double bond over here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And I'll draw in that lone pair of electrons on that top oxygen here. Should only be one bond now to this carbon. I'll put in my hydrogen. And now actually, we have a double bond over here. So let me draw everything in. And let me show the movement of electrons. So first, let's start with these electrons, which I will make magenta."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And now actually, we have a double bond over here. So let me draw everything in. And let me show the movement of electrons. So first, let's start with these electrons, which I will make magenta. Those electrons moved into here to form a double bond. And then let's highlight these electrons here in red. So these electrons came off onto our oxygen right here."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So first, let's start with these electrons, which I will make magenta. Those electrons moved into here to form a double bond. And then let's highlight these electrons here in red. So these electrons came off onto our oxygen right here. And now we have a plus 1 formal charge on this oxygen. So the product of nucleophilic attack by the carbonyl oxygen is resonance stabilized. And then to go to our product, just think about a base coming along, something like water."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So these electrons came off onto our oxygen right here. And now we have a plus 1 formal charge on this oxygen. So the product of nucleophilic attack by the carbonyl oxygen is resonance stabilized. And then to go to our product, just think about a base coming along, something like water. And we could take this proton. And then these electrons would be left behind. These electrons would be left behind on that oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And then to go to our product, just think about a base coming along, something like water. And we could take this proton. And then these electrons would be left behind. These electrons would be left behind on that oxygen. So let's draw in our product. We would have a benzene ring. So here's our benzene ring."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "These electrons would be left behind on that oxygen. So let's draw in our product. We would have a benzene ring. So here's our benzene ring. We would have an oxygen here. And then we would have our carbonyl and then our hydrogen. So this oxygen has two lone pairs of electrons."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So here's our benzene ring. We would have an oxygen here. And then we would have our carbonyl and then our hydrogen. So this oxygen has two lone pairs of electrons. And so does this one. So we could highlight some of those electrons. Let me make them green."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So this oxygen has two lone pairs of electrons. And so does this one. So we could highlight some of those electrons. Let me make them green. So these electrons in green here came off onto this oxygen. And that gives us our product. Notice that in our product, we have a chiral center."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "Let me make them green. So these electrons in green here came off onto this oxygen. And that gives us our product. Notice that in our product, we have a chiral center. So this carbon right here is a chiral center. We have four different groups attached to it. So we would expect to get a racemic mixture."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "Notice that in our product, we have a chiral center. So this carbon right here is a chiral center. We have four different groups attached to it. So we would expect to get a racemic mixture. We should get both enantiomers here. Because remember, going back to our carbocation, this carbocation is planar. And our nucleophile can attack from either side."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So we would expect to get a racemic mixture. We should get both enantiomers here. Because remember, going back to our carbocation, this carbocation is planar. And our nucleophile can attack from either side. If you're wondering why this oxygen is not as nucleophilic as the carbonyl oxygen, let's show the result of what would happen if this oxygen attacked our positively charged carbon. So let me draw in our benzene ring. So let's sketch that in."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And our nucleophile can attack from either side. If you're wondering why this oxygen is not as nucleophilic as the carbonyl oxygen, let's show the result of what would happen if this oxygen attacked our positively charged carbon. So let me draw in our benzene ring. So let's sketch that in. And we would form a bond between the oxygen and, let me highlight our carbon, which I made red below. So this is our carbon in red. And I'm going to show this lone pair of electrons forming a bond with that carbon."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's sketch that in. And we would form a bond between the oxygen and, let me highlight our carbon, which I made red below. So this is our carbon in red. And I'm going to show this lone pair of electrons forming a bond with that carbon. So that's bonded to this oxygen. And the oxygen is bonded to our carbonyl, which is bonded to our hydrogen. And our oxygen still has a hydrogen on it and a lone pair of electrons."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And I'm going to show this lone pair of electrons forming a bond with that carbon. So that's bonded to this oxygen. And the oxygen is bonded to our carbonyl, which is bonded to our hydrogen. And our oxygen still has a hydrogen on it and a lone pair of electrons. So there's still a hydrogen on it. And there's still a lone pair of electrons, which gives this oxygen a plus 1 formal charge. Notice this positively charged oxygen is right next to this carbonyl carbon."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And our oxygen still has a hydrogen on it and a lone pair of electrons. So there's still a hydrogen on it. And there's still a lone pair of electrons, which gives this oxygen a plus 1 formal charge. Notice this positively charged oxygen is right next to this carbonyl carbon. And we know that this carbonyl carbon is partially positive because this carbonyl oxygen withdraws electron density from it. So you have a positively charged oxygen next to a partially positively charged carbon. And we know that like charges repel."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "Notice this positively charged oxygen is right next to this carbonyl carbon. And we know that this carbonyl carbon is partially positive because this carbonyl oxygen withdraws electron density from it. So you have a positively charged oxygen next to a partially positively charged carbon. And we know that like charges repel. So having these two positive charges next to each other would destabilize this structure. And so that's the reason why this oxygen is not the nucleophile. The carbonyl oxygen is."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And we know that like charges repel. So having these two positive charges next to each other would destabilize this structure. And so that's the reason why this oxygen is not the nucleophile. The carbonyl oxygen is. Let's look at one final example. So for this alkyl halide, this is a tertiary alkyl halide and a tertiary substrate means, think about an SN1 mechanism. So our first step here would be loss of a leaving group."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "The carbonyl oxygen is. Let's look at one final example. So for this alkyl halide, this is a tertiary alkyl halide and a tertiary substrate means, think about an SN1 mechanism. So our first step here would be loss of a leaving group. These electrons come off to form the iodide anion. And we're taking a bond away from this carbon in red to form a carbocation. Let's get some more room."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So our first step here would be loss of a leaving group. These electrons come off to form the iodide anion. And we're taking a bond away from this carbon in red to form a carbocation. Let's get some more room. Let's go down here. And let's draw our carbocation. So we have a six-membered ring."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "Let's get some more room. Let's go down here. And let's draw our carbocation. So we have a six-membered ring. So let me draw in our six-membered ring here. And our carbon in red is this carbon. So that carbon gets a plus 1 formal charge."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So we have a six-membered ring. So let me draw in our six-membered ring here. And our carbon in red is this carbon. So that carbon gets a plus 1 formal charge. So we have a tertiary carbocation, which is relatively stable for a carbocation. And in our next step, our nucleophile attacks our electrophile. And our nucleophile is our solvent, which is methanol."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So that carbon gets a plus 1 formal charge. So we have a tertiary carbocation, which is relatively stable for a carbocation. And in our next step, our nucleophile attacks our electrophile. And our nucleophile is our solvent, which is methanol. So our nucleophile is going to attack our electrophile. Our oxygen is going to form a bond with this carbon in red. So let's draw the result of our nucleophilic attack."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And our nucleophile is our solvent, which is methanol. So our nucleophile is going to attack our electrophile. Our oxygen is going to form a bond with this carbon in red. So let's draw the result of our nucleophilic attack. So put in this ring here. Move the methyl group over to make room for the oxygen. So here's that oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's draw the result of our nucleophilic attack. So put in this ring here. Move the methyl group over to make room for the oxygen. So here's that oxygen. Let's highlight the carbon in red. And let's also highlight some electrons. So these electrons in magenta form the bond between the oxygen and the carbon in red."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So here's that oxygen. Let's highlight the carbon in red. And let's also highlight some electrons. So these electrons in magenta form the bond between the oxygen and the carbon in red. The oxygen is still bonded to a hydrogen and a methyl group. So let's put those in. So here is our hydrogen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So these electrons in magenta form the bond between the oxygen and the carbon in red. The oxygen is still bonded to a hydrogen and a methyl group. So let's put those in. So here is our hydrogen. And here is our methyl group. We still have a lone pair of electrons on the oxygen. So let's put in that lone pair of electrons."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So here is our hydrogen. And here is our methyl group. We still have a lone pair of electrons on the oxygen. So let's put in that lone pair of electrons. And that gives the oxygen a plus 1 formal charge. To get to our product, we need a neutral product. So another molecule of methanol comes along."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So let's put in that lone pair of electrons. And that gives the oxygen a plus 1 formal charge. To get to our product, we need a neutral product. So another molecule of methanol comes along. But this time, instead of acting like a nucleophile, it's going to act as a base and take off that proton. So here is our molecule of methanol. And we're going to take this proton and leave these electrons behind on the oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "So another molecule of methanol comes along. But this time, instead of acting like a nucleophile, it's going to act as a base and take off that proton. So here is our molecule of methanol. And we're going to take this proton and leave these electrons behind on the oxygen. So let's draw in our final product up here. So here is our ring. And let's move up a little bit."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And we're going to take this proton and leave these electrons behind on the oxygen. So let's draw in our final product up here. So here is our ring. And let's move up a little bit. So let's move up. On our ring, we have this methyl group here. And we have our oxygen and another methyl group, two lone pairs of electrons on the oxygen."}, {"video_title": "Sn1 and Sn2 summary.mp3", "Sentence": "And let's move up a little bit. So let's move up. On our ring, we have this methyl group here. And we have our oxygen and another methyl group, two lone pairs of electrons on the oxygen. So let's show those electrons. So these electrons in here in blue come off onto the oxygen to form our final product. Notice that for our final product, we don't have any chiral centers to worry about."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So we're going from the alkyl halide on the left to the alkene on the right. The first step is loss of our leaving groups. The electrons in this bond come off onto the chlorine to form the chloride anion, and we know that the chloride anion is a good leaving group. We're taking a bond away from the carbon that I just marked in red, so that's where our carbocation will form. So let's draw out our carbon skeleton, and we'll put in these methyl groups right here. And the carbon in red is this carbon here. So that would be a secondary carbocation, so it has a plus one formal charge, and it's secondary because the carbon in red is directly bonded to two other carbons."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "We're taking a bond away from the carbon that I just marked in red, so that's where our carbocation will form. So let's draw out our carbon skeleton, and we'll put in these methyl groups right here. And the carbon in red is this carbon here. So that would be a secondary carbocation, so it has a plus one formal charge, and it's secondary because the carbon in red is directly bonded to two other carbons. So we have a secondary carbocation, and if you look at it, there's a possibility for a rearrangement. We could take one of the methyl groups on the carbon in magenta on the right, and we could move it over to the positively charged carbon. So we get a methyl shift here, and the reason why we would get a methyl shift is that's gonna create a more stable carbocation."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So that would be a secondary carbocation, so it has a plus one formal charge, and it's secondary because the carbon in red is directly bonded to two other carbons. So we have a secondary carbocation, and if you look at it, there's a possibility for a rearrangement. We could take one of the methyl groups on the carbon in magenta on the right, and we could move it over to the positively charged carbon. So we get a methyl shift here, and the reason why we would get a methyl shift is that's gonna create a more stable carbocation. So now there's only one methyl group on this carbon. There's a methyl group over here to begin with, and we just moved a methyl group to the carbon in red. So let me highlight the carbon in red here."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So we get a methyl shift here, and the reason why we would get a methyl shift is that's gonna create a more stable carbocation. So now there's only one methyl group on this carbon. There's a methyl group over here to begin with, and we just moved a methyl group to the carbon in red. So let me highlight the carbon in red here. We moved a methyl group to that carbon, so let me draw that in. So that means we took a bond away from this carbon, so that's this carbon, and so that's where our positive charge is now. So we have a plus one formal charge on the carbon that I've now labeled in light blue, and this is a tertiary carbocation."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So let me highlight the carbon in red here. We moved a methyl group to that carbon, so let me draw that in. So that means we took a bond away from this carbon, so that's this carbon, and so that's where our positive charge is now. So we have a plus one formal charge on the carbon that I've now labeled in light blue, and this is a tertiary carbocation. So let me highlight those carbons here. So the carbon in light blue is directly bonded to three other carbons, so this carbon right here, this one, and this one. So this is a tertiary carbocation, which we know is much more stable than a secondary carbocation."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So we have a plus one formal charge on the carbon that I've now labeled in light blue, and this is a tertiary carbocation. So let me highlight those carbons here. So the carbon in light blue is directly bonded to three other carbons, so this carbon right here, this one, and this one. So this is a tertiary carbocation, which we know is much more stable than a secondary carbocation. Our carbocation is further stabilized by the fact that we have a polar protic solvent here, so this is a very stable carbocation. At this point, you could have two possible reaction paths. You have water, which could function as a nucleophile, and if that happened, you would get a substitution product, but now we know we need an elimination product."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So this is a tertiary carbocation, which we know is much more stable than a secondary carbocation. Our carbocation is further stabilized by the fact that we have a polar protic solvent here, so this is a very stable carbocation. At this point, you could have two possible reaction paths. You have water, which could function as a nucleophile, and if that happened, you would get a substitution product, but now we know we need an elimination product. We have an alkene on the right here, so water's gonna function as a base, and water's going to take a proton from one of the carbons next door to our carbocation, so one of the carbons in magenta. And if we look at our product here, the double bond formed between this carbon and this carbon, so that means we need to take a proton from the carbon in magenta on the top right. So let me highlight that carbon here."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "You have water, which could function as a nucleophile, and if that happened, you would get a substitution product, but now we know we need an elimination product. We have an alkene on the right here, so water's gonna function as a base, and water's going to take a proton from one of the carbons next door to our carbocation, so one of the carbons in magenta. And if we look at our product here, the double bond formed between this carbon and this carbon, so that means we need to take a proton from the carbon in magenta on the top right. So let me highlight that carbon here. So we need to take a proton from this carbon. So let me draw in one, because we know there's one hydrogen bonded to that carbon, and water's gonna function as our weak base here, so let me draw in H2O, and water's gonna take that proton. So a lone pair of electrons on the oxygen take this proton right here, and then these electrons would move in to form our double bond, and that gives us our product."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So let me highlight that carbon here. So we need to take a proton from this carbon. So let me draw in one, because we know there's one hydrogen bonded to that carbon, and water's gonna function as our weak base here, so let me draw in H2O, and water's gonna take that proton. So a lone pair of electrons on the oxygen take this proton right here, and then these electrons would move in to form our double bond, and that gives us our product. So let me highlight those electrons. So the electrons that form our double bond, I'll make them dark blue right here, so those electrons move in here to form our alkene. And we've shown a mechanism for this reaction."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So a lone pair of electrons on the oxygen take this proton right here, and then these electrons would move in to form our double bond, and that gives us our product. So let me highlight those electrons. So the electrons that form our double bond, I'll make them dark blue right here, so those electrons move in here to form our alkene. And we've shown a mechanism for this reaction. Let's look at another E1 reaction, and let's say our goal was to draw all of the products from this elimination reaction. On the left we have our alcohol, and we're reacting our alcohol with sulfuric acid, and we're heating our reaction mixture. We've already seen from earlier videos the first step when you have an alcohol is a proton transfer."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "And we've shown a mechanism for this reaction. Let's look at another E1 reaction, and let's say our goal was to draw all of the products from this elimination reaction. On the left we have our alcohol, and we're reacting our alcohol with sulfuric acid, and we're heating our reaction mixture. We've already seen from earlier videos the first step when you have an alcohol is a proton transfer. The alcohol functions as a base, and sulfuric acid donates a proton. So I'll draw an H plus here. We're gonna protonate our alcohol first."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "We've already seen from earlier videos the first step when you have an alcohol is a proton transfer. The alcohol functions as a base, and sulfuric acid donates a proton. So I'll draw an H plus here. We're gonna protonate our alcohol first. So one of the lone pairs of electrons on oxygen picks up this proton, which came from sulfuric acid. So let's draw in what we would have. So here's our carbon chain."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "We're gonna protonate our alcohol first. So one of the lone pairs of electrons on oxygen picks up this proton, which came from sulfuric acid. So let's draw in what we would have. So here's our carbon chain. We would have two methyl groups on that carbon, and the oxygen now has two bonds to hydrogen. It still has one lone pair of electrons, but now the oxygen has a plus one formal charge. So our electrons in magenta pick up our proton from sulfuric acid to form this bond."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So here's our carbon chain. We would have two methyl groups on that carbon, and the oxygen now has two bonds to hydrogen. It still has one lone pair of electrons, but now the oxygen has a plus one formal charge. So our electrons in magenta pick up our proton from sulfuric acid to form this bond. Now the reason why this step happens first when you have an alcohol is it forms a better leaving group for your E1 mechanism. Water is a much better leaving group than the hydroxide anion. So next, the electrons in this bond come off onto the oxygen, and water leaves."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So our electrons in magenta pick up our proton from sulfuric acid to form this bond. Now the reason why this step happens first when you have an alcohol is it forms a better leaving group for your E1 mechanism. Water is a much better leaving group than the hydroxide anion. So next, the electrons in this bond come off onto the oxygen, and water leaves. So when we do that, we take a bond away from this carbon, the carbon in red. So that's where our carbocation forms. If I draw in my carbon skeleton here, so I have methyl groups coming off of this carbon."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So next, the electrons in this bond come off onto the oxygen, and water leaves. So when we do that, we take a bond away from this carbon, the carbon in red. So that's where our carbocation forms. If I draw in my carbon skeleton here, so I have methyl groups coming off of this carbon. The carbon in red is this carbon, and that's a secondary carbocation, because the carbon in red is directly bonded to two other carbons, which I've just marked there in magenta. So this is a secondary carbocation. And let me draw in the plus one formal charge on the carbon in red."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "If I draw in my carbon skeleton here, so I have methyl groups coming off of this carbon. The carbon in red is this carbon, and that's a secondary carbocation, because the carbon in red is directly bonded to two other carbons, which I've just marked there in magenta. So this is a secondary carbocation. And let me draw in the plus one formal charge on the carbon in red. Next, we think about the possibility of a rearrangement. So can we do anything to form a more stable carbocation? Well, just like the previous example, we could have a methyl shift."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "And let me draw in the plus one formal charge on the carbon in red. Next, we think about the possibility of a rearrangement. So can we do anything to form a more stable carbocation? Well, just like the previous example, we could have a methyl shift. Let me go ahead and leave some more room in there. So we could take, let's say, this methyl group, and we could move this methyl group over to the carbon in red. So let's show a methyl shift."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "Well, just like the previous example, we could have a methyl shift. Let me go ahead and leave some more room in there. So we could take, let's say, this methyl group, and we could move this methyl group over to the carbon in red. So let's show a methyl shift. So now there's only one methyl group on this carbon, and a methyl group moved over here to this carbon. So this is the carbon in red. So the carbon in red no longer has a plus one formal charge."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So let's show a methyl shift. So now there's only one methyl group on this carbon, and a methyl group moved over here to this carbon. So this is the carbon in red. So the carbon in red no longer has a plus one formal charge. The plus one formal charge goes to the carbon in magenta that just lost a bond. So now, our plus one formal charge is on this carbon. Let me highlight it."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So the carbon in red no longer has a plus one formal charge. The plus one formal charge goes to the carbon in magenta that just lost a bond. So now, our plus one formal charge is on this carbon. Let me highlight it. So this carbon right here in magenta. And that is a tertiary carbocation, and we know tertiary carbocations are more stable than secondary carbocations. Next, since this is an E1 mechanism, we know that a weak base comes along and takes a proton."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "Let me highlight it. So this carbon right here in magenta. And that is a tertiary carbocation, and we know tertiary carbocations are more stable than secondary carbocations. Next, since this is an E1 mechanism, we know that a weak base comes along and takes a proton. So in this case, we could take a proton from a few different places. We could take a proton from this carbon. So we think about our weak base coming along, and we draw this in here."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "Next, since this is an E1 mechanism, we know that a weak base comes along and takes a proton. So in this case, we could take a proton from a few different places. We could take a proton from this carbon. So we think about our weak base coming along, and we draw this in here. And our base is gonna take this proton, which means that these electrons would move in here to get rid of our formal charge and to form an alkene as our product. So let's draw in the product up here. Our double bond would form here."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So we think about our weak base coming along, and we draw this in here. And our base is gonna take this proton, which means that these electrons would move in here to get rid of our formal charge and to form an alkene as our product. So let's draw in the product up here. Our double bond would form here. And let's draw in everything else. So that is one of the possible products. Let's highlight our electrons."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "Our double bond would form here. And let's draw in everything else. So that is one of the possible products. Let's highlight our electrons. I'll use light blue. So the electrons in this bond moved in to form our alkene. All right, what about if I took a proton from a different place?"}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "Let's highlight our electrons. I'll use light blue. So the electrons in this bond moved in to form our alkene. All right, what about if I took a proton from a different place? So let me just redraw that carbocation here just so things don't get too busy. So I'm gonna redraw our carbocation. So the carbon in magenta is this one."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "All right, what about if I took a proton from a different place? So let me just redraw that carbocation here just so things don't get too busy. So I'm gonna redraw our carbocation. So the carbon in magenta is this one. So that has a plus one formal charge. Let me draw that in. And the carbon in red is this carbon."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So the carbon in magenta is this one. So that has a plus one formal charge. Let me draw that in. And the carbon in red is this carbon. So what happens if we took a proton from the carbon in red? Let me draw in a proton here. So our weak base comes along and takes this proton, let's say."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "And the carbon in red is this carbon. So what happens if we took a proton from the carbon in red? Let me draw in a proton here. So our weak base comes along and takes this proton, let's say. And so these electrons would move in to here to form our double bond, to form another alkene. So let's draw that product. So we have our carbon skeleton here."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So our weak base comes along and takes this proton, let's say. And so these electrons would move in to here to form our double bond, to form another alkene. So let's draw that product. So we have our carbon skeleton here. And this time the double bond forms between the carbons in magenta and red. So the double bond would form in here. So let's highlight those electrons in light blue."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So we have our carbon skeleton here. And this time the double bond forms between the carbons in magenta and red. So the double bond would form in here. So let's highlight those electrons in light blue. These electrons would move in to here to form our double bond. All right, there's actually one more product for this reaction. And let's go back to our secondary carbocation over here."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So let's highlight those electrons in light blue. These electrons would move in to here to form our double bond. All right, there's actually one more product for this reaction. And let's go back to our secondary carbocation over here. So if it doesn't rearrange, you could actually take a proton from this secondary carbocation and form our last product. So let's think about the carbon in red right here. So we're looking at the carbons next door to the carbon in red."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "And let's go back to our secondary carbocation over here. So if it doesn't rearrange, you could actually take a proton from this secondary carbocation and form our last product. So let's think about the carbon in red right here. So we're looking at the carbons next door to the carbon in red. Well, the carbon in magenta to the left doesn't have any protons. So we can't take a proton from that one. But the carbon to the right in magenta does."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So we're looking at the carbons next door to the carbon in red. Well, the carbon in magenta to the left doesn't have any protons. So we can't take a proton from that one. But the carbon to the right in magenta does. So if I squeeze in a hydrogen in here, finally our base could come along and take this proton. And if that happened, then these electrons would move in to here. So that gives us our final product."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "But the carbon to the right in magenta does. So if I squeeze in a hydrogen in here, finally our base could come along and take this proton. And if that happened, then these electrons would move in to here. So that gives us our final product. So let me draw in our skeleton here. And then our double bond would form right, actually let me just redraw that double bond so things are getting, let me just redraw the whole thing. I think I have a little bit more space than I thought I did."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So that gives us our final product. So let me draw in our skeleton here. And then our double bond would form right, actually let me just redraw that double bond so things are getting, let me just redraw the whole thing. I think I have a little bit more space than I thought I did. So let's sketch in our carbon skeleton here. Our double bond would form here. And then we have our methyl groups coming off of that carbon."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "I think I have a little bit more space than I thought I did. So let's sketch in our carbon skeleton here. Our double bond would form here. And then we have our methyl groups coming off of that carbon. So the electrons in light blue moved in to form our double bond. So we'd form three products. Three products from this E1 reaction."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "And then we have our methyl groups coming off of that carbon. So the electrons in light blue moved in to form our double bond. So we'd form three products. Three products from this E1 reaction. If we think about which one would be the major product, let's look at degrees of substitution. So let's go over to the alkene on the right. So when we think about the two carbons across our double bonds, there are one, two, three, four alkyl groups."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "Three products from this E1 reaction. If we think about which one would be the major product, let's look at degrees of substitution. So let's go over to the alkene on the right. So when we think about the two carbons across our double bonds, there are one, two, three, four alkyl groups. So this is a tetra-substituted alkene. So this should be the major product. This is the most stable alkene."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So when we think about the two carbons across our double bonds, there are one, two, three, four alkyl groups. So this is a tetra-substituted alkene. So this should be the major product. This is the most stable alkene. Next, let's look at this one. So here's a carbon and here's a carbon across our double bond. We can see this time we have only two alkyl groups."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "This is the most stable alkene. Next, let's look at this one. So here's a carbon and here's a carbon across our double bond. We can see this time we have only two alkyl groups. So this is a di-substituted alkene. And then finally, over here on the left, this would be a mono-substituted alkene. So here are two carbons."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "We can see this time we have only two alkyl groups. So this is a di-substituted alkene. And then finally, over here on the left, this would be a mono-substituted alkene. So here are two carbons. We have one alkyl group, so a mono-substituted alkene. And this one came from the secondary carbocation from no rearrangement. So this one's not gonna be a major product."}, {"video_title": "E1 mechanism carbocations and rearrangements.mp3", "Sentence": "So here are two carbons. We have one alkyl group, so a mono-substituted alkene. And this one came from the secondary carbocation from no rearrangement. So this one's not gonna be a major product. So only a very small percentage of your product would be this mono-substituted alkene. Most of your product is gonna be your di- and your tetra-substituted alkene. So with your tetra-substituted alkene being your major product, since it is the most stable."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we start with bromobenzene here. And to bromobenzene, we add some sodium amide, which is a strong base, and some liquid ammonia. And you can see that we have substituted an amino group for our halogen on our ring to form aniline as our product. Let's go ahead and look at the mechanism here. And so we're going to start with the elimination part. So let me go ahead and write down here, elimination. And the sodium amide functions as a base."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and look at the mechanism here. And so we're going to start with the elimination part. So let me go ahead and write down here, elimination. And the sodium amide functions as a base. So we can go ahead and draw the amide anion here with a negative 1 formal charge on this nitrogen. And so it's going to take this aromatic proton right here, which leaves these electrons behind on this carbon. So let's go ahead and show the result of our acid-base reaction."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the sodium amide functions as a base. So we can go ahead and draw the amide anion here with a negative 1 formal charge on this nitrogen. And so it's going to take this aromatic proton right here, which leaves these electrons behind on this carbon. So let's go ahead and show the result of our acid-base reaction. So we have our ring. We still have our leaving group for the moment. And now we have a negative 1 formal charge on this carbon."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the result of our acid-base reaction. So we have our ring. We still have our leaving group for the moment. And now we have a negative 1 formal charge on this carbon. So we form a carbanion. Let me go ahead and highlight these electrons. So these electrons in here are now on this carbon, forming a carbanion that's ortho to our leaving group, so an ortho carbanion."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now we have a negative 1 formal charge on this carbon. So we form a carbanion. Let me go ahead and highlight these electrons. So these electrons in here are now on this carbon, forming a carbanion that's ortho to our leaving group, so an ortho carbanion. So when these electrons in magenta move into here, that allows these electrons to kick off onto our halogen. So it leaves as an anion. And we create what's called the benzyne molecule, so a triple bond in our molecule like that."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here are now on this carbon, forming a carbanion that's ortho to our leaving group, so an ortho carbanion. So when these electrons in magenta move into here, that allows these electrons to kick off onto our halogen. So it leaves as an anion. And we create what's called the benzyne molecule, so a triple bond in our molecule like that. And so these electrons in magenta have moved in here to form a triple bond. This is a little bit different from usual triple bonds. This happens to be an unstable intermediate."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we create what's called the benzyne molecule, so a triple bond in our molecule like that. And so these electrons in magenta have moved in here to form a triple bond. This is a little bit different from usual triple bonds. This happens to be an unstable intermediate. So this benzyne molecule turns out to be very reactive. And so in the next step, we'll call this the addition portion of our mechanism. Our benzyne molecule is right here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This happens to be an unstable intermediate. So this benzyne molecule turns out to be very reactive. And so in the next step, we'll call this the addition portion of our mechanism. Our benzyne molecule is right here. And it's going to react with amides, the amide anion. But this time, the anion is going to function as a nucleophile. And so when I go ahead and show my amide anion like that, I'm going to show it functioning as a nucleophile attacking our triple bond, so attacking one of the carbons, so I'm going to say on that side, which would push these electrons back onto this carbon."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Our benzyne molecule is right here. And it's going to react with amides, the amide anion. But this time, the anion is going to function as a nucleophile. And so when I go ahead and show my amide anion like that, I'm going to show it functioning as a nucleophile attacking our triple bond, so attacking one of the carbons, so I'm going to say on that side, which would push these electrons back onto this carbon. And so we can go ahead and show that. So we have our ring. Now we have our NH2 attached to our ring."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so when I go ahead and show my amide anion like that, I'm going to show it functioning as a nucleophile attacking our triple bond, so attacking one of the carbons, so I'm going to say on that side, which would push these electrons back onto this carbon. And so we can go ahead and show that. So we have our ring. Now we have our NH2 attached to our ring. And we had these electrons move out onto here, onto this carbon. So once again, the electrons in magenta have moved out here to form a carb anion, again, an ortho carb anion. Our last step would be to protonate our ring."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now we have our NH2 attached to our ring. And we had these electrons move out onto here, onto this carbon. So once again, the electrons in magenta have moved out here to form a carb anion, again, an ortho carb anion. Our last step would be to protonate our ring. And so we have ammonia. So ammonia comes along. It's going to function as an acid and donate a proton."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Our last step would be to protonate our ring. And so we have ammonia. So ammonia comes along. It's going to function as an acid and donate a proton. So these electrons in magenta are going to pick up a proton here, leaving these electrons behind. And we protonate our ring. And we're done with our mechanism."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's going to function as an acid and donate a proton. So these electrons in magenta are going to pick up a proton here, leaving these electrons behind. And we protonate our ring. And we're done with our mechanism. We have formed aniline as our product like that. All right, so that's the mechanism for elimination addition. Let's see if we can go ahead and do a practice problem here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're done with our mechanism. We have formed aniline as our product like that. All right, so that's the mechanism for elimination addition. Let's see if we can go ahead and do a practice problem here. And here we have a disubstituted ring. And our two substituents are para to each other. So we're going to add our strong base, so sodium amide, so negative 1 formal charge here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's see if we can go ahead and do a practice problem here. And here we have a disubstituted ring. And our two substituents are para to each other. So we're going to add our strong base, so sodium amide, so negative 1 formal charge here. First step is elimination. So first step is an acid-base reaction where the amide takes a proton that's next to our halogen. And so we have two choices here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to add our strong base, so sodium amide, so negative 1 formal charge here. First step is elimination. So first step is an acid-base reaction where the amide takes a proton that's next to our halogen. And so we have two choices here. We could take this proton that's next to our halogen. Or we could take this proton that's next to our halogen. In this case, it doesn't matter because they are equivalent, pretty much the same thing if you think about the symmetry of the molecule."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we have two choices here. We could take this proton that's next to our halogen. Or we could take this proton that's next to our halogen. In this case, it doesn't matter because they are equivalent, pretty much the same thing if you think about the symmetry of the molecule. And so this is going to take, I'm just going to say it's going to take this proton to make our lives easier here. And then these electrons would remain behind on this carbon. So let's go ahead and show our acid-base reaction here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In this case, it doesn't matter because they are equivalent, pretty much the same thing if you think about the symmetry of the molecule. And so this is going to take, I'm just going to say it's going to take this proton to make our lives easier here. And then these electrons would remain behind on this carbon. So let's go ahead and show our acid-base reaction here. So we have our ring. We have our methyl group here on our ring. We have our halogen here on our ring."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show our acid-base reaction here. So we have our ring. We have our methyl group here on our ring. We have our halogen here on our ring. And now we have a negative 1 formal charge on this carbon. So we have electrons on that carbon, negative 1 formal charge. These are going to move into here to form a triple bond."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our halogen here on our ring. And now we have a negative 1 formal charge on this carbon. So we have electrons on that carbon, negative 1 formal charge. These are going to move into here to form a triple bond. And these electrons kick off onto our leaving group. So the chloride anion leaves. And we form benzyne."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "These are going to move into here to form a triple bond. And these electrons kick off onto our leaving group. So the chloride anion leaves. And we form benzyne. So we form our benzyne intermediate. And so I can go ahead and show that right here. So there's benzyne."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we form benzyne. So we form our benzyne intermediate. And so I can go ahead and show that right here. So there's benzyne. There's still a methyl group attached to our ring. Now once again, I chose to show this proton participating in the acid-base reaction. It could have been, I could have showed it in this one."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there's benzyne. There's still a methyl group attached to our ring. Now once again, I chose to show this proton participating in the acid-base reaction. It could have been, I could have showed it in this one. But it's just helpful the way I drew my benzene ring. I happen to show my pi bond right here. And so it's easier to show it there."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It could have been, I could have showed it in this one. But it's just helpful the way I drew my benzene ring. I happen to show my pi bond right here. And so it's easier to show it there. But you only get one benzyne intermediate here. And so here is our benzyne intermediate, which next is going to react with the amide anion, which is now going to function as a nucleophile. So I have a negative 1 formal charge on my nitrogen."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it's easier to show it there. But you only get one benzyne intermediate here. And so here is our benzyne intermediate, which next is going to react with the amide anion, which is now going to function as a nucleophile. So I have a negative 1 formal charge on my nitrogen. And it's going to function as a nucleophile. And it could attack either side of that triple bond. I'm going to go ahead and show it attacking this side of our triple bond."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I have a negative 1 formal charge on my nitrogen. And it's going to function as a nucleophile. And it could attack either side of that triple bond. I'm going to go ahead and show it attacking this side of our triple bond. And so these electrons would come off onto there. So let's go ahead and show the result of that. And so I have my ring."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and show it attacking this side of our triple bond. And so these electrons would come off onto there. So let's go ahead and show the result of that. And so I have my ring. And I'm going to add on my NH2 to this top carbon this time. And my electrons went off over here onto this carbon. So I still have a methyl group like that."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I have my ring. And I'm going to add on my NH2 to this top carbon this time. And my electrons went off over here onto this carbon. So I still have a methyl group like that. Now I could have showed the amide anion attacking the other side of the triple bond. That's possible too. So I could have thought about my nucleophile attacking over here, which would push these electrons off onto that carbon."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I still have a methyl group like that. Now I could have showed the amide anion attacking the other side of the triple bond. That's possible too. So I could have thought about my nucleophile attacking over here, which would push these electrons off onto that carbon. So let's go ahead and show that possibility as well. So once again, we have our ring. So we have these electrons on our ring."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could have thought about my nucleophile attacking over here, which would push these electrons off onto that carbon. So let's go ahead and show that possibility as well. So once again, we have our ring. So we have these electrons on our ring. We have our methyl group. This time, the NH2 added on to the right side of where our triple bond used to be. And these electrons ended up over here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have these electrons on our ring. We have our methyl group. This time, the NH2 added on to the right side of where our triple bond used to be. And these electrons ended up over here. And so of course, the final step, which I won't show the mechanism for, we just need to protonate that anion. And we end up with our two final products. And so let me go ahead and draw those."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And these electrons ended up over here. And so of course, the final step, which I won't show the mechanism for, we just need to protonate that anion. And we end up with our two final products. And so let me go ahead and draw those. So we have our two groups. And in our top example, our para to each other. And then our other products, our two groups on our ring are meta to each other."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so let me go ahead and draw those. So we have our two groups. And in our top example, our para to each other. And then our other products, our two groups on our ring are meta to each other. And so you get a mixture of products here. So thinking about your benzyne mechanism. So let's look at one more type of elimination of a reaction using the benzyne mechanism."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then our other products, our two groups on our ring are meta to each other. And so you get a mixture of products here. So thinking about your benzyne mechanism. So let's look at one more type of elimination of a reaction using the benzyne mechanism. So another famous one. And you can see this time, instead of using sodium amide, we're using sodium hydroxide. But we know that this is a strong base."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at one more type of elimination of a reaction using the benzyne mechanism. So another famous one. And you can see this time, instead of using sodium amide, we're using sodium hydroxide. But we know that this is a strong base. And so that will work. You have to heat it up a lot and add some water, some proton sources. And you can see that eventually, you're going to get an OH group substituting in for your halogen."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But we know that this is a strong base. And so that will work. You have to heat it up a lot and add some water, some proton sources. And you can see that eventually, you're going to get an OH group substituting in for your halogen. So this is a very famous reaction. Let's see if we can do this problem down here, where we have a methyl group on our ring. And we have our chlorine like that."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you can see that eventually, you're going to get an OH group substituting in for your halogen. So this is a very famous reaction. Let's see if we can do this problem down here, where we have a methyl group on our ring. And we have our chlorine like that. So this is a little bit different from the one that we just did. So these two are meta to each other. And those other ones were para."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have our chlorine like that. So this is a little bit different from the one that we just did. So these two are meta to each other. And those other ones were para. So once again, we're going to think about the hydroxide anion functioning as a strong base. And it's going to take a proton off of our ring. Remember, it's going to take a proton off next to our leaving group."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And those other ones were para. So once again, we're going to think about the hydroxide anion functioning as a strong base. And it's going to take a proton off of our ring. Remember, it's going to take a proton off next to our leaving group. And so it could take a proton off from here. Or it could take a proton off from here. And those are two different positions this time."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Remember, it's going to take a proton off next to our leaving group. And so it could take a proton off from here. Or it could take a proton off from here. And those are two different positions this time. So we're going to get different benzyne intermediates. So we need to think about two different benzynes. So that's different from the previous example."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And those are two different positions this time. So we're going to get different benzyne intermediates. So we need to think about two different benzynes. So that's different from the previous example. We only had to think about one benzyne. So first, I'm going to think about the proton on the right. So if this proton leaves, you can think about these electrons in here moving in to form your triple bond."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that's different from the previous example. We only had to think about one benzyne. So first, I'm going to think about the proton on the right. So if this proton leaves, you can think about these electrons in here moving in to form your triple bond. So let's go ahead and draw that one first. So the elimination of your halogen would give you your benzyne intermediate. And your benzyne intermediate would look like this with your methyl group being right here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if this proton leaves, you can think about these electrons in here moving in to form your triple bond. So let's go ahead and draw that one first. So the elimination of your halogen would give you your benzyne intermediate. And your benzyne intermediate would look like this with your methyl group being right here. And so that's one possible benzyne intermediate. Of course, we could have taken off this proton. So sodium hydroxide could have taken off that proton."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And your benzyne intermediate would look like this with your methyl group being right here. And so that's one possible benzyne intermediate. Of course, we could have taken off this proton. So sodium hydroxide could have taken off that proton. And your halogen could have left in the elimination step to form another benzyne. And this time, it would be between this hydrogen and the chlorine. So it would be over here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So sodium hydroxide could have taken off that proton. And your halogen could have left in the elimination step to form another benzyne. And this time, it would be between this hydrogen and the chlorine. So it would be over here. So benzyne would have to be over on that side. And so we could go ahead and draw our benzyne like that. Now, notice I'm showing a different resonance form of the pi electrons in the ring."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it would be over here. So benzyne would have to be over on that side. And so we could go ahead and draw our benzyne like that. Now, notice I'm showing a different resonance form of the pi electrons in the ring. But we know that that's OK. I just happened to start out with my pi electrons in my benzene ring like this. But I could have just as easily have started out the opposite way."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, notice I'm showing a different resonance form of the pi electrons in the ring. But we know that that's OK. I just happened to start out with my pi electrons in my benzene ring like this. But I could have just as easily have started out the opposite way. So what I drew here is not incorrect. It's just a different way of showing it. And it's necessary for us to show the benzyne intermediate this way."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But I could have just as easily have started out the opposite way. So what I drew here is not incorrect. It's just a different way of showing it. And it's necessary for us to show the benzyne intermediate this way. So now we have two benzyne intermediates. And in the next step, hydroxide would function as a nucleophile. And it could attack either side."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it's necessary for us to show the benzyne intermediate this way. So now we have two benzyne intermediates. And in the next step, hydroxide would function as a nucleophile. And it could attack either side. So let's go ahead and think about the hydroxide anion functioning as a nucleophile. And it could attack. Let's go ahead and draw the possible products."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it could attack either side. So let's go ahead and think about the hydroxide anion functioning as a nucleophile. And it could attack. Let's go ahead and draw the possible products. I could think about hydroxide attacking this side of the triple bond, in which case our final product would look like this. We would have our ring. We would have our methyl group here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw the possible products. I could think about hydroxide attacking this side of the triple bond, in which case our final product would look like this. We would have our ring. We would have our methyl group here. And then we would have our OH group para to that methyl group. So that's one of our possible products. So let me go ahead and use a different color so we can see that better."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We would have our methyl group here. And then we would have our OH group para to that methyl group. So that's one of our possible products. So let me go ahead and use a different color so we can see that better. So nucleophilic attack here would give us this product. Nucleophilic attack here would give us a different product. So let's go ahead and draw that."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and use a different color so we can see that better. So nucleophilic attack here would give us this product. Nucleophilic attack here would give us a different product. So let's go ahead and draw that. We would show our ring. And we would show our pi electrons. We would show our OH adding here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that. We would show our ring. And we would show our pi electrons. We would show our OH adding here. And we would still have this methyl group. And so you can see these are different products. So we have two different products so far."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We would show our OH adding here. And we would still have this methyl group. And so you can see these are different products. So we have two different products so far. And once again, we could go over to this molecule. And we could show the same thing. We show hydroxide attacking the left side of our triple bond."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have two different products so far. And once again, we could go over to this molecule. And we could show the same thing. We show hydroxide attacking the left side of our triple bond. So let's go ahead and draw that product. So if hydroxide attacked the left side, it would add on to the left side of where our triple bond used to be. And now we would have this product."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We show hydroxide attacking the left side of our triple bond. So let's go ahead and draw that product. So if hydroxide attacked the left side, it would add on to the left side of where our triple bond used to be. And now we would have this product. So that's another possible product. And now, of course, we could have attack on the right side of our triple bond for this one. So let's go ahead and draw the fourth possibility here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now we would have this product. So that's another possible product. And now, of course, we could have attack on the right side of our triple bond for this one. So let's go ahead and draw the fourth possibility here. And so now we have our benzene ring. We have our OH group is added onto here. And then we have a methyl group right here."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the fourth possibility here. And so now we have our benzene ring. We have our OH group is added onto here. And then we have a methyl group right here. And so if you look at it, you can see that these two are actually the same molecule. Because these benzene rings just draw on the pi electrons in different places. But we know that those are just resonance structures of each other."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a methyl group right here. And so if you look at it, you can see that these two are actually the same molecule. Because these benzene rings just draw on the pi electrons in different places. But we know that those are just resonance structures of each other. And so we get a total of three products for this reaction. So this would just be one molecule. And then this would be two."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But we know that those are just resonance structures of each other. And so we get a total of three products for this reaction. So this would just be one molecule. And then this would be two. And then this would be three. So this would be my approach to answering this question on an exam. You don't have time to draw out the full mechanism."}, {"video_title": "Nucleophilic aromatic substitution II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then this would be two. And then this would be three. So this would be my approach to answering this question on an exam. You don't have time to draw out the full mechanism. So think about eliminating your halogen to form the benzyne intermediate. Think about how many benzyne intermediates are possible. And then finally, think about adding the nucleophile to either side of your triple bond."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And if we wanted to name ethers, there are a couple different ways to do it. We'll start off with the common way of naming ethers, which follows the pattern of first naming one of the alkyl groups attached to your oxygen, and then naming the other alkyl group, and then followed by the word ether like that. So alkyl, alkyl, ether. Let's look at an example of common nomenclature. And we'll start with this molecule here. So if I wanted to name this ether, I would first focus in on what sort of alkyl groups do we have attached to our oxygen. Over here on the left, that's a methyl group."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an example of common nomenclature. And we'll start with this molecule here. So if I wanted to name this ether, I would first focus in on what sort of alkyl groups do we have attached to our oxygen. Over here on the left, that's a methyl group. And over here on the right, that's an ethyl group. So we need to think about the alphabet rule. And we know that E comes before M. So we're going to write ethyl before methyl."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Over here on the left, that's a methyl group. And over here on the right, that's an ethyl group. So we need to think about the alphabet rule. And we know that E comes before M. So we're going to write ethyl before methyl. So it will be called ethyl methyl ether like that. So let's do another one. So our goal is to name this ether."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we know that E comes before M. So we're going to write ethyl before methyl. So it will be called ethyl methyl ether like that. So let's do another one. So our goal is to name this ether. And we first, of course, take a look at the alkyl groups that we have attached to our oxygen. So on the left here, we recognize this as being a tert-butyl group. And on the right, we have a methyl group."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So our goal is to name this ether. And we first, of course, take a look at the alkyl groups that we have attached to our oxygen. So on the left here, we recognize this as being a tert-butyl group. And on the right, we have a methyl group. So in thinking about using the alphabet rule, B comes before M. So the tert-butyl group is going to come before the methyl group in the name. So it should be called tert-butyl methyl ether like that. So the tert part isn't a part of the alphabet."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And on the right, we have a methyl group. So in thinking about using the alphabet rule, B comes before M. So the tert-butyl group is going to come before the methyl group in the name. So it should be called tert-butyl methyl ether like that. So the tert part isn't a part of the alphabet. You're comparing the butyl versus the methyl like that. Now, tert-butyl methyl ether might be the correct name. Although that's probably not what you will hear this molecule called in the laboratory."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the tert part isn't a part of the alphabet. You're comparing the butyl versus the methyl like that. Now, tert-butyl methyl ether might be the correct name. Although that's probably not what you will hear this molecule called in the laboratory. Usually, you'll hear this called methyl tert-butyl ether or MTBE. So MTBE is a very common organic solvent. It works very well for a number of things."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Although that's probably not what you will hear this molecule called in the laboratory. Usually, you'll hear this called methyl tert-butyl ether or MTBE. So MTBE is a very common organic solvent. It works very well for a number of things. And again, MTBE is what you will hear. But technically, that's not the correct name. Let's look at an ether that has alkyl groups that are the same."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It works very well for a number of things. And again, MTBE is what you will hear. But technically, that's not the correct name. Let's look at an ether that has alkyl groups that are the same. So if we have this as our ether, we have two ethyl groups for this one. So we're going to call this diethyl ether. So diethyl ether."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an ether that has alkyl groups that are the same. So if we have this as our ether, we have two ethyl groups for this one. So we're going to call this diethyl ether. So diethyl ether. So you can use the prefix of di here. Now, diethyl ether, of course, is the famous one. This is the one that everyone thinks of when they think about ethers."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So diethyl ether. So you can use the prefix of di here. Now, diethyl ether, of course, is the famous one. This is the one that everyone thinks of when they think about ethers. Let's look at another way to name ethers. And this is the official IUPAC way of naming them. Now, the common way of naming ethers is so common that it's accepted by IUPAC."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This is the one that everyone thinks of when they think about ethers. Let's look at another way to name ethers. And this is the official IUPAC way of naming them. Now, the common way of naming ethers is so common that it's accepted by IUPAC. It's a common nomenclature. But there is an IUPAC way of doing it for more complicated molecules. And that is to name your ether as a substituent, which we call an alkoxy."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now, the common way of naming ethers is so common that it's accepted by IUPAC. It's a common nomenclature. But there is an IUPAC way of doing it for more complicated molecules. And that is to name your ether as a substituent, which we call an alkoxy. So we're going to name ethers as a substituent and then have a parent alkane. So we're going to call these alkoxy alkanes, like that. So what you would do is if you had just this generic ether here like this, ROR prime, you would find the larger alkyl group."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And that is to name your ether as a substituent, which we call an alkoxy. So we're going to name ethers as a substituent and then have a parent alkane. So we're going to call these alkoxy alkanes, like that. So what you would do is if you had just this generic ether here like this, ROR prime, you would find the larger alkyl group. And that would be your parent name. So let's say the R prime group was my longer carbon chain. That would be my parent alkane name, like that."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So what you would do is if you had just this generic ether here like this, ROR prime, you would find the larger alkyl group. And that would be your parent name. So let's say the R prime group was my longer carbon chain. That would be my parent alkane name, like that. And then you would name this portion of the molecule as an alkoxy substituent on your alkane. So let's look at an example of IUPAC nomenclature. So let's go ahead and we'll name that first ether that we did a few minutes ago."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That would be my parent alkane name, like that. And then you would name this portion of the molecule as an alkoxy substituent on your alkane. So let's look at an example of IUPAC nomenclature. So let's go ahead and we'll name that first ether that we did a few minutes ago. So the very first one we did, we were doing common nomenclature. We called this ethyl methyl ether. So let's go ahead and name it using IUPAC nomenclature here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and we'll name that first ether that we did a few minutes ago. So the very first one we did, we were doing common nomenclature. We called this ethyl methyl ether. So let's go ahead and name it using IUPAC nomenclature here. So what we would do is find the larger group. And that's going to be my parent name. So my larger group would be this over here on the right."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and name it using IUPAC nomenclature here. So what we would do is find the larger group. And that's going to be my parent name. So my larger group would be this over here on the right. And if we were to number that, you could say that's number one and number two right here. Number one and number two, if you wanted to. You don't really have to for this example."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So my larger group would be this over here on the right. And if we were to number that, you could say that's number one and number two right here. Number one and number two, if you wanted to. You don't really have to for this example. But just to get you thinking about longest carbon chain, right, that would be ethane right there. So we'll go ahead and write ethane as our parent alkane name. And then what do we have coming off of the ethane portion of the molecule here?"}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "You don't really have to for this example. But just to get you thinking about longest carbon chain, right, that would be ethane right there. So we'll go ahead and write ethane as our parent alkane name. And then what do we have coming off of the ethane portion of the molecule here? So what is our substituent? Well, this is the ether portion. We're going to name it as an alkoxy."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then what do we have coming off of the ethane portion of the molecule here? So what is our substituent? Well, this is the ether portion. We're going to name it as an alkoxy. And since we have one carbon to deal with, we know that our root is meth. And as an alkoxy, it'd be methoxy. So the complete IUPAC name for this molecule would be methoxyethane, like that."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We're going to name it as an alkoxy. And since we have one carbon to deal with, we know that our root is meth. And as an alkoxy, it'd be methoxy. So the complete IUPAC name for this molecule would be methoxyethane, like that. And we don't have to worry about numbers, since we have only two carbons on our parent chain here. So methoxyethane would be the IUPAC name for this molecule. The common name would be ethylmethyl ether."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the complete IUPAC name for this molecule would be methoxyethane, like that. And we don't have to worry about numbers, since we have only two carbons on our parent chain here. So methoxyethane would be the IUPAC name for this molecule. The common name would be ethylmethyl ether. Both are acceptable names. Let's do another one where we have a very similar looking molecule, except in this case, we're going to add on another carbon there. So if I'm thinking about my parent alkane over here on the right, there would be three carbons for my parent alkane."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "The common name would be ethylmethyl ether. Both are acceptable names. Let's do another one where we have a very similar looking molecule, except in this case, we're going to add on another carbon there. So if I'm thinking about my parent alkane over here on the right, there would be three carbons for my parent alkane. So if I wanted to number that, it'd be 1, 2, and 3. A three-carbon alkane is, of course, propane. So I go ahead and write propane right here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm thinking about my parent alkane over here on the right, there would be three carbons for my parent alkane. So if I wanted to number that, it'd be 1, 2, and 3. A three-carbon alkane is, of course, propane. So I go ahead and write propane right here. And I look at my ether substituent once again. And I look and see how many carbons I have. And there's one carbon."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So I go ahead and write propane right here. And I look at my ether substituent once again. And I look and see how many carbons I have. And there's one carbon. There's one carbon on the substituent. So once again, it would be methoxy. So I'll go ahead and write methoxy in here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And there's one carbon. There's one carbon on the substituent. So once again, it would be methoxy. So I'll go ahead and write methoxy in here. So methoxypropane. OK, this time I need to put a number on there. And that group is coming off of carbon 2."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and write methoxy in here. So methoxypropane. OK, this time I need to put a number on there. And that group is coming off of carbon 2. So the complete IUPAC name would be 2-methoxypropane, like that. Let's do a much more complicated one that has a little bit of stereochemistry in it. So if this was the molecule that I was trying to name."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And that group is coming off of carbon 2. So the complete IUPAC name would be 2-methoxypropane, like that. Let's do a much more complicated one that has a little bit of stereochemistry in it. So if this was the molecule that I was trying to name. And let's go ahead and put a bromine here like that. All right, so for this one, once again, I have to think about the larger group as my parent name. So if I look at those two alkyl groups, the alkyl group on the left looks like the longest one to me."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if this was the molecule that I was trying to name. And let's go ahead and put a bromine here like that. All right, so for this one, once again, I have to think about the larger group as my parent name. So if I look at those two alkyl groups, the alkyl group on the left looks like the longest one to me. And I want to number to give my substituents the lowest number possible. So if I look over here on the left, I can see that there are four carbons in my larger substituent. So four carbons is going to be my parent name here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at those two alkyl groups, the alkyl group on the left looks like the longest one to me. And I want to number to give my substituents the lowest number possible. So if I look over here on the left, I can see that there are four carbons in my larger substituent. So four carbons is going to be my parent name here. So I'm going to call this butane. So let's go ahead and start naming it with butane right here. So this would be butane so far."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So four carbons is going to be my parent name here. So I'm going to call this butane. So let's go ahead and start naming it with butane right here. So this would be butane so far. Now when I number that butane, I want to get the lowest numbers possible to my substituents. So I could start from the left or I could start from the right. And starting from the right makes more sense because I have a substituent coming off of carbon 1."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this would be butane so far. Now when I number that butane, I want to get the lowest numbers possible to my substituents. So I could start from the left or I could start from the right. And starting from the right makes more sense because I have a substituent coming off of carbon 1. I have a substituent coming off of carbon 2, and then 3, and 4, like that. So if I'm thinking about those two substituents, let's think about how I would name them. So over here on the right for my alkoxy substituent, this time there are two carbons in my alkoxy substituent right here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And starting from the right makes more sense because I have a substituent coming off of carbon 1. I have a substituent coming off of carbon 2, and then 3, and 4, like that. So if I'm thinking about those two substituents, let's think about how I would name them. So over here on the right for my alkoxy substituent, this time there are two carbons in my alkoxy substituent right here. So two carbons would be eth, so that would be ethoxy. And ethoxy is coming off of carbon 1. So I can go ahead and write that in here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the right for my alkoxy substituent, this time there are two carbons in my alkoxy substituent right here. So two carbons would be eth, so that would be ethoxy. And ethoxy is coming off of carbon 1. So I can go ahead and write that in here. So 1-ethoxybutane is what I have so far. And I also have a bromine coming off of carbon 2. So it would be 2-bromo."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and write that in here. So 1-ethoxybutane is what I have so far. And I also have a bromine coming off of carbon 2. So it would be 2-bromo. So I can go ahead and put in 2-bromo, 1-ethoxy. And that follows the alphabet because B comes before E. So 2-bromo, 1-ethoxybutane. And then we have to worry about the absolute configuration at this carbon right here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it would be 2-bromo. So I can go ahead and put in 2-bromo, 1-ethoxy. And that follows the alphabet because B comes before E. So 2-bromo, 1-ethoxybutane. And then we have to worry about the absolute configuration at this carbon right here. So carbon 2 is a chirality center. So we need to think about how to assign priority to those four groups. So if I think about the atoms directly attached to my chirality center, first let's go ahead and identify my chirality center."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then we have to worry about the absolute configuration at this carbon right here. So carbon 2 is a chirality center. So we need to think about how to assign priority to those four groups. So if I think about the atoms directly attached to my chirality center, first let's go ahead and identify my chirality center. That would be this one right here. Four different groups attached to it because there's also a hydrogen going away from me in space. And I think about atomic numbers."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I think about the atoms directly attached to my chirality center, first let's go ahead and identify my chirality center. That would be this one right here. Four different groups attached to it because there's also a hydrogen going away from me in space. And I think about atomic numbers. So I have carbon versus carbon versus bromine right here. So bromine, of course, has the highest atomic number. It gets highest priority like that."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And I think about atomic numbers. So I have carbon versus carbon versus bromine right here. So bromine, of course, has the highest atomic number. It gets highest priority like that. Now my hydrogen, of course, is going to get lowest priority. So that's priority number four. And now I have two groups to worry about."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It gets highest priority like that. Now my hydrogen, of course, is going to get lowest priority. So that's priority number four. And now I have two groups to worry about. So I have two carbons to worry about. Let's go ahead and mark those carbons again. So which one of these carbons is going to get higher priority?"}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And now I have two groups to worry about. So I have two carbons to worry about. Let's go ahead and mark those carbons again. So which one of these carbons is going to get higher priority? Well, it's all about what they're attached to. So the carbon on the left is attached to another carbon and two hydrogens. The carbon on the right is attached to an oxygen and two hydrogens."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So which one of these carbons is going to get higher priority? Well, it's all about what they're attached to. So the carbon on the left is attached to another carbon and two hydrogens. The carbon on the right is attached to an oxygen and two hydrogens. So in terms of atomic number, carbon versus oxygen, the oxygen will win. And this substituent on the right would get the highest priority. So this would get a 2 over here."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the right is attached to an oxygen and two hydrogens. So in terms of atomic number, carbon versus oxygen, the oxygen will win. And this substituent on the right would get the highest priority. So this would get a 2 over here. All this stuff over there would get a 2. And over here on the left, this would be a 3. So we have 1, 2, 3 going around this way, going around counterclockwise, which is the S absolute configuration."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this would get a 2 over here. All this stuff over there would get a 2. And over here on the left, this would be a 3. So we have 1, 2, 3 going around this way, going around counterclockwise, which is the S absolute configuration. So this is S 2 bromo 1 ethoxy butane for the final name. Let's do one more example of naming an ether here. So let's go ahead and look at one that has a ring."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we have 1, 2, 3 going around this way, going around counterclockwise, which is the S absolute configuration. So this is S 2 bromo 1 ethoxy butane for the final name. Let's do one more example of naming an ether here. So let's go ahead and look at one that has a ring. And we'll put a double bond in our ring like that. And then we'll have our ether over here on the right. So if I wanted to name this molecule, I would think about my two alkyl groups and think about which one is the larger one."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and look at one that has a ring. And we'll put a double bond in our ring like that. And then we'll have our ether over here on the right. So if I wanted to name this molecule, I would think about my two alkyl groups and think about which one is the larger one. And of course, all this stuff on the left is going to be my parent name. And then this is going to be my alkoxy substituent like that. So on the left, I know what that molecule is."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I wanted to name this molecule, I would think about my two alkyl groups and think about which one is the larger one. And of course, all this stuff on the left is going to be my parent name. And then this is going to be my alkoxy substituent like that. So on the left, I know what that molecule is. I know that's called cyclohexene from an earlier video. So this is cyclohexene as my parent name. So cyclohexene."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, I know what that molecule is. I know that's called cyclohexene from an earlier video. So this is cyclohexene as my parent name. So cyclohexene. I now need to number my ring to give my alkoxy substituent the lowest number possible. So if I wanted to number my ring to give my alkoxy substituent the lowest number possible, I should start here and make that 1, 2, 3, and 4 like that. So we think about what is that alkoxy substituent."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So cyclohexene. I now need to number my ring to give my alkoxy substituent the lowest number possible. So if I wanted to number my ring to give my alkoxy substituent the lowest number possible, I should start here and make that 1, 2, 3, and 4 like that. So we think about what is that alkoxy substituent. It is an ethoxy group, because an ethoxy substituent, because I have two carbons right there. So I have an ethoxy coming off of carbon 4. So this would be 4-ethoxy."}, {"video_title": "Ether nomenclature Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we think about what is that alkoxy substituent. It is an ethoxy group, because an ethoxy substituent, because I have two carbons right there. So I have an ethoxy coming off of carbon 4. So this would be 4-ethoxy. And if you wanted to, you could put the 1 in here, 1-cyclohexene. You could leave it out if you wanted to. So it doesn't really matter."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you start with an aldehyde or a ketone and add a catalytic amount of acid or base, you'll find the aldehyde or ketone is going to be in equilibrium with this product over here on the right, which we call an enol. So the name enol comes from the fact that we have a double bond in the molecule, so that's where the ene part comes in. And we also have an alcohol, right? You can see the OH over here, so that's where the OL comes in. So this is the enol form. And then over here, this is the keto form. So the keto form and the enol form."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can see the OH over here, so that's where the OL comes in. So this is the enol form. And then over here, this is the keto form. So the keto form and the enol form. And these are different molecules. They're isomers of each other, so we call them tautomers. And they're in equilibrium with each other."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the keto form and the enol form. And these are different molecules. They're isomers of each other, so we call them tautomers. And they're in equilibrium with each other. They're not different resonance structures. Let's see if we can analyze our aldehyde or ketone to see how to form our enol. And so if we look at the carbon that's next to the carbonyl carbon, we call this the alpha carbon."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And they're in equilibrium with each other. They're not different resonance structures. Let's see if we can analyze our aldehyde or ketone to see how to form our enol. And so if we look at the carbon that's next to the carbonyl carbon, we call this the alpha carbon. And there are two hydrogens attached to the alpha carbon in this case, so let me go ahead and draw those in. And those are called the alpha protons. And so if we think about transferring one of those alpha protons from the alpha carbon to the oxygen, even though it's most likely not the same proton, it just helps to think about doing that."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so if we look at the carbon that's next to the carbonyl carbon, we call this the alpha carbon. And there are two hydrogens attached to the alpha carbon in this case, so let me go ahead and draw those in. And those are called the alpha protons. And so if we think about transferring one of those alpha protons from the alpha carbon to the oxygen, even though it's most likely not the same proton, it just helps to think about doing that. We can also think about moving the double bond, right? So over here on the left, the double bond is between the carbon and the oxygen, and we're moving that double bond over here between the two carbons. So transferring one alpha proton and shifting your double bond converts the keto form into the enol form."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so if we think about transferring one of those alpha protons from the alpha carbon to the oxygen, even though it's most likely not the same proton, it just helps to think about doing that. We can also think about moving the double bond, right? So over here on the left, the double bond is between the carbon and the oxygen, and we're moving that double bond over here between the two carbons. So transferring one alpha proton and shifting your double bond converts the keto form into the enol form. And then we also have a hydrogen, right? So over here, we still have a hydrogen left on this carbon. So let me go ahead and draw in that hydrogen."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So transferring one alpha proton and shifting your double bond converts the keto form into the enol form. And then we also have a hydrogen, right? So over here, we still have a hydrogen left on this carbon. So let me go ahead and draw in that hydrogen. So that's this hydrogen in blue here. So that's how to think about converting a keto tautomer into an enol one. Let's look at the acid catalyzed mechanism for this."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw in that hydrogen. So that's this hydrogen in blue here. So that's how to think about converting a keto tautomer into an enol one. Let's look at the acid catalyzed mechanism for this. So if we start with our aldehyde or ketone and add H3O+, the first thing that's gonna happen is protonation of our carbonyl. And so a lone pair of electrons picks up this proton like that. So we can go ahead and draw that."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the acid catalyzed mechanism for this. So if we start with our aldehyde or ketone and add H3O+, the first thing that's gonna happen is protonation of our carbonyl. And so a lone pair of electrons picks up this proton like that. So we can go ahead and draw that. We would protonate our carbonyl. So now our oxygen would have a plus one formal charge. Let me just go ahead and draw in those electrons here."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and draw that. We would protonate our carbonyl. So now our oxygen would have a plus one formal charge. Let me just go ahead and draw in those electrons here. And let's say we started with an aldehyde, so we'll make this an H. So the lone pair of electrons on our oxygen, right, picked up a proton like that. We can draw a resonance structure for this, right? We could move these electrons off onto our oxygen."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me just go ahead and draw in those electrons here. And let's say we started with an aldehyde, so we'll make this an H. So the lone pair of electrons on our oxygen, right, picked up a proton like that. We can draw a resonance structure for this, right? We could move these electrons off onto our oxygen. So let's go ahead and show a resonance structure. So we would have our R group, right? And now we would have our oxygen with two lone pairs of electrons."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could move these electrons off onto our oxygen. So let's go ahead and show a resonance structure. So we would have our R group, right? And now we would have our oxygen with two lone pairs of electrons. Let me go ahead and draw in those two lone pairs of electrons on our oxygen. And then we took a bond away from carbon, right? So we took a bond away from this carbon."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And now we would have our oxygen with two lone pairs of electrons. Let me go ahead and draw in those two lone pairs of electrons on our oxygen. And then we took a bond away from carbon, right? So we took a bond away from this carbon. So this carbon right here. So plus one formal charge on that carbon. And then we could show the movement of those electrons."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we took a bond away from this carbon. So this carbon right here. So plus one formal charge on that carbon. And then we could show the movement of those electrons. So these electrons right here, I'm saying moving out onto the oxygen like that. And so this is our intermediate here. All right, so we know that our alpha carbon has two protons on it."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we could show the movement of those electrons. So these electrons right here, I'm saying moving out onto the oxygen like that. And so this is our intermediate here. All right, so we know that our alpha carbon has two protons on it. So once again, let's find our alpha carbon. Here it is right here. We know we have two protons attached to it, two alpha protons, if you will."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we know that our alpha carbon has two protons on it. So once again, let's find our alpha carbon. Here it is right here. We know we have two protons attached to it, two alpha protons, if you will. And so in the next step of our mechanism, we're gonna get a molecule of water acting as a base. Let me go ahead and show a molecule of water here. And the water's gonna take one of those alpha protons, right, let's say once again it takes this alpha proton, and leave these electrons behind."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know we have two protons attached to it, two alpha protons, if you will. And so in the next step of our mechanism, we're gonna get a molecule of water acting as a base. Let me go ahead and show a molecule of water here. And the water's gonna take one of those alpha protons, right, let's say once again it takes this alpha proton, and leave these electrons behind. They're gonna move in here to form our double bonds. So let's go ahead and draw our product, right? So we would have our R group here."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the water's gonna take one of those alpha protons, right, let's say once again it takes this alpha proton, and leave these electrons behind. They're gonna move in here to form our double bonds. So let's go ahead and draw our product, right? So we would have our R group here. And now we would have our double bond form between our two carbons, and then we would have our oxygen, and then we would have two lone pairs of electrons on our oxygen. We would have our hydrogen, and then we would have another hydrogen right here. So let's go ahead and follow some of those electrons."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our R group here. And now we would have our double bond form between our two carbons, and then we would have our oxygen, and then we would have two lone pairs of electrons on our oxygen. We would have our hydrogen, and then we would have another hydrogen right here. So let's go ahead and follow some of those electrons. All right, so let's go ahead and make these electrons in here blue. All right, so these electrons are gonna move in here. It doesn't really matter which one you say it is."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and follow some of those electrons. All right, so let's go ahead and make these electrons in here blue. All right, so these electrons are gonna move in here. It doesn't really matter which one you say it is. Let's just say it's that one, right, to form our double bond. And then the electrons in red moved off onto this oxygen, and then we said that these electrons were in magenta. And so you can see that we have formed our enol here, right?"}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't really matter which one you say it is. Let's just say it's that one, right, to form our double bond. And then the electrons in red moved off onto this oxygen, and then we said that these electrons were in magenta. And so you can see that we have formed our enol here, right? So this is our enol, right? And then we started with our keto form like that. So keto enol tautomerization."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so you can see that we have formed our enol here, right? So this is our enol, right? And then we started with our keto form like that. So keto enol tautomerization. Let's look at the base catalyzed version, right? So once again, we start with our aldehyde or ketone. But this time, we're going to add a base, so something like hydroxide."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So keto enol tautomerization. Let's look at the base catalyzed version, right? So once again, we start with our aldehyde or ketone. But this time, we're going to add a base, so something like hydroxide. And so we find our alpha carbon. So here's our alpha carbon, once again with two alpha protons. So I'm gonna go ahead and draw in those two protons here."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But this time, we're going to add a base, so something like hydroxide. And so we find our alpha carbon. So here's our alpha carbon, once again with two alpha protons. So I'm gonna go ahead and draw in those two protons here. And the base is gonna take one of those protons. Let's say it takes this one over here on the right. That leaves these electrons behind on this carbon."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna go ahead and draw in those two protons here. And the base is gonna take one of those protons. Let's say it takes this one over here on the right. That leaves these electrons behind on this carbon. So let's go ahead and draw the resulting anion here. So we would have our carbonyl like that, right? And once again, let's say we started with an aldehyde."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That leaves these electrons behind on this carbon. So let's go ahead and draw the resulting anion here. So we would have our carbonyl like that, right? And once again, let's say we started with an aldehyde. And then we would have a lone pair of electrons on this carbon, the carbon in red here. So let me go ahead and identify those electrons. So these electrons in here in magenta have moved off onto this carbon like that, which gives that carbon a negative one formal charge."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And once again, let's say we started with an aldehyde. And then we would have a lone pair of electrons on this carbon, the carbon in red here. So let me go ahead and identify those electrons. So these electrons in here in magenta have moved off onto this carbon like that, which gives that carbon a negative one formal charge. It's a carb anion. There's still a hydrogen attached to that carbon in red. This hydrogen right here is still attached to it."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here in magenta have moved off onto this carbon like that, which gives that carbon a negative one formal charge. It's a carb anion. There's still a hydrogen attached to that carbon in red. This hydrogen right here is still attached to it. I'm just not drawing it in so we can see a little bit better. All right, so this is one form of the anion that we could have. We could draw a resonance structure to show the other form."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This hydrogen right here is still attached to it. I'm just not drawing it in so we can see a little bit better. All right, so this is one form of the anion that we could have. We could draw a resonance structure to show the other form. So if we move these electrons in magenta into here and push these electrons off onto the oxygen, let's draw the resonance structure. So we would have our R group here. We would have a double bond."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could draw a resonance structure to show the other form. So if we move these electrons in magenta into here and push these electrons off onto the oxygen, let's draw the resonance structure. So we would have our R group here. We would have a double bond. And then our oxygen would have three lone pairs of electrons, giving it a negative one formal charge. And then we would have our hydrogen over here. So the electrons in magenta moved in here to form our pi bond."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We would have a double bond. And then our oxygen would have three lone pairs of electrons, giving it a negative one formal charge. And then we would have our hydrogen over here. So the electrons in magenta moved in here to form our pi bond. And then we can say that these electrons in here moved off onto our oxygen. So we could go ahead and show that. And so let me just go ahead and put the other bracket on here."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta moved in here to form our pi bond. And then we can say that these electrons in here moved off onto our oxygen. So we could go ahead and show that. And so let me just go ahead and put the other bracket on here. And so we have two forms of this anion. This is called the enolate anion. So this is the enolate anion."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so let me just go ahead and put the other bracket on here. And so we have two forms of this anion. This is called the enolate anion. So this is the enolate anion. This is going to be extremely important in future reactions. And you can see the enolate anion has two resonance structures. One where we've shown the negative charge on the carbon."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is the enolate anion. This is going to be extremely important in future reactions. And you can see the enolate anion has two resonance structures. One where we've shown the negative charge on the carbon. All right, so that would be this one right over here. So the negative charge in the carbon. So this is our carbanion form."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "One where we've shown the negative charge on the carbon. All right, so that would be this one right over here. So the negative charge in the carbon. So this is our carbanion form. So carbanion. And then we also have a resonance structure where the negative charge is on the oxygen. So we could call this the oxyanion."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is our carbanion form. So carbanion. And then we also have a resonance structure where the negative charge is on the oxygen. So we could call this the oxyanion. And if you think about which one contributes more to the overall hybrid, oxygen is more electronegative than carbon. And so it's better able to have a negative one formal charge on it. So the oxyanion contributes more to the resonance hybrid."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we could call this the oxyanion. And if you think about which one contributes more to the overall hybrid, oxygen is more electronegative than carbon. And so it's better able to have a negative one formal charge on it. So the oxyanion contributes more to the resonance hybrid. All right, let's think about the last step in our mechanism, right, to form our enol. If we think about our oxyanion, all we'd have to do is protonate that oxygen here. So we could just go ahead and draw a water molecule."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the oxyanion contributes more to the resonance hybrid. All right, let's think about the last step in our mechanism, right, to form our enol. If we think about our oxyanion, all we'd have to do is protonate that oxygen here. So we could just go ahead and draw a water molecule. All right, so we have a water molecule. This time water's going to function as an acid. It's going to donate a proton."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we could just go ahead and draw a water molecule. All right, so we have a water molecule. This time water's going to function as an acid. It's going to donate a proton. So let's say these electrons in blue, right, take this proton, leave these electrons behind. And so from our oxyanion, we can go ahead and draw our enol product. So we have our R group here."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's going to donate a proton. So let's say these electrons in blue, right, take this proton, leave these electrons behind. And so from our oxyanion, we can go ahead and draw our enol product. So we have our R group here. We would have our double bond. We would have our oxygen, right, now protonated like this to form our enol product. So let me just go ahead and show those electrons in blue, right, picked up a proton here to form our enol."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have our R group here. We would have our double bond. We would have our oxygen, right, now protonated like this to form our enol product. So let me just go ahead and show those electrons in blue, right, picked up a proton here to form our enol. So that's how to get there using base catalyzed. And once again, we will talk much more about the enolate anion in future videos here. So let's look at a situation where the alpha carbon is a chiral center."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me just go ahead and show those electrons in blue, right, picked up a proton here to form our enol. So that's how to get there using base catalyzed. And once again, we will talk much more about the enolate anion in future videos here. So let's look at a situation where the alpha carbon is a chiral center. Okay, so let's look at this right here. So here's our alpha carbon, right, and it's a chiral, let's just say it's a chiral center. So if R and R double prime are different from each other, we would have four different things attached to this carbon, right?"}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a situation where the alpha carbon is a chiral center. Okay, so let's look at this right here. So here's our alpha carbon, right, and it's a chiral, let's just say it's a chiral center. So if R and R double prime are different from each other, we would have four different things attached to this carbon, right? And so the alpha carbon here is sp3 hybridized with tetrahedral geometry. So let's say it's either the R or the S in the antemer. So it doesn't really matter which one."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if R and R double prime are different from each other, we would have four different things attached to this carbon, right? And so the alpha carbon here is sp3 hybridized with tetrahedral geometry. So let's say it's either the R or the S in the antemer. So it doesn't really matter which one. But you can see now we have only one alpha proton, right, only one alpha proton. But because there is an alpha proton, right, we can form an enol. So in either an acid or a base catalyzed mechanism, we could think about the proton here in red, you could think about transferring one to this oxygen and moving your double bond, and then we form our enol."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it doesn't really matter which one. But you can see now we have only one alpha proton, right, only one alpha proton. But because there is an alpha proton, right, we can form an enol. So in either an acid or a base catalyzed mechanism, we could think about the proton here in red, you could think about transferring one to this oxygen and moving your double bond, and then we form our enol. So here is our enol. Now let's look and see what happened to the carbon in red right here. So on the left, the alpha carbon was sp3 hybridized with tetrahedral geometry."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in either an acid or a base catalyzed mechanism, we could think about the proton here in red, you could think about transferring one to this oxygen and moving your double bond, and then we form our enol. So here is our enol. Now let's look and see what happened to the carbon in red right here. So on the left, the alpha carbon was sp3 hybridized with tetrahedral geometry. Now this carbon is sp2 hybridized with trigonal planar geometry. And so whatever stereochemical information we had over here on the left, right, whether it was the R or the S enantiomer, it's been lost now that we formed the enol. The enol is achiral, it's flat, it's planar."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, the alpha carbon was sp3 hybridized with tetrahedral geometry. Now this carbon is sp2 hybridized with trigonal planar geometry. And so whatever stereochemical information we had over here on the left, right, whether it was the R or the S enantiomer, it's been lost now that we formed the enol. The enol is achiral, it's flat, it's planar. And so when we reform the ketoform, right, so one of the possibilities is to form the enantiomer that we started with, but the other possibility is to form the other enantiomer, and so you can see that's what I've shown here. I've shown the hydrogen now going away from us and our R double prime group coming out at us. So this is the enantiomer."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The enol is achiral, it's flat, it's planar. And so when we reform the ketoform, right, so one of the possibilities is to form the enantiomer that we started with, but the other possibility is to form the other enantiomer, and so you can see that's what I've shown here. I've shown the hydrogen now going away from us and our R double prime group coming out at us. So this is the enantiomer. And so because we formed the enol, right, we can get a mixture of enantiomers, so enolization can lead to racemization. We can get a mixture of enantiomers, and if we wait long enough, right, we can get an equal mixture of these guys. So this one and this one, right, would be in equilibrium with our enol form."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is the enantiomer. And so because we formed the enol, right, we can get a mixture of enantiomers, so enolization can lead to racemization. We can get a mixture of enantiomers, and if we wait long enough, right, we can get an equal mixture of these guys. So this one and this one, right, would be in equilibrium with our enol form. And so that's something to think about if you have a chiral center at your alpha carbon. Let's look at two quick examples of keto and enol forms. And so over here on the left we have cyclohexanone, and then on the right would be the enol version of it, right?"}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this one and this one, right, would be in equilibrium with our enol form. And so that's something to think about if you have a chiral center at your alpha carbon. Let's look at two quick examples of keto and enol forms. And so over here on the left we have cyclohexanone, and then on the right would be the enol version of it, right? So you can think about one of these as being your alpha carbon, right, and you can move these electrons in here and push those electrons off, and you can see how that would give you this enol form. So it turns out that the keto form is favored, right? So the equilibrium is actually far to the left, favoring formation of the keto form, so even under just normal conditions, so not acid or base catalyzed."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so over here on the left we have cyclohexanone, and then on the right would be the enol version of it, right? So you can think about one of these as being your alpha carbon, right, and you can move these electrons in here and push those electrons off, and you can see how that would give you this enol form. So it turns out that the keto form is favored, right? So the equilibrium is actually far to the left, favoring formation of the keto form, so even under just normal conditions, so not acid or base catalyzed. And so there's only a trace amount of the enol present. However, there are some cases where the enol is extra stabilized, and that's the case for this example down here. So we have the keto form, and we have the enol form."}, {"video_title": "Keto-enol tautomerization (by Jay) Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the equilibrium is actually far to the left, favoring formation of the keto form, so even under just normal conditions, so not acid or base catalyzed. And so there's only a trace amount of the enol present. However, there are some cases where the enol is extra stabilized, and that's the case for this example down here. So we have the keto form, and we have the enol form. So once again, you can think about these electrons moving in here, pushing those electrons off, giving you your enol form. This is a specially stabilized enol, right? So this is phenol right here, and we know that phenol has an aromatic ring, and so the formation of the enol form is extra stabilized because of this aromatic ring."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "For this Diels-Alder reaction, I've added on an electron donating group to the diene. So here's the diene, and notice there's a methoxy group attached to this carbon. That means there are two possible regiochemical outcomes for this Diels-Alder reaction. So we could form this product, or we could form this product. So which product is favored? To figure that out, we need to draw resonance structures for both the diene and the dienophile. And let's start with the diene."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we could form this product, or we could form this product. So which product is favored? To figure that out, we need to draw resonance structures for both the diene and the dienophile. And let's start with the diene. We could take this lone pair of electrons on this oxygen and move them into here, and then push these electrons off onto this carbon. So let's draw that resonance structure. So now this oxygen would have a double bond to this carbon, and then this carbon would have a lone pair of electrons on it."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's start with the diene. We could take this lone pair of electrons on this oxygen and move them into here, and then push these electrons off onto this carbon. So let's draw that resonance structure. So now this oxygen would have a double bond to this carbon, and then this carbon would have a lone pair of electrons on it. Let me go ahead and finish drawing in the rest of these bonds. And this oxygen still has one lone pair of electrons on it, which gives this oxygen a plus one formal charge. And this carbon would get a negative one formal charge."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So now this oxygen would have a double bond to this carbon, and then this carbon would have a lone pair of electrons on it. Let me go ahead and finish drawing in the rest of these bonds. And this oxygen still has one lone pair of electrons on it, which gives this oxygen a plus one formal charge. And this carbon would get a negative one formal charge. Let me highlight that carbon. So this carbon right here gets a negative one formal charge. Let me draw in my resonance brackets here."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon would get a negative one formal charge. Let me highlight that carbon. So this carbon right here gets a negative one formal charge. Let me draw in my resonance brackets here. Let's look at the dienophile next. So we know that this oxygen is electronegative, so electron density is going to flow towards that oxygen. I could take these electrons and move them into here, and these electrons come off onto the oxygen."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw in my resonance brackets here. Let's look at the dienophile next. So we know that this oxygen is electronegative, so electron density is going to flow towards that oxygen. I could take these electrons and move them into here, and these electrons come off onto the oxygen. So let's draw that resonance structure. We would have a double bond here, and our oxygen would have three lone pairs of electrons around it, which gives that oxygen a negative one formal charge. Let me put in my resonance brackets."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I could take these electrons and move them into here, and these electrons come off onto the oxygen. So let's draw that resonance structure. We would have a double bond here, and our oxygen would have three lone pairs of electrons around it, which gives that oxygen a negative one formal charge. Let me put in my resonance brackets. And notice we took a bond away from this carbon. So this carbon down here gets a plus one formal charge. So our diene has a carbon that is electron rich, and our dienophile has a carbon that is electron poor."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me put in my resonance brackets. And notice we took a bond away from this carbon. So this carbon down here gets a plus one formal charge. So our diene has a carbon that is electron rich, and our dienophile has a carbon that is electron poor. And we know that opposite charges attract. So we could just line up these two carbons, and that allows us to predict the regiochemistry for this reaction. Technically, the Diels-Alder reaction is not an ionic reaction."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So our diene has a carbon that is electron rich, and our dienophile has a carbon that is electron poor. And we know that opposite charges attract. So we could just line up these two carbons, and that allows us to predict the regiochemistry for this reaction. Technically, the Diels-Alder reaction is not an ionic reaction. It's a pericyclic reaction. But this trick does allow you to predict the product. So let's go ahead and use it."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Technically, the Diels-Alder reaction is not an ionic reaction. It's a pericyclic reaction. But this trick does allow you to predict the product. So let's go ahead and use it. Let's get some more space down here, and let's redraw our dienophile first. So let me draw this in. So we had our double bond, and then we had our carbonyl."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and use it. Let's get some more space down here, and let's redraw our dienophile first. So let me draw this in. So we had our double bond, and then we had our carbonyl. We know that this carbon down here is the one that is electron poor. So I'm saying that's partially positive. So we need to line up that electron poor carbon with the electron rich carbon on the diene."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we had our double bond, and then we had our carbonyl. We know that this carbon down here is the one that is electron poor. So I'm saying that's partially positive. So we need to line up that electron poor carbon with the electron rich carbon on the diene. So we want to make this carbon the electron rich one. And that means the methoxy group must come off of this carbon. And now we can see the regiochemistry for our Diels-Alder reaction."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we need to line up that electron poor carbon with the electron rich carbon on the diene. So we want to make this carbon the electron rich one. And that means the methoxy group must come off of this carbon. And now we can see the regiochemistry for our Diels-Alder reaction. Remember, the Diels-Alder reaction is a concerted movement of six pi electrons. So these pi electrons are going to move into here to form a bond between these two carbons. These pi electrons move into here to form a bond, and these pi electrons move down."}, {"video_title": "Diels-Alder regiochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And now we can see the regiochemistry for our Diels-Alder reaction. Remember, the Diels-Alder reaction is a concerted movement of six pi electrons. So these pi electrons are going to move into here to form a bond between these two carbons. These pi electrons move into here to form a bond, and these pi electrons move down. So we form our cyclohexene ring, and then we would have our methoxy group coming off of this carbon, and we would have our ketone coming off of this carbon. So following our electrons, so these electrons in red formed this bond, and then our electrons in blue formed this bond, and our electrons in magenta formed this bond. So this turns out to be the product of our Diels-Alder reaction, and let me go back up to here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we'll see how to add them on the same side, or a syn addition. So we start with R-alkene. And to R-alkene, we add osmium tetroxide. So this is the OSO4 here. And we can also add water and tert-butanol. And what that does is that forms your diol over here, adding your two OH groups on the same side for a syn addition. You don't have to use water and tert-butanol."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is the OSO4 here. And we can also add water and tert-butanol. And what that does is that forms your diol over here, adding your two OH groups on the same side for a syn addition. You don't have to use water and tert-butanol. There are several other ways to do it. I've seen hydrogen peroxide and aqueous sodium bisulfite. So just do whatever your professor has for your class."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You don't have to use water and tert-butanol. There are several other ways to do it. I've seen hydrogen peroxide and aqueous sodium bisulfite. So just do whatever your professor has for your class. So let's look at the mechanism and figure out why this is a syn addition of the OH. So as you can see over here, we have our alkene and our osmium tetroxide right up here. And in this mechanism, we're going to get a concerted six-electron movement."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So just do whatever your professor has for your class. So let's look at the mechanism and figure out why this is a syn addition of the OH. So as you can see over here, we have our alkene and our osmium tetroxide right up here. And in this mechanism, we're going to get a concerted six-electron movement. So six electrons are going to move at the same time. So these electrons in here are going to bond to this carbon. That's going to push the electrons in this pi bond off to bond to this oxygen."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And in this mechanism, we're going to get a concerted six-electron movement. So six electrons are going to move at the same time. So these electrons in here are going to bond to this carbon. That's going to push the electrons in this pi bond off to bond to this oxygen. And then the electrons in this bond are going to move into here onto the osmium. So let's look at what that would give us. So here we have our osmium right here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's going to push the electrons in this pi bond off to bond to this oxygen. And then the electrons in this bond are going to move into here onto the osmium. So let's look at what that would give us. So here we have our osmium right here. And then that's double bonded to these oxygens. And then down here, there used to be two bonds to this oxygen. Now there's only one because the other electrons moved in here to form the bond between the oxygen and the carbon."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here we have our osmium right here. And then that's double bonded to these oxygens. And then down here, there used to be two bonds to this oxygen. Now there's only one because the other electrons moved in here to form the bond between the oxygen and the carbon. So now we have, let's see, we'll go ahead and put a wedge and a dash here for this guy. And then this carbon over here on the right, it has a wedge and a dash bonded to something else. And then it is now bonded to this oxygen on the right here like that."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now there's only one because the other electrons moved in here to form the bond between the oxygen and the carbon. So now we have, let's see, we'll go ahead and put a wedge and a dash here for this guy. And then this carbon over here on the right, it has a wedge and a dash bonded to something else. And then it is now bonded to this oxygen on the right here like that. And then our lone pair of electrons moved in here on the osmium. So let's color coordinate so we can follow along here. So these electrons in here were the ones that moved in here to form our bond."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then it is now bonded to this oxygen on the right here like that. And then our lone pair of electrons moved in here on the osmium. So let's color coordinate so we can follow along here. So these electrons in here were the ones that moved in here to form our bond. And the electrons in this bond right in here, those moved to form this bond. And then finally, these electrons in here, these are the ones that are now lone pair of electrons on the osmium like that. So we formed an osmate ester right here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here were the ones that moved in here to form our bond. And the electrons in this bond right in here, those moved to form this bond. And then finally, these electrons in here, these are the ones that are now lone pair of electrons on the osmium like that. So we formed an osmate ester right here. And we can hydrolyze our osmate ester with the addition of water. So if we add some water to it, so H2O, or like I said, there are several other things that you can add instead. The water's going to hydrolyze this osmate ester and give you back your diol up here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we formed an osmate ester right here. And we can hydrolyze our osmate ester with the addition of water. So if we add some water to it, so H2O, or like I said, there are several other things that you can add instead. The water's going to hydrolyze this osmate ester and give you back your diol up here. It'll give you back this with your two OH groups on the same side. And the reason why those two OH groups end up on the same side is because of this osmate ester intermediate. The way those two oxygens add, the way these two oxygens add here, they add on the same side."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The water's going to hydrolyze this osmate ester and give you back your diol up here. It'll give you back this with your two OH groups on the same side. And the reason why those two OH groups end up on the same side is because of this osmate ester intermediate. The way those two oxygens add, the way these two oxygens add here, they add on the same side. They add in a syn fashion. All right, so what's the other product we get? Well, after this osmate ester is hydrolyzed, the osmium is going to be right here bonded to our two oxygens."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The way those two oxygens add, the way these two oxygens add here, they add on the same side. They add in a syn fashion. All right, so what's the other product we get? Well, after this osmate ester is hydrolyzed, the osmium is going to be right here bonded to our two oxygens. And it's going to form an OH on either side here like that. So if you're looking at oxidation states, so if we check out the oxidation state of osmium over here on the left, so this osmium, if you remember from general chemistry, doing your oxidation states, you'll see that osmium is a plus 8 stage over here. So it's plus 8 over here on the left."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, after this osmate ester is hydrolyzed, the osmium is going to be right here bonded to our two oxygens. And it's going to form an OH on either side here like that. So if you're looking at oxidation states, so if we check out the oxidation state of osmium over here on the left, so this osmium, if you remember from general chemistry, doing your oxidation states, you'll see that osmium is a plus 8 stage over here. So it's plus 8 over here on the left. And after the osmate ester is hydrolyzed, we get this structure over here. And again, from general chemistry, if you assign your oxidation states to that osmium, you're going to get a plus 6. So we have a plus 8 to plus 6, so a decrease or a reduction in the oxidation number."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's plus 8 over here on the left. And after the osmate ester is hydrolyzed, we get this structure over here. And again, from general chemistry, if you assign your oxidation states to that osmium, you're going to get a plus 6. So we have a plus 8 to plus 6, so a decrease or a reduction in the oxidation number. So osmium is reduced while those two carbons are being oxidized, while your alkene is oxidized to a diol. So that's a redox reaction, of course. Let's look at how you would do this in practice."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have a plus 8 to plus 6, so a decrease or a reduction in the oxidation number. So osmium is reduced while those two carbons are being oxidized, while your alkene is oxidized to a diol. So that's a redox reaction, of course. Let's look at how you would do this in practice. Because the problem with osmium is it's very toxic. It's also very expensive. So you don't want to use very much of it."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at how you would do this in practice. Because the problem with osmium is it's very toxic. It's also very expensive. So you don't want to use very much of it. You want to use as little of the osmium as you possibly can. So how do you get around the toxicity and also the cost? So let's look at cyclohexene."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you don't want to use very much of it. You want to use as little of the osmium as you possibly can. So how do you get around the toxicity and also the cost? So let's look at cyclohexene. If we were to add osmium tetroxide to cyclohexene, it'd be OSO4. We want to add only a catalytic amount, the smallest amount that we absolutely need. So we're going to add a catalytic amount of this."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at cyclohexene. If we were to add osmium tetroxide to cyclohexene, it'd be OSO4. We want to add only a catalytic amount, the smallest amount that we absolutely need. So we're going to add a catalytic amount of this. So catalytic amount, just a very, very small amount of osmium tetroxide. And we're going to add something called NMO, which is N-methylmorpholine N-oxide. And what this is going to do is it's going to regenerate your OSO4 by oxidizing the osmium plus 6 back into osmium plus 8."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to add a catalytic amount of this. So catalytic amount, just a very, very small amount of osmium tetroxide. And we're going to add something called NMO, which is N-methylmorpholine N-oxide. And what this is going to do is it's going to regenerate your OSO4 by oxidizing the osmium plus 6 back into osmium plus 8. So the NMO is going to take this one over here, the osmium that's in the plus 6, and it's going to oxidize it back into the plus 8 stage so that we can reuse our OSO4 in the reaction. And that way, we can use very, very small amounts of it. So that's what the presence of NMO does."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And what this is going to do is it's going to regenerate your OSO4 by oxidizing the osmium plus 6 back into osmium plus 8. So the NMO is going to take this one over here, the osmium that's in the plus 6, and it's going to oxidize it back into the plus 8 stage so that we can reuse our OSO4 in the reaction. And that way, we can use very, very small amounts of it. So that's what the presence of NMO does. So let's see our solvents. We use water and tert-butanol, just like we did above. So I'll just write out the abbreviation this time."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's what the presence of NMO does. So let's see our solvents. We use water and tert-butanol, just like we did above. So I'll just write out the abbreviation this time. T-B-U-O-H. And we're going to get a syn addition of OHs. So we're going to get an OH, OHs that add on the same side. So an OH, OH that adds on the same side, OH that adds on the same side."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll just write out the abbreviation this time. T-B-U-O-H. And we're going to get a syn addition of OHs. So we're going to get an OH, OHs that add on the same side. So an OH, OH that adds on the same side, OH that adds on the same side. And you might think, oh, well, wouldn't we get another product here? I could draw two dashes instead of two wedges. Wouldn't that be a different product?"}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So an OH, OH that adds on the same side, OH that adds on the same side. And you might think, oh, well, wouldn't we get another product here? I could draw two dashes instead of two wedges. Wouldn't that be a different product? Well, in reality, these are the exact same molecule. So this is actually a meso compound. So there's a plane of symmetry."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Wouldn't that be a different product? Well, in reality, these are the exact same molecule. So this is actually a meso compound. So there's a plane of symmetry. So there's a plane of symmetry right here. And you can superimpose these two molecules on each other, therefore, the same thing. So we're only going to get one product."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So there's a plane of symmetry. So there's a plane of symmetry right here. And you can superimpose these two molecules on each other, therefore, the same thing. So we're only going to get one product. So only one product for this reaction, where we add our two OHs on the same side. Let's look at another way to achieve syn dihydroxylation here. So let's look at a different way to do it."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're only going to get one product. So only one product for this reaction, where we add our two OHs on the same side. Let's look at another way to achieve syn dihydroxylation here. So let's look at a different way to do it. In this case, we're going to use permanganate. So we take our alkene, and then we take permanganate, MnO4 minus, or something like potassium permanganate. And usually you do this reaction in cold sodium hydroxide solution, a cold aqueous solution of sodium hydroxide."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a different way to do it. In this case, we're going to use permanganate. So we take our alkene, and then we take permanganate, MnO4 minus, or something like potassium permanganate. And usually you do this reaction in cold sodium hydroxide solution, a cold aqueous solution of sodium hydroxide. So water's in there as well. And what you're going to get is a syn addition of your OHs. You're going to get your two OH groups adding on to the same side, just like we did before."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And usually you do this reaction in cold sodium hydroxide solution, a cold aqueous solution of sodium hydroxide. So water's in there as well. And what you're going to get is a syn addition of your OHs. You're going to get your two OH groups adding on to the same side, just like we did before. And your other product would be MnO2. So you can follow this reaction pretty easily, because we know that MnO4 minus is purple. And the oxidation state of manganese on the left here is plus 7 when you assign your oxidation states."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You're going to get your two OH groups adding on to the same side, just like we did before. And your other product would be MnO2. So you can follow this reaction pretty easily, because we know that MnO4 minus is purple. And the oxidation state of manganese on the left here is plus 7 when you assign your oxidation states. And over here on the right, it goes to plus 4. So there's a reduction. Manganese has been reduced while those two carbons have been oxidized for this redox reaction."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the oxidation state of manganese on the left here is plus 7 when you assign your oxidation states. And over here on the right, it goes to plus 4. So there's a reduction. Manganese has been reduced while those two carbons have been oxidized for this redox reaction. MnO4 minus is purple. MnO2 is kind of a really, really dark brown kind of color. So when you do this reaction, it's pretty much instantaneous."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Manganese has been reduced while those two carbons have been oxidized for this redox reaction. MnO4 minus is purple. MnO2 is kind of a really, really dark brown kind of color. So when you do this reaction, it's pretty much instantaneous. Everything turns from purple to brown, and you get your diol. And there's a fast way of figuring out oxidation, if something is an oxidation or a reduction. You could assign oxidation states like we did in the previous video, like we did in one of the earlier videos."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when you do this reaction, it's pretty much instantaneous. Everything turns from purple to brown, and you get your diol. And there's a fast way of figuring out oxidation, if something is an oxidation or a reduction. You could assign oxidation states like we did in the previous video, like we did in one of the earlier videos. Another way to do it is to just count the number of bonds to oxygen. So over here on the left, you're alkene, zero bonds of carbon to oxygen. Over here on the right, each of your two carbons has one bond of carbon to oxygen."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You could assign oxidation states like we did in the previous video, like we did in one of the earlier videos. Another way to do it is to just count the number of bonds to oxygen. So over here on the left, you're alkene, zero bonds of carbon to oxygen. Over here on the right, each of your two carbons has one bond of carbon to oxygen. So you can tell it's been oxidized that way. That's a fast way of doing it. Why is this a syn dihydroxylation?"}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Over here on the right, each of your two carbons has one bond of carbon to oxygen. So you can tell it's been oxidized that way. That's a fast way of doing it. Why is this a syn dihydroxylation? Why do our two H's add on on the same side? Well, if we draw out the mechanism really fast, so we have MnO4 minus. So let's draw our permanganate anion."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Why is this a syn dihydroxylation? Why do our two H's add on on the same side? Well, if we draw out the mechanism really fast, so we have MnO4 minus. So let's draw our permanganate anion. So here is the dot structure for our permanganate anion, so negative charge in this oxygen because it has three lone pairs of electrons around it. So if we can fit in those three lone pairs of electrons. So it's a very, very similar mechanism to before."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw our permanganate anion. So here is the dot structure for our permanganate anion, so negative charge in this oxygen because it has three lone pairs of electrons around it. So if we can fit in those three lone pairs of electrons. So it's a very, very similar mechanism to before. We take our alkene. We're going to get a concerted six electron movement. So these electrons are going to move into here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's a very, very similar mechanism to before. We take our alkene. We're going to get a concerted six electron movement. So these electrons are going to move into here. These electrons are going to form a bond here. And these electrons are going to bond right here. So it's just like before."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons are going to move into here. These electrons are going to form a bond here. And these electrons are going to bond right here. So it's just like before. It's just like with the osmium. So we get our manganese. And these two top oxygens don't do anything."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's just like before. It's just like with the osmium. So we get our manganese. And these two top oxygens don't do anything. So we'll just go ahead and leave them like that. This oxygen here gets bonded to a carbon. Carbon, oxygen, like that."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And these two top oxygens don't do anything. So we'll just go ahead and leave them like that. This oxygen here gets bonded to a carbon. Carbon, oxygen, like that. And then we have our wedge and our dash like that. And then our lone pair of electrons and our manganese. So it's the exact same thing as before."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Carbon, oxygen, like that. And then we have our wedge and our dash like that. And then our lone pair of electrons and our manganese. So it's the exact same thing as before. We can hydrolyze it to get our diol like that. Exact same mechanism, which is why our two OH's add syn like that. The problem with using permanganate is permanganate is such a strong oxidizing agent."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's the exact same thing as before. We can hydrolyze it to get our diol like that. Exact same mechanism, which is why our two OH's add syn like that. The problem with using permanganate is permanganate is such a strong oxidizing agent. It's hard to stop it at this first oxidation. So if you do everything really cold with ice, and then you stop your reaction immediately, it's not too hard to get your diol out. But if you heat things up or let things go for a long time, you won't be able to stop the oxidation at one oxidation."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The problem with using permanganate is permanganate is such a strong oxidizing agent. It's hard to stop it at this first oxidation. So if you do everything really cold with ice, and then you stop your reaction immediately, it's not too hard to get your diol out. But if you heat things up or let things go for a long time, you won't be able to stop the oxidation at one oxidation. It'll keep going. So let's go ahead and draw what would happen if we added some heat to our oxidation with permanganate. So we add potassium permanganate, so KMnO4."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But if you heat things up or let things go for a long time, you won't be able to stop the oxidation at one oxidation. It'll keep going. So let's go ahead and draw what would happen if we added some heat to our oxidation with permanganate. So we add potassium permanganate, so KMnO4. And we add some heat to it this time. So what's going to happen if you add heat to it? It's not going to stop at the diol."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we add potassium permanganate, so KMnO4. And we add some heat to it this time. So what's going to happen if you add heat to it? It's not going to stop at the diol. It's going to keep oxidizing. So instead of one bond of carbon to oxygen, you're actually going to get two bonds of carbon to oxygen. So let me redraw my alkene here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's not going to stop at the diol. It's going to keep oxidizing. So instead of one bond of carbon to oxygen, you're actually going to get two bonds of carbon to oxygen. So let me redraw my alkene here. And let me show you a clever way of doing alkene cleavage. So if I just make this a little bit longer than usual. So here's my alkene."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me redraw my alkene here. And let me show you a clever way of doing alkene cleavage. So if I just make this a little bit longer than usual. So here's my alkene. The result is going to cleave this bond between my two carbons. So I'm going to actually break it. That's alkene cleavage."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here's my alkene. The result is going to cleave this bond between my two carbons. So I'm going to actually break it. That's alkene cleavage. So I'm going to go like that. And I'll shorten it a little bit right here like that. And I'm going to put auctions in there."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That's alkene cleavage. So I'm going to go like that. And I'll shorten it a little bit right here like that. And I'm going to put auctions in there. So to draw my product for this alkene cleavage, I'm going to put an auction here and an auction here. So there are now two bonds. One of carbon to oxygen instead of one bond like we did up there."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to put auctions in there. So to draw my product for this alkene cleavage, I'm going to put an auction here and an auction here. So there are now two bonds. One of carbon to oxygen instead of one bond like we did up there. So it's been oxidized. And this could be a ketone. This could be a carboxylic acid."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "One of carbon to oxygen instead of one bond like we did up there. So it's been oxidized. And this could be a ketone. This could be a carboxylic acid. This could be CO2. And it all depends on what kind of reaction that you're doing. So let's do a quick example."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This could be a carboxylic acid. This could be CO2. And it all depends on what kind of reaction that you're doing. So let's do a quick example. So let's start over here. Let's start with this as our starting reactant. We'll put a methyl group there, a hydrogen here, a methyl group here, and a methyl group here."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's do a quick example. So let's start over here. Let's start with this as our starting reactant. We'll put a methyl group there, a hydrogen here, a methyl group here, and a methyl group here. So we add potassium permanganate and heat. And I'm going to go ahead and redraw my reactant so we can do that little trick. So I'm going to redraw my reactant."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'll put a methyl group there, a hydrogen here, a methyl group here, and a methyl group here. So we add potassium permanganate and heat. And I'm going to go ahead and redraw my reactant so we can do that little trick. So I'm going to redraw my reactant. This time I'm going to make these a little bit skinnier. And I'll make this a CH3 and make this a hydrogen and put our two methyl groups over here like that. So alkene cleavage."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to redraw my reactant. This time I'm going to make these a little bit skinnier. And I'll make this a CH3 and make this a hydrogen and put our two methyl groups over here like that. So alkene cleavage. So you heat it up. You can't stop it at the first oxidation. Permanganate is such a strong oxidizer."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So alkene cleavage. So you heat it up. You can't stop it at the first oxidation. Permanganate is such a strong oxidizer. So it's going to keep on going. And we're going to get alkene cleavage. We're going to break the bond between our two carbons."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Permanganate is such a strong oxidizer. So it's going to keep on going. And we're going to get alkene cleavage. We're going to break the bond between our two carbons. And then each of these carbons is going to get an oxygen. So now there are two bonds of carbon to oxygen. So I'll go ahead and do that."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to break the bond between our two carbons. And then each of these carbons is going to get an oxygen. So now there are two bonds of carbon to oxygen. So I'll go ahead and do that. Let's put my oxygen in there like that. And it doesn't even stop here. So the molecule on the left, our aldehyde here, we haven't talked about aldehydes in this course."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and do that. Let's put my oxygen in there like that. And it doesn't even stop here. So the molecule on the left, our aldehyde here, we haven't talked about aldehydes in this course. But aldehydes are easily oxidized, especially in the presence of something like permanganate. So this aldehyde right here is going to get oxidized again. So we have acid aldehyde."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the molecule on the left, our aldehyde here, we haven't talked about aldehydes in this course. But aldehydes are easily oxidized, especially in the presence of something like permanganate. So this aldehyde right here is going to get oxidized again. So we have acid aldehyde. It's going to get oxidized. And if we oxidize an aldehyde, two bonds of carbon to oxygen, we would oxidize it to something that has three bonds of carbon to oxygen. So that's our three bonds of carbon to oxygen like that."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have acid aldehyde. It's going to get oxidized. And if we oxidize an aldehyde, two bonds of carbon to oxygen, we would oxidize it to something that has three bonds of carbon to oxygen. So that's our three bonds of carbon to oxygen like that. So acid aldehyde gets oxidized to acetic acid. So let's just emphasize that. Over here, two bonds of carbon to oxygen."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's our three bonds of carbon to oxygen like that. So acid aldehyde gets oxidized to acetic acid. So let's just emphasize that. Over here, two bonds of carbon to oxygen. Over here, three bonds of carbon to oxygen. So it has been oxidized. Once again, you could assign oxidation states."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Over here, two bonds of carbon to oxygen. Over here, three bonds of carbon to oxygen. So it has been oxidized. Once again, you could assign oxidation states. We just don't have time in this video to do that. So you're going to get acetic acid as one of your products. And then we had acetone over here as our other product."}, {"video_title": "Syn dihydroxylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Once again, you could assign oxidation states. We just don't have time in this video to do that. So you're going to get acetic acid as one of your products. And then we had acetone over here as our other product. So alkene cleavage. This reaction isn't very useful because of these side reactions. It's hard to stop the oxidation."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "What I want to do in this video is to try to figure out what type of reaction or reactions might occur if we have, what is this? This is 1, 2, 3, 4, 5, it's in a cycle. This is bromocyclopentane. If we have some bromocyclopentane dissolved in our solvent is dimethylformamide. Sometimes you'll see that just written as DMF and I've actually drawn the formula for it here so we can think about what type of a solvent it is. Also in our solution we have the methoxide ion. We also have the methoxide ion right here."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "If we have some bromocyclopentane dissolved in our solvent is dimethylformamide. Sometimes you'll see that just written as DMF and I've actually drawn the formula for it here so we can think about what type of a solvent it is. Also in our solution we have the methoxide ion. We also have the methoxide ion right here. Let's think about what type of reaction might occur. Just to narrow things down, we'll think about it in the context of the last four types of reactions we've looked at. This might be an SN2 reaction, an SN1 reaction, an E2 reaction, or an E1 reaction."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "We also have the methoxide ion right here. Let's think about what type of reaction might occur. Just to narrow things down, we'll think about it in the context of the last four types of reactions we've looked at. This might be an SN2 reaction, an SN1 reaction, an E2 reaction, or an E1 reaction. We're going to look at all the clues and figure out what's likely to occur and then actually draw the mechanism for it occurring. The first thing, since they gave us the solvent and other things that are in the solvent, let's think about how those might affect the reaction. If we look at this solvent right here, whenever you look at any of these reactions, when you look at the solvent you just want to think about is it protic or not."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "This might be an SN2 reaction, an SN1 reaction, an E2 reaction, or an E1 reaction. We're going to look at all the clues and figure out what's likely to occur and then actually draw the mechanism for it occurring. The first thing, since they gave us the solvent and other things that are in the solvent, let's think about how those might affect the reaction. If we look at this solvent right here, whenever you look at any of these reactions, when you look at the solvent you just want to think about is it protic or not. Protic means that it has hydrogens that can be released or that the electrons can be nabbed off and these protons can just float around. If we look over here, we do have hydrogens, but all of the hydrogens are bonded to carbon. Carbon is unlikely to just steal a hydrogen's electrons and let the hydrogen float around."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "If we look at this solvent right here, whenever you look at any of these reactions, when you look at the solvent you just want to think about is it protic or not. Protic means that it has hydrogens that can be released or that the electrons can be nabbed off and these protons can just float around. If we look over here, we do have hydrogens, but all of the hydrogens are bonded to carbon. Carbon is unlikely to just steal a hydrogen's electrons and let the hydrogen float around. Carbon is not that electronegative. If you had hydrogens bonded to an oxygen, that would be a different question. Then you would have a protic solvent."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Carbon is unlikely to just steal a hydrogen's electrons and let the hydrogen float around. Carbon is not that electronegative. If you had hydrogens bonded to an oxygen, that would be a different question. Then you would have a protic solvent. In this case, all the hydrogens bonded to carbons are not likely to get their electrons nabbed off and float around as free protons. This is an aprotic solvent. We've gone over this a little bit with SN2 and SN1, but the same idea applies."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Then you would have a protic solvent. In this case, all the hydrogens bonded to carbons are not likely to get their electrons nabbed off and float around as free protons. This is an aprotic solvent. We've gone over this a little bit with SN2 and SN1, but the same idea applies. In order to have an SN2 or an E2 reaction, you have to have either a strong nucleophile or a strong base. The same thing could actually be both, although they're not always correlated. We've seen that before."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "We've gone over this a little bit with SN2 and SN1, but the same idea applies. In order to have an SN2 or an E2 reaction, you have to have either a strong nucleophile or a strong base. The same thing could actually be both, although they're not always correlated. We've seen that before. If you had a protic solvent, it would stabilize the strong base or the strong nucleophile. The protons would react with them. They would take the electrons from that strong base or that strong nucleophile."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "We've seen that before. If you had a protic solvent, it would stabilize the strong base or the strong nucleophile. The protons would react with them. They would take the electrons from that strong base or that strong nucleophile. In order to have an SN2 or an E2, you have to have no protons flying around. You need an aprotic solvent. This aprotic solvent will favor SN2 or an E2 reaction."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "They would take the electrons from that strong base or that strong nucleophile. In order to have an SN2 or an E2, you have to have no protons flying around. You need an aprotic solvent. This aprotic solvent will favor SN2 or an E2 reaction. Our mind is already thinking in SN2 or E2. Let's think about the reactants themselves. Over here we have the methoxide ion."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "This aprotic solvent will favor SN2 or an E2 reaction. Our mind is already thinking in SN2 or E2. Let's think about the reactants themselves. Over here we have the methoxide ion. Let's think about whether it's a strong or weak nucleophile. It's actually a pretty strong nucleophile. That would put us in the direction of an SN2."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Over here we have the methoxide ion. Let's think about whether it's a strong or weak nucleophile. It's actually a pretty strong nucleophile. That would put us in the direction of an SN2. We have two data points for SN2. Remember, it has to go in there and be active. It's not too big of a molecule, so it's not going to be hindered."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "That would put us in the direction of an SN2. We have two data points for SN2. Remember, it has to go in there and be active. It's not too big of a molecule, so it's not going to be hindered. It's also an extremely strong base, even stronger than hydroxide. It's also an extremely strong base, which might lead us, or that does imply that we're going to have an E2 reaction. The last thing we need to think about is the carbon where the leaving group might leave from."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "It's not too big of a molecule, so it's not going to be hindered. It's also an extremely strong base, even stronger than hydroxide. It's also an extremely strong base, which might lead us, or that does imply that we're going to have an E2 reaction. The last thing we need to think about is the carbon where the leaving group might leave from. Immediately when you look at the bromocyclopentane, there's only one functional group attached to the chain, and that is the bromo group right here. It is attached to this carbon. We can call that the alpha carbon."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "The last thing we need to think about is the carbon where the leaving group might leave from. Immediately when you look at the bromocyclopentane, there's only one functional group attached to the chain, and that is the bromo group right here. It is attached to this carbon. We can call that the alpha carbon. It is a secondary carbon. This carbon right here is bonded to one, two other carbons. This alpha carbon is a secondary carbon."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "We can call that the alpha carbon. It is a secondary carbon. This carbon right here is bonded to one, two other carbons. This alpha carbon is a secondary carbon. That kind of makes it neutral in this mix. If it was a methyl or primary carbon, it would favor SN2, actually. Methyl, the only thing you could have is an SN2."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "This alpha carbon is a secondary carbon. That kind of makes it neutral in this mix. If it was a methyl or primary carbon, it would favor SN2, actually. Methyl, the only thing you could have is an SN2. If it was a tertiary carbon, it would favor SN1 or E1, because it would favor a stable carbocation. The leaving group could just leave, and if this guy was bonded to another carbon, it would be very stable. In this situation, it's a secondary carbon bonded to two carbons."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Methyl, the only thing you could have is an SN2. If it was a tertiary carbon, it would favor SN1 or E1, because it would favor a stable carbocation. The leaving group could just leave, and if this guy was bonded to another carbon, it would be very stable. In this situation, it's a secondary carbon bonded to two carbons. It's a little bit neutral. Any of these reactions might occur. When we look at all of the other data points, they're pointing at both SN2 or E2."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "In this situation, it's a secondary carbon bonded to two carbons. It's a little bit neutral. Any of these reactions might occur. When we look at all of the other data points, they're pointing at both SN2 or E2. We have strong nucleophile slash base. We have an aprotic solvent. It's going to be SN2 or an E2 reaction."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "When we look at all of the other data points, they're pointing at both SN2 or E2. We have strong nucleophile slash base. We have an aprotic solvent. It's going to be SN2 or an E2 reaction. Let's actually draw the reaction. Let me do the SN2 first. Let me do it in orange."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "It's going to be SN2 or an E2 reaction. Let's actually draw the reaction. Let me do the SN2 first. Let me do it in orange. If we were to have an SN2 reaction, let me redraw the molecule. Let me draw the cyclopentane part. I want to make sure."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Let me do it in orange. If we were to have an SN2 reaction, let me redraw the molecule. Let me draw the cyclopentane part. I want to make sure. Let me draw it the same way I had it drawn up there. The pentagon is facing upwards. Then we have our bromo group right there."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "I want to make sure. Let me draw it the same way I had it drawn up there. The pentagon is facing upwards. Then we have our bromo group right there. We have our methoxide ion right over here. CH3O minus. Another way we could view it is that this oxygen has 1, 2, 3, 4, 5, 6, 7 valence electrons with a negative charge."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Then we have our bromo group right there. We have our methoxide ion right over here. CH3O minus. Another way we could view it is that this oxygen has 1, 2, 3, 4, 5, 6, 7 valence electrons with a negative charge. One of these electrons right over here, this can attack the substrate right over there, that carbon. Right when that happens, simultaneously, this bromine is going to be able to nab an electron from that same carbon. Then we are going to be left with the bromine now becomes the bromide anion."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Another way we could view it is that this oxygen has 1, 2, 3, 4, 5, 6, 7 valence electrons with a negative charge. One of these electrons right over here, this can attack the substrate right over there, that carbon. Right when that happens, simultaneously, this bromine is going to be able to nab an electron from that same carbon. Then we are going to be left with the bromine now becomes the bromide anion. It now has 1, 2, 3, 4, 5, 6, 7 valence electrons. 1, 2, 3, 4, 5, 6, 7. Now it nabbed one more electron, making it bromide."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Then we are going to be left with the bromine now becomes the bromide anion. It now has 1, 2, 3, 4, 5, 6, 7 valence electrons. 1, 2, 3, 4, 5, 6, 7. Now it nabbed one more electron, making it bromide. Now it has a negative charge. If we were to draw the chain, it would look like this. We could draw it on this side, so it's attacking from the other side."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Now it nabbed one more electron, making it bromide. Now it has a negative charge. If we were to draw the chain, it would look like this. We could draw it on this side, so it's attacking from the other side. This isn't a chiral substrate, so we don't have to be too particular about how we draw the connections to the carbon. We're not even showing anything popping in or out. We would have the methoxide ion, now it's bonded, so it's no longer an ion."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "We could draw it on this side, so it's attacking from the other side. This isn't a chiral substrate, so we don't have to be too particular about how we draw the connections to the carbon. We're not even showing anything popping in or out. We would have the methoxide ion, now it's bonded, so it's no longer an ion. It's OCH3. Just like that, has bonded to this carbon. Implicitly, this carbon had another hydrogen that we are not showing."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "We would have the methoxide ion, now it's bonded, so it's no longer an ion. It's OCH3. Just like that, has bonded to this carbon. Implicitly, this carbon had another hydrogen that we are not showing. That quickly, that was the SN2 reaction. That is the mechanism. Now let's think about what the E2 reaction."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Implicitly, this carbon had another hydrogen that we are not showing. That quickly, that was the SN2 reaction. That is the mechanism. Now let's think about what the E2 reaction. To do the E2 properly, to give it justice, we're going to have to draw some of the hydrogens. On the E2 reaction, let me draw that in blue. The E2 reaction, let me draw the cyclopentane part."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Now let's think about what the E2 reaction. To do the E2 properly, to give it justice, we're going to have to draw some of the hydrogens. On the E2 reaction, let me draw that in blue. The E2 reaction, let me draw the cyclopentane part. Let me draw it big. Actually, over here, it's less important to draw it too big. Let me draw the pentagon, just like that."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "The E2 reaction, let me draw the cyclopentane part. Let me draw it big. Actually, over here, it's less important to draw it too big. Let me draw the pentagon, just like that. That is the bromine, 3, 4, 5, 6, and it has a 7th valence electron right over here. This is the alpha carbon. That right there is the alpha carbon."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Let me draw the pentagon, just like that. That is the bromine, 3, 4, 5, 6, and it has a 7th valence electron right over here. This is the alpha carbon. That right there is the alpha carbon. Then there are two beta carbons. There are two beta carbons right over there and there. They each have two hydrogens on them."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "That right there is the alpha carbon. Then there are two beta carbons. There are two beta carbons right over there and there. They each have two hydrogens on them. I know it's becoming a little hard to read. They each have two hydrogens on them. In an E2 reaction, the strong base will react."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "They each have two hydrogens on them. I know it's becoming a little hard to read. They each have two hydrogens on them. In an E2 reaction, the strong base will react. Let me make it a little cleaner than that. Let me get rid of the beta. The beta makes it a little dirty."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "In an E2 reaction, the strong base will react. Let me make it a little cleaner than that. Let me get rid of the beta. The beta makes it a little dirty. They each have two hydrogens on them. In an E2 reaction, the strong base, over here, the methoxide ion was acting as a strong nucleophile. E2 is going to act as a strong base."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "The beta makes it a little dirty. They each have two hydrogens on them. In an E2 reaction, the strong base, over here, the methoxide ion was acting as a strong nucleophile. E2 is going to act as a strong base. It's going to nab off a hydrogen off of one of the beta carbons. You might say, which one? Let's look at Zaitsev's rule."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "E2 is going to act as a strong base. It's going to nab off a hydrogen off of one of the beta carbons. You might say, which one? Let's look at Zaitsev's rule. It doesn't matter. These are symmetric. They are both bonded to two other carbons."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Let's look at Zaitsev's rule. It doesn't matter. These are symmetric. They are both bonded to two other carbons. They both are bonded to the same number of hydrogens. It doesn't matter. It's actually going to be random, which one."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "They are both bonded to two other carbons. They both are bonded to the same number of hydrogens. It doesn't matter. It's actually going to be random, which one. You actually won't be able to tell the difference because it's symmetric. Let's just draw it like this. Let me draw the methoxide ion."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "It's actually going to be random, which one. You actually won't be able to tell the difference because it's symmetric. Let's just draw it like this. Let me draw the methoxide ion. 1, 2, 3, 4, or anion, maybe I should say, 5, 6. Then it has one bond to the CH3. It has a negative charge."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Let me draw the methoxide ion. 1, 2, 3, 4, or anion, maybe I should say, 5, 6. Then it has one bond to the CH3. It has a negative charge. Very, very, very strong base. It can go over here and nab the hydrogen and leave hydrogen's electron behind. It can, maybe I'll take a color."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "It has a negative charge. Very, very, very strong base. It can go over here and nab the hydrogen and leave hydrogen's electron behind. It can, maybe I'll take a color. This electron can be given to the hydrogen so that it forms a bond with it. Hydrogen's electron, let me do this in a suitably different color. Hydrogen's electron that is sitting right over there can now be given to the alpha carbon to form a double bond."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "It can, maybe I'll take a color. This electron can be given to the hydrogen so that it forms a bond with it. Hydrogen's electron, let me do this in a suitably different color. Hydrogen's electron that is sitting right over there can now be given to the alpha carbon to form a double bond. Now that the alpha carbon is getting that electron, now the bromo group can leave. It's a decent leaving group, and that was another thing that we should think about in our equation. A good leaving group actually favors all of the reactions."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Hydrogen's electron that is sitting right over there can now be given to the alpha carbon to form a double bond. Now that the alpha carbon is getting that electron, now the bromo group can leave. It's a decent leaving group, and that was another thing that we should think about in our equation. A good leaving group actually favors all of the reactions. SN2, E2, SN1, E1. There the carbon's getting the electron, and then the bromine can then take this carbon's electron. Just in one step, that's what's distinctive about the E2 and the SN2 reactions."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "A good leaving group actually favors all of the reactions. SN2, E2, SN1, E1. There the carbon's getting the electron, and then the bromine can then take this carbon's electron. Just in one step, that's what's distinctive about the E2 and the SN2 reactions. All of the reactions are involved in the rate determining step, and there really is only one step. Just like that, after that happens, what we're left with is the methoxide anion takes the hydrogen, so it becomes methanol. Let me draw that."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Just in one step, that's what's distinctive about the E2 and the SN2 reactions. All of the reactions are involved in the rate determining step, and there really is only one step. Just like that, after that happens, what we're left with is the methoxide anion takes the hydrogen, so it becomes methanol. Let me draw that. It becomes methanol. It had one, two, three, four, and then five. That's this one right there, but then this guy goes and bonds with the hydrogen."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Let me draw that. It becomes methanol. It had one, two, three, four, and then five. That's this one right there, but then this guy goes and bonds with the hydrogen. This guy goes and bonds with the hydrogen just like that. Hydrogen leaves its electron behind. Let me draw the cyclopentane part now."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "That's this one right there, but then this guy goes and bonds with the hydrogen. This guy goes and bonds with the hydrogen just like that. Hydrogen leaves its electron behind. Let me draw the cyclopentane part now. The cyclopentane looked like this before. If I just focus on the ring. This guy was bonded to a hydrogen."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Let me draw the cyclopentane part now. The cyclopentane looked like this before. If I just focus on the ring. This guy was bonded to a hydrogen. He was bonded to this hydrogen over here, but now that electron is going to be used to form a bond with this alpha carbon right over here. Let me draw the alpha carbon. The alpha carbon is right over there."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "This guy was bonded to a hydrogen. He was bonded to this hydrogen over here, but now that electron is going to be used to form a bond with this alpha carbon right over here. Let me draw the alpha carbon. The alpha carbon is right over there. Obviously, implicitly, at every one of these edges, we have a carbon, but now a double bond is going to form with that alpha carbon. We could just draw it like that, a double bond. Obviously, there's another carbon here."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "The alpha carbon is right over there. Obviously, implicitly, at every one of these edges, we have a carbon, but now a double bond is going to form with that alpha carbon. We could just draw it like that, a double bond. Obviously, there's another carbon here. Another carbon over there. Now this double bond will form. Now the bromide has left."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Obviously, there's another carbon here. Another carbon over there. Now this double bond will form. Now the bromide has left. It's taken an electron with it from that carbon. Now that the carbon doesn't need it, it was already starting to hog it because it's so electronegative. That's bromine."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Now the bromide has left. It's taken an electron with it from that carbon. Now that the carbon doesn't need it, it was already starting to hog it because it's so electronegative. That's bromine. Takes that orange electron. Now it is bromide, and we're done. Just to go back to the original question here, which reaction is likely to occur, which mechanism, it's actually both SN2 and E2."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "That's bromine. Takes that orange electron. Now it is bromide, and we're done. Just to go back to the original question here, which reaction is likely to occur, which mechanism, it's actually both SN2 and E2. You would see a mix of both of these occurring because you have all of the environmental factors that would enable both. You would have both of these mechanisms. Let me separate them out."}, {"video_title": "Comparing E2 E1 Sn2 Sn1 Reactions.mp3", "Sentence": "Just to go back to the original question here, which reaction is likely to occur, which mechanism, it's actually both SN2 and E2. You would see a mix of both of these occurring because you have all of the environmental factors that would enable both. You would have both of these mechanisms. Let me separate them out. Here's the SN2 reaction. You would have the SN2 reaction occurring in your vial or your pot or whatever you're making all this stuff occur in. You would also have your E2 reaction."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So these first two right here, they actually look like completely different molecules. So your gut impulse might be, see, these are completely different molecules. And it wouldn't be completely off. But if we look a little bit closer, you see that this guy on the left has one, two, three, four carbons. And so does this guy on the right. He has one, two, three, four carbons. This guy on the left has two, four, six, seven, eight hydrogens."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "But if we look a little bit closer, you see that this guy on the left has one, two, three, four carbons. And so does this guy on the right. He has one, two, three, four carbons. This guy on the left has two, four, six, seven, eight hydrogens. This guy on the right has two, four, six, eight hydrogens. And they both have one oxygen. So both of the molecular formulas for both of these things are four carbons, eight hydrogens, and one oxygen."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "This guy on the left has two, four, six, seven, eight hydrogens. This guy on the right has two, four, six, eight hydrogens. And they both have one oxygen. So both of the molecular formulas for both of these things are four carbons, eight hydrogens, and one oxygen. They're both C4H8O. So they have the same molecular formula. They're made up of the same thing."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So both of the molecular formulas for both of these things are four carbons, eight hydrogens, and one oxygen. They're both C4H8O. So they have the same molecular formula. They're made up of the same thing. So these are going to be isomers. And they're a special type of isomers. In this situation, we don't have the same bonds."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "They're made up of the same thing. So these are going to be isomers. And they're a special type of isomers. In this situation, we don't have the same bonds. We're made up of the same things, but the bonds, what is connected to what is different. So we call this a constitutional isomer. So we are essentially made up of the same things, but we are actually two different molecules."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "In this situation, we don't have the same bonds. We're made up of the same things, but the bonds, what is connected to what is different. So we call this a constitutional isomer. So we are essentially made up of the same things, but we are actually two different molecules. Actually, two very different molecules here. Now let's look at this next guy over here. So if we look at this molecule, it does look like this carbon is chiral."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So we are essentially made up of the same things, but we are actually two different molecules. Actually, two very different molecules here. Now let's look at this next guy over here. So if we look at this molecule, it does look like this carbon is chiral. It is an asymmetric carbon. It is bonded to four different groups, fluorine, bromine, hydrogen, and then a methyl group. And so is this one."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So if we look at this molecule, it does look like this carbon is chiral. It is an asymmetric carbon. It is bonded to four different groups, fluorine, bromine, hydrogen, and then a methyl group. And so is this one. And they're both made up of the same things. You have the carbon. And not only are they made up of the same things, but the bonding is the same."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And so is this one. And they're both made up of the same things. You have the carbon. And not only are they made up of the same things, but the bonding is the same. So carbon to a fluorine, carbon to a fluorine, carbon to a bromine, carbon to bromine, carbon to hydrogen in both, and then carbon to the methyl group in both. But they don't look quite the same. Are they mirror images?"}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And not only are they made up of the same things, but the bonding is the same. So carbon to a fluorine, carbon to a fluorine, carbon to a bromine, carbon to bromine, carbon to hydrogen in both, and then carbon to the methyl group in both. But they don't look quite the same. Are they mirror images? Well, no. This guy's mirror image would have the fluorine popping out here. The hydrogen going back here."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Are they mirror images? Well, no. This guy's mirror image would have the fluorine popping out here. The hydrogen going back here. And then would have the bromine pointing out here. Let's see if I can somehow get from this guy to that guy. Let me flip this guy first."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "The hydrogen going back here. And then would have the bromine pointing out here. Let's see if I can somehow get from this guy to that guy. Let me flip this guy first. So a good thing to do would be to just flip. Let me see the fastest way I could potentially get there. Well, let me just flip it like this."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Let me flip this guy first. So a good thing to do would be to just flip. Let me see the fastest way I could potentially get there. Well, let me just flip it like this. So I'm going to flip out of the page, you can imagine. I'm going to flip it like this. So I'm going to take this methyl group and then put it on the right-hand side."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Well, let me just flip it like this. So I'm going to flip out of the page, you can imagine. I'm going to flip it like this. So I'm going to take this methyl group and then put it on the right-hand side. And you can imagine I'm going to turn it so it would come out of the page and then go back down. So if I did that, what would it look like? I would have the carbon."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So I'm going to take this methyl group and then put it on the right-hand side. And you can imagine I'm going to turn it so it would come out of the page and then go back down. So if I did that, what would it look like? I would have the carbon. This carbon here, I would have the methyl group on that side now. And then since I flipped it over, the bromine was in the plane of the page. It'll still be in the plane of the page."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "I would have the carbon. This carbon here, I would have the methyl group on that side now. And then since I flipped it over, the bromine was in the plane of the page. It'll still be in the plane of the page. But since I flipped it over, the hydrogen, which was in the back, will now be in the front. And the fluorine will now be in the back, because I flipped it over. So the fluorine is now in the back."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It'll still be in the plane of the page. But since I flipped it over, the hydrogen, which was in the back, will now be in the front. And the fluorine will now be in the back, because I flipped it over. So the fluorine is now in the back. Now, how does this compare to that? Let's see if I can somehow get there. Well, if I take this fluorine and I rotate it to where the hydrogen is, and I take the hydrogen and rotate it to where, it's all going to happen at once, to where the bromine is, and I take the bromine and rotate it to where the fluorine is, I get that."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So the fluorine is now in the back. Now, how does this compare to that? Let's see if I can somehow get there. Well, if I take this fluorine and I rotate it to where the hydrogen is, and I take the hydrogen and rotate it to where, it's all going to happen at once, to where the bromine is, and I take the bromine and rotate it to where the fluorine is, I get that. So I can flip it and then I can rotate it around this bond axis right there, and I would get to that molecule there. So even though they look pretty different, with a flip and a rotation, you actually see that these are the same molecule. Next one."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Well, if I take this fluorine and I rotate it to where the hydrogen is, and I take the hydrogen and rotate it to where, it's all going to happen at once, to where the bromine is, and I take the bromine and rotate it to where the fluorine is, I get that. So I can flip it and then I can rotate it around this bond axis right there, and I would get to that molecule there. So even though they look pretty different, with a flip and a rotation, you actually see that these are the same molecule. Next one. So let's see, what do we have here? Let me switch colors. So over here, this part of both of these molecules look the same."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Next one. So let's see, what do we have here? Let me switch colors. So over here, this part of both of these molecules look the same. We have the carbons on both of them. This carbon looks like a chiral center. It's bonded to three different groups."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So over here, this part of both of these molecules look the same. We have the carbons on both of them. This carbon looks like a chiral center. It's bonded to three different groups. You might say, oh, it's two carbons, but this is a methyl group and then this side has all this business over it. So this is definitely a chiral carbon. And over here, same thing."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It's bonded to three different groups. You might say, oh, it's two carbons, but this is a methyl group and then this side has all this business over it. So this is definitely a chiral carbon. And over here, same thing. It's a chiral carbon. And this has the same thing. It's bonded to four different things."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And over here, same thing. It's a chiral carbon. And this has the same thing. It's bonded to four different things. So each of these molecules have two chiral carbons, and it looks like they're made up of the same things. And not only are they made up of the same things, but the bonds are made in the same way. So this carbon is bonded to a hydrogen and a fluorine, and then two other carbons, same thing."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It's bonded to four different things. So each of these molecules have two chiral carbons, and it looks like they're made up of the same things. And not only are they made up of the same things, but the bonds are made in the same way. So this carbon is bonded to a hydrogen and a fluorine, and then two other carbons, same thing. A hydrogen and a fluorine. Carbon, it looks like it's bonded to a hydrogen and a chlorine. So it's made up of the same constituents, and they're bonded in the same way."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So this carbon is bonded to a hydrogen and a fluorine, and then two other carbons, same thing. A hydrogen and a fluorine. Carbon, it looks like it's bonded to a hydrogen and a chlorine. So it's made up of the same constituents, and they're bonded in the same way. So these look like, but the bonding is a little bit different. Over here on this one on the left, the hydrogen goes in the back, and over here, the hydrogen's in the front. And over here, the chlorine's in back, and over here, the chlorine's in front."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So it's made up of the same constituents, and they're bonded in the same way. So these look like, but the bonding is a little bit different. Over here on this one on the left, the hydrogen goes in the back, and over here, the hydrogen's in the front. And over here, the chlorine's in back, and over here, the chlorine's in front. So these look like stereoisomers. You saw earlier in this video, you saw structural isomers. Made up of the same things, but the connections are all different."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And over here, the chlorine's in back, and over here, the chlorine's in front. So these look like stereoisomers. You saw earlier in this video, you saw structural isomers. Made up of the same things, but the connections are all different. Stereoisomers, they're made up of the same thing. The connections are the same, but their three-dimensional configuration is a little bit different. For example, here on this carbon, it's connected to the same things as this carbon, but over here, the fluorine's out front, and over here, the fluorine's backwards."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Made up of the same things, but the connections are all different. Stereoisomers, they're made up of the same thing. The connections are the same, but their three-dimensional configuration is a little bit different. For example, here on this carbon, it's connected to the same things as this carbon, but over here, the fluorine's out front, and over here, the fluorine's backwards. And the same thing for the chlorine here. It's back here, and it's front here. Now, let's see if they're related, I guess, in a more nuanced way."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "For example, here on this carbon, it's connected to the same things as this carbon, but over here, the fluorine's out front, and over here, the fluorine's backwards. And the same thing for the chlorine here. It's back here, and it's front here. Now, let's see if they're related, I guess, in a more nuanced way. Well, you could imagine putting a mirror behind this molecule. If you put a mirror behind this molecule, what would its reflection look like? So if you put a mirror behind it, in the image of the mirror, this hydrogen would now, since the mirror's behind this whole molecule, this hydrogen's actually closer to the mirror."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Now, let's see if they're related, I guess, in a more nuanced way. Well, you could imagine putting a mirror behind this molecule. If you put a mirror behind this molecule, what would its reflection look like? So if you put a mirror behind it, in the image of the mirror, this hydrogen would now, since the mirror's behind this whole molecule, this hydrogen's actually closer to the mirror. So in the mirror image, you would have a hydrogen that's pointed out, and then you would have the carbon, and then you would have the fluorine being further away. And same thing in the mirror image here. You would have the chlorine coming closer, since this chlorine is further back, closer to the mirror, and then you would have the hydrogen pointing outwards like that."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So if you put a mirror behind it, in the image of the mirror, this hydrogen would now, since the mirror's behind this whole molecule, this hydrogen's actually closer to the mirror. So in the mirror image, you would have a hydrogen that's pointed out, and then you would have the carbon, and then you would have the fluorine being further away. And same thing in the mirror image here. You would have the chlorine coming closer, since this chlorine is further back, closer to the mirror, and then you would have the hydrogen pointing outwards like that. And then obviously, the rest of the molecule would look exactly the same. And so this mirror image that I just thought about in white is exactly what this molecule is. Hydrogen pointing out in front, hydrogen pointing out in front."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "You would have the chlorine coming closer, since this chlorine is further back, closer to the mirror, and then you would have the hydrogen pointing outwards like that. And then obviously, the rest of the molecule would look exactly the same. And so this mirror image that I just thought about in white is exactly what this molecule is. Hydrogen pointing out in front, hydrogen pointing out in front. You might say, wait, this hydrogen is on the right, this one's on the left. It doesn't matter. This is actually saying that the hydrogen's pointing out front, fluorine's pointing out back, hydrogen out front, fluorine back, chlorine out front, hydrogen back."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Hydrogen pointing out in front, hydrogen pointing out in front. You might say, wait, this hydrogen is on the right, this one's on the left. It doesn't matter. This is actually saying that the hydrogen's pointing out front, fluorine's pointing out back, hydrogen out front, fluorine back, chlorine out front, hydrogen back. So these are actually mirror images, but they're not the easy mirror images that we've done in the past, where the mirror was just like that in between the two. This one is a mirror image where you'd place the mirror either on top of or behind one of the molecules. So this is a class of stereoisomers, and we've brought up this word before."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "This is actually saying that the hydrogen's pointing out front, fluorine's pointing out back, hydrogen out front, fluorine back, chlorine out front, hydrogen back. So these are actually mirror images, but they're not the easy mirror images that we've done in the past, where the mirror was just like that in between the two. This one is a mirror image where you'd place the mirror either on top of or behind one of the molecules. So this is a class of stereoisomers, and we've brought up this word before. We call this enantiomers. So each of these are an enantiomer. I'll say they are enantiomers of each other."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So this is a class of stereoisomers, and we've brought up this word before. We call this enantiomers. So each of these are an enantiomer. I'll say they are enantiomers of each other. They're stereoisomers. They're made up of the same molecules, so they have the same constituents. They also have the same connections."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "I'll say they are enantiomers of each other. They're stereoisomers. They're made up of the same molecules, so they have the same constituents. They also have the same connections. And not only do they have the same connections, that so far gets us a stereoisomer, but they're a special kind of stereoisomer called an enantiomer, where they are actual mirror images of each other. Now what is this one over here in blue? Just like the last one, it looks like it's made up of the same things."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "They also have the same connections. And not only do they have the same connections, that so far gets us a stereoisomer, but they're a special kind of stereoisomer called an enantiomer, where they are actual mirror images of each other. Now what is this one over here in blue? Just like the last one, it looks like it's made up of the same things. You have these carbons, these carbons, these carbons, and hydrogens up there. Same thing over there. You have a hydrogen, bromine, hydrogen and a bromine, hydrogen and chlorine, hydrogen and chlorine, hydrogen and chlorine, hydrogen and chlorine."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Just like the last one, it looks like it's made up of the same things. You have these carbons, these carbons, these carbons, and hydrogens up there. Same thing over there. You have a hydrogen, bromine, hydrogen and a bromine, hydrogen and chlorine, hydrogen and chlorine, hydrogen and chlorine, hydrogen and chlorine. So it's made up of the same things. They're connected in the same way. So they're definitely stereoisomers."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "You have a hydrogen, bromine, hydrogen and a bromine, hydrogen and chlorine, hydrogen and chlorine, hydrogen and chlorine, hydrogen and chlorine. So it's made up of the same things. They're connected in the same way. So they're definitely stereoisomers. Well, let's make sure they're not the same molecule first. Here hydrogen's in the front. There hydrogen's in the back."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So they're definitely stereoisomers. Well, let's make sure they're not the same molecule first. Here hydrogen's in the front. There hydrogen's in the back. Here hydrogen is in the back. Here hydrogen's in the front. So they're not the same molecule."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "There hydrogen's in the back. Here hydrogen is in the back. Here hydrogen's in the front. So they're not the same molecule. They have a different three-dimensional configuration, although the bond connections are the same. So these are stereoisomers. Let's see if they're enantiomers."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So they're not the same molecule. They have a different three-dimensional configuration, although the bond connections are the same. So these are stereoisomers. Let's see if they're enantiomers. So if we look at it like this, you put a mirror here, you wouldn't get this guy over here. Then you would have a chlorine out front and a hydrogen. So you won't get it if you get a mirror over there."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Let's see if they're enantiomers. So if we look at it like this, you put a mirror here, you wouldn't get this guy over here. Then you would have a chlorine out front and a hydrogen. So you won't get it if you get a mirror over there. But if we do the same exercise that we did in the last pair, if you put a mirror behind this guy, and I'm just going to focus on stuff that's just forward and back, because that's what's relevant if the mirror is sitting behind the molecule. So if the mirror is sitting behind the molecule, this bromine is actually closer to the mirror than that hydrogen. So the bromine will now be out front, and then the hydrogen will be in back."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So you won't get it if you get a mirror over there. But if we do the same exercise that we did in the last pair, if you put a mirror behind this guy, and I'm just going to focus on stuff that's just forward and back, because that's what's relevant if the mirror is sitting behind the molecule. So if the mirror is sitting behind the molecule, this bromine is actually closer to the mirror than that hydrogen. So the bromine will now be out front, and then the hydrogen will be in back. This hydrogen will be in the back. I'm trying to do kind of a mirror image if it's hard to conceptualize, and then that would all look the same. And then this chlorine will now be out front, and this hydrogen will now be in the back in our mirror image, if you can visualize it."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So the bromine will now be out front, and then the hydrogen will be in back. This hydrogen will be in the back. I'm trying to do kind of a mirror image if it's hard to conceptualize, and then that would all look the same. And then this chlorine will now be out front, and this hydrogen will now be in the back in our mirror image, if you can visualize it. And then we have another one, and then this chlorine is closer to the mirror that it's sitting on top of. So in the mirror image, it would be pointing out, and then this hydrogen would be pointing back. Now let's see, is our mirror image the same as this?"}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And then this chlorine will now be out front, and this hydrogen will now be in the back in our mirror image, if you can visualize it. And then we have another one, and then this chlorine is closer to the mirror that it's sitting on top of. So in the mirror image, it would be pointing out, and then this hydrogen would be pointing back. Now let's see, is our mirror image the same as this? So in the mirror image, our bromine is pointing in the front, hydrogen in the back there. Then we have hydrogen in our mirror image. We have the hydrogen in back, chlorine in front, same there."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Now let's see, is our mirror image the same as this? So in the mirror image, our bromine is pointing in the front, hydrogen in the back there. Then we have hydrogen in our mirror image. We have the hydrogen in back, chlorine in front, same there. So so far, it's looking like a mirror image. And then in this last carbon over here, chlorine in front, hydrogen in back. But here we have chlorine in the back, hydrogen in front."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "We have the hydrogen in back, chlorine in front, same there. So so far, it's looking like a mirror image. And then in this last carbon over here, chlorine in front, hydrogen in back. But here we have chlorine in the back, hydrogen in front. So this part, you could think of it this way. This is the mirror image of this. This is the mirror image of this part."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "But here we have chlorine in the back, hydrogen in front. So this part, you could think of it this way. This is the mirror image of this. This is the mirror image of this part. But this is not the mirror image of that part. So when you have a stereoisomer that is not a mirror, when you have two stereoisomers that aren't mirror images of each other, we call them diastereomers. I always have trouble saying that."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "This is the mirror image of this part. But this is not the mirror image of that part. So when you have a stereoisomer that is not a mirror, when you have two stereoisomers that aren't mirror images of each other, we call them diastereomers. I always have trouble saying that. Let me write it. These are diastereomers, which is essentially saying it's a stereoisomer that is not an enantiomer. That's all it means."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "I always have trouble saying that. Let me write it. These are diastereomers, which is essentially saying it's a stereoisomer that is not an enantiomer. That's all it means. Stereoisomer, not enantiomer. A stereoisomer is either going to be an enantiomer or a diastereomer. Now let's do this last one."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "That's all it means. Stereoisomer, not enantiomer. A stereoisomer is either going to be an enantiomer or a diastereomer. Now let's do this last one. Let's see. We have this kind of cyclohexane ring, and you have a bromo on, I guess, the number 1 and the number 2 group, depending how you think about it. It looks like they are mirror images of each other."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Now let's do this last one. Let's see. We have this kind of cyclohexane ring, and you have a bromo on, I guess, the number 1 and the number 2 group, depending how you think about it. It looks like they are mirror images of each other. We could put a mirror right there, and they definitely look like mirror images. And this is a chiral carbon here. It's bonded to one carbon group that is different than this carbon group."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It looks like they are mirror images of each other. We could put a mirror right there, and they definitely look like mirror images. And this is a chiral carbon here. It's bonded to one carbon group that is different than this carbon group. This carbon group has a bromine. This carbon group just has a bunch of hydrogens on it, if you kind of go in that direction. So it's hydrogen and then a bromine."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It's bonded to one carbon group that is different than this carbon group. This carbon group has a bromine. This carbon group just has a bunch of hydrogens on it, if you kind of go in that direction. So it's hydrogen and then a bromine. So that is chiral. And then same argument. That is also chiral."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So it's hydrogen and then a bromine. So that is chiral. And then same argument. That is also chiral. And obviously this one is chiral and that is chiral. But if you think about it, they are mirror images of each other, and they each have two chiral centers or two chiral carbons. But if you think about it, all you have to do is flip this guy over, and you will get this molecule."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "That is also chiral. And obviously this one is chiral and that is chiral. But if you think about it, they are mirror images of each other, and they each have two chiral centers or two chiral carbons. But if you think about it, all you have to do is flip this guy over, and you will get this molecule. These are the same molecules. So it is the same molecule. So this is interesting."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "But if you think about it, all you have to do is flip this guy over, and you will get this molecule. These are the same molecules. So it is the same molecule. So this is interesting. And we saw this when we first learned about chirality. Even though we have two chiral centers, this is not a chiral molecule. It is the same thing as its mirror image."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "So this is interesting. And we saw this when we first learned about chirality. Even though we have two chiral centers, this is not a chiral molecule. It is the same thing as its mirror image. It is superimposable on its mirror image. So even though it has chiral carbons in it, it is not a chiral molecule. And we call these mesocompounds."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It is the same thing as its mirror image. It is superimposable on its mirror image. So even though it has chiral carbons in it, it is not a chiral molecule. And we call these mesocompounds. And we could point to one of them, because they really are the same compound. This is a mesocompound. It has chiral centers."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And we call these mesocompounds. And we could point to one of them, because they really are the same compound. This is a mesocompound. It has chiral centers. It has chiral carbons, I guess you could say it. But it is not a chiral compound. And the way to spot these, fairly straightforward, is that you have chiral centers, but there is a line of symmetry here."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "It has chiral centers. It has chiral carbons, I guess you could say it. But it is not a chiral compound. And the way to spot these, fairly straightforward, is that you have chiral centers, but there is a line of symmetry here. There is a line of symmetry right here. These two sides of the compound are mirror images of each other. Now, these would not be the same molecule if I changed that to a fluorine."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "And the way to spot these, fairly straightforward, is that you have chiral centers, but there is a line of symmetry here. There is a line of symmetry right here. These two sides of the compound are mirror images of each other. Now, these would not be the same molecule if I changed that to a fluorine. Then all of a sudden, you do not have this symmetry. These are mirror images, but they would not be superimposable. So if that was a fluorine, these would actually be enantiomers."}, {"video_title": "Stereoisomers, enantiomers, diastereomers, constitutional isomers and meso compounds Khan Academy.mp3", "Sentence": "Now, these would not be the same molecule if I changed that to a fluorine. Then all of a sudden, you do not have this symmetry. These are mirror images, but they would not be superimposable. So if that was a fluorine, these would actually be enantiomers. And this would not be only one mesocompound. It would be two different enantiomers. And one of them would have a r direction, and one of them would have an s direction, if we go with the naming conventions that we learned in the last few videos."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here I have a box, so here we have a box of mass m, so this box is mass m here, and there's no friction between the box and the ground here, but the box is attached to a wall right here by a spring, and so if you grab this box and you pull this box to the right, so you're gonna apply a force to the right, and so you're going to stretch the spring, so you move the box over to here, and the spring would obviously stretch out to there, like that. Let's say you moved the box a distance of delta x, so from here to here, we move the box a distance delta x. The box is gonna feel a force pulling it to the left, so if you're just holding it there, you're holding it because there's a force of the spring that wants to pull the box back to the left, so this is the force of the spring, and according to Hooke's Law, the force of the spring is equal to negative kx, and the negative sign just means a restoring force, so this negative sign here means a restoring force, meaning the force of the spring is back towards this original position right here. K is called the spring constant, so this is the spring constant right here, and it depends on how strong or weak the spring is, so if you have a very stiff spring or a strong spring, you have an increased value for k. If there's a loose spring or a weak spring, a decreased value for k, so if there's a strong, let me go ahead and write that down, strong spring or stiff spring, increased value for k. If you have a weak or a loose spring, a decreased value for k. X refers to the displacement from the original position here, so we displaced the box delta x here, so that's what we're talking about here, so let's think about what these terms mean for the force of the spring. If you have a strong spring, the box is going to experience a stronger force. According to Hooke's Law, if you increase k, you increase the force of the spring, and also the more you stretch it, so if you increase the value for delta x, also the stronger the force of the spring would be. And so if you get into the physics of it, we're not gonna get too far into it, but you could set this equal to mass times acceleration, because negative kx, f is equal to ma, that's equal to ma, and the acceleration is the second derivative of the position, so you could write the acceleration as being the second derivative of the position."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "K is called the spring constant, so this is the spring constant right here, and it depends on how strong or weak the spring is, so if you have a very stiff spring or a strong spring, you have an increased value for k. If there's a loose spring or a weak spring, a decreased value for k, so if there's a strong, let me go ahead and write that down, strong spring or stiff spring, increased value for k. If you have a weak or a loose spring, a decreased value for k. X refers to the displacement from the original position here, so we displaced the box delta x here, so that's what we're talking about here, so let's think about what these terms mean for the force of the spring. If you have a strong spring, the box is going to experience a stronger force. According to Hooke's Law, if you increase k, you increase the force of the spring, and also the more you stretch it, so if you increase the value for delta x, also the stronger the force of the spring would be. And so if you get into the physics of it, we're not gonna get too far into it, but you could set this equal to mass times acceleration, because negative kx, f is equal to ma, that's equal to ma, and the acceleration is the second derivative of the position, so you could write the acceleration as being the second derivative of the position. And I won't get into the physics of it, but eventually you can solve for the frequency of oscillation for this spring mass system. And when you do that math, let me go ahead and write down what you would get, you're going to get the frequency of oscillation is equal to one over two pi square root of k over m. So let's think about what the frequency of oscillation is referring to. So if you, once again, you pull the box to the right, so we're starting with our box at this position right here, and then if you release it, think about what happens, what's the motion of the box?"}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if you get into the physics of it, we're not gonna get too far into it, but you could set this equal to mass times acceleration, because negative kx, f is equal to ma, that's equal to ma, and the acceleration is the second derivative of the position, so you could write the acceleration as being the second derivative of the position. And I won't get into the physics of it, but eventually you can solve for the frequency of oscillation for this spring mass system. And when you do that math, let me go ahead and write down what you would get, you're going to get the frequency of oscillation is equal to one over two pi square root of k over m. So let's think about what the frequency of oscillation is referring to. So if you, once again, you pull the box to the right, so we're starting with our box at this position right here, and then if you release it, think about what happens, what's the motion of the box? Well, the force of the spring is going to cause the box to go towards this direction, it's gonna keep going because of energy, and it's going to go all the way over to here. And so now it's going to compress the spring. And now there's a force of the spring back towards this direction, like that."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you, once again, you pull the box to the right, so we're starting with our box at this position right here, and then if you release it, think about what happens, what's the motion of the box? Well, the force of the spring is going to cause the box to go towards this direction, it's gonna keep going because of energy, and it's going to go all the way over to here. And so now it's going to compress the spring. And now there's a force of the spring back towards this direction, like that. And of course, the spring is now compressed, all the energy is stored in the compression of the spring, it pushes on the box, and the box therefore goes back towards the equilibrium position, the center, but it's gonna keep on going until it reaches this position where we started, so the original position. So that's like one oscillation. And the time it takes for one oscillation is called the period."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And now there's a force of the spring back towards this direction, like that. And of course, the spring is now compressed, all the energy is stored in the compression of the spring, it pushes on the box, and the box therefore goes back towards the equilibrium position, the center, but it's gonna keep on going until it reaches this position where we started, so the original position. So that's like one oscillation. And the time it takes for one oscillation is called the period. So that would be the period, the period is measured in seconds. So the period is measured in seconds, the time it takes for one oscillation. Alright, so one over the period, one over the period is equal to the frequency, so you could write frequency like that, or you could write frequency like this."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the time it takes for one oscillation is called the period. So that would be the period, the period is measured in seconds. So the period is measured in seconds, the time it takes for one oscillation. Alright, so one over the period, one over the period is equal to the frequency, so you could write frequency like that, or you could write frequency like this. And so the frequency would be equal, the units would be one over seconds, and this is talking about the number of oscillations per second. So frequency is number of oscillations, number of oscillations per second. So what affects the frequency?"}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Alright, so one over the period, one over the period is equal to the frequency, so you could write frequency like that, or you could write frequency like this. And so the frequency would be equal, the units would be one over seconds, and this is talking about the number of oscillations per second. So frequency is number of oscillations, number of oscillations per second. So what affects the frequency? Well, once again, the spring constant affects the frequency, so k. So if you increase the value for k, you're going to increase the frequency. If you have a strong spring, it's gonna cause that mass to oscillate faster. And, so let me write that down, you increase k, increase the strength of the spring, you increase the frequency."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So what affects the frequency? Well, once again, the spring constant affects the frequency, so k. So if you increase the value for k, you're going to increase the frequency. If you have a strong spring, it's gonna cause that mass to oscillate faster. And, so let me write that down, you increase k, increase the strength of the spring, you increase the frequency. What about the mass? If you increase the mass, what happens to the frequency? If you increase this number, just think about it mathematically, that would decrease this number."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And, so let me write that down, you increase k, increase the strength of the spring, you increase the frequency. What about the mass? If you increase the mass, what happens to the frequency? If you increase this number, just think about it mathematically, that would decrease this number. So an increase in the mass would decrease the frequency. You wouldn't get as many oscillations per second, you'd get a slower oscillation. And so this is what we're thinking about when we're treating this bond as a spring."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you increase this number, just think about it mathematically, that would decrease this number. So an increase in the mass would decrease the frequency. You wouldn't get as many oscillations per second, you'd get a slower oscillation. And so this is what we're thinking about when we're treating this bond as a spring. So let's go back up here to this diagram, where we have the bond as a spring. So let's think about keeping the carbon stationary for now. And so we keep the carbon stationary, and we're gonna pull on the hydrogen here."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this is what we're thinking about when we're treating this bond as a spring. So let's go back up here to this diagram, where we have the bond as a spring. So let's think about keeping the carbon stationary for now. And so we keep the carbon stationary, and we're gonna pull on the hydrogen here. So we're gonna pull to the right. So let me go ahead and draw a line. So we're gonna pull to the right and stretch the spring out."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so we keep the carbon stationary, and we're gonna pull on the hydrogen here. So we're gonna pull to the right. So let me go ahead and draw a line. So we're gonna pull to the right and stretch the spring out. So we're gonna pull the hydrogen here to the right. So the force that we apply is in this direction. And so the hydrogen feels a restoring force in this direction, and that's the force of the spring."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna pull to the right and stretch the spring out. So we're gonna pull the hydrogen here to the right. So the force that we apply is in this direction. And so the hydrogen feels a restoring force in this direction, and that's the force of the spring. And so once again, if you have a really strong bond, so a really strong bond, that means you have an increased value for K. So a stronger bond means a stronger spring constant, if you will. And so you're gonna have an increased frequency of oscillation. So after you release this hydrogen, it's going to oscillate in a way analogous to this spring mass system here."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the hydrogen feels a restoring force in this direction, and that's the force of the spring. And so once again, if you have a really strong bond, so a really strong bond, that means you have an increased value for K. So a stronger bond means a stronger spring constant, if you will. And so you're gonna have an increased frequency of oscillation. So after you release this hydrogen, it's going to oscillate in a way analogous to this spring mass system here. So once again, if you increase the strength of the spring, you increase the frequency. What happens if you change the mass? So what happens if you, let me use a different color here."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So after you release this hydrogen, it's going to oscillate in a way analogous to this spring mass system here. So once again, if you increase the strength of the spring, you increase the frequency. What happens if you change the mass? So what happens if you, let me use a different color here. So what happens if you change the mass? So instead of hydrogen, if you change it to carbon or oxygen or something that has more mass than hydrogen, what would happen to the frequency of oscillation? If you increase the mass, you decrease the frequency of oscillation."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So what happens if you, let me use a different color here. So what happens if you change the mass? So instead of hydrogen, if you change it to carbon or oxygen or something that has more mass than hydrogen, what would happen to the frequency of oscillation? If you increase the mass, you decrease the frequency of oscillation. And so this is a pretty good model if you think about it this way. However, we're talking about only the hydrogen moving this time, but we know that when we're talking about a stretching vibration, both of these are moving. So let's get some more space and let's deal with that next."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you increase the mass, you decrease the frequency of oscillation. And so this is a pretty good model if you think about it this way. However, we're talking about only the hydrogen moving this time, but we know that when we're talking about a stretching vibration, both of these are moving. So let's get some more space and let's deal with that next. So we have this situation where we have two masses. I'm just gonna generalize it now. So we have M1 and then we have M2 over here and the bond between them."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's get some more space and let's deal with that next. So we have this situation where we have two masses. I'm just gonna generalize it now. So we have M1 and then we have M2 over here and the bond between them. So M1 and M2 are the masses of the nuclei of the atoms that we're talking about. So both masses are actually moving in our situation. So we have to amend the equation for frequency slightly."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have M1 and then we have M2 over here and the bond between them. So M1 and M2 are the masses of the nuclei of the atoms that we're talking about. So both masses are actually moving in our situation. So we have to amend the equation for frequency slightly. So let's go back up here and let's look at this equation again. So the frequency of oscillation is equal to one over two pi square root of K over M. But this is assuming that only one mass is moving. We have both masses moving when we're talking about a stretching vibration."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have to amend the equation for frequency slightly. So let's go back up here and let's look at this equation again. So the frequency of oscillation is equal to one over two pi square root of K over M. But this is assuming that only one mass is moving. We have both masses moving when we're talking about a stretching vibration. And so we have to use something different for M. We use what's called the reduced mass. So let me go ahead and write it down here. So the frequency is equal to one over two pi square root of K over M. But we can't write M anymore because we have a slightly different situation."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have both masses moving when we're talking about a stretching vibration. And so we have to use something different for M. We use what's called the reduced mass. So let me go ahead and write it down here. So the frequency is equal to one over two pi square root of K over M. But we can't write M anymore because we have a slightly different situation. We're gonna use that symbol to represent the reduced mass. So the reduced mass is equal to the M1 times M2 divided by M1 plus M2. And we're talking about the mass of the nuclei in AMUs."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the frequency is equal to one over two pi square root of K over M. But we can't write M anymore because we have a slightly different situation. We're gonna use that symbol to represent the reduced mass. So the reduced mass is equal to the M1 times M2 divided by M1 plus M2. And we're talking about the mass of the nuclei in AMUs. We'll worry more about units in the next video. And you can use the atomic mass to get an approximate value for the mass of the nuclei. So let's think about a carbon-hydrogen bond."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we're talking about the mass of the nuclei in AMUs. We'll worry more about units in the next video. And you can use the atomic mass to get an approximate value for the mass of the nuclei. So let's think about a carbon-hydrogen bond. So a carbon-hydrogen bond. So that would be like M1 is equal to carbon and M2 is equal to hydrogen. What is the reduced mass equal to?"}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about a carbon-hydrogen bond. So a carbon-hydrogen bond. So that would be like M1 is equal to carbon and M2 is equal to hydrogen. What is the reduced mass equal to? So the reduced mass is equal to, well, the atomic mass of carbon is 12. So it'd be 12 times the atomic mass of hydrogen which is one. Divide that by 12 plus one."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What is the reduced mass equal to? So the reduced mass is equal to, well, the atomic mass of carbon is 12. So it'd be 12 times the atomic mass of hydrogen which is one. Divide that by 12 plus one. So let's do that math. So let's go ahead and do that really quickly. So 12 times one is 12 divided by 12 plus one which is 13."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Divide that by 12 plus one. So let's do that math. So let's go ahead and do that really quickly. So 12 times one is 12 divided by 12 plus one which is 13. And so we get.923. So we get.923. So that's the reduced mass of our system."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 12 times one is 12 divided by 12 plus one which is 13. And so we get.923. So we get.923. So that's the reduced mass of our system. Let's do another one. Let's do carbon-carbon this time. So carbon-carbon."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's the reduced mass of our system. Let's do another one. Let's do carbon-carbon this time. So carbon-carbon. So the reduced mass is equal to 12 times 12 over 12 plus 12. So what is the reduced mass equal to now? So 12 times 12 is 144."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So carbon-carbon. So the reduced mass is equal to 12 times 12 over 12 plus 12. So what is the reduced mass equal to now? So 12 times 12 is 144. Divide that by 12 plus 12 which is 24. And so we get six. So the reduced mass is equal to six here."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 12 times 12 is 144. Divide that by 12 plus 12 which is 24. And so we get six. So the reduced mass is equal to six here. So let's think about what that does to the frequency of vibration. So if we increase the reduced mass, so if we're going from a reduced mass of.923 to a reduced mass of six, we're increasing this. We're increasing the reduced mass."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the reduced mass is equal to six here. So let's think about what that does to the frequency of vibration. So if we increase the reduced mass, so if we're going from a reduced mass of.923 to a reduced mass of six, we're increasing this. We're increasing the reduced mass. What happens to the frequency of vibration? Well, if we increase this value, of course we're going to decrease the frequency of vibration. So we would expect a carbon-carbon single bond to have a lower frequency of vibration than a carbon-hydrogen single bond."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're increasing the reduced mass. What happens to the frequency of vibration? Well, if we increase this value, of course we're going to decrease the frequency of vibration. So we would expect a carbon-carbon single bond to have a lower frequency of vibration than a carbon-hydrogen single bond. So let's do this idea one more time except let's think about a double bond. So this is a carbon-carbon single bond. What about a carbon-carbon double bond?"}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we would expect a carbon-carbon single bond to have a lower frequency of vibration than a carbon-hydrogen single bond. So let's do this idea one more time except let's think about a double bond. So this is a carbon-carbon single bond. What about a carbon-carbon double bond? Well, the reduced mass would be the same. It would still be six. But what happens to the spring constant?"}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What about a carbon-carbon double bond? Well, the reduced mass would be the same. It would still be six. But what happens to the spring constant? What happens to the force constant? Well, for a carbon-carbon double bond, we can pretend like this is twice as strong as a single bond. So if the value for the single bond, if the spring constant was K for the single bond, for the double bond, it would be 2K."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But what happens to the spring constant? What happens to the force constant? Well, for a carbon-carbon double bond, we can pretend like this is twice as strong as a single bond. So if the value for the single bond, if the spring constant was K for the single bond, for the double bond, it would be 2K. It would be twice that. And so we're increasing the value of the spring constant. We're increasing K because a double bond, we're assuming a double bond is twice as strong as a single bond."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if the value for the single bond, if the spring constant was K for the single bond, for the double bond, it would be 2K. It would be twice that. And so we're increasing the value of the spring constant. We're increasing K because a double bond, we're assuming a double bond is twice as strong as a single bond. So what happens to the frequency of vibration if you increase K? If you increase this value, you're going to increase this value. So if you increase K, you're going to increase the frequency of vibration because you have a stronger bond."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're increasing K because a double bond, we're assuming a double bond is twice as strong as a single bond. So what happens to the frequency of vibration if you increase K? If you increase this value, you're going to increase this value. So if you increase K, you're going to increase the frequency of vibration because you have a stronger bond. And so the bottom line is, the things you have to remember are stronger bonds vibrate faster. So that's what this is saying right here. A stronger bond, increase K, you get an increased frequency of vibration."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you increase K, you're going to increase the frequency of vibration because you have a stronger bond. And so the bottom line is, the things you have to remember are stronger bonds vibrate faster. So that's what this is saying right here. A stronger bond, increase K, you get an increased frequency of vibration. So a stronger bond vibrates faster. And what about a lighter atom here? So if we're thinking about a hydrogen, so a hydrogen, a lighter atom, that has a smaller value for the reduced mass."}, {"video_title": "Bonds as springs Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "A stronger bond, increase K, you get an increased frequency of vibration. So a stronger bond vibrates faster. And what about a lighter atom here? So if we're thinking about a hydrogen, so a hydrogen, a lighter atom, that has a smaller value for the reduced mass. So that's going to vibrate faster than a heavier atom. So stronger bonds vibrate faster and lighter atoms vibrate faster too. And this is what we gather from thinking about bonds and springs."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Here we have the dot structure for menthol, which is a famous compound found in various mint oils, like peppermint. Our goal is to draw both chair conformations for menthol, and then we're gonna choose the more stable one. The first thing we have to do is assign numbers to our substituents on the ring. And how you assign numbers does not have to follow IUPAC nomenclature. For example, I'm gonna call this carbon one, this carbon two, this carbon three, and this carbon four. And if you already know how to name this compound, you'll know that's not how to number it, according to IUPAC nomenclature. But this numbering system is just to help us draw our chair conformations."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And how you assign numbers does not have to follow IUPAC nomenclature. For example, I'm gonna call this carbon one, this carbon two, this carbon three, and this carbon four. And if you already know how to name this compound, you'll know that's not how to number it, according to IUPAC nomenclature. But this numbering system is just to help us draw our chair conformations. So let's start with one chair conformation. We've already seen how to draw chair conformations. You start with two parallel lines, one that's offset from the other."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "But this numbering system is just to help us draw our chair conformations. So let's start with one chair conformation. We've already seen how to draw chair conformations. You start with two parallel lines, one that's offset from the other. So here is our first parallel line, and then here is the other one. Next we draw two horizontal lines. So this horizontal line intersects with the top point on the top line, so something like that."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "You start with two parallel lines, one that's offset from the other. So here is our first parallel line, and then here is the other one. Next we draw two horizontal lines. So this horizontal line intersects with the top point on the top line, so something like that. So it comes close to this point. And then we draw another one down here that intersects with the bottom point on the bottom line, like that. Next we draw a line from the top dotted line down to the bottom."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this horizontal line intersects with the top point on the top line, so something like that. So it comes close to this point. And then we draw another one down here that intersects with the bottom point on the bottom line, like that. Next we draw a line from the top dotted line down to the bottom. And then we draw another line over here in parallel with the one that we just drew. So something like that. Next you put in your last set of parallel lines."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next we draw a line from the top dotted line down to the bottom. And then we draw another line over here in parallel with the one that we just drew. So something like that. Next you put in your last set of parallel lines. So from this point to this point, and then from this point to this point. So now we have our carbon skeleton. Now we need to put in our bonds."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next you put in your last set of parallel lines. So from this point to this point, and then from this point to this point. So now we have our carbon skeleton. Now we need to put in our bonds. And we call this carbon one. And we know we start axial up at carbon one. So there's axial up, carbon two is axial down."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now we need to put in our bonds. And we call this carbon one. And we know we start axial up at carbon one. So there's axial up, carbon two is axial down. We just keep alternating. Carbon three is axial up, carbon four would be down, carbon five is axial up, and then carbon six is axial down. Next we put in equatorial."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So there's axial up, carbon two is axial down. We just keep alternating. Carbon three is axial up, carbon four would be down, carbon five is axial up, and then carbon six is axial down. Next we put in equatorial. So at carbon one, this would be equatorial down. Carbon two would be up, carbon three would be down, carbon four would be up, carbon five down, and finally carbon six would be up. Let's look at our groups on the ring."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next we put in equatorial. So at carbon one, this would be equatorial down. Carbon two would be up, carbon three would be down, carbon four would be up, carbon five down, and finally carbon six would be up. Let's look at our groups on the ring. So we have a methyl group at carbon one. And since this is a wedge, that means it's going up in space. So we're gonna put the methyl group going up relative to the plane of the ring."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at our groups on the ring. So we have a methyl group at carbon one. And since this is a wedge, that means it's going up in space. So we're gonna put the methyl group going up relative to the plane of the ring. So that must mean the methyl group is axial, because axial's the only one that's going up at carbon one. Next we look at the OH. The OH is going up at carbon three."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna put the methyl group going up relative to the plane of the ring. So that must mean the methyl group is axial, because axial's the only one that's going up at carbon one. Next we look at the OH. The OH is going up at carbon three. So we need to, let's go ahead and number our ring here. So this would be carbon one, carbon two, and then carbon three. So we need to put the OH going up at carbon three."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "The OH is going up at carbon three. So we need to, let's go ahead and number our ring here. So this would be carbon one, carbon two, and then carbon three. So we need to put the OH going up at carbon three. And the only way we can do that is by putting the OH on axial. So we're gonna put the OH here on carbon three, so going up in space. And then finally we have our isopropyl group at carbon four."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we need to put the OH going up at carbon three. And the only way we can do that is by putting the OH on axial. So we're gonna put the OH here on carbon three, so going up in space. And then finally we have our isopropyl group at carbon four. And this is a dash, so this is going away from us in space, or down relative to the plane of the ring. So over here, let me go ahead and number carbon four. And we have two choices."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then finally we have our isopropyl group at carbon four. And this is a dash, so this is going away from us in space, or down relative to the plane of the ring. So over here, let me go ahead and number carbon four. And we have two choices. So where do we put our isopropyl group? It must be going down relative to the plane of the ring, which must mean it's axial. Again, that's the only one that's going down."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And we have two choices. So where do we put our isopropyl group? It must be going down relative to the plane of the ring, which must mean it's axial. Again, that's the only one that's going down. So let's draw in our isopropyl group at carbon four. So there's one chair conformation for menthol. Alright, we know this chair conformation is in equilibrium with our other chair conformations."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Again, that's the only one that's going down. So let's draw in our isopropyl group at carbon four. So there's one chair conformation for menthol. Alright, we know this chair conformation is in equilibrium with our other chair conformations. Let's go ahead and draw that. So two parallel lines that are offset from each other, so something like that. And then we draw in our dotted lines right here, just as guidelines to help us as we're drawing our chair conformation."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Alright, we know this chair conformation is in equilibrium with our other chair conformations. Let's go ahead and draw that. So two parallel lines that are offset from each other, so something like that. And then we draw in our dotted lines right here, just as guidelines to help us as we're drawing our chair conformation. We go from this point down to our bottom line, we go from our bottom line up to our top line, and then we need to connect the dots to finish our chair conformation. We know this is carbon one now, so we start axial down at carbon one, so it's the opposite of the other chair conformation. At carbon two would be this point, that's axial up."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we draw in our dotted lines right here, just as guidelines to help us as we're drawing our chair conformation. We go from this point down to our bottom line, we go from our bottom line up to our top line, and then we need to connect the dots to finish our chair conformation. We know this is carbon one now, so we start axial down at carbon one, so it's the opposite of the other chair conformation. At carbon two would be this point, that's axial up. Carbon three would be axial down. Carbon four would be axial up. Carbon five would be down, and six would be up."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "At carbon two would be this point, that's axial up. Carbon three would be axial down. Carbon four would be axial up. Carbon five would be down, and six would be up. Next, equatorial. So at carbon one, this would be up, and then at carbon two, it would be down, so we just keep alternating. At carbon three, it would be up."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Carbon five would be down, and six would be up. Next, equatorial. So at carbon one, this would be up, and then at carbon two, it would be down, so we just keep alternating. At carbon three, it would be up. At carbon four, down. Carbon five, up. And carbon six, down."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "At carbon three, it would be up. At carbon four, down. Carbon five, up. And carbon six, down. So let me redo that carbon six one, that wasn't very good, so let me draw that one in again. Now, carbon one, this time we know when this undergoes a ring flip, this is carbon one. And we'll see this in the video that I'll show you in a few minutes."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And carbon six, down. So let me redo that carbon six one, that wasn't very good, so let me draw that one in again. Now, carbon one, this time we know when this undergoes a ring flip, this is carbon one. And we'll see this in the video that I'll show you in a few minutes. So that's carbon one, this is carbon two, this must be carbon three, and then this is carbon four. So let's put in our groups. We know at carbon one, we have a methyl group, and this methyl group is up relative to the plane of the ring."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And we'll see this in the video that I'll show you in a few minutes. So that's carbon one, this is carbon two, this must be carbon three, and then this is carbon four. So let's put in our groups. We know at carbon one, we have a methyl group, and this methyl group is up relative to the plane of the ring. We know when this undergoes a ring flip, the methyl group has to stay up. So the methyl group goes up axial to up equatorial when this undergoes a ring flip. Next, let's look at the OH."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We know at carbon one, we have a methyl group, and this methyl group is up relative to the plane of the ring. We know when this undergoes a ring flip, the methyl group has to stay up. So the methyl group goes up axial to up equatorial when this undergoes a ring flip. Next, let's look at the OH. So the OH is also up relative to the plane of the ring. So it's gonna go on carbon three, it's gonna go up relative to the plane of the ring. So it's up axial for the chair conformation on the left, and it's up equatorial for the chair conformation on the right."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at the OH. So the OH is also up relative to the plane of the ring. So it's gonna go on carbon three, it's gonna go up relative to the plane of the ring. So it's up axial for the chair conformation on the left, and it's up equatorial for the chair conformation on the right. Finally, we look at carbon four, where we have our isopropyl group, which is down relative to the plane of the ring, so it's gonna stay down, but it's gonna go from axial to equatorial, because that makes it down right here, so there's our isopropyl group going down relative to the plane of the ring here. Sometimes it's really hard to tell if an equatorial bond is up or down relative to the plane, and the way to tell is to look at the axial one. So this axial one at carbon four here is very obviously up, which must mean that this one is going down."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So it's up axial for the chair conformation on the left, and it's up equatorial for the chair conformation on the right. Finally, we look at carbon four, where we have our isopropyl group, which is down relative to the plane of the ring, so it's gonna stay down, but it's gonna go from axial to equatorial, because that makes it down right here, so there's our isopropyl group going down relative to the plane of the ring here. Sometimes it's really hard to tell if an equatorial bond is up or down relative to the plane, and the way to tell is to look at the axial one. So this axial one at carbon four here is very obviously up, which must mean that this one is going down. So that's a little trick to help you if you're stuck with those bonds. Alright, let me go ahead and put in hydrogens. So I went ahead and drew in every bond, so I might as well put in hydrogens in all of these chair conformations."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this axial one at carbon four here is very obviously up, which must mean that this one is going down. So that's a little trick to help you if you're stuck with those bonds. Alright, let me go ahead and put in hydrogens. So I went ahead and drew in every bond, so I might as well put in hydrogens in all of these chair conformations. So I'll leave off the hydrogens on the isopropyl group, so we put those in. You don't have to put in these hydrogens when you're drawing your chair conformations. Actually, it's probably easier to see if you leave them out, but if it helps you, if it helps you to get it in the drawing chair frame of mind, you can go ahead and do that."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So I went ahead and drew in every bond, so I might as well put in hydrogens in all of these chair conformations. So I'll leave off the hydrogens on the isopropyl group, so we put those in. You don't have to put in these hydrogens when you're drawing your chair conformations. Actually, it's probably easier to see if you leave them out, but if it helps you, if it helps you to get it in the drawing chair frame of mind, you can go ahead and do that. So now we have both chair conformations, and let's analyze them in terms of stability, but first let's check out the video to make sure that we drew the proper chair conformations. So here we have one chair conformation for menthol, and at carbon one, you can see we have a methyl group going up axial. If that's carbon one, then this is carbon two right here, and then at carbon three, we have an OH going up axial."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Actually, it's probably easier to see if you leave them out, but if it helps you, if it helps you to get it in the drawing chair frame of mind, you can go ahead and do that. So now we have both chair conformations, and let's analyze them in terms of stability, but first let's check out the video to make sure that we drew the proper chair conformations. So here we have one chair conformation for menthol, and at carbon one, you can see we have a methyl group going up axial. If that's carbon one, then this is carbon two right here, and then at carbon three, we have an OH going up axial. Then at carbon four, we have our isopropyl group, so down axial. If we undergo a ring flip, so if I rotate this carbon up in space, and then if I rotate this other carbon down, and if we turn it a little bit, we'll be able to see our chair conformation better. You can see all those groups we just talked about went from axial to equatorial, which is the more stable place for these relatively bulky substituents."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "If that's carbon one, then this is carbon two right here, and then at carbon three, we have an OH going up axial. Then at carbon four, we have our isopropyl group, so down axial. If we undergo a ring flip, so if I rotate this carbon up in space, and then if I rotate this other carbon down, and if we turn it a little bit, we'll be able to see our chair conformation better. You can see all those groups we just talked about went from axial to equatorial, which is the more stable place for these relatively bulky substituents. Let's compare the chair conformations that we drew with what we saw in the video. So in the video, this is carbon one, and you can see this methyl group is up axial, and that's what we have here. Our methyl group is up axial."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "You can see all those groups we just talked about went from axial to equatorial, which is the more stable place for these relatively bulky substituents. Let's compare the chair conformations that we drew with what we saw in the video. So in the video, this is carbon one, and you can see this methyl group is up axial, and that's what we have here. Our methyl group is up axial. This would be carbon two, this is carbon three, we have our OH up axial, just like we have it here, so up axial. Then at carbon four, we should have an isopropyl group going down in space, and this is carbon four. You can't really see the bond, but you can see that this isopropyl group is going down."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Our methyl group is up axial. This would be carbon two, this is carbon three, we have our OH up axial, just like we have it here, so up axial. Then at carbon four, we should have an isopropyl group going down in space, and this is carbon four. You can't really see the bond, but you can see that this isopropyl group is going down. When this chair conformation undergoes a ring flip, all of those axial bonds go equatorial. So on the right, we can see that. This would be carbon one now, and the methyl is still up relative to the plane of the ring, but it is equatorial."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "You can't really see the bond, but you can see that this isopropyl group is going down. When this chair conformation undergoes a ring flip, all of those axial bonds go equatorial. So on the right, we can see that. This would be carbon one now, and the methyl is still up relative to the plane of the ring, but it is equatorial. The hydrogen that was equatorial right here in our drawing, you can see now it's gone axial, just like we have here. This is carbon two, this is carbon three. So at carbon three, we have our OH up equatorial, just like we drew it up here, and then at carbon four, we have our isopropyl group down equatorial."}, {"video_title": "Polysubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This would be carbon one now, and the methyl is still up relative to the plane of the ring, but it is equatorial. The hydrogen that was equatorial right here in our drawing, you can see now it's gone axial, just like we have here. This is carbon two, this is carbon three. So at carbon three, we have our OH up equatorial, just like we drew it up here, and then at carbon four, we have our isopropyl group down equatorial. So here it is, down. We know that the equatorial position is the more stable position for a relatively bulky substituent. So if all three substituents are out to the side, this is the more stable conformation."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's think a little bit about the different clues that have led us to conclude that we have these lithospheric plates moving relative to each other. Now the first clue, and this is something that I think many students, even in elementary school, first experience when they first learn about geography, is it looks like the continents could kind of fit into each other. And the most obvious one of these is when you look at this kind of little pointy part of South America. And if you have a more detailed map, it really is amazing how well it seems to fit into the Nigerian basin right here in Africa. It looks like at one time this little pointy part was nudged into this part of Africa, that they were actually connected. And if you're a little bit more creative, there are other parts of the world that you can kind of start to see how they might have fit in with each other in the past. And that by itself, that's just a very small clue, but it kind of hints at, well maybe if at one time they were next to each other, or if this was kind of connected, then they've had to move the part at some time."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if you have a more detailed map, it really is amazing how well it seems to fit into the Nigerian basin right here in Africa. It looks like at one time this little pointy part was nudged into this part of Africa, that they were actually connected. And if you're a little bit more creative, there are other parts of the world that you can kind of start to see how they might have fit in with each other in the past. And that by itself, that's just a very small clue, but it kind of hints at, well maybe if at one time they were next to each other, or if this was kind of connected, then they've had to move the part at some time. Although it doesn't tell us that it's still moving, or what might have caused the movement. And it definitely doesn't definitively tell us that they even moved. Maybe this is just a coincidence that this coast of South America looks very similar to this coast right here of Africa."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that by itself, that's just a very small clue, but it kind of hints at, well maybe if at one time they were next to each other, or if this was kind of connected, then they've had to move the part at some time. Although it doesn't tell us that it's still moving, or what might have caused the movement. And it definitely doesn't definitively tell us that they even moved. Maybe this is just a coincidence that this coast of South America looks very similar to this coast right here of Africa. Now the next clues, which really came over, I would say about the last 60 or 70 years. The first clue is that, okay, if you go to the mid-Atlantic ridge right here, so if you look at the Atlantic Ocean, let me look at this photograph right over here. So this is, it's a little, you don't normally see the oceans highlighted like this, so let me make it very clear to you."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe this is just a coincidence that this coast of South America looks very similar to this coast right here of Africa. Now the next clues, which really came over, I would say about the last 60 or 70 years. The first clue is that, okay, if you go to the mid-Atlantic ridge right here, so if you look at the Atlantic Ocean, let me look at this photograph right over here. So this is, it's a little, you don't normally see the oceans highlighted like this, so let me make it very clear to you. This right here is South America. This is South America. This right here is Africa."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is, it's a little, you don't normally see the oceans highlighted like this, so let me make it very clear to you. This right here is South America. This is South America. This right here is Africa. This right here is North America. So if you actually look at the elevations in the middle of the ocean, people noticed in the middle of the 20th century that, gee, there's a ridge in the middle of the Atlantic Ocean. There's kind of a mountain ridge that goes straight up the middle of the Atlantic Ocean."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right here is Africa. This right here is North America. So if you actually look at the elevations in the middle of the ocean, people noticed in the middle of the 20th century that, gee, there's a ridge in the middle of the Atlantic Ocean. There's kind of a mountain ridge that goes straight up the middle of the Atlantic Ocean. So that by itself doesn't tell you that you have these plates that are moving apart, but it is kind of a curious thing to look at. And not only is there a ridge, there's a lot of underwater volcanic activity. You have magma flowing out and lava flowing into the water, and it's kind of forming this ridge that really goes across the whole Atlantic Ocean."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's kind of a mountain ridge that goes straight up the middle of the Atlantic Ocean. So that by itself doesn't tell you that you have these plates that are moving apart, but it is kind of a curious thing to look at. And not only is there a ridge, there's a lot of underwater volcanic activity. You have magma flowing out and lava flowing into the water, and it's kind of forming this ridge that really goes across the whole Atlantic Ocean. There are other ridges in the world like that, underwater ridges. You have one over here in the Pacific Ocean. You have these here in the Indian Ocean."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have magma flowing out and lava flowing into the water, and it's kind of forming this ridge that really goes across the whole Atlantic Ocean. There are other ridges in the world like that, underwater ridges. You have one over here in the Pacific Ocean. You have these here in the Indian Ocean. So that by itself, that's just a little clue, but that by itself doesn't explain, doesn't tell you that these plates are actually moving apart at the ridge. The more conclusive, this is just the beginning of the clue, but what made this conclusive is, one, the separate discovery. This is what's interesting, is that you have these separate discoveries in different domains that eventually let you kind of come to a pretty neat conclusion."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have these here in the Indian Ocean. So that by itself, that's just a little clue, but that by itself doesn't explain, doesn't tell you that these plates are actually moving apart at the ridge. The more conclusive, this is just the beginning of the clue, but what made this conclusive is, one, the separate discovery. This is what's interesting, is that you have these separate discoveries in different domains that eventually let you kind of come to a pretty neat conclusion. So you've had a separate discovery that if you look at different eras of magnetic rock, or maybe I should say different magnetic rock from different periods in geologic time, and you can tell where they are in geologic time by how they're layered. So this would be newer rock, this would be newer, and then this would be a little bit older, and then this would be even older. Geologists noticed something interesting."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is what's interesting, is that you have these separate discoveries in different domains that eventually let you kind of come to a pretty neat conclusion. So you've had a separate discovery that if you look at different eras of magnetic rock, or maybe I should say different magnetic rock from different periods in geologic time, and you can tell where they are in geologic time by how they're layered. So this would be newer rock, this would be newer, and then this would be a little bit older, and then this would be even older. Geologists noticed something interesting. If I were to take magnetic rock, and if it was molten lava, and if it were to harden, remember, it's magnetic rock, so it would want to align with the poles the same way a compass would. So if I had a bunch of magnetic, so let's say this is some lava right here, and so the molecules can align themselves. Since when it's liquid and they can align themselves, they're going to naturally want to align with the poles."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Geologists noticed something interesting. If I were to take magnetic rock, and if it was molten lava, and if it were to harden, remember, it's magnetic rock, so it would want to align with the poles the same way a compass would. So if I had a bunch of magnetic, so let's say this is some lava right here, and so the molecules can align themselves. Since when it's liquid and they can align themselves, they're going to naturally want to align with the poles. So they'll naturally all want to align in one direction because of Earth's magnetic field. And so when that lava hardens into actual rock, that alignment will kind of be frozen. Now, if Earth's magnetic field was constant over time, then when you look at magnetic rocks from any period, you would expect them all to be aligned in the same direction."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Since when it's liquid and they can align themselves, they're going to naturally want to align with the poles. So they'll naturally all want to align in one direction because of Earth's magnetic field. And so when that lava hardens into actual rock, that alignment will kind of be frozen. Now, if Earth's magnetic field was constant over time, then when you look at magnetic rocks from any period, you would expect them all to be aligned in the same direction. So since we're taking a cross-section of rock here, let's say an alignment towards the North Pole looks like this. I draw it like that, that's kind of an arrow pointing into our screen. And let's say an alignment pointing to the South Pole would look like this."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, if Earth's magnetic field was constant over time, then when you look at magnetic rocks from any period, you would expect them all to be aligned in the same direction. So since we're taking a cross-section of rock here, let's say an alignment towards the North Pole looks like this. I draw it like that, that's kind of an arrow pointing into our screen. And let's say an alignment pointing to the South Pole would look like this. This would be an arrow pointing out of our screen. So what you would expect is the newer rock, that kind of the alignment of the rock would go into the screen. And then the older rock, it would still go older into the screen."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say an alignment pointing to the South Pole would look like this. This would be an arrow pointing out of our screen. So what you would expect is the newer rock, that kind of the alignment of the rock would go into the screen. And then the older rock, it would still go older into the screen. So if I were to draw a top view, let me draw it like this, just so I make sure that everyone is on the same page. Let me just draw a cross-section like this. So we know what we're talking about."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then the older rock, it would still go older into the screen. So if I were to draw a top view, let me draw it like this, just so I make sure that everyone is on the same page. Let me just draw a cross-section like this. So we know what we're talking about. Let me draw a cross-section like this. So this is the surface up here. This up here is the surface."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we know what we're talking about. Let me draw a cross-section like this. So this is the surface up here. This up here is the surface. When I talk about going into the page, that means that the magnetic rock would be aligned in that direction. And when I talk about going out of the page, that means that the magnetic rock would be aligned in that direction. Now, like I said, if the magnetic field of Earth never changed, then lava that essentially turns into hard, cools down into non-lava rock, or you can say freezes into rock, it would all point in the same direction regardless of when it hardened."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This up here is the surface. When I talk about going into the page, that means that the magnetic rock would be aligned in that direction. And when I talk about going out of the page, that means that the magnetic rock would be aligned in that direction. Now, like I said, if the magnetic field of Earth never changed, then lava that essentially turns into hard, cools down into non-lava rock, or you can say freezes into rock, it would all point in the same direction regardless of when it hardened. This would be the situation in a constant magnetic field. But what we've seen is that that's not the case. When you look at older magnetic rock, and depending on how old you go, you have the newer rock that's aligned with our current magnetic field."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, like I said, if the magnetic field of Earth never changed, then lava that essentially turns into hard, cools down into non-lava rock, or you can say freezes into rock, it would all point in the same direction regardless of when it hardened. This would be the situation in a constant magnetic field. But what we've seen is that that's not the case. When you look at older magnetic rock, and depending on how old you go, you have the newer rock that's aligned with our current magnetic field. You go a little bit older, and right now we think it's about 780,000 years ago. Roughly, you have to find rock of that age, magnetic rock that hardened at that time. It's actually in the opposite direction."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When you look at older magnetic rock, and depending on how old you go, you have the newer rock that's aligned with our current magnetic field. You go a little bit older, and right now we think it's about 780,000 years ago. Roughly, you have to find rock of that age, magnetic rock that hardened at that time. It's actually in the opposite direction. So it's actually the magnetic rock has hardened in a way so that it's pointing. It's as if the North Pole was at the South Pole now, the magnetic North Pole. So it's aligned in the opposite direction."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's actually in the opposite direction. So it's actually the magnetic rock has hardened in a way so that it's pointing. It's as if the North Pole was at the South Pole now, the magnetic North Pole. So it's aligned in the opposite direction. So it's kind of pointing out of the page here. And if you get even older rock, it's more aligned with our traditional direction. So it's more aligned than that."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's aligned in the opposite direction. So it's kind of pointing out of the page here. And if you get even older rock, it's more aligned with our traditional direction. So it's more aligned than that. And so the only conclusion, the only reasonable conclusion that we can draw from this is that Earth's magnetic field has actually fluctuated over time. Earth's magnetic field fluctuates. Now you're probably thinking, Sal, how is this relevant to plate tectonics?"}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's more aligned than that. And so the only conclusion, the only reasonable conclusion that we can draw from this is that Earth's magnetic field has actually fluctuated over time. Earth's magnetic field fluctuates. Now you're probably thinking, Sal, how is this relevant to plate tectonics? Well, once you accept that magnetic fields fluctuate over the history of the Earth, there's another interesting observation you can make about the rock that's kind of at the basin of the ocean floor. So not only do you have this mid-Atlantic ridge, do you have these volcanoes spewing kind of new rock into the ocean, creating this kind of underwater mountain ridge, but it also turns out that the rock that forms the sea floor also contains a lot of magnetite, which is magnetic. And what's really interesting about that, so let me draw, so let's say that this is, so we have a top view, just like we have over here."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now you're probably thinking, Sal, how is this relevant to plate tectonics? Well, once you accept that magnetic fields fluctuate over the history of the Earth, there's another interesting observation you can make about the rock that's kind of at the basin of the ocean floor. So not only do you have this mid-Atlantic ridge, do you have these volcanoes spewing kind of new rock into the ocean, creating this kind of underwater mountain ridge, but it also turns out that the rock that forms the sea floor also contains a lot of magnetite, which is magnetic. And what's really interesting about that, so let me draw, so let's say that this is, so we have a top view, just like we have over here. So let's say that this is the mid-Atlantic ridge right here. So mid-Atlantic ridge. Now this is really cool."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's really interesting about that, so let me draw, so let's say that this is, so we have a top view, just like we have over here. So let's say that this is the mid-Atlantic ridge right here. So mid-Atlantic ridge. Now this is really cool. So when they look at rocks that are very close to the mid-Atlantic ridge, they're aligned, and once again we're looking at rocks at the floor of the ocean, they're aligned in a way that you would expect with the current magnetic field. They're aligned just like that, the way that you would expect when you're looking at the magnetic rock that's close to the ridge. But if you go a little bit further, and when I say a little bit, I'm talking about thousands of miles, but when you go further out from that, you have stripes of other, you have other magnetic rock that is going in the opposite direction."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now this is really cool. So when they look at rocks that are very close to the mid-Atlantic ridge, they're aligned, and once again we're looking at rocks at the floor of the ocean, they're aligned in a way that you would expect with the current magnetic field. They're aligned just like that, the way that you would expect when you're looking at the magnetic rock that's close to the ridge. But if you go a little bit further, and when I say a little bit, I'm talking about thousands of miles, but when you go further out from that, you have stripes of other, you have other magnetic rock that is going in the opposite direction. It's going in the opposite direction. It's going like this. And what's even cooler than the idea that it's switched directions depending on how far you've gone from the rift, is that there's a symmetric stripe of magnetic rock on exactly the same distance, or roughly the same distance away from the rift, that's also pointing in that same direction."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if you go a little bit further, and when I say a little bit, I'm talking about thousands of miles, but when you go further out from that, you have stripes of other, you have other magnetic rock that is going in the opposite direction. It's going in the opposite direction. It's going like this. And what's even cooler than the idea that it's switched directions depending on how far you've gone from the rift, is that there's a symmetric stripe of magnetic rock on exactly the same distance, or roughly the same distance away from the rift, that's also pointing in that same direction. And you go a little bit further out, and you'll find some rock that's pointing in the original direction. And even better, you go on the symmetric other side of the actual ridge, and you find another set of rocks that's doing the exact same thing. So if you accept that Earth's magnetic field has kind of been flip-flopping over time, the only reasonable conclusion, at least that I could think of, or that the geologists can think of, is that, sure, all of this was formed at a similar period in time."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's even cooler than the idea that it's switched directions depending on how far you've gone from the rift, is that there's a symmetric stripe of magnetic rock on exactly the same distance, or roughly the same distance away from the rift, that's also pointing in that same direction. And you go a little bit further out, and you'll find some rock that's pointing in the original direction. And even better, you go on the symmetric other side of the actual ridge, and you find another set of rocks that's doing the exact same thing. So if you accept that Earth's magnetic field has kind of been flip-flopping over time, the only reasonable conclusion, at least that I could think of, or that the geologists can think of, is that, sure, all of this was formed at a similar period in time. This came out as lava, magnetic lava, and then it all aligned with Earth's magnetic field, and that's why it looks similar. You fast-forward in time some, and the only way that these could have formed, and they could have been so similar, so if we rewind in time, the only way that these purple magnetic rocks could have aligned this way, in exactly the same way, at exactly the same distance, is that at some point, they were much closer to each other, if they were actually connected. So if we rewind in time, maybe at the mid-Atlantic rift, you had all of the purple rock coming out from those underwater volcanoes, and at that time, Earth's magnetic field was the opposite as it is right now."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you accept that Earth's magnetic field has kind of been flip-flopping over time, the only reasonable conclusion, at least that I could think of, or that the geologists can think of, is that, sure, all of this was formed at a similar period in time. This came out as lava, magnetic lava, and then it all aligned with Earth's magnetic field, and that's why it looks similar. You fast-forward in time some, and the only way that these could have formed, and they could have been so similar, so if we rewind in time, the only way that these purple magnetic rocks could have aligned this way, in exactly the same way, at exactly the same distance, is that at some point, they were much closer to each other, if they were actually connected. So if we rewind in time, maybe at the mid-Atlantic rift, you had all of the purple rock coming out from those underwater volcanoes, and at that time, Earth's magnetic field was the opposite as it is right now. And then, of course, you had this blue rock that is looking like that. And so this seems like a reasonable explanation. This rock and this rock were at some point touching."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we rewind in time, maybe at the mid-Atlantic rift, you had all of the purple rock coming out from those underwater volcanoes, and at that time, Earth's magnetic field was the opposite as it is right now. And then, of course, you had this blue rock that is looking like that. And so this seems like a reasonable explanation. This rock and this rock were at some point touching. They were actually formed at the exact place and at the exact same time. And so if this is the case, if at one point, this purple rock was all together, and they formed at the same time at the mid-Atlantic rift, we're assuming all the rock was... Well, we don't have to make that assumption, but if you assume that they formed at the same time, and that based on the pattern, it really does look like they do, and it's a symmetric distance away from that rift, then the only reasonable conclusion I can think of is that the rift has had to move apart. The rift has had to move apart from this period to that period."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This rock and this rock were at some point touching. They were actually formed at the exact place and at the exact same time. And so if this is the case, if at one point, this purple rock was all together, and they formed at the same time at the mid-Atlantic rift, we're assuming all the rock was... Well, we don't have to make that assumption, but if you assume that they formed at the same time, and that based on the pattern, it really does look like they do, and it's a symmetric distance away from that rift, then the only reasonable conclusion I can think of is that the rift has had to move apart. The rift has had to move apart from this period to that period. And there was a time when all this blue rock was together. So that by itself, that frankly, is the most definitive evidence in the 1960s where it kind of became conclusive that you did have these plates that were moving away from each other. You did have these plates that were moving away from each other."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The rift has had to move apart from this period to that period. And there was a time when all this blue rock was together. So that by itself, that frankly, is the most definitive evidence in the 1960s where it kind of became conclusive that you did have these plates that were moving away from each other. You did have these plates that were moving away from each other. And obviously, if the plates are moving away from each other at some point, and that means that, you know, just based on the way the map looks, at some point, they're also going to be moving into each other. We could talk more about that in future videos. But, you know, right, at certain points, they're actually moving on one plate, is moving under another."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You did have these plates that were moving away from each other. And obviously, if the plates are moving away from each other at some point, and that means that, you know, just based on the way the map looks, at some point, they're also going to be moving into each other. We could talk more about that in future videos. But, you know, right, at certain points, they're actually moving on one plate, is moving under another. And we'll talk about how that might partially explain, and we'll talk about all of the explanations for why we think the plates might actually be moving. But now if we fast forward to more present times, now that we have GPS satellites and all the rest, we can actually measure the movement of the plates. This is actually an image from NASA showing the vector of the movements at different points on the surface of the planet."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But, you know, right, at certain points, they're actually moving on one plate, is moving under another. And we'll talk about how that might partially explain, and we'll talk about all of the explanations for why we think the plates might actually be moving. But now if we fast forward to more present times, now that we have GPS satellites and all the rest, we can actually measure the movement of the plates. This is actually an image from NASA showing the vector of the movements at different points on the surface of the planet. And you can see we've gotten a lot of vectors from the United States, so it's almost hard to read since it's so chock full of vectors. But you can see right over here in Hawaii, the Pacific plate at that point is moving in this northwest direction, as measured by GPS satellites. And I want to make clear, this movement is relatively slow."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is actually an image from NASA showing the vector of the movements at different points on the surface of the planet. And you can see we've gotten a lot of vectors from the United States, so it's almost hard to read since it's so chock full of vectors. But you can see right over here in Hawaii, the Pacific plate at that point is moving in this northwest direction, as measured by GPS satellites. And I want to make clear, this movement is relatively slow. It's roughly the speed at which fingernails grow. But if you do it over millions of years, that actually amounts to thousands of miles. So we're talking on the order of about a centimeter a year for most of the plates."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I want to make clear, this movement is relatively slow. It's roughly the speed at which fingernails grow. But if you do it over millions of years, that actually amounts to thousands of miles. So we're talking on the order of about a centimeter a year for most of the plates. Some of the plates might be moving a little bit faster, maybe close to 10 or 15 centimeters. But most are moving about a centimeter a year, at the same rate your fingernails are going. But this is fascinating because we can actually measure it because GPS is so accurate."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're talking on the order of about a centimeter a year for most of the plates. Some of the plates might be moving a little bit faster, maybe close to 10 or 15 centimeters. But most are moving about a centimeter a year, at the same rate your fingernails are going. But this is fascinating because we can actually measure it because GPS is so accurate. Over here it looks like the North American plate is kind of rotating generally in that direction. We have the Nazca plate right here is moving roughly in that direction, moving into the South American plate. I'll leave you there right now."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is fascinating because we can actually measure it because GPS is so accurate. Over here it looks like the North American plate is kind of rotating generally in that direction. We have the Nazca plate right here is moving roughly in that direction, moving into the South American plate. I'll leave you there right now. Actually, before I leave you there, this is another thing that I got off of Wikipedia that shows that same magnetic striping. It's maybe a slightly neater drawing. I don't know which one might be more helpful for you."}, {"video_title": "Plate tectonics Evidence of plate movement Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll leave you there right now. Actually, before I leave you there, this is another thing that I got off of Wikipedia that shows that same magnetic striping. It's maybe a slightly neater drawing. I don't know which one might be more helpful for you. But I'll leave you there in this video. In the next video, we'll think about some of the theories. We know now that the plates are moving."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we start with nitrobenzene. And if our goal is to nitrate nitrobenzene, we need to add something like fuming nitric acid, sulfuric acid as our catalyst. And we also need to heat it up. So we need to force this reaction to occur because nitrobenzene turns out to be not as reactive as benzene itself. And so the end result is to add a nitro group onto the meta position. So we added a nitro group onto the meta position on our benzene ring. And so the meta product dominates here, not the ortho-para."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we need to force this reaction to occur because nitrobenzene turns out to be not as reactive as benzene itself. And so the end result is to add a nitro group onto the meta position. So we added a nitro group onto the meta position on our benzene ring. And so the meta product dominates here, not the ortho-para. Let's see if we can figure out why by looking at some resonance structures for this mechanism. And so we'll start with an ortho attack. So in our first example here, we'll do an ortho attack where we add the nitro group onto the ortho position."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the meta product dominates here, not the ortho-para. Let's see if we can figure out why by looking at some resonance structures for this mechanism. And so we'll start with an ortho attack. So in our first example here, we'll do an ortho attack where we add the nitro group onto the ortho position. So remember, the function of the nitric acid and sulfuric acid is to generate your electrophile, which is the nitronium ion right here with a plus 1 formal charge on the nitrogen. And so that's our electrophile. The benzene ring is going to function as the nucleophile."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So in our first example here, we'll do an ortho attack where we add the nitro group onto the ortho position. So remember, the function of the nitric acid and sulfuric acid is to generate your electrophile, which is the nitronium ion right here with a plus 1 formal charge on the nitrogen. And so that's our electrophile. The benzene ring is going to function as the nucleophile. And so these pi electrons here are going to attack this nitrogen, pushing these electrons off onto the oxygen. So let's go ahead and look at the results of our nucleophilic attack. We have a nitro group right here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The benzene ring is going to function as the nucleophile. And so these pi electrons here are going to attack this nitrogen, pushing these electrons off onto the oxygen. So let's go ahead and look at the results of our nucleophilic attack. We have a nitro group right here. I want to show an ortho attack. So I'm going to show the NO2 adding onto the ortho position. So here's the ortho position with an NO2."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have a nitro group right here. I want to show an ortho attack. So I'm going to show the NO2 adding onto the ortho position. So here's the ortho position with an NO2. That carbon also has a hydrogen bonded to it. Let's go ahead and highlight some electrons. So these pi electrons right here are forming a bond between the nitrogen and the carbon on our ring."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So here's the ortho position with an NO2. That carbon also has a hydrogen bonded to it. Let's go ahead and highlight some electrons. So these pi electrons right here are forming a bond between the nitrogen and the carbon on our ring. So that takes a bond away from this carbon. And so that's where our plus 1 formal charge is going to go. And then, of course, we still have these pi electrons in our ring like that."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons right here are forming a bond between the nitrogen and the carbon on our ring. So that takes a bond away from this carbon. And so that's where our plus 1 formal charge is going to go. And then, of course, we still have these pi electrons in our ring like that. Let's go ahead and draw a resonance structure for this ion here. So we have a pi bond next to a positive charge. So we could take these pi electrons and move them over to here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then, of course, we still have these pi electrons in our ring like that. Let's go ahead and draw a resonance structure for this ion here. So we have a pi bond next to a positive charge. So we could take these pi electrons and move them over to here. And let's go ahead and draw that resonance structure. So we have our ring, top nitro group. Over here, we have the other nitro group in the ortho position."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we could take these pi electrons and move them over to here. And let's go ahead and draw that resonance structure. So we have our ring, top nitro group. Over here, we have the other nitro group in the ortho position. We have these pi electrons at the top. We have these pi electrons over here now. So let's highlight those."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Over here, we have the other nitro group in the ortho position. We have these pi electrons at the top. We have these pi electrons over here now. So let's highlight those. So the electrons in the pi electrons in blue have moved over to here, taking a bond away from this carbon. So that carbon gets a plus 1 formal charge like that. So there's a resonance structure."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's highlight those. So the electrons in the pi electrons in blue have moved over to here, taking a bond away from this carbon. So that carbon gets a plus 1 formal charge like that. So there's a resonance structure. Let's go ahead and draw one more. So I could show, once again, these pi electrons moving down to here. And we have our ring."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there's a resonance structure. Let's go ahead and draw one more. So I could show, once again, these pi electrons moving down to here. And we have our ring. We have this top nitro group, which I'm actually going to go ahead and draw in the formal charges. And you'll see why in a second. So this oxygen is a negative 1 formal charge."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have our ring. We have this top nitro group, which I'm actually going to go ahead and draw in the formal charges. And you'll see why in a second. So this oxygen is a negative 1 formal charge. And this nitrogen has a plus 1 formal charge like that. So over here, we have our other nitro group in the ortho position. We have pi electrons here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen is a negative 1 formal charge. And this nitrogen has a plus 1 formal charge like that. So over here, we have our other nitro group in the ortho position. We have pi electrons here. And we just moved some pi electrons over to this position. So let's go ahead and highlight those too. So over here in red, these pi electrons have moved over to here, taking a bond away from this top carbon here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have pi electrons here. And we just moved some pi electrons over to this position. So let's go ahead and highlight those too. So over here in red, these pi electrons have moved over to here, taking a bond away from this top carbon here. So that's where our plus 1 charge goes now. So we have a plus 1 formal charge on this top carbon here. And this is a destabilizing resonance structure."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So over here in red, these pi electrons have moved over to here, taking a bond away from this top carbon here. So that's where our plus 1 charge goes now. So we have a plus 1 formal charge on this top carbon here. And this is a destabilizing resonance structure. And we know that because we have a positive 1 formal charge in this nitrogen and a positive 1 formal charge on this carbon on our ring. And so like charges repel and therefore destabilize this resonance structure. So we have a destabilizing resonance structure."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this is a destabilizing resonance structure. And we know that because we have a positive 1 formal charge in this nitrogen and a positive 1 formal charge on this carbon on our ring. And so like charges repel and therefore destabilize this resonance structure. So we have a destabilizing resonance structure. And remember, it's actually really a hybrid of our resonance structures for our sigma complex. But one of those resonance structures is destabilizing, which means that this sigma complex is not very likely to form. So let's go ahead and look at a meta attack."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a destabilizing resonance structure. And remember, it's actually really a hybrid of our resonance structures for our sigma complex. But one of those resonance structures is destabilizing, which means that this sigma complex is not very likely to form. So let's go ahead and look at a meta attack. And you'll see that we will not have a destabilizing resonance structure when we do a meta attack. So let's go ahead and once again show our nitrobenzene and our nitronium ion. And this time, we will do a meta attack."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and look at a meta attack. And you'll see that we will not have a destabilizing resonance structure when we do a meta attack. So let's go ahead and once again show our nitrobenzene and our nitronium ion. And this time, we will do a meta attack. So if I want to show a nitro group adding on to the meta position, I would once again use these pi electrons. So nucleophilic attack pushes these electrons off. And so we're going to once again show the resulting carbocation here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this time, we will do a meta attack. So if I want to show a nitro group adding on to the meta position, I would once again use these pi electrons. So nucleophilic attack pushes these electrons off. And so we're going to once again show the resulting carbocation here. So we have a nitro group right here. And this time, we're showing the nitro group adding on meta. And once again, there's a hydrogen attached to our ring."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to once again show the resulting carbocation here. So we have a nitro group right here. And this time, we're showing the nitro group adding on meta. And once again, there's a hydrogen attached to our ring. And these pi electrons here are forming the bond between this carbon and our nitrogen, taking a bond away from this carbon. So that carbon gets a plus 1 formal charge. And we still have, of course, pi electrons in our ring."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, there's a hydrogen attached to our ring. And these pi electrons here are forming the bond between this carbon and our nitrogen, taking a bond away from this carbon. So that carbon gets a plus 1 formal charge. And we still have, of course, pi electrons in our ring. And so that's our first resonance structure. We can draw another one. I could take these pi electrons and move them over to here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we still have, of course, pi electrons in our ring. And so that's our first resonance structure. We can draw another one. I could take these pi electrons and move them over to here. So let's go ahead and show the next resonance structure with our ring, our nitro group here, our nitro group in the meta position, and hydrogen also attached to that carbon, pi electrons here. And pi electrons have moved over to here. So let me highlight those."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I could take these pi electrons and move them over to here. So let's go ahead and show the next resonance structure with our ring, our nitro group here, our nitro group in the meta position, and hydrogen also attached to that carbon, pi electrons here. And pi electrons have moved over to here. So let me highlight those. So these pi electrons have moved over to here, taking a bond away from this carbon. So we get a plus 1 formal charge here. We can draw another resonance structure, taking these pi electrons, moving them over to here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me highlight those. So these pi electrons have moved over to here, taking a bond away from this carbon. So we get a plus 1 formal charge here. We can draw another resonance structure, taking these pi electrons, moving them over to here. So let's go ahead and do that. We have our ring once again. We have a nitro group in the top carbon."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We can draw another resonance structure, taking these pi electrons, moving them over to here. So let's go ahead and do that. We have our ring once again. We have a nitro group in the top carbon. We have a nitro group in the meta position. Once again, we have hydrogen. We have these pi electrons."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have a nitro group in the top carbon. We have a nitro group in the meta position. Once again, we have hydrogen. We have these pi electrons. And we now have moved the pi electrons over to here. So let me highlight those. So in red, these pi electrons have moved over to this position."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have these pi electrons. And we now have moved the pi electrons over to here. So let me highlight those. So in red, these pi electrons have moved over to this position. Taking a bond away from this carbon. So we get a plus 1 formal charge on that carbon. And so these are the three resonance structures that you have for a meta attack."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So in red, these pi electrons have moved over to this position. Taking a bond away from this carbon. So we get a plus 1 formal charge on that carbon. And so these are the three resonance structures that you have for a meta attack. Notice, we don't have a destabilizing one. So in our three resonance structures, none of them have the two positive charges right next to each other, as we saw in the previous example. So it's not so much that the sigma complex for a meta attack is extra stable."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so these are the three resonance structures that you have for a meta attack. Notice, we don't have a destabilizing one. So in our three resonance structures, none of them have the two positive charges right next to each other, as we saw in the previous example. So it's not so much that the sigma complex for a meta attack is extra stable. It's just that the sigma complex for a meta attack doesn't have any destabilizing-like charges repelling each other. And so because there is no destabilization, the meta sigma complex becomes the most stable one and the one that's most likely to form in your mechanism. And so that's why a nitro group on your ring is going to function as a meta director."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's not so much that the sigma complex for a meta attack is extra stable. It's just that the sigma complex for a meta attack doesn't have any destabilizing-like charges repelling each other. And so because there is no destabilization, the meta sigma complex becomes the most stable one and the one that's most likely to form in your mechanism. And so that's why a nitro group on your ring is going to function as a meta director. Let's go ahead and do a para attack so that you can see that a para attack is going to give you the same situation as an ortho attack. So if I wanted to add this nitro group onto the para position, I would have to use these pi electrons over here this time. So nucleophilic attack."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's why a nitro group on your ring is going to function as a meta director. Let's go ahead and do a para attack so that you can see that a para attack is going to give you the same situation as an ortho attack. So if I wanted to add this nitro group onto the para position, I would have to use these pi electrons over here this time. So nucleophilic attack. So the nucleophile attacks the electrophile, once again pushing those electrons off. And so we have our nitro group here. We have these pi electrons."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So nucleophilic attack. So the nucleophile attacks the electrophile, once again pushing those electrons off. And so we have our nitro group here. We have these pi electrons. We are now going to show the nitro group in the para position. So if I'm saying these electrons in magenta are going to form a bond with this nitrogen right here, taking a bond away from that carbon. So that gets our plus 1 formal charge right here."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have these pi electrons. We are now going to show the nitro group in the para position. So if I'm saying these electrons in magenta are going to form a bond with this nitrogen right here, taking a bond away from that carbon. So that gets our plus 1 formal charge right here. So a resonant structure would be to take these pi electrons and move them over to here. So let's go ahead and show the next one. So we have our ring."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that gets our plus 1 formal charge right here. So a resonant structure would be to take these pi electrons and move them over to here. So let's go ahead and show the next one. So we have our ring. This time I'm going to draw out the formal charges on my nitro group. So I have an oxygen that's negatively charged. This nitrogen has a plus 1 formal charge on it."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our ring. This time I'm going to draw out the formal charges on my nitro group. So I have an oxygen that's negatively charged. This nitrogen has a plus 1 formal charge on it. And my pi electrons, I have pi electrons here. I have my nitro group, once again, in the para position. And I have some pi electrons over here as well."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This nitrogen has a plus 1 formal charge on it. And my pi electrons, I have pi electrons here. I have my nitro group, once again, in the para position. And I have some pi electrons over here as well. So let's highlight those. So the pi electrons in blue have moved over to here, taking a bond away from this top carbon. And so I can go ahead and draw a plus 1 formal charge on this top carbon."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I have some pi electrons over here as well. So let's highlight those. So the pi electrons in blue have moved over to here, taking a bond away from this top carbon. And so I can go ahead and draw a plus 1 formal charge on this top carbon. And once again, this is our destabilizing resonant structure. I have a positively charged carbon on my ring right next to my positively charged nitrogen. So like charges repel, this is the destabilizing resonant structure, which, of course, destabilizes the sigma complex for a para attack."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I can go ahead and draw a plus 1 formal charge on this top carbon. And once again, this is our destabilizing resonant structure. I have a positively charged carbon on my ring right next to my positively charged nitrogen. So like charges repel, this is the destabilizing resonant structure, which, of course, destabilizes the sigma complex for a para attack. Just to be complete, I can draw another resonant structure. So I could take these pi electrons over to here. So let's go ahead and draw our third one."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So like charges repel, this is the destabilizing resonant structure, which, of course, destabilizes the sigma complex for a para attack. Just to be complete, I can draw another resonant structure. So I could take these pi electrons over to here. So let's go ahead and draw our third one. So I have my nitro group. I have pi electrons here, pi electrons here. And I have a nitro in the para position."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw our third one. So I have my nitro group. I have pi electrons here, pi electrons here. And I have a nitro in the para position. And let me go ahead and highlight those electrons one last time. So I have these electrons in red have moved over to here, taking a bond away from this carbon. And so that will get our plus 1 formal charge for our last resonant structure."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I have a nitro in the para position. And let me go ahead and highlight those electrons one last time. So I have these electrons in red have moved over to here, taking a bond away from this carbon. And so that will get our plus 1 formal charge for our last resonant structure. So once again, remember that the sigma complex is a hybrid of these resonant structures. But since we can draw a destabilizing resonant structure, this sigma complex is not the one that's most likely to form. So the meta attack is preferred."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that will get our plus 1 formal charge for our last resonant structure. So once again, remember that the sigma complex is a hybrid of these resonant structures. But since we can draw a destabilizing resonant structure, this sigma complex is not the one that's most likely to form. So the meta attack is preferred. So one more quick thing about metadirectors, so an easy way to recognize a metadirector. So we've seen in this example, our substituent has an atom has a positive charge right next to our benzene ring. And so one way to look for a metadirector would be here's just a generic substituent, y, directly bonded to our benzene ring."}, {"video_title": "Meta directors I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the meta attack is preferred. So one more quick thing about metadirectors, so an easy way to recognize a metadirector. So we've seen in this example, our substituent has an atom has a positive charge right next to our benzene ring. And so one way to look for a metadirector would be here's just a generic substituent, y, directly bonded to our benzene ring. So the atom directly bonded to your benzene ring, if there's a plus 1 formal charge on it, we've just seen that resonant structures are destabilized for an ortho para attack. Therefore, a meta attack is preferred. So you could look for a plus 1 formal charge, or you could look for a partial positive."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "The concept of steric number is very useful because it tells us the number of hybridized orbitals that we have. So to find the steric number, you add up the number of sigma bonds, or single bonds, and to that, you add the number of lone pairs of electrons. So let's go ahead and do it for methane. So if I wanted to find the steric number, steric number is equal to the number of sigma bonds. So I look at around my carbon here, and I see one, two, three, and four sigma, or single bonds. So I have four sigma bonds. I have zero lone pairs of electrons around that carbon."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So if I wanted to find the steric number, steric number is equal to the number of sigma bonds. So I look at around my carbon here, and I see one, two, three, and four sigma, or single bonds. So I have four sigma bonds. I have zero lone pairs of electrons around that carbon. So four plus zero gives me a steric number of four. In the last video, we saw that sp3 hybridized situation, we get four hybrid orbitals. And that's how many we need."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "I have zero lone pairs of electrons around that carbon. So four plus zero gives me a steric number of four. In the last video, we saw that sp3 hybridized situation, we get four hybrid orbitals. And that's how many we need. The steric number tells us we need four hybridized orbitals. So we took one s orbital and three p orbitals, and that gave us four sp3 hybrid orbitals. So this carbon must be sp3 hybridized."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And that's how many we need. The steric number tells us we need four hybridized orbitals. So we took one s orbital and three p orbitals, and that gave us four sp3 hybrid orbitals. So this carbon must be sp3 hybridized. So let's go ahead and draw that in here. This carbon is sp3 hybridized. And in the last video, we also drew everything out."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So this carbon must be sp3 hybridized. So let's go ahead and draw that in here. This carbon is sp3 hybridized. And in the last video, we also drew everything out. So we drew in those four sp3 hybrid orbitals for that carbon. And we had one valence electron in each of those four sp3 hybrid orbitals. And then hydrogen had one valence electron in an unhybridized s orbital."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And in the last video, we also drew everything out. So we drew in those four sp3 hybrid orbitals for that carbon. And we had one valence electron in each of those four sp3 hybrid orbitals. And then hydrogen had one valence electron in an unhybridized s orbital. So we drew in our hydrogens and the one valence electron like that. This head-on overlap, this is, of course, a sigma bond. So again, we talked about this in the last video."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And then hydrogen had one valence electron in an unhybridized s orbital. So we drew in our hydrogens and the one valence electron like that. This head-on overlap, this is, of course, a sigma bond. So again, we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs. So these electron pairs are going to repel each other, like charges repel. And so the idea of VSEPR theory tells us these electron pairs are going to repel and try to get as far away from each other as they possibly can in space."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So again, we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs. So these electron pairs are going to repel each other, like charges repel. And so the idea of VSEPR theory tells us these electron pairs are going to repel and try to get as far away from each other as they possibly can in space. And that means that the arrangement of those electron pairs ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon like that."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And so the idea of VSEPR theory tells us these electron pairs are going to repel and try to get as far away from each other as they possibly can in space. And that means that the arrangement of those electron pairs ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon like that. When we think about the molecular geometry, so that's like electron group geometry, when I think about the geometry of the entire molecule, I could think about drawing in those electrons, those bonding electrons like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then these lines mean in the plane of the page. And so we can go ahead and draw in our hydrogens."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So we have a tetrahedral arrangement of electron pairs around our carbon like that. When we think about the molecular geometry, so that's like electron group geometry, when I think about the geometry of the entire molecule, I could think about drawing in those electrons, those bonding electrons like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then these lines mean in the plane of the page. And so we can go ahead and draw in our hydrogens. And this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral. So let's go ahead and write that."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And so we can go ahead and draw in our hydrogens. And this is just one way to represent the methane molecule, which attempts to show the geometry of the entire molecule. So the arrangement of the atoms turns out to also be tetrahedral. So let's go ahead and write that. So tetrahedral. And let's see if we can see that four-sided figure. So a tetrahedron is a four-sided figure."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that. So tetrahedral. And let's see if we can see that four-sided figure. So a tetrahedron is a four-sided figure. So we can think about this being one face. And then let's go ahead and draw a second face. If I draw a line back here, that gives us four faces to our tetrahedron."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So a tetrahedron is a four-sided figure. So we can think about this being one face. And then let's go ahead and draw a second face. If I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral. The molecular geometry of methane is tetrahedral. And then we also have a bond angle."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "If I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral. The molecular geometry of methane is tetrahedral. And then we also have a bond angle. Let me go ahead and draw that in. So a bond angle, this hydrogen-carbon-hydrogen bond angle in here is approximately 109.5 degrees. Let's go ahead and do the same type of analysis for a different molecule here."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And then we also have a bond angle. Let me go ahead and draw that in. So a bond angle, this hydrogen-carbon-hydrogen bond angle in here is approximately 109.5 degrees. Let's go ahead and do the same type of analysis for a different molecule here. So let's do it for ammonia next. So we have NH3. If I want to find the steric number, the steric number is equal to the number of sigma bonds."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and do the same type of analysis for a different molecule here. So let's do it for ammonia next. So we have NH3. If I want to find the steric number, the steric number is equal to the number of sigma bonds. That's 1, 2, 3. So three sigma bonds plus number of lone pairs of electrons. So I have one lone pair of electrons here."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "If I want to find the steric number, the steric number is equal to the number of sigma bonds. That's 1, 2, 3. So three sigma bonds plus number of lone pairs of electrons. So I have one lone pair of electrons here. So 3 plus 1 gives me a steric number of 4. So I need four hybridized orbitals. And once again, when I need four hybridized orbitals, I know that this nitrogen must be SP3 hybridized, because SP3 hybridization gives us four hybrid orbitals."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So I have one lone pair of electrons here. So 3 plus 1 gives me a steric number of 4. So I need four hybridized orbitals. And once again, when I need four hybridized orbitals, I know that this nitrogen must be SP3 hybridized, because SP3 hybridization gives us four hybrid orbitals. And so let's go ahead and draw those four hybrid orbitals. We would have nitrogen. And let's go ahead and draw in all four of those."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And once again, when I need four hybridized orbitals, I know that this nitrogen must be SP3 hybridized, because SP3 hybridization gives us four hybrid orbitals. And so let's go ahead and draw those four hybrid orbitals. We would have nitrogen. And let's go ahead and draw in all four of those. So 1, 2, 3, and 4. Those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and draw in all four of those. So 1, 2, 3, and 4. Those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron. And then you'd have two in this one, like that. And then you go ahead and put in your hydrogens. Once again, each hydrogen has one electron in an unhybridized s orbital."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron. And then you'd have two in this one, like that. And then you go ahead and put in your hydrogens. Once again, each hydrogen has one electron in an unhybridized s orbital. So we go ahead and draw in those hydrogens. So our overlap of orbitals. So here's a sigma bond."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "Once again, each hydrogen has one electron in an unhybridized s orbital. So we go ahead and draw in those hydrogens. So our overlap of orbitals. So here's a sigma bond. Here's a sigma bond. And here's a sigma bond. So three sigma bonds in ammonia."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So here's a sigma bond. Here's a sigma bond. And here's a sigma bond. So three sigma bonds in ammonia. And then we have this lone pair up here. So the arrangement of these electron pairs is just what we talked about before. So we have this tetrahedral arrangement of electron pairs or electron groups."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So three sigma bonds in ammonia. And then we have this lone pair up here. So the arrangement of these electron pairs is just what we talked about before. So we have this tetrahedral arrangement of electron pairs or electron groups. So VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule. So if I go ahead and draw in another picture over here to talk about the molecular geometry, I go ahead and draw in the bonding electrons like that."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So we have this tetrahedral arrangement of electron pairs or electron groups. So VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule. So if I go ahead and draw in another picture over here to talk about the molecular geometry, I go ahead and draw in the bonding electrons like that. And then I'll put in my non-bonding electrons up here, this lone pair right here, housed in an sp3 hybridized orbital. So the arrangement of the atoms turns out not to be tetrahedral. And that has to do with this lone pair of electrons up here at the top."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So if I go ahead and draw in another picture over here to talk about the molecular geometry, I go ahead and draw in the bonding electrons like that. And then I'll put in my non-bonding electrons up here, this lone pair right here, housed in an sp3 hybridized orbital. So the arrangement of the atoms turns out not to be tetrahedral. And that has to do with this lone pair of electrons up here at the top. So this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example. And because it's going to repel those electrons a little bit more strongly, you're not going to get a bond angle of 109.5. It's going to decrease the bond angle."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And that has to do with this lone pair of electrons up here at the top. So this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example. And because it's going to repel those electrons a little bit more strongly, you're not going to get a bond angle of 109.5. It's going to decrease the bond angle. So let me go ahead and use the same color we used before. So this bond angle is not 109.5. It goes down a little bit because of the extra repulsion."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "It's going to decrease the bond angle. So let me go ahead and use the same color we used before. So this bond angle is not 109.5. It goes down a little bit because of the extra repulsion. So it turns out to be approximately 107 degrees. And in terms of the shape of the molecule, we don't say tetrahedral. We say trigonal pyramidal."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "It goes down a little bit because of the extra repulsion. So it turns out to be approximately 107 degrees. And in terms of the shape of the molecule, we don't say tetrahedral. We say trigonal pyramidal. So let me go ahead and write that here. So the geometry of the ammonia molecule is trigonal pyramidal. And let's analyze that a little bit."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "We say trigonal pyramidal. So let me go ahead and write that here. So the geometry of the ammonia molecule is trigonal pyramidal. And let's analyze that a little bit. So trigonal refers to the fact that nitrogen is bonded to three atoms here. So nitrogen's bonded to three hydrogens. So that takes care of the trigonal part."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And let's analyze that a little bit. So trigonal refers to the fact that nitrogen is bonded to three atoms here. So nitrogen's bonded to three hydrogens. So that takes care of the trigonal part. The pyramidal part comes in because when you're doing molecular geometry, you ignore lone pairs of electrons. So if you ignore that lone pair of electrons and just look at this nitrogen at the top of this pyramid right here, so that's where the pyramidal term comes in. So bonded to three other atoms like this, this, and this for our pyramid."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So that takes care of the trigonal part. The pyramidal part comes in because when you're doing molecular geometry, you ignore lone pairs of electrons. So if you ignore that lone pair of electrons and just look at this nitrogen at the top of this pyramid right here, so that's where the pyramidal term comes in. So bonded to three other atoms like this, this, and this for our pyramid. So trigonal pyramidal is the geometry of the ammonia molecule. But the nitrogen is sp3 hybridized. All right, let's do one more example."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So bonded to three other atoms like this, this, and this for our pyramid. So trigonal pyramidal is the geometry of the ammonia molecule. But the nitrogen is sp3 hybridized. All right, let's do one more example. Let's do water. So first we calculate the steric number. So the steric number is equal to the number of sigma bonds."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "All right, let's do one more example. Let's do water. So first we calculate the steric number. So the steric number is equal to the number of sigma bonds. So that's 1, 2, so two sigma bonds, plus numbers of lone pairs of electrons. So here's a lone pair, here's a lone pair. So we have 2 plus 2, which is equal to 4."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So the steric number is equal to the number of sigma bonds. So that's 1, 2, so two sigma bonds, plus numbers of lone pairs of electrons. So here's a lone pair, here's a lone pair. So we have 2 plus 2, which is equal to 4. So we need four hybridized orbitals. As we've seen in the previous two examples, when you need four hybridized orbitals, that's an sp3 hybridization situation. You have four sp3 hybridized orbitals."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So we have 2 plus 2, which is equal to 4. So we need four hybridized orbitals. As we've seen in the previous two examples, when you need four hybridized orbitals, that's an sp3 hybridization situation. You have four sp3 hybridized orbitals. So this oxygen is sp3 hybridized. So I'll go ahead and write that in here. So oxygen is sp3 hybridized."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "You have four sp3 hybridized orbitals. So this oxygen is sp3 hybridized. So I'll go ahead and write that in here. So oxygen is sp3 hybridized. So we can draw that out, showing oxygen with its four sp3 hybrid orbitals. So there's four of them. So I'm going to go ahead and draw in all four."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So oxygen is sp3 hybridized. So we can draw that out, showing oxygen with its four sp3 hybrid orbitals. So there's four of them. So I'm going to go ahead and draw in all four. In terms of electrons, this orbital gets one, this orbital gets one, and these orbitals are going to get two like that. So that takes care of oxygen's six valence electrons. When you're drawing in your hydrogens, so let's go ahead and put in the hydrogen here."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and draw in all four. In terms of electrons, this orbital gets one, this orbital gets one, and these orbitals are going to get two like that. So that takes care of oxygen's six valence electrons. When you're drawing in your hydrogens, so let's go ahead and put in the hydrogen here. So once again, each hydrogen with one electron in an unhybridized s orbital like that. So in terms of overlap of bonds, here's one sigma bond, and here's another sigma bond. So that's our two sigma bonds for water."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "When you're drawing in your hydrogens, so let's go ahead and put in the hydrogen here. So once again, each hydrogen with one electron in an unhybridized s orbital like that. So in terms of overlap of bonds, here's one sigma bond, and here's another sigma bond. So that's our two sigma bonds for water. Once again, the arrangement of these electron pairs is tetrahedral. So VSEPR theory says the electrons repel. And so the electron group geometry, you could say, is tetrahedral."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So that's our two sigma bonds for water. Once again, the arrangement of these electron pairs is tetrahedral. So VSEPR theory says the electrons repel. And so the electron group geometry, you could say, is tetrahedral. But that's not the geometry of the entire molecule. So that's just thinking about electron groups and these hybrid orbitals. The geometry of the molecule is different."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And so the electron group geometry, you could say, is tetrahedral. But that's not the geometry of the entire molecule. So that's just thinking about electron groups and these hybrid orbitals. The geometry of the molecule is different. So we'll go ahead and draw that over here. So we have our water molecule. And draw in our bonding electrons."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "The geometry of the molecule is different. So we'll go ahead and draw that over here. So we have our water molecule. And draw in our bonding electrons. And now let's put in our non-bonding electrons like that. So we have a different situation than with ammonia. With ammonia, we had one lone pair of electrons repelling these bonding electrons up here."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And draw in our bonding electrons. And now let's put in our non-bonding electrons like that. So we have a different situation than with ammonia. With ammonia, we had one lone pair of electrons repelling these bonding electrons up here. For water, we have two lone pairs of electrons repelling these bonding electrons. And so that's going to change the bond angle. It's going to shorten even more than in the previous example."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "With ammonia, we had one lone pair of electrons repelling these bonding electrons up here. For water, we have two lone pairs of electrons repelling these bonding electrons. And so that's going to change the bond angle. It's going to shorten even more than in the previous example. So the bond angle decreases. So this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So thinking about the molecular geometry, or the shape of the water molecule, so we actually call this bent or angular."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "It's going to shorten even more than in the previous example. So the bond angle decreases. So this bond angle in here decreases to approximately 105 degrees, rounded up a little bit. So thinking about the molecular geometry, or the shape of the water molecule, so we actually call this bent or angular. So this is a bent geometry. Because you ignore the lone pairs of electrons. And that would just give you this oxygen here, and then this angle."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "So thinking about the molecular geometry, or the shape of the water molecule, so we actually call this bent or angular. So this is a bent geometry. Because you ignore the lone pairs of electrons. And that would just give you this oxygen here, and then this angle. So you could also call this angular. So we have this bent molecular geometry like that, or angular. And once again, for molecular geometry, ignore your lone pairs of electrons."}, {"video_title": "Steric number AP Chemistry Khan Academy.mp3", "Sentence": "And that would just give you this oxygen here, and then this angle. So you could also call this angular. So we have this bent molecular geometry like that, or angular. And once again, for molecular geometry, ignore your lone pairs of electrons. So these are examples of three molecules. And the central atom in all three of these molecules is sp3 hybridized. And so this is one way to figure out your overall molecular geometry, and to think about bond angles, and to think about how those hybrid orbitals affect the structure of these molecules."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We know the direction they're moving in, what does that tell us about where they came from? So let's just do the thought experiment. Right now South America and Africa are moving away from each other because of new plate material being created at the mid-Atlantic rift. Let's rewind it, let's bring them back together. We know that India is jamming into the Eurasian plate right now, causing the Himalayas to get higher and higher. What if we rewind that? Let's bring India back down towards Antarctica."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's rewind it, let's bring them back together. We know that India is jamming into the Eurasian plate right now, causing the Himalayas to get higher and higher. What if we rewind that? Let's bring India back down towards Antarctica. Same thing with Australia. We have new plate material being formed between Australia and Antarctica that's making the continents move apart. Let's bring them back together."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's bring India back down towards Antarctica. Same thing with Australia. We have new plate material being formed between Australia and Antarctica that's making the continents move apart. Let's bring them back together. Let's rewind the clock. Even North America, it's not as obvious from this diagram, but if you actually look at the GPS data, it becomes pretty obvious. That North America right now is kind of moving in a counter-clockwise rotation."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's bring them back together. Let's rewind the clock. Even North America, it's not as obvious from this diagram, but if you actually look at the GPS data, it becomes pretty obvious. That North America right now is kind of moving in a counter-clockwise rotation. So let's rewind it. Let's go back, moving it in a clockwise direction. Instead of Eurasia going further away from North America, let's bring it back together."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That North America right now is kind of moving in a counter-clockwise rotation. So let's rewind it. Let's go back, moving it in a clockwise direction. Instead of Eurasia going further away from North America, let's bring it back together. So what you can imagine is a reality where India and Australia are jammed down into Antarctica, South America and Africa are jammed together, North America is jammed in there, and essentially Eurasia is also jammed in there. So it looks like they all would clump together if you go back a few hundred million years. Based on just that thought experiment, you can imagine at one point all of the continents on the world were merged into one supercontinent."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Instead of Eurasia going further away from North America, let's bring it back together. So what you can imagine is a reality where India and Australia are jammed down into Antarctica, South America and Africa are jammed together, North America is jammed in there, and essentially Eurasia is also jammed in there. So it looks like they all would clump together if you go back a few hundred million years. Based on just that thought experiment, you can imagine at one point all of the continents on the world were merged into one supercontinent. That supercontinent is called Pangea. Pan for entire or whole, and Gea for coming from Gea, for the world. It turns out that all of the evidence we've seen actually does make us believe that there was a supercontinent called Pangea."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Based on just that thought experiment, you can imagine at one point all of the continents on the world were merged into one supercontinent. That supercontinent is called Pangea. Pan for entire or whole, and Gea for coming from Gea, for the world. It turns out that all of the evidence we've seen actually does make us believe that there was a supercontinent called Pangea. Obviously there probably weren't things on the planet calling it anything back then. Well, there were things back then, but not things that would actually go and try to label continents that we know of. But all of the evidence tells us that Pangea existed about 200 to 300 million years ago, roughly 250 million, give or take, years ago."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It turns out that all of the evidence we've seen actually does make us believe that there was a supercontinent called Pangea. Obviously there probably weren't things on the planet calling it anything back then. Well, there were things back then, but not things that would actually go and try to label continents that we know of. But all of the evidence tells us that Pangea existed about 200 to 300 million years ago, roughly 250 million, give or take, years ago. I want to be clear, this was not the first supercontinent. To a large degree, it's kind of the most recent supercontinent that's easiest for us to construct because it was the most recent one. But we believe that there were other supercontinents before this."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But all of the evidence tells us that Pangea existed about 200 to 300 million years ago, roughly 250 million, give or take, years ago. I want to be clear, this was not the first supercontinent. To a large degree, it's kind of the most recent supercontinent that's easiest for us to construct because it was the most recent one. But we believe that there were other supercontinents before this. If you rewind even more, you would have to break up Pangea and it would reform, but we're not going back in time. There were several supercontinents in the past that broke up, reformed, broke up, reformed, and the last time that we had a supercontinent was Pangea, about 250 million years ago. Now it's broken up into our current day geography."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we believe that there were other supercontinents before this. If you rewind even more, you would have to break up Pangea and it would reform, but we're not going back in time. There were several supercontinents in the past that broke up, reformed, broke up, reformed, and the last time that we had a supercontinent was Pangea, about 250 million years ago. Now it's broken up into our current day geography. I won't go into all of the detail why we believe that there was a Pangea about 250 million years ago, or this diagram tells us about 225 million years ago, give or take. But I'll go into some of the interesting evidence. On a very high level, you have a lot of rock commonalities between things that would have had to combine during Pangea."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now it's broken up into our current day geography. I won't go into all of the detail why we believe that there was a Pangea about 250 million years ago, or this diagram tells us about 225 million years ago, give or take. But I'll go into some of the interesting evidence. On a very high level, you have a lot of rock commonalities between things that would have had to combine during Pangea. Probably the most interesting thing is the fossil evidence. There's a whole bunch of fossils, and here are examples of it, from species that were around between 200 and 300 million years ago. Their fossils are found in a very specific place, this animal right here, Sinognathus."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "On a very high level, you have a lot of rock commonalities between things that would have had to combine during Pangea. Probably the most interesting thing is the fossil evidence. There's a whole bunch of fossils, and here are examples of it, from species that were around between 200 and 300 million years ago. Their fossils are found in a very specific place, this animal right here, Sinognathus. This animal's fossils are only found in this area of South America, a nice clean band here, and this part of Africa. Not only does South America look like it fits very nicely into Africa, but the fossil evidence also makes it look like there was a nice clean band where this animal lived and where we find the fossils. It really makes it seem like these were connected at least when this animal lived, maybe on the order of 250 million years ago."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Their fossils are found in a very specific place, this animal right here, Sinognathus. This animal's fossils are only found in this area of South America, a nice clean band here, and this part of Africa. Not only does South America look like it fits very nicely into Africa, but the fossil evidence also makes it look like there was a nice clean band where this animal lived and where we find the fossils. It really makes it seem like these were connected at least when this animal lived, maybe on the order of 250 million years ago. This species right over here, its fossils are found in this area. Let me do it in a color that has more contrast. In this area right over here."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It really makes it seem like these were connected at least when this animal lived, maybe on the order of 250 million years ago. This species right over here, its fossils are found in this area. Let me do it in a color that has more contrast. In this area right over here. This plant, its fossils. Now this starts to connect a lot of dots between a lot of content. Its fossils are found in this entire area, across South America, Africa, Antarctica, India, and Australia."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In this area right over here. This plant, its fossils. Now this starts to connect a lot of dots between a lot of content. Its fossils are found in this entire area, across South America, Africa, Antarctica, India, and Australia. Not only does it look like the continents kind of fit together in a puzzle piece, not only do we get it to a configuration like this, we essentially just rewind the movement that we're seeing now, but the fossil evidence also kind of confirms that they fit together in this way. This animal right here, we find fossils on this nice stripe that goes from Africa through India all the way to Antarctica. Now this only gives us evidence of kind of the more southern hemisphere of Pangea, but there is other evidence."}, {"video_title": "Pangaea Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Its fossils are found in this entire area, across South America, Africa, Antarctica, India, and Australia. Not only does it look like the continents kind of fit together in a puzzle piece, not only do we get it to a configuration like this, we essentially just rewind the movement that we're seeing now, but the fossil evidence also kind of confirms that they fit together in this way. This animal right here, we find fossils on this nice stripe that goes from Africa through India all the way to Antarctica. Now this only gives us evidence of kind of the more southern hemisphere of Pangea, but there is other evidence. We find kind of continuing mountain chains between North America and Europe. We find rock evidence where just the way we see the fossils kind of line up nicely, there's a common rock that lines up nicely between South America and Africa and other continents that were at once connected. So all of the evidence, as far as we can tell now, does make us think that there at one time was a Pangea."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So far we've dealt with simple chains and rings of carbons, but let's think about what happens when things get a little bit more complex. Let me just draw a molecule here, and we can think about how we'll name it. So let me draw it like that. So here we don't have a simple chain. It branches off at some point. So what we do in this situation is we find the longest chain in this molecule. And let's think about what the longest chain is."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So here we don't have a simple chain. It branches off at some point. So what we do in this situation is we find the longest chain in this molecule. And let's think about what the longest chain is. If we start here, we get 1, 2, 3, 4, 5, 6, 7, 8 carbons. If we start here, we get 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. So what we do is we look at the longest chain."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And let's think about what the longest chain is. If we start here, we get 1, 2, 3, 4, 5, 6, 7, 8 carbons. If we start here, we get 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. So what we do is we look at the longest chain. The longest chain is this one right here. Let me do this in a different color. So this is 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So what we do is we look at the longest chain. The longest chain is this one right here. Let me do this in a different color. So this is 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. And so that will be kind of the core of our naming. So it's 9 carbons. So that right here, longest chain, has 9 carbons."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. And so that will be kind of the core of our naming. So it's 9 carbons. So that right here, longest chain, has 9 carbons. You can kind of view this as the backbone of our molecule and also the backbone of our naming. 9 carbons. So we're dealing with a nonane."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So that right here, longest chain, has 9 carbons. You can kind of view this as the backbone of our molecule and also the backbone of our naming. 9 carbons. So we're dealing with a nonane. Remember, 9, non for 9. And then ane because we're dealing with an alkane. We have all single bonds over here."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So we're dealing with a nonane. Remember, 9, non for 9. And then ane because we're dealing with an alkane. We have all single bonds over here. Now, what do we do with this thing over here? Well, what's kind of sticking off of this chain? Well, we essentially have one carbon here that's attached to the chain."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "We have all single bonds over here. Now, what do we do with this thing over here? Well, what's kind of sticking off of this chain? Well, we essentially have one carbon here that's attached to the chain. This carbon is bonded to a carbon on the chain. If we wanted to draw the entire molecule, we could draw a carbon here, and it would be bonded to 3 hydrogens. But that is all implicit."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Well, we essentially have one carbon here that's attached to the chain. This carbon is bonded to a carbon on the chain. If we wanted to draw the entire molecule, we could draw a carbon here, and it would be bonded to 3 hydrogens. But that is all implicit. But what is this thing right here? Well, it's one carbon. So you might be tempted to say it's methane."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "But that is all implicit. But what is this thing right here? Well, it's one carbon. So you might be tempted to say it's methane. But it's not methane because it's attached to other things. But it would be right to use the prefix meth. So this right here, you would use the prefix meth."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So you might be tempted to say it's methane. But it's not methane because it's attached to other things. But it would be right to use the prefix meth. So this right here, you would use the prefix meth. So this is one carbon. So you would want to use the prefix meth. But because it isn't the main chain, it is added to something else, we don't write methane, we write methyl."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this right here, you would use the prefix meth. So this is one carbon. So you would want to use the prefix meth. But because it isn't the main chain, it is added to something else, we don't write methane, we write methyl. So let me write this right here. Methyl. So this means that it branches off longer chain."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "But because it isn't the main chain, it is added to something else, we don't write methane, we write methyl. So let me write this right here. Methyl. So this means that it branches off longer chain. So this molecule right here would be called, and actually there's one more thing that we have to think about. This thing could have been, we could just call it methyl nonane, let me just write that down right now. So we could just call this right from the get-go, we could call it methyl nonane."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this means that it branches off longer chain. So this molecule right here would be called, and actually there's one more thing that we have to think about. This thing could have been, we could just call it methyl nonane, let me just write that down right now. So we could just call this right from the get-go, we could call it methyl nonane. But there's one problem with just calling it methyl nonane. This is methyl nonane, but so let me draw it. So this thing I just drew here, this is 1, 2, 3, 4, 5, 6, 7, 8, 9, so that's the nonane, and the one I drew here is 1, 2, 3, the methyl is right over here."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So we could just call this right from the get-go, we could call it methyl nonane. But there's one problem with just calling it methyl nonane. This is methyl nonane, but so let me draw it. So this thing I just drew here, this is 1, 2, 3, 4, 5, 6, 7, 8, 9, so that's the nonane, and the one I drew here is 1, 2, 3, the methyl is right over here. That's what I just drew. That is, let me do it the same color. So this is what I just drew."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this thing I just drew here, this is 1, 2, 3, 4, 5, 6, 7, 8, 9, so that's the nonane, and the one I drew here is 1, 2, 3, the methyl is right over here. That's what I just drew. That is, let me do it the same color. So this is what I just drew. But maybe, what if it was something like this? What if it was 1, 2, 3, 4, 5, 6, 7, 8, 9, and let's say that the methyl group was, instead of being right here, let's say the methyl group was right here. So how would you name these two things differently?"}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is what I just drew. But maybe, what if it was something like this? What if it was 1, 2, 3, 4, 5, 6, 7, 8, 9, and let's say that the methyl group was, instead of being right here, let's say the methyl group was right here. So how would you name these two things differently? This name right now does not differentiate between the methyl group being on this carbon versus the methyl group being on this carbon over here or this carbon over there. And what you do in this case is you number the carbons on the longest chain, on the main backbone, and you number them so that the methyl group is attached to the lowest possible number. So there's two possible ways to number this chain."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So how would you name these two things differently? This name right now does not differentiate between the methyl group being on this carbon versus the methyl group being on this carbon over here or this carbon over there. And what you do in this case is you number the carbons on the longest chain, on the main backbone, and you number them so that the methyl group is attached to the lowest possible number. So there's two possible ways to number this chain. You could start here as 1, 2, 3, 4, 5, 6, 7, then you would say that this would be the methyl group is attached to the 7-carbon, or you could start numbering from this end of the chain. Would be 1, 2, 3. Or so it could also be attached to the third, 3, 4, 5, 6, 7, 8, 9."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So there's two possible ways to number this chain. You could start here as 1, 2, 3, 4, 5, 6, 7, then you would say that this would be the methyl group is attached to the 7-carbon, or you could start numbering from this end of the chain. Would be 1, 2, 3. Or so it could also be attached to the third, 3, 4, 5, 6, 7, 8, 9. So you want to number it so you're closest to the methyl group, so you want to start here, 1, 2, 3, 4, 5, 6, 7, 8, 9. So in this case, this would not just be methyl nonane, this thing right here would be 3. Let me make this very clear."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Or so it could also be attached to the third, 3, 4, 5, 6, 7, 8, 9. So you want to number it so you're closest to the methyl group, so you want to start here, 1, 2, 3, 4, 5, 6, 7, 8, 9. So in this case, this would not just be methyl nonane, this thing right here would be 3. Let me make this very clear. It would be 3-methyl nonane. Because the methyl group, that 1-carbon, that 1CH3, is attached to the third carbon on our nonane backbone. So that right there is 3-methyl nonane."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Let me make this very clear. It would be 3-methyl nonane. Because the methyl group, that 1-carbon, that 1CH3, is attached to the third carbon on our nonane backbone. So that right there is 3-methyl nonane. Since I drew this out, what is this right here? Well, once again, we have this nonane backbone, 1, 2, 3, 4, 5, 6, 7, 8, 9. And you want to number it so that this methyl group is at the lowest number carbon as possible."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So that right there is 3-methyl nonane. Since I drew this out, what is this right here? Well, once again, we have this nonane backbone, 1, 2, 3, 4, 5, 6, 7, 8, 9. And you want to number it so that this methyl group is at the lowest number carbon as possible. So it's closer to this end than that end. So it's 1, 2, 3, 4. I'll do that in magenta."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And you want to number it so that this methyl group is at the lowest number carbon as possible. So it's closer to this end than that end. So it's 1, 2, 3, 4. I'll do that in magenta. And you can keep numbering, 5, 6, 7, 8, 9. So this one is going to be 4. And then you have your methyl group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "I'll do that in magenta. And you can keep numbering, 5, 6, 7, 8, 9. So this one is going to be 4. And then you have your methyl group. It's just 1-carbon sitting right there. So 4-methyl nonane. Now let's go the other way around."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And then you have your methyl group. It's just 1-carbon sitting right there. So 4-methyl nonane. Now let's go the other way around. Let's say we start with a name and we want to figure out its formula. And hopefully this gives you a good understanding how things will get more complex. And over the next few videos, it'll get more and more complex."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Now let's go the other way around. Let's say we start with a name and we want to figure out its formula. And hopefully this gives you a good understanding how things will get more complex. And over the next few videos, it'll get more and more complex. But you'll see it's all at least reasonably logical. So let's say I were to give you, let me think of one. Let's say 2-propylheptane."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And over the next few videos, it'll get more and more complex. But you'll see it's all at least reasonably logical. So let's say I were to give you, let me think of one. Let's say 2-propylheptane. So how do we take this apart? So the first thing you see, it is an alkane. It's all going to be single bonds."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Let's say 2-propylheptane. So how do we take this apart? So the first thing you see, it is an alkane. It's all going to be single bonds. Heptane, this is kind of the core. So the longest chain here, hept is the prefix for 7. So this tells us, let me make this clear, this tells us that we're dealing with all single bonds."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "It's all going to be single bonds. Heptane, this is kind of the core. So the longest chain here, hept is the prefix for 7. So this tells us, let me make this clear, this tells us that we're dealing with all single bonds. That's what the ane tells us. If we had double bonds, it would be ene. Triple bonds, eine."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this tells us, let me make this clear, this tells us that we're dealing with all single bonds. That's what the ane tells us. If we had double bonds, it would be ene. Triple bonds, eine. We're going to see that in the future, but let's stay simple right now. The hept tells us that we're dealing with 7 carbons. And then what does this 2-propyl means?"}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Triple bonds, eine. We're going to see that in the future, but let's stay simple right now. The hept tells us that we're dealing with 7 carbons. And then what does this 2-propyl means? Well, propyl tells us, what was the prop prefix? If methyl is 1, ethyl is 2, propyl is 3, butyl is 4. So this is a 3-carbon group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "And then what does this 2-propyl means? Well, propyl tells us, what was the prop prefix? If methyl is 1, ethyl is 2, propyl is 3, butyl is 4. So this is a 3-carbon group. And it's going to be attached to the second carbon on the heptane. So let me draw the heptane chain. So we have 7 carbons."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is a 3-carbon group. And it's going to be attached to the second carbon on the heptane. So let me draw the heptane chain. So we have 7 carbons. So 1, 2, 3, 4, 5, 6, 7. And then on the second carbon, so let me pick a good color here, on the second carbon, so if we just number it 1, 2, 3, 4, 5, 6, 7, on the second carbon right there, we have a propyl group. But there's a 3-carbon group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So we have 7 carbons. So 1, 2, 3, 4, 5, 6, 7. And then on the second carbon, so let me pick a good color here, on the second carbon, so if we just number it 1, 2, 3, 4, 5, 6, 7, on the second carbon right there, we have a propyl group. But there's a 3-carbon group. So let me draw the propyl group. So we have a 1, 2, 3. 3-carbon group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "But there's a 3-carbon group. So let me draw the propyl group. So we have a 1, 2, 3. 3-carbon group. And it's attached at the second carbon right there. So this molecule, 2-propylheptane, looks just like this. Let's do another one, maybe one that seems a little bit more difficult."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "3-carbon group. And it's attached at the second carbon right there. So this molecule, 2-propylheptane, looks just like this. Let's do another one, maybe one that seems a little bit more difficult. Let's say that we have 6-butyl tetradecane. And all of these might seem a little daunting when you see it at first, but if you really just break it up, it actually is pretty logical. So once again, we have the ane there, so it's all going to be single bonds."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one, maybe one that seems a little bit more difficult. Let's say that we have 6-butyl tetradecane. And all of these might seem a little daunting when you see it at first, but if you really just break it up, it actually is pretty logical. So once again, we have the ane there, so it's all going to be single bonds. What is tetradec? What is that prefix? Well, dec is 10, and then we have tetra."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we have the ane there, so it's all going to be single bonds. What is tetradec? What is that prefix? Well, dec is 10, and then we have tetra. That's 4 plus 10. That's 14. Tetradec is 14."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Well, dec is 10, and then we have tetra. That's 4 plus 10. That's 14. Tetradec is 14. So this is telling us that we're dealing with a 14-carbon chain. So let's draw that. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 carbon chain, and if we number them, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, this tells us that the sixth carbon will be a butyl group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Tetradec is 14. So this is telling us that we're dealing with a 14-carbon chain. So let's draw that. So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 carbon chain, and if we number them, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, this tells us that the sixth carbon will be a butyl group. So let's see, the sixth carbon is right here. And butyl, and you should memorize this, methyl is 1, ethyl is 2, propyl is 3, butyl is 4. So we have a 4-carbon group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 carbon chain, and if we number them, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, this tells us that the sixth carbon will be a butyl group. So let's see, the sixth carbon is right here. And butyl, and you should memorize this, methyl is 1, ethyl is 2, propyl is 3, butyl is 4. So we have a 4-carbon group. So let me write this in pink maybe. So this is a 4-carbon group. It's actually what you call an alkyl group."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So we have a 4-carbon group. So let me write this in pink maybe. So this is a 4-carbon group. It's actually what you call an alkyl group. It's attached to something else. So on the sixth carbon, we have a 4-carbon group. So 1, let me draw it this way."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "It's actually what you call an alkyl group. It's attached to something else. So on the sixth carbon, we have a 4-carbon group. So 1, let me draw it this way. No, this is fine. Let's say we have 1, 2, 3, 4, so that is our butyl right there, and it's attached to the sixth carbon right there. So this is what our molecule would look like."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So 1, let me draw it this way. No, this is fine. Let's say we have 1, 2, 3, 4, so that is our butyl right there, and it's attached to the sixth carbon right there. So this is what our molecule would look like. Now, let me ask you a question. Would you ever see a 9-butyl tetradecane? So something where, so let me redraw it like this."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is what our molecule would look like. Now, let me ask you a question. Would you ever see a 9-butyl tetradecane? So something where, so let me redraw it like this. So a 9-butyl tetradecane, so it would have a butyl over here, 1, 2, 3, 4. Would you ever see a 9-butyl tetradecane? Would you ever see that written?"}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "So something where, so let me redraw it like this. So a 9-butyl tetradecane, so it would have a butyl over here, 1, 2, 3, 4. Would you ever see a 9-butyl tetradecane? Would you ever see that written? You might say, oh yeah, Sal, you just drew it. And the reason why you won't ever see that written is because there's a better way to number it if it's like this. Instead of starting over here at 1 at this end, you'd want to start 1 at that end."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Would you ever see that written? You might say, oh yeah, Sal, you just drew it. And the reason why you won't ever see that written is because there's a better way to number it if it's like this. Instead of starting over here at 1 at this end, you'd want to start 1 at that end. So instead of saying 9-butyl tetradecane, you should number it the other way. You should say this is 1, 2, 3, 4, 5, 6. So this is actually also 6-butyl tetradecane."}, {"video_title": "Naming alkanes with alkyl groups Organic chemistry Khan Academy.mp3", "Sentence": "Instead of starting over here at 1 at this end, you'd want to start 1 at that end. So instead of saying 9-butyl tetradecane, you should number it the other way. You should say this is 1, 2, 3, 4, 5, 6. So this is actually also 6-butyl tetradecane. Make sure I got that right. 1, 2, 3, 4, 5, 6. So instead of starting from the left and making this the ninth carbon, you always want to start numbering from the direction that has the lowest number for the first group."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So why would something, or even a star that's two or three solar masses, its gravity isn't so strong that it keeps light from escaping. Why would a black hole that has the same mass, why would that keep light from escaping? And to understand that, let's just think a little bit about, and I'll just do Newtonian classical physics right here. I won't get into the whole general relativity of things. And this really will just give us the intuition of why a smaller, denser thing of the same mass can exert a stronger gravitational pull. So let's imagine, so let's take two examples. Let's say I have some star here."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I won't get into the whole general relativity of things. And this really will just give us the intuition of why a smaller, denser thing of the same mass can exert a stronger gravitational pull. So let's imagine, so let's take two examples. Let's say I have some star here. Let's just call that mass m1. And let's say that its radius, let's just call this r. And let's say that I have some other mass right at the surface of this star. It's somehow able to survive those surface temperatures."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say I have some star here. Let's just call that mass m1. And let's say that its radius, let's just call this r. And let's say that I have some other mass right at the surface of this star. It's somehow able to survive those surface temperatures. And this mass over here has a mass of m1. This mass over here has a mass of m2. The universal law of gravitation tells us that the force between these two masses is going to be equal to the gravitational constant times the product of the masses."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's somehow able to survive those surface temperatures. And this mass over here has a mass of m1. This mass over here has a mass of m2. The universal law of gravitation tells us that the force between these two masses is going to be equal to the gravitational constant times the product of the masses. So m1 times m2, all of that over the square of the distance. Now let me be very clear. You might say, wait, this magenta mass right here is touching this larger mass."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The universal law of gravitation tells us that the force between these two masses is going to be equal to the gravitational constant times the product of the masses. So m1 times m2, all of that over the square of the distance. Now let me be very clear. You might say, wait, this magenta mass right here is touching this larger mass. Isn't the distance 0? And you have to be very careful. This is the distance between their center of masses."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You might say, wait, this magenta mass right here is touching this larger mass. Isn't the distance 0? And you have to be very careful. This is the distance between their center of masses. So the center of mass of this large mass over here is r away from this mass that's on the surface. Now with that said, let's take another example. Let's say that this large, massive star, or whatever it might be, eventually condenses into something 1,000 times smaller."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the distance between their center of masses. So the center of mass of this large mass over here is r away from this mass that's on the surface. Now with that said, let's take another example. Let's say that this large, massive star, or whatever it might be, eventually condenses into something 1,000 times smaller. So let me draw it like this. And obviously I'm not drawing it to scale. So let's say we have another case like this."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say that this large, massive star, or whatever it might be, eventually condenses into something 1,000 times smaller. So let me draw it like this. And obviously I'm not drawing it to scale. So let's say we have another case like this. And I'm not drawing it to scale. So this object, maybe it's the same object, or maybe it's a different object, that has the exact same mass as this larger object, but now it has a much smaller radius. It now has a much smaller radius."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say we have another case like this. And I'm not drawing it to scale. So this object, maybe it's the same object, or maybe it's a different object, that has the exact same mass as this larger object, but now it has a much smaller radius. It now has a much smaller radius. So that radius now, the radius is 1 over, let's just say it's 1,000th of this radius over here. So it's 1, maybe I'll just call it r over 1,000. So if this had a million kilometer radius, so that would make it roughly about twice the radius of the sun."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It now has a much smaller radius. So that radius now, the radius is 1 over, let's just say it's 1,000th of this radius over here. So it's 1, maybe I'll just call it r over 1,000. So if this had a million kilometer radius, so that would make it roughly about twice the radius of the sun. If this was a million kilometer radius right over here, this would be 1,000 kilometer radius. So maybe we're talking about something that's approaching a neutron star. But we don't have to think about what it actually is."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if this had a million kilometer radius, so that would make it roughly about twice the radius of the sun. If this was a million kilometer radius right over here, this would be 1,000 kilometer radius. So maybe we're talking about something that's approaching a neutron star. But we don't have to think about what it actually is. Let's just think about the thought experiment here. So let's say I have this thing over here. And let's say I have something on the surface of this."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we don't have to think about what it actually is. Let's just think about the thought experiment here. So let's say I have this thing over here. And let's say I have something on the surface of this. So let's say I have that same mass, it's on the surface of this thing. So this is m2 right over here. So what is going to be the force between these two masses?"}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say I have something on the surface of this. So let's say I have that same mass, it's on the surface of this thing. So this is m2 right over here. So what is going to be the force between these two masses? How strong are they going to want to, what's the force pulling them together? So let's just do the universal law of gravitation again. The force, let's just call this force 1, and let's call this force 2."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what is going to be the force between these two masses? How strong are they going to want to, what's the force pulling them together? So let's just do the universal law of gravitation again. The force, let's just call this force 1, and let's call this force 2. Once again, it's going to be the gravitational constant times the product of their masses. So the big m1 times the smaller mass, m2, all of that over this distance squared, this radius squared. Remember, it's the distance to the center of masses."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The force, let's just call this force 1, and let's call this force 2. Once again, it's going to be the gravitational constant times the product of their masses. So the big m1 times the smaller mass, m2, all of that over this distance squared, this radius squared. Remember, it's the distance to the center of masses. This center of mass here, we're considering m2 to kind of be just a point mass right over there. So what's the radius squared? It's going to be r over 1,000 squared."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, it's the distance to the center of masses. This center of mass here, we're considering m2 to kind of be just a point mass right over there. So what's the radius squared? It's going to be r over 1,000 squared. Or if we simplify this, what will this be? This is the same thing, and I'll just write it in one color just because it takes less time. Gravitational constant m1, m2 over r squared over 1,000 squared, or over 1 million."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's going to be r over 1,000 squared. Or if we simplify this, what will this be? This is the same thing, and I'll just write it in one color just because it takes less time. Gravitational constant m1, m2 over r squared over 1,000 squared, or over 1 million. Over 1 million. That's just 1,000 squared. Or we can multiply the numerator and the denominator by a million, and this is going to be equal to 1 million, I'm going to write it out, 1 million, let me scroll to the right a little bit, times the gravitational constant times m1, m2, all of that over r squared."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Gravitational constant m1, m2 over r squared over 1,000 squared, or over 1 million. Over 1 million. That's just 1,000 squared. Or we can multiply the numerator and the denominator by a million, and this is going to be equal to 1 million, I'm going to write it out, 1 million, let me scroll to the right a little bit, times the gravitational constant times m1, m2, all of that over r squared. Now what is this thing right over here? That's the same thing as this F1. So this is going to be 1 million times F1."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or we can multiply the numerator and the denominator by a million, and this is going to be equal to 1 million, I'm going to write it out, 1 million, let me scroll to the right a little bit, times the gravitational constant times m1, m2, all of that over r squared. Now what is this thing right over here? That's the same thing as this F1. So this is going to be 1 million times F1. So even though the masses involved are the same, this yellow object right here is the same mass as this larger object over here. It's able to exert a million times the gravitational force on this point mass, and actually vice versa. They're both being attracted."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is going to be 1 million times F1. So even though the masses involved are the same, this yellow object right here is the same mass as this larger object over here. It's able to exert a million times the gravitational force on this point mass, and actually vice versa. They're both being attracted. They're both exerting this on each other. And the reality is, is because this thing is smaller, because this m1 on the right here, this one I'm coloring in, because this one is smaller and denser, this particle is able to get closer to its center of mass. Now you might be saying, OK, well I can buy that."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're both being attracted. They're both exerting this on each other. And the reality is, is because this thing is smaller, because this m1 on the right here, this one I'm coloring in, because this one is smaller and denser, this particle is able to get closer to its center of mass. Now you might be saying, OK, well I can buy that. This just comes straight from the universal law of gravitation. But wouldn't something closer to this center of mass experience that same thing? If this was a star, wouldn't photons that are over here, wouldn't this experience the same force?"}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now you might be saying, OK, well I can buy that. This just comes straight from the universal law of gravitation. But wouldn't something closer to this center of mass experience that same thing? If this was a star, wouldn't photons that are over here, wouldn't this experience the same force? If this distance right here is r over 1,000, wouldn't some photon here, or atom here, or molecule, or whatever it's over here, wouldn't that experience the same force, this million times the force as this thing? And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening? It no longer has the entire mass is no longer pulling on it in that direction."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If this was a star, wouldn't photons that are over here, wouldn't this experience the same force? If this distance right here is r over 1,000, wouldn't some photon here, or atom here, or molecule, or whatever it's over here, wouldn't that experience the same force, this million times the force as this thing? And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening? It no longer has the entire mass is no longer pulling on it in that direction. It's no longer pulling it in that inward direction. You now have all of this mass over here. Let me think of the best way of doing it."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It no longer has the entire mass is no longer pulling on it in that direction. It's no longer pulling it in that inward direction. You now have all of this mass over here. Let me think of the best way of doing it. So you could think of it all of this mass over here is pulling it in an outward direction. It's not telling you what that mass out there is doing, since that mass itself is being pulled inward. It is pushing down on this."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me think of the best way of doing it. So you could think of it all of this mass over here is pulling it in an outward direction. It's not telling you what that mass out there is doing, since that mass itself is being pulled inward. It is pushing down on this. It is exerting pressure on that point. But the actual gravitational force that that point is experiencing is actually going to be less. It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is pushing down on this. It is exerting pressure on that point. But the actual gravitational force that that point is experiencing is actually going to be less. It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction. And so you can imagine if you were in the center of a really massive object, there would be no net gravitational force being pulled on you, because you're at its center of mass. The rest of the mass is outward. So at every point, it will be pulling you outward."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction. And so you can imagine if you were in the center of a really massive object, there would be no net gravitational force being pulled on you, because you're at its center of mass. The rest of the mass is outward. So at every point, it will be pulling you outward. And so that's why if you were to enter the core of a star, if you were to get a lot closer to its center of mass, it's not going to be pulling on you with this type of force. And the only way you can get these type of forces is if the entire mass is contained in a very dense region, in a very small region. And that's why a black hole is able to exert such strong gravity that not even light can escape."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So you would just break this up the way we've done it in the last several videos. It is a suffix is ane, so it is an alkane, all single bonds. So single bonds. It's pentane, so we're dealing with five carbons on kind of the base or on kind of the backbone. So this is five carbons. And it's a cyclopentane, so it's five carbons in a ring. So it's five carbon ring is the backbone."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "It's pentane, so we're dealing with five carbons on kind of the base or on kind of the backbone. So this is five carbons. And it's a cyclopentane, so it's five carbons in a ring. So it's five carbon ring is the backbone. And then we have a butyl group added to that five carbon ring. Now you might say, hey Sal, how do I know which carbon to add it to? When you're dealing with a ring and you only have one group on the ring, it doesn't matter."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So it's five carbon ring is the backbone. And then we have a butyl group added to that five carbon ring. Now you might say, hey Sal, how do I know which carbon to add it to? When you're dealing with a ring and you only have one group on the ring, it doesn't matter. Let me just show you what I mean. So let's draw the five carbon ring. Let's draw the cyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "When you're dealing with a ring and you only have one group on the ring, it doesn't matter. Let me just show you what I mean. So let's draw the five carbon ring. Let's draw the cyclopentane. So it'll just be a pentagon, so 1, 2, 3, 4, 5. And then it's a ring, so you can connect them. 1, 2, 3, 4, 5."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw the cyclopentane. So it'll just be a pentagon, so 1, 2, 3, 4, 5. And then it's a ring, so you can connect them. 1, 2, 3, 4, 5. Now, it doesn't matter where I draw the butyl group. It's all symmetric around there. We just have a ring and it's connected to a butyl group at some point."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5. Now, it doesn't matter where I draw the butyl group. It's all symmetric around there. We just have a ring and it's connected to a butyl group at some point. It'll start to matter once we add more than one group. So we can just pick any of these carbons to add the butyl group to. Now, just as a review, the but- prefix, that refers to, remember, methyl ethyl propyl or meth-eth-prop-bute."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "We just have a ring and it's connected to a butyl group at some point. It'll start to matter once we add more than one group. So we can just pick any of these carbons to add the butyl group to. Now, just as a review, the but- prefix, that refers to, remember, methyl ethyl propyl or meth-eth-prop-bute. This is four carbons. This is a four-carbon alkyl group. So let me just add it here."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Now, just as a review, the but- prefix, that refers to, remember, methyl ethyl propyl or meth-eth-prop-bute. This is four carbons. This is a four-carbon alkyl group. So let me just add it here. I could have added it to any of these carbons around this cyclopentane ring. So if I just add it right here, I'm going to have four carbons. So 1, 2, 3, 4."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So let me just add it here. I could have added it to any of these carbons around this cyclopentane ring. So if I just add it right here, I'm going to have four carbons. So 1, 2, 3, 4. That is the butyl part of this whole thing. And then let me just attach them up. So you might be tempted to just draw this right there."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4. That is the butyl part of this whole thing. And then let me just attach them up. So you might be tempted to just draw this right there. And actually, this would be right. This is butyl cyclopentane. But a question might rise."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So you might be tempted to just draw this right there. And actually, this would be right. This is butyl cyclopentane. But a question might rise. I just happened to connect the cyclopentane to the butyl at this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "But a question might rise. I just happened to connect the cyclopentane to the butyl at this first carbon on the butyl right there. I could have just as easily done it like this. I could have just as easily had it like this. Where, let me draw my butyl again. So I have 1, 2, 3, 4. So once again, this is a butyl."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "I could have just as easily had it like this. Where, let me draw my butyl again. So I have 1, 2, 3, 4. So once again, this is a butyl. But instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Maybe, let me do it with that yellow color. Maybe it's bonded right here."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So once again, this is a butyl. But instead of being bonded to the cyclopentane on my first carbon, maybe it's bonded right here. Maybe, let me do it with that yellow color. Maybe it's bonded right here. This seems like maybe this could also be butyl cyclopentane. It looks like we have a butyl group. This is butyl right here."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Maybe it's bonded right here. This seems like maybe this could also be butyl cyclopentane. It looks like we have a butyl group. This is butyl right here. I drew a butyl group right over here. And I also drew a butyl group right over here. But these are fundamentally two different molecular structures."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "This is butyl right here. I drew a butyl group right over here. And I also drew a butyl group right over here. But these are fundamentally two different molecular structures. I'm touching the first carbon here. I'm touching the second carbon over here. Now there's two ways to differentiate this."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "But these are fundamentally two different molecular structures. I'm touching the first carbon here. I'm touching the second carbon over here. Now there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Now there's two ways to differentiate this. One is the common naming and one is the systematic naming. So let me differentiate between the two. So in the common naming, and this can get a little bit involved. And this, frankly, is probably the most complicated or a part of naming organic compounds. Systematic is often more complicated, but it's easier to, I guess, systematically come up with it. So there's a common and then there's a systematic."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So in the common naming, and this can get a little bit involved. And this, frankly, is probably the most complicated or a part of naming organic compounds. Systematic is often more complicated, but it's easier to, I guess, systematically come up with it. So there's a common and then there's a systematic. Systematic. So the common way of doing it is if you just say butyl cyclopentane, that implies that you are bonding to the first, or depending how you view it, the last carbon in the chain. So this right here is butyl cyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So there's a common and then there's a systematic. Systematic. So the common way of doing it is if you just say butyl cyclopentane, that implies that you are bonding to the first, or depending how you view it, the last carbon in the chain. So this right here is butyl cyclopentane. This right here is not just butyl cyclopentane. What you would do is you definitely have a cyclopentane ring. So this would definitely be a cyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So this right here is butyl cyclopentane. This right here is not just butyl cyclopentane. What you would do is you definitely have a cyclopentane ring. So this would definitely be a cyclopentane. Let me put some space here. This is definitely going to be a cyclopentane. And you do have a butyl group on it."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So this would definitely be a cyclopentane. Let me put some space here. This is definitely going to be a cyclopentane. And you do have a butyl group on it. So we do have a butyl group. But because we aren't bonded to the first carbon, we're bonded to a carbon that is bonded to two other carbons. We call this sec-butyl cyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "And you do have a butyl group on it. So we do have a butyl group. But because we aren't bonded to the first carbon, we're bonded to a carbon that is bonded to two other carbons. We call this sec-butyl cyclopentane. Let me, so this is sec. And everything I'm doing is obviously freehand. If you were to see this in a book, the sec would be italicized."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "We call this sec-butyl cyclopentane. Let me, so this is sec. And everything I'm doing is obviously freehand. If you were to see this in a book, the sec would be italicized. Or sometimes it would be written as s, as s-butyl cyclopentane. And this sec means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group and say, well, which of these carbons is attached to two others?"}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "If you were to see this in a book, the sec would be italicized. Or sometimes it would be written as s, as s-butyl cyclopentane. And this sec means that we have attached to a carbon that is touching two other carbons. So you look at the butyl group and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same molecular structure. So that's what you do when you're attached to that guy right over there."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So you look at the butyl group and say, well, which of these carbons is attached to two others? It's either that one or that one. And regardless of whether you're attached to this or this, if you think about it, it's fundamentally the same molecular structure. So that's what you do when you're attached to that guy right over there. But what about the situation, we're dealing with just the common names right here. What about the situation where it looks like this? So we have our cyclopentane right there."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So that's what you do when you're attached to that guy right over there. But what about the situation, we're dealing with just the common names right here. What about the situation where it looks like this? So we have our cyclopentane right there. And we have a, I guess we could call it a butyl group. It'll have four carbons in it. But let's say that the four carbons look something like this."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So we have our cyclopentane right there. And we have a, I guess we could call it a butyl group. It'll have four carbons in it. But let's say that the four carbons look something like this. Let's say our four carbons, so we have one, two, three, four. One, two, three, four carbons. And we're bonded to this one right over here."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "But let's say that the four carbons look something like this. Let's say our four carbons, so we have one, two, three, four. One, two, three, four carbons. And we're bonded to this one right over here. So whenever you're bonded to, I guess, one end of the four carbon group, and it branches off at the other end, and it seems a little complicated. This only deals for alkyl groups below, really, five or six carbons. This we call an isobutyl group."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "And we're bonded to this one right over here. So whenever you're bonded to, I guess, one end of the four carbon group, and it branches off at the other end, and it seems a little complicated. This only deals for alkyl groups below, really, five or six carbons. This we call an isobutyl group. So let me write this down. So this right here is secbutyl or s-butyl, sometimes for short. This right here is isobutyl."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "This we call an isobutyl group. So let me write this down. So this right here is secbutyl or s-butyl, sometimes for short. This right here is isobutyl. That is an isobutyl group. And then the last thing to worry about when you're dealing with butyl groups is something like this. So you could also draw four carbons like this."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "This right here is isobutyl. That is an isobutyl group. And then the last thing to worry about when you're dealing with butyl groups is something like this. So you could also draw four carbons like this. You have one carbon, one, two, three, four. One, two, three, four carbons, and you're attached over here. Now this naming, this group right here, you're going to see the systematic naming is much easier for these compounds."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So you could also draw four carbons like this. You have one carbon, one, two, three, four. One, two, three, four carbons, and you're attached over here. Now this naming, this group right here, you're going to see the systematic naming is much easier for these compounds. This group right here, over here, the carbon you're attached to is attached to two other carbons. So it is secbutyl. When you're attached to three, it is t-butyl or tertbutyl."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Now this naming, this group right here, you're going to see the systematic naming is much easier for these compounds. This group right here, over here, the carbon you're attached to is attached to two other carbons. So it is secbutyl. When you're attached to three, it is t-butyl or tertbutyl. So this right here is a tertbutyl group, or sometimes called a t-butyl. And I really want you to understand the difference here. The common naming, it's easier to say and easier to spell, but it's sometimes a little confusing."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "When you're attached to three, it is t-butyl or tertbutyl. So this right here is a tertbutyl group, or sometimes called a t-butyl. And I really want you to understand the difference here. The common naming, it's easier to say and easier to spell, but it's sometimes a little confusing. This is just straight up butyl. So you would call this butylcyclopentane. This is secbutyl because this guy's connected to two carbons, that's where the sec comes from."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "The common naming, it's easier to say and easier to spell, but it's sometimes a little confusing. This is just straight up butyl. So you would call this butylcyclopentane. This is secbutyl because this guy's connected to two carbons, that's where the sec comes from. Sometimes it'll be s-butyl. So this could be called secbutylcyclopentane or s-butylcyclopentane. This, because we're attached to the end away from this branching off, is still a butyl group since we have four carbons."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "This is secbutyl because this guy's connected to two carbons, that's where the sec comes from. Sometimes it'll be s-butyl. So this could be called secbutylcyclopentane or s-butylcyclopentane. This, because we're attached to the end away from this branching off, is still a butyl group since we have four carbons. But since we're attached here, this is isobutyl. So this is isobutylcyclopentane. And then finally, since the carbon we're attaching to is attached to one, two, three other carbons, it is a tertbutyl or a t-butyl group."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "This, because we're attached to the end away from this branching off, is still a butyl group since we have four carbons. But since we're attached here, this is isobutyl. So this is isobutylcyclopentane. And then finally, since the carbon we're attaching to is attached to one, two, three other carbons, it is a tertbutyl or a t-butyl group. So this is t-butylcyclopentane. That's the common naming. So maybe I should clear out systematic here just so it's clear to you that everything we've done here is common naming."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, since the carbon we're attaching to is attached to one, two, three other carbons, it is a tertbutyl or a t-butyl group. So this is t-butylcyclopentane. That's the common naming. So maybe I should clear out systematic here just so it's clear to you that everything we've done here is common naming. So let me write down. It won't hurt to write them down again. Because the more familiar you are with these, the better."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So maybe I should clear out systematic here just so it's clear to you that everything we've done here is common naming. So let me write down. It won't hurt to write them down again. Because the more familiar you are with these, the better. So this is just butylcyclopentane. This is s- or secbutylcyclopentane. And this is isobutylcyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Because the more familiar you are with these, the better. So this is just butylcyclopentane. This is s- or secbutylcyclopentane. And this is isobutylcyclopentane. And then finally, this is tertbutyl or t-butylcyclopentane. Now, I said these are the common naming. What are the systematic naming?"}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "And this is isobutylcyclopentane. And then finally, this is tertbutyl or t-butylcyclopentane. Now, I said these are the common naming. What are the systematic naming? Well, in the systematic, this is still butylcyclopentane. So let me write this down. Systematic."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "What are the systematic naming? Well, in the systematic, this is still butylcyclopentane. So let me write this down. Systematic. This is still butylcyclopentane, which makes sense, this is very clearly a cyclopentane, this is very clearly a butyl group. But the systematic naming, what we try to do is we try to name this group right here just as we would name a traditional chain, but we end it with an il. So if you look at this right here, what we do is we just consider the chain where we attach."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Systematic. This is still butylcyclopentane, which makes sense, this is very clearly a cyclopentane, this is very clearly a butyl group. But the systematic naming, what we try to do is we try to name this group right here just as we would name a traditional chain, but we end it with an il. So if you look at this right here, what we do is we just consider the chain where we attach. So if you look at, we attached over here, so the longest chain from that point is there and there. So if you look at it like that, it looks like you have one, two, three carbons, and you have one carbon attached on the beginning. So this little group right here in the systematic naming, this looks like a one, two, three, three carbons, that's the prope prefix."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at this right here, what we do is we just consider the chain where we attach. So if you look at, we attached over here, so the longest chain from that point is there and there. So if you look at it like that, it looks like you have one, two, three carbons, and you have one carbon attached on the beginning. So this little group right here in the systematic naming, this looks like a one, two, three, three carbons, that's the prope prefix. So we're dealing with a prope, and it's all going to be one group, so it's a propyl group. This is a propyl group, but it has a methyl, remember, meth is one carbon, it has a methyl group attached on the first carbon. So this is one methyl, one methylpropyl."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So this little group right here in the systematic naming, this looks like a one, two, three, three carbons, that's the prope prefix. So we're dealing with a prope, and it's all going to be one group, so it's a propyl group. This is a propyl group, but it has a methyl, remember, meth is one carbon, it has a methyl group attached on the first carbon. So this is one methyl, one methylpropyl. Now that describes just the group. One methylpropyl describes just this part right here. That describes just that right over there."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So this is one methyl, one methylpropyl. Now that describes just the group. One methylpropyl describes just this part right here. That describes just that right over there. And then to have the whole compound, to describe the whole compound, you'd put this in parentheses, so this is the systematic naming. So one methyl, I put an l there, we'll put it in the same color, one methyl, because you're starting where you're attaching. So one methyl, you have a methyl group right there on that first carbon, it's a propyl chain, one, two, three, propyl, and then you would say cyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "That describes just that right over there. And then to have the whole compound, to describe the whole compound, you'd put this in parentheses, so this is the systematic naming. So one methyl, I put an l there, we'll put it in the same color, one methyl, because you're starting where you're attaching. So one methyl, you have a methyl group right there on that first carbon, it's a propyl chain, one, two, three, propyl, and then you would say cyclopentane. That's the systematic name for that. Now if you look at this one right here, what in the common name is isobutyl, what you do is you look at where we attach is one, two, three carbons. So once again, let me do it in that same, one, two, three carbons, so once again, this is a propyl, prop is for three, but with a methyl group now is attached to the second carbon."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So one methyl, you have a methyl group right there on that first carbon, it's a propyl chain, one, two, three, propyl, and then you would say cyclopentane. That's the systematic name for that. Now if you look at this one right here, what in the common name is isobutyl, what you do is you look at where we attach is one, two, three carbons. So once again, let me do it in that same, one, two, three carbons, so once again, this is a propyl, prop is for three, but with a methyl group now is attached to the second carbon. So this is two methyl, let me make some space here, this is two methyl, so that describes this group right here, that describes this entire group, cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as isobutyl cyclopentane or 2-methylpropylcyclopentane."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So once again, let me do it in that same, one, two, three carbons, so once again, this is a propyl, prop is for three, but with a methyl group now is attached to the second carbon. So this is two methyl, let me make some space here, this is two methyl, so that describes this group right here, that describes this entire group, cyclopentane. Remember, this is a systematic name. You might sometimes see this referred to as isobutyl cyclopentane or 2-methylpropylcyclopentane. This is actually a yl, spelled it wrong. And then finally, do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon, and the longest chain I can do starting with that carbon is just one chain right there, so we just have a two-carbon chain, one, two."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "You might sometimes see this referred to as isobutyl cyclopentane or 2-methylpropylcyclopentane. This is actually a yl, spelled it wrong. And then finally, do the same exact idea here, but it becomes a little bit more interesting. Over here, we are attached to this carbon, and the longest chain I can do starting with that carbon is just one chain right there, so we just have a two-carbon chain, one, two. The prefix for two-carbon is ethyl, or eth, and since it's a group, ethyl, and then we have two methyl groups. We have two methyl groups attached right over there, and it's attached on the one carbon. This is one, and this is two."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "Over here, we are attached to this carbon, and the longest chain I can do starting with that carbon is just one chain right there, so we just have a two-carbon chain, one, two. The prefix for two-carbon is ethyl, or eth, and since it's a group, ethyl, and then we have two methyl groups. We have two methyl groups attached right over there, and it's attached on the one carbon. This is one, and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write 1, 1 to show that we have two groups attached to the first carbon, and both of them are methyl. So we write 1, 1 dimethyl, di for two, dimethyl."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "This is one, and this is two. We call the one carbon where we are attached to the broader chain. So what this is going to be, you would actually write 1, 1 to show that we have two groups attached to the first carbon, and both of them are methyl. So we write 1, 1 dimethyl, di for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming, is 1, 1. We have two groups attached to this first carbon. 1, 1 dimethyl ethyl."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "So we write 1, 1 dimethyl, di for two, dimethyl. So this entire group right here, which we also called t-butyl, in systematic naming, is 1, 1. We have two groups attached to this first carbon. 1, 1 dimethyl ethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached there. That's why we write 1, 1."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "1, 1 dimethyl ethyl. That's this whole thing. This is the ethyl, and then we have two methyls attached there. That's why we write 1, 1. They're both attached to the one. We have two of them. That's why we wrote di over there."}, {"video_title": "Common and systematic naming iso-, sec-, and tert- prefixes Organic chemistry Khan Academy.mp3", "Sentence": "That's why we write 1, 1. They're both attached to the one. We have two of them. That's why we wrote di over there. So it's 1, 1 dimethyl ethyl, and then finally cyclopentane. So hopefully that doesn't confuse you too much. I think if you watch the video over and over and try to practice it with your own problems, you'll see that the systematic name way is actually pretty, pretty logical."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we also hint at the fact that it's actually caused by the tilt of the Earth. And so in this video, I want to show you how the tilt of the Earth causes the seasons to happen. So let's draw as many diagrams as possible here, because at least for my brain, they help me visualize what's actually going on. So we could imagine a top view first. So let's have a top view. That is the sun right over there. And let me draw Earth's orbit."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we could imagine a top view first. So let's have a top view. That is the sun right over there. And let me draw Earth's orbit. So Earth's orbit maybe looks something like that. It is almost circular, so I'll draw it as something that's pretty close to a circle right over here. And I'm going to draw Earth at different points in its orbit."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me draw Earth's orbit. So Earth's orbit maybe looks something like that. It is almost circular, so I'll draw it as something that's pretty close to a circle right over here. And I'm going to draw Earth at different points in its orbit. And I'm going to try to depict the tilt of its rotational axis. And obviously, this is not drawn anywhere near close to scale. Earth is much further away from the sun and much, much smaller than the sun as well."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm going to draw Earth at different points in its orbit. And I'm going to try to depict the tilt of its rotational axis. And obviously, this is not drawn anywhere near close to scale. Earth is much further away from the sun and much, much smaller than the sun as well. So I'll draw the Earth at that point. And at this point, the Earth will be tilted away from the sun. So Earth's tilt does not change, if you think about the direction, or at least over the course of a year, if we think about relatively small periods of time, it does not change relative to the direction that it's pointing at in the universe."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Earth is much further away from the sun and much, much smaller than the sun as well. So I'll draw the Earth at that point. And at this point, the Earth will be tilted away from the sun. So Earth's tilt does not change, if you think about the direction, or at least over the course of a year, if we think about relatively small periods of time, it does not change relative to the direction that it's pointing at in the universe. And we'll talk about that in a second. But let's say right over here, we are pointed away from the sun. So we're up and out of this page."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So Earth's tilt does not change, if you think about the direction, or at least over the course of a year, if we think about relatively small periods of time, it does not change relative to the direction that it's pointing at in the universe. And we'll talk about that in a second. But let's say right over here, we are pointed away from the sun. So we're up and out of this page. So we are up and out. So if I wanted to put some perspective on an arrow, it would be up and out. It would be more like up and out of this page."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're up and out of this page. So we are up and out. So if I wanted to put some perspective on an arrow, it would be up and out. It would be more like up and out of this page. So that's the direction if you were to come straight out of the North Pole. And if you were to go straight out of the South Pole, you would go below that circle right over there. And if I wanted to draw the same position, but if we're looking sideways along the plane, the orbital plane, or the plane of Earth's orbit."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would be more like up and out of this page. So that's the direction if you were to come straight out of the North Pole. And if you were to go straight out of the South Pole, you would go below that circle right over there. And if I wanted to draw the same position, but if we're looking sideways along the plane, the orbital plane, or the plane of Earth's orbit. So if we're looking at it from that direction, so let me do it this way. If we're looking at it directly sideways, this is the sun right over here. This is the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if I wanted to draw the same position, but if we're looking sideways along the plane, the orbital plane, or the plane of Earth's orbit. So if we're looking at it from that direction, so let me do it this way. If we're looking at it directly sideways, this is the sun right over here. This is the sun. That is the sun. And this is Earth at that position. This is Earth right over there."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the sun. That is the sun. And this is Earth at that position. This is Earth right over there. If I were to draw an arrow pointing straight out of the North Pole, it would look something like this. So this arrow and this arrow, they are both popping straight out of the North Pole. And so when we talk about the tilt of the Earth, we're talking about the tilt of its orbital axis, kind of this pole that could go straight between the South Pole and the North Pole."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is Earth right over there. If I were to draw an arrow pointing straight out of the North Pole, it would look something like this. So this arrow and this arrow, they are both popping straight out of the North Pole. And so when we talk about the tilt of the Earth, we're talking about the tilt of its orbital axis, kind of this pole that could go straight between the South Pole and the North Pole. The angle between that and a pole that would actually be at a 90 degree angle, or perpendicular to the plane of its orbit. And so compared to if it was just straight up and down, relative to the plane of the orbit. So this right here is the angle of Earth's tilt."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so when we talk about the tilt of the Earth, we're talking about the tilt of its orbital axis, kind of this pole that could go straight between the South Pole and the North Pole. The angle between that and a pole that would actually be at a 90 degree angle, or perpendicular to the plane of its orbit. And so compared to if it was just straight up and down, relative to the plane of the orbit. So this right here is the angle of Earth's tilt. Let me draw that a little bit bigger, just so it becomes a little bit clearer. So if this is the plane of the orbit, we're looking sideways along the plane of the orbit. And this is Earth right over here."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right here is the angle of Earth's tilt. Let me draw that a little bit bigger, just so it becomes a little bit clearer. So if this is the plane of the orbit, we're looking sideways along the plane of the orbit. And this is Earth right over here. So this is Earth. My best attempt to draw a circle. That is Earth."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is Earth right over here. So this is Earth. My best attempt to draw a circle. That is Earth. Earth does not rotate. Its axis of rotation is not perpendicular to the plane of the orbit. So this is how Earth would orbit."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is Earth. Earth does not rotate. Its axis of rotation is not perpendicular to the plane of the orbit. So this is how Earth would orbit. This is how Earth would rotate if it was. Earth rotates. Earth's rotational axis is at an angle to that vertical relative to the plane of its orbit, I guess you could say it."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is how Earth would orbit. This is how Earth would rotate if it was. Earth rotates. Earth's rotational axis is at an angle to that vertical relative to the plane of its orbit, I guess you could say it. It rotates at an angle like this. So this would be the North Pole. That is the South Pole."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Earth's rotational axis is at an angle to that vertical relative to the plane of its orbit, I guess you could say it. It rotates at an angle like this. So this would be the North Pole. That is the South Pole. This is the South Pole. And so it rotates like this. And that angle relative to being vertical with respect to the orbital plane, this angle right here for Earth right now is 23.4 degrees."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is the South Pole. This is the South Pole. And so it rotates like this. And that angle relative to being vertical with respect to the orbital plane, this angle right here for Earth right now is 23.4 degrees. And if we're talking about relatively short periods of time like our lifespans, that is constant. But it is actually changing over long periods of time. That is changing between, and these are rough numbers, it is changing between 22.1 degrees and 24.5 degrees, if my sources are correct."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that angle relative to being vertical with respect to the orbital plane, this angle right here for Earth right now is 23.4 degrees. And if we're talking about relatively short periods of time like our lifespans, that is constant. But it is actually changing over long periods of time. That is changing between, and these are rough numbers, it is changing between 22.1 degrees and 24.5 degrees, if my sources are correct. But that gives a rough estimate of what is changing between. But I want to make it clear, this is not happening overnight. The period for it to go from roughly a 22 degree angle to a 24.5 degree angle and back to a 22 degree angle is 41,000 years."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is changing between, and these are rough numbers, it is changing between 22.1 degrees and 24.5 degrees, if my sources are correct. But that gives a rough estimate of what is changing between. But I want to make it clear, this is not happening overnight. The period for it to go from roughly a 22 degree angle to a 24.5 degree angle and back to a 22 degree angle is 41,000 years. And this long-term change in the tilt, this might play into some of the long-term climactic change. Maybe it might contribute on some level to some of the ice ages that have formed over Earth's past. But for the sake of thinking about our annual seasons, you don't have to worry too much, or you don't have to worry at all really about this variation."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The period for it to go from roughly a 22 degree angle to a 24.5 degree angle and back to a 22 degree angle is 41,000 years. And this long-term change in the tilt, this might play into some of the long-term climactic change. Maybe it might contribute on some level to some of the ice ages that have formed over Earth's past. But for the sake of thinking about our annual seasons, you don't have to worry too much, or you don't have to worry at all really about this variation. You really just have to know that it is tilted. And right now it is tilted at an angle of 23.4 degrees. Now, you might say, OK, I understand what the tilt is, but how does that change the seasons in either the northern or the southern hemisphere?"}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But for the sake of thinking about our annual seasons, you don't have to worry too much, or you don't have to worry at all really about this variation. You really just have to know that it is tilted. And right now it is tilted at an angle of 23.4 degrees. Now, you might say, OK, I understand what the tilt is, but how does that change the seasons in either the northern or the southern hemisphere? And to do that, I'm going to imagine the Earth when the northern hemisphere is most tilted away from the sun and when it is most tilted towards the sun. So remember, this tilt, the direction this arrow points into relative to the rest of the universe, if we assume that this tilt is at 23.4%, it's not changing throughout the year. But depending on where it is in the orbit, it's either going to be tilting away from the sun, as it is in this example right over here, or it will be tilting towards the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, you might say, OK, I understand what the tilt is, but how does that change the seasons in either the northern or the southern hemisphere? And to do that, I'm going to imagine the Earth when the northern hemisphere is most tilted away from the sun and when it is most tilted towards the sun. So remember, this tilt, the direction this arrow points into relative to the rest of the universe, if we assume that this tilt is at 23.4%, it's not changing throughout the year. But depending on where it is in the orbit, it's either going to be tilting away from the sun, as it is in this example right over here, or it will be tilting towards the sun. I'll do the towards the sun in this magenta color. Or it would be tilting towards the sun. So six months later, when the Earth is over here, it's going to, relative to the rest of the universe, it will be tilted in that same direction."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But depending on where it is in the orbit, it's either going to be tilting away from the sun, as it is in this example right over here, or it will be tilting towards the sun. I'll do the towards the sun in this magenta color. Or it would be tilting towards the sun. So six months later, when the Earth is over here, it's going to, relative to the rest of the universe, it will be tilted in that same direction. It will be tilted in that same direction up and up, out of this page and to the right. So out of this page and to the right again, just like it was over here. But now that it's on the other side of the sun, that makes it tilt a little bit more towards the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So six months later, when the Earth is over here, it's going to, relative to the rest of the universe, it will be tilted in that same direction. It will be tilted in that same direction up and up, out of this page and to the right. So out of this page and to the right again, just like it was over here. But now that it's on the other side of the sun, that makes it tilt a little bit more towards the sun. If I were to draw it right over here, it is now tilted towards the sun. It is now tilted towards the sun. And what I want to think about is, how much sunlight will different parts of the planet receive?"}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But now that it's on the other side of the sun, that makes it tilt a little bit more towards the sun. If I were to draw it right over here, it is now tilted towards the sun. It is now tilted towards the sun. And what I want to think about is, how much sunlight will different parts of the planet receive? And I'll focus on the northern hemisphere, but you can make a similar argument for the southern hemisphere. I want to think about how much sunlight they receive when it's tilted away or tilted towards the sun. And so let's think about those two situations."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what I want to think about is, how much sunlight will different parts of the planet receive? And I'll focus on the northern hemisphere, but you can make a similar argument for the southern hemisphere. I want to think about how much sunlight they receive when it's tilted away or tilted towards the sun. And so let's think about those two situations. So first of all, let's think about this situation here where we are tilted away from the sun. So let me zoom in a little bit. So this is the situation where we're tilted away from the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so let's think about those two situations. So first of all, let's think about this situation here where we are tilted away from the sun. So let me zoom in a little bit. So this is the situation where we're tilted away from the sun. So if this is the vertical, so let me draw it. I could actually just use this diagram, but let me make it. So we're tilted away from the sun like this."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the situation where we're tilted away from the sun. So if this is the vertical, so let me draw it. I could actually just use this diagram, but let me make it. So we're tilted away from the sun like this. Let me do this in a different color. So if we have an arrow coming straight out of the North Pole, it would look like this. If we have an arrow coming straight out of the North Pole and we are rotating around like that."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're tilted away from the sun like this. Let me do this in a different color. So if we have an arrow coming straight out of the North Pole, it would look like this. If we have an arrow coming straight out of the North Pole and we are rotating around like that. So we're out of the page on the left-hand side and then into the page on the right-hand side. So we're rotating towards the east constantly. So this arrow is in the direction of that arrow is in the direction of the east."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we have an arrow coming straight out of the North Pole and we are rotating around like that. So we're out of the page on the left-hand side and then into the page on the right-hand side. So we're rotating towards the east constantly. So this arrow is in the direction of that arrow is in the direction of the east. So when we are at this point in Earth's orbit, and actually let me copy and paste this, and I'm going to use the same exact diagram for the different seasons. So let me copy and then let me paste this exact diagram. I'll do it over here for two different points."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this arrow is in the direction of that arrow is in the direction of the east. So when we are at this point in Earth's orbit, and actually let me copy and paste this, and I'm going to use the same exact diagram for the different seasons. So let me copy and then let me paste this exact diagram. I'll do it over here for two different points. So when we are here in Earth's orbit, where is the sunlight coming from? Well, it's going to be coming from the left, at least the way I've drawn the diagram right over here. So the sunlight is coming from the left."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll do it over here for two different points. So when we are here in Earth's orbit, where is the sunlight coming from? Well, it's going to be coming from the left, at least the way I've drawn the diagram right over here. So the sunlight is coming from the left. Sunlight is coming from the left in this situation. And so if you think about it, what half of the Earth is being, or what part of the Earth is being lit by sunlight? Or what part of the Earth is in daylight, the way I've drawn it right over here?"}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the sunlight is coming from the left. Sunlight is coming from the left in this situation. And so if you think about it, what half of the Earth is being, or what part of the Earth is being lit by sunlight? Or what part of the Earth is in daylight, the way I've drawn it right over here? Well, the part that is facing the sun. So all of this right over here is going to be in daylight. As we rotate, whatever part of the surface of the Earth enters into this yellow part right over here will be in daylight."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or what part of the Earth is in daylight, the way I've drawn it right over here? Well, the part that is facing the sun. So all of this right over here is going to be in daylight. As we rotate, whatever part of the surface of the Earth enters into this yellow part right over here will be in daylight. But let's think about what's happening at different parts of the Earth. So let me draw the equator, which separates our northern and southern hemispheres. So this is the equator."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As we rotate, whatever part of the surface of the Earth enters into this yellow part right over here will be in daylight. But let's think about what's happening at different parts of the Earth. So let me draw the equator, which separates our northern and southern hemispheres. So this is the equator. And then let me go into the northern hemisphere. And I want to show you why, when the North Pole is pointed away from the sun, why this is our winter. So when we're pointed away from the sun, well, if we go to the Arctic Circle."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the equator. And then let me go into the northern hemisphere. And I want to show you why, when the North Pole is pointed away from the sun, why this is our winter. So when we're pointed away from the sun, well, if we go to the Arctic Circle. So let me go right over here. Let me go to some point in the Arctic Circle. As it goes, as the Earth rotates every 24 hours, this point on the globe will just rotate around just like that."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when we're pointed away from the sun, well, if we go to the Arctic Circle. So let me go right over here. Let me go to some point in the Arctic Circle. As it goes, as the Earth rotates every 24 hours, this point on the globe will just rotate around just like that. It will just keep rotating around just like that. And so my question is, that point in the Arctic Circle as it rotates, will it ever see sunlight? Well, no."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As it goes, as the Earth rotates every 24 hours, this point on the globe will just rotate around just like that. It will just keep rotating around just like that. And so my question is, that point in the Arctic Circle as it rotates, will it ever see sunlight? Well, no. It will never see sunlight, because the North Pole is tilted away from the sun. So what I'm shading here in purple, that part of the Earth, when it's completely tilted away, will never see sunlight. Or at least it won't see sunlight while it's tilted away, while it's in this position, or in this position in the orbit."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, no. It will never see sunlight, because the North Pole is tilted away from the sun. So what I'm shading here in purple, that part of the Earth, when it's completely tilted away, will never see sunlight. Or at least it won't see sunlight while it's tilted away, while it's in this position, or in this position in the orbit. Never. I won't say never, because once it becomes summer, they will be able to see it. So no sunlight, no day, I guess you could say."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or at least it won't see sunlight while it's tilted away, while it's in this position, or in this position in the orbit. Never. I won't say never, because once it becomes summer, they will be able to see it. So no sunlight, no day, I guess you could say. No daylight. If you go to slightly more southern latitudes, so let's say you go over here. So maybe that's the latitude of something like, I don't know, New York or San Francisco or something like that."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So no sunlight, no day, I guess you could say. No daylight. If you go to slightly more southern latitudes, so let's say you go over here. So maybe that's the latitude of something like, I don't know, New York or San Francisco or something like that. Let's think about what it would see as the Earth rotates every 24 hours. So this would be daylight, daylight, daylight, daylight. Then nighttime, nighttime, nighttime, nighttime."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So maybe that's the latitude of something like, I don't know, New York or San Francisco or something like that. Let's think about what it would see as the Earth rotates every 24 hours. So this would be daylight, daylight, daylight, daylight. Then nighttime, nighttime, nighttime, nighttime. This is now going behind the globe. Nighttime, nighttime, nighttime, nighttime, nighttime. Daylight, daylight, daylight, daylight."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then nighttime, nighttime, nighttime, nighttime. This is now going behind the globe. Nighttime, nighttime, nighttime, nighttime, nighttime. Daylight, daylight, daylight, daylight. So if you just compare this, so let me do the daylight in orange. So daylight is in orange. And then nighttime I will do in this bluish, purplish color."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Daylight, daylight, daylight, daylight. So if you just compare this, so let me do the daylight in orange. So daylight is in orange. And then nighttime I will do in this bluish, purplish color. So nighttime over here. So if you go to really northern latitudes, like the Arctic Circle, they don't get any daylight when we're tilted away from the Earth. And if we go to slightly still northern latitudes, but not as north as the Arctic Circle, it does get daylight, but it gets a lot less daylight."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then nighttime I will do in this bluish, purplish color. So nighttime over here. So if you go to really northern latitudes, like the Arctic Circle, they don't get any daylight when we're tilted away from the Earth. And if we go to slightly still northern latitudes, but not as north as the Arctic Circle, it does get daylight, but it gets a lot less daylight. It spends a lot less time in the daylight than in the nighttime. So notice, if you say that this circumference represents the positions over 24 hours, it spends much less time in the daylight than it does in the nighttime. So because while the northern hemisphere is tilted away from the Earth, the latitudes in the northern hemisphere are getting less daylight."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if we go to slightly still northern latitudes, but not as north as the Arctic Circle, it does get daylight, but it gets a lot less daylight. It spends a lot less time in the daylight than in the nighttime. So notice, if you say that this circumference represents the positions over 24 hours, it spends much less time in the daylight than it does in the nighttime. So because while the northern hemisphere is tilted away from the Earth, the latitudes in the northern hemisphere are getting less daylight. So they're getting less daylight. They're also getting less energy from the sun. And so that's what leads to winter, or just being generally colder."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So because while the northern hemisphere is tilted away from the Earth, the latitudes in the northern hemisphere are getting less daylight. So they're getting less daylight. They're also getting less energy from the sun. And so that's what leads to winter, or just being generally colder. And to see what happens in the summer, let's just go the other side. So now we're going to the other side of our orbit around the sun. This is going to be six months later."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that's what leads to winter, or just being generally colder. And to see what happens in the summer, let's just go the other side. So now we're going to the other side of our orbit around the sun. This is going to be six months later. And notice, the actual direction relative to the rest of the universe has not changed. We're still pointed in that same direction. We still have a 23.4 degree tilt relative to, I guess, being straight up and down."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is going to be six months later. And notice, the actual direction relative to the rest of the universe has not changed. We're still pointed in that same direction. We still have a 23.4 degree tilt relative to, I guess, being straight up and down. But now, once we're over here, the light from the sun is going to be coming from the right. So the light from the sun is going to be coming from the right, just like that. And now, if on this diagram at least, this is the side of the Earth that is going to be getting the sunlight."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We still have a 23.4 degree tilt relative to, I guess, being straight up and down. But now, once we're over here, the light from the sun is going to be coming from the right. So the light from the sun is going to be coming from the right, just like that. And now, if on this diagram at least, this is the side of the Earth that is going to be getting the sunlight. And let me draw the equator again, or my best attempt to draw the equator. I'll draw the equator in that same color, actually, in that green color. So this separates the northern and the southern hemisphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now, if on this diagram at least, this is the side of the Earth that is going to be getting the sunlight. And let me draw the equator again, or my best attempt to draw the equator. I'll draw the equator in that same color, actually, in that green color. So this separates the northern and the southern hemisphere. And now, let's think about the Arctic Circle. So let's say I'm sitting here in the Arctic Circle. As the day goes on, as 24 hours go around, I'll keep rotating around here."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this separates the northern and the southern hemisphere. And now, let's think about the Arctic Circle. So let's say I'm sitting here in the Arctic Circle. As the day goes on, as 24 hours go around, I'll keep rotating around here. But notice, the whole time I am inside of the sun. I'm getting no night time. There is no night in the Arctic Circle while we are tilted towards the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As the day goes on, as 24 hours go around, I'll keep rotating around here. But notice, the whole time I am inside of the sun. I'm getting no night time. There is no night in the Arctic Circle while we are tilted towards the sun. And if we still do that fairly northern latitude, but not as far as the Arctic Circle, maybe in San Francisco or New York or something like that, if we go to that latitude, notice how much time we spend in the sun. So maybe we just enter. So this is right at sunrise."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There is no night in the Arctic Circle while we are tilted towards the sun. And if we still do that fairly northern latitude, but not as far as the Arctic Circle, maybe in San Francisco or New York or something like that, if we go to that latitude, notice how much time we spend in the sun. So maybe we just enter. So this is right at sunrise. And then as the day goes on, we are in sunlight, sunlight, sunlight, sunlight, sunlight, sunlight, sunlight, sunlight. Then we hit sunset. Then we hit night time, night time."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is right at sunrise. And then as the day goes on, we are in sunlight, sunlight, sunlight, sunlight, sunlight, sunlight, sunlight, sunlight. Then we hit sunset. Then we hit night time, night time. Then we hit night time. And then we get sunrise again. And so when you look at the amount of time that something in the northern hemisphere spends in the daylight versus sunlight, you'll see it spends a lot more time in the daylight when the northern hemisphere is tilted towards the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then we hit night time, night time. Then we hit night time. And then we get sunrise again. And so when you look at the amount of time that something in the northern hemisphere spends in the daylight versus sunlight, you'll see it spends a lot more time in the daylight when the northern hemisphere is tilted towards the sun. So this is more day, less night. So it is getting more energy from the sun. So when it's tilted towards the sun, it is getting more energy from the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so when you look at the amount of time that something in the northern hemisphere spends in the daylight versus sunlight, you'll see it spends a lot more time in the daylight when the northern hemisphere is tilted towards the sun. So this is more day, less night. So it is getting more energy from the sun. So when it's tilted towards the sun, it is getting more energy from the sun. So things will generally be warmer. And so you are now talking about summer in the northern hemisphere. And the arguments for the southern hemisphere are identical."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when it's tilted towards the sun, it is getting more energy from the sun. So things will generally be warmer. And so you are now talking about summer in the northern hemisphere. And the arguments for the southern hemisphere are identical. You could even play it right over here. When the northern hemisphere is tilted away from the sun, then the southern hemisphere is tilted towards the sun. And so for example, the South Pole will have all daylight and no night time."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the arguments for the southern hemisphere are identical. You could even play it right over here. When the northern hemisphere is tilted away from the sun, then the southern hemisphere is tilted towards the sun. And so for example, the South Pole will have all daylight and no night time. And southern latitudes will have more daylight than night time. And so the south will have summer. So this is summer in the south, in the southern hemisphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so for example, the South Pole will have all daylight and no night time. And southern latitudes will have more daylight than night time. And so the south will have summer. So this is summer in the south, in the southern hemisphere. And it's winter in the north. And then down here, the southern hemisphere is pointed away from the sun. So this is winter in the southern hemisphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is summer in the south, in the southern hemisphere. And it's winter in the north. And then down here, the southern hemisphere is pointed away from the sun. So this is winter in the southern hemisphere. And you might be saying, hey, Sal, what about, you haven't talked a lot about spring and fall. Well, let's think about it. Well, if we're talking about the northern hemisphere, this over here we just tided was winter in the northern hemisphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is winter in the southern hemisphere. And you might be saying, hey, Sal, what about, you haven't talked a lot about spring and fall. Well, let's think about it. Well, if we're talking about the northern hemisphere, this over here we just tided was winter in the northern hemisphere. And we are going to rotate around the sun. And at some point, we're going to get over here. And then because of this tilt, we aren't pointed away or towards the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, if we're talking about the northern hemisphere, this over here we just tided was winter in the northern hemisphere. And we are going to rotate around the sun. And at some point, we're going to get over here. And then because of this tilt, we aren't pointed away or towards the sun. We're kind of pointed sideways relative to the direction of the sun. But this doesn't favor one hemisphere over the other. So when we're over here in, and this will actually be the spring now, both hemispheres are getting the equal amount of daylight and sunlight."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then because of this tilt, we aren't pointed away or towards the sun. We're kind of pointed sideways relative to the direction of the sun. But this doesn't favor one hemisphere over the other. So when we're over here in, and this will actually be the spring now, both hemispheres are getting the equal amount of daylight and sunlight. Or for a given latitude above or below the equator, they're getting the same amount. And the same thing is true over here when we get to, so this is the spring, this is the summer in the northern hemisphere. Now this will be the fall in the northern hemisphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when we're over here in, and this will actually be the spring now, both hemispheres are getting the equal amount of daylight and sunlight. Or for a given latitude above or below the equator, they're getting the same amount. And the same thing is true over here when we get to, so this is the spring, this is the summer in the northern hemisphere. Now this will be the fall in the northern hemisphere. And once again, we're tilted in this direction. And so the northern hemisphere isn't tilted away or towards the sun. And so both hemispheres are going to get the same amount of radiation from the sun."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now this will be the fall in the northern hemisphere. And once again, we're tilted in this direction. And so the northern hemisphere isn't tilted away or towards the sun. And so both hemispheres are going to get the same amount of radiation from the sun. So you really see the extremes in the winters and the summers. Now one thing I do want to make clear, and I started off with just the length of day and nighttime, because frankly that's maybe a little bit, or at least in my brain, a little bit easier to visualize. But that by itself does not account for all of the difference between summer and winter."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so both hemispheres are going to get the same amount of radiation from the sun. So you really see the extremes in the winters and the summers. Now one thing I do want to make clear, and I started off with just the length of day and nighttime, because frankly that's maybe a little bit, or at least in my brain, a little bit easier to visualize. But that by itself does not account for all of the difference between summer and winter. Another cause, and actually this is probably the biggest cause, is if you think about the total amount of sun. So let's talk about the northern hemisphere winter. And let's say there's a certain amount of sunlight that is reaching the Earth."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But that by itself does not account for all of the difference between summer and winter. Another cause, and actually this is probably the biggest cause, is if you think about the total amount of sun. So let's talk about the northern hemisphere winter. And let's say there's a certain amount of sunlight that is reaching the Earth. So there is a certain amount of sunlight that is reaching the Earth. So this is the total amount of sunlight that's reaching the Earth at any point in time. You see that much more of that is hitting the southern hemisphere than the northern hemisphere here."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say there's a certain amount of sunlight that is reaching the Earth. So there is a certain amount of sunlight that is reaching the Earth. So this is the total amount of sunlight that's reaching the Earth at any point in time. You see that much more of that is hitting the southern hemisphere than the northern hemisphere here. If you imagine it, all of these rays right over here are hitting the southern hemisphere. So a majority of the rays are hitting the southern hemisphere and much fewer are hitting the northern hemisphere. So actually a smaller amount of the radiation period, at even a given period in time, not even talking about the amount of time you're facing the sun, but at any given moment in time, more energy is hitting the southern hemisphere than the northern."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You see that much more of that is hitting the southern hemisphere than the northern hemisphere here. If you imagine it, all of these rays right over here are hitting the southern hemisphere. So a majority of the rays are hitting the southern hemisphere and much fewer are hitting the northern hemisphere. So actually a smaller amount of the radiation period, at even a given period in time, not even talking about the amount of time you're facing the sun, but at any given moment in time, more energy is hitting the southern hemisphere than the northern. And the opposite is true when the tilt is then towards the sun. Now a disproportionate amount of the sun's energy is hitting the northern hemisphere. So if you just think that this is all of the energy from the sun, most of it, all of these rays up here, are hitting the northern hemisphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So actually a smaller amount of the radiation period, at even a given period in time, not even talking about the amount of time you're facing the sun, but at any given moment in time, more energy is hitting the southern hemisphere than the northern. And the opposite is true when the tilt is then towards the sun. Now a disproportionate amount of the sun's energy is hitting the northern hemisphere. So if you just think that this is all of the energy from the sun, most of it, all of these rays up here, are hitting the northern hemisphere. And only these down here are hitting the southern hemisphere. And on top of that, what makes it even more extreme is that the actual angle that the, and of course this is to some degree is due to the fact of where the angle of the sun relative to the horizon or where you are on Earth. But even more than that, if you are on, let's say that this is the land."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you just think that this is all of the energy from the sun, most of it, all of these rays up here, are hitting the northern hemisphere. And only these down here are hitting the southern hemisphere. And on top of that, what makes it even more extreme is that the actual angle that the, and of course this is to some degree is due to the fact of where the angle of the sun relative to the horizon or where you are on Earth. But even more than that, if you are on, let's say that this is the land. And we're talking about the winter in the northern hemisphere. So let's say you're talking about, let's say we're up over here at this northern latitude. And this is just what, and we're looking, we're just looking at the sun here."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But even more than that, if you are on, let's say that this is the land. And we're talking about the winter in the northern hemisphere. So let's say you're talking about, let's say we're up over here at this northern latitude. And this is just what, and we're looking, we're just looking at the sun here. And over here you could see even when we are closest to the sun, the sun is not directly overhead. When we're closest to the sun, the sun still is pretty low on the horizon. So maybe right over here when we're closest to the sun in the winter, the sun might be right over here."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is just what, and we're looking, we're just looking at the sun here. And over here you could see even when we are closest to the sun, the sun is not directly overhead. When we're closest to the sun, the sun still is pretty low on the horizon. So maybe right over here when we're closest to the sun in the winter, the sun might be right over here. But if you look at that same latitude in the summer, when it is closest to the sun, the sun is more close to being directly overhead. It still won't be directly overhead because we're still at a relatively northern latitude. But the sun is going to be much higher in the sky."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So maybe right over here when we're closest to the sun in the winter, the sun might be right over here. But if you look at that same latitude in the summer, when it is closest to the sun, the sun is more close to being directly overhead. It still won't be directly overhead because we're still at a relatively northern latitude. But the sun is going to be much higher in the sky. And these are all related to each other. It's kind of connected with this idea that more energy is hitting one hemisphere or the other. But also when you have a, I guess you could say, a steeper angle from the rays of the sun with the earth, it's actually going to be dissipated less by the atmosphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the sun is going to be much higher in the sky. And these are all related to each other. It's kind of connected with this idea that more energy is hitting one hemisphere or the other. But also when you have a, I guess you could say, a steeper angle from the rays of the sun with the earth, it's actually going to be dissipated less by the atmosphere. And let me just make it clear how this is. So in the summer, so let's say that that's the land. And let's say that, let me draw the atmosphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But also when you have a, I guess you could say, a steeper angle from the rays of the sun with the earth, it's actually going to be dissipated less by the atmosphere. And let me just make it clear how this is. So in the summer, so let's say that that's the land. And let's say that, let me draw the atmosphere. I'll draw the atmosphere in white. So all of this area right over here, this is the atmosphere. And obviously, there's not a hard boundary for the atmosphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say that, let me draw the atmosphere. I'll draw the atmosphere in white. So all of this area right over here, this is the atmosphere. And obviously, there's not a hard boundary for the atmosphere. But let's just say this is the densest part of the atmosphere. In the summer, when the sun is higher in the sky, the rays from the sun are dissipated by less atmosphere. So they have to get through this much atmosphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And obviously, there's not a hard boundary for the atmosphere. But let's just say this is the densest part of the atmosphere. In the summer, when the sun is higher in the sky, the rays from the sun are dissipated by less atmosphere. So they have to get through this much atmosphere. And they're bounced off. And they heat some of that atmosphere. And they're absorbed before they get to the ground."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they have to get through this much atmosphere. And they're bounced off. And they heat some of that atmosphere. And they're absorbed before they get to the ground. In the winter, when the sun is lower in the sky, so maybe the sun is out here. Let me draw it a little bit. So when the sun is lower in the sky relative to this point, you see that the rays of sunlight have to travel through a lot more atmosphere."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they're absorbed before they get to the ground. In the winter, when the sun is lower in the sky, so maybe the sun is out here. Let me draw it a little bit. So when the sun is lower in the sky relative to this point, you see that the rays of sunlight have to travel through a lot more atmosphere. So they get dissipated much more before they get to this point on the planet. So all in all, it is the tilt that is causing the changes in the season. But it's causing it for multiple reasons."}, {"video_title": "How earth's tilt causes seasons Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when the sun is lower in the sky relative to this point, you see that the rays of sunlight have to travel through a lot more atmosphere. So they get dissipated much more before they get to this point on the planet. So all in all, it is the tilt that is causing the changes in the season. But it's causing it for multiple reasons. One is when you're tilted towards the sun, you're getting more absolute hours of daylight. Not only are you getting more absolute hours of daylight, but at any given moment, most or more of the sun's total rays that are hitting the Earth are hitting the northern hemisphere as opposed to the southern hemisphere. And the stuff that's hitting the places that have summer, it has to go through less atmosphere."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we called these points black holes. And we learned there's an event horizon around these black holes, and if anything gets closer or goes within the boundary of that event horizon, there's no way they can ever escape from the black hole. All it can do is get closer and closer to the black hole, and that includes light, and that's why it's called a black hole. So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, I'll do it in purple, because it's supposed to be black, this entire thing will appear black. It will emit no light. Now, these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So even though all of the mass is at the central point, this entire area, or the entire surface of the event horizon, I'll do it in purple, because it's supposed to be black, this entire thing will appear black. It will emit no light. Now, these type of black holes that we described, we call those stellar black holes. And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we've observed are on the order of 33 solar masses, give or take. So very massive to begin with. Let's just be clear, and this is what the remnant of the star has to be."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's because they're formed from collapsing massive stars. And the largest stellar black holes that we've observed are on the order of 33 solar masses, give or take. So very massive to begin with. Let's just be clear, and this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae, plural of supernova. Now, there's another class of black holes here, and these are somewhat mysterious. And they're called supermassive black holes."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's just be clear, and this is what the remnant of the star has to be. So a lot more of the original star's mass might have been pushed off in supernovae, plural of supernova. Now, there's another class of black holes here, and these are somewhat mysterious. And they're called supermassive black holes. Supermassive black holes. And to some degree, the word super isn't big enough. Supermassive black holes."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they're called supermassive black holes. Supermassive black holes. And to some degree, the word super isn't big enough. Supermassive black holes. Because they're not just a little bit more massive than stellar black holes, they're a lot more massive. They're on the order of hundreds of thousands to billions of solar masses. 100,000s to billions times the mass of our sun."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Supermassive black holes. Because they're not just a little bit more massive than stellar black holes, they're a lot more massive. They're on the order of hundreds of thousands to billions of solar masses. 100,000s to billions times the mass of our sun. And what's interesting about these, other than the fact that they're super huge, is that there doesn't seem to be black holes in between. Or at least we haven't observed black holes in between. The largest stellar black hole's 33 solar masses."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "100,000s to billions times the mass of our sun. And what's interesting about these, other than the fact that they're super huge, is that there doesn't seem to be black holes in between. Or at least we haven't observed black holes in between. The largest stellar black hole's 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The largest stellar black hole's 33 solar masses. And then there are these supermassive black holes that we think exist. And we think they mainly exist in the centers of galaxies. And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question. If all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form is that you have a regular stellar black hole in an area that has a lot of matter that it can accrete around it."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we think most, if not all, centers of galaxies actually have one of these supermassive black holes. But it's kind of an interesting question. If all black holes were formed from collapsing stars, wouldn't we see things in between? So one theory of how these really massive black holes form is that you have a regular stellar black hole in an area that has a lot of matter that it can accrete around it. So let's imagine you have a regular, so I'll draw the, this is the event horizon around it. The actual black hole's going to be in the center of that, or the mass of the black hole will be in the center of it. And then over time, you have just more and more mass just falling into this black hole."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So one theory of how these really massive black holes form is that you have a regular stellar black hole in an area that has a lot of matter that it can accrete around it. So let's imagine you have a regular, so I'll draw the, this is the event horizon around it. The actual black hole's going to be in the center of that, or the mass of the black hole will be in the center of it. And then over time, you have just more and more mass just falling into this black hole. Just more and more stuff just keeps falling into this black hole, and then it just keeps growing. And so this could be a plausible reason, or at least the mass in the center keeps growing, and so the event horizon will also keep growing in radius. Now, this is a plausible explanation based on our current understanding."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then over time, you have just more and more mass just falling into this black hole. Just more and more stuff just keeps falling into this black hole, and then it just keeps growing. And so this could be a plausible reason, or at least the mass in the center keeps growing, and so the event horizon will also keep growing in radius. Now, this is a plausible explanation based on our current understanding. But the reason why this one doesn't gel that well is if this was the explanation for supermassive black holes, you expect to see more black holes in between, maybe black holes with 100 solar masses, or 1,000 solar masses, or 10,000 solar masses. But we're not seeing those right now. We just see the stellar black holes, and we see the supermassive black holes."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, this is a plausible explanation based on our current understanding. But the reason why this one doesn't gel that well is if this was the explanation for supermassive black holes, you expect to see more black holes in between, maybe black holes with 100 solar masses, or 1,000 solar masses, or 10,000 solar masses. But we're not seeing those right now. We just see the stellar black holes, and we see the supermassive black holes. So another possible explanation, this is where my inclinations lean towards this one because it kind of explains the gap, is that these supermassive black holes actually formed shortly after the Big Bang. These are primordial black holes. These started near the beginning of our universe."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We just see the stellar black holes, and we see the supermassive black holes. So another possible explanation, this is where my inclinations lean towards this one because it kind of explains the gap, is that these supermassive black holes actually formed shortly after the Big Bang. These are primordial black holes. These started near the beginning of our universe. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer and closer together, and they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures to really collapse into what we think is a single point."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These started near the beginning of our universe. Now remember, what do you need to have a black hole? You need to have an amazingly dense amount of matter or a dense amount of mass. If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer and closer together, and they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you have a lot of mass in a very small volume, then their gravitational pull will pull them closer and closer together, and they'll be able to overcome all of the electron degeneracy pressures, and the neutron degeneracy pressures, and the quark degeneracy pressures to really collapse into what we think is a single point. I want to be clear here, too. We don't know it's a single point. We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least as we understand it right now, have everything colliding into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely dense point in space, or dense amount of matter."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We've never gone into the center of a black hole. Just the mathematics of the black holes, or at least as we understand it right now, have everything colliding into a single point where the math starts to break down. So we're really not sure what happens at that very small center point. But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter was in the universe, was in a much denser space, because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now, what we've talked about before when we talked about cosmic background radiation is at that point, the universe was relatively uniform."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But needless to say, it will be an unbelievably, maybe infinite, maybe almost infinitely dense point in space, or dense amount of matter. And the reason why I kind of favor this primordial black hole and why this would make sense is right after the formation of the universe, all of the matter was in the universe, was in a much denser space, because the universe was smaller. So let's say that this is right after the Big Bang, some period of time after the Big Bang. Now, what we've talked about before when we talked about cosmic background radiation is at that point, the universe was relatively uniform. It was super, super dense, but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes, because if you look at a point here, sure, there's a ton of mass very close to it, but it's very close to it in every direction. So it would be pulled, the gravitational force would be the same in every direction, if it was completely uniform."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, what we've talked about before when we talked about cosmic background radiation is at that point, the universe was relatively uniform. It was super, super dense, but it was relatively uniform. So a universe like this, there's no reason why anything would collapse into black holes, because if you look at a point here, sure, there's a ton of mass very close to it, but it's very close to it in every direction. So it would be pulled, the gravitational force would be the same in every direction, if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly non-uniform. So let's say it becomes slightly non-uniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly non-uniform."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it would be pulled, the gravitational force would be the same in every direction, if it was completely uniform. But if you go shortly after the Big Bang, maybe because of slight quantum fluctuation effects, it becomes slightly non-uniform. So let's say it becomes slightly non-uniform, but it still is unbelievably dense. So let's say it looks something like this, where you have areas that are denser, but it's slightly non-uniform. But extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say it looks something like this, where you have areas that are denser, but it's slightly non-uniform. But extremely dense. So here, all of a sudden, you have the type of densities necessary for a black hole. And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you can imagine, in that primordial universe, that very shortly after the Big Bang, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And where you have higher densities, where it's less uniform, here, all of a sudden, you will have inward force. The gravitational pull from things outside of this area are going to be less than the gravitational pull towards those areas. And the more things get pulled towards it, the less uniform it's going to get. So you can imagine, in that primordial universe, that very shortly after the Big Bang, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough so that you could have this snowballing effect. So that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the sun. And this is maybe the even more interesting part."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you can imagine, in that primordial universe, that very shortly after the Big Bang, when things were very dense and closely packed together, we may have had the conditions where these supermassive black holes could have formed. Where we had so much mass in such a small volume, and it was just not uniform enough so that you could have this snowballing effect. So that more and more mass would collect into these supermassive black holes that are hundreds of thousands to billions of times the mass of the sun. And this is maybe the even more interesting part. Those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And then not everything would go into a black hole."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is maybe the even more interesting part. Those black holes would become the centers of future galaxies. So you have these black holes forming, these supermassive black holes forming. And then not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going past it fast enough, it'll just start going in orbit around the black hole. And so you can imagine that this is how the early galaxies, or even our galaxy, formed."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then not everything would go into a black hole. Only if it didn't have a lot of angular velocity, then it might go into the black hole. But if it's going past it fast enough, it'll just start going in orbit around the black hole. And so you can imagine that this is how the early galaxies, or even our galaxy, formed. And so you might be wondering, well, what about the galaxy at the center of the Milky Way? Or sorry, what about the black hole at the center of the Milky Way? And we think there is one."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you can imagine that this is how the early galaxies, or even our galaxy, formed. And so you might be wondering, well, what about the galaxy at the center of the Milky Way? Or sorry, what about the black hole at the center of the Milky Way? And we think there is one. Because we've observed stars orbiting very quickly around something at the center of the universe. I'm sorry, at the center of our Milky Way. I want to be very clear."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we think there is one. Because we've observed stars orbiting very quickly around something at the center of the universe. I'm sorry, at the center of our Milky Way. I want to be very clear. Not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the sun."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I want to be very clear. Not at the center of the universe. And the only plausible explanation for it orbiting so quickly around something is that it has to have a density of either a black hole or something that will eventually turn into a black hole. And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are, and somehow they grow into supermassive black holes."}, {"video_title": "Supermassive black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when you do the math for the middle of our galaxy, the center of the Milky Way, our supermassive black hole is on the order of 4 million times the mass of the sun. So hopefully that gives you a little bit of food for thought. There aren't just only stellar collapsed black holes. Or maybe there are, and somehow they grow into supermassive black holes. And that everything in between we just can't observe. Or that there really are a different class of black holes that are actually formed different ways. Maybe they formed near the beginning of the actual universe, when the density of things was a little ununiform, things condensed into each other."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the hydrohalogenation reaction of alkynes. We've seen this reaction before with alkenes, and there are some similarities and some differences. So let's look and see what happens here. We start with our alkyne, our triple bond, and we add our hydrogen halide. And we add either one or two molar equivalents of our hydrogen halide. If we add one equivalent of our hydrogen halide, the hydrogen and the halogen are going to add from opposite sides. And the halogen is going to add to the most substituted carbon."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We start with our alkyne, our triple bond, and we add our hydrogen halide. And we add either one or two molar equivalents of our hydrogen halide. If we add one equivalent of our hydrogen halide, the hydrogen and the halogen are going to add from opposite sides. And the halogen is going to add to the most substituted carbon. So this halogen here is going to add to the most substituted carbon, which in terms of regiochemistry, that was called Markovnikov's rule. So Markovnikov's rule, we're going to add the halogen to the most substituted carbon in the reaction. So with two molar equivalents, you're going to end up with two halogens on the same carbon like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the halogen is going to add to the most substituted carbon. So this halogen here is going to add to the most substituted carbon, which in terms of regiochemistry, that was called Markovnikov's rule. So Markovnikov's rule, we're going to add the halogen to the most substituted carbon in the reaction. So with two molar equivalents, you're going to end up with two halogens on the same carbon like that. So let's take a look at one of the proposed mechanisms for the hydrohalogenation of alkynes. And it closely parallels the hydrohalogenation of alkenes. And this isn't considered to be the perfect mechanism for alkynes, but we're going to start with that just to show why this is Markovnikov's in terms of regiochemistry."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So with two molar equivalents, you're going to end up with two halogens on the same carbon like that. So let's take a look at one of the proposed mechanisms for the hydrohalogenation of alkynes. And it closely parallels the hydrohalogenation of alkenes. And this isn't considered to be the perfect mechanism for alkynes, but we're going to start with that just to show why this is Markovnikov's in terms of regiochemistry. So let's say this was the mechanism. We start with our alkyne. And we have our hydrogen halide, so our hydrogen halide like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this isn't considered to be the perfect mechanism for alkynes, but we're going to start with that just to show why this is Markovnikov's in terms of regiochemistry. So let's say this was the mechanism. We start with our alkyne. And we have our hydrogen halide, so our hydrogen halide like that. And when we did this mechanism for alkenes, we said the first thing that happens was these electrons in this bond are going to form a bond with this proton here. And then these electrons are going to kick off onto your halogen like that. So let's see what we can do."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have our hydrogen halide, so our hydrogen halide like that. And when we did this mechanism for alkenes, we said the first thing that happens was these electrons in this bond are going to form a bond with this proton here. And then these electrons are going to kick off onto your halogen like that. So let's see what we can do. So let's say the hydrogen adds on to the right side. So now there's only a double bond between my two carbons. And the hydrogen added on to the right carbon like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's see what we can do. So let's say the hydrogen adds on to the right side. So now there's only a double bond between my two carbons. And the hydrogen added on to the right carbon like that. And so now this right carbon has a bond going down like this. And now this carbon on the left has only three bonds to it. So that gives it a plus 1 formal charge like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the hydrogen added on to the right carbon like that. And so now this right carbon has a bond going down like this. And now this carbon on the left has only three bonds to it. So that gives it a plus 1 formal charge like that. And then our halogen is going to have a negative 1 formal charge. And so our halogen would act as our nucleophile. And our positively charged carbon right here would act as our electrophile."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that gives it a plus 1 formal charge like that. And then our halogen is going to have a negative 1 formal charge. And so our halogen would act as our nucleophile. And our positively charged carbon right here would act as our electrophile. And so you would get the halogen adding on to that carbon. And since a carbocation was involved in the mechanism, that's the reason why Markovnikov's rule is seen. Because you want to form the most stable carbocation possible."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And our positively charged carbon right here would act as our electrophile. And so you would get the halogen adding on to that carbon. And since a carbocation was involved in the mechanism, that's the reason why Markovnikov's rule is seen. Because you want to form the most stable carbocation possible. So the more substituted carbocation is the more stable one. So this really isn't quite the correct mechanism. And the reason why it's not quite the correct mechanism is because if this was the right mechanism, the rate of this reaction would depend on the concentration of two molecules, bimolecular."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Because you want to form the most stable carbocation possible. So the more substituted carbocation is the more stable one. So this really isn't quite the correct mechanism. And the reason why it's not quite the correct mechanism is because if this was the right mechanism, the rate of this reaction would depend on the concentration of two molecules, bimolecular. So if we were to write the rate law for this mechanism, we would say, oh, OK, I expect the rate of reaction for hydrohalogenation of alkynes. We have a rate constant in there to be proportional to the concentration of your alkyne and then the concentration of your hydrogen halide. Two things, because that's what we started in this mechanism."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why it's not quite the correct mechanism is because if this was the right mechanism, the rate of this reaction would depend on the concentration of two molecules, bimolecular. So if we were to write the rate law for this mechanism, we would say, oh, OK, I expect the rate of reaction for hydrohalogenation of alkynes. We have a rate constant in there to be proportional to the concentration of your alkyne and then the concentration of your hydrogen halide. Two things, because that's what we started in this mechanism. We had our hydrogen halide right here. We had our alkyne. And that is not observed experimentally."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Two things, because that's what we started in this mechanism. We had our hydrogen halide right here. We had our alkyne. And that is not observed experimentally. So this mechanism is not quite right. So this mechanism is not quite what happens. But it is kind of important to think about this carbon as being the positively charged one and Markovnikov's rule applying."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that is not observed experimentally. So this mechanism is not quite right. So this mechanism is not quite what happens. But it is kind of important to think about this carbon as being the positively charged one and Markovnikov's rule applying. So it helps to think about this mechanism. But this mechanism isn't quite right, because this is not the experimental rate law. The experimental rate law turns out to be second order with respect to your hydrogen halide and third order overall."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But it is kind of important to think about this carbon as being the positively charged one and Markovnikov's rule applying. So it helps to think about this mechanism. But this mechanism isn't quite right, because this is not the experimental rate law. The experimental rate law turns out to be second order with respect to your hydrogen halide and third order overall. So if you put a superscript 2 there, it's actually dependent upon two molecules of your hydrogen halide and one molecule of your alkyne. So since it's actually third order overall, we need to come up with a different mechanism. So let's look at a mechanism that is currently proposed as the mechanism for the hydrohalogenation of alkynes."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The experimental rate law turns out to be second order with respect to your hydrogen halide and third order overall. So if you put a superscript 2 there, it's actually dependent upon two molecules of your hydrogen halide and one molecule of your alkyne. So since it's actually third order overall, we need to come up with a different mechanism. So let's look at a mechanism that is currently proposed as the mechanism for the hydrohalogenation of alkynes. We start with our alkyne here. And we need two molecules of our hydrogen halide. So here's one molecule of our hydrogen halide like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a mechanism that is currently proposed as the mechanism for the hydrohalogenation of alkynes. We start with our alkyne here. And we need two molecules of our hydrogen halide. So here's one molecule of our hydrogen halide like that. And let's go ahead and draw the other one down here. So let's put in those lone pairs of electrons like that. So in this mechanism, all three of these molecules are reacting at the same time."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here's one molecule of our hydrogen halide like that. And let's go ahead and draw the other one down here. So let's put in those lone pairs of electrons like that. So in this mechanism, all three of these molecules are reacting at the same time. So as these electrons in here from the bond between the hydrogen and the halogen, the halogen is going to take those electrons and they're going to move and form a new bond with this carbon. At the same time that's happening, these two electrons are going to form a bond with this proton. And then these electrons are going to kick off onto your halogen like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in this mechanism, all three of these molecules are reacting at the same time. So as these electrons in here from the bond between the hydrogen and the halogen, the halogen is going to take those electrons and they're going to move and form a new bond with this carbon. At the same time that's happening, these two electrons are going to form a bond with this proton. And then these electrons are going to kick off onto your halogen like that. So if we were to draw the transition state of all this stuff happening at the same time, let's go ahead and put in our brackets here like that. So what's going to happen? Well, we're going to have carbon double bonded to another carbon like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons are going to kick off onto your halogen like that. So if we were to draw the transition state of all this stuff happening at the same time, let's go ahead and put in our brackets here like that. So what's going to happen? Well, we're going to have carbon double bonded to another carbon like that. And then we're going to have a lot of partial bonds in here. So let's use a different color for partial bonds. We have our hydrogen and our halogen."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, we're going to have carbon double bonded to another carbon like that. And then we're going to have a lot of partial bonds in here. So let's use a different color for partial bonds. We have our hydrogen and our halogen. And then we have our hydrogen and our halogen up here. And then let's use blue for partial bonds. So this halogen is forming a bond with this carbon here."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have our hydrogen and our halogen. And then we have our hydrogen and our halogen up here. And then let's use blue for partial bonds. So this halogen is forming a bond with this carbon here. At the same time, this bond is breaking in here like that. And we started with a triple bond, but that triple bond is leaving to form a bond over here with this proton like that. At the same time, this proton's bond is breaking with this halogen like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this halogen is forming a bond with this carbon here. At the same time, this bond is breaking in here like that. And we started with a triple bond, but that triple bond is leaving to form a bond over here with this proton like that. At the same time, this proton's bond is breaking with this halogen like that. So this is all one giant transition state. And if you think about what's happening, the reason why I wanted to show this mechanism up here is to show you if this happened stepwise, then this carbon on the left will get your full positive charge. Let me just highlight that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "At the same time, this proton's bond is breaking with this halogen like that. So this is all one giant transition state. And if you think about what's happening, the reason why I wanted to show this mechanism up here is to show you if this happened stepwise, then this carbon on the left will get your full positive charge. Let me just highlight that. This carbon left gets your full positive charge. This one right here. And so if we think about what's happening in this mechanism down here, it's not quite the same thing."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me just highlight that. This carbon left gets your full positive charge. This one right here. And so if we think about what's happening in this mechanism down here, it's not quite the same thing. But since the bond is leaving this carbon on the left, the triple bond is leaving this carbon on the left, this is the one that's going to get some partial carbocationic character. So it's going to be partially positive. And it's hard to see in this mechanism how this carbon could be partially positive."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so if we think about what's happening in this mechanism down here, it's not quite the same thing. But since the bond is leaving this carbon on the left, the triple bond is leaving this carbon on the left, this is the one that's going to get some partial carbocationic character. So it's going to be partially positive. And it's hard to see in this mechanism how this carbon could be partially positive. But up here, it's easy to see because the triple bond is breaking from the carbon on the left, and it's going over here to the carbon on the right. So that's the reason why it's important to think about this mechanism. This gives you your partial carbocationic character, which explains the Markovnikov's regiochemistry."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And it's hard to see in this mechanism how this carbon could be partially positive. But up here, it's easy to see because the triple bond is breaking from the carbon on the left, and it's going over here to the carbon on the right. So that's the reason why it's important to think about this mechanism. This gives you your partial carbocationic character, which explains the Markovnikov's regiochemistry. So and now we're pretty much done. Once those electrons finish moving, we're going to get our alkene. So we're going to form our alkene like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This gives you your partial carbocationic character, which explains the Markovnikov's regiochemistry. So and now we're pretty much done. Once those electrons finish moving, we're going to get our alkene. So we're going to form our alkene like that. And let's see, we formed a bond with our halogen down here, and then we formed a bond up here with that proton. And so that is going to be the product of our mechanism here like that. So let's look at a couple reactions."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to form our alkene like that. And let's see, we formed a bond with our halogen down here, and then we formed a bond up here with that proton. And so that is going to be the product of our mechanism here like that. So let's look at a couple reactions. So let's look at the hydrohalogenation of alkynes here. So we'll start with an alkyne. So we'll go ahead and draw it like that."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a couple reactions. So let's look at the hydrohalogenation of alkynes here. So we'll start with an alkyne. So we'll go ahead and draw it like that. And remember, you want to make it linear around your alkynes since they are linear. So in the first reaction, we'll just add one molar equivalent of our hydrogen halide like that. So I know I'm going to add on hydrogen and halogen across my triple bond."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we'll go ahead and draw it like that. And remember, you want to make it linear around your alkynes since they are linear. So in the first reaction, we'll just add one molar equivalent of our hydrogen halide like that. So I know I'm going to add on hydrogen and halogen across my triple bond. I know that's going to form a double bond. So the first thing I'm going to do is count how many carbons I have. Let's see, 1, 2, 3, 4, and 5."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I know I'm going to add on hydrogen and halogen across my triple bond. I know that's going to form a double bond. So the first thing I'm going to do is count how many carbons I have. Let's see, 1, 2, 3, 4, and 5. So I know that I'm going to turn a 5-carbon alkyne into a 5-carbon alkene. So let's go ahead and draw our 5-carbon alkene. So 1, 2, 3, 4, and 5."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's see, 1, 2, 3, 4, and 5. So I know that I'm going to turn a 5-carbon alkyne into a 5-carbon alkene. So let's go ahead and draw our 5-carbon alkene. So 1, 2, 3, 4, and 5. And the alkene is going to be between these first two carbons over here on the right like that. So that's part of my product. But now I have to figure out, OK, which one of these two carbons do I add my halogen to?"}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, and 5. And the alkene is going to be between these first two carbons over here on the right like that. So that's part of my product. But now I have to figure out, OK, which one of these two carbons do I add my halogen to? Do I add it to this carbon over here on the left side of my triple bond? Or do I add that halogen over here to the carbon on the right side of my triple bond? And to do that, you need to think about Markovnikov's rule."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But now I have to figure out, OK, which one of these two carbons do I add my halogen to? Do I add it to this carbon over here on the left side of my triple bond? Or do I add that halogen over here to the carbon on the right side of my triple bond? And to do that, you need to think about Markovnikov's rule. So if I added it to the right one, that would give me a primary carbocation. I mean, you have to think about which one will give you the most stable carbocation. And the most stable carbocation would be your secondary one over here on the left."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And to do that, you need to think about Markovnikov's rule. So if I added it to the right one, that would give me a primary carbocation. I mean, you have to think about which one will give you the most stable carbocation. And the most stable carbocation would be your secondary one over here on the left. And so that is the one that your halogen is going to add. Your halogen is going to add over here on the left side. So again, watch those earlier videos for much more in terms of detail about carbocations and stability and Markovnikov's rule."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the most stable carbocation would be your secondary one over here on the left. And so that is the one that your halogen is going to add. Your halogen is going to add over here on the left side. So again, watch those earlier videos for much more in terms of detail about carbocations and stability and Markovnikov's rule. So let's do the addition of two equivalents of hydrogen halide or just make it in excess here. So in excess. So you have to think about the fact that, OK, so once again, I'm starting with five carbons."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So again, watch those earlier videos for much more in terms of detail about carbocations and stability and Markovnikov's rule. So let's do the addition of two equivalents of hydrogen halide or just make it in excess here. So in excess. So you have to think about the fact that, OK, so once again, I'm starting with five carbons. So I'm going to get five carbons for my product here. And go back up here to our original reaction. You can see if you have two molar equivalents, you're going to end up adding your two halogens to the same carbon."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you have to think about the fact that, OK, so once again, I'm starting with five carbons. So I'm going to get five carbons for my product here. And go back up here to our original reaction. You can see if you have two molar equivalents, you're going to end up adding your two halogens to the same carbon. And I probably should have drawn this a little bit differently on the general reaction. But this also is going to exhibit Markovnikov's Ferrigio chemistry. So when you're trying to figure out which carbon that is, think about which one would be the most stable carbocation again."}, {"video_title": "Hydrohalogenation of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You can see if you have two molar equivalents, you're going to end up adding your two halogens to the same carbon. And I probably should have drawn this a little bit differently on the general reaction. But this also is going to exhibit Markovnikov's Ferrigio chemistry. So when you're trying to figure out which carbon that is, think about which one would be the most stable carbocation again. So once again, we have two choices. The carbon on the left side, the carbon on the right side, the carbon on the left side gives us more stable. So we're going to add on two equivalents of our halogen here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Before we get into nuclear shielding, we need to review some physics. So let's say we have current in a loop of wire. So on the left is our loop of wire, and let's say that current is going in this direction. So in physics, you represent current by I. And let's say we're looking down on this loop of wire. And so over here, this would be the top view. And if we're looking down, current would be going in a clockwise fashion, so around this loop."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in physics, you represent current by I. And let's say we're looking down on this loop of wire. And so over here, this would be the top view. And if we're looking down, current would be going in a clockwise fashion, so around this loop. In physics, current is thought of as being moving positive charges. So even though that's not really what's happening, but the moving charge creates a magnetic field. And so the current's going to create a magnetic field."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if we're looking down, current would be going in a clockwise fashion, so around this loop. In physics, current is thought of as being moving positive charges. So even though that's not really what's happening, but the moving charge creates a magnetic field. And so the current's going to create a magnetic field. We can figure out the direction of the magnetic field by using a variation of the right-hand rule. So if we think about our right hand being right here on our loop, we point our thumb in the direction of the current. So the current's going to the left at this point, so we point our thumb to the left, and this is going to be the back of my right hand here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the current's going to create a magnetic field. We can figure out the direction of the magnetic field by using a variation of the right-hand rule. So if we think about our right hand being right here on our loop, we point our thumb in the direction of the current. So the current's going to the left at this point, so we point our thumb to the left, and this is going to be the back of my right hand here. So if using your right hand, there's only one direction for your fingers to curl. And in this loop, your fingers would curl down. So in this loop, your fingers curl down."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the current's going to the left at this point, so we point our thumb to the left, and this is going to be the back of my right hand here. So if using your right hand, there's only one direction for your fingers to curl. And in this loop, your fingers would curl down. So in this loop, your fingers curl down. And that's the direction of the magnetic field created by the current. And so from a top view, the magnetic field is going into the page. And if you're looking at it from this orientation, the magnetic field would be going down."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in this loop, your fingers curl down. And that's the direction of the magnetic field created by the current. And so from a top view, the magnetic field is going into the page. And if you're looking at it from this orientation, the magnetic field would be going down. So a magnetic field, represented by B here, is created by current in our loop of wire. In reality, it's the electrons that are moving. And since the electrons are negatively charged, the electrons move in an opposite direction from the current."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if you're looking at it from this orientation, the magnetic field would be going down. So a magnetic field, represented by B here, is created by current in our loop of wire. In reality, it's the electrons that are moving. And since the electrons are negatively charged, the electrons move in an opposite direction from the current. So the electrons are actually going around this way. So if you look at a top view, the electrons would be going around counterclockwise. And so this is important, the idea of moving electrons creating a magnetic field."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And since the electrons are negatively charged, the electrons move in an opposite direction from the current. So the electrons are actually going around this way. So if you look at a top view, the electrons would be going around counterclockwise. And so this is important, the idea of moving electrons creating a magnetic field. Now let's look at a situation where we have a proton involved, so proton NMR. In the last video, I talked about how in proton NMR, you apply an external magnetic field. So this vector here represents our external magnetic field, B0."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this is important, the idea of moving electrons creating a magnetic field. Now let's look at a situation where we have a proton involved, so proton NMR. In the last video, I talked about how in proton NMR, you apply an external magnetic field. So this vector here represents our external magnetic field, B0. And in the presence of an external magnetic field, electron density around our proton circulates. So if you think about this as being a proton, and you think about some electron density going around the proton. So here's some electron density that's circulating."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this vector here represents our external magnetic field, B0. And in the presence of an external magnetic field, electron density around our proton circulates. So if you think about this as being a proton, and you think about some electron density going around the proton. So here's some electron density that's circulating. The electron density that's circulating creates an induced magnetic field. So if the electrons are moving this way, you could think about this situation here. And the induced magnetic field would go down."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's some electron density that's circulating. The electron density that's circulating creates an induced magnetic field. So if the electrons are moving this way, you could think about this situation here. And the induced magnetic field would go down. So the induced magnetic field opposes the applied magnetic field. So here's the induced magnetic field. Let me use a different color here for that."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the induced magnetic field would go down. So the induced magnetic field opposes the applied magnetic field. So here's the induced magnetic field. Let me use a different color here for that. So this is the induced magnetic field, which is in a direction, this vector is in a direction opposite to this magnetic field. This is an effect called diamagnetism. And so the proton, the proton right here experiences a smaller overall magnetic field."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me use a different color here for that. So this is the induced magnetic field, which is in a direction, this vector is in a direction opposite to this magnetic field. This is an effect called diamagnetism. And so the proton, the proton right here experiences a smaller overall magnetic field. So let's think about that. So if we have an applied magnetic field of a certain magnitude, so B0, and the circulating electron density produces an induced magnetic field that opposes the applied field, the proton is going to feel an overall smaller magnetic field. So let me go ahead and draw that in here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the proton, the proton right here experiences a smaller overall magnetic field. So let's think about that. So if we have an applied magnetic field of a certain magnitude, so B0, and the circulating electron density produces an induced magnetic field that opposes the applied field, the proton is going to feel an overall smaller magnetic field. So let me go ahead and draw that in here. So the proton experiences a smaller magnetic field, which I will call B effective. So the effective magnetic field that the proton experiences. Or you could think about it like this."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that in here. So the proton experiences a smaller magnetic field, which I will call B effective. So the effective magnetic field that the proton experiences. Or you could think about it like this. If you start, the effective magnetic field experienced by the proton would be equal to the original magnetic field, the applied magnetic field, minus the induced magnetic field. And so this proton, this proton, this nucleus is shielded from the external magnetic field by electrons. So this proton here is said to be shielded."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Or you could think about it like this. If you start, the effective magnetic field experienced by the proton would be equal to the original magnetic field, the applied magnetic field, minus the induced magnetic field. And so this proton, this proton, this nucleus is shielded from the external magnetic field by electrons. So this proton here is said to be shielded. And if you increase the electron density around a proton, you would therefore increase the shielding of that proton. So shielding, shielding has the effect of lowering the effective, the effective magnetic field experienced by the proton. So let's think about two examples now."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this proton here is said to be shielded. And if you increase the electron density around a proton, you would therefore increase the shielding of that proton. So shielding, shielding has the effect of lowering the effective, the effective magnetic field experienced by the proton. So let's think about two examples now. So first let's start with, let's start with just a bare proton. So over here we have just a proton all by itself. It's completely deshielded."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about two examples now. So first let's start with, let's start with just a bare proton. So over here we have just a proton all by itself. It's completely deshielded. There are no electrons around it. Let me go ahead and write that. So we have a completely deshielded proton here because there are no electrons."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's completely deshielded. There are no electrons around it. Let me go ahead and write that. So we have a completely deshielded proton here because there are no electrons. Therefore this deshielded proton is going to experience the full effect of the applied magnetic field. Alright, so, and we know from the previous video that the applied magnetic field, the external magnetic field, right, causes your alpha and your beta spin states to be separated by a certain distance here. So here's the alpha spin state and here's the beta spin state."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have a completely deshielded proton here because there are no electrons. Therefore this deshielded proton is going to experience the full effect of the applied magnetic field. Alright, so, and we know from the previous video that the applied magnetic field, the external magnetic field, right, causes your alpha and your beta spin states to be separated by a certain distance here. So here's the alpha spin state and here's the beta spin state. And this would be a certain energy difference between our two spin states. So this is an energy difference right here. Now let's move to the example on the right."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's the alpha spin state and here's the beta spin state. And this would be a certain energy difference between our two spin states. So this is an energy difference right here. Now let's move to the example on the right. So the example on the right, this proton here is a proton in a molecule. It's shielded. There's electron density around this proton."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Now let's move to the example on the right. So the example on the right, this proton here is a proton in a molecule. It's shielded. There's electron density around this proton. Alright, so this is a shielded proton. Let me go ahead and write that, shielded proton. And we've just talked about what that means."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "There's electron density around this proton. Alright, so this is a shielded proton. Let me go ahead and write that, shielded proton. And we've just talked about what that means. A shielded proton, right, has circulating electron density that creates a magnetic field that opposes the applied magnetic field. And so the proton feels a smaller, a smaller effective magnetic field. So we decrease, we decrease the magnetic field experienced by this proton."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we've just talked about what that means. A shielded proton, right, has circulating electron density that creates a magnetic field that opposes the applied magnetic field. And so the proton feels a smaller, a smaller effective magnetic field. So we decrease, we decrease the magnetic field experienced by this proton. In the previous video, I talked about what happens when you have a decreased magnetic field. The magnetic field strength corresponds to the energy difference between the alpha and the beta states. And so if we're decreasing the magnetic field compared to the example on the left, we're going to decrease the energy."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we decrease, we decrease the magnetic field experienced by this proton. In the previous video, I talked about what happens when you have a decreased magnetic field. The magnetic field strength corresponds to the energy difference between the alpha and the beta states. And so if we're decreasing the magnetic field compared to the example on the left, we're going to decrease the energy. So decreasing the magnetic field decreases the energy difference between the alpha and the beta states. So I can go ahead and write that. I can show the alpha and the beta states here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if we're decreasing the magnetic field compared to the example on the left, we're going to decrease the energy. So decreasing the magnetic field decreases the energy difference between the alpha and the beta states. So I can go ahead and write that. I can show the alpha and the beta states here. And I can show a smaller gap between them. So there's a decrease in energy. And we know that that energy difference, E is equal to h nu."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I can show the alpha and the beta states here. And I can show a smaller gap between them. So there's a decrease in energy. And we know that that energy difference, E is equal to h nu. So if we decrease the energy, we're going to decrease the frequency. So the energy and the frequency are directly proportional. So if you decrease the energy difference, you decrease the frequency."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we know that that energy difference, E is equal to h nu. So if we decrease the energy, we're going to decrease the frequency. So the energy and the frequency are directly proportional. So if you decrease the energy difference, you decrease the frequency. And so therefore, a shielded proton absorbs at a lower frequency than a deshielded proton. So a deshielded proton, the energy difference would correspond to a higher frequency because there's a larger difference in energy. And so that's what we need to think about when we're looking at an NMR spectrum."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you decrease the energy difference, you decrease the frequency. And so therefore, a shielded proton absorbs at a lower frequency than a deshielded proton. So a deshielded proton, the energy difference would correspond to a higher frequency because there's a larger difference in energy. And so that's what we need to think about when we're looking at an NMR spectrum. And so I just went ahead and drew a just generic NMR. This isn't a real NMR. We're just trying to think about this example of these two protons here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that's what we need to think about when we're looking at an NMR spectrum. And so I just went ahead and drew a just generic NMR. This isn't a real NMR. We're just trying to think about this example of these two protons here. So we have one spectrum up here. So this would be, let me go ahead and mark this. So this is the deshielded spectrum."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're just trying to think about this example of these two protons here. So we have one spectrum up here. So this would be, let me go ahead and mark this. So this is the deshielded spectrum. And then this one down here represents the shielded spectrum. Again, not a real NMR spectrum, just helping to think about what's happening here. And for the example on the left, for the deshielded protons, let's think about this really fast."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is the deshielded spectrum. And then this one down here represents the shielded spectrum. Again, not a real NMR spectrum, just helping to think about what's happening here. And for the example on the left, for the deshielded protons, let's think about this really fast. So as you go to the left on an NMR spectrum, you get more and more deshielded. And if you're more and more deshielded, you experience a greater magnetic field. So a greater magnetic field."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And for the example on the left, for the deshielded protons, let's think about this really fast. So as you go to the left on an NMR spectrum, you get more and more deshielded. And if you're more and more deshielded, you experience a greater magnetic field. So a greater magnetic field. A greater magnetic field corresponds to a greater difference in energy. And a greater difference in energy corresponds to a higher frequency absorbed. So a higher frequency absorbed here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a greater magnetic field. A greater magnetic field corresponds to a greater difference in energy. And a greater difference in energy corresponds to a higher frequency absorbed. So a higher frequency absorbed here. And so therefore, as we go to the left, we're talking about an increasingly deshielded proton. And this signal that appears at your NMR right here, so this is the signal for this deshielded proton. We're talking about a high frequency signal."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a higher frequency absorbed here. And so therefore, as we go to the left, we're talking about an increasingly deshielded proton. And this signal that appears at your NMR right here, so this is the signal for this deshielded proton. We're talking about a high frequency signal. So moving to the left on an NMR spectrum, we're talking about higher frequency signal. Alright, let's think about the shielded proton over here on the right. So we're thinking about the shielded proton now."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're talking about a high frequency signal. So moving to the left on an NMR spectrum, we're talking about higher frequency signal. Alright, let's think about the shielded proton over here on the right. So we're thinking about the shielded proton now. And as you move to the right on your NMR spectrum, so we're moving to the right on our NMR spectrum, we're getting more and more shielded. So this signal is the signal for this proton, so it's more shielded than the one on the left. So if you move to the right, you're talking about increasing shielding."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're thinking about the shielded proton now. And as you move to the right on your NMR spectrum, so we're moving to the right on our NMR spectrum, we're getting more and more shielded. So this signal is the signal for this proton, so it's more shielded than the one on the left. So if you move to the right, you're talking about increasing shielding. And increasing shielding decreases the effective magnetic field. Decreasing the effective magnetic field decreases the energy difference between the alpha and the beta states, and therefore decreases the frequency absorbed. Alright, so as you move to the right, you're talking about a lower frequency signal."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you move to the right, you're talking about increasing shielding. And increasing shielding decreases the effective magnetic field. Decreasing the effective magnetic field decreases the energy difference between the alpha and the beta states, and therefore decreases the frequency absorbed. Alright, so as you move to the right, you're talking about a lower frequency signal. So as you move to the right, you're talking about a lower frequency. Alright, so this is the idea of FT NMR, which I briefly introduced in the previous video. So in FT NMR, you're holding the external magnetic field constant, and you're hitting the sample with a short pulse that contains a range of frequencies."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Alright, so as you move to the right, you're talking about a lower frequency signal. So as you move to the right, you're talking about a lower frequency. Alright, so this is the idea of FT NMR, which I briefly introduced in the previous video. So in FT NMR, you're holding the external magnetic field constant, and you're hitting the sample with a short pulse that contains a range of frequencies. And so these frequencies correspond to the energy differences. So one frequency might correspond to this energy difference, and when the proton goes back to the lower energy state, the NMR machine gives you this signal. Another frequency might correspond to this energy difference, and once again, the NMR would give you this signal."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in FT NMR, you're holding the external magnetic field constant, and you're hitting the sample with a short pulse that contains a range of frequencies. And so these frequencies correspond to the energy differences. So one frequency might correspond to this energy difference, and when the proton goes back to the lower energy state, the NMR machine gives you this signal. Another frequency might correspond to this energy difference, and once again, the NMR would give you this signal. And so that's the idea about FT NMR. You do all this at once, and the NMR machine gives you your NMR spectrum. For older NMRs, you would hold the frequency constant and vary the strength of the magnetic field."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Another frequency might correspond to this energy difference, and once again, the NMR would give you this signal. And so that's the idea about FT NMR. You do all this at once, and the NMR machine gives you your NMR spectrum. For older NMRs, you would hold the frequency constant and vary the strength of the magnetic field. And for older NMRs, it turns out that as you go to the right, you needed a higher, you needed a higher magnetic field strength, and so we called this upfield. So this would be a shift upfield, if you will. And as you go to the left on an NMR spectrum, you needed a lower magnetic field strength, and so this is called downfield."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For older NMRs, you would hold the frequency constant and vary the strength of the magnetic field. And for older NMRs, it turns out that as you go to the right, you needed a higher, you needed a higher magnetic field strength, and so we called this upfield. So this would be a shift upfield, if you will. And as you go to the left on an NMR spectrum, you needed a lower magnetic field strength, and so this is called downfield. So upfield and downfield are two terms that you might hear, and they're older terminology that relate to an older kind of NMR, but you'll still hear them, and I'm sure I will use those terms sometime as well. So in this video, we've talked about two protons with different amounts of shielding. So a completely bare proton, completely deshielded, and a shielded proton here."}, {"video_title": "Nuclear shielding Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And as you go to the left on an NMR spectrum, you needed a lower magnetic field strength, and so this is called downfield. So upfield and downfield are two terms that you might hear, and they're older terminology that relate to an older kind of NMR, but you'll still hear them, and I'm sure I will use those terms sometime as well. So in this video, we've talked about two protons with different amounts of shielding. So a completely bare proton, completely deshielded, and a shielded proton here. So two protons with different amounts of shielding are in two different environments, and we get two different signals, right? Two different signals in two different, having different frequencies on our NMR. So if you have two protons in the same environment, you should only get one signal, and we'll talk more about this in the next video."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The second way was to first form a halohydrin using bromine water, and then using sodium hydroxide to start an intramolecular Williamson ether synthesis to form our epoxide. In this video, we'll look at the stereochemistry of epoxide formation for either of these two reactions. And if we start with a cis alkene, so our hydrogen's on the same side of the double bond, or you could think about the R groups as being on the same side. For the product, the R groups are still going to be on the same side of where the double bond used to be. If you're looking at a trans alkene, so I know this is trans, since my hydrogens are on opposite sides, you think about the R groups being on opposite sides of the double bond. And in the product, the R groups are still going to be on opposite sides of where the double bond used to be. So let's look at a reaction involving stereochemistry."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "For the product, the R groups are still going to be on the same side of where the double bond used to be. If you're looking at a trans alkene, so I know this is trans, since my hydrogens are on opposite sides, you think about the R groups being on opposite sides of the double bond. And in the product, the R groups are still going to be on opposite sides of where the double bond used to be. So let's look at a reaction involving stereochemistry. So if I start out with an alkene, and I'm going to put a methyl group here, and I'm going to put an ethyl group over here, and then I'm going to have my hydrogens like that. And if I react this alkene with either of those two reagents, so I'll just go ahead and put ditto marks again. Either of those two are going to give me the same products."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at a reaction involving stereochemistry. So if I start out with an alkene, and I'm going to put a methyl group here, and I'm going to put an ethyl group over here, and then I'm going to have my hydrogens like that. And if I react this alkene with either of those two reagents, so I'll just go ahead and put ditto marks again. Either of those two are going to give me the same products. If I think about the formation of an epoxide, so I know I'm going to make an epoxide, and I know I need to have those R groups on opposite sides of where the double bond used to be. So one possible product would be to have the ethyl group coming out at me in space, and then the methyl group would therefore need to be going away from me in space like that. So if that's my product, at this carbon over here on the left, I must have my hydrogen going away from me in space."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Either of those two are going to give me the same products. If I think about the formation of an epoxide, so I know I'm going to make an epoxide, and I know I need to have those R groups on opposite sides of where the double bond used to be. So one possible product would be to have the ethyl group coming out at me in space, and then the methyl group would therefore need to be going away from me in space like that. So if that's my product, at this carbon over here on the left, I must have my hydrogen going away from me in space. And at this carbon over here on the right, I must have my hydrogen coming out at me in space like that. So that's one possible product. I could actually draw another product where my R groups are on opposite sides of where the double bond used to be."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if that's my product, at this carbon over here on the left, I must have my hydrogen going away from me in space. And at this carbon over here on the right, I must have my hydrogen coming out at me in space like that. So that's one possible product. I could actually draw another product where my R groups are on opposite sides of where the double bond used to be. So I could have my ethyl group going away from me in space at the carbon on the left. And I could therefore have my methyl group coming out at me on the carbon on the right like that. And so this would have a hydrogen coming out at me, and then this carbon would have a hydrogen going away from me like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I could actually draw another product where my R groups are on opposite sides of where the double bond used to be. So I could have my ethyl group going away from me in space at the carbon on the left. And I could therefore have my methyl group coming out at me on the carbon on the right like that. And so this would have a hydrogen coming out at me, and then this carbon would have a hydrogen going away from me like that. So I have two possible products. And these are actually enantiomers of each other. So you're actually going to create a racemic mixture, 50% of one enantiomer and 50% of the other one as well, because this reaction creates two new chirality centers."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this would have a hydrogen coming out at me, and then this carbon would have a hydrogen going away from me like that. So I have two possible products. And these are actually enantiomers of each other. So you're actually going to create a racemic mixture, 50% of one enantiomer and 50% of the other one as well, because this reaction creates two new chirality centers. So if I take a look at this molecule, these two carbons are chirality centers. So there's stereochemistry at both of those carbons. So when you're drawing the enantiomers, if I start with that molecule on the left, I have my ethyl group coming out at me."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So you're actually going to create a racemic mixture, 50% of one enantiomer and 50% of the other one as well, because this reaction creates two new chirality centers. So if I take a look at this molecule, these two carbons are chirality centers. So there's stereochemistry at both of those carbons. So when you're drawing the enantiomers, if I start with that molecule on the left, I have my ethyl group coming out at me. All I have to do is reverse that absolute configuration to make the ethyl group going away from me for the enantiomer on the right. If I'm looking at this chirality center over here, I have my methyl group going away from me. All I have to do is reverse that absolute configuration, have the methyl group coming out at me."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So when you're drawing the enantiomers, if I start with that molecule on the left, I have my ethyl group coming out at me. All I have to do is reverse that absolute configuration to make the ethyl group going away from me for the enantiomer on the right. If I'm looking at this chirality center over here, I have my methyl group going away from me. All I have to do is reverse that absolute configuration, have the methyl group coming out at me. So these two are enantiomers of each other. Let's look at the details of the formation of these two enantiomers, and then we'll also name them using stereochemistry. So if we start with the alkene over here on the left, so if we're starting with that alkene, and I think about these two carbons, these two carbons are sp2 hybridized, meaning the molecule will be planar at those two carbons."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "All I have to do is reverse that absolute configuration, have the methyl group coming out at me. So these two are enantiomers of each other. Let's look at the details of the formation of these two enantiomers, and then we'll also name them using stereochemistry. So if we start with the alkene over here on the left, so if we're starting with that alkene, and I think about these two carbons, these two carbons are sp2 hybridized, meaning the molecule will be planar at those two carbons. So I could think about that alkene as being planar at those two carbons. So if I'm going to go ahead and draw that molecule here, so I'm going to put my methyl group over here on the left, and hydrogen over here, and then I'm going to have a hydrogen over here, and then an ethyl group like that. So it's the same molecule as this one over here on the left."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we start with the alkene over here on the left, so if we're starting with that alkene, and I think about these two carbons, these two carbons are sp2 hybridized, meaning the molecule will be planar at those two carbons. So I could think about that alkene as being planar at those two carbons. So if I'm going to go ahead and draw that molecule here, so I'm going to put my methyl group over here on the left, and hydrogen over here, and then I'm going to have a hydrogen over here, and then an ethyl group like that. So it's the same molecule as this one over here on the left. It's just rotated a little bit. And I know that those two carbons are on either side of my double bond, or sp2 hybridized, meaning that that portion of the molecule is planar. So I can think about that portion of the molecule as planar."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's the same molecule as this one over here on the left. It's just rotated a little bit. And I know that those two carbons are on either side of my double bond, or sp2 hybridized, meaning that that portion of the molecule is planar. So I can think about that portion of the molecule as planar. And I could think, therefore, about the oxygen adding from either side of that plane. The oxygen could add from the top of the plane, or the oxygen could add below the plane. So let's go ahead and draw the results of those two different ways of adding the oxygen."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I can think about that portion of the molecule as planar. And I could think, therefore, about the oxygen adding from either side of that plane. The oxygen could add from the top of the plane, or the oxygen could add below the plane. So let's go ahead and draw the results of those two different ways of adding the oxygen. So over here on the left, I'm going to think about what happens to the molecule when the oxygen adds to the top of that plane. So if the oxygen adds to the top of the plane, I'm going to go ahead and keep my methyl group and my ethyl group on opposite sides of where the double bond used to be. And therefore, the two hydrogens are also going to be on opposite sides of where the double bond used to be."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the results of those two different ways of adding the oxygen. So over here on the left, I'm going to think about what happens to the molecule when the oxygen adds to the top of that plane. So if the oxygen adds to the top of the plane, I'm going to go ahead and keep my methyl group and my ethyl group on opposite sides of where the double bond used to be. And therefore, the two hydrogens are also going to be on opposite sides of where the double bond used to be. And I'm going to show the oxygen adding to the top of the plane. So therefore, my epoxide would look like that. So that's one possible way to form your epoxide."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, the two hydrogens are also going to be on opposite sides of where the double bond used to be. And I'm going to show the oxygen adding to the top of the plane. So therefore, my epoxide would look like that. So that's one possible way to form your epoxide. So let's go ahead and think about what would the epoxide look like if the oxygen added from below that plane. So if the oxygen added from below the plane, once again, we need to think about keeping our methyl group and our ethyl group on opposite sides of where the double bond used to be like that. And the hydrogens are also going to be on opposite sides of where the double bond used to be like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's one possible way to form your epoxide. So let's go ahead and think about what would the epoxide look like if the oxygen added from below that plane. So if the oxygen added from below the plane, once again, we need to think about keeping our methyl group and our ethyl group on opposite sides of where the double bond used to be like that. And the hydrogens are also going to be on opposite sides of where the double bond used to be like that. And then we're going to show our oxygen adding to form our epoxide from below the plane. So our epoxide might look like that. So these are my two possible products."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the hydrogens are also going to be on opposite sides of where the double bond used to be like that. And then we're going to show our oxygen adding to form our epoxide from below the plane. So our epoxide might look like that. So these are my two possible products. And these are enantiomers of each other, but it's kind of hard to see that as we've drawn them right here. So let's see if we can get a different vantage point on our epoxide product. So I'm going to put my eye right here."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these are my two possible products. And these are enantiomers of each other, but it's kind of hard to see that as we've drawn them right here. So let's see if we can get a different vantage point on our epoxide product. So I'm going to put my eye right here. And I'm going to stare at this epoxide like that. So if I'm staring at that epoxide, I could redraw my epoxide here. So I'm going to put that portion of the molecule like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to put my eye right here. And I'm going to stare at this epoxide like that. So if I'm staring at that epoxide, I could redraw my epoxide here. So I'm going to put that portion of the molecule like that. Now, if I'm staring at it that way, then I'm going to first focus in on the carbon on my left. So that would be this carbon down here. And I can see that there's an ethyl group coming out at me."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to put that portion of the molecule like that. Now, if I'm staring at it that way, then I'm going to first focus in on the carbon on my left. So that would be this carbon down here. And I can see that there's an ethyl group coming out at me. So I'm going to go ahead and draw an ethyl group coming out at me in space. And I can see that there's a hydrogen going away from me in space. Hopefully it's obvious this hydrogen is going away from me in space."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I can see that there's an ethyl group coming out at me. So I'm going to go ahead and draw an ethyl group coming out at me in space. And I can see that there's a hydrogen going away from me in space. Hopefully it's obvious this hydrogen is going away from me in space. So I can represent that. So I'll put my dash in here to show my hydrogen going away. And now we're going to take a look at the carbon on the right side."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Hopefully it's obvious this hydrogen is going away from me in space. So I can represent that. So I'll put my dash in here to show my hydrogen going away. And now we're going to take a look at the carbon on the right side. So this carbon on the other side right here. This time, I can see that the hydrogen is coming out at me in space. So I'm going to go ahead and draw the hydrogen coming out at me in space like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And now we're going to take a look at the carbon on the right side. So this carbon on the other side right here. This time, I can see that the hydrogen is coming out at me in space. So I'm going to go ahead and draw the hydrogen coming out at me in space like that. And therefore, this methyl group back here, this methyl group is going away from me in space. So I can go ahead and show that methyl group going away like that. So that's one enantiomer."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and draw the hydrogen coming out at me in space like that. And therefore, this methyl group back here, this methyl group is going away from me in space. So I can go ahead and show that methyl group going away like that. So that's one enantiomer. Let's go ahead and try to redraw the one on the right here. So again, if I'm staring at this epoxide like that, what do I see? Well, I will see the epoxide as being upside down."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's one enantiomer. Let's go ahead and try to redraw the one on the right here. So again, if I'm staring at this epoxide like that, what do I see? Well, I will see the epoxide as being upside down. So the oxygen would be down here. And I can see that if I'm looking at the carbon on the left side of my vantage point, so that would be this carbon down here. I can see there's an ethyl group coming out at me in space."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, I will see the epoxide as being upside down. So the oxygen would be down here. And I can see that if I'm looking at the carbon on the left side of my vantage point, so that would be this carbon down here. I can see there's an ethyl group coming out at me in space. So here is my ethyl group coming out at me in space. And this hydrogen would be going away from me. So I'll represent that hydrogen going away from me like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I can see there's an ethyl group coming out at me in space. So here is my ethyl group coming out at me in space. And this hydrogen would be going away from me. So I'll represent that hydrogen going away from me like that. And then if we move to the carbon on the right side here, I can see that this hydrogen is coming out at me in space. So I can go ahead and draw the hydrogen coming out at me in space. And the methyl group back here would be going away from me in space."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'll represent that hydrogen going away from me like that. And then if we move to the carbon on the right side here, I can see that this hydrogen is coming out at me in space. So I can go ahead and draw the hydrogen coming out at me in space. And the methyl group back here would be going away from me in space. So I can have the methyl group as a dash like that. So this is one of my products. And again, it's still not obvious that these are enantiomers."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the methyl group back here would be going away from me in space. So I can have the methyl group as a dash like that. So this is one of my products. And again, it's still not obvious that these are enantiomers. So this is where the model set comes in handy. So if you make this molecule on the right, and then you hold this oxygen and you rotate this oxygen up, much easier to see with the model set right in front of you, you will find that this molecule is the exact same molecule as the one that we're going to draw right here. So if we rotate it so the oxygen is now pointing upwards, that's actually going to take this hydrogen back here and move it to the front."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And again, it's still not obvious that these are enantiomers. So this is where the model set comes in handy. So if you make this molecule on the right, and then you hold this oxygen and you rotate this oxygen up, much easier to see with the model set right in front of you, you will find that this molecule is the exact same molecule as the one that we're going to draw right here. So if we rotate it so the oxygen is now pointing upwards, that's actually going to take this hydrogen back here and move it to the front. So when you're drawing that molecule rotated, that hydrogen is going to move to the front, which pushes the ethyl group to the back. So the ethyl group is actually going to the back here. Same thing with this methyl group."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we rotate it so the oxygen is now pointing upwards, that's actually going to take this hydrogen back here and move it to the front. So when you're drawing that molecule rotated, that hydrogen is going to move to the front, which pushes the ethyl group to the back. So the ethyl group is actually going to the back here. Same thing with this methyl group. The methyl group was in the back. When you rotate it that way, the methyl group is going to end up coming out at you like that. And then that means the hydrogen is going to go away from you in space."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Same thing with this methyl group. The methyl group was in the back. When you rotate it that way, the methyl group is going to end up coming out at you like that. And then that means the hydrogen is going to go away from you in space. So now we can see, hopefully, that these two are enantiomers to each other. And let's go ahead and think about naming this product. So if I were going to name these two molecules, I have to think about how to do it."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then that means the hydrogen is going to go away from you in space. So now we can see, hopefully, that these two are enantiomers to each other. And let's go ahead and think about naming this product. So if I were going to name these two molecules, I have to think about how to do it. So I need to find my longest carbon chain. And I want to give my epoxy substituent the lowest number possible. So if I'm going to number my carbon chain, I would make this carbon number 1, this carbon number 2, 3, 4, and 5, like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I were going to name these two molecules, I have to think about how to do it. So I need to find my longest carbon chain. And I want to give my epoxy substituent the lowest number possible. So if I'm going to number my carbon chain, I would make this carbon number 1, this carbon number 2, 3, 4, and 5, like that. So I'm going to name it as a pentane base. So this would be pentane, like that. And I can see my epoxy occurs between carbons 2 and 3."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm going to number my carbon chain, I would make this carbon number 1, this carbon number 2, 3, 4, and 5, like that. So I'm going to name it as a pentane base. So this would be pentane, like that. And I can see my epoxy occurs between carbons 2 and 3. So it would be 2, 3-epoxypentane, like that. So this enantiomer would be 2, 3-epoxypentane. This one would be 2."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I can see my epoxy occurs between carbons 2 and 3. So it would be 2, 3-epoxypentane, like that. So this enantiomer would be 2, 3-epoxypentane. This one would be 2. So I can go ahead and write 2, 3-epoxypentane, like that. But now I have to think about the stereochemistry at carbons 2 and 3. So that makes things a little bit trickier."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This one would be 2. So I can go ahead and write 2, 3-epoxypentane, like that. But now I have to think about the stereochemistry at carbons 2 and 3. So that makes things a little bit trickier. So I'm going to look at the enantiomer on the left. And I'm going to redraw the enantiomer on the left. So I'm going to redraw this one right down here."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that makes things a little bit trickier. So I'm going to look at the enantiomer on the left. And I'm going to redraw the enantiomer on the left. So I'm going to redraw this one right down here. And let's see if we can start to assign some stereochemistry. So I have my hydrogen coming out at me. I have my methyl group going away from me, like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to redraw this one right down here. And let's see if we can start to assign some stereochemistry. So I have my hydrogen coming out at me. I have my methyl group going away from me, like that. This ethyl group is coming out at me. This hydrogen is going away from me. So let's start with this carbon."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I have my methyl group going away from me, like that. This ethyl group is coming out at me. This hydrogen is going away from me. So let's start with this carbon. If I wanted to figure out the absolute configuration at that carbon, I have to think about what atoms are directly connected to that carbon. So let me go ahead and draw a carbon in here. So that is one of the atoms directly attached to that carbon in blue."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with this carbon. If I wanted to figure out the absolute configuration at that carbon, I have to think about what atoms are directly connected to that carbon. So let me go ahead and draw a carbon in here. So that is one of the atoms directly attached to that carbon in blue. So the carbon in blue is directly attached to this carbon, this oxygen, this hydrogen, and this carbon right down here. So those would be the four atoms. So if I'm trying to determine priority, I think about the atomic number."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that is one of the atoms directly attached to that carbon in blue. So the carbon in blue is directly attached to this carbon, this oxygen, this hydrogen, and this carbon right down here. So those would be the four atoms. So if I'm trying to determine priority, I think about the atomic number. And I know that oxygen has the highest atomic number out of those four atoms. So the oxygen is going to get a number 1. Hydrogen has the lowest priority, so hydrogen gets a number 4."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm trying to determine priority, I think about the atomic number. And I know that oxygen has the highest atomic number out of those four atoms. So the oxygen is going to get a number 1. Hydrogen has the lowest priority, so hydrogen gets a number 4. So now I have to think about these two carbons. If there's a tie, I have to think about what those two carbons are attached to. Well, the carbon on the right, so this carbon right here, this carbon right here is attached to an oxygen."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Hydrogen has the lowest priority, so hydrogen gets a number 4. So now I have to think about these two carbons. If there's a tie, I have to think about what those two carbons are attached to. Well, the carbon on the right, so this carbon right here, this carbon right here is attached to an oxygen. So let me go ahead and write this here. The carbon on the right is directly attached to an oxygen. It's directly attached, right over here, this oxygen, is directly attached to a carbon."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, the carbon on the right, so this carbon right here, this carbon right here is attached to an oxygen. So let me go ahead and write this here. The carbon on the right is directly attached to an oxygen. It's directly attached, right over here, this oxygen, is directly attached to a carbon. And it's directly attached to a hydrogen. So oxygen, carbon, hydrogen. Let's go ahead and look at the carbon on the left."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's directly attached, right over here, this oxygen, is directly attached to a carbon. And it's directly attached to a hydrogen. So oxygen, carbon, hydrogen. Let's go ahead and look at the carbon on the left. So this carbon right here, what is that carbon attached to? It's attached to another carbon over here, and two hydrogens, so CHH. So the oxygen is going to beat the carbon in terms of atomic number."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and look at the carbon on the left. So this carbon right here, what is that carbon attached to? It's attached to another carbon over here, and two hydrogens, so CHH. So the oxygen is going to beat the carbon in terms of atomic number. So this carbon on the right is going to get highest priority. So this carbon is going to get a number 2 here, and that means this carbon is going to get a number 3. So for absolute configuration, my lowest priority group is going away from me, and I'm traveling around this way."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the oxygen is going to beat the carbon in terms of atomic number. So this carbon on the right is going to get highest priority. So this carbon is going to get a number 2 here, and that means this carbon is going to get a number 3. So for absolute configuration, my lowest priority group is going away from me, and I'm traveling around this way. So I'm going clockwise, so it's an R absolute configuration at that carbon. OK, let's go ahead, and since that drawing's really busy, let's go ahead and draw it one more time, and we'll figure out the absolute configuration at the other carbon. So let's see, we still have my ethyl group coming out at me."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So for absolute configuration, my lowest priority group is going away from me, and I'm traveling around this way. So I'm going clockwise, so it's an R absolute configuration at that carbon. OK, let's go ahead, and since that drawing's really busy, let's go ahead and draw it one more time, and we'll figure out the absolute configuration at the other carbon. So let's see, we still have my ethyl group coming out at me. Still have the hydrogen going away from me. Still have this hydrogen coming out at me. Still have this methyl group going away from me, like that."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's see, we still have my ethyl group coming out at me. Still have the hydrogen going away from me. Still have this hydrogen coming out at me. Still have this methyl group going away from me, like that. So if I'm trying to find the absolute configuration for this carbon, we're going to approach it the same way. Look at the atom directly attached to that carbon, so it would be oxygen, carbon, carbon, hydrogen. So priorities, the oxygen gets highest priority."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Still have this methyl group going away from me, like that. So if I'm trying to find the absolute configuration for this carbon, we're going to approach it the same way. Look at the atom directly attached to that carbon, so it would be oxygen, carbon, carbon, hydrogen. So priorities, the oxygen gets highest priority. The hydrogen gets lowest priority. And then this carbon over here is directly attached to an oxygen, so this is going to get second highest priority, and that makes this methyl group over here the third highest priority. So the 1, 2, 3 is going around this way, which is counterclockwise, which makes you think it might be S, but remember that trick that I told you about in an earlier video."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So priorities, the oxygen gets highest priority. The hydrogen gets lowest priority. And then this carbon over here is directly attached to an oxygen, so this is going to get second highest priority, and that makes this methyl group over here the third highest priority. So the 1, 2, 3 is going around this way, which is counterclockwise, which makes you think it might be S, but remember that trick that I told you about in an earlier video. It looks S, but this hydrogen is coming out at me, so all I have to do is switch that, and that takes care of the fact that my lowest priority group is not pointing away from me. So it looks S, but since the hydrogen is coming out at me, I can say with confidence that it's R for an absolute configuration. So we can go ahead and finalize the name."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the 1, 2, 3 is going around this way, which is counterclockwise, which makes you think it might be S, but remember that trick that I told you about in an earlier video. It looks S, but this hydrogen is coming out at me, so all I have to do is switch that, and that takes care of the fact that my lowest priority group is not pointing away from me. So it looks S, but since the hydrogen is coming out at me, I can say with confidence that it's R for an absolute configuration. So we can go ahead and finalize the name. So at carbon 2 and at carbon 3, we have an R absolute configuration, so the name of this enantiomer would be 2R, 3R, 2, 3-epoxypentane. And for this enantiomer over here on the right, I know that it's the enantiomer, so I just have to switch the absolute configuration. So this one is going to be 2S and 3S, 2, 3-epoxypentane."}, {"video_title": "Preparation of epoxides Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and finalize the name. So at carbon 2 and at carbon 3, we have an R absolute configuration, so the name of this enantiomer would be 2R, 3R, 2, 3-epoxypentane. And for this enantiomer over here on the right, I know that it's the enantiomer, so I just have to switch the absolute configuration. So this one is going to be 2S and 3S, 2, 3-epoxypentane. And we don't have time in this video to go ahead and assign absolute configuration to the enantiomer on the right, but you can go ahead and do so for practice, and you should get 2S, 3S for its absolute configuration. Now, this is a problem, because if you're forming a racemic mixture of your epoxide, that's not always what you want to do in organic chemistry. So there are ways to use chiral catalysts that allow you to select out for one of these enantiomers."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we're asked to draw all of the structural isomers that have the molecular formula C5H12. The word isomer means same parts, and so we're talking about the same number of atoms. All of our structural isomers are gonna have five carbons and 12 hydrogens. Our isomers are gonna differ in how those atoms are connected to each other, so they differ in terms of their structure, and that's why we call them structural isomers. You can also call them constitutional isomers. We need five carbons, so for our first isomer, we could just draw five carbons in a chain. Here are my five carbons in a chain."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Our isomers are gonna differ in how those atoms are connected to each other, so they differ in terms of their structure, and that's why we call them structural isomers. You can also call them constitutional isomers. We need five carbons, so for our first isomer, we could just draw five carbons in a chain. Here are my five carbons in a chain. You should have already seen the video on bond line structures before you watched this one. Let's draw out those five carbons, and let's double-check and make sure we have the correct number of hydrogens. The carbon on the far left has three hydrogens, so here we have our three hydrogens."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Here are my five carbons in a chain. You should have already seen the video on bond line structures before you watched this one. Let's draw out those five carbons, and let's double-check and make sure we have the correct number of hydrogens. The carbon on the far left has three hydrogens, so here we have our three hydrogens. Next carbon has two. Same with the next carbon, so two for this one, two for the next carbon, and finally, three hydrogens for the last carbon. Let's count up everything and make sure we have the correct molecular formula."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the far left has three hydrogens, so here we have our three hydrogens. Next carbon has two. Same with the next carbon, so two for this one, two for the next carbon, and finally, three hydrogens for the last carbon. Let's count up everything and make sure we have the correct molecular formula. We have one, two, three, four, five carbons, so that's C5, and then we should have 12 hydrogens. Here's three, plus two gives us five, plus two gives us seven, plus two gives us nine, and then we have three more for a total of 12. C5H12 is the molecular formula for this compound."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's count up everything and make sure we have the correct molecular formula. We have one, two, three, four, five carbons, so that's C5, and then we should have 12 hydrogens. Here's three, plus two gives us five, plus two gives us seven, plus two gives us nine, and then we have three more for a total of 12. C5H12 is the molecular formula for this compound. Let's draw another structural isomer that has the same molecular formula. Instead of drawing five carbons in a chain, now we have to draw four. Let's start by drawing four carbons."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "C5H12 is the molecular formula for this compound. Let's draw another structural isomer that has the same molecular formula. Instead of drawing five carbons in a chain, now we have to draw four. Let's start by drawing four carbons. We need a total of five carbons, so we need to show the fifth carbon branching off of our chain. We could show the fifth carbon branching off of our chain here. Let's draw in those five carbons."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's start by drawing four carbons. We need a total of five carbons, so we need to show the fifth carbon branching off of our chain. We could show the fifth carbon branching off of our chain here. Let's draw in those five carbons. Here we have our five carbons. Let's count up hydrogens. Carbon on the left has three, so three hydrogens here, three hydrogens on this top carbon."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw in those five carbons. Here we have our five carbons. Let's count up hydrogens. Carbon on the left has three, so three hydrogens here, three hydrogens on this top carbon. There's only one hydrogen on this carbon, two hydrogens on this one, and finally, three hydrogens on this carbon. Let's count up our atoms. Let's use red for this one."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Carbon on the left has three, so three hydrogens here, three hydrogens on this top carbon. There's only one hydrogen on this carbon, two hydrogens on this one, and finally, three hydrogens on this carbon. Let's count up our atoms. Let's use red for this one. We have one, two, three, four, five carbons, so that's C5, and then for hydrogens, we have three here, plus three gives us six, plus one gives us seven, plus two gives us nine, and three more for a total of 12. C5H12 is the molecular formula for this compound. These two drawings represent two different molecules."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's use red for this one. We have one, two, three, four, five carbons, so that's C5, and then for hydrogens, we have three here, plus three gives us six, plus one gives us seven, plus two gives us nine, and three more for a total of 12. C5H12 is the molecular formula for this compound. These two drawings represent two different molecules. Both of these molecules have the molecular formula C5H12, but they differ in terms of how those atoms are connected. They differ in terms of their structure. We call them structural isomers of each other."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "These two drawings represent two different molecules. Both of these molecules have the molecular formula C5H12, but they differ in terms of how those atoms are connected. They differ in terms of their structure. We call them structural isomers of each other. To draw another structural isomer, some students might say, oh, well, we could start with four carbons in our chain again, and this time, instead of showing a branch off of this carbon, we could show a branch off of this carbon. A student might draw this structure and say, okay, there's a different structural isomer, but actually, these are just two different ways to represent the same molecule. If you analyze that second structure that we just drew, the connections are the same."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We call them structural isomers of each other. To draw another structural isomer, some students might say, oh, well, we could start with four carbons in our chain again, and this time, instead of showing a branch off of this carbon, we could show a branch off of this carbon. A student might draw this structure and say, okay, there's a different structural isomer, but actually, these are just two different ways to represent the same molecule. If you analyze that second structure that we just drew, the connections are the same. We have a CH right here bonded to a CH3, bonded to a CH3, and bonded to a CH2, and the CH2 is bonded to a CH3. That's the same structure as what we drew out over here. It looks like it's a different structure."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If you analyze that second structure that we just drew, the connections are the same. We have a CH right here bonded to a CH3, bonded to a CH3, and bonded to a CH2, and the CH2 is bonded to a CH3. That's the same structure as what we drew out over here. It looks like it's a different structure. It's a different drawing than the one up here, but actually, this is just two different ways to represent the same molecule. We have two structural isomers so far. Let's think about one more."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It looks like it's a different structure. It's a different drawing than the one up here, but actually, this is just two different ways to represent the same molecule. We have two structural isomers so far. Let's think about one more. We can no longer do four carbons in our chain, so we go down to three carbons. We start with three carbons in our chain. We know we need a total of five carbons, so we need to show two more carbons added to our chain."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about one more. We can no longer do four carbons in our chain, so we go down to three carbons. We start with three carbons in our chain. We know we need a total of five carbons, so we need to show two more carbons added to our chain. We would have to add those two carbons to our central carbon like that. Let's draw out all of our carbons here. Let's add in our hydrogens."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We know we need a total of five carbons, so we need to show two more carbons added to our chain. We would have to add those two carbons to our central carbon like that. Let's draw out all of our carbons here. Let's add in our hydrogens. This carbon would have three hydrogens. Same with this carbon, and the same with this one, and finally, the same for this carbon. The carbon in the center, this carbon in the center here already has four bonds, so it doesn't have any hydrogens on it."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's add in our hydrogens. This carbon would have three hydrogens. Same with this carbon, and the same with this one, and finally, the same for this carbon. The carbon in the center, this carbon in the center here already has four bonds, so it doesn't have any hydrogens on it. Let's count up everything. Let's count our carbons first. One, two, three, four, five carbons, so C5, and then we have three hydrogens, plus three is six, plus three is nine, plus three is 12."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The carbon in the center, this carbon in the center here already has four bonds, so it doesn't have any hydrogens on it. Let's count up everything. Let's count our carbons first. One, two, three, four, five carbons, so C5, and then we have three hydrogens, plus three is six, plus three is nine, plus three is 12. C5H12 is the molecular formula for this compound. This is another structural isomer, so it's a different molecule from the other two. We have a total of three structural isomers that have the molecular formula C5H12."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five carbons, so C5, and then we have three hydrogens, plus three is six, plus three is nine, plus three is 12. C5H12 is the molecular formula for this compound. This is another structural isomer, so it's a different molecule from the other two. We have a total of three structural isomers that have the molecular formula C5H12. Now let's draw all of the structural isomers that have the molecular formula C3H8O. We'll start with the molecule we talked about in the bond line structure videos. That molecule looked like this."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have a total of three structural isomers that have the molecular formula C5H12. Now let's draw all of the structural isomers that have the molecular formula C3H8O. We'll start with the molecule we talked about in the bond line structure videos. That molecule looked like this. We had three carbons, and then we had an OH coming off of the central carbon. Let's expand that out and make sure that this has the correct molecular formula. We have our three carbons, and on the middle carbon we have an OH, so an oxygen bonded to a hydrogen."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That molecule looked like this. We had three carbons, and then we had an OH coming off of the central carbon. Let's expand that out and make sure that this has the correct molecular formula. We have our three carbons, and on the middle carbon we have an OH, so an oxygen bonded to a hydrogen. I'll go ahead and put lone pairs of electrons on this oxygen. How many hydrogens do we need to add to the carbon on the left? We need to add three hydrogens, so we'll go ahead and draw in those three hydrogens."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have our three carbons, and on the middle carbon we have an OH, so an oxygen bonded to a hydrogen. I'll go ahead and put lone pairs of electrons on this oxygen. How many hydrogens do we need to add to the carbon on the left? We need to add three hydrogens, so we'll go ahead and draw in those three hydrogens. The carbon in the center already has three bonds, so it needs one more, so we add one hydrogen to that carbon. The carbon on the right needs three hydrogens. Let's count everything up now."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We need to add three hydrogens, so we'll go ahead and draw in those three hydrogens. The carbon in the center already has three bonds, so it needs one more, so we add one hydrogen to that carbon. The carbon on the right needs three hydrogens. Let's count everything up now. We'll start with our carbons. We have one, two, three carbons, so that's C3. We have three hydrogens here, and three here, so that's six plus one is seven."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's count everything up now. We'll start with our carbons. We have one, two, three carbons, so that's C3. We have three hydrogens here, and three here, so that's six plus one is seven. Don't forget about the hydrogen on the oxygen for eight. We have eight hydrogens, and obviously we have one oxygen here. I went ahead and put in lone pairs of electrons on that oxygen."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have three hydrogens here, and three here, so that's six plus one is seven. Don't forget about the hydrogen on the oxygen for eight. We have eight hydrogens, and obviously we have one oxygen here. I went ahead and put in lone pairs of electrons on that oxygen. The molecular formula for this molecule is C3H8O. If I numbered this, if I said this was carbon one, and this was carbon two, and this was carbon three, that helps us to draw the next structural isomer because we can think about instead of that OH group coming off of carbon two, what if that OH group came off of carbon one? Let's draw out our three carbons here."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I went ahead and put in lone pairs of electrons on that oxygen. The molecular formula for this molecule is C3H8O. If I numbered this, if I said this was carbon one, and this was carbon two, and this was carbon three, that helps us to draw the next structural isomer because we can think about instead of that OH group coming off of carbon two, what if that OH group came off of carbon one? Let's draw out our three carbons here. Now we put our OH group coming off of carbon one. Let's expand this out and draw the Lewis dot structure and make sure that this has the correct molecular formula. We have three carbons again in a row, and then the carbon on the left is bonded to the oxygen."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw out our three carbons here. Now we put our OH group coming off of carbon one. Let's expand this out and draw the Lewis dot structure and make sure that this has the correct molecular formula. We have three carbons again in a row, and then the carbon on the left is bonded to the oxygen. The oxygen is bonded to a hydrogen. I'll put in lone pairs of electrons on the oxygen. Now we need to add in carbon-hydrogen bonds."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have three carbons again in a row, and then the carbon on the left is bonded to the oxygen. The oxygen is bonded to a hydrogen. I'll put in lone pairs of electrons on the oxygen. Now we need to add in carbon-hydrogen bonds. This carbon needs two. The next carbon also needs two, and the carbon on the end would need three. That's one, two, and three."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now we need to add in carbon-hydrogen bonds. This carbon needs two. The next carbon also needs two, and the carbon on the end would need three. That's one, two, and three. When we add everything up, let's use blue for that. That's one, two, three carbons, so we have C3. We have three hydrogens here, plus two is five, plus two is seven, and one here is eight."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That's one, two, and three. When we add everything up, let's use blue for that. That's one, two, three carbons, so we have C3. We have three hydrogens here, plus two is five, plus two is seven, and one here is eight. C3H8, and then of course our oxygen. C3H8O is the molecular formula. Next, some students might think, okay, we put an OH coming off of carbon one, but what if I put an OH on the other side?"}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have three hydrogens here, plus two is five, plus two is seven, and one here is eight. C3H8, and then of course our oxygen. C3H8O is the molecular formula. Next, some students might think, okay, we put an OH coming off of carbon one, but what if I put an OH on the other side? Over here on the other side. Let's see what would that give us. If I put an OH coming off of that carbon, hopefully it's obvious that these two represent the same molecule."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, some students might think, okay, we put an OH coming off of carbon one, but what if I put an OH on the other side? Over here on the other side. Let's see what would that give us. If I put an OH coming off of that carbon, hopefully it's obvious that these two represent the same molecule. There's no difference in terms of how those two are connected structurally. This is the same molecule, so two different ways to draw the same one. This is not a new structural isomer, just a new way of looking at this molecule."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If I put an OH coming off of that carbon, hopefully it's obvious that these two represent the same molecule. There's no difference in terms of how those two are connected structurally. This is the same molecule, so two different ways to draw the same one. This is not a new structural isomer, just a new way of looking at this molecule. Now, let's draw one more. We can't put the OH on the other carbon, so now we have to figure out something else that we can do. We could this time put two carbons in a row, and put an oxygen in between, so putting oxygen to break up our carbon chain."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This is not a new structural isomer, just a new way of looking at this molecule. Now, let's draw one more. We can't put the OH on the other carbon, so now we have to figure out something else that we can do. We could this time put two carbons in a row, and put an oxygen in between, so putting oxygen to break up our carbon chain. Now, this would be carbon bonded to carbon, bonded to oxygen, bonded to carbon, and then we fill in our hydrogen, so there'd be three on this carbon. There would be two on this carbon. There would be three on this carbon, and I could put in lone pairs of electrons on the oxygen like that, and count everything up."}, {"video_title": "Structural (constitutional) isomers Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We could this time put two carbons in a row, and put an oxygen in between, so putting oxygen to break up our carbon chain. Now, this would be carbon bonded to carbon, bonded to oxygen, bonded to carbon, and then we fill in our hydrogen, so there'd be three on this carbon. There would be two on this carbon. There would be three on this carbon, and I could put in lone pairs of electrons on the oxygen like that, and count everything up. We have one, two, three carbons, so that's C3. We have three hydrogens, plus two is five, plus three is eight, so we have the H8, and then of course the one oxygen, so this is another structural isomer. Again, some students might say, well, we could go like this, and this would be yet another structural isomer like that, but really, this is just another way to draw this molecule, so it's not a new structural isomer."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes one dot structure is not enough to completely describe a molecule or an ion. Sometimes you need two or more. And here's an example. This is the acetate anion, and this dot structure does not completely describe the acetate anion. We need to draw a resonance structure, another resonance structure. And so what we're gonna do is take a lone pair of electrons from this oxygen and move that lone pair of electrons in here to form a double bond between this carbon and that oxygen. At the same time, we're gonna take these two pi electrons here and move those pi electrons out onto the top oxygen."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is the acetate anion, and this dot structure does not completely describe the acetate anion. We need to draw a resonance structure, another resonance structure. And so what we're gonna do is take a lone pair of electrons from this oxygen and move that lone pair of electrons in here to form a double bond between this carbon and that oxygen. At the same time, we're gonna take these two pi electrons here and move those pi electrons out onto the top oxygen. So let's go ahead and draw a resonance double-headed arrow here. And when you're drawing resonance structures, you usually put in brackets. Let's go ahead and draw the other resonance structure."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "At the same time, we're gonna take these two pi electrons here and move those pi electrons out onto the top oxygen. So let's go ahead and draw a resonance double-headed arrow here. And when you're drawing resonance structures, you usually put in brackets. Let's go ahead and draw the other resonance structure. So now there would be a double bond between this carbon and this oxygen here. This oxygen on the bottom right used to have three lone pairs of electrons around it. Now it only has two, because one of those lone pairs moved in to form that pi bond."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw the other resonance structure. So now there would be a double bond between this carbon and this oxygen here. This oxygen on the bottom right used to have three lone pairs of electrons around it. Now it only has two, because one of those lone pairs moved in to form that pi bond. The oxygen on the top used to have a double bond. Right now it has only a single bond to it. And it used to have two lone pairs of electrons, and now it has three lone pairs of electrons."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now it only has two, because one of those lone pairs moved in to form that pi bond. The oxygen on the top used to have a double bond. Right now it has only a single bond to it. And it used to have two lone pairs of electrons, and now it has three lone pairs of electrons. That gives the top oxygen a negative one formal charge. And make sure you understand formal charges before you get into drawing resonance structures. So it's extremely important to understand that."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And it used to have two lone pairs of electrons, and now it has three lone pairs of electrons. That gives the top oxygen a negative one formal charge. And make sure you understand formal charges before you get into drawing resonance structures. So it's extremely important to understand that. All right, so next, let's follow those electrons just to make sure we know what happened here. So these electrons in magenta moved in here to form our pi bond like that. And the electrons over here in blue moved out onto the top oxygen."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's extremely important to understand that. All right, so next, let's follow those electrons just to make sure we know what happened here. So these electrons in magenta moved in here to form our pi bond like that. And the electrons over here in blue moved out onto the top oxygen. So let's say those electrons in blue are these electrons like that. So we have two resonance structures for the acetate anion. And neither of these structures completely describes the acetate anion."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons over here in blue moved out onto the top oxygen. So let's say those electrons in blue are these electrons like that. So we have two resonance structures for the acetate anion. And neither of these structures completely describes the acetate anion. We need to draw a hybrid of these two. And so if we take a look at, let's say the oxygen on the bottom right here, we can see there's a single bond between this carbon and this oxygen. If we look at this one over here, we see there's now a double bond between that carbon and the oxygen."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And neither of these structures completely describes the acetate anion. We need to draw a hybrid of these two. And so if we take a look at, let's say the oxygen on the bottom right here, we can see there's a single bond between this carbon and this oxygen. If we look at this one over here, we see there's now a double bond between that carbon and the oxygen. So if we think about a hybrid of these two resonance structures, let's go ahead and draw it in here, we can't just draw a single bond between the carbon and that oxygen. There's some partial double bond character there. So it's a hybrid of the two structures above."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we look at this one over here, we see there's now a double bond between that carbon and the oxygen. So if we think about a hybrid of these two resonance structures, let's go ahead and draw it in here, we can't just draw a single bond between the carbon and that oxygen. There's some partial double bond character there. So it's a hybrid of the two structures above. So let's go ahead and draw in a partial bond here like that. The exact same thing for the top oxygen. Here we have a double bond, and then over here we have a single bond."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's a hybrid of the two structures above. So let's go ahead and draw in a partial bond here like that. The exact same thing for the top oxygen. Here we have a double bond, and then over here we have a single bond. So somewhere in between, right? So in between is going to be our hybrid. So let's go ahead and draw that in."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here we have a double bond, and then over here we have a single bond. So somewhere in between, right? So in between is going to be our hybrid. So let's go ahead and draw that in. So we can't just draw a single bond in our hybrid. We have to show some partial double bond character, drawing the dotted line in there like that. And also charge, right?"}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that in. So we can't just draw a single bond in our hybrid. We have to show some partial double bond character, drawing the dotted line in there like that. And also charge, right? So if we think about charge, the negative charge is on the oxygen on the bottom right, and then over here the negative charge is on the top oxygen. And so that negative charge is actually delocalized. So it's not localized to one oxygen, it's delocalized, it's distributed evenly over both of those oxygens here."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And also charge, right? So if we think about charge, the negative charge is on the oxygen on the bottom right, and then over here the negative charge is on the top oxygen. And so that negative charge is actually delocalized. So it's not localized to one oxygen, it's delocalized, it's distributed evenly over both of those oxygens here. And so this is just one way to represent the hybrid here. And studies have shown that the hybrid is closer to what the actual anion looks like. So studies have been done on these bond lengths here, and the bonding between this carbon and this oxygen, it turns out to be the exact same bond length as the bond between the carbon and this oxygen."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's not localized to one oxygen, it's delocalized, it's distributed evenly over both of those oxygens here. And so this is just one way to represent the hybrid here. And studies have shown that the hybrid is closer to what the actual anion looks like. So studies have been done on these bond lengths here, and the bonding between this carbon and this oxygen, it turns out to be the exact same bond length as the bond between the carbon and this oxygen. So it's the exact same bond length, and so the hybrid, again, is a better picture of what the anion actually looks like. Let's think about what would happen if we just moved the electrons in magenta in. So if I go back to the very first thing I talked about, and you're like, well, why didn't we just stop after moving these electrons in magenta?"}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So studies have been done on these bond lengths here, and the bonding between this carbon and this oxygen, it turns out to be the exact same bond length as the bond between the carbon and this oxygen. So it's the exact same bond length, and so the hybrid, again, is a better picture of what the anion actually looks like. Let's think about what would happen if we just moved the electrons in magenta in. So if I go back to the very first thing I talked about, and you're like, well, why didn't we just stop after moving these electrons in magenta? Let's go ahead and draw what we would have if we stopped after moving the electrons in magenta. So we would have this. So the electrons in magenta moved in here to form our double bond."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I go back to the very first thing I talked about, and you're like, well, why didn't we just stop after moving these electrons in magenta? Let's go ahead and draw what we would have if we stopped after moving the electrons in magenta. So we would have this. So the electrons in magenta moved in here to form our double bond. And if we don't push off those electrons in blue, this might be our resonance structure. The problem with this one is, of course, the fact that this carbon here has five bonds to it. So one, two, three, four, five."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta moved in here to form our double bond. And if we don't push off those electrons in blue, this might be our resonance structure. The problem with this one is, of course, the fact that this carbon here has five bonds to it. So one, two, three, four, five. So five bonds, so 10 electrons around it. We know that carbon can't exceed the octet of electrons because of its position on the periodic table. So this is not a valid structure."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So one, two, three, four, five. So five bonds, so 10 electrons around it. We know that carbon can't exceed the octet of electrons because of its position on the periodic table. So this is not a valid structure. And so this is one of the patterns that we're gonna be talking about in the next video. So the pattern is a lone pair of electrons, so next to a pi bond, which is the example we see here for the acetate anion. And so these are the two resonance structures."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is not a valid structure. And so this is one of the patterns that we're gonna be talking about in the next video. So the pattern is a lone pair of electrons, so next to a pi bond, which is the example we see here for the acetate anion. And so these are the two resonance structures. The problem with the word resonance is, when you're a student, you might think that the anion will resonate back and forth between this one and this one. That's just kind of what the name seems to imply. And that's not actually what's happening."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so these are the two resonance structures. The problem with the word resonance is, when you're a student, you might think that the anion will resonate back and forth between this one and this one. That's just kind of what the name seems to imply. And that's not actually what's happening. It's just that we can't draw, if we're just drawing one dot structure, this is not an accurate description. And so the electrons are actually delocalized, so it's not resonating back and forth. When you draw resonance structures in your head, think about what that means for the hybrid and how the resonance structures would contribute to the overall hybrid."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And that's not actually what's happening. It's just that we can't draw, if we're just drawing one dot structure, this is not an accurate description. And so the electrons are actually delocalized, so it's not resonating back and forth. When you draw resonance structures in your head, think about what that means for the hybrid and how the resonance structures would contribute to the overall hybrid. So don't forget about your brackets and your double-headed arrows and also your formal charges. So you have to put those in when you're drawing your resonance structures. All right, let's look at an application of the acetate anion here and the resonance structures that we can draw."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "When you draw resonance structures in your head, think about what that means for the hybrid and how the resonance structures would contribute to the overall hybrid. So don't forget about your brackets and your double-headed arrows and also your formal charges. So you have to put those in when you're drawing your resonance structures. All right, let's look at an application of the acetate anion here and the resonance structures that we can draw. So if we look at the acetate anion, so we just talked about the fact that one of these lone pairs here, so this is not localized to the oxygen, it's delocalized, so we can move those electrons in here, we push those electrons off onto the oxygen, we can draw a resonance structure. And so this negative one formal charge is not localized to this oxygen, it's delocalized. And so because we can spread out some of that negative charge, that increases the stability of the anion here."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at an application of the acetate anion here and the resonance structures that we can draw. So if we look at the acetate anion, so we just talked about the fact that one of these lone pairs here, so this is not localized to the oxygen, it's delocalized, so we can move those electrons in here, we push those electrons off onto the oxygen, we can draw a resonance structure. And so this negative one formal charge is not localized to this oxygen, it's delocalized. And so because we can spread out some of that negative charge, that increases the stability of the anion here. So this is relatively stable, so increased stability due to delocalization. All right, so the fact that we can draw an extra resonance structure means that the anion has been stabilized. And so this is called pushing electrons, right?"}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so because we can spread out some of that negative charge, that increases the stability of the anion here. So this is relatively stable, so increased stability due to delocalization. All right, so the fact that we can draw an extra resonance structure means that the anion has been stabilized. And so this is called pushing electrons, right? So we're moving electrons around and it's extremely important to feel comfortable with moving electrons around and being able to follow them. So the only way to do this, to get good at this, is to do a lot of practice problems, so please do that. Do lots of practice problems in your textbook."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this is called pushing electrons, right? So we're moving electrons around and it's extremely important to feel comfortable with moving electrons around and being able to follow them. So the only way to do this, to get good at this, is to do a lot of practice problems, so please do that. Do lots of practice problems in your textbook. If we compare that to the ethoxide anion, so over here, all right, if we try to do the same thing, if we try to take a lone pair of electrons on this oxygen and move it into here, we can't do that because this carbon right here already has four bonds, right? So it's already bonded to two hydrogens and then we have this bond and this bond. And so moving those electrons in, trying to delocalize those electrons would give us five bonds to carbon, so we can't do that."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Do lots of practice problems in your textbook. If we compare that to the ethoxide anion, so over here, all right, if we try to do the same thing, if we try to take a lone pair of electrons on this oxygen and move it into here, we can't do that because this carbon right here already has four bonds, right? So it's already bonded to two hydrogens and then we have this bond and this bond. And so moving those electrons in, trying to delocalize those electrons would give us five bonds to carbon, so we can't do that. We can't draw a resonance structure for the ethoxide anion. So those electrons are localized to this oxygen and so this oxygen has a full negative one formal charge. And since we can't spread out that negative charge, it's going to destabilize this anion."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so moving those electrons in, trying to delocalize those electrons would give us five bonds to carbon, so we can't do that. We can't draw a resonance structure for the ethoxide anion. So those electrons are localized to this oxygen and so this oxygen has a full negative one formal charge. And since we can't spread out that negative charge, it's going to destabilize this anion. So this is not as stable, so decreased stability compared to the anion on the left because we can't draw a resonance structure. If we think about the conjugate acids to these bases, so the conjugate acid to the acetate anion would be, of course, acetic acid, so we go ahead and draw in acetic acid like that. The conjugate acid to the ethoxide anion would, of course, be ethanol, so we go ahead and draw in ethanol."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And since we can't spread out that negative charge, it's going to destabilize this anion. So this is not as stable, so decreased stability compared to the anion on the left because we can't draw a resonance structure. If we think about the conjugate acids to these bases, so the conjugate acid to the acetate anion would be, of course, acetic acid, so we go ahead and draw in acetic acid like that. The conjugate acid to the ethoxide anion would, of course, be ethanol, so we go ahead and draw in ethanol. And we think about which one of those is more acidic. We know that acetic acid is more acidic. It's more likely to donate a proton because the conjugate base is more stable because of, you could think about resonance, or delocalization of electrons."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The conjugate acid to the ethoxide anion would, of course, be ethanol, so we go ahead and draw in ethanol. And we think about which one of those is more acidic. We know that acetic acid is more acidic. It's more likely to donate a proton because the conjugate base is more stable because of, you could think about resonance, or delocalization of electrons. If you're looking at ethanol, ethanol's not as likely to donate its proton because the conjugate base, the ethoxide anion, is not as stable because you can't draw any resonance structures for it. The negative charge is not able to be delocalized. It's localized to that oxygen."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's more likely to donate a proton because the conjugate base is more stable because of, you could think about resonance, or delocalization of electrons. If you're looking at ethanol, ethanol's not as likely to donate its proton because the conjugate base, the ethoxide anion, is not as stable because you can't draw any resonance structures for it. The negative charge is not able to be delocalized. It's localized to that oxygen. So this is just one application of thinking about resonance structures. And again, do lots of practice. In the next video, we'll talk about different patterns that you can look for."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's localized to that oxygen. So this is just one application of thinking about resonance structures. And again, do lots of practice. In the next video, we'll talk about different patterns that you can look for. And we talked about one in this video. We took a lone pair of electrons, so right here in green, right? And we noticed this lone pair of electrons was next to a pi bond, and so we were able to draw another resonance structure for it."}, {"video_title": "Resonance structure Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "In the next video, we'll talk about different patterns that you can look for. And we talked about one in this video. We took a lone pair of electrons, so right here in green, right? And we noticed this lone pair of electrons was next to a pi bond, and so we were able to draw another resonance structure for it. We don't have that situation with ethoxide. We have a lone pair of electrons, but we don't have a pi bond next to it. And so again, more in the next video on that."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we've drawn cyclohexane like that. Now we know from the last several videos that all of the bonds for a carbon don't sit in the same plane. If we take the example of methane, that's the simplest example, you have your carbon sitting in the middle. You'll have kind of a hydrogen popping out like that, another hydrogen that's in the plane of the screen, another one that's behind the screen, and another one that is straight up. So you kind of have this tetrahedral structure. And in the case of methane, you have that 109.5 degree bond angles. Carbon likes to form bonds of this shape."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "You'll have kind of a hydrogen popping out like that, another hydrogen that's in the plane of the screen, another one that's behind the screen, and another one that is straight up. So you kind of have this tetrahedral structure. And in the case of methane, you have that 109.5 degree bond angles. Carbon likes to form bonds of this shape. It won't always be 109.5 degrees. It'll be something close to it, depending on what the different atoms or molecules are that it is bonded to. So given that, what would a cyclohexane molecule actually look like if we try to visualize it in 3 dimensions?"}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Carbon likes to form bonds of this shape. It won't always be 109.5 degrees. It'll be something close to it, depending on what the different atoms or molecules are that it is bonded to. So given that, what would a cyclohexane molecule actually look like if we try to visualize it in 3 dimensions? So to think about that, let's think about these two bonds first. I'll try my best to draw it in one of its 3-dimensional shapes. So those bonds right there, I will draw like that."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So given that, what would a cyclohexane molecule actually look like if we try to visualize it in 3 dimensions? So to think about that, let's think about these two bonds first. I'll try my best to draw it in one of its 3-dimensional shapes. So those bonds right there, I will draw like that. And then this down here in orange, I will draw like this. And then this up here in magenta, I will draw like that. And then, let me see, in purple, I'll do these two right over here, and I'll draw them like this."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So those bonds right there, I will draw like that. And then this down here in orange, I will draw like this. And then this up here in magenta, I will draw like that. And then, let me see, in purple, I'll do these two right over here, and I'll draw them like this. So you have that and like that. So hopefully it makes clear that over there is that end over there. This end over here is this end over here."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then, let me see, in purple, I'll do these two right over here, and I'll draw them like this. So you have that and like that. So hopefully it makes clear that over there is that end over there. This end over here is this end over here. And this way that I've drawn the cyclohexane is called a chair configuration, chair shape. And it might be obvious. It looks like a chair."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This end over here is this end over here. And this way that I've drawn the cyclohexane is called a chair configuration, chair shape. And it might be obvious. It looks like a chair. That's the back of the chair. This is where you would sit down on the chair, and I guess the back of your calves would go against here. Your knees would sit on it someplace like that."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It looks like a chair. That's the back of the chair. This is where you would sit down on the chair, and I guess the back of your calves would go against here. Your knees would sit on it someplace like that. That's called the chair configuration. Now another configuration that it could be in is called the boat configuration. And so if I were to put this exact one in the boat configuration, if I take it from a slightly different perspective, if I'm looking at it kind of head on, it would look something like this."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Your knees would sit on it someplace like that. That's called the chair configuration. Now another configuration that it could be in is called the boat configuration. And so if I were to put this exact one in the boat configuration, if I take it from a slightly different perspective, if I'm looking at it kind of head on, it would look something like this. Now the first thing you're probably saying is, Sal, you said that the reason why it looks like this is because carbon likes to form these kind of tetrahedral or this tripod-shaped bonds. I don't see the tripod-shaped bonds either here or here. Let me draw that boat a little bit, at least this end of the boat, a little bit better."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And so if I were to put this exact one in the boat configuration, if I take it from a slightly different perspective, if I'm looking at it kind of head on, it would look something like this. Now the first thing you're probably saying is, Sal, you said that the reason why it looks like this is because carbon likes to form these kind of tetrahedral or this tripod-shaped bonds. I don't see the tripod-shaped bonds either here or here. Let me draw that boat a little bit, at least this end of the boat, a little bit better. There you go. And you say, well, I don't see that tripod shape over there. And to see the tripod shape, you just have to draw the hydrogens."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw that boat a little bit, at least this end of the boat, a little bit better. There you go. And you say, well, I don't see that tripod shape over there. And to see the tripod shape, you just have to draw the hydrogens. So let me draw some hydrogens here. So let me draw a hydrogen here that will go straight down like this, a hydrogen that goes straight down over here, a hydrogen that goes straight up over here, straight up over here, straight down over here, straight up over there. I've now drawn one hydrogen on every carbon."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And to see the tripod shape, you just have to draw the hydrogens. So let me draw some hydrogens here. So let me draw a hydrogen here that will go straight down like this, a hydrogen that goes straight down over here, a hydrogen that goes straight up over here, straight up over here, straight down over here, straight up over there. I've now drawn one hydrogen on every carbon. And now let me draw some hydrogens. Let me draw a hydrogen here that goes straight up, not up, really to the side over here. So a hydrogen there."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "I've now drawn one hydrogen on every carbon. And now let me draw some hydrogens. Let me draw a hydrogen here that goes straight up, not up, really to the side over here. So a hydrogen there. Let me draw a hydrogen over here that does the same thing. So those guys have their hydrogens. A hydrogen right over here."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So a hydrogen there. Let me draw a hydrogen over here that does the same thing. So those guys have their hydrogens. A hydrogen right over here. And then, let's see, this guy needs his hydrogens still. So he'll have a hydrogen that goes down like that and a hydrogen that goes like that. And then this guy will have a hydrogen that goes like that."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "A hydrogen right over here. And then, let's see, this guy needs his hydrogens still. So he'll have a hydrogen that goes down like that and a hydrogen that goes like that. And then this guy will have a hydrogen that goes like that. And when you see it like this, if you look at any one carbon on this molecule, if you look at any one carbon, you can see that it's forming the same tetrahedral shape, that it has a tripod at every one. Over here, you have that close to, roughly, 109, 110 degree angle between each of the constituents that are bonding to the carbon. Now, I've drawn the different hydrogens that are coming off of these carbons in different colors."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then this guy will have a hydrogen that goes like that. And when you see it like this, if you look at any one carbon on this molecule, if you look at any one carbon, you can see that it's forming the same tetrahedral shape, that it has a tripod at every one. Over here, you have that close to, roughly, 109, 110 degree angle between each of the constituents that are bonding to the carbon. Now, I've drawn the different hydrogens that are coming off of these carbons in different colors. And I've done it for a purpose. The ones that are going straight up or straight down, we call those axial hydrogens. And the ones I drew in orange that are kind of going to the side in some level, we call these equatorial hydrogens."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now, I've drawn the different hydrogens that are coming off of these carbons in different colors. And I've done it for a purpose. The ones that are going straight up or straight down, we call those axial hydrogens. And the ones I drew in orange that are kind of going to the side in some level, we call these equatorial hydrogens. And there's a reason why it's useful to know that name, is when we talk about the different configurations, the different chair and boats, whether something is equatorial or axial can change if this were to flip up or vice versa and things like that. And we'll talk more about that in the next video. And the reason why they're called equatorial is, if you think about it, and it's sometimes hard to visualize, this bond right here is parallel to this bond right over there."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And the ones I drew in orange that are kind of going to the side in some level, we call these equatorial hydrogens. And there's a reason why it's useful to know that name, is when we talk about the different configurations, the different chair and boats, whether something is equatorial or axial can change if this were to flip up or vice versa and things like that. And we'll talk more about that in the next video. And the reason why they're called equatorial is, if you think about it, and it's sometimes hard to visualize, this bond right here is parallel to this bond right over there. And this bond right over here is parallel to this. The equatorial bonds are parallel to some part of the ring. So that one is parallel to that right over there."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why they're called equatorial is, if you think about it, and it's sometimes hard to visualize, this bond right here is parallel to this bond right over there. And this bond right over here is parallel to this. The equatorial bonds are parallel to some part of the ring. So that one is parallel to that right over there. Actually, I could even color code it that. I don't want to use that same color. This is parallel to this."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So that one is parallel to that right over there. Actually, I could even color code it that. I don't want to use that same color. This is parallel to this. And this is parallel to that. And we could do it for all the equatorial bonds. So for example, I'm running out of colors here."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This is parallel to this. And this is parallel to that. And we could do it for all the equatorial bonds. So for example, I'm running out of colors here. So this right here is parallel to this and this and that over there. So we could keep doing it for all of them. I could do it for the other set right here."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So for example, I'm running out of colors here. So this right here is parallel to this and this and that over there. So we could keep doing it for all of them. I could do it for the other set right here. This guy right here is parallel to that guy over there. I didn't quite draw it like that, but hopefully it makes the idea clear. And I'll do one more of these just to show what's parallel to what."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "I could do it for the other set right here. This guy right here is parallel to that guy over there. I didn't quite draw it like that, but hopefully it makes the idea clear. And I'll do one more of these just to show what's parallel to what. This bond is parallel to that. So the ones that are parallel to some part of the ring, we're calling equatorial. And the ones that kind of jump out of the ring, that aren't parallel to any other part of the ring, we're calling those axial."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And I'll do one more of these just to show what's parallel to what. This bond is parallel to that. So the ones that are parallel to some part of the ring, we're calling equatorial. And the ones that kind of jump out of the ring, that aren't parallel to any other part of the ring, we're calling those axial. And the way I've drawn it here, the axials are the ones that point straight up and point straight down. We could do the same thing on a boat configuration. Now one question you might ask is, well, there's these two configurations."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And the ones that kind of jump out of the ring, that aren't parallel to any other part of the ring, we're calling those axial. And the way I've drawn it here, the axials are the ones that point straight up and point straight down. We could do the same thing on a boat configuration. Now one question you might ask is, well, there's these two configurations. Both of these would result in tetrahedral type shapes at each of the carbons. In fact, let me draw it for you. So this axial hydrogen is pointing straight down."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now one question you might ask is, well, there's these two configurations. Both of these would result in tetrahedral type shapes at each of the carbons. In fact, let me draw it for you. So this axial hydrogen is pointing straight down. This one is pointing straight down. Here, this hydrogen is actually going to point straight down because we've flipped it up. And then over here, you would have a hydrogen that points straight up and then one that's kind of pointing down."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this axial hydrogen is pointing straight down. This one is pointing straight down. Here, this hydrogen is actually going to point straight down because we've flipped it up. And then over here, you would have a hydrogen that points straight up and then one that's kind of pointing down. This gives a tripod there. To have the tripod over here, you'll have to have a hydrogen that points a little bit like that, one that points a little bit like that, one that's pointing a little bit like that along. Well, you can kind of view it along the same plane as this guy would be parallel."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then over here, you would have a hydrogen that points straight up and then one that's kind of pointing down. This gives a tripod there. To have the tripod over here, you'll have to have a hydrogen that points a little bit like that, one that points a little bit like that, one that's pointing a little bit like that along. Well, you can kind of view it along the same plane as this guy would be parallel. It's hard to see it in this, but he would actually be parallel to that. This guy would be out like this. And then this guy would have an axial hydrogen."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Well, you can kind of view it along the same plane as this guy would be parallel. It's hard to see it in this, but he would actually be parallel to that. This guy would be out like this. And then this guy would have an axial hydrogen. And then he would have one equatorial one just like that. So you could draw the tripod shapes in either the chair or boat configuration. But one question is, well, what's more stable?"}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then this guy would have an axial hydrogen. And then he would have one equatorial one just like that. So you could draw the tripod shapes in either the chair or boat configuration. But one question is, well, what's more stable? That's actually one of the main points of being able to visually think about the three-dimensional structure of any of these hydrocarbons, or in this case, cyclohexane. So in this situation, we know from past videos that all of these carbons with their hydrogens around them, these bonds, these have electron clouds around them. The electron clouds are negative."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "But one question is, well, what's more stable? That's actually one of the main points of being able to visually think about the three-dimensional structure of any of these hydrocarbons, or in this case, cyclohexane. So in this situation, we know from past videos that all of these carbons with their hydrogens around them, these bonds, these have electron clouds around them. The electron clouds are negative. And so they want to get as far away from each other as possible. In this chair configuration, you have this carbon up here, the CH2 we could consider. It has two hydrogens and it's connected to the rest of the ring."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "The electron clouds are negative. And so they want to get as far away from each other as possible. In this chair configuration, you have this carbon up here, the CH2 we could consider. It has two hydrogens and it's connected to the rest of the ring. It's as far as possible from this CH2 as possible. So in that situation, we have a lower potential energy, or it is a more stable shape or more stable configuration. In the boat configuration, this CH2 up here is much closer to this CH2."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It has two hydrogens and it's connected to the rest of the ring. It's as far as possible from this CH2 as possible. So in that situation, we have a lower potential energy, or it is a more stable shape or more stable configuration. In the boat configuration, this CH2 up here is much closer to this CH2. I mean, that's really the main difference between the two. And they want to get away from each other. They want to repel each other."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "In the boat configuration, this CH2 up here is much closer to this CH2. I mean, that's really the main difference between the two. And they want to get away from each other. They want to repel each other. So this one will have higher potential energy, or it will be less stable. So this is just a starting point of how to visualize cyclic hydrocarbons. And we'll use this information in the next video to think a little bit more about maybe the different chair configurations that a molecule could have and what could be more stable."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "They want to repel each other. So this one will have higher potential energy, or it will be less stable. So this is just a starting point of how to visualize cyclic hydrocarbons. And we'll use this information in the next video to think a little bit more about maybe the different chair configurations that a molecule could have and what could be more stable. In this situation, in the case of just cyclohexane, the two chair configurations are equally stable. And let me just touch on that a second. So you have, well, I don't have to, actually let me see."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And we'll use this information in the next video to think a little bit more about maybe the different chair configurations that a molecule could have and what could be more stable. In this situation, in the case of just cyclohexane, the two chair configurations are equally stable. And let me just touch on that a second. So you have, well, I don't have to, actually let me see. I won't copy and paste it. I'll just redraw the other chair configuration for this guy. Actually, let me just do it separately over here because I've made the colors here so confusing."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So you have, well, I don't have to, actually let me see. I won't copy and paste it. I'll just redraw the other chair configuration for this guy. Actually, let me just do it separately over here because I've made the colors here so confusing. Let me draw the same cyclohexane, but in two different chair configurations that it could be equilibrium in. So you could have this one. So this could be one chair configuration."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me just do it separately over here because I've made the colors here so confusing. Let me draw the same cyclohexane, but in two different chair configurations that it could be equilibrium in. So you could have this one. So this could be one chair configuration. And I'll draw it like this. And then the same cyclohexane could be in equilibrium with this other chair configuration that looks like this. Let me have a little more space here."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this could be one chair configuration. And I'll draw it like this. And then the same cyclohexane could be in equilibrium with this other chair configuration that looks like this. Let me have a little more space here. So it looks like this. Let me do the pink. Goes up like that, like that."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let me have a little more space here. So it looks like this. Let me do the pink. Goes up like that, like that. This pink guy goes like this. And then the blue guy is going to be just like this. So notice, in this situation, this carbon appears kind of at the top of the chair, and this carbon is at the bottom."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Goes up like that, like that. This pink guy goes like this. And then the blue guy is going to be just like this. So notice, in this situation, this carbon appears kind of at the top of the chair, and this carbon is at the bottom. And then they've flipped, but these are equally stable configurations. But one thing to think about is all of the axial guys on this carbon here turned into equatorial on this carbon and vice versa on the two. Let me show it to you."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So notice, in this situation, this carbon appears kind of at the top of the chair, and this carbon is at the bottom. And then they've flipped, but these are equally stable configurations. But one thing to think about is all of the axial guys on this carbon here turned into equatorial on this carbon and vice versa on the two. Let me show it to you. Let me just draw the hydrogens on this carbon. So if I were to draw this carbon's hydrogens, it has an axial hydrogen and has an equatorial hydrogen whose bond would be parallel to that, just like that. And this guy would have an equatorial hydrogen whose bond is parallel to actually both of these guys, and an axial hydrogen."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let me show it to you. Let me just draw the hydrogens on this carbon. So if I were to draw this carbon's hydrogens, it has an axial hydrogen and has an equatorial hydrogen whose bond would be parallel to that, just like that. And this guy would have an equatorial hydrogen whose bond is parallel to actually both of these guys, and an axial hydrogen. But when it flips, and I'm just drawing those guys' hydrogens, but when this structure flips like that, what happens? Well, this hydrogen over here goes into this position, and this yellow hydrogen over here goes into this position. So over here it was equatorial, and now it becomes axial, and the same argument can be made over here."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And this guy would have an equatorial hydrogen whose bond is parallel to actually both of these guys, and an axial hydrogen. But when it flips, and I'm just drawing those guys' hydrogens, but when this structure flips like that, what happens? Well, this hydrogen over here goes into this position, and this yellow hydrogen over here goes into this position. So over here it was equatorial, and now it becomes axial, and the same argument can be made over here. This equatorial hydrogen, when this whole blue part flips down, now becomes axial, and this axial hydrogen, when you flip it down, becomes equatorial. And you can actually do that for all of the hydrogens. Over here you have an axial hydrogen."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So over here it was equatorial, and now it becomes axial, and the same argument can be made over here. This equatorial hydrogen, when this whole blue part flips down, now becomes axial, and this axial hydrogen, when you flip it down, becomes equatorial. And you can actually do that for all of the hydrogens. Over here you have an axial hydrogen. Once you flip it, let's see, you have an axial hydrogen, and then you have an equatorial hydrogen and an equatorial hydrogen. When you flip it, these two equatorial hydrogens become axial. So they become axial, and then both of these guys become equatorial."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Over here you have an axial hydrogen. Once you flip it, let's see, you have an axial hydrogen, and then you have an equatorial hydrogen and an equatorial hydrogen. When you flip it, these two equatorial hydrogens become axial. So they become axial, and then both of these guys become equatorial. So let me do that in yellow. Both this guy and this guy become equatorial. So this and that become equatorial."}, {"video_title": "Chair and boat shapes for cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So they become axial, and then both of these guys become equatorial. So let me do that in yellow. Both this guy and this guy become equatorial. So this and that become equatorial. They become parallel to the other end. And you could do it for these two hydrogens as well. So that's another interesting thing to think about, and this is really just practice on visualizing what's going on when we visualize these molecules in three dimensions."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And here's the signal for the protons on methane. So this signal occurs at approximately one part per million. And remember from the first few videos on proton NMR, what that signal is talking about, it's talking about the energy difference between the alpha and the beta spin states. So this is the alpha spin state and this is the beta spin state. There's an energy difference between those two spin states. And this energy difference corresponds to a frequency because E is equal to h nu. And the energy difference also corresponds to the effective magnetic field felt by a proton."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is the alpha spin state and this is the beta spin state. There's an energy difference between those two spin states. And this energy difference corresponds to a frequency because E is equal to h nu. And the energy difference also corresponds to the effective magnetic field felt by a proton. So if I draw in a magnetic field here, so the effective magnetic field controls the energy difference. So let's think about this. If I have a certain effective magnetic field, I get a certain difference in energy between the alpha and the beta spin states."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the energy difference also corresponds to the effective magnetic field felt by a proton. So if I draw in a magnetic field here, so the effective magnetic field controls the energy difference. So let's think about this. If I have a certain effective magnetic field, I get a certain difference in energy between the alpha and the beta spin states. The energy corresponds to a frequency that's absorbed. And so this, this signal is a certain frequency. We said before this was a lower frequency."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If I have a certain effective magnetic field, I get a certain difference in energy between the alpha and the beta spin states. The energy corresponds to a frequency that's absorbed. And so this, this signal is a certain frequency. We said before this was a lower frequency. This is a lower frequency signal right here. So this is a lower frequency signal. And in an earlier video, we talked about how to compare frequency to chemical shift."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We said before this was a lower frequency. This is a lower frequency signal right here. So this is a lower frequency signal. And in an earlier video, we talked about how to compare frequency to chemical shift. So a low frequency gives a low chemical shift. So one is a low chemical shift here. And so the protons in methane are shielded compared to the protons in chloromethane."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And in an earlier video, we talked about how to compare frequency to chemical shift. So a low frequency gives a low chemical shift. So one is a low chemical shift here. And so the protons in methane are shielded compared to the protons in chloromethane. So let's look at chloromethane next down here. So for chloromethane, we have three equivalent protons, so one signal on our NMR spectrum. And this signal occurs just past three, so approximately 3.1 parts per million."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the protons in methane are shielded compared to the protons in chloromethane. So let's look at chloromethane next down here. So for chloromethane, we have three equivalent protons, so one signal on our NMR spectrum. And this signal occurs just past three, so approximately 3.1 parts per million. And let's see if we can understand why this occurs. So we now have an electronegative atom, right? Chlorine is much more electronegative than carbon."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this signal occurs just past three, so approximately 3.1 parts per million. And let's see if we can understand why this occurs. So we now have an electronegative atom, right? Chlorine is much more electronegative than carbon. So chlorine is going to withdraw some electron density. And so the chlorine gets partially negative. We give the carbon a partial positive here."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Chlorine is much more electronegative than carbon. So chlorine is going to withdraw some electron density. And so the chlorine gets partially negative. We give the carbon a partial positive here. So the chlorine's withdrawing electron density from these protons. So these protons are deshielded from the applied magnetic field. So here we have deshielded protons."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We give the carbon a partial positive here. So the chlorine's withdrawing electron density from these protons. So these protons are deshielded from the applied magnetic field. So here we have deshielded protons. So deshielding protons, if the protons are deshielded from the applied magnetic field, that means those protons experience a greater effective magnetic field. So let me go ahead and draw this in. I'm just gonna exaggerate to get the point across."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here we have deshielded protons. So deshielding protons, if the protons are deshielded from the applied magnetic field, that means those protons experience a greater effective magnetic field. So let me go ahead and draw this in. I'm just gonna exaggerate to get the point across. So a greater effective magnetic field for a proton means a greater difference in energy between your alpha and your beta spin states. So the alpha and beta spin states here. Now, since we have a deshielded proton, right, we have a greater difference in energy between our spin states."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I'm just gonna exaggerate to get the point across. So a greater effective magnetic field for a proton means a greater difference in energy between your alpha and your beta spin states. So the alpha and beta spin states here. Now, since we have a deshielded proton, right, we have a greater difference in energy between our spin states. And energy corresponds to frequency. So a greater effective magnetic field means a greater energy difference, which means a larger frequency, right? So a higher frequency absorbed."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Now, since we have a deshielded proton, right, we have a greater difference in energy between our spin states. And energy corresponds to frequency. So a greater effective magnetic field means a greater energy difference, which means a larger frequency, right? So a higher frequency absorbed. And so this, right, this would be a higher frequency compared to the previous example. So everything is relative here. So a higher frequency signal compared to the protons in methane."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a higher frequency absorbed. And so this, right, this would be a higher frequency compared to the previous example. So everything is relative here. So a higher frequency signal compared to the protons in methane. And therefore, we get a higher value for the chemical shift. So let's just sum this up really quickly. So a shielded protons, right, are gonna give you a lower frequency signal, and therefore a lower value for the chemical shift."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a higher frequency signal compared to the protons in methane. And therefore, we get a higher value for the chemical shift. So let's just sum this up really quickly. So a shielded protons, right, are gonna give you a lower frequency signal, and therefore a lower value for the chemical shift. A deshielded proton is gonna give you a higher frequency signal and a higher chemical shift. Alright, so once again, just comparing these two things, that's what electronegativity does, right? So the more, here we have an electronegative atom that's deshielding the protons, giving a higher chemical shift."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a shielded protons, right, are gonna give you a lower frequency signal, and therefore a lower value for the chemical shift. A deshielded proton is gonna give you a higher frequency signal and a higher chemical shift. Alright, so once again, just comparing these two things, that's what electronegativity does, right? So the more, here we have an electronegative atom that's deshielding the protons, giving a higher chemical shift. So that's the idea. And let's apply this to a chart that has a bunch of different functional groups here. And let's think about the different chemical shifts for protons in different environments."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the more, here we have an electronegative atom that's deshielding the protons, giving a higher chemical shift. So that's the idea. And let's apply this to a chart that has a bunch of different functional groups here. And let's think about the different chemical shifts for protons in different environments. Alright, so we just said that if you deshield a proton, you're gonna get a higher frequency signal, and therefore a higher chemical shift. And this is called downfield. So the left side of this NMR spectrum, these are more deshielded protons."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's think about the different chemical shifts for protons in different environments. Alright, so we just said that if you deshield a proton, you're gonna get a higher frequency signal, and therefore a higher chemical shift. And this is called downfield. So the left side of this NMR spectrum, these are more deshielded protons. To the right side of the NMR spectrum, we're talking about more shielded protons, therefore a lower frequency signal, therefore a lower chemical shift, and you could use the older term upfield if you wanted to as well. So we just talked about methane, right? So we're talking about an alkane-type environment here."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the left side of this NMR spectrum, these are more deshielded protons. To the right side of the NMR spectrum, we're talking about more shielded protons, therefore a lower frequency signal, therefore a lower chemical shift, and you could use the older term upfield if you wanted to as well. So we just talked about methane, right? So we're talking about an alkane-type environment here. So the proton on a carbon in an alkane-type environment, the chemical shift, this is a shielded proton, right? So we would expect a low frequency signal or a low chemical shift. So somewhere in the range of.5 to two is where we'd expect the signal for a proton in an alkane-type environment, so somewhere in that range."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're talking about an alkane-type environment here. So the proton on a carbon in an alkane-type environment, the chemical shift, this is a shielded proton, right? So we would expect a low frequency signal or a low chemical shift. So somewhere in the range of.5 to two is where we'd expect the signal for a proton in an alkane-type environment, so somewhere in that range. Alright, so those are more shielded. Next, we talked about chloromethane, right? In chloromethane, we had an electronegative atom on a carbon that was bonded to our proton."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So somewhere in the range of.5 to two is where we'd expect the signal for a proton in an alkane-type environment, so somewhere in that range. Alright, so those are more shielded. Next, we talked about chloromethane, right? In chloromethane, we had an electronegative atom on a carbon that was bonded to our proton. So that's this situation. Let me use yellow for this. So here we have Y as an electronegative atom."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "In chloromethane, we had an electronegative atom on a carbon that was bonded to our proton. So that's this situation. Let me use yellow for this. So here we have Y as an electronegative atom. So you could think about something like chlorine or fluorine, so a halogen, or you could think about oxygen, also electronegative, right? So if Y is an electronegative atom, Y withdraws electron density from this carbon, and that deshields this proton that's directly on that carbon. So deshielding the proton gives you a higher chemical shift, and so you'd expect this shift to be approximately 2.5 to 4.5."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here we have Y as an electronegative atom. So you could think about something like chlorine or fluorine, so a halogen, or you could think about oxygen, also electronegative, right? So if Y is an electronegative atom, Y withdraws electron density from this carbon, and that deshields this proton that's directly on that carbon. So deshielding the proton gives you a higher chemical shift, and so you'd expect this shift to be approximately 2.5 to 4.5. So if you see a signal in 2.5 to 4.5 range, it could be a proton that's on a carbon that's directly bonded to an electronegative atom, like a halogen or like oxygen. All right, so in between those two examples, right, so the signal for this proton right here, right, this proton would show up approximately 2 to 2.5. And so this proton is directly bonded to a carbon, but this carbon is not directly bonded to an electronegative atom."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So deshielding the proton gives you a higher chemical shift, and so you'd expect this shift to be approximately 2.5 to 4.5. So if you see a signal in 2.5 to 4.5 range, it could be a proton that's on a carbon that's directly bonded to an electronegative atom, like a halogen or like oxygen. All right, so in between those two examples, right, so the signal for this proton right here, right, this proton would show up approximately 2 to 2.5. And so this proton is directly bonded to a carbon, but this carbon is not directly bonded to an electronegative atom. But it is bonded to this carbon, which is a carbonyl here. So this oxygen, right, is more electronegative. This oxygen withdraws some electron density."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this proton is directly bonded to a carbon, but this carbon is not directly bonded to an electronegative atom. But it is bonded to this carbon, which is a carbonyl here. So this oxygen, right, is more electronegative. This oxygen withdraws some electron density. But not quite as much as in this example with this electronegative atom directly on this carbon. And so the signal, the chemical shift is in between here. So a proton that's on a carbon, that's next to a carbonyl, look for that approximately 2 to 2.5."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen withdraws some electron density. But not quite as much as in this example with this electronegative atom directly on this carbon. And so the signal, the chemical shift is in between here. So a proton that's on a carbon, that's next to a carbonyl, look for that approximately 2 to 2.5. Again, all of these are just approximate ranges here. So I tried to give nice, easy numbers to remember. Next, let's look at the proton on alcohol."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a proton that's on a carbon, that's next to a carbonyl, look for that approximately 2 to 2.5. Again, all of these are just approximate ranges here. So I tried to give nice, easy numbers to remember. Next, let's look at the proton on alcohol. So right here. Well, alcohols have hydrogen bonding. And hydrogen bonding has a deshielding effect."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at the proton on alcohol. So right here. Well, alcohols have hydrogen bonding. And hydrogen bonding has a deshielding effect. So increased hydrogen bonding, increased deshielding. The problem is the amount of hydrogen bonding depends on things like concentration and temperature. And since those things can vary, right, you get different amounts of hydrogen bonding, you get different amounts of deshielding, you get a different range, a pretty broad range here for your possible signal."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And hydrogen bonding has a deshielding effect. So increased hydrogen bonding, increased deshielding. The problem is the amount of hydrogen bonding depends on things like concentration and temperature. And since those things can vary, right, you get different amounts of hydrogen bonding, you get different amounts of deshielding, you get a different range, a pretty broad range here for your possible signal. So approximately 2 to 5 for the signal on an alcohol. But it might not even be in that range. So just think about 2 to 5 for the proton on an alcohol as an approximate region."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And since those things can vary, right, you get different amounts of hydrogen bonding, you get different amounts of deshielding, you get a different range, a pretty broad range here for your possible signal. So approximately 2 to 5 for the signal on an alcohol. But it might not even be in that range. So just think about 2 to 5 for the proton on an alcohol as an approximate region. Next, let's look at the proton on a double bond here. So proton, right, bonded to a carbon. Proton on a double bond, the shift is approximately 4.5 to 6.5."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So just think about 2 to 5 for the proton on an alcohol as an approximate region. Next, let's look at the proton on a double bond here. So proton, right, bonded to a carbon. Proton on a double bond, the shift is approximately 4.5 to 6.5. So let's see if we can understand why. One way to think about it is using electronegativity. And so if we think about this carbon here, this carbon is sp2 hybridized."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Proton on a double bond, the shift is approximately 4.5 to 6.5. So let's see if we can understand why. One way to think about it is using electronegativity. And so if we think about this carbon here, this carbon is sp2 hybridized. And if we compare that carbon to this carbon, right, this carbon is sp3 hybridized. Remember from hybridization videos that an sp2 hybrid orbital has more s character than an sp3 hybridized orbital. Therefore, the electrons are held closer to the nucleus."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if we think about this carbon here, this carbon is sp2 hybridized. And if we compare that carbon to this carbon, right, this carbon is sp3 hybridized. Remember from hybridization videos that an sp2 hybrid orbital has more s character than an sp3 hybridized orbital. Therefore, the electrons are held closer to the nucleus. So you can say that an sp2 hybridized carbon is more electronegative than an sp3 hybridized carbon. So if you wanna think about it that way, that's one way to think about it. And so this sp2 hybridized carbon is withdrawing more electron density from this proton."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, the electrons are held closer to the nucleus. So you can say that an sp2 hybridized carbon is more electronegative than an sp3 hybridized carbon. So if you wanna think about it that way, that's one way to think about it. And so this sp2 hybridized carbon is withdrawing more electron density from this proton. Let me use a different color here. So this sp2 hybridized carbon is withdrawing more electron density, deshielding this proton, and giving you a higher chemical shift than for a proton bonded to an sp3 hybridized carbon. So that's one way to explain this."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this sp2 hybridized carbon is withdrawing more electron density from this proton. Let me use a different color here. So this sp2 hybridized carbon is withdrawing more electron density, deshielding this proton, and giving you a higher chemical shift than for a proton bonded to an sp3 hybridized carbon. So that's one way to explain this. But it doesn't, that line of reasoning isn't exactly, that doesn't hold up completely. Because if we next look at a proton on a triple bond here, so if I draw a triple bond in this proton right here, so you might think, okay, well, this carbon is sp hybridized. And I know that sp hybridized carbon, an sp hybridized orbital has even more s character than an sp2 hybridized orbital."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's one way to explain this. But it doesn't, that line of reasoning isn't exactly, that doesn't hold up completely. Because if we next look at a proton on a triple bond here, so if I draw a triple bond in this proton right here, so you might think, okay, well, this carbon is sp hybridized. And I know that sp hybridized carbon, an sp hybridized orbital has even more s character than an sp2 hybridized orbital. So therefore, you could think about an sp hybridized carbon being more electronegative, and these electrons are closer to this carbon. And so you might think, oh, that's going to deshield, right, that's going to deshield this proton, and we would expect a signal that's an even higher chemical shift than for this proton. And that's not what we observe."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And I know that sp hybridized carbon, an sp hybridized orbital has even more s character than an sp2 hybridized orbital. So therefore, you could think about an sp hybridized carbon being more electronegative, and these electrons are closer to this carbon. And so you might think, oh, that's going to deshield, right, that's going to deshield this proton, and we would expect a signal that's an even higher chemical shift than for this proton. And that's not what we observe. So the proton on a triple bond, actually, this shows up somewhere in this range, so somewhere around two to 2.5 approximately. And so it's not just electronegativity that you have to think about. So there's another effect that's causing the chemical shift for this proton that we'll talk about in the next video."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that's not what we observe. So the proton on a triple bond, actually, this shows up somewhere in this range, so somewhere around two to 2.5 approximately. And so it's not just electronegativity that you have to think about. So there's another effect that's causing the chemical shift for this proton that we'll talk about in the next video. And it's the same thing, it's actually the same thing for the proton on a benzene ring. So we'll save that discussion for the next video. So we'll talk about this, and we'll talk about this in the next video."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So there's another effect that's causing the chemical shift for this proton that we'll talk about in the next video. And it's the same thing, it's actually the same thing for the proton on a benzene ring. So we'll save that discussion for the next video. So we'll talk about this, and we'll talk about this in the next video. If we move on to an aldehyde, right, so for an aldehyde, we have this carbonyl here. The oxygen's withdrawing electron density, right, away from the proton on the aldehyde. And so it's deshielding that proton, right, therefore we'd expect the signal for that proton to occur at a higher chemical shift, so somewhere around nine to 10 is where we'd expect the shift for this proton."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we'll talk about this, and we'll talk about this in the next video. If we move on to an aldehyde, right, so for an aldehyde, we have this carbonyl here. The oxygen's withdrawing electron density, right, away from the proton on the aldehyde. And so it's deshielding that proton, right, therefore we'd expect the signal for that proton to occur at a higher chemical shift, so somewhere around nine to 10 is where we'd expect the shift for this proton. Finally, let's look at a carboxylic acid. So the signal for this proton, approximately 10 to 12. So once again, we have this carbonyl here withdrawing electron density."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so it's deshielding that proton, right, therefore we'd expect the signal for that proton to occur at a higher chemical shift, so somewhere around nine to 10 is where we'd expect the shift for this proton. Finally, let's look at a carboxylic acid. So the signal for this proton, approximately 10 to 12. So once again, we have this carbonyl here withdrawing electron density. We have another oxygen here withdrawing some electron density. So you could think about electronegativity effects. You could also think about resonance effects."}, {"video_title": "Electronegativity and chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we have this carbonyl here withdrawing electron density. We have another oxygen here withdrawing some electron density. So you could think about electronegativity effects. You could also think about resonance effects. And you could also think about there's some hydrogen bonding effects. So there's all kinds of things going on here with the carboxylic acid. And pretty much, if you're just looking in the 10 to 12 region and you see a signal, think the proton on a carboxylic acid."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Or another way of saying that, formal charge is equal to the number of valence electrons the atom is supposed to have minus the number of valence electrons that the atom actually has in the drawing. So let's assign a formal charge to the nitrogen in this molecule. And remember that each bond represents two electrons. So I'm gonna draw in the electrons in this bond so it's easier for us to assign a formal charge to the nitrogen. So formal charge is equal to the number of valence electrons that nitrogen is supposed to have. We know that nitrogen is supposed to have five valence electrons because of its position on the periodic table. So this is five."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna draw in the electrons in this bond so it's easier for us to assign a formal charge to the nitrogen. So formal charge is equal to the number of valence electrons that nitrogen is supposed to have. We know that nitrogen is supposed to have five valence electrons because of its position on the periodic table. So this is five. And from that, we subtract the number of valence electrons that nitrogen has in our drawing. So let's go back over here to the dot structure and let's look at these bonds. We know that from this bond here on the left, nitrogen gets one of those electrons and from this bond on the right, nitrogen gets one of those electrons and hydrogen gets the other."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is five. And from that, we subtract the number of valence electrons that nitrogen has in our drawing. So let's go back over here to the dot structure and let's look at these bonds. We know that from this bond here on the left, nitrogen gets one of those electrons and from this bond on the right, nitrogen gets one of those electrons and hydrogen gets the other. And same for this nitrogen-hydrogen bond, nitrogen gets one of the electrons and hydrogen gets the other. So how many electrons do we have around nitrogen in our drawing? Let's count them up."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that from this bond here on the left, nitrogen gets one of those electrons and from this bond on the right, nitrogen gets one of those electrons and hydrogen gets the other. And same for this nitrogen-hydrogen bond, nitrogen gets one of the electrons and hydrogen gets the other. So how many electrons do we have around nitrogen in our drawing? Let's count them up. This would be one, two, three, and then we have a lone pair of electrons on the nitrogen. So that's four and five. So in our drawing, nitrogen is surrounded by five valence electrons."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's count them up. This would be one, two, three, and then we have a lone pair of electrons on the nitrogen. So that's four and five. So in our drawing, nitrogen is surrounded by five valence electrons. So we put five minus five, which is equal to zero. So nitrogen has a formal charge of zero. Let me go ahead and redraw that."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in our drawing, nitrogen is surrounded by five valence electrons. So we put five minus five, which is equal to zero. So nitrogen has a formal charge of zero. Let me go ahead and redraw that. So we had our nitrogen here with our two hydrogens and a lone pair of electrons on the nitrogen. We found the nitrogen to have a formal charge of zero. So we have a pattern."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and redraw that. So we had our nitrogen here with our two hydrogens and a lone pair of electrons on the nitrogen. We found the nitrogen to have a formal charge of zero. So we have a pattern. Every time that you see nitrogen with three bonds, let me draw these in here, one, two, three. So three bonds and one lone pair of electrons, the formal charge is equal to zero. So when nitrogen has three bonds and one lone pair of electrons, the formal charge is equal to zero."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have a pattern. Every time that you see nitrogen with three bonds, let me draw these in here, one, two, three. So three bonds and one lone pair of electrons, the formal charge is equal to zero. So when nitrogen has three bonds and one lone pair of electrons, the formal charge is equal to zero. And sometimes you don't want to draw in lone pairs of electrons, so you can just leave those off. You could just say, all right, well if I just draw this, and you know the formal charge on nitrogen is zero, then it's assumed you also know there's a lone pair of electrons on that nitrogen. So this is just another way of representing the same molecule, leaving off the lone pair, because you should know it's there."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So when nitrogen has three bonds and one lone pair of electrons, the formal charge is equal to zero. And sometimes you don't want to draw in lone pairs of electrons, so you can just leave those off. You could just say, all right, well if I just draw this, and you know the formal charge on nitrogen is zero, then it's assumed you also know there's a lone pair of electrons on that nitrogen. So this is just another way of representing the same molecule, leaving off the lone pair, because you should know it's there. Let's look at other examples where nitrogen has a formal charge of zero. So we'll start with the example on the left here. And if we look at this nitrogen, and we know it has a formal charge of zero, let's see how many bonds it has."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is just another way of representing the same molecule, leaving off the lone pair, because you should know it's there. Let's look at other examples where nitrogen has a formal charge of zero. So we'll start with the example on the left here. And if we look at this nitrogen, and we know it has a formal charge of zero, let's see how many bonds it has. Let's use red here. So here's one bond, two bonds, and then three bonds. So three bonds, and with a formal charge of zero, we know there should be a lone pair of electrons on that nitrogen."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if we look at this nitrogen, and we know it has a formal charge of zero, let's see how many bonds it has. Let's use red here. So here's one bond, two bonds, and then three bonds. So three bonds, and with a formal charge of zero, we know there should be a lone pair of electrons on that nitrogen. So you could leave it off, and just know it's there, or you could draw them in. So I'll go ahead and draw in the lone pair of electrons on the nitrogen, so formal charge of zero. Let's look at the one on the right."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So three bonds, and with a formal charge of zero, we know there should be a lone pair of electrons on that nitrogen. So you could leave it off, and just know it's there, or you could draw them in. So I'll go ahead and draw in the lone pair of electrons on the nitrogen, so formal charge of zero. Let's look at the one on the right. So if we assume that nitrogen has a formal charge of zero, let's see how many bonds we have here. So here's one, two, and three. So we have three bonds."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the one on the right. So if we assume that nitrogen has a formal charge of zero, let's see how many bonds we have here. So here's one, two, and three. So we have three bonds. So we'd still need one lone pair of electrons. So if you wanted to show the lone pair of electrons, you could put them in there like that. Notice this gives nitrogen an octet of electrons around it."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have three bonds. So we'd still need one lone pair of electrons. So if you wanted to show the lone pair of electrons, you could put them in there like that. Notice this gives nitrogen an octet of electrons around it. So count those up. Here's two, four, six, and eight. So nitrogen would have an octet."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Notice this gives nitrogen an octet of electrons around it. So count those up. Here's two, four, six, and eight. So nitrogen would have an octet. And remember, you could just leave off that lone pair of electrons, and it's assumed, if we know nitrogen has a formal charge of zero, that there is a lone pair. And we just didn't want to take the time to draw them in. Let's assign formal charge to another nitrogen, so down here."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So nitrogen would have an octet. And remember, you could just leave off that lone pair of electrons, and it's assumed, if we know nitrogen has a formal charge of zero, that there is a lone pair. And we just didn't want to take the time to draw them in. Let's assign formal charge to another nitrogen, so down here. So what is the formal charge of nitrogen now? Let's draw in our electrons. So each bond is two electrons."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's assign formal charge to another nitrogen, so down here. So what is the formal charge of nitrogen now? Let's draw in our electrons. So each bond is two electrons. So I draw those in there. And the formal charge on nitrogen is equal to the number of valence electrons that nitrogen is supposed to have, which we already know is five. So we put a five in here."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So each bond is two electrons. So I draw those in there. And the formal charge on nitrogen is equal to the number of valence electrons that nitrogen is supposed to have, which we already know is five. So we put a five in here. And from that, we subtract the number of valence electrons that nitrogen actually has in our drawing. So for these bonds, hydrogen gets one electron, and nitrogen gets one for each of these bonds. So that allows us to see there are four electrons around nitrogen."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we put a five in here. And from that, we subtract the number of valence electrons that nitrogen actually has in our drawing. So for these bonds, hydrogen gets one electron, and nitrogen gets one for each of these bonds. So that allows us to see there are four electrons around nitrogen. So here's one, two, three, and four. So in our drawing, nitrogen only has four electrons around it. So this would be five minus four, which gives us a formal charge of plus one."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that allows us to see there are four electrons around nitrogen. So here's one, two, three, and four. So in our drawing, nitrogen only has four electrons around it. So this would be five minus four, which gives us a formal charge of plus one. So it's like nitrogen lost a valence electron, right? It's supposed to have five, and here we see only four around it. So it's as if it lost a valence electron."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this would be five minus four, which gives us a formal charge of plus one. So it's like nitrogen lost a valence electron, right? It's supposed to have five, and here we see only four around it. So it's as if it lost a valence electron. So it's plus one for the formal charge. All right, let me redraw that. So we have our nitrogen with four bonds to hydrogen."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's as if it lost a valence electron. So it's plus one for the formal charge. All right, let me redraw that. So we have our nitrogen with four bonds to hydrogen. And the nitrogen has a plus one formal charge. You should recognize this as being the ammonium ion from general chemistry. So this has a formal charge of plus one."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have our nitrogen with four bonds to hydrogen. And the nitrogen has a plus one formal charge. You should recognize this as being the ammonium ion from general chemistry. So this has a formal charge of plus one. So we have another pattern to think about here. So let's draw that in. We have one, two, three, four bonds, and zero lone pairs of electrons."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this has a formal charge of plus one. So we have another pattern to think about here. So let's draw that in. We have one, two, three, four bonds, and zero lone pairs of electrons. So when nitrogen has four bonds, four bonds and zero lone pairs, zero lone pairs of electrons, we've already seen the formal charge should be equal to plus one. So let's look at some examples where nitrogen has a formal charge of plus one. So the example on the left, we can see there are four bonds, and there are no lone pairs on that nitrogen."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have one, two, three, four bonds, and zero lone pairs of electrons. So when nitrogen has four bonds, four bonds and zero lone pairs, zero lone pairs of electrons, we've already seen the formal charge should be equal to plus one. So let's look at some examples where nitrogen has a formal charge of plus one. So the example on the left, we can see there are four bonds, and there are no lone pairs on that nitrogen. So that's a plus one formal charge. Over here on the right, same idea. Here's one bond, two bond, three bonds, and four bonds, and no lone pairs, so a plus one formal charge on the nitrogen."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the example on the left, we can see there are four bonds, and there are no lone pairs on that nitrogen. So that's a plus one formal charge. Over here on the right, same idea. Here's one bond, two bond, three bonds, and four bonds, and no lone pairs, so a plus one formal charge on the nitrogen. All right, finally, one more nitrogen to assign a formal charge to. So let's look at this one. Let's draw in the electrons in the bonds."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here's one bond, two bond, three bonds, and four bonds, and no lone pairs, so a plus one formal charge on the nitrogen. All right, finally, one more nitrogen to assign a formal charge to. So let's look at this one. Let's draw in the electrons in the bonds. So here's two electrons, and here's two electrons. What is the formal charge on nitrogen? Formal charge is equal to number of valence electrons nitrogen is supposed to have, which we know is five, and from that we subtract the number of valence electrons nitrogen actually has in our dot structure."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw in the electrons in the bonds. So here's two electrons, and here's two electrons. What is the formal charge on nitrogen? Formal charge is equal to number of valence electrons nitrogen is supposed to have, which we know is five, and from that we subtract the number of valence electrons nitrogen actually has in our dot structure. So again, we go over here, and we look at this bond, and we give one electron to nitrogen and one electron to the other atom, and over here we give one electron to nitrogen and one electron to the other atom, and now we have two lone pairs of electrons on the nitrogen. So how many is that total? This would be one, two, three, four, five, and six."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Formal charge is equal to number of valence electrons nitrogen is supposed to have, which we know is five, and from that we subtract the number of valence electrons nitrogen actually has in our dot structure. So again, we go over here, and we look at this bond, and we give one electron to nitrogen and one electron to the other atom, and over here we give one electron to nitrogen and one electron to the other atom, and now we have two lone pairs of electrons on the nitrogen. So how many is that total? This would be one, two, three, four, five, and six. So six electrons around our nitrogen, so five minus six gives us negative one, so a formal charge of negative one. Let me go ahead and redraw that so I can draw it out here. So nitrogen with two lone pairs of electrons we just found has a formal charge of negative one."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This would be one, two, three, four, five, and six. So six electrons around our nitrogen, so five minus six gives us negative one, so a formal charge of negative one. Let me go ahead and redraw that so I can draw it out here. So nitrogen with two lone pairs of electrons we just found has a formal charge of negative one. If I wanted to leave off the lone pairs of electrons, I could do that. I could just write NH here and put a negative one formal charge, and because of this pattern, you should know there are two lone pairs of electrons on that nitrogen. Let me just clarify the pattern here."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So nitrogen with two lone pairs of electrons we just found has a formal charge of negative one. If I wanted to leave off the lone pairs of electrons, I could do that. I could just write NH here and put a negative one formal charge, and because of this pattern, you should know there are two lone pairs of electrons on that nitrogen. Let me just clarify the pattern here. The pattern for a formal charge of negative one on nitrogen would be two bonds, here are the two bonds, and two lone pairs of electrons. So when nitrogen has two bonds and two lone pairs of electrons, nitrogen should have a formal charge of negative one. Let's look at some examples of that."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me just clarify the pattern here. The pattern for a formal charge of negative one on nitrogen would be two bonds, here are the two bonds, and two lone pairs of electrons. So when nitrogen has two bonds and two lone pairs of electrons, nitrogen should have a formal charge of negative one. Let's look at some examples of that. So down here we have nitrogen. So here's nitrogen with no lone pairs of electrons drawn in, but you know this nitrogen has a negative one formal charge because it's telling you that right here. How many bonds do we have?"}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at some examples of that. So down here we have nitrogen. So here's nitrogen with no lone pairs of electrons drawn in, but you know this nitrogen has a negative one formal charge because it's telling you that right here. How many bonds do we have? Well, here's one bond, and here's the other bond. So we have our two bonds, but we don't have our two lone pairs drawn in, so you could just know that they are there, or I'll go ahead and add them in here. So here's one lone pair of electrons, and here's the other lone pair of electrons on that nitrogen."}, {"video_title": "Formal charge on nitrogen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "How many bonds do we have? Well, here's one bond, and here's the other bond. So we have our two bonds, but we don't have our two lone pairs drawn in, so you could just know that they are there, or I'll go ahead and add them in here. So here's one lone pair of electrons, and here's the other lone pair of electrons on that nitrogen. Notice that gives that nitrogen an octet of electrons. Over here on the right, let's do the same thing. You know this nitrogen has a negative one formal charge, so you know it must have two bonds and two lone pairs of electrons."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's our dot structure for dibutylamine, and we start by drawing a line about 3,000, and we know to the right of that, we expect to find the signal for the carbon-hydrogen bond stretch, where we're talking about an sp3 hybridized carbon. And so this is the bond to hydrogen region on the IR spectrum. And notice we have another signal right here, so a signal that's at a higher wave number than the carbon-hydrogen bond stretch. So if we drop down here, and we can estimate the wave number, so it's approximately 31, 32, 33. So at approximately 3,300 wave numbers, we get another signal, so another bond of an atom to hydrogen. And this is the bond stretch for nitrogen-hydrogen, so that's this bond stretching right here. So in magenta, this is the nitrogen-hydrogen bond stretch."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we drop down here, and we can estimate the wave number, so it's approximately 31, 32, 33. So at approximately 3,300 wave numbers, we get another signal, so another bond of an atom to hydrogen. And this is the bond stretch for nitrogen-hydrogen, so that's this bond stretching right here. So in magenta, this is the nitrogen-hydrogen bond stretch. Let's compare the strength of that bond to a carbon-hydrogen bond, where the carbon is sp3 hybridized. We know that the wave number is dependent on two things from an earlier video. We know it's dependent upon the force constant, k, or the spring constant, k, and the reduced mass."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in magenta, this is the nitrogen-hydrogen bond stretch. Let's compare the strength of that bond to a carbon-hydrogen bond, where the carbon is sp3 hybridized. We know that the wave number is dependent on two things from an earlier video. We know it's dependent upon the force constant, k, or the spring constant, k, and the reduced mass. Well, the reduced mass for these two bonds is approximately the same. So if you did a calculation for the reduced mass for nitrogen-hydrogen and for carbon-hydrogen, you're going to get approximately the same value for the reduced mass. And so that's not what's affecting the different wave numbers for these signals here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We know it's dependent upon the force constant, k, or the spring constant, k, and the reduced mass. Well, the reduced mass for these two bonds is approximately the same. So if you did a calculation for the reduced mass for nitrogen-hydrogen and for carbon-hydrogen, you're going to get approximately the same value for the reduced mass. And so that's not what's affecting the different wave numbers for these signals here. So it must be the force constant, must be k. And since the nitrogen-hydrogen bond has a greater weight, the signal shows up at a higher wave number, that must mean it's a stronger bond. Because if you increase the force constant, increase the strength of the bond, you increase the wave number. You increase the frequency."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that's not what's affecting the different wave numbers for these signals here. So it must be the force constant, must be k. And since the nitrogen-hydrogen bond has a greater weight, the signal shows up at a higher wave number, that must mean it's a stronger bond. Because if you increase the force constant, increase the strength of the bond, you increase the wave number. You increase the frequency. And so the nitrogen-hydrogen bond is stronger than the carbon-hydrogen bond, where the carbon is sp3 hybridized. And if it's stronger, it takes more energy to cause that bond to stretch. And so let's really quickly talk about energy."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "You increase the frequency. And so the nitrogen-hydrogen bond is stronger than the carbon-hydrogen bond, where the carbon is sp3 hybridized. And if it's stronger, it takes more energy to cause that bond to stretch. And so let's really quickly talk about energy. So energy is equal to h, which is Planck's constant, times the frequency. So when you're talking about the energy of a photon, it's equal to h nu. And nu is our frequency, and we know that relates to wave number."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so let's really quickly talk about energy. So energy is equal to h, which is Planck's constant, times the frequency. So when you're talking about the energy of a photon, it's equal to h nu. And nu is our frequency, and we know that relates to wave number. So the frequency is equal to the wave number times the speed of light. And so we talked about this in an earlier video. So if you take this and plug it into here, you can see the energy is directly proportional to wave number."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And nu is our frequency, and we know that relates to wave number. So the frequency is equal to the wave number times the speed of light. And so we talked about this in an earlier video. So if you take this and plug it into here, you can see the energy is directly proportional to wave number. So this would be E is equal to h times the wave number times the speed of light. And this is one of the reasons why you see IR spectrum done in wave numbers, because you can also think about energy. So if you increase the wave number, if you're talking about an increased wave number, you're talking about increased energy."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you take this and plug it into here, you can see the energy is directly proportional to wave number. So this would be E is equal to h times the wave number times the speed of light. And this is one of the reasons why you see IR spectrum done in wave numbers, because you can also think about energy. So if you increase the wave number, if you're talking about an increased wave number, you're talking about increased energy. So as you go this way, so as you increase in wave number, you're also talking about increasing in energy. And so you could think about it takes more energy. More energy is needed to stretch a stronger bond."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you increase the wave number, if you're talking about an increased wave number, you're talking about increased energy. So as you go this way, so as you increase in wave number, you're also talking about increasing in energy. And so you could think about it takes more energy. More energy is needed to stretch a stronger bond. And so it takes more energy to stretch this nitrogen-hydrogen bond. So again, think about a bond as a spring. If you have a really stiff or strong spring, it takes more energy to stretch that spring out, as compared to a looser spring."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "More energy is needed to stretch a stronger bond. And so it takes more energy to stretch this nitrogen-hydrogen bond. So again, think about a bond as a spring. If you have a really stiff or strong spring, it takes more energy to stretch that spring out, as compared to a looser spring. And so that's thinking about energy, and also looking at a typical IR spectrum here for a secondary amine. So this nitrogen here is bonded to two carbons. So this is a secondary amine."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you have a really stiff or strong spring, it takes more energy to stretch that spring out, as compared to a looser spring. And so that's thinking about energy, and also looking at a typical IR spectrum here for a secondary amine. So this nitrogen here is bonded to two carbons. So this is a secondary amine. And in a secondary amine, you're going to get one signal, approximately 3,300 here. So let's compare this IR spectrum of a secondary amine with another amine. So this is a primary amine."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is a secondary amine. And in a secondary amine, you're going to get one signal, approximately 3,300 here. So let's compare this IR spectrum of a secondary amine with another amine. So this is a primary amine. So let's compare it to a butyl amine. So over here, this is a primary amine. The nitrogen is bonded to one carbon."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is a primary amine. So let's compare it to a butyl amine. So over here, this is a primary amine. The nitrogen is bonded to one carbon. So we're talking about a primary amine now. And let's analyze the IR spectrum. So once again, we're going to draw a line around 3,000."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The nitrogen is bonded to one carbon. So we're talking about a primary amine now. And let's analyze the IR spectrum. So once again, we're going to draw a line around 3,000. And we know that this in here is talking about the carbon-hydrogen bond stretch for an sp3 hybridized carbon. Once again, let's look at just past that, in the bond-to-hydrogen region. And we get two signals this time."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we're going to draw a line around 3,000. And we know that this in here is talking about the carbon-hydrogen bond stretch for an sp3 hybridized carbon. Once again, let's look at just past that, in the bond-to-hydrogen region. And we get two signals this time. So if we look over here, there are two signals. This signal, let's drop down. This is approximately 3,300."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we get two signals this time. So if we look over here, there are two signals. This signal, let's drop down. This is approximately 3,300. So we have one signal, approximately 3,300. And then we have another signal. Let me go ahead and make that green here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is approximately 3,300. So we have one signal, approximately 3,300. And then we have another signal. Let me go ahead and make that green here. So we get another signal right here, which is a little bit higher in terms of a wave number. So if we drop down, this signal is approximately 3,400. So we know this is where we would expect to find the nitrogen-hydrogen bond stretch."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and make that green here. So we get another signal right here, which is a little bit higher in terms of a wave number. So if we drop down, this signal is approximately 3,400. So we know this is where we would expect to find the nitrogen-hydrogen bond stretch. We get two signals. And we need to figure out what's going on here. Well, this has to do with symmetric and asymmetric stretching."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we know this is where we would expect to find the nitrogen-hydrogen bond stretch. We get two signals. And we need to figure out what's going on here. Well, this has to do with symmetric and asymmetric stretching. So let's look at two generic amines here. And let's talk about what the difference is between symmetric and asymmetric stretching. So if you have symmetric stretching, so you have these bonds are stretching in phase, if you will."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, this has to do with symmetric and asymmetric stretching. So let's look at two generic amines here. And let's talk about what the difference is between symmetric and asymmetric stretching. So if you have symmetric stretching, so you have these bonds are stretching in phase, if you will. So you can think about the hydrogens stretching away from the nitrogen at the same time. So this is called symmetric stretching. So this is symmetric stretching."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you have symmetric stretching, so you have these bonds are stretching in phase, if you will. So you can think about the hydrogens stretching away from the nitrogen at the same time. So this is called symmetric stretching. So this is symmetric stretching. And this one over here, let me go ahead and draw what's happening over here. So this time, these two nitrogen-hydrogen bonds are stretching out of phase. So if that hydrogen is stretching this way, this hydrogen might be contracting here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is symmetric stretching. And this one over here, let me go ahead and draw what's happening over here. So this time, these two nitrogen-hydrogen bonds are stretching out of phase. So if that hydrogen is stretching this way, this hydrogen might be contracting here. So that's an asymmetric stretch. Let me go ahead and write that. So we're talking about an asymmetric stretch here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if that hydrogen is stretching this way, this hydrogen might be contracting here. So that's an asymmetric stretch. Let me go ahead and write that. So we're talking about an asymmetric stretch here. So this is what's happening. This is why we get these two different signals. It turns out it takes less energy to do the symmetric stretching."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're talking about an asymmetric stretch here. So this is what's happening. This is why we get these two different signals. It turns out it takes less energy to do the symmetric stretching. So if it takes less energy to do the symmetric stretching, this is the one that we find at a lower wave number. So remember, wave numbers correspond to energy. So it takes less energy to do a symmetric stretching."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It turns out it takes less energy to do the symmetric stretching. So if it takes less energy to do the symmetric stretching, this is the one that we find at a lower wave number. So remember, wave numbers correspond to energy. So it takes less energy to do a symmetric stretching. And so that's this signal. It takes a little more energy to do asymmetric stretch. And so that's this signal right up here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it takes less energy to do a symmetric stretching. And so that's this signal. It takes a little more energy to do asymmetric stretch. And so that's this signal right up here. So we get two different signals here for our primary amine. So two signals. And it's tempting to say, oh, we get two signals because we have two nitrogen-hydrogen bonds."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that's this signal right up here. So we get two different signals here for our primary amine. So two signals. And it's tempting to say, oh, we get two signals because we have two nitrogen-hydrogen bonds. So here's a nitrogen-hydrogen bond, and here's a nitrogen-hydrogen bond. But that's not really what's happening. Some of the molecules are having a symmetric stretch, and some of the molecules are having an asymmetric stretch."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it's tempting to say, oh, we get two signals because we have two nitrogen-hydrogen bonds. So here's a nitrogen-hydrogen bond, and here's a nitrogen-hydrogen bond. But that's not really what's happening. Some of the molecules are having a symmetric stretch, and some of the molecules are having an asymmetric stretch. And so that's why you see these two different signals. And so once again, let's just really quickly compare these two different amines. So a secondary amine is going to give you only one signal on your IR spectrum, whereas a primary amine is going to give you two signals, these two different signals here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Some of the molecules are having a symmetric stretch, and some of the molecules are having an asymmetric stretch. And so that's why you see these two different signals. And so once again, let's just really quickly compare these two different amines. So a secondary amine is going to give you only one signal on your IR spectrum, whereas a primary amine is going to give you two signals, these two different signals here. So that's something to look out for. Also, we can think about the carbon-hydrogen sp3 stretching here. For example, if you have a CH2 here, so a CH2, let me go ahead and draw two different situations here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a secondary amine is going to give you only one signal on your IR spectrum, whereas a primary amine is going to give you two signals, these two different signals here. So that's something to look out for. Also, we can think about the carbon-hydrogen sp3 stretching here. For example, if you have a CH2 here, so a CH2, let me go ahead and draw two different situations here. So a CH2 in a molecule, you can have the same thing. You can have a symmetric stretch and an asymmetric stretch. So if these hydrogens are both stretching at the same time, that's a symmetric stretch."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For example, if you have a CH2 here, so a CH2, let me go ahead and draw two different situations here. So a CH2 in a molecule, you can have the same thing. You can have a symmetric stretch and an asymmetric stretch. So if these hydrogens are both stretching at the same time, that's a symmetric stretch. You could also have an asymmetric stretch like this. And once again, you'll find the asymmetric stretch, so this one, actually takes a little bit more energy. So you're going to find this signal at a slightly higher wave number."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if these hydrogens are both stretching at the same time, that's a symmetric stretch. You could also have an asymmetric stretch like this. And once again, you'll find the asymmetric stretch, so this one, actually takes a little bit more energy. So you're going to find this signal at a slightly higher wave number. So it's a pretty small difference, but it is a slightly higher wave number for this stretch. And that's one of the reasons why you get such a hard-to-interpret signal in here. So less than 3,000, there's a lot of stuff going on."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So you're going to find this signal at a slightly higher wave number. So it's a pretty small difference, but it is a slightly higher wave number for this stretch. And that's one of the reasons why you get such a hard-to-interpret signal in here. So less than 3,000, there's a lot of stuff going on. And it's too difficult to worry about in great detail, just to understand that that's where you would expect to find your carbon-hydrogen bond stretch, where you're talking about sp3 hybridized carbon. Finally, let's look at one more example of a symmetric and asymmetric stretch, and that's an acid anhydride. So let's look at an IR spectrum, just a generic IR spectrum for an acid anhydride."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So less than 3,000, there's a lot of stuff going on. And it's too difficult to worry about in great detail, just to understand that that's where you would expect to find your carbon-hydrogen bond stretch, where you're talking about sp3 hybridized carbon. Finally, let's look at one more example of a symmetric and asymmetric stretch, and that's an acid anhydride. So let's look at an IR spectrum, just a generic IR spectrum for an acid anhydride. So we draw a line around 1,500. To divide our two regions, we draw a line around 3,000 here. So we know this is our carbon-hydrogen in here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at an IR spectrum, just a generic IR spectrum for an acid anhydride. So we draw a line around 1,500. To divide our two regions, we draw a line around 3,000 here. So we know this is our carbon-hydrogen in here. And then we get these two very intense signals. So let's figure out where these are, approximately. So this signal right in here, let me use a different color for right here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we know this is our carbon-hydrogen in here. And then we get these two very intense signals. So let's figure out where these are, approximately. So this signal right in here, let me use a different color for right here. We drop down. So where is that approximately? Well, if this is 1,500, 1,600, 1,700, so that's pretty close to, let's say, 1,760 here for this signal."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this signal right in here, let me use a different color for right here. We drop down. So where is that approximately? Well, if this is 1,500, 1,600, 1,700, so that's pretty close to, let's say, 1,760 here for this signal. So 1,760 wave numbers for that signal. And then this other signal, we drop down to here. This is a tiny bit past 1,800."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, if this is 1,500, 1,600, 1,700, so that's pretty close to, let's say, 1,760 here for this signal. So 1,760 wave numbers for that signal. And then this other signal, we drop down to here. This is a tiny bit past 1,800. So we'll say approximately 1,800 here. So we get these two different strong signals for an acid anhydride. And once again, we're talking about symmetric and asymmetric stretching."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is a tiny bit past 1,800. So we'll say approximately 1,800 here. So we get these two different strong signals for an acid anhydride. And once again, we're talking about symmetric and asymmetric stretching. So a symmetric stretch, let me go down here. So this carbonyl, of course, is what we're talking about. We're talking about this really strong absorbance in the double bond region."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we're talking about symmetric and asymmetric stretching. So a symmetric stretch, let me go down here. So this carbonyl, of course, is what we're talking about. We're talking about this really strong absorbance in the double bond region. So in here is the double bond region on our IR spectrum. We get these two strong signals. And the signal at a lower wave number is due to symmetric stretching."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're talking about this really strong absorbance in the double bond region. So in here is the double bond region on our IR spectrum. We get these two strong signals. And the signal at a lower wave number is due to symmetric stretching. So this carbonyl could be stretching in phase with this carbonyl. So that's our symmetric stretch signal. So that's this one down here."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the signal at a lower wave number is due to symmetric stretching. So this carbonyl could be stretching in phase with this carbonyl. So that's our symmetric stretch signal. So that's this one down here. And then once again, we could get an asymmetric stretch. So we could have one of these. We could have this one stretching and this one contracting here for our spring."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's this one down here. And then once again, we could get an asymmetric stretch. So we could have one of these. We could have this one stretching and this one contracting here for our spring. It takes more energy to do an asymmetric stretch. And so that's this higher signal here. And so if you see an IR spectrum and you see these two really intense signals, we're talking about the carbonyl."}, {"video_title": "Symmetric and asymmetric stretching Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We could have this one stretching and this one contracting here for our spring. It takes more energy to do an asymmetric stretch. And so that's this higher signal here. And so if you see an IR spectrum and you see these two really intense signals, we're talking about the carbonyl. So right in here, this is our double bond region. We're talking about the carbonyl stretch with a large dipole moment, so a large change in dipole moment when it stretches. So that's why we see such an intense signal here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you look at the red and blue protons, they're both attached to this carbon. And if we see this double bond here, we have these different groups attached to this double bond. And since there's no rotation around the double bond, the red and the blue protons are locked into different environments. Therefore, they are not chemically equivalent. And since those protons are not equivalent, they can couple together. And since this is occurring on the same carbon, we call this geminal coupling. So geminal coupling here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, they are not chemically equivalent. And since those protons are not equivalent, they can couple together. And since this is occurring on the same carbon, we call this geminal coupling. So geminal coupling here. So geminal referring to the fact that both protons are in the same carbon. And coupling can occur. So those protons are close enough where they can affect each other."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So geminal coupling here. So geminal referring to the fact that both protons are in the same carbon. And coupling can occur. So those protons are close enough where they can affect each other. So let's think about first the NMR spectrum with no coupling. So we would expect one signal for the blue proton and one signal for the red proton. So here's the spectrum with no coupling."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So those protons are close enough where they can affect each other. So let's think about first the NMR spectrum with no coupling. So we would expect one signal for the blue proton and one signal for the red proton. So here's the spectrum with no coupling. But we know that the red proton's magnetic moment can align either with the external magnetic field or against the external magnetic field. And that causes the signal for the blue proton to be split into two. So if I go down here, so we actually see a doublet for the signal for the blue proton."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's the spectrum with no coupling. But we know that the red proton's magnetic moment can align either with the external magnetic field or against the external magnetic field. And that causes the signal for the blue proton to be split into two. So if I go down here, so we actually see a doublet for the signal for the blue proton. Same thing for the blue proton. The magnetic moment can be aligned either with the external magnetic field or against it. And that splits the signal for the red proton into a doublet."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if I go down here, so we actually see a doublet for the signal for the blue proton. Same thing for the blue proton. The magnetic moment can be aligned either with the external magnetic field or against it. And that splits the signal for the red proton into a doublet. So two peaks for the signal for the red proton. I went into much more detail about this in the spin-spin splitting, spin-spin coupling video. In this video, we're more concerned with the idea of the coupling constant."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that splits the signal for the red proton into a doublet. So two peaks for the signal for the red proton. I went into much more detail about this in the spin-spin splitting, spin-spin coupling video. In this video, we're more concerned with the idea of the coupling constant. And the coupling constant refers to the distance between the peaks of a signal. So if you think about the distance between the two peaks of the signal, that is the coupling constant. And the coupling constant is the same for both of these signals because these protons are splitting each other."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we're more concerned with the idea of the coupling constant. And the coupling constant refers to the distance between the peaks of a signal. So if you think about the distance between the two peaks of the signal, that is the coupling constant. And the coupling constant is the same for both of these signals because these protons are splitting each other. They are coupled together. The coupling constant is measured in hertz. So it turns out to be 1.4 hertz."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the coupling constant is the same for both of these signals because these protons are splitting each other. They are coupled together. The coupling constant is measured in hertz. So it turns out to be 1.4 hertz. And if it's 1.4 hertz for this one, it must be 1.4 hertz for this one because those protons are coupled together. The reason why we use hertz is because it's the same coupling constant no matter what NMR spectrometer you're using. So it doesn't matter what the operating frequency is."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it turns out to be 1.4 hertz. And if it's 1.4 hertz for this one, it must be 1.4 hertz for this one because those protons are coupled together. The reason why we use hertz is because it's the same coupling constant no matter what NMR spectrometer you're using. So it doesn't matter what the operating frequency is. You're going to get the same coupling constant. If we look at the actual NMR spectrum, over here is a zoom in of the actual NMR spectrum. The signal for the red proton is right here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it doesn't matter what the operating frequency is. You're going to get the same coupling constant. If we look at the actual NMR spectrum, over here is a zoom in of the actual NMR spectrum. The signal for the red proton is right here. And the signal for the blue proton is over here. So when I looked at the spectrum with interaction, the spectrum with coupling between the protons, we just assumed that the heights of these two peaks were the same. But if I look at the actual NMR spectrum, they're not quite the same."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The signal for the red proton is right here. And the signal for the blue proton is over here. So when I looked at the spectrum with interaction, the spectrum with coupling between the protons, we just assumed that the heights of these two peaks were the same. But if I look at the actual NMR spectrum, they're not quite the same. So this one right here is a little bit higher. And if you draw an arrow pointing towards the higher peak, that arrow points towards the signal of the proton that's causing the splitting. So that arrow is pointing to the right."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But if I look at the actual NMR spectrum, they're not quite the same. So this one right here is a little bit higher. And if you draw an arrow pointing towards the higher peak, that arrow points towards the signal of the proton that's causing the splitting. So that arrow is pointing to the right. And that's where we find the signal for the red proton, which is causing the splitting of the blue proton. So the doublet points towards the proton with which it is coupled. And the same thing for this signal."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that arrow is pointing to the right. And that's where we find the signal for the red proton, which is causing the splitting of the blue proton. So the doublet points towards the proton with which it is coupled. And the same thing for this signal. So this peak's a little bit higher. So we draw an arrow pointing towards the higher peak. And so the doublet points toward the proton with which it is coupled."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the same thing for this signal. So this peak's a little bit higher. So we draw an arrow pointing towards the higher peak. And so the doublet points toward the proton with which it is coupled. And so you get this situation where you get these doublets with a roof over their head. So if you can imagine this roof over them like that. So sometimes you'll see this on NMR spectrum."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the doublet points toward the proton with which it is coupled. And so you get this situation where you get these doublets with a roof over their head. So if you can imagine this roof over them like that. So sometimes you'll see this on NMR spectrum. And if you think about that they're pointing towards the proton with which it is coupled, sometimes it can help you when you're trying to understand what's going on in your NMR spectrum. All right, let's look at another example for a coupling constant. So let's look at this molecule."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So sometimes you'll see this on NMR spectrum. And if you think about that they're pointing towards the proton with which it is coupled, sometimes it can help you when you're trying to understand what's going on in your NMR spectrum. All right, let's look at another example for a coupling constant. So let's look at this molecule. And let's focus in on the ethyl group. So over here, this carbon has two protons. We expect one signal for those protons."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this molecule. And let's focus in on the ethyl group. So over here, this carbon has two protons. We expect one signal for those protons. And then over here, this carbon has three protons. So we would expect another signal for these protons. Let's focus in on the protons in blue."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We expect one signal for those protons. And then over here, this carbon has three protons. So we would expect another signal for these protons. Let's focus in on the protons in blue. So how many neighboring protons do we have? Well, those protons in blue are attached to this carbon. The next door carbon is this one."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's focus in on the protons in blue. So how many neighboring protons do we have? Well, those protons in blue are attached to this carbon. The next door carbon is this one. So how many neighbors? One, two, three. So three neighboring protons."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The next door carbon is this one. So how many neighbors? One, two, three. So three neighboring protons. So n is equal to 3. And using the n plus 1 rule, we expect n plus 1 peaks. So 3 plus 1 is equal to 4."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So three neighboring protons. So n is equal to 3. And using the n plus 1 rule, we expect n plus 1 peaks. So 3 plus 1 is equal to 4. So we would expect a signal with four peaks. So we'd expect a quartet. So let me go ahead and draw that down here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 3 plus 1 is equal to 4. So we would expect a signal with four peaks. So we'd expect a quartet. So let me go ahead and draw that down here. So we would expect a quartet for that signal. So this is supposed to represent what you would see on an NMR spectrum. Next, let's do the protons in red here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that down here. So we would expect a quartet for that signal. So this is supposed to represent what you would see on an NMR spectrum. Next, let's do the protons in red here. So how many neighboring protons do they have? Well, they're all attached to this carbon. And the next door carbon is here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's do the protons in red here. So how many neighboring protons do they have? Well, they're all attached to this carbon. And the next door carbon is here. And we have two protons on the next door carbon. So we have two neighbors. So n is equal to 2."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the next door carbon is here. And we have two protons on the next door carbon. So we have two neighbors. So n is equal to 2. And so we expect 2 plus 1 peaks. So three peaks or a triplet. So let me see if I can draw in a triplet here."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So n is equal to 2. And so we expect 2 plus 1 peaks. So three peaks or a triplet. So let me see if I can draw in a triplet here. So this would be the signal for these protons. Since the red and the blue protons are splitting each other, the coupling constant should be the same. So the distance between the peaks should be the same."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me see if I can draw in a triplet here. So this would be the signal for these protons. Since the red and the blue protons are splitting each other, the coupling constant should be the same. So the distance between the peaks should be the same. So it turns out to be 7 hertz. So this distance should be 7 hertz. Same with this distance."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the distance between the peaks should be the same. So it turns out to be 7 hertz. So this distance should be 7 hertz. Same with this distance. So we just have to pretend like they're all equivalent here. And the same for this one. So a coupling constant of 7 hertz."}, {"video_title": "Coupling constant Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Same with this distance. So we just have to pretend like they're all equivalent here. And the same for this one. So a coupling constant of 7 hertz. Same for this signal. So this distance should be 7 hertz. And also for this one."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Well, we start with the longest carbon chain. So there are seven carbons in my longest carbon chain. So I would call this heptane. And I number it to give the substituent the lowest number possible. So in this example, it doesn't really matter if I start from the left or from the right. In both examples, you would end up with a 4 for your substituent there. Now, this substituent looks different from ones we've seen before."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And I number it to give the substituent the lowest number possible. So in this example, it doesn't really matter if I start from the left or from the right. In both examples, you would end up with a 4 for your substituent there. Now, this substituent looks different from ones we've seen before. There are three carbons in it, but those carbons are not in a straight chain alkyl group. So if I look at it, there are three carbons, but they're not going in a straight chain. They're branching of branching here."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Now, this substituent looks different from ones we've seen before. There are three carbons in it, but those carbons are not in a straight chain alkyl group. So if I look at it, there are three carbons, but they're not going in a straight chain. They're branching of branching here. So this is kind of weird. How do we name this substituent? Well, down here, I have the same substituent."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "They're branching of branching here. So this is kind of weird. How do we name this substituent? Well, down here, I have the same substituent. I'm going to draw this little zigzag line to indicate that that substituent is coming off of some straight chain alkane. And when you're naming a complex substituent like this, you actually use the same rules that you would use for a straight chain alkane. So you first identify the longest carbon chain, which in this case is only two carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Well, down here, I have the same substituent. I'm going to draw this little zigzag line to indicate that that substituent is coming off of some straight chain alkane. And when you're naming a complex substituent like this, you actually use the same rules that you would use for a straight chain alkane. So you first identify the longest carbon chain, which in this case is only two carbons. So that would be an ethyl group coming off my carbon chain. So I'm going to go ahead and name that as an ethyl group. I'm going to go ahead and number it to give my branching group there the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So you first identify the longest carbon chain, which in this case is only two carbons. So that would be an ethyl group coming off my carbon chain. So I'm going to go ahead and name that as an ethyl group. I'm going to go ahead and number it to give my branching group there the lowest number possible. So I go 1 and 2. So what is my substituent coming off of my ethyl group? Well, that's a methyl group coming off of carbon 1."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and number it to give my branching group there the lowest number possible. So I go 1 and 2. So what is my substituent coming off of my ethyl group? Well, that's a methyl group coming off of carbon 1. So I name it as 1-methyl ethyl. So now that complex substituent is named as 1-methyl ethyl. So I could go ahead and put that into my name."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Well, that's a methyl group coming off of carbon 1. So I name it as 1-methyl ethyl. So now that complex substituent is named as 1-methyl ethyl. So I could go ahead and put that into my name. So coming off of carbon 4, I have 1-methyl ethyl. And I'm going to put that in parentheses. And all of that is coming off of carbon 4 for my molecule."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So I could go ahead and put that into my name. So coming off of carbon 4, I have 1-methyl ethyl. And I'm going to put that in parentheses. And all of that is coming off of carbon 4 for my molecule. So 4-1-methyl ethyl heptane would be an acceptable IUPAC way of naming that molecule. So if you're naming your complex substituent as 1-methyl ethyl, that's the official IUPAC way. But there are also common names for these complex substituents."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And all of that is coming off of carbon 4 for my molecule. So 4-1-methyl ethyl heptane would be an acceptable IUPAC way of naming that molecule. So if you're naming your complex substituent as 1-methyl ethyl, that's the official IUPAC way. But there are also common names for these complex substituents. So the common name for 1-methyl ethyl is isopropyl. So isopropyl is the common name. And isopropyl is used so frequently that it's perfectly acceptable to use isopropyl for the name of this molecule as well."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "But there are also common names for these complex substituents. So the common name for 1-methyl ethyl is isopropyl. So isopropyl is the common name. And isopropyl is used so frequently that it's perfectly acceptable to use isopropyl for the name of this molecule as well. So you could have said, oh, this is 4-isopropyl heptane. And you would have been absolutely correct. So that's yet another IUPAC name."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And isopropyl is used so frequently that it's perfectly acceptable to use isopropyl for the name of this molecule as well. So you could have said, oh, this is 4-isopropyl heptane. And you would have been absolutely correct. So that's yet another IUPAC name. So iso means same. And it probably comes from the fact that you have these two methyl groups giving you this y shape that are the same. So that's one complex substituent, one that has three carbons on it."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So that's yet another IUPAC name. So iso means same. And it probably comes from the fact that you have these two methyl groups giving you this y shape that are the same. So that's one complex substituent, one that has three carbons on it. Let's look at a bunch of complex substituents that have a total of four carbons on them. So all of these guys have a total of four carbons. And let's do that same trick with the zigzag line so we can ignore the rest of the molecule and just think about them as being alkyl groups."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So that's one complex substituent, one that has three carbons on it. Let's look at a bunch of complex substituents that have a total of four carbons on them. So all of these guys have a total of four carbons. And let's do that same trick with the zigzag line so we can ignore the rest of the molecule and just think about them as being alkyl groups. So how do I name these? The same steps. You find your longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And let's do that same trick with the zigzag line so we can ignore the rest of the molecule and just think about them as being alkyl groups. So how do I name these? The same steps. You find your longest carbon chain. So for this one, the longest carbon chain would be 3. It's an alkyl group, so it's propyl. And when you number it, this would get a 1, this would get a 2, and this would get a 3."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "You find your longest carbon chain. So for this one, the longest carbon chain would be 3. It's an alkyl group, so it's propyl. And when you number it, this would get a 1, this would get a 2, and this would get a 3. So you have a methyl group coming off of carbon 1 here. So it would be 1-methylpropyl as the name of that complex substituent. The common name for that is secbutyl."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And when you number it, this would get a 1, this would get a 2, and this would get a 3. So you have a methyl group coming off of carbon 1 here. So it would be 1-methylpropyl as the name of that complex substituent. The common name for that is secbutyl. So butyl because there are a total of four carbons. Let's do the next one. So 1, 2, 3 again."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "The common name for that is secbutyl. So butyl because there are a total of four carbons. Let's do the next one. So 1, 2, 3 again. So when I number it 1, 2, 3, I can see that it would be propyl once again. So I go ahead and write propyl here. And what is coming off of that group?"}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3 again. So when I number it 1, 2, 3, I can see that it would be propyl once again. So I go ahead and write propyl here. And what is coming off of that group? Well, I have a methyl group coming off of carbon 2 this time. So it would be 2-methylpropyl. So 2-methylpropyl for this complex substituent."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And what is coming off of that group? Well, I have a methyl group coming off of carbon 2 this time. So it would be 2-methylpropyl. So 2-methylpropyl for this complex substituent. The common name for this is isobutyl. So butyl again because there are four total carbons in this complex substituent. Iso because once again, you have these two methyl groups."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So 2-methylpropyl for this complex substituent. The common name for this is isobutyl. So butyl again because there are four total carbons in this complex substituent. Iso because once again, you have these two methyl groups. So they're like the same. So you get that Y formation. So that's isobutyl."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Iso because once again, you have these two methyl groups. So they're like the same. So you get that Y formation. So that's isobutyl. The next one, longest carbon chain. There are two carbons in my longest carbon chain. So that would be ethyl."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So that's isobutyl. The next one, longest carbon chain. There are two carbons in my longest carbon chain. So that would be ethyl. And when I number my longest carbon chain, I can see that I have two methyl groups. And each of those methyl groups is coming off of carbon 1. So I would say this would be 1, 1-dimethyl ethyl."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So that would be ethyl. And when I number my longest carbon chain, I can see that I have two methyl groups. And each of those methyl groups is coming off of carbon 1. So I would say this would be 1, 1-dimethyl ethyl. So 1, 1-dimethyl ethyl would be the IUPAC name for this complex substituent. You will also see tert-butyl. So tert-butyl is probably used even more frequently."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So I would say this would be 1, 1-dimethyl ethyl. So 1, 1-dimethyl ethyl would be the IUPAC name for this complex substituent. You will also see tert-butyl. So tert-butyl is probably used even more frequently. So tert-butyl, again butyl because there are a total of four carbons here. So those are the three possibilities for a complex substituent with a total of four carbons. Let's look at just a few of the possibilities for complex substituents that have five carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So tert-butyl is probably used even more frequently. So tert-butyl, again butyl because there are a total of four carbons here. So those are the three possibilities for a complex substituent with a total of four carbons. Let's look at just a few of the possibilities for complex substituents that have five carbons. There are actually much more than this, but these are the ones that are most commonly used. So let's just focus in on these two. So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight chain alkane."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at just a few of the possibilities for complex substituents that have five carbons. There are actually much more than this, but these are the ones that are most commonly used. So let's just focus in on these two. So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight chain alkane. And once again, we find our longest carbon chain, 1, 2, 3, 4. So that would be butyl. And when I number that carbon chain, 1, 2, 3, 4, I can see that I have a methyl group coming off of carbon 3."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we'll draw our zigzag line to represent the fact that this is actually connected to some straight chain alkane. And once again, we find our longest carbon chain, 1, 2, 3, 4. So that would be butyl. And when I number that carbon chain, 1, 2, 3, 4, I can see that I have a methyl group coming off of carbon 3. So it would be 3-methylbutyl for the IUPAC name. And this is also called isopentyl. So you could say isopentyl since there are five carbons now."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And when I number that carbon chain, 1, 2, 3, 4, I can see that I have a methyl group coming off of carbon 3. So it would be 3-methylbutyl for the IUPAC name. And this is also called isopentyl. So you could say isopentyl since there are five carbons now. And iso because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So you could say isopentyl since there are five carbons now. And iso because again, you have this methyl group and this methyl group looking like a Y. They're like the same thing. Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right, longest carbon chain? 1, 2, 3."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Or I've seen this called isoamyl before. So isoamyl or isopentyl are acceptable IUPAC names as well. What about this one on the right, longest carbon chain? 1, 2, 3. So that would be propyl. And numbering it, 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2-2-dimethylpropyl, otherwise known as neopentyl since once again, you have five carbons for these."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3. So that would be propyl. And numbering it, 1, 2, 3, immediately it is obvious that you have two methyl groups coming off of carbon 2. So it would be 2-2-dimethylpropyl, otherwise known as neopentyl since once again, you have five carbons for these. So again, there are many more. And we'll stop with those. And so that gives you an idea about how to approach naming complex substituents."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So it would be 2-2-dimethylpropyl, otherwise known as neopentyl since once again, you have five carbons for these. So again, there are many more. And we'll stop with those. And so that gives you an idea about how to approach naming complex substituents. And of course, when you name complex substituents, you have to use them when you're naming straight chain or cycloalkane molecules. So let's look at a cycloalkane molecule. And let's see how to name this guy."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And so that gives you an idea about how to approach naming complex substituents. And of course, when you name complex substituents, you have to use them when you're naming straight chain or cycloalkane molecules. So let's look at a cycloalkane molecule. And let's see how to name this guy. Well, I have four carbons in my ring. And I have four carbons in this group. So tie goes to the cycloalkane."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And let's see how to name this guy. Well, I have four carbons in my ring. And I have four carbons in this group. So tie goes to the cycloalkane. So remember from the last video, if you have an equal number of carbons in your ring as with your chain, you're going to name it as an alkyl cycloalkane. The cycloalkane wins the tie. So there are four carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So tie goes to the cycloalkane. So remember from the last video, if you have an equal number of carbons in your ring as with your chain, you're going to name it as an alkyl cycloalkane. The cycloalkane wins the tie. So there are four carbons. So this would be cyclobutane. So let's go ahead and write cyclobutane here. And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So there are four carbons. So this would be cyclobutane. So let's go ahead and write cyclobutane here. And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3. So that would be propyl. And then when you number that complex substituent, 1, 2, 3, obviously there is a methyl group coming off of carbon 1. So you would write 1-methylpropyl."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And once you've determined that you're going to name it as a cycloalkane, then you have to look at this complex substituent and say, OK, well, that's 1, 2, 3. So that would be propyl. And then when you number that complex substituent, 1, 2, 3, obviously there is a methyl group coming off of carbon 1. So you would write 1-methylpropyl. So 1-methylpropyl. And if you wanted to, you could identify that 1-methylpropyl is coming off of carbon 1 of your cyclobutane. So you could put this in parentheses and write 1-methylpropylcyclobutane."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So you would write 1-methylpropyl. So 1-methylpropyl. And if you wanted to, you could identify that 1-methylpropyl is coming off of carbon 1 of your cyclobutane. So you could put this in parentheses and write 1-methylpropylcyclobutane. Or you could just leave the 1 off and say 1-methylpropylcyclobutane, because again, it is implied. What is the common name for this complex substituent? So 1-methylpropyl, we go back up here, and we find 1-methylpropyl was also called secbutyl."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So you could put this in parentheses and write 1-methylpropylcyclobutane. Or you could just leave the 1 off and say 1-methylpropylcyclobutane, because again, it is implied. What is the common name for this complex substituent? So 1-methylpropyl, we go back up here, and we find 1-methylpropyl was also called secbutyl. So we could have also have named this molecule secbutylcyclobutane. So let's go ahead and write that. So secbutylcyclobutane is a perfectly acceptable IUPAC name as well."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So 1-methylpropyl, we go back up here, and we find 1-methylpropyl was also called secbutyl. So we could have also have named this molecule secbutylcyclobutane. So let's go ahead and write that. So secbutylcyclobutane is a perfectly acceptable IUPAC name as well. So it just depends. Do you want to do it the official IUPAC way, or do you want to memorize some of these common names? Let's do one more thing in this video."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So secbutylcyclobutane is a perfectly acceptable IUPAC name as well. So it just depends. Do you want to do it the official IUPAC way, or do you want to memorize some of these common names? Let's do one more thing in this video. Let's classify carbons. So we're going to do something called classification of carbons. This is a topic that comes up over and over again throughout an organic chemistry course."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more thing in this video. Let's classify carbons. So we're going to do something called classification of carbons. This is a topic that comes up over and over again throughout an organic chemistry course. So the earlier you learn this concept, the better off you are. So classification of carbons. So let's say I have carbon connected to three hydrogens, and then I also have it connected to one other carbon in some R group."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "This is a topic that comes up over and over again throughout an organic chemistry course. So the earlier you learn this concept, the better off you are. So classification of carbons. So let's say I have carbon connected to three hydrogens, and then I also have it connected to one other carbon in some R group. I want to know how to classify this carbon. So this carbon is connected to one other carbon, so therefore we say it is primary. So that is a primary carbon right there."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So let's say I have carbon connected to three hydrogens, and then I also have it connected to one other carbon in some R group. I want to know how to classify this carbon. So this carbon is connected to one other carbon, so therefore we say it is primary. So that is a primary carbon right there. Let's take off one of the hydrogens, and let's put on another R group. So I'll make it R prime to distinguish it from the first R group. So this time, if I wanted to know the classification of this carbon, it's connected to two other carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So that is a primary carbon right there. Let's take off one of the hydrogens, and let's put on another R group. So I'll make it R prime to distinguish it from the first R group. So this time, if I wanted to know the classification of this carbon, it's connected to two other carbons. So it is said to be secondary. So it is a secondary carbon like that. And I'm going to, once again, take off one of the hydrogens."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So this time, if I wanted to know the classification of this carbon, it's connected to two other carbons. So it is said to be secondary. So it is a secondary carbon like that. And I'm going to, once again, take off one of the hydrogens. So I'll make it an R double prime group. And now, if I wanted to classify my central carbon, now this is connected to one, two, three other carbons. So it is said to be tertiary."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to, once again, take off one of the hydrogens. So I'll make it an R double prime group. And now, if I wanted to classify my central carbon, now this is connected to one, two, three other carbons. So it is said to be tertiary. So that is a tertiary carbon like that. And finally, I have one more example, of course. I take off the last hydrogen."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So it is said to be tertiary. So that is a tertiary carbon like that. And finally, I have one more example, of course. I take off the last hydrogen. So now I have R R prime, R double prime, and R triple prime. So what is the classification of this carbon now connected to four other carbons? So it is said to be quaternary."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "I take off the last hydrogen. So now I have R R prime, R double prime, and R triple prime. So what is the classification of this carbon now connected to four other carbons? So it is said to be quaternary. So that is a quaternary carbon right here. So quaternary. All right, so if I'm trying to think about where some of these common names come from, I can see, oh, well, right here I have carbons bonded to two other carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So it is said to be quaternary. So that is a quaternary carbon right here. So quaternary. All right, so if I'm trying to think about where some of these common names come from, I can see, oh, well, right here I have carbons bonded to two other carbons. Well, that would be secondary, right? So SEC for my prefix. So let's go back up here and let's see if we can find those examples."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "All right, so if I'm trying to think about where some of these common names come from, I can see, oh, well, right here I have carbons bonded to two other carbons. Well, that would be secondary, right? So SEC for my prefix. So let's go back up here and let's see if we can find those examples. So here I have this carbon bonded to two other carbons. So this carbon was said to be secondary. So I think that's where this comes from."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So let's go back up here and let's see if we can find those examples. So here I have this carbon bonded to two other carbons. So this carbon was said to be secondary. So I think that's where this comes from. I've never seen that explained in a textbook or anywhere, but it just makes sense. So it's ignoring the fact that this carbon is actually attached to a ring. It's saying this carbon on my complex substituent is bonded to two other carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So I think that's where this comes from. I've never seen that explained in a textbook or anywhere, but it just makes sense. So it's ignoring the fact that this carbon is actually attached to a ring. It's saying this carbon on my complex substituent is bonded to two other carbons. So it is secondary on that complex substituent. What about tertiary? So carbon bonded to three other carbons is said to be tertiary."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "It's saying this carbon on my complex substituent is bonded to two other carbons. So it is secondary on that complex substituent. What about tertiary? So carbon bonded to three other carbons is said to be tertiary. So if I go back up here again, I can say, well, that would make sense because if I look at this carbon, it's bonded to three other carbons. So I could say that is a tertiary carbon. And once again, I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So carbon bonded to three other carbons is said to be tertiary. So if I go back up here again, I can say, well, that would make sense because if I look at this carbon, it's bonded to three other carbons. So I could say that is a tertiary carbon. And once again, I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring. So if you just look at the complex substituent, that carbon is said to be tertiary, which I think is where the name comes from. Let's do one more example of assigning classification of carbons to this molecule. So let's look at this carbon right here."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "And once again, I'm ignoring the fact that this carbon is actually bonded to another carbon on the ring. So if you just look at the complex substituent, that carbon is said to be tertiary, which I think is where the name comes from. Let's do one more example of assigning classification of carbons to this molecule. So let's look at this carbon right here. This carbon is bonded to one other carbon and three hydrogens. So this carbon is said to be primary. This carbon right here is bonded to two other carbons."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this carbon right here. This carbon is bonded to one other carbon and three hydrogens. So this carbon is said to be primary. This carbon right here is bonded to two other carbons. So it is said to be secondary. This carbon right here is bonded to three other carbons, so it is tertiary. This carbon is bond to one other carbon, so it is primary."}, {"video_title": "Alkane and cycloalkane nomenclature III Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here is bonded to two other carbons. So it is said to be secondary. This carbon right here is bonded to three other carbons, so it is tertiary. This carbon is bond to one other carbon, so it is primary. This carbon is bonded to three other carbons, so it is tertiary. And all the carbons on the ring right here are bonded to two other carbons, so they are all said to be secondary. So that's a very important skill to develop, classifying your carbons."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Just so we get a little practice of naming. Let's see, this is one, two, three, four carbons. So it has but- as a prefix. No double bonds or triple bonds, so it is butane. And we have a chloro group here. So if we start numbering at the side closest to it, one, two, so it's two chloro butane. So let's think about what might happen here."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "No double bonds or triple bonds, so it is butane. And we have a chloro group here. So if we start numbering at the side closest to it, one, two, so it's two chloro butane. So let's think about what might happen here. So the first thing, let me just redraw the molecule right there. The first thing you need to realize is this sodium methoxide, it is a salt. When it's not dissolved, it's made up of a positive sodium cation, let me draw them right here."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about what might happen here. So the first thing, let me just redraw the molecule right there. The first thing you need to realize is this sodium methoxide, it is a salt. When it's not dissolved, it's made up of a positive sodium cation, let me draw them right here. A positive sodium cation and a negative methoxide anion. So let me draw the methoxide part right here. So a negative methoxide anion."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "When it's not dissolved, it's made up of a positive sodium cation, let me draw them right here. A positive sodium cation and a negative methoxide anion. So let me draw the methoxide part right here. So a negative methoxide anion. And then it has, so normally it would have just two pairs, but now we have three pairs here. The oxygen has an extra electron. The oxygen has an extra electron."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So a negative methoxide anion. And then it has, so normally it would have just two pairs, but now we have three pairs here. The oxygen has an extra electron. The oxygen has an extra electron. Actually, let me draw another electron as the extra electron. This will be useful for our mechanism. So it could be any of them."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen has an extra electron. Actually, let me draw another electron as the extra electron. This will be useful for our mechanism. So it could be any of them. But the oxygen has an extra electron, it has a negative charge, in solid form when they're not dissolved, they form an ionic bond and they form a crystal-like structure. It's a salt. But when you dissolve it in something, in a solution, in this case we have methanol as the solution, they will dissociate from each other."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it could be any of them. But the oxygen has an extra electron, it has a negative charge, in solid form when they're not dissolved, they form an ionic bond and they form a crystal-like structure. It's a salt. But when you dissolve it in something, in a solution, in this case we have methanol as the solution, they will dissociate from each other. And what you have right here is this methoxide. This is a very, very, very strong base. So this right here is a strong base."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But when you dissolve it in something, in a solution, in this case we have methanol as the solution, they will dissociate from each other. And what you have right here is this methoxide. This is a very, very, very strong base. So this right here is a strong base. And if you use the Lewis definition of a base, that means it really, really, really wants to give away this electron to something else. If you use the Bronsted-Lowry definition, this means that it really, really, really wants to take a proton off of something else. And in this situation, that is exactly what it will do."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this right here is a strong base. And if you use the Lewis definition of a base, that means it really, really, really wants to give away this electron to something else. If you use the Bronsted-Lowry definition, this means that it really, really, really wants to take a proton off of something else. And in this situation, that is exactly what it will do. And I'll just give you one. I'll actually give you the most likely reaction to occur here. And we'll talk about other reactions and why this is the most likely reaction in future videos."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And in this situation, that is exactly what it will do. And I'll just give you one. I'll actually give you the most likely reaction to occur here. And we'll talk about other reactions and why this is the most likely reaction in future videos. So it wants to nab a proton. It is a strong base. It wants to give away this electron."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And we'll talk about other reactions and why this is the most likely reaction in future videos. So it wants to nab a proton. It is a strong base. It wants to give away this electron. So let's say that it gives away this electron to this hydrogen right over here. Now, this hydrogen already had an electron. If it's getting an electron from the methoxide anion, the methoxide base, then it can give away its electron to the rest of the molecule."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It wants to give away this electron. So let's say that it gives away this electron to this hydrogen right over here. Now, this hydrogen already had an electron. If it's getting an electron from the methoxide anion, the methoxide base, then it can give away its electron to the rest of the molecule. So now it can give away this electron to the rest of the molecule. Now, carbon won't need the electron. Carbon doesn't want to have a negative charge."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "If it's getting an electron from the methoxide anion, the methoxide base, then it can give away its electron to the rest of the molecule. So now it can give away this electron to the rest of the molecule. Now, carbon won't need the electron. Carbon doesn't want to have a negative charge. Maybe simultaneously, that electron goes to that carbon right over there. But once again, this carbon doesn't want it. It only has four bonds."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Carbon doesn't want to have a negative charge. Maybe simultaneously, that electron goes to that carbon right over there. But once again, this carbon doesn't want it. It only has four bonds. But what we see is we have this chloro here. This is a highly electronegative, I guess you could call it a group right now. The chlorine is very electronegative."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It only has four bonds. But what we see is we have this chloro here. This is a highly electronegative, I guess you could call it a group right now. The chlorine is very electronegative. So the whole time, the chlorine was already tugging on this electron right here. So now, all of a sudden, this is all happening simultaneously. When an electron becomes available to this carbon, then now this chlorine, this carbon doesn't need this electron anymore."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The chlorine is very electronegative. So the whole time, the chlorine was already tugging on this electron right here. So now, all of a sudden, this is all happening simultaneously. When an electron becomes available to this carbon, then now this chlorine, this carbon doesn't need this electron anymore. The chlorine already wanted it. So now the chlorine can take the electron. And just like that, in exactly one step, let's think about what happened."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "When an electron becomes available to this carbon, then now this chlorine, this carbon doesn't need this electron anymore. The chlorine already wanted it. So now the chlorine can take the electron. And just like that, in exactly one step, let's think about what happened. If all of this whole chain reaction, and I shouldn't call it a chain reaction, this simultaneous reaction occurred, what are we left with? So let me redraw my 2-chlorobutane. But we have to change it now."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And just like that, in exactly one step, let's think about what happened. If all of this whole chain reaction, and I shouldn't call it a chain reaction, this simultaneous reaction occurred, what are we left with? So let me redraw my 2-chlorobutane. But we have to change it now. So now, the chlorine, this has disappeared. So let me clear it. That's disappeared."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But we have to change it now. So now, the chlorine, this has disappeared. So let me clear it. That's disappeared. The chlorine has now left. This chlorine is now up here. It is left."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "That's disappeared. The chlorine has now left. This chlorine is now up here. It is left. It had this electron right over there, which is right over there. And then the electron it was paired with that was forming a bond is now also on the chlorine. So now it becomes a chloride anion."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It is left. It had this electron right over there, which is right over there. And then the electron it was paired with that was forming a bond is now also on the chlorine. So now it becomes a chloride anion. So now it is a chloride anion. If I want to draw the rest of the valence electrons, 1, 2, 3, 4, 5, 6, and it had the seventh one right here. 1, 2, 3, 4, 5, 6, 7, 8 now."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So now it becomes a chloride anion. So now it is a chloride anion. If I want to draw the rest of the valence electrons, 1, 2, 3, 4, 5, 6, and it had the seventh one right here. 1, 2, 3, 4, 5, 6, 7, 8 now. So it now has a negative charge. Now, this electron up here, let me clear this part out as well. So let me clear that out as well."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7, 8 now. So it now has a negative charge. Now, this electron up here, let me clear this part out as well. So let me clear that out as well. Now, this electron right here, this magenta electron, this is now given to this carbon. So let me draw it here. So that magenta electron is now given to that carbon."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So let me clear that out as well. Now, this electron right here, this magenta electron, this is now given to this carbon. So let me draw it here. So that magenta electron is now given to that carbon. And if we look at the other end, the one that was paired with, that was bonded with, it still is bonded with it. So that green electron is now still on that carbon, and now they are bonded. And now they're forming a double bond."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So that magenta electron is now given to that carbon. And if we look at the other end, the one that was paired with, that was bonded with, it still is bonded with it. So that green electron is now still on that carbon, and now they are bonded. And now they're forming a double bond. This all of a sudden has become an alkene. And now the methoxide took the hydrogen. So let me redraw the methoxide."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And now they're forming a double bond. This all of a sudden has become an alkene. And now the methoxide took the hydrogen. So let me redraw the methoxide. So OCH3, the oxygen had 1, 2, 3, 4, 5, it had 6, well, actually it had 7 valence electrons, one with the carbon and then all of these 6 unpaired ones. Neutral oxygen has 6. But now it gave one of them away to a hydrogen."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So let me redraw the methoxide. So OCH3, the oxygen had 1, 2, 3, 4, 5, it had 6, well, actually it had 7 valence electrons, one with the carbon and then all of these 6 unpaired ones. Neutral oxygen has 6. But now it gave one of them away to a hydrogen. It gave that green electron over there to the hydrogen. I'll make this hydrogen the same color. To this hydrogen right over here."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But now it gave one of them away to a hydrogen. It gave that green electron over there to the hydrogen. I'll make this hydrogen the same color. To this hydrogen right over here. So now this hydrogen is now bonded with it. So what are we left with? We have a chloride anion."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "To this hydrogen right over here. So now this hydrogen is now bonded with it. So what are we left with? We have a chloride anion. We have a chloride anion. So this is chloride anion. We now have methanol."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We have a chloride anion. We have a chloride anion. So this is chloride anion. We now have methanol. This was a strong base. Now it has become its conjugate acid. So now we have methanol, which is the same as the solvent, same as solution."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We now have methanol. This was a strong base. Now it has become its conjugate acid. So now we have methanol, which is the same as the solvent, same as solution. So it's now mixed in. And now we're left off with 1, 2, 3, 4, still 4, still we still have 4 carbons. But now it's an alkene."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So now we have methanol, which is the same as the solvent, same as solution. So it's now mixed in. And now we're left off with 1, 2, 3, 4, still 4, still we still have 4 carbons. But now it's an alkene. We have a double bond. So we could call this butyrene or sometimes called 2-butene. So let's think about what happened here."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But now it's an alkene. We have a double bond. So we could call this butyrene or sometimes called 2-butene. So let's think about what happened here. It all happened simultaneously. Both reactants were present. There was actually only one step."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about what happened here. It all happened simultaneously. Both reactants were present. There was actually only one step. So this is the rate determining step. So if we were to try to name it, we'd probably have a 2 in it someplace. And something got eliminated."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "There was actually only one step. So this is the rate determining step. So if we were to try to name it, we'd probably have a 2 in it someplace. And something got eliminated. The chloride, or I guess you could call it the chloro group, got eliminated. It left. It was a leaving group in this situation."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And something got eliminated. The chloride, or I guess you could call it the chloro group, got eliminated. It left. It was a leaving group in this situation. So this was eliminated. And this type of reaction, where something is eliminated and both of the reactants are participating in the rate determining step, and we only had one step here, so that was the rate determining step, is called an E2 reaction. And once again, E stands for elimination."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It was a leaving group in this situation. So this was eliminated. And this type of reaction, where something is eliminated and both of the reactants are participating in the rate determining step, and we only had one step here, so that was the rate determining step, is called an E2 reaction. And once again, E stands for elimination. And the 2 stands for that both reactants involved in the rate determining step. We had one reactant and two reactants. They were both involved in the rate determining step."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And once again, E stands for elimination. And the 2 stands for that both reactants involved in the rate determining step. We had one reactant and two reactants. They were both involved in the rate determining step. E2, just like SN2. I mean, it's obviously not the same reaction, but SN2 had substitution, with leaving group left, nucleophile came. It was substitution using a nucleophile, so that's the S and the N. And then the 2 was we had both reactants."}, {"video_title": "E2 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "They were both involved in the rate determining step. E2, just like SN2. I mean, it's obviously not the same reaction, but SN2 had substitution, with leaving group left, nucleophile came. It was substitution using a nucleophile, so that's the S and the N. And then the 2 was we had both reactants. It was all happening simultaneously. I'll leave you there. We're going to go into more detail on the different types of E2 reactions, which ones are favored, and then we'll talk a little bit about E1."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "If we start with an aldehyde or a ketone and we add an excess of alcohol in an acidic environment, we can form our acetal. We talked about ways to increase the formation of your acetal by removing, in this case, water from your reaction to drive the equilibrium to the right. But let's say you increase the concentration of water. So let's say you increase the concentration of this. You would push the equilibrium to the left and you would hydrolyze your acetal and convert it back into your original aldehyde or ketone and your alcohol. So this can actually be a useful reaction if you want to use an acetal as a protecting group. And so let's look at an example of hydrolyzing an acetal."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's say you increase the concentration of this. You would push the equilibrium to the left and you would hydrolyze your acetal and convert it back into your original aldehyde or ketone and your alcohol. So this can actually be a useful reaction if you want to use an acetal as a protecting group. And so let's look at an example of hydrolyzing an acetal. So if we have this acetal over here on the left, which is the same one that we made in the previous video, let's see how to analyze the products that we would make by hydrolyzing it. So we add HCl as our acid catalyst and we add an excess of water. And so we're going to get back our original aldehyde and alcohol here."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so let's look at an example of hydrolyzing an acetal. So if we have this acetal over here on the left, which is the same one that we made in the previous video, let's see how to analyze the products that we would make by hydrolyzing it. So we add HCl as our acid catalyst and we add an excess of water. And so we're going to get back our original aldehyde and alcohol here. And so let's analyze the structure of this. So this carbon right here is the important one. So let me go ahead and change color here."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to get back our original aldehyde and alcohol here. And so let's analyze the structure of this. So this carbon right here is the important one. So let me go ahead and change color here. So this carbon right here in magenta, we go back up to here, that's this carbon we're talking about here. And so we know that we have a carbon attached to that. So that would be this R group right here."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and change color here. So this carbon right here in magenta, we go back up to here, that's this carbon we're talking about here. And so we know that we have a carbon attached to that. So that would be this R group right here. And then we also have a hydrogen coming off of that carbon. So that would be this hydrogen right here. And so immediately, we can work backwards and see we have an R group, we have a hydrogen."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So that would be this R group right here. And then we also have a hydrogen coming off of that carbon. So that would be this hydrogen right here. And so immediately, we can work backwards and see we have an R group, we have a hydrogen. We're dealing with an aldehyde right here, specifically a two-carbon aldehyde. So there's two carbons on this. So acetaldehyde is one of our products."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so immediately, we can work backwards and see we have an R group, we have a hydrogen. We're dealing with an aldehyde right here, specifically a two-carbon aldehyde. So there's two carbons on this. So acetaldehyde is one of our products. So let's go ahead and draw acetaldehyde here, a two-carbon aldehyde. And then our other product we know is going to be an alcohol. So if we analyze the structure of our acetal, we should be able to figure out what kind of alcohol it is."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So acetaldehyde is one of our products. So let's go ahead and draw acetaldehyde here, a two-carbon aldehyde. And then our other product we know is going to be an alcohol. So if we analyze the structure of our acetal, we should be able to figure out what kind of alcohol it is. So if we look at that, that's like this portion right here, which we know came from our alcohol. So all we have to do is add on a hydrogen, and then we have the structure of our alcohol product here. So it would be a four-carbon alcohol."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So if we analyze the structure of our acetal, we should be able to figure out what kind of alcohol it is. So if we look at that, that's like this portion right here, which we know came from our alcohol. So all we have to do is add on a hydrogen, and then we have the structure of our alcohol product here. So it would be a four-carbon alcohol. So one, two, three, and four. And the exact same thing down here. So our other product would be butanol."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So it would be a four-carbon alcohol. So one, two, three, and four. And the exact same thing down here. So our other product would be butanol. And we have two equivalents of it. So that would be one, two, three, four carbons. So by looking at our acetal and thinking about hydrolyzing it and thinking about where those portions came from, we can easily come up with the products of this hydrolysis of this acetal."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So our other product would be butanol. And we have two equivalents of it. So that would be one, two, three, four carbons. So by looking at our acetal and thinking about hydrolyzing it and thinking about where those portions came from, we can easily come up with the products of this hydrolysis of this acetal. So let's see how we could use this reaction if we were trying to synthesize this molecule over here on the right. And so if we're trying to synthesize this molecule from this molecule, first you might be able to think you could figure it out by looking at the functional groups. So we have a ketone over here on the left, and we want a ketone over here for our product."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So by looking at our acetal and thinking about hydrolyzing it and thinking about where those portions came from, we can easily come up with the products of this hydrolysis of this acetal. So let's see how we could use this reaction if we were trying to synthesize this molecule over here on the right. And so if we're trying to synthesize this molecule from this molecule, first you might be able to think you could figure it out by looking at the functional groups. So we have a ketone over here on the left, and we want a ketone over here for our product. We also have a carboxylic acid over here on the left, and then we have an alcohol over here on the right. So if you're thinking about how to complete this kind of transformation, you might think to reduce the carboxylic acid to form your alcohol. And so you could do that with something like lithium aluminum hydride."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So we have a ketone over here on the left, and we want a ketone over here for our product. We also have a carboxylic acid over here on the left, and then we have an alcohol over here on the right. So if you're thinking about how to complete this kind of transformation, you might think to reduce the carboxylic acid to form your alcohol. And so you could do that with something like lithium aluminum hydride. So you might think you could just add some lithium aluminum hydride here in the first step. And in the second step, add a source of protons to protonate your alkoxide anion to form your alcohol as your product. The only problem with trying to do this in one step is lithium aluminum hydride is also going to reduce your ketone here to form a secondary alcohol."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so you could do that with something like lithium aluminum hydride. So you might think you could just add some lithium aluminum hydride here in the first step. And in the second step, add a source of protons to protonate your alkoxide anion to form your alcohol as your product. The only problem with trying to do this in one step is lithium aluminum hydride is also going to reduce your ketone here to form a secondary alcohol. And so this will not work. The first thing you have to do is protect the ketone, and then you can use your lithium aluminum hydride. So we're going to protect our ketone using an acetal, of course, and we're going to react it with ethylene glycol."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "The only problem with trying to do this in one step is lithium aluminum hydride is also going to reduce your ketone here to form a secondary alcohol. And so this will not work. The first thing you have to do is protect the ketone, and then you can use your lithium aluminum hydride. So we're going to protect our ketone using an acetal, of course, and we're going to react it with ethylene glycol. So if we react our starting compound here with ethylene glycol and we use an acid catalyst, we're going to form an acetal. So the ethylene glycol is going to react with the ketone portion of the molecule to form an acetal, specifically a cyclic acetal. So over here on the left, we still have our carboxylic acid."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to protect our ketone using an acetal, of course, and we're going to react it with ethylene glycol. So if we react our starting compound here with ethylene glycol and we use an acid catalyst, we're going to form an acetal. So the ethylene glycol is going to react with the ketone portion of the molecule to form an acetal, specifically a cyclic acetal. So over here on the left, we still have our carboxylic acid. And over here on the right, now we're going to have an oxygen bonded to this carbon, another oxygen bonded to this carbon, and they are, of course, connected. And so that's our cyclic acetal, which came from this oxygen, this carbon, this carbon, and this oxygen. So that's this oxygen, this carbon, this carbon, and this oxygen."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the left, we still have our carboxylic acid. And over here on the right, now we're going to have an oxygen bonded to this carbon, another oxygen bonded to this carbon, and they are, of course, connected. And so that's our cyclic acetal, which came from this oxygen, this carbon, this carbon, and this oxygen. So that's this oxygen, this carbon, this carbon, and this oxygen. So acetals are stable in basic conditions. And so now we can add our lithium aluminum hydride. So we add our lithium aluminum hydride here in the first step."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So that's this oxygen, this carbon, this carbon, and this oxygen. So acetals are stable in basic conditions. And so now we can add our lithium aluminum hydride. So we add our lithium aluminum hydride here in the first step. And that is going to reduce our carboxylic acid. And so in the second step, we can add a source of protons and we can add some excess water. And so we can protonate the alkoxide anions to form our alcohol product."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So we add our lithium aluminum hydride here in the first step. And that is going to reduce our carboxylic acid. And so in the second step, we can add a source of protons and we can add some excess water. And so we can protonate the alkoxide anions to form our alcohol product. And also, we can add an excess of water in the acidic environment is going to hydrolyze our acetal. So we're also going to hydrolyze our cyclic acetal here and get back our original ketone. And so we've now been able to make our desired product."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so we can protonate the alkoxide anions to form our alcohol product. And also, we can add an excess of water in the acidic environment is going to hydrolyze our acetal. So we're also going to hydrolyze our cyclic acetal here and get back our original ketone. And so we've now been able to make our desired product. And so that's one way to use an acetal as a protecting group. Let's look at another type of reaction here. So this one's formation of a thioacetal."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so we've now been able to make our desired product. And so that's one way to use an acetal as a protecting group. Let's look at another type of reaction here. So this one's formation of a thioacetal. So this is completely analogous to the formation of an acetal. So you start off with an aldehyde or a ketone. And this time, instead of using an alcohol, you're going to use a thiol."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So this one's formation of a thioacetal. So this is completely analogous to the formation of an acetal. So you start off with an aldehyde or a ketone. And this time, instead of using an alcohol, you're going to use a thiol. So instead of having an oxygen here, you have a sulfur. So you need acidic conditions. And you can even use something like boron trifluoride here, which can function as a Lewis acid catalyst for this reaction."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And this time, instead of using an alcohol, you're going to use a thiol. So instead of having an oxygen here, you have a sulfur. So you need acidic conditions. And you can even use something like boron trifluoride here, which can function as a Lewis acid catalyst for this reaction. And you form a thioacetal as your product. So once again, you have your R double prime group coming from your thiol. And then you have a sulfur here instead of an oxygen."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And you can even use something like boron trifluoride here, which can function as a Lewis acid catalyst for this reaction. And you form a thioacetal as your product. So once again, you have your R double prime group coming from your thiol. And then you have a sulfur here instead of an oxygen. And then you have two of these to form your thioacetal. And so one reason you might want to form a thioacetal instead of an acetal is thioacetals have an additional reaction that they undergo. And we can use it in this transformation."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And then you have a sulfur here instead of an oxygen. And then you have two of these to form your thioacetal. And so one reason you might want to form a thioacetal instead of an acetal is thioacetals have an additional reaction that they undergo. And we can use it in this transformation. So let's say our goal was to go from the product on the left to the product on the right. And so you can see that we have reduced our ketone here to form this cyclohexane portion. So reducing your ketone."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And we can use it in this transformation. So let's say our goal was to go from the product on the left to the product on the right. And so you can see that we have reduced our ketone here to form this cyclohexane portion. So reducing your ketone. So the first thing we could do to do this transformation is to form a thioacetal. So it's going to be, again, analogous to the formation of an acetal. This time, we're going to use, instead of a diol, we're going to use something with two SH groups instead here."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So reducing your ketone. So the first thing we could do to do this transformation is to form a thioacetal. So it's going to be, again, analogous to the formation of an acetal. This time, we're going to use, instead of a diol, we're going to use something with two SH groups instead here. So a dithiol here. So we can use an acidic environment here. And we're going to form a cyclic thioacetal."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "This time, we're going to use, instead of a diol, we're going to use something with two SH groups instead here. So a dithiol here. So we can use an acidic environment here. And we're going to form a cyclic thioacetal. So very similar to what we did before. The ring is here. The carboxylic acid is going to be untouched."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to form a cyclic thioacetal. So very similar to what we did before. The ring is here. The carboxylic acid is going to be untouched. And instead of an oxygen directly bonded to our ring, we're going to have a sulfur. And the same thing on this side, a sulfur, and then our two carbons. So once again, let's follow those atoms here."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "The carboxylic acid is going to be untouched. And instead of an oxygen directly bonded to our ring, we're going to have a sulfur. And the same thing on this side, a sulfur, and then our two carbons. So once again, let's follow those atoms here. So a sulfur, two carbons, two carbons, and a sulfur. So here's your sulfur, two carbons, two carbons, and then another sulfur. So a cyclic thioacetal."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So once again, let's follow those atoms here. So a sulfur, two carbons, two carbons, and a sulfur. So here's your sulfur, two carbons, two carbons, and then another sulfur. So a cyclic thioacetal. And then there's actually a reaction for converting this compound into our target compound. And this involves a special kind of nickel. So if we use rainy nickel here, it's a special kind of finely divided nickel that has already adsorbed some hydrogens."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So a cyclic thioacetal. And then there's actually a reaction for converting this compound into our target compound. And this involves a special kind of nickel. So if we use rainy nickel here, it's a special kind of finely divided nickel that has already adsorbed some hydrogens. And what it's going to do is add some hydrogens to this carbon right here. So we're going to add two hydrogens to this carbon. And you can see that we have reduced the thioacetal and we have formed our desired product."}, {"video_title": "Acetals as protecting groups and thioacetals Organic chemistry Khan Academy.mp3", "Sentence": "So if we use rainy nickel here, it's a special kind of finely divided nickel that has already adsorbed some hydrogens. And what it's going to do is add some hydrogens to this carbon right here. So we're going to add two hydrogens to this carbon. And you can see that we have reduced the thioacetal and we have formed our desired product. So this is another way to reduce a ketone here to this alkane portion of the molecule. And so there are other ways to do it. And this just gives you another tool that you can use for synthesis problems."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's really interesting about the simulation, besides the fact that it's just mesmerizing and extremely cool, is it shows the particles collide. Once they get to a certain critical mass, you see that they get colored yellow, maybe to indicate that they are now a star. Fusion can now occur. And you can zoom in at different levels to really see how the different particles or the different masses are interacting. And then you can actually rotate that to see a little bit clearer. This is if I'm looking kind of right on top of it to see how they're interacting. And it's a three-dimensional simulation, so it's a very rich way of thinking about these."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you can zoom in at different levels to really see how the different particles or the different masses are interacting. And then you can actually rotate that to see a little bit clearer. This is if I'm looking kind of right on top of it to see how they're interacting. And it's a three-dimensional simulation, so it's a very rich way of thinking about these. And what's exciting for me is it's highly dependent on what the initial conditions are. In an earlier version of Peter's simulation, he did not give a net angular momentum to the system. And so you did not have as much of kind of the planet satellite or as much of the disk structures forming, although right here we don't have too much of a disk structure, although there does seem to be a dominant plane in this scenario."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's a three-dimensional simulation, so it's a very rich way of thinking about these. And what's exciting for me is it's highly dependent on what the initial conditions are. In an earlier version of Peter's simulation, he did not give a net angular momentum to the system. And so you did not have as much of kind of the planet satellite or as much of the disk structures forming, although right here we don't have too much of a disk structure, although there does seem to be a dominant plane in this scenario. And what's exciting is here we have a binary system. Sometimes you restart it. You might not have a binary system."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you did not have as much of kind of the planet satellite or as much of the disk structures forming, although right here we don't have too much of a disk structure, although there does seem to be a dominant plane in this scenario. And what's exciting is here we have a binary system. Sometimes you restart it. You might not have a binary system. Depending on the initial conditions, you might have something that starts to look like the Milky Way. Sometimes you might have something that looks very different than the Milky Way. And it really gives us clues of why we see such diversity, especially when we're looking at galaxies, the structure of galaxies, that it's highly dependent on initial conditions."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You might not have a binary system. Depending on the initial conditions, you might have something that starts to look like the Milky Way. Sometimes you might have something that looks very different than the Milky Way. And it really gives us clues of why we see such diversity, especially when we're looking at galaxies, the structure of galaxies, that it's highly dependent on initial conditions. One can argue that our own solar system did have some net initial angular momentum because the current theory, what really catalyzed it was a nearby supernova that sent a shock wave and allowed the dust that would form our solar system to reach a critical mass and start to condense into the sun and the planets. And so this isn't, at least in my mind, too unrealistic of a scenario. And it's really cool to look at."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it really gives us clues of why we see such diversity, especially when we're looking at galaxies, the structure of galaxies, that it's highly dependent on initial conditions. One can argue that our own solar system did have some net initial angular momentum because the current theory, what really catalyzed it was a nearby supernova that sent a shock wave and allowed the dust that would form our solar system to reach a critical mass and start to condense into the sun and the planets. And so this isn't, at least in my mind, too unrealistic of a scenario. And it's really cool to look at. And it really gives you a sense of things. You already see you have a binary star. They're kind of orbiting around each other or orbiting around the center of mass, which kind of looks like around each other."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's really cool to look at. And it really gives you a sense of things. You already see you have a binary star. They're kind of orbiting around each other or orbiting around the center of mass, which kind of looks like around each other. And then this star right over here has its own kind of captive planet that is just rotating around it. We can see it a little bit clearer. If we had a very, at least from this perspective, a very close range, we can zoom in a little bit more to see it a little bit better."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're kind of orbiting around each other or orbiting around the center of mass, which kind of looks like around each other. And then this star right over here has its own kind of captive planet that is just rotating around it. We can see it a little bit clearer. If we had a very, at least from this perspective, a very close range, we can zoom in a little bit more to see it a little bit better. This has a satellite, but then they're also kind of dancing around each other. So it's a really fascinating simulation. I could really stare at this and play with it for days."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we had a very, at least from this perspective, a very close range, we can zoom in a little bit more to see it a little bit better. This has a satellite, but then they're also kind of dancing around each other. So it's a really fascinating simulation. I could really stare at this and play with it for days. And I encourage you to play with it, restart it, see how the initial conditions or what type of solar systems or galaxies you might end up with, whether they form disks, whether you have binary systems or not, whether you have planets with satellites. And then if you are more advanced, actually play with the code and see if you can really change the initial conditions, the starting velocities of things, the number of particles of things, the distribution of mass that you start off with, the angular momentum that you start off with, and see how that might change the structure of the universes that you create. And I'm going to add an annotation to this video that links directly to this simulation."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I could really stare at this and play with it for days. And I encourage you to play with it, restart it, see how the initial conditions or what type of solar systems or galaxies you might end up with, whether they form disks, whether you have binary systems or not, whether you have planets with satellites. And then if you are more advanced, actually play with the code and see if you can really change the initial conditions, the starting velocities of things, the number of particles of things, the distribution of mass that you start off with, the angular momentum that you start off with, and see how that might change the structure of the universes that you create. And I'm going to add an annotation to this video that links directly to this simulation. And I'll also put the link inside of the description. So have fun. I could literally spend hours with this."}, {"video_title": "Accreting mass due to gravity simulation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm going to add an annotation to this video that links directly to this simulation. And I'll also put the link inside of the description. So have fun. I could literally spend hours with this. It's a fascinating, fascinating module that he's created. We zoom in and out. And I really thank Peter Kohlingridge for this incredible contribution."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This 1.5 to 3 times is the mass once it's shed off a lot of the, I guess, outer material of the star, and this is really the mass of the remnant of the star, kind of the core of the star. But that remnant, once it stops fusing, once it stops having outward pressure, once it has enough density, this, we saw in the last video, will cause a supernova, it will cause a shockwave to move out through the rest of the material and essentially cause it to blow up, and this will condense into a neutron star. Now in this video, what I want to talk about is what if we're starting with a star that has a mass more than, this is give or take, we don't know the actual firm boundaries here, but what if we have a star that is more than 20 times the mass of the sun, and this is kind of the original mass before the star burns itself out. Or, when that star has kind of reached this old age, once it has that iron core, it has more than, so I could say the remnant, the dense remnant has more than 3 to 4 times the mass of the sun. And remember, it's going to have 3 to 4 times the mass of the sun, but it's going to be far denser, it's just going to be a core, it's going to be an iron-nickel core that's no longer fusing. So what happens to these stars? So it turns out that these are so massive that even the neutron degeneracy pressure will not be enough to keep the mass from imploding."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or, when that star has kind of reached this old age, once it has that iron core, it has more than, so I could say the remnant, the dense remnant has more than 3 to 4 times the mass of the sun. And remember, it's going to have 3 to 4 times the mass of the sun, but it's going to be far denser, it's just going to be a core, it's going to be an iron-nickel core that's no longer fusing. So what happens to these stars? So it turns out that these are so massive that even the neutron degeneracy pressure will not be enough to keep the mass from imploding. And these stars, all of the mass in these stars, will just keep imploding. So in the neutron, so we imagine the first, in kind of sun-like stars, things would collapse into white dwarfs. So maybe I should draw that in white."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it turns out that these are so massive that even the neutron degeneracy pressure will not be enough to keep the mass from imploding. And these stars, all of the mass in these stars, will just keep imploding. So in the neutron, so we imagine the first, in kind of sun-like stars, things would collapse into white dwarfs. So maybe I should draw that in white. So they would collapse into white dwarfs. No, that's not white either. There you go."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So maybe I should draw that in white. So they would collapse into white dwarfs. No, that's not white either. There you go. They would collapse into white dwarfs eventually. So this is a white dwarf. And here, the pressure that's keeping this from collapsing further is electron degeneracy pressure."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There you go. They would collapse into white dwarfs eventually. So this is a white dwarf. And here, the pressure that's keeping this from collapsing further is electron degeneracy pressure. The atoms are squeezed so much that the electrons are essentially keeping them from squeezing anymore. But if the pressure gets large enough, then you have the neutron star. So you have even more mass and even a smaller, and I'm not drawing this to scale, neutron stars are tiny."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And here, the pressure that's keeping this from collapsing further is electron degeneracy pressure. The atoms are squeezed so much that the electrons are essentially keeping them from squeezing anymore. But if the pressure gets large enough, then you have the neutron star. So you have even more mass and even a smaller, and I'm not drawing this to scale, neutron stars are tiny. White dwarf stars are on the scale of an Earth-like planet. Neutron stars, we learned in the last video, are on the scale of a city. So these are super dense, super tiny, and this has more mass than this over here."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have even more mass and even a smaller, and I'm not drawing this to scale, neutron stars are tiny. White dwarf stars are on the scale of an Earth-like planet. Neutron stars, we learned in the last video, are on the scale of a city. So these are super dense, super tiny, and this has more mass than this over here. In fact, maybe I should just draw it as a dot, just so you have a sense of how dense it is. It's really just like one big atomic nucleus. Well, it's still small, but it's the size of a city."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So these are super dense, super tiny, and this has more mass than this over here. In fact, maybe I should just draw it as a dot, just so you have a sense of how dense it is. It's really just like one big atomic nucleus. Well, it's still small, but it's the size of a city. It's like a nucleus the size of a city. But this right here is a neutron star. And what's unintuitive about what I'm drawing is each of these smaller things have more mass."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, it's still small, but it's the size of a city. It's like a nucleus the size of a city. But this right here is a neutron star. And what's unintuitive about what I'm drawing is each of these smaller things have more mass. This overcame the electron degeneracy pressure to collapse even further. But if the mass is large enough, and this is what we're talking about in this video, even the neutron degeneracy pressure will not be able to keep that mass from collapsing. And there's even theoretical quark stars where the quark degeneracy pressure, but if you get even beyond that, it all collapses into a single point, and I'm simplifying here, but it collapses into a single point of infinite density."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's unintuitive about what I'm drawing is each of these smaller things have more mass. This overcame the electron degeneracy pressure to collapse even further. But if the mass is large enough, and this is what we're talking about in this video, even the neutron degeneracy pressure will not be able to keep that mass from collapsing. And there's even theoretical quark stars where the quark degeneracy pressure, but if you get even beyond that, it all collapses into a single point, and I'm simplifying here, but it collapses into a single point of infinite density. Infinite mass density. And this is really the mass of a black hole. And I'm calling it the mass of a black hole because there's different ways how you can view where a black hole starts and ends or what exactly is the black hole."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there's even theoretical quark stars where the quark degeneracy pressure, but if you get even beyond that, it all collapses into a single point, and I'm simplifying here, but it collapses into a single point of infinite density. Infinite mass density. And this is really the mass of a black hole. And I'm calling it the mass of a black hole because there's different ways how you can view where a black hole starts and ends or what exactly is the black hole. So this is all the mass of the black hole. Or we could say of the original star. So when we're talking about that remnant being 3 to 4 solar masses, all of that mass is now being contained."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm calling it the mass of a black hole because there's different ways how you can view where a black hole starts and ends or what exactly is the black hole. So this is all the mass of the black hole. Or we could say of the original star. So when we're talking about that remnant being 3 to 4 solar masses, all of that mass is now being contained. Well, not all of it. Some of it was released as energy during the supernova, and that was also true of the neutron star. But most of that mass is now being contained in this infinitely small point."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when we're talking about that remnant being 3 to 4 solar masses, all of that mass is now being contained. Well, not all of it. Some of it was released as energy during the supernova, and that was also true of the neutron star. But most of that mass is now being contained in this infinitely small point. And you'll hear physicists and mathematicians talk about singularities. And singularities are really points, even in mathematics, where everything breaks down, where nothing starts to make sense anymore, where the mathematical equations don't give you a defined answer. And this is a singularity because you have a ton of mass in an infinitely small space."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But most of that mass is now being contained in this infinitely small point. And you'll hear physicists and mathematicians talk about singularities. And singularities are really points, even in mathematics, where everything breaks down, where nothing starts to make sense anymore, where the mathematical equations don't give you a defined answer. And this is a singularity because you have a ton of mass in an infinitely small space. You essentially have an infinite density right here. And this is hard to visualize, but you have kind of an infinite curvature in space-time right here, and I can't visualize that. Maybe we'll think about that in more videos."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is a singularity because you have a ton of mass in an infinitely small space. You essentially have an infinite density right here. And this is hard to visualize, but you have kind of an infinite curvature in space-time right here, and I can't visualize that. Maybe we'll think about that in more videos. But the reason why I said that there's different ways to think about where a black hole is or where it starts and ends is that this is where the mass is. And if there was any other mass that was over here, it would obviously be attracted to this mass and then become part of that singularity. It would add to that mass, that already huge mass that's in an infinitely small point in space."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe we'll think about that in more videos. But the reason why I said that there's different ways to think about where a black hole is or where it starts and ends is that this is where the mass is. And if there was any other mass that was over here, it would obviously be attracted to this mass and then become part of that singularity. It would add to that mass, that already huge mass that's in an infinitely small point in space. But the reason why the boundary is hard to define is because there's some point in space around that singularity at which no matter what that thing is, no matter how much energy that thing has, it will not be able to escape the gravitational influence of the black hole, of that ultra-dense mass. So even if it was electromagnetic radiation, even if it was light, and even if it's light that's shown away from the mass, it will eventually have to go back. It will not be able to escape the gravitational influence."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would add to that mass, that already huge mass that's in an infinitely small point in space. But the reason why the boundary is hard to define is because there's some point in space around that singularity at which no matter what that thing is, no matter how much energy that thing has, it will not be able to escape the gravitational influence of the black hole, of that ultra-dense mass. So even if it was electromagnetic radiation, even if it was light, and even if it's light that's shown away from the mass, it will eventually have to go back. It will not be able to escape the gravitational influence. And so the boundary, where if you're within that boundary, that's really a sphere, so that boundary around the singularity, where if you're within the boundary, no matter what you do, no matter if you're electromagnetic radiation, you're still going to, you're never going to be able to escape the black hole. If you are beyond that boundary, you might be able to escape the black hole. So this guy could escape."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It will not be able to escape the gravitational influence. And so the boundary, where if you're within that boundary, that's really a sphere, so that boundary around the singularity, where if you're within the boundary, no matter what you do, no matter if you're electromagnetic radiation, you're still going to, you're never going to be able to escape the black hole. If you are beyond that boundary, you might be able to escape the black hole. So this guy could escape. This guy over here, no matter what he does, is going to have to go back into the black hole. This boundary right here is called the event horizon. Another word used in a lot of science fiction movies, and for good reason, because it's fascinating."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this guy could escape. This guy over here, no matter what he does, is going to have to go back into the black hole. This boundary right here is called the event horizon. Another word used in a lot of science fiction movies, and for good reason, because it's fascinating. We'll actually learn in future videos, hopefully about Hawking radiation, we'll see that that is not radiation from the black hole itself. It's the byproduct of quantum effects that are occurring at the event horizon. But the event horizon is just this kind of point in space, or this sphere in space, or this boundary in space."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Another word used in a lot of science fiction movies, and for good reason, because it's fascinating. We'll actually learn in future videos, hopefully about Hawking radiation, we'll see that that is not radiation from the black hole itself. It's the byproduct of quantum effects that are occurring at the event horizon. But the event horizon is just this kind of point in space, or this sphere in space, or this boundary in space. Anything closer than or within the event horizon has to eventually end up in the singularity, contributing to that mass. Anything on the outside has a chance of escaping. So what would a black hole look like?"}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the event horizon is just this kind of point in space, or this sphere in space, or this boundary in space. Anything closer than or within the event horizon has to eventually end up in the singularity, contributing to that mass. Anything on the outside has a chance of escaping. So what would a black hole look like? Well, not even light can escape from it, so it will be black. It will be black in the purest sense. It will not emit any type of radiation from the black hole itself, from that mass."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what would a black hole look like? Well, not even light can escape from it, so it will be black. It will be black in the purest sense. It will not emit any type of radiation from the black hole itself, from that mass. So here are some depictions I got from NASA of black holes. So just to be clear, what's happening here, what you're seeing here is black. You can view that as the black hole."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It will not emit any type of radiation from the black hole itself, from that mass. So here are some depictions I got from NASA of black holes. So just to be clear, what's happening here, what you're seeing here is black. You can view that as the black hole. When people talk about the black hole, that's often what they're talking about. But there's a point of infinite density at the center of this black sphere right here. And what you see is that black sphere, that really is the boundary of the event horizon."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can view that as the black hole. When people talk about the black hole, that's often what they're talking about. But there's a point of infinite density at the center of this black sphere right here. And what you see is that black sphere, that really is the boundary of the event horizon. So this right here is the boundary of the event horizon. And what we're seeing right here is the accretion disk around the black hole. As all of this matter gets closer and closer to it, it's being squeezed more and more, it's moving faster and faster, and getting hotter and hotter."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what you see is that black sphere, that really is the boundary of the event horizon. So this right here is the boundary of the event horizon. And what we're seeing right here is the accretion disk around the black hole. As all of this matter gets closer and closer to it, it's being squeezed more and more, it's moving faster and faster, and getting hotter and hotter. And that's why the way this art is depicted, it looks like this stuff over here is redder and hotter than the stuff further out. It's just accelerating as it approaches that event horizon. Once it's in the event horizon, we cannot even see the light that it's emitting, even though it's starting to become unbelievably energetic."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As all of this matter gets closer and closer to it, it's being squeezed more and more, it's moving faster and faster, and getting hotter and hotter. And that's why the way this art is depicted, it looks like this stuff over here is redder and hotter than the stuff further out. It's just accelerating as it approaches that event horizon. Once it's in the event horizon, we cannot even see the light that it's emitting, even though it's starting to become unbelievably energetic. Here's some other pictures. This is a picture of a star being ripped apart. Not a picture, this is actually an artist's depiction."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once it's in the event horizon, we cannot even see the light that it's emitting, even though it's starting to become unbelievably energetic. Here's some other pictures. This is a picture of a star being ripped apart. Not a picture, this is actually an artist's depiction. We never were able to get such good pictures of actual action occurring near black holes. These are artist's depictions. But this is a star being ripped apart by a black hole."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Not a picture, this is actually an artist's depiction. We never were able to get such good pictures of actual action occurring near black holes. These are artist's depictions. But this is a star being ripped apart by a black hole. So this star is getting pretty close to this black hole. Already out here, where the star is, it's very strong gravitational attraction. So any mass that's being emitted from the star in that direction is slowly being pulled into the black hole."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is a star being ripped apart by a black hole. So this star is getting pretty close to this black hole. Already out here, where the star is, it's very strong gravitational attraction. So any mass that's being emitted from the star in that direction is slowly being pulled into the black hole. So the star is kind of being ripped apart by the black hole. This is maybe a better depiction of it. This is the star at first, and once it becomes under the influence of the black hole's gravitation, it starts to kind of elongate and gets ripped apart, and its matter starts spiraling in closer and closer to that black hole."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So any mass that's being emitted from the star in that direction is slowly being pulled into the black hole. So the star is kind of being ripped apart by the black hole. This is maybe a better depiction of it. This is the star at first, and once it becomes under the influence of the black hole's gravitation, it starts to kind of elongate and gets ripped apart, and its matter starts spiraling in closer and closer to that black hole. Once it's in the event horizon, we won't even see it anymore, because even the light from that matter, that intensely hot matter that's entering into the black hole, cannot even escape the black hole itself. Anyway, hopefully you found that interesting. And I want to be clear."}, {"video_title": "Black holes Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the star at first, and once it becomes under the influence of the black hole's gravitation, it starts to kind of elongate and gets ripped apart, and its matter starts spiraling in closer and closer to that black hole. Once it's in the event horizon, we won't even see it anymore, because even the light from that matter, that intensely hot matter that's entering into the black hole, cannot even escape the black hole itself. Anyway, hopefully you found that interesting. And I want to be clear. We still don't understand a lot about black holes. In fact, this whole notion of a singularity, the fact that all the math and all the theory breaks down at the singularity is a pretty good sign that our theory isn't complete. Because if our theory is complete, we would maybe get something a little bit more sensical than just all of our equations not making sense at that infinitely dense point."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we're gonna think about doing things in reverse. So we're gonna start with a retroaldol reaction and see how that way of thinking can apply to retrosynthesis. And so let's start with cinnamaldehyde right here. And if we do a retroaldol reaction, the mechanism is pretty much the exact reverse of an aldol condensation. And so if we had a base like sodium carbonate, we're gonna form benzaldehyde and acetaldehyde. Cinnamaldehyde is the molecule that gives cinnamon its smell and benzaldehyde smells like almonds. And so this is a pretty cool experiment to do."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if we do a retroaldol reaction, the mechanism is pretty much the exact reverse of an aldol condensation. And so if we had a base like sodium carbonate, we're gonna form benzaldehyde and acetaldehyde. Cinnamaldehyde is the molecule that gives cinnamon its smell and benzaldehyde smells like almonds. And so this is a pretty cool experiment to do. You can start with a molecule that smells like cinnamon and end with a molecule that smells like almonds. Let's analyze our cinnamaldehyde starting compound here to see how we would form those products. We know the carbon next to our carbonyl is our alpha carbon, and we know the carbon next to that is the beta carbon."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this is a pretty cool experiment to do. You can start with a molecule that smells like cinnamon and end with a molecule that smells like almonds. Let's analyze our cinnamaldehyde starting compound here to see how we would form those products. We know the carbon next to our carbonyl is our alpha carbon, and we know the carbon next to that is the beta carbon. And so looking at the structure, we know that there's a hydrogen attached to our alpha carbon like that. So if we think about breaking this double bond, we can see this two carbon set up over here on the right for acetaldehyde. So those are the two carbons, and then this is the hydrogen that's bonded to that alpha carbon there."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know the carbon next to our carbonyl is our alpha carbon, and we know the carbon next to that is the beta carbon. And so looking at the structure, we know that there's a hydrogen attached to our alpha carbon like that. So if we think about breaking this double bond, we can see this two carbon set up over here on the right for acetaldehyde. So those are the two carbons, and then this is the hydrogen that's bonded to that alpha carbon there. Looking at the structure again on the right, we know that there are two more hydrogens bonded to that alpha carbon. And so therefore, if we're doing things in reverse, we can think about adding two hydrogens to this alpha carbon. So let me go ahead and draw two hydrogens right here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So those are the two carbons, and then this is the hydrogen that's bonded to that alpha carbon there. Looking at the structure again on the right, we know that there are two more hydrogens bonded to that alpha carbon. And so therefore, if we're doing things in reverse, we can think about adding two hydrogens to this alpha carbon. So let me go ahead and draw two hydrogens right here. So I'm just gonna write H2. And then we can think about adding an oxygen to the beta carbon. So we can think about adding H2O."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw two hydrogens right here. So I'm just gonna write H2. And then we can think about adding an oxygen to the beta carbon. So we can think about adding H2O. I know there's already a hydrogen on that beta carbon, so I could think about this being the hydrogen on that carbon. And then if I add an oxygen to that carbon, that would be this oxygen right here. So just a way of thinking in reverse, adding water, breaking your double bond, and adding an oxygen and two hydrogens is one way of thinking about forming your product here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can think about adding H2O. I know there's already a hydrogen on that beta carbon, so I could think about this being the hydrogen on that carbon. And then if I add an oxygen to that carbon, that would be this oxygen right here. So just a way of thinking in reverse, adding water, breaking your double bond, and adding an oxygen and two hydrogens is one way of thinking about forming your product here. Or you could think about the reverse. You could think about losing water from this portion and sticking these two fragments together to give you your aldol condensation product. So once again, this way of thinking can be very useful if you're trying to retro-synthesize something."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So just a way of thinking in reverse, adding water, breaking your double bond, and adding an oxygen and two hydrogens is one way of thinking about forming your product here. Or you could think about the reverse. You could think about losing water from this portion and sticking these two fragments together to give you your aldol condensation product. So once again, this way of thinking can be very useful if you're trying to retro-synthesize something. So if you're thinking in reverse. So for example, let's say a question on a test was, show how you could synthesize this enone here. So once again, we're gonna do the same kind of analysis."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So once again, this way of thinking can be very useful if you're trying to retro-synthesize something. So if you're thinking in reverse. So for example, let's say a question on a test was, show how you could synthesize this enone here. So once again, we're gonna do the same kind of analysis. We're gonna find our alpha carbon right here. So this is our alpha carbon, and this is our beta carbon. All right, so attached to our alpha carbon, we know that there's a hydrogen right here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we're gonna do the same kind of analysis. We're gonna find our alpha carbon right here. So this is our alpha carbon, and this is our beta carbon. All right, so attached to our alpha carbon, we know that there's a hydrogen right here. So we're gonna think about breaking this double bond. So let's go ahead and draw a retro-synthesis arrow here. So we're gonna break that double bond."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, so attached to our alpha carbon, we know that there's a hydrogen right here. So we're gonna think about breaking this double bond. So let's go ahead and draw a retro-synthesis arrow here. So we're gonna break that double bond. So on the left, we're gonna have our benzene ring. We're gonna have our carbonyl here. And then we're gonna have a hydrogen attached to this carbon, attached to our alpha carbon."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we're gonna break that double bond. So on the left, we're gonna have our benzene ring. We're gonna have our carbonyl here. And then we're gonna have a hydrogen attached to this carbon, attached to our alpha carbon. And remember, we're gonna add two hydrogens to our alpha carbon. So we're gonna add two hydrogens to our alpha carbon, and that would give us the ketone that we would need. All right, for our beta carbon, we know there's already a hydrogen attached right here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we're gonna have a hydrogen attached to this carbon, attached to our alpha carbon. And remember, we're gonna add two hydrogens to our alpha carbon. So we're gonna add two hydrogens to our alpha carbon, and that would give us the ketone that we would need. All right, for our beta carbon, we know there's already a hydrogen attached right here. And so we're going to add an oxygen to this beta carbon on the left, so it's just our way of thinking about it. So over here on the right, we're gonna have our carbonyl. All right, and then we're adding an oxygen."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, for our beta carbon, we know there's already a hydrogen attached right here. And so we're going to add an oxygen to this beta carbon on the left, so it's just our way of thinking about it. So over here on the right, we're gonna have our carbonyl. All right, and then we're adding an oxygen. All right, we're adding an oxygen right here. And then we had that hydrogen, so I'm gonna go ahead and draw in that hydrogen in blue. And then we can go ahead and draw our ring."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, and then we're adding an oxygen. All right, we're adding an oxygen right here. And then we had that hydrogen, so I'm gonna go ahead and draw in that hydrogen in blue. And then we can go ahead and draw our ring. So we have our ring right here. And then we have our nitro group coming off over here. So just to check ourselves and make sure that this is the right way of thinking about it in reverse, if we take water from this portion, right, and stick those two fragments together, then that would give us our conjugated enone product over here on the left."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we can go ahead and draw our ring. So we have our ring right here. And then we have our nitro group coming off over here. So just to check ourselves and make sure that this is the right way of thinking about it in reverse, if we take water from this portion, right, and stick those two fragments together, then that would give us our conjugated enone product over here on the left. And so that's, again, one way of thinking about how to do these sorts of problems. And so if they wanted you to give some reaction conditions here, you could go ahead and draw what you would need. All right, so you'd say, well, I need to start with this ketone, right, like this."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So just to check ourselves and make sure that this is the right way of thinking about it in reverse, if we take water from this portion, right, and stick those two fragments together, then that would give us our conjugated enone product over here on the left. And so that's, again, one way of thinking about how to do these sorts of problems. And so if they wanted you to give some reaction conditions here, you could go ahead and draw what you would need. All right, so you'd say, well, I need to start with this ketone, right, like this. And then I would also need an aldehyde. So I'm redrawing the 4-nitrobenzaldehyde product. All right, so we have 4-nitrobenzaldehyde like this."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, so you'd say, well, I need to start with this ketone, right, like this. And then I would also need an aldehyde. So I'm redrawing the 4-nitrobenzaldehyde product. All right, so we have 4-nitrobenzaldehyde like this. All right, so this compound, this ketone, of course, is this compound right here. And then 4-nitrobenzaldehyde here is the same one that we drew right here. So if you take those two, right, and you put them together, you should form your conjugated enone because remember, thinking about your alpha protons, right, there are no alpha protons on our 4-nitrobenzaldehyde, right?"}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we have 4-nitrobenzaldehyde like this. All right, so this compound, this ketone, of course, is this compound right here. And then 4-nitrobenzaldehyde here is the same one that we drew right here. So if you take those two, right, and you put them together, you should form your conjugated enone because remember, thinking about your alpha protons, right, there are no alpha protons on our 4-nitrobenzaldehyde, right? So this carbon right here doesn't have any alpha protons. So the only source for alpha protons would be this carbon right here. This is your only alpha carbon with alpha protons on it."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if you take those two, right, and you put them together, you should form your conjugated enone because remember, thinking about your alpha protons, right, there are no alpha protons on our 4-nitrobenzaldehyde, right? So this carbon right here doesn't have any alpha protons. So the only source for alpha protons would be this carbon right here. This is your only alpha carbon with alpha protons on it. So in terms of what else we would need, we would need a base, right, to do our aldol condensation. So we could add something like sodium hydroxide as our base. And we could create a solution of sodium hydroxide and water and then also ethanol."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is your only alpha carbon with alpha protons on it. So in terms of what else we would need, we would need a base, right, to do our aldol condensation. So we could add something like sodium hydroxide as our base. And we could create a solution of sodium hydroxide and water and then also ethanol. So if you took this ketone, right, and this aldehyde and used these reaction conditions, you should form this conjugated enone as your major product. And so that's how to synthesize it. So once again, thinking in reverse."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we could create a solution of sodium hydroxide and water and then also ethanol. So if you took this ketone, right, and this aldehyde and used these reaction conditions, you should form this conjugated enone as your major product. And so that's how to synthesize it. So once again, thinking in reverse. Let's do another one. This one's a little bit different because we don't have an enone or an enal as our target compound. We have an aldol as our target compound."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So once again, thinking in reverse. Let's do another one. This one's a little bit different because we don't have an enone or an enal as our target compound. We have an aldol as our target compound. But we can use the same kind of thinking. We can find our carbonyl and identify the carbon next to that as being our alpha carbon. And then thinking about this as being our beta carbon."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have an aldol as our target compound. But we can use the same kind of thinking. We can find our carbonyl and identify the carbon next to that as being our alpha carbon. And then thinking about this as being our beta carbon. And we know that this is the bond that forms. So we can think about breaking this bond. And we know that we have a hydrogen on this carbon right here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then thinking about this as being our beta carbon. And we know that this is the bond that forms. So we can think about breaking this bond. And we know that we have a hydrogen on this carbon right here. So I'm just going to see if I can stick it into there. So when I think about retrosynthesis, go ahead and draw my retrosynthesis arrow in here. So on the left, I can see on the left that I'm going to have this carbon, two, three, four, and five."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we know that we have a hydrogen on this carbon right here. So I'm just going to see if I can stick it into there. So when I think about retrosynthesis, go ahead and draw my retrosynthesis arrow in here. So on the left, I can see on the left that I'm going to have this carbon, two, three, four, and five. So I have these five carbons right here. So I'm going to go ahead and draw those in. So we have our carbonyl."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, I can see on the left that I'm going to have this carbon, two, three, four, and five. So I have these five carbons right here. So I'm going to go ahead and draw those in. So we have our carbonyl. And then we have our five carbons. And then I already have this hydrogen in magenta. And we're only going to add one hydrogen this time."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have our carbonyl. And then we have our five carbons. And then I already have this hydrogen in magenta. And we're only going to add one hydrogen this time. So let me go ahead and draw in one hydrogen in green because this isn't the conjugated product. We stopped at the aldol if we're thinking about it in reverse. And so we're not going to add two hydrogens."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we're only going to add one hydrogen this time. So let me go ahead and draw in one hydrogen in green because this isn't the conjugated product. We stopped at the aldol if we're thinking about it in reverse. And so we're not going to add two hydrogens. We're only going to add one here. And so what else would we need? So over here on the right, let's think about the fact that there is hydrogen attached to this carbon right here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we're not going to add two hydrogens. We're only going to add one here. And so what else would we need? So over here on the right, let's think about the fact that there is hydrogen attached to this carbon right here. And then how many carbons would we have for our other compound? One, two, three, and four. So we need a four-carbon carbonyl."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the right, let's think about the fact that there is hydrogen attached to this carbon right here. And then how many carbons would we have for our other compound? One, two, three, and four. So we need a four-carbon carbonyl. And there's a hydrogen attached to it. So a four-carbon carbonyl. So let me go ahead and draw that in."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we need a four-carbon carbonyl. And there's a hydrogen attached to it. So a four-carbon carbonyl. So let me go ahead and draw that in. One, two, three, four. And then we have this hydrogen in blue to it. And so now we've identified, thinking about it in reverse, what compounds we would need to form this aldol product."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that in. One, two, three, four. And then we have this hydrogen in blue to it. And so now we've identified, thinking about it in reverse, what compounds we would need to form this aldol product. We would need this ketone and this aldehyde. But in terms of actually synthesizing the aldol compound on the left, we have to be a little bit careful about how to do it. We need to do a directed aldol addition here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so now we've identified, thinking about it in reverse, what compounds we would need to form this aldol product. We would need this ketone and this aldehyde. But in terms of actually synthesizing the aldol compound on the left, we have to be a little bit careful about how to do it. We need to do a directed aldol addition here. So let's think about starting with this ketone. So think about starting with this ketone here. So let me go ahead and redraw it."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We need to do a directed aldol addition here. So let's think about starting with this ketone. So think about starting with this ketone here. So let me go ahead and redraw it. I'm going to draw it a slightly different way. So that's this ketone right here. And to that ketone, we're going to add LDA."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and redraw it. I'm going to draw it a slightly different way. So that's this ketone right here. And to that ketone, we're going to add LDA. So we're going to form a lithium enolate here. And once we've formed our lithium enolate, then we can add our aldehyde. So in the second step, we can add our aldehyde here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And to that ketone, we're going to add LDA. So we're going to form a lithium enolate here. And once we've formed our lithium enolate, then we can add our aldehyde. So in the second step, we can add our aldehyde here. So we talked about this in an earlier video. So we add our aldehyde. It's a four-carbon aldehyde."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in the second step, we can add our aldehyde here. So we talked about this in an earlier video. So we add our aldehyde. It's a four-carbon aldehyde. So we're going to add butanol. And then that's going to give us a lithium alkoxide. So we need to protonate the alkoxide."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's a four-carbon aldehyde. So we're going to add butanol. And then that's going to give us a lithium alkoxide. So we need to protonate the alkoxide. So in the workup, we can add some water here. And that would give us our aldol product. So that would give us this."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we need to protonate the alkoxide. So in the workup, we can add some water here. And that would give us our aldol product. So that would give us this. So that is the way to synthesize our aldol products. Let me go ahead and draw it in here. So we would form this."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that would give us this. So that is the way to synthesize our aldol products. Let me go ahead and draw it in here. So we would form this. So a directed aldol addition. Let's do one more of this retrosynthesis approach. And let's think about how to make this molecule."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we would form this. So a directed aldol addition. Let's do one more of this retrosynthesis approach. And let's think about how to make this molecule. So this is a little bit different than before. But we can do our same steps. We can identify our alpha carbon, the one next to our carbonyl."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's think about how to make this molecule. So this is a little bit different than before. But we can do our same steps. We can identify our alpha carbon, the one next to our carbonyl. And then our beta carbon is the one next to that. So we could think about breaking this double bond and adding two hydrogens to our alpha carbon. So let's go ahead and do that."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can identify our alpha carbon, the one next to our carbonyl. And then our beta carbon is the one next to that. So we could think about breaking this double bond and adding two hydrogens to our alpha carbon. So let's go ahead and do that. So we're going to break that double bond and add two hydrogens to the alpha carbon. So thinking about this in terms of retrosynthesis, we have a ring here. And then let me go ahead and draw this over here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and do that. So we're going to break that double bond and add two hydrogens to the alpha carbon. So thinking about this in terms of retrosynthesis, we have a ring here. And then let me go ahead and draw this over here. We would have two hydrogens over here. Let me go ahead and get some more room. So probably easier to think about adding your oxygen to your beta carbon first."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then let me go ahead and draw this over here. We would have two hydrogens over here. Let me go ahead and get some more room. So probably easier to think about adding your oxygen to your beta carbon first. So let's do that. We know we're going to add an oxygen to our beta carbon. So let's go ahead and put in our oxygen on our beta carbon."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So probably easier to think about adding your oxygen to your beta carbon first. So let's do that. We know we're going to add an oxygen to our beta carbon. So let's go ahead and put in our oxygen on our beta carbon. I'll do that in green. And then we're going to think about adding two hydrogens to our alpha carbon. So we're going to draw in our alpha carbon like that."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and put in our oxygen on our beta carbon. I'll do that in green. And then we're going to think about adding two hydrogens to our alpha carbon. So we're going to draw in our alpha carbon like that. And we're going to add two hydrogens. I'm going to put those in green like this. That just gives us a little bit more room."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to draw in our alpha carbon like that. And we're going to add two hydrogens. I'm going to put those in green like this. That just gives us a little bit more room. And then we would have our carbonyl like that. And then we would have, let's make sure to count your carbons when you're doing these. So let's go ahead and show those carbons."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That just gives us a little bit more room. And then we would have our carbonyl like that. And then we would have, let's make sure to count your carbons when you're doing these. So let's go ahead and show those carbons. So this carbon right here is this carbon. And then this carbon is this one right here. And then this carbon, our carbonyl one, is right here."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show those carbons. So this carbon right here is this carbon. And then this carbon is this one right here. And then this carbon, our carbonyl one, is right here. And so you can see this is actually an intramolecular aldol condensation. So if you think about losing water right here and sticking those two fragments together, you would form this compound on the left. And so this is a pretty cool reaction."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then this carbon, our carbonyl one, is right here. And so you can see this is actually an intramolecular aldol condensation. So if you think about losing water right here and sticking those two fragments together, you would form this compound on the left. And so this is a pretty cool reaction. We could redraw this. So on an exam, you could draw it like that. But most of the time, you would see it in a different conformation."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this is a pretty cool reaction. We could redraw this. So on an exam, you could draw it like that. But most of the time, you would see it in a different conformation. So you would see the compound looking like this. So let me go ahead and draw that. So identify your carbons."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But most of the time, you would see it in a different conformation. So you would see the compound looking like this. So let me go ahead and draw that. So identify your carbons. Carbon in magenta is the one right here. Carbon in blue is the one right here. And the carbon in red is the one right here, which has your carbonyl."}, {"video_title": "Retro-aldol and retrosynthesis Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So identify your carbons. Carbon in magenta is the one right here. Carbon in blue is the one right here. And the carbon in red is the one right here, which has your carbonyl. And then we'll go ahead and make our alpha carbon right here green, so this carbon right here. So this must have been your starting compound. And so if you add a base, you could convert this molecule into your target compound."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "Here's another way to do a mixed or crossed aldol condensation, so this time using a lithium enolate. So if we took this ketone and this aldehyde and just mixed them together with some base, we would get a mixture of products. We wouldn't get our desired product, this conjugated enone over here. So if we do this stepwise, we can do a directed aldol condensation, and so if we take this ketone and add LDA to it, and then we add our aldehyde, and then in our workup, we can add water and toluene sulfonic acid as an acid source, we can get our conjugated enone in a decent yield. And so let's look at each of these steps one by one. So we'll first start off with the addition of LDA, which we know is a strong base. So the strong base is going to take a proton from our ketone, and so let's analyze our ketone here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So if we do this stepwise, we can do a directed aldol condensation, and so if we take this ketone and add LDA to it, and then we add our aldehyde, and then in our workup, we can add water and toluene sulfonic acid as an acid source, we can get our conjugated enone in a decent yield. And so let's look at each of these steps one by one. So we'll first start off with the addition of LDA, which we know is a strong base. So the strong base is going to take a proton from our ketone, and so let's analyze our ketone here. So the alpha carbons are the ones next to the carbonyl, and so this is an alpha carbon on our ketone, and this is an alpha carbon on our ketone. So which one of those alpha carbons will our strong base deprotonate? So we saw in a previous video that LDA is going to form the kinetic enolate, so it's going to take a proton from the least sterically hindered side, and so it's going to take a proton from this right side over here to form the kinetic enolate."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So the strong base is going to take a proton from our ketone, and so let's analyze our ketone here. So the alpha carbons are the ones next to the carbonyl, and so this is an alpha carbon on our ketone, and this is an alpha carbon on our ketone. So which one of those alpha carbons will our strong base deprotonate? So we saw in a previous video that LDA is going to form the kinetic enolate, so it's going to take a proton from the least sterically hindered side, and so it's going to take a proton from this right side over here to form the kinetic enolate. If it took one from the left, that would be the thermodynamic enolate. So I can go ahead and draw in an alpha proton on our alpha carbon right here. So let's say this is the one that LDA is going to take."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So we saw in a previous video that LDA is going to form the kinetic enolate, so it's going to take a proton from the least sterically hindered side, and so it's going to take a proton from this right side over here to form the kinetic enolate. If it took one from the left, that would be the thermodynamic enolate. So I can go ahead and draw in an alpha proton on our alpha carbon right here. So let's say this is the one that LDA is going to take. So let's go ahead and draw in LDA. So I'm going to go ahead and draw my nitrogen in here, and then I have my isopropyl groups on my nitrogen, and then I'm going to have the nitrogen bonded to lithium right here like that. So when this deprotonation occurs, it's actually a cyclic mechanism."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's say this is the one that LDA is going to take. So let's go ahead and draw in LDA. So I'm going to go ahead and draw my nitrogen in here, and then I have my isopropyl groups on my nitrogen, and then I'm going to have the nitrogen bonded to lithium right here like that. So when this deprotonation occurs, it's actually a cyclic mechanism. So the oxygen is going to be forming a bond with this lithium. The nitrogen is going to be forming a bond with this proton here. So let's go ahead and show the cyclic mechanism."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So when this deprotonation occurs, it's actually a cyclic mechanism. So the oxygen is going to be forming a bond with this lithium. The nitrogen is going to be forming a bond with this proton here. So let's go ahead and show the cyclic mechanism. So if these electrons right here on the nitrogen anion move in and take this proton, we're going to form a bond right here. These electrons would move in here to form a carbon-carbon double bond, and then these electrons would kick off onto your oxygen to form a bond with lithium. So let's go ahead and show the product, and then we'll show the movement of those electrons here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the cyclic mechanism. So if these electrons right here on the nitrogen anion move in and take this proton, we're going to form a bond right here. These electrons would move in here to form a carbon-carbon double bond, and then these electrons would kick off onto your oxygen to form a bond with lithium. So let's go ahead and show the product, and then we'll show the movement of those electrons here. So we have our oxygen. It's now going to be bonded to lithium like that. And now we have a carbon-carbon double bond."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the product, and then we'll show the movement of those electrons here. So we have our oxygen. It's now going to be bonded to lithium like that. And now we have a carbon-carbon double bond. And then we would have nitrogen bonded to hydrogen now, and we have an amine over here on the right. So let's go ahead and follow some of those electrons. So if I'm saying these electrons in blue here on the nitrogen anion are going to pick up this proton to form our amine like that."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And now we have a carbon-carbon double bond. And then we would have nitrogen bonded to hydrogen now, and we have an amine over here on the right. So let's go ahead and follow some of those electrons. So if I'm saying these electrons in blue here on the nitrogen anion are going to pick up this proton to form our amine like that. I could say that these electrons in here, once it's deprotonated, they could form our double bond right in here. And then I could also point out that these electrons right here could move out to bond with the lithium to form our lithium enolate. And so once again, this is the kinetic enolate, the one that's formed the fastest."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm saying these electrons in blue here on the nitrogen anion are going to pick up this proton to form our amine like that. I could say that these electrons in here, once it's deprotonated, they could form our double bond right in here. And then I could also point out that these electrons right here could move out to bond with the lithium to form our lithium enolate. And so once again, this is the kinetic enolate, the one that's formed the fastest. And the use of LDA helps us to get our kinetic enolate, because these bulky isopropyl groups would prevent deprotonation at the alpha carbon on the left. And so now we've formed our lithium enolate, and so let's go to the next step. So in the next step, we're going to add our aldehyde right here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, this is the kinetic enolate, the one that's formed the fastest. And the use of LDA helps us to get our kinetic enolate, because these bulky isopropyl groups would prevent deprotonation at the alpha carbon on the left. And so now we've formed our lithium enolate, and so let's go to the next step. So in the next step, we're going to add our aldehyde right here. So we're going to add butanol. So let's go ahead and look at the formation of our lithium enolate, and then let's add butanol to that. So let's go ahead and draw in butanol."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So in the next step, we're going to add our aldehyde right here. So we're going to add butanol. So let's go ahead and look at the formation of our lithium enolate, and then let's add butanol to that. So let's go ahead and draw in butanol. So here we have our carbonyl, and then we would have four carbons, so two, three, four, like that. We know that aldehydes can function as electrophiles. So in the second step, we're going to add an electrophilic carbonyl compound."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in butanol. So here we have our carbonyl, and then we would have four carbons, so two, three, four, like that. We know that aldehydes can function as electrophiles. So in the second step, we're going to add an electrophilic carbonyl compound. We know this is electrophilic, because the oxygen is partially negative, and this carbonyl carbon right here is partially positive, like that. And so in the second step, it's going to be a nucleophile attacking an electrophile. So the enolate is going to function as a nucleophile."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So in the second step, we're going to add an electrophilic carbonyl compound. We know this is electrophilic, because the oxygen is partially negative, and this carbonyl carbon right here is partially positive, like that. And so in the second step, it's going to be a nucleophile attacking an electrophile. So the enolate is going to function as a nucleophile. So we're going to form a bond, this time between lithium and oxygen over here, and then we're going to form a carbon-carbon bond down here. So let's show the movement of those electrons. So the lithium enolate functions as our nucleophile."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So the enolate is going to function as a nucleophile. So we're going to form a bond, this time between lithium and oxygen over here, and then we're going to form a carbon-carbon bond down here. So let's show the movement of those electrons. So the lithium enolate functions as our nucleophile. The aldehyde functions as our electrophile. So if these electrons move into here, then these electrons are going to bond with that carbon, and then these electrons can bond in here. So another cyclic mechanism."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So the lithium enolate functions as our nucleophile. The aldehyde functions as our electrophile. So if these electrons move into here, then these electrons are going to bond with that carbon, and then these electrons can bond in here. So another cyclic mechanism. And so let's go ahead and draw the result of that. So we would have these carbons. We would form a carbonyl right here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So another cyclic mechanism. And so let's go ahead and draw the result of that. So we would have these carbons. We would form a carbonyl right here. Now we'd form a new carbon-carbon bond right here, and then our oxygen would now be bonded to our lithium, and then we'll draw in the rest of our carbon. So we form a lithium alkoxide product. Following those electrons, so these electrons in here in red moved into here to form our carbonyl."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "We would form a carbonyl right here. Now we'd form a new carbon-carbon bond right here, and then our oxygen would now be bonded to our lithium, and then we'll draw in the rest of our carbon. So we form a lithium alkoxide product. Following those electrons, so these electrons in here in red moved into here to form our carbonyl. The electrons in here in magenta, these are the ones that attacked our carbonyl carbon here to form our new carbon-carbon bond. And then finally, I could say make these electrons in here blue, and then forming a bond between oxygen and lithium to form our lithium alkoxide intermediate here. And so two cyclic mechanisms form your carbon-carbon bond here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "Following those electrons, so these electrons in here in red moved into here to form our carbonyl. The electrons in here in magenta, these are the ones that attacked our carbonyl carbon here to form our new carbon-carbon bond. And then finally, I could say make these electrons in here blue, and then forming a bond between oxygen and lithium to form our lithium alkoxide intermediate here. And so two cyclic mechanisms form your carbon-carbon bond here. So first you would deprotonate. We talked about this cyclic mechanism up here. And then once you add your aldehyde, once you've formed your lithium enolate, another cyclic mechanism will give you your lithium alkoxide."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And so two cyclic mechanisms form your carbon-carbon bond here. So first you would deprotonate. We talked about this cyclic mechanism up here. And then once you add your aldehyde, once you've formed your lithium enolate, another cyclic mechanism will give you your lithium alkoxide. So let's look at the next step. The third step, you're going to add water in your workup. And so here I've drawn the lithium alkoxide a little bit differently."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And then once you add your aldehyde, once you've formed your lithium enolate, another cyclic mechanism will give you your lithium alkoxide. So let's look at the next step. The third step, you're going to add water in your workup. And so here I've drawn the lithium alkoxide a little bit differently. So the electrons in blue that I had over here in the lithium-oxygen bond, I'm just going to put those on the oxygen this time. We know the oxygen has two other lone pairs of electrons, which give it a negative 1 formal charge, and then lithium would be a plus 1 charge. And so this is just another way to represent our lithium alkoxide."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And so here I've drawn the lithium alkoxide a little bit differently. So the electrons in blue that I had over here in the lithium-oxygen bond, I'm just going to put those on the oxygen this time. We know the oxygen has two other lone pairs of electrons, which give it a negative 1 formal charge, and then lithium would be a plus 1 charge. And so this is just another way to represent our lithium alkoxide. This is how we've usually done it in our videos here. And so in the next step, in the workup, if you do an aqueous workup, so you add some water here, you would protonate your alkoxide anion. So let's go ahead and show that."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And so this is just another way to represent our lithium alkoxide. This is how we've usually done it in our videos here. And so in the next step, in the workup, if you do an aqueous workup, so you add some water here, you would protonate your alkoxide anion. So let's go ahead and show that. So we take a proton here from water. And let's go ahead and draw the product. So once again, we would have our carbonyl."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. So we take a proton here from water. And let's go ahead and draw the product. So once again, we would have our carbonyl. And then we would protonate the oxygen to form our aldol. And so here is our aldol product. So once you form your aldol, let's go ahead and look at the next step here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we would have our carbonyl. And then we would protonate the oxygen to form our aldol. And so here is our aldol product. So once you form your aldol, let's go ahead and look at the next step here. We added some toluenesulfonic acid. So we added a source of protons. And so in the final step, to get to our enone, we need to dehydrate our aldol."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So once you form your aldol, let's go ahead and look at the next step here. We added some toluenesulfonic acid. So we added a source of protons. And so in the final step, to get to our enone, we need to dehydrate our aldol. And so we've seen how to form our enone using base. And this time, we're going to do an acid-catalyzed dehydration. So if you add a source of protons here, we know we still have two lone pairs of electrons on our oxygen."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And so in the final step, to get to our enone, we need to dehydrate our aldol. And so we've seen how to form our enone using base. And this time, we're going to do an acid-catalyzed dehydration. So if you add a source of protons here, we know we still have two lone pairs of electrons on our oxygen. One of those lone pairs could pick up that proton. So we protonate. So let's go ahead and draw what we would make here."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So if you add a source of protons here, we know we still have two lone pairs of electrons on our oxygen. One of those lone pairs could pick up that proton. So we protonate. So let's go ahead and draw what we would make here. So we have our carbonyl. And we have all of these carbons. And now we would have this oxygen, right, which still has one lone pair of electrons."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw what we would make here. So we have our carbonyl. And we have all of these carbons. And now we would have this oxygen, right, which still has one lone pair of electrons. So let's go ahead and show those electrons. These electrons in magenta picked up a proton right here. So we have one lone pair of electrons left, which gives that oxygen a plus 1 formal charge."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And now we would have this oxygen, right, which still has one lone pair of electrons. So let's go ahead and show those electrons. These electrons in magenta picked up a proton right here. So we have one lone pair of electrons left, which gives that oxygen a plus 1 formal charge. And so we have an excellent leaving group here. If we think about these electrons moving off onto the oxygen, we have water as a leaving group. So loss of water at this step would yield a cation."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So we have one lone pair of electrons left, which gives that oxygen a plus 1 formal charge. And so we have an excellent leaving group here. If we think about these electrons moving off onto the oxygen, we have water as a leaving group. So loss of water at this step would yield a cation. So let's go ahead and draw the cation that would form. So we have our carbonyl. And we have all of these carbons."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So loss of water at this step would yield a cation. So let's go ahead and draw the cation that would form. So we have our carbonyl. And we have all of these carbons. And so we lost a bond to this carbon right here. So this is the carbon that's going to form our cation. So let's go ahead and draw a plus 1 formal charge on our cation like that."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And we have all of these carbons. And so we lost a bond to this carbon right here. So this is the carbon that's going to form our cation. So let's go ahead and draw a plus 1 formal charge on our cation like that. And so in the next step, a base is going to come along. And it's going to take a proton from our alpha carbon. So once again, we think about where our carbonyl is."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw a plus 1 formal charge on our cation like that. And so in the next step, a base is going to come along. And it's going to take a proton from our alpha carbon. So once again, we think about where our carbonyl is. The carbon next to it is our alpha carbon. And so there's a proton on here. So we can deprotonate it with a base."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we think about where our carbonyl is. The carbon next to it is our alpha carbon. And so there's a proton on here. So we can deprotonate it with a base. So I'm just going to write a generic base. But it could be something like the water that just left in the previous step. So a base is going to come along and take this proton."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So we can deprotonate it with a base. So I'm just going to write a generic base. But it could be something like the water that just left in the previous step. So a base is going to come along and take this proton. And so these electrons are going to move in here to form your double bond. That's going to take away your plus 1 formal charge. So let's go ahead and draw our product."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So a base is going to come along and take this proton. And so these electrons are going to move in here to form your double bond. That's going to take away your plus 1 formal charge. So let's go ahead and draw our product. So we would have our carbonyl. And then we would now have a double bond right here. And then we draw in the rest of our carbons like that."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw our product. So we would have our carbonyl. And then we would now have a double bond right here. And then we draw in the rest of our carbons like that. So if we show those electrons, let's make them blue this time. So these electrons in here. So the base takes the proton."}, {"video_title": "Mixed (crossed) aldol condensation using a lithium enolate Organic chemistry Khan Academy.mp3", "Sentence": "And then we draw in the rest of our carbons like that. So if we show those electrons, let's make them blue this time. So these electrons in here. So the base takes the proton. And the electrons move in to give us our double bond. And we now have formed our enone as our product. So once again, this is an example of a directed aldol condensation."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we're going to use the concept of retrosynthesis, so working backwards and thinking about what could be an immediate precursor to this molecule here. To do that, let's analyze the groups that are attached to our ring. Right here, we have an amino group. And we know that's an ortho para director. We know that because of the lone pair of electrons on this nitrogen right next to our ring. And then over here, we have a carboxylic acid functional group attached to our ring. And we know this carbonyl carbon here is partially positive, which makes this a meta director."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we know that's an ortho para director. We know that because of the lone pair of electrons on this nitrogen right next to our ring. And then over here, we have a carboxylic acid functional group attached to our ring. And we know this carbonyl carbon here is partially positive, which makes this a meta director. Now, we haven't covered any reactions that would install an amino group meta to a carboxylic acid group. But we have talked about how to put a nitro group meta to a carboxylic acid group. So let's turn that amino group into a nitro group."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we know this carbonyl carbon here is partially positive, which makes this a meta director. Now, we haven't covered any reactions that would install an amino group meta to a carboxylic acid group. But we have talked about how to put a nitro group meta to a carboxylic acid group. So let's turn that amino group into a nitro group. So let's go ahead and draw our benzene ring here. And we'll go ahead and put our carboxylic acid down here. And instead of an amino group, let's draw a nitro group like that."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's turn that amino group into a nitro group. So let's go ahead and draw our benzene ring here. And we'll go ahead and put our carboxylic acid down here. And instead of an amino group, let's draw a nitro group like that. So all we need to do now is think about a way to reduce our nitro group to our amino group. And there are many ways to do it. One thing you could use would be a metal like iron and some hydrochloric acid."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And instead of an amino group, let's draw a nitro group like that. So all we need to do now is think about a way to reduce our nitro group to our amino group. And there are many ways to do it. One thing you could use would be a metal like iron and some hydrochloric acid. You could have used tin. You could have also used hydrogen and a metal catalyst like nickel. So there are many ways to reduce a nitro group to an amino group on a benzene ring."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "One thing you could use would be a metal like iron and some hydrochloric acid. You could have used tin. You could have also used hydrogen and a metal catalyst like nickel. So there are many ways to reduce a nitro group to an amino group on a benzene ring. And now I know how to put that nitro group on meta to my carboxylic acid. So we have a meta director right here. And so if I take that nitro group off, I can come up with the precursor to this molecule, which would be just benzoic acid."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there are many ways to reduce a nitro group to an amino group on a benzene ring. And now I know how to put that nitro group on meta to my carboxylic acid. So we have a meta director right here. And so if I take that nitro group off, I can come up with the precursor to this molecule, which would be just benzoic acid. So go ahead and draw benzoic acid here. And I have to think about the reagents necessary for a nitration reaction, which of course would be concentrated nitric acid and concentrated sulfuric acid. And so once again, because my carbonyl here on my carboxylic acid is meta directing, the nitro group ends up in the meta position."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if I take that nitro group off, I can come up with the precursor to this molecule, which would be just benzoic acid. So go ahead and draw benzoic acid here. And I have to think about the reagents necessary for a nitration reaction, which of course would be concentrated nitric acid and concentrated sulfuric acid. And so once again, because my carbonyl here on my carboxylic acid is meta directing, the nitro group ends up in the meta position. So now I have benzoic acid. And I need to make a benzoic acid from benzene. And if you think about one of the reactions we did in the benzylic position video, we were able to create a benzoic acid molecule from an alkyl benzene."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, because my carbonyl here on my carboxylic acid is meta directing, the nitro group ends up in the meta position. So now I have benzoic acid. And I need to make a benzoic acid from benzene. And if you think about one of the reactions we did in the benzylic position video, we were able to create a benzoic acid molecule from an alkyl benzene. And so the immediate precursor to this molecule must be some sort of alkyl benzene, so a benzene ring with an alkyl group on it. So just to make things easy, I'll just make it a methyl group here, so a methyl group like that. So toluene as our immediate precursor to benzoic acid."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if you think about one of the reactions we did in the benzylic position video, we were able to create a benzoic acid molecule from an alkyl benzene. And so the immediate precursor to this molecule must be some sort of alkyl benzene, so a benzene ring with an alkyl group on it. So just to make things easy, I'll just make it a methyl group here, so a methyl group like that. So toluene as our immediate precursor to benzoic acid. To oxidize the alkyl side chain, I need to add something like sodium dichromate, so Na2Cr2O7. A source of protons, or something like sulfuric acid, and some heat. And so this is oxidation of an alkyl benzene like that."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So toluene as our immediate precursor to benzoic acid. To oxidize the alkyl side chain, I need to add something like sodium dichromate, so Na2Cr2O7. A source of protons, or something like sulfuric acid, and some heat. And so this is oxidation of an alkyl benzene like that. You could have also done this reaction with permanganate and heat as well. And you could have chosen any number of carbons on your alkyl side chain. So I just chose one to make it easy."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this is oxidation of an alkyl benzene like that. You could have also done this reaction with permanganate and heat as well. And you could have chosen any number of carbons on your alkyl side chain. So I just chose one to make it easy. So this carbon, when this group gets oxidized, would turn into that carbon right here. Now we would need to convert benzene into toluene. And so we could do that using a Friedel-Crafts alkylation reaction."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I just chose one to make it easy. So this carbon, when this group gets oxidized, would turn into that carbon right here. Now we would need to convert benzene into toluene. And so we could do that using a Friedel-Crafts alkylation reaction. And so we need one carbon. So for our alkyl chloride, we would need a one carbon alkyl chloride, so CH3Cl. And our catalyst for Friedel-Crafts alkylation would be aluminum chloride like that."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we could do that using a Friedel-Crafts alkylation reaction. And so we need one carbon. So for our alkyl chloride, we would need a one carbon alkyl chloride, so CH3Cl. And our catalyst for Friedel-Crafts alkylation would be aluminum chloride like that. So our synthesis is complete. Let's go ahead and run through the reactions very quickly. And let's make sure that everything makes sense."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And our catalyst for Friedel-Crafts alkylation would be aluminum chloride like that. So our synthesis is complete. Let's go ahead and run through the reactions very quickly. And let's make sure that everything makes sense. So we start with benzene. And we do a Friedel-Crafts alkylation to put a methyl group onto our benzene ring to form toluene. Next, we oxidize that alkyl side chain with sodium dichromate to convert it into a carboxylic acid, which is now a metadirector."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let's make sure that everything makes sense. So we start with benzene. And we do a Friedel-Crafts alkylation to put a methyl group onto our benzene ring to form toluene. Next, we oxidize that alkyl side chain with sodium dichromate to convert it into a carboxylic acid, which is now a metadirector. So when you nitrate your ring, the nitrate group gets put on in a position meta to your carboxylic acid. Finally, you reduce your nitro group to turn it into an amino group. And you are done."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Next, we oxidize that alkyl side chain with sodium dichromate to convert it into a carboxylic acid, which is now a metadirector. So when you nitrate your ring, the nitrate group gets put on in a position meta to your carboxylic acid. Finally, you reduce your nitro group to turn it into an amino group. And you are done. So let's do one more synthesis problem here. And so we can see that this time our target molecule is over here on the right. And immediately, I see a bromine at the benzylic position, the carbon right next to our benzene ring."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you are done. So let's do one more synthesis problem here. And so we can see that this time our target molecule is over here on the right. And immediately, I see a bromine at the benzylic position, the carbon right next to our benzene ring. And so I know a reaction that will put a bromine in the benzylic position. If you remember, that's the free radical bromination reaction. And so I can go ahead and draw the precursor to that molecule, which would be a benzene ring."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And immediately, I see a bromine at the benzylic position, the carbon right next to our benzene ring. And so I know a reaction that will put a bromine in the benzylic position. If you remember, that's the free radical bromination reaction. And so I can go ahead and draw the precursor to that molecule, which would be a benzene ring. This bromine over here is on our ring. And we would have just an ethyl group now like that. And so I would need some NBS, some heat."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I can go ahead and draw the precursor to that molecule, which would be a benzene ring. This bromine over here is on our ring. And we would have just an ethyl group now like that. And so I would need some NBS, some heat. I could use carbon tetrachloride as a solvent, a peroxide to initiate the free radical mechanism. And this is a free radical bromination reaction, which puts a bromine at the benzylic position. All right, trying to figure out a precursor to this molecule."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I would need some NBS, some heat. I could use carbon tetrachloride as a solvent, a peroxide to initiate the free radical mechanism. And this is a free radical bromination reaction, which puts a bromine at the benzylic position. All right, trying to figure out a precursor to this molecule. Let's analyze what sort of groups we have on our ring. So this ethyl group is an alkyl group. And remember, that's an ortho para director."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "All right, trying to figure out a precursor to this molecule. Let's analyze what sort of groups we have on our ring. So this ethyl group is an alkyl group. And remember, that's an ortho para director. This bromine here, of course, is also an ortho para director because of the lone pairs of electrons on it. So this is an ortho para director. We have two ortho para directors on our ring."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And remember, that's an ortho para director. This bromine here, of course, is also an ortho para director because of the lone pairs of electrons on it. So this is an ortho para director. We have two ortho para directors on our ring. But those ortho para directors are meta to each other. So we need to think about a way to turn one of those ortho para directors into a meta director. And a good way of doing that would be to turn our alkyl group into a group that has a carbonyl on it, so an acyl group."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have two ortho para directors on our ring. But those ortho para directors are meta to each other. So we need to think about a way to turn one of those ortho para directors into a meta director. And a good way of doing that would be to turn our alkyl group into a group that has a carbonyl on it, so an acyl group. So thinking about the precursor molecule here, I draw my benzene ring. I'm going to leave my bromine here. And instead of an alkyl group, we're going to have an acyl group now because our acyl group is a meta director."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And a good way of doing that would be to turn our alkyl group into a group that has a carbonyl on it, so an acyl group. So thinking about the precursor molecule here, I draw my benzene ring. I'm going to leave my bromine here. And instead of an alkyl group, we're going to have an acyl group now because our acyl group is a meta director. So this carbon right here is partially positive. So we need to think about how to reduce an acyl group to an alkyl group. And one way of doing that would be to do a Clemenson reduction."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And instead of an alkyl group, we're going to have an acyl group now because our acyl group is a meta director. So this carbon right here is partially positive. So we need to think about how to reduce an acyl group to an alkyl group. And one way of doing that would be to do a Clemenson reduction. So you take amalgamated zinc. So we take some zinc here, so an amalgam with mercury, and also some HCl. And so that will reduce our acyl group to an alkyl group like that."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And one way of doing that would be to do a Clemenson reduction. So you take amalgamated zinc. So we take some zinc here, so an amalgam with mercury, and also some HCl. And so that will reduce our acyl group to an alkyl group like that. Now that we have an acyl group on there, which is a meta director, that would allow us to install the bromine meta to that acyl group. So when I think about the precursor to this molecule, I just take off that bromine. And so I have my ring with an acyl group on that ring."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that will reduce our acyl group to an alkyl group like that. Now that we have an acyl group on there, which is a meta director, that would allow us to install the bromine meta to that acyl group. So when I think about the precursor to this molecule, I just take off that bromine. And so I have my ring with an acyl group on that ring. And once again, since this is a meta director here, all I need to do is a bromination reaction. So once again, I would need some bromine. And I would need our catalyst, so FeBr3."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I have my ring with an acyl group on that ring. And once again, since this is a meta director here, all I need to do is a bromination reaction. So once again, I would need some bromine. And I would need our catalyst, so FeBr3. And that reaction will work. Finally, I have to go from benzene to this molecule. And once again, I can do that with a Friedel-Crafts acylation reaction."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I would need our catalyst, so FeBr3. And that reaction will work. Finally, I have to go from benzene to this molecule. And once again, I can do that with a Friedel-Crafts acylation reaction. I would need two carbons to do my Friedel-Crafts acylation reaction. So for my acyl chloride, I need to make sure I have two carbons. And once again, my catalyst would be something like aluminum chloride for this Friedel-Crafts acylation."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, I can do that with a Friedel-Crafts acylation reaction. I would need two carbons to do my Friedel-Crafts acylation reaction. So for my acyl chloride, I need to make sure I have two carbons. And once again, my catalyst would be something like aluminum chloride for this Friedel-Crafts acylation. And so this would be the complete synthesis. Let's once again run through each step. Starting with benzene, we do a Friedel-Crafts acylation to install an acyl group on our ring like that, which is a meta director."}, {"video_title": "Synthesis of substituted benzene rings II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, my catalyst would be something like aluminum chloride for this Friedel-Crafts acylation. And so this would be the complete synthesis. Let's once again run through each step. Starting with benzene, we do a Friedel-Crafts acylation to install an acyl group on our ring like that, which is a meta director. So if we follow that with a bromination, the bromine is installed meta to our acyl group. Our acyl group is then reduced using a zinc amalgam and converted into an ethyl group right here. And then finally, we do a free radical bromination to put a bromine at the benzylic position, and we are done."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And she made, a little over 100 years ago, this is in the early 1900s, while working for Edward Charles Pickering, who was a Harvard astronomer, while working for his observatory, she made what is arguably, well, definitely one of the most important discoveries in all of astronomy. Probably, and I would say it ranks their top three, because it really enabled people like Hubble to start realizing that the universe is expanding. Or even being able to think about how to measure distances to objects in space well beyond the reach of our tools with parallax. We saw with parallax, you have to have extremely sensitive instruments just to even measure distances to stars relatively close to us. Very sensitive instruments to get to stars maybe further out into our galaxy. And we don't have the instruments even today to measure things beyond our galaxy. But because of Henrietta Swan Leavitt, we were able to approximate or get good senses of the distance to objects beyond our galaxy."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We saw with parallax, you have to have extremely sensitive instruments just to even measure distances to stars relatively close to us. Very sensitive instruments to get to stars maybe further out into our galaxy. And we don't have the instruments even today to measure things beyond our galaxy. But because of Henrietta Swan Leavitt, we were able to approximate or get good senses of the distance to objects beyond our galaxy. So let's just think about what she did. So her job was literally to classify stars in the large Magellanic, I have trouble saying that, Magellanic cloud and the small Magellanic clouds. And this is what they look like from the southern hemisphere."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But because of Henrietta Swan Leavitt, we were able to approximate or get good senses of the distance to objects beyond our galaxy. So let's just think about what she did. So her job was literally to classify stars in the large Magellanic, I have trouble saying that, Magellanic cloud and the small Magellanic clouds. And this is what they look like from the southern hemisphere. This is the large right over here. And this is the small right over here. And remember, this is before Hubble realized or showed the world that there are stars beyond our galaxy, that there are galaxies beyond our galaxy."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is what they look like from the southern hemisphere. This is the large right over here. And this is the small right over here. And remember, this is before Hubble realized or showed the world that there are stars beyond our galaxy, that there are galaxies beyond our galaxy. So at this point in time, people didn't even fully appreciate that these were separate galaxies. We just said, hey, these are kind of these blobs or these clusters of stars that we see in the southern hemisphere. And just to get a sense of where they are relative to our galaxy, the Milky Way galaxy, this is obviously not an actual picture."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And remember, this is before Hubble realized or showed the world that there are stars beyond our galaxy, that there are galaxies beyond our galaxy. So at this point in time, people didn't even fully appreciate that these were separate galaxies. We just said, hey, these are kind of these blobs or these clusters of stars that we see in the southern hemisphere. And just to get a sense of where they are relative to our galaxy, the Milky Way galaxy, this is obviously not an actual picture. We can't take a picture from this vantage point. This would have to be very, very far away. But this is the Milky Way right here."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just to get a sense of where they are relative to our galaxy, the Milky Way galaxy, this is obviously not an actual picture. We can't take a picture from this vantage point. This would have to be very, very far away. But this is the Milky Way right here. And this is the small Magellanic cloud. And this is the large Magellanic cloud. I'm getting better at saying it."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is the Milky Way right here. And this is the small Magellanic cloud. And this is the large Magellanic cloud. I'm getting better at saying it. So her job was literally just to classify the different stars that she saw. But while she was classifying, she looked at these things called variables. It turns out what she was looking at were a class of stars called Cepheid or Cepheid variable stars."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'm getting better at saying it. So her job was literally just to classify the different stars that she saw. But while she was classifying, she looked at these things called variables. It turns out what she was looking at were a class of stars called Cepheid or Cepheid variable stars. And what's interesting about them is two things. They're super duper bright. They're up to 30,000 times as luminous as the sun."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It turns out what she was looking at were a class of stars called Cepheid or Cepheid variable stars. And what's interesting about them is two things. They're super duper bright. They're up to 30,000 times as luminous as the sun. And they're 5 to 20 times more massive than the sun. 5 to 20 times the sun's mass. But what makes them interesting is one, they're really bright."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're up to 30,000 times as luminous as the sun. And they're 5 to 20 times more massive than the sun. 5 to 20 times the sun's mass. But what makes them interesting is one, they're really bright. So you can see them from really far away. You can see these Cepheid variable stars in other galaxies. In fact, we can see it well beyond even the small Magellanic cloud or the large Magellanic cloud."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what makes them interesting is one, they're really bright. So you can see them from really far away. You can see these Cepheid variable stars in other galaxies. In fact, we can see it well beyond even the small Magellanic cloud or the large Magellanic cloud. But you can see these stars in other galaxies. And what's even more interesting about them is that their intensity is variable, that they become brighter and dimmer with a well-defined period. So if you're looking at a Cepheid variable star, and this is just kind of a simulation, a very cheap simulation, it might look like this."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, we can see it well beyond even the small Magellanic cloud or the large Magellanic cloud. But you can see these stars in other galaxies. And what's even more interesting about them is that their intensity is variable, that they become brighter and dimmer with a well-defined period. So if you're looking at a Cepheid variable star, and this is just kind of a simulation, a very cheap simulation, it might look like this. And then over the course of the next three, four days, it might reduce in intensity to something like this. And then after three, four days again, it will look like this. And then it'll look like this again."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you're looking at a Cepheid variable star, and this is just kind of a simulation, a very cheap simulation, it might look like this. And then over the course of the next three, four days, it might reduce in intensity to something like this. And then after three, four days again, it will look like this. And then it'll look like this again. So its actual intensity is going up and down with a well-defined period. So if this takes three days and this is another three days, then the period, one entire cycle of its going from low intensity back to high intensity is going to be six days. So this is a six-day period."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then it'll look like this again. So its actual intensity is going up and down with a well-defined period. So if this takes three days and this is another three days, then the period, one entire cycle of its going from low intensity back to high intensity is going to be six days. So this is a six-day period. And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she plotted, she assumed that things, everything in each of these clouds are roughly the same distance away. Everything in the large Magellanic cloud is roughly the same distance away. And it's obviously not exact."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is a six-day period. And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she plotted, she assumed that things, everything in each of these clouds are roughly the same distance away. Everything in the large Magellanic cloud is roughly the same distance away. And it's obviously not exact. This is an entire galaxy, so you have obviously things further away in that galaxy and things closer up. You have stars here and here, and their distance isn't going to be exactly the same to us, that we're sitting maybe over here someplace. But it's going to be close."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's obviously not exact. This is an entire galaxy, so you have obviously things further away in that galaxy and things closer up. You have stars here and here, and their distance isn't going to be exactly the same to us, that we're sitting maybe over here someplace. But it's going to be close. It wasn't a bad approximation. And by making that assumption, she saw something pretty neat. So let me plot this right over here."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's going to be close. It wasn't a bad approximation. And by making that assumption, she saw something pretty neat. So let me plot this right over here. So she plotted on the horizontal axis, she plotted the relative luminosity. So really, the only way that she could measure this is just how bright did they look to her? And she's assuming that they're same distance."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me plot this right over here. So she plotted on the horizontal axis, she plotted the relative luminosity. So really, the only way that she could measure this is just how bright did they look to her? And she's assuming that they're same distance. So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer. So if you assume that they're all roughly the same distance, then how bright it is will tell you how bright it is at the actual star. So she plotted relative luminosity of a star on one axis, and on the other axis, she plotted the period of these variable stars."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And she's assuming that they're same distance. So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer. So if you assume that they're all roughly the same distance, then how bright it is will tell you how bright it is at the actual star. So she plotted relative luminosity of a star on one axis, and on the other axis, she plotted the period of these variable stars. And what I'm going to do is I'm going to do this on a logarithmic scale. So let's say that this is in days. So this is one day, this is 10 days, this is 100 days, right over here."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So she plotted relative luminosity of a star on one axis, and on the other axis, she plotted the period of these variable stars. And what I'm going to do is I'm going to do this on a logarithmic scale. So let's say that this is in days. So this is one day, this is 10 days, this is 100 days, right over here. It's a logarithmic scale because I'm going up in powers of 10. I could say that if we take the log of these, this would be 0, this would be 1, this would be 2. And so that's what I'm using as a scale."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is one day, this is 10 days, this is 100 days, right over here. It's a logarithmic scale because I'm going up in powers of 10. I could say that if we take the log of these, this would be 0, this would be 1, this would be 2. And so that's what I'm using as a scale. So I'm using the log of the period, or I'm just marking them as 1, 10, 100, but I'm giving each of these factors of 10 an equal spacing. When you plot it on this scale, the relative luminosity versus the period, she got a plot that looks something like this. This is obviously not exact."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that's what I'm using as a scale. So I'm using the log of the period, or I'm just marking them as 1, 10, 100, but I'm giving each of these factors of 10 an equal spacing. When you plot it on this scale, the relative luminosity versus the period, she got a plot that looks something like this. This is obviously not exact. She got a plot that looks something like this. It was a fairly linear relationship when you plot the relative luminosity against the log of the period. So this is obviously a logarithmic scale over here."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is obviously not exact. She got a plot that looks something like this. It was a fairly linear relationship when you plot the relative luminosity against the log of the period. So this is obviously a logarithmic scale over here. And so you could fit a line. And why I'd argue, and I think most people would argue, this is one of the most important discoveries in astronomy, is if you know, because think about what the problem here is. We can look at all of these stars in space."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is obviously a logarithmic scale over here. And so you could fit a line. And why I'd argue, and I think most people would argue, this is one of the most important discoveries in astronomy, is if you know, because think about what the problem here is. We can look at all of these stars in space. Let's say you look at a fraction of the sky and you look at something that looks like that. So it's really bright. And then you see something dim that looks like that."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We can look at all of these stars in space. Let's say you look at a fraction of the sky and you look at something that looks like that. So it's really bright. And then you see something dim that looks like that. So if you have a very superficial understanding, you say, oh, this star is brighter. You would say that this is a fundamentally brighter star. But how do you know that?"}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then you see something dim that looks like that. So if you have a very superficial understanding, you say, oh, this star is brighter. You would say that this is a fundamentally brighter star. But how do you know that? Maybe instead of being brighter, maybe it's just a dimmer, closer star. Maybe this is an entire galaxy, but it's so far away that you can't even tell. But all of a sudden, by the work that Henrietta Leavitt did, if you see one of these Cepheid variable stars in another galaxy, you know its relative brightness compared to other Cepheid variable stars."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But how do you know that? Maybe instead of being brighter, maybe it's just a dimmer, closer star. Maybe this is an entire galaxy, but it's so far away that you can't even tell. But all of a sudden, by the work that Henrietta Leavitt did, if you see one of these Cepheid variable stars in another galaxy, you know its relative brightness compared to other Cepheid variable stars. And so if you can place just one of these Cepheid variable stars, if you know exactly the distance to one of them, and then you know its absolute luminosity, you then know the absolute luminosity of any other Cepheid variable stars. So let's say using parallax, which is our other tool, we find, let's say there's some star in our galaxy. And let's say using parallax, we're able to come up with a pretty good measure that it is, I don't know, let's say it's 100 light years away."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But all of a sudden, by the work that Henrietta Leavitt did, if you see one of these Cepheid variable stars in another galaxy, you know its relative brightness compared to other Cepheid variable stars. And so if you can place just one of these Cepheid variable stars, if you know exactly the distance to one of them, and then you know its absolute luminosity, you then know the absolute luminosity of any other Cepheid variable stars. So let's say using parallax, which is our other tool, we find, let's say there's some star in our galaxy. And let's say using parallax, we're able to come up with a pretty good measure that it is, I don't know, let's say it's 100 light years away. And this star is a Cepheid variable star. And let's say its period is one day. So we now know something interesting."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say using parallax, we're able to come up with a pretty good measure that it is, I don't know, let's say it's 100 light years away. And this star is a Cepheid variable star. And let's say its period is one day. So we now know something interesting. We know variable stars with a period of one day at 100 light years away will look like this. Will look like this drawing right over here. So if we later on see a Cepheid variable star with a period of one day, so it gets brighter and dim over the course of one day, and maybe it's redshifted as well, but maybe it looks a little bit dimmer, it looks like this."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we now know something interesting. We know variable stars with a period of one day at 100 light years away will look like this. Will look like this drawing right over here. So if we later on see a Cepheid variable star with a period of one day, so it gets brighter and dim over the course of one day, and maybe it's redshifted as well, but maybe it looks a little bit dimmer, it looks like this. We now know that if it was 100 light years away, it would have this luminosity. So based on how much dimmer it is, we can then figure out how much further away this Cepheid variable star is. If that confuses you a little bit, I'll do a little bit more details in the next few videos so we can get a closer sense of how the math would work."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we later on see a Cepheid variable star with a period of one day, so it gets brighter and dim over the course of one day, and maybe it's redshifted as well, but maybe it looks a little bit dimmer, it looks like this. We now know that if it was 100 light years away, it would have this luminosity. So based on how much dimmer it is, we can then figure out how much further away this Cepheid variable star is. If that confuses you a little bit, I'll do a little bit more details in the next few videos so we can get a closer sense of how the math would work. But this was a big discovery, just discovering this class of stars, this Cepheid variable class. She wasn't the one who discovered them. People knew before her that there were these stars that got brighter and dimmer."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If that confuses you a little bit, I'll do a little bit more details in the next few videos so we can get a closer sense of how the math would work. But this was a big discovery, just discovering this class of stars, this Cepheid variable class. She wasn't the one who discovered them. People knew before her that there were these stars that got brighter and dimmer. But what her big discovery was is seeing this linear relationship between the relative luminosity of these stars and their period. Because then, if we see Cepheid variable stars in completely different galaxies or galactic clusters, by looking at their period, we know what their real relative luminosity is. And then we could guess how far those things really are."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these are these super giant stars, as much as 30,000 times as bright as the sun, as a mass 20 times the mass of the sun. And what's neat about them is, one, because they're so large and so bright, you can see them really, really far away. And what's even neater about them is that they're variable, that they pulsate. And because their pulsations are related to their actual luminosity, you know if you see a Cepheid variable star in some distant galaxy, you know what its luminosity actually is if you were kind of at the star, because you can see how its period of pulsation. And so if you know its actual luminosity, and then you know, obviously, its apparent luminosity, you know how much it's gotten dim. And the more dim it's gotten from its actual state, you know the farther away it is. So that's the real value of them."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And because their pulsations are related to their actual luminosity, you know if you see a Cepheid variable star in some distant galaxy, you know what its luminosity actually is if you were kind of at the star, because you can see how its period of pulsation. And so if you know its actual luminosity, and then you know, obviously, its apparent luminosity, you know how much it's gotten dim. And the more dim it's gotten from its actual state, you know the farther away it is. So that's the real value of them. What I want to do in this video is to try to explain why they're variable, why they pulsate. And to do that, what we're going to think about is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw neutral helium."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's the real value of them. What I want to do in this video is to try to explain why they're variable, why they pulsate. And to do that, what we're going to think about is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw neutral helium. Neutral helium's got two protons, two neutrons, and then two electrons. And obviously, this is not drawn to scale. So this is neutral helium right over here."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just to review, helium, so neutral helium, let me draw neutral helium. Neutral helium's got two protons, two neutrons, and then two electrons. And obviously, this is not drawn to scale. So this is neutral helium right over here. Now, if you singly ionize helium, you knock off one of these electrons. And these type of things happen in stars, when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is neutral helium right over here. Now, if you singly ionize helium, you knock off one of these electrons. And these type of things happen in stars, when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off, so now you only have one electron. And now you have a net positive charge."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off, so now you only have one electron. And now you have a net positive charge. So here, let me do this in a different color. This helium now has a net charge. We could write 1 plus here, but if you just write a plus, you implicitly mean a positive charge of 1."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now you have a net positive charge. So here, let me do this in a different color. This helium now has a net charge. We could write 1 plus here, but if you just write a plus, you implicitly mean a positive charge of 1. Now, you can also doubly ionize helium if the environment is hot enough. You can doubly ionize helium. And doubly ionizing helium is essentially knocking off both of the electrons."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We could write 1 plus here, but if you just write a plus, you implicitly mean a positive charge of 1. Now, you can also doubly ionize helium if the environment is hot enough. You can doubly ionize helium. And doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus like this. This right here is doubly ionized helium. Now, I just said, in order to do this, you have to have a hotter environment."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus like this. This right here is doubly ionized helium. Now, I just said, in order to do this, you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both. This electron really doesn't want to leave. To take an electron off of something that's already positive is difficult."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, I just said, in order to do this, you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both. This electron really doesn't want to leave. To take an electron off of something that's already positive is difficult. You have to have a lot of, really, pressure and temperature. This is cooler. And this is all relative."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "To take an electron off of something that's already positive is difficult. You have to have a lot of, really, pressure and temperature. This is cooler. And this is all relative. We're talking about the insides of stars. So this is a hotter part of the star versus a cooler part of the star, I guess, is the way you think about it. It's a very hot environment by our traditional everyday standards."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is all relative. We're talking about the insides of stars. So this is a hotter part of the star versus a cooler part of the star, I guess, is the way you think about it. It's a very hot environment by our traditional everyday standards. Now, the other thing about the doubly ionized helium is that it is more opaque, which means it doesn't allow light to go through it. It actually absorbs light. It is more opaque."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's a very hot environment by our traditional everyday standards. Now, the other thing about the doubly ionized helium is that it is more opaque, which means it doesn't allow light to go through it. It actually absorbs light. It is more opaque. It absorbs light. Or another way, it absorbs that light energy. That energy will make it even hotter."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is more opaque. It absorbs light. Or another way, it absorbs that light energy. That energy will make it even hotter. So that's just something to think about. Now, the singly ionized helium is more transparent. It allows the light to pass through it."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That energy will make it even hotter. So that's just something to think about. Now, the singly ionized helium is more transparent. It allows the light to pass through it. So it doesn't get heated as much by photons that are kind of going near it or through it or whatever. It allows them to go through it. Here, the photons are going to actually heat up the ion."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It allows the light to pass through it. So it doesn't get heated as much by photons that are kind of going near it or through it or whatever. It allows them to go through it. Here, the photons are going to actually heat up the ion. So let's think about how this might cause a Cepheid variable to pulsate. So assuming that Cepheid variables have large enough quantities, I should say, of these ions, we can imagine that when a Cepheid variable is dim. So let me draw a dim Cepheid variable."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Here, the photons are going to actually heat up the ion. So let's think about how this might cause a Cepheid variable to pulsate. So assuming that Cepheid variables have large enough quantities, I should say, of these ions, we can imagine that when a Cepheid variable is dim. So let me draw a dim Cepheid variable. So I'll draw that like I'll draw this in a dim color. So this is a dim Cepheid variable right here. In its dim state, just like this, you have a lot of the doubly ionized helium in the star, at least kind of the outer surface of the star."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me draw a dim Cepheid variable. So I'll draw that like I'll draw this in a dim color. So this is a dim Cepheid variable right here. In its dim state, just like this, you have a lot of the doubly ionized helium in the star, at least kind of the outer surface of the star. And so this does not allow a lot of light to pass through. So this is the dim part of the pulsation of the Cepheid variable. Now, because this doubly ionized helium is opaque, it is absorbing the light."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In its dim state, just like this, you have a lot of the doubly ionized helium in the star, at least kind of the outer surface of the star. And so this does not allow a lot of light to pass through. So this is the dim part of the pulsation of the Cepheid variable. Now, because this doubly ionized helium is opaque, it is absorbing the light. It is getting heated. It is getting heated. And because it's getting heated, it'll cause the star to expand."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, because this doubly ionized helium is opaque, it is absorbing the light. It is getting heated. It is getting heated. And because it's getting heated, it'll cause the star to expand. So because it's getting heated, it'll become more energetic, and the star will actually expand. Now, as the star expands, because this doubly ionized helium is getting heated, what's going to happen? The further away you are from the core of the star, the cooler it gets."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And because it's getting heated, it'll cause the star to expand. So because it's getting heated, it'll become more energetic, and the star will actually expand. Now, as the star expands, because this doubly ionized helium is getting heated, what's going to happen? The further away you are from the core of the star, the cooler it gets. So this expanded because it was getting heated, but then because it expanded, the outer layers of the star become cooler. And since they're cooler, helium won't be doubly ionized anymore. It'll get an electron from each helium atom."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The further away you are from the core of the star, the cooler it gets. So this expanded because it was getting heated, but then because it expanded, the outer layers of the star become cooler. And since they're cooler, helium won't be doubly ionized anymore. It'll get an electron from each helium atom. It can now get an electron from the plasma, I guess we can say, to become singly ionized helium. So now we have singly ionized helium. And now the star is going to be more transparent."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It'll get an electron from each helium atom. It can now get an electron from the plasma, I guess we can say, to become singly ionized helium. So now we have singly ionized helium. And now the star is going to be more transparent. It's going to allow more light to pass through it. So now this is the bright part of the pulsation. It's going to allow more light through."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now the star is going to be more transparent. It's going to allow more light to pass through it. So now this is the bright part of the pulsation. It's going to allow more light through. So now it is bright. The star is bright. But what's happening now?"}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's going to allow more light through. So now it is bright. The star is bright. But what's happening now? Because the light is no longer, or it's not being absorbed as well by the helium when it was a doubly ionized helium, now it's letting most of the light, or a lot more of the light, get through. It's not going to get heated as much. And so it won't have the kinetic energy to kind of keep pushing out, to keep moving outward."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's happening now? Because the light is no longer, or it's not being absorbed as well by the helium when it was a doubly ionized helium, now it's letting most of the light, or a lot more of the light, get through. It's not going to get heated as much. And so it won't have the kinetic energy to kind of keep pushing out, to keep moving outward. And so it'll collapse back into the star. And so then this will cool down and collapse back in. And when it collapses back in, what's going to happen?"}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so it won't have the kinetic energy to kind of keep pushing out, to keep moving outward. And so it'll collapse back into the star. And so then this will cool down and collapse back in. And when it collapses back in, what's going to happen? When it collapses back in, when these helium atoms get closer to the center of the star, to the core of the star, they're going to be heated again, because they're closer now to the core. And when they get heated, they're going to become doubly ionized. So then we have doubly ionized helium again."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when it collapses back in, what's going to happen? When it collapses back in, when these helium atoms get closer to the center of the star, to the core of the star, they're going to be heated again, because they're closer now to the core. And when they get heated, they're going to become doubly ionized. So then we have doubly ionized helium again. And then the cycle will go again. It is now opaque. It will now absorb more energy."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So then we have doubly ionized helium again. And then the cycle will go again. It is now opaque. It will now absorb more energy. That'll cause it to have more kinetic energy to expand. Once it expands, it'll get cool again, and transparent, and bright. And so this is the current best theory of why Cepheid variable stars are variable to begin with."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "For example, let's look at this substituted cyclohexane on the left here, and let's first try to solve this problem without looking at a chair conformation. So I would start by saying my chlorine is attached to my alpha carbon, and the carbons directly bonded to the alpha carbon would be the beta carbons. So there's a beta carbon on the right, which I call beta one, and there's a beta carbon on the left, which I will call beta two. I know that sodium ethoxide will be my strong base. The ethoxide anion will take a proton from one of those beta carbons, and a double bond would form between the alpha carbon and one of those beta carbons, and the chlorine would leave as the chloride anion. So a double bond forms between the alpha and the beta one carbon, so I'll draw in a double bond here, and the chlorine is now gone, but I would still have this methyl group going away from us in space, so CH3, so that's one possible product. And you might think, oh, I could also form a double bond between the alpha and the beta two carbons, so let me draw a double bond in here, and so my methyl group would just be on a straight line this time, and so you might think that would be one of the products."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "I know that sodium ethoxide will be my strong base. The ethoxide anion will take a proton from one of those beta carbons, and a double bond would form between the alpha carbon and one of those beta carbons, and the chlorine would leave as the chloride anion. So a double bond forms between the alpha and the beta one carbon, so I'll draw in a double bond here, and the chlorine is now gone, but I would still have this methyl group going away from us in space, so CH3, so that's one possible product. And you might think, oh, I could also form a double bond between the alpha and the beta two carbons, so let me draw a double bond in here, and so my methyl group would just be on a straight line this time, and so you might think that would be one of the products. But when you actually think about chair conformations, you'll realize that you would not make this product, so this product is not observed, so you would only get this one. And let's look at why. So first we need to number our cyclohexane ring, and remember, when you number a cyclohexane ring for a chair conformation, you're not necessarily doing IUPAC nomenclature, you're just using whatever numbers you need to help you draw a proper chair conformation."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "And you might think, oh, I could also form a double bond between the alpha and the beta two carbons, so let me draw a double bond in here, and so my methyl group would just be on a straight line this time, and so you might think that would be one of the products. But when you actually think about chair conformations, you'll realize that you would not make this product, so this product is not observed, so you would only get this one. And let's look at why. So first we need to number our cyclohexane ring, and remember, when you number a cyclohexane ring for a chair conformation, you're not necessarily doing IUPAC nomenclature, you're just using whatever numbers you need to help you draw a proper chair conformation. So I'm gonna start with the chlorine being number one, and then I'm gonna go around to the right, so two, three, carbon four, carbon five, and carbon six. So we're not naming this, we're just trying to get a chair conformation. So on my chair conformation down here, I always say this is carbon one, and so I need to have the chlorine up at carbon one, and so I only have one choice, the chlorine has to be up axial."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "So first we need to number our cyclohexane ring, and remember, when you number a cyclohexane ring for a chair conformation, you're not necessarily doing IUPAC nomenclature, you're just using whatever numbers you need to help you draw a proper chair conformation. So I'm gonna start with the chlorine being number one, and then I'm gonna go around to the right, so two, three, carbon four, carbon five, and carbon six. So we're not naming this, we're just trying to get a chair conformation. So on my chair conformation down here, I always say this is carbon one, and so I need to have the chlorine up at carbon one, and so I only have one choice, the chlorine has to be up axial. And so if I go around to carbon six, so this would be carbon one here, two, three, four, five, and six, I have a methyl group going away from me in space, so this would be going down, so I'm gonna draw in an ME for a methyl group right here, so it's down axial. So now let's draw in some hydrogens on our beta carbons. So let me highlight our beta carbons here, I'll use red, so this would be, what I've marked as being beta one, so I have two hydrogens on that carbon, so I'll draw those in here."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "So on my chair conformation down here, I always say this is carbon one, and so I need to have the chlorine up at carbon one, and so I only have one choice, the chlorine has to be up axial. And so if I go around to carbon six, so this would be carbon one here, two, three, four, five, and six, I have a methyl group going away from me in space, so this would be going down, so I'm gonna draw in an ME for a methyl group right here, so it's down axial. So now let's draw in some hydrogens on our beta carbons. So let me highlight our beta carbons here, I'll use red, so this would be, what I've marked as being beta one, so I have two hydrogens on that carbon, so I'll draw those in here. And then my other beta carbon, which I called beta two up here, so I only have, so this is beta two, I have only one hydrogen, and it is equatorial. So let's go to a video so we can analyze which one of these protons will participate in our E two mechanism. Here's our chair conformation, and you can see at the alpha carbon we have the yellow chlorine up and axial."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "So let me highlight our beta carbons here, I'll use red, so this would be, what I've marked as being beta one, so I have two hydrogens on that carbon, so I'll draw those in here. And then my other beta carbon, which I called beta two up here, so I only have, so this is beta two, I have only one hydrogen, and it is equatorial. So let's go to a video so we can analyze which one of these protons will participate in our E two mechanism. Here's our chair conformation, and you can see at the alpha carbon we have the yellow chlorine up and axial. When we go to the beta one carbon, the hydrogen in green is the only one that's antiperiplanar to the halogen. If I turn to the side here, it's easier to see that we have all four of those atoms in the same plane, so the green proton is antiperiplanar to the chlorine. The other hydrogen, the one in white, is not antiperiplanar, so it will not participate in our E two mechanism."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Here's our chair conformation, and you can see at the alpha carbon we have the yellow chlorine up and axial. When we go to the beta one carbon, the hydrogen in green is the only one that's antiperiplanar to the halogen. If I turn to the side here, it's easier to see that we have all four of those atoms in the same plane, so the green proton is antiperiplanar to the chlorine. The other hydrogen, the one in white, is not antiperiplanar, so it will not participate in our E two mechanism. We go to the beta two carbon, and this hydrogen in white is not antiperiplanar, and when we look at the down axial position, it's occupied by a methyl group, so that's where a hydrogen would need to be if it were to participate in an E two mechanism. For E two elimination in cyclohexanes, the halogen must be axial. Here is our halogen that's axial, and when the halogen is axial in this chair conformation, the only hydrogen that's antiperiplanar is this one in green as we saw in the video."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "The other hydrogen, the one in white, is not antiperiplanar, so it will not participate in our E two mechanism. We go to the beta two carbon, and this hydrogen in white is not antiperiplanar, and when we look at the down axial position, it's occupied by a methyl group, so that's where a hydrogen would need to be if it were to participate in an E two mechanism. For E two elimination in cyclohexanes, the halogen must be axial. Here is our halogen that's axial, and when the halogen is axial in this chair conformation, the only hydrogen that's antiperiplanar is this one in green as we saw in the video. If a strong base comes along and takes the proton in green, the electrons in here would move in to form our double bond, and these electrons come off onto the chlorine. A double bond forms between the alpha and the beta one carbons, which would give us this as our only product. We don't get a double bond forming between our alpha and our beta two carbon because we would need to have a hydrogen where our methyl group is."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Here is our halogen that's axial, and when the halogen is axial in this chair conformation, the only hydrogen that's antiperiplanar is this one in green as we saw in the video. If a strong base comes along and takes the proton in green, the electrons in here would move in to form our double bond, and these electrons come off onto the chlorine. A double bond forms between the alpha and the beta one carbons, which would give us this as our only product. We don't get a double bond forming between our alpha and our beta two carbon because we would need to have a hydrogen where our methyl group is. If we did have a hydrogen here, this hydrogen would be antiperiplanar to our halogen, but in this case, we get only one product. This is the only product observed, and we figured that out because we drew our chair conformation. Let's do another E2 mechanism for a substituted cyclohexane, and I'll start by numbering my cyclohexane."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "We don't get a double bond forming between our alpha and our beta two carbon because we would need to have a hydrogen where our methyl group is. If we did have a hydrogen here, this hydrogen would be antiperiplanar to our halogen, but in this case, we get only one product. This is the only product observed, and we figured that out because we drew our chair conformation. Let's do another E2 mechanism for a substituted cyclohexane, and I'll start by numbering my cyclohexane. If that's carbon one, then this is carbon two, this is carbon three, and this is carbon four. Again, the numbering is not for nomenclature. It's just to help me out with my chair conformation."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Let's do another E2 mechanism for a substituted cyclohexane, and I'll start by numbering my cyclohexane. If that's carbon one, then this is carbon two, this is carbon three, and this is carbon four. Again, the numbering is not for nomenclature. It's just to help me out with my chair conformation. If I call this carbon one, I need a methyl group going up, which means the methyl group must be up axial. At carbon three, I need a chlorine that's going down, so this would be carbon two, this would be carbon three, which means the chlorine is down equatorial. Then at carbon four, I have this isopropyl group going down, so this would be carbon four, and going down means the isopropyl group is down axial."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "It's just to help me out with my chair conformation. If I call this carbon one, I need a methyl group going up, which means the methyl group must be up axial. At carbon three, I need a chlorine that's going down, so this would be carbon two, this would be carbon three, which means the chlorine is down equatorial. Then at carbon four, I have this isopropyl group going down, so this would be carbon four, and going down means the isopropyl group is down axial. For an E2 elimination in cyclohexanes, the halogen needs to be axial, but here the halogen is equatorial. Next, we need to think about a ring flip. On the right would be our other chair conformation, and this is carbon one, so if I have my methyl group up axial for the chair conformation on the left, it must be up equatorial for the chair conformation on the right."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Then at carbon four, I have this isopropyl group going down, so this would be carbon four, and going down means the isopropyl group is down axial. For an E2 elimination in cyclohexanes, the halogen needs to be axial, but here the halogen is equatorial. Next, we need to think about a ring flip. On the right would be our other chair conformation, and this is carbon one, so if I have my methyl group up axial for the chair conformation on the left, it must be up equatorial for the chair conformation on the right. This would be carbon two, and this would be carbon three. I need a chlorine that's down, so on the left, the chlorine is down equatorial. On the right for this chair conformation, it would be down axial."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "On the right would be our other chair conformation, and this is carbon one, so if I have my methyl group up axial for the chair conformation on the left, it must be up equatorial for the chair conformation on the right. This would be carbon two, and this would be carbon three. I need a chlorine that's down, so on the left, the chlorine is down equatorial. On the right for this chair conformation, it would be down axial. Then at carbon four, I need to put an isopropyl group going down. On the left, the isopropyl group was down axial. On the right, it'll be down equatorial."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "On the right for this chair conformation, it would be down axial. Then at carbon four, I need to put an isopropyl group going down. On the left, the isopropyl group was down axial. On the right, it'll be down equatorial. The chlorine is now axial, and so this is my alpha carbon right here. Now I can look for beta carbons, so the carbons next door to the alpha carbon. Here's a beta carbon, which I'll mark as being beta one, and then here's a beta carbon, which I'll mark as being beta two."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "On the right, it'll be down equatorial. The chlorine is now axial, and so this is my alpha carbon right here. Now I can look for beta carbons, so the carbons next door to the alpha carbon. Here's a beta carbon, which I'll mark as being beta one, and then here's a beta carbon, which I'll mark as being beta two. Let's look for hydrogens that are antiperiplanar on those beta carbons. Here's a hydrogen on beta one that's antiperiplanar, and then on beta two, here's a hydrogen that is antiperiplanar. Those are the protons that will be lost in our mechanism."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Here's a beta carbon, which I'll mark as being beta one, and then here's a beta carbon, which I'll mark as being beta two. Let's look for hydrogens that are antiperiplanar on those beta carbons. Here's a hydrogen on beta one that's antiperiplanar, and then on beta two, here's a hydrogen that is antiperiplanar. Those are the protons that will be lost in our mechanism. Before we draw the products, let's think about which one of these conformations is more stable. We saw in earlier videos that you want to put the bulky groups equatorial out to the side. The bulkiest group is this isopropyl group right here, and it's more stable when it's out to the side."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Those are the protons that will be lost in our mechanism. Before we draw the products, let's think about which one of these conformations is more stable. We saw in earlier videos that you want to put the bulky groups equatorial out to the side. The bulkiest group is this isopropyl group right here, and it's more stable when it's out to the side. When this isopropyl group is axial, there's a lot of interaction with other groups on the ring. The conformation on the right is the more stable conformation. Let's take a look at that in a video."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "The bulkiest group is this isopropyl group right here, and it's more stable when it's out to the side. When this isopropyl group is axial, there's a lot of interaction with other groups on the ring. The conformation on the right is the more stable conformation. Let's take a look at that in a video. At carbon one, I have a methyl group that's up axial. This is carbon two, and this is carbon three with my chlorine down equatorial. At carbon four, I have my isopropyl group down axial, but you can see all of the steric hindrance."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Let's take a look at that in a video. At carbon one, I have a methyl group that's up axial. This is carbon two, and this is carbon three with my chlorine down equatorial. At carbon four, I have my isopropyl group down axial, but you can see all of the steric hindrance. That destabilizes this conformation. This is the less stable conformation. We do a ring flip, and it's hard to do with this model set, but I'm going to approximate the other chair conformation."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "At carbon four, I have my isopropyl group down axial, but you can see all of the steric hindrance. That destabilizes this conformation. This is the less stable conformation. We do a ring flip, and it's hard to do with this model set, but I'm going to approximate the other chair conformation. At carbon one, the methyl group is now up equatorial. At carbon three, the halogen is down axial. At carbon four, the bulky isopropyl group is equatorial and out to the side."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "We do a ring flip, and it's hard to do with this model set, but I'm going to approximate the other chair conformation. At carbon one, the methyl group is now up equatorial. At carbon three, the halogen is down axial. At carbon four, the bulky isopropyl group is equatorial and out to the side. The carbon bond to the halogen is my alpha carbon, and next to that would be a beta carbon. This beta hydrogen is antiperiplanar to this halogen. We have another beta carbon over here with another hydrogen that's antiperiplanar to this halogen."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "At carbon four, the bulky isopropyl group is equatorial and out to the side. The carbon bond to the halogen is my alpha carbon, and next to that would be a beta carbon. This beta hydrogen is antiperiplanar to this halogen. We have another beta carbon over here with another hydrogen that's antiperiplanar to this halogen. Finally, let's draw our two products. Let's take a proton from the beta one position first. Our base, let me draw it in here."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "We have another beta carbon over here with another hydrogen that's antiperiplanar to this halogen. Finally, let's draw our two products. Let's take a proton from the beta one position first. Our base, let me draw it in here. Our strong base is going to take this proton. These electrons move into here to form our double bond, and these electrons come off onto the chlorine to form the chloride anion as our leaving group. This would form a double bond between what I called carbons two and three, because this is carbon one, this is carbon two, this is carbon three, and this is carbon four."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Our base, let me draw it in here. Our strong base is going to take this proton. These electrons move into here to form our double bond, and these electrons come off onto the chlorine to form the chloride anion as our leaving group. This would form a double bond between what I called carbons two and three, because this is carbon one, this is carbon two, this is carbon three, and this is carbon four. We have a methyl group that's up at carbon one. Let me draw in our methyl group up at carbon one. We put that on a wedge."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "This would form a double bond between what I called carbons two and three, because this is carbon one, this is carbon two, this is carbon three, and this is carbon four. We have a methyl group that's up at carbon one. Let me draw in our methyl group up at carbon one. We put that on a wedge. If that's carbon one, then this is carbon two, and this is carbon three, and that's where our double bond forms. The double bond forms between carbons two and three. Then at carbon four, we have this isopropyl group going down."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "We put that on a wedge. If that's carbon one, then this is carbon two, and this is carbon three, and that's where our double bond forms. The double bond forms between carbons two and three. Then at carbon four, we have this isopropyl group going down. Let me go ahead and put that in going away from us. We put that on a dash. That's one product."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Then at carbon four, we have this isopropyl group going down. Let me go ahead and put that in going away from us. We put that on a dash. That's one product. If our base took this proton, then the electrons would move into here, and these electrons come off onto the chlorine. If we took a proton from the beta two carbon, we would form a double bond between carbons three and four. Here's carbons three and four."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "That's one product. If our base took this proton, then the electrons would move into here, and these electrons come off onto the chlorine. If we took a proton from the beta two carbon, we would form a double bond between carbons three and four. Here's carbons three and four. I put a double bond in there. I still have a methyl group that's going up at what I called carbon one. My isopropyl group is at carbon four, but since now this carbon is sp two hybridized, I need to draw in this isopropyl group on a straight line."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Here's carbons three and four. I put a double bond in there. I still have a methyl group that's going up at what I called carbon one. My isopropyl group is at carbon four, but since now this carbon is sp two hybridized, I need to draw in this isopropyl group on a straight line. Sometimes students will put this isopropyl group in on a wedge or a dash, but you're trying to show the planar geometry around this carbon, so a straight line is what you need. Those are the two products. Let's do one more."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "My isopropyl group is at carbon four, but since now this carbon is sp two hybridized, I need to draw in this isopropyl group on a straight line. Sometimes students will put this isopropyl group in on a wedge or a dash, but you're trying to show the planar geometry around this carbon, so a straight line is what you need. Those are the two products. Let's do one more. You can see this substrate is very similar to the one we did in the previous example. The only difference is this time the chlorine is on a wedge instead of a dash. If I number my ring one, two, three, four, I've already put in both chair conformations to save time."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "Let's do one more. You can see this substrate is very similar to the one we did in the previous example. The only difference is this time the chlorine is on a wedge instead of a dash. If I number my ring one, two, three, four, I've already put in both chair conformations to save time. That's carbon one. This is carbon two. This is carbon three, and this is carbon four."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "If I number my ring one, two, three, four, I've already put in both chair conformations to save time. That's carbon one. This is carbon two. This is carbon three, and this is carbon four. On the other chair conformation, this is carbon one, two, three, and four. Notice for the chair conformation on the left, we have the chlorine in the axial position. This would be the alpha carbon, and the carbons next to the alpha carbon would be the beta carbon."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "This is carbon three, and this is carbon four. On the other chair conformation, this is carbon one, two, three, and four. Notice for the chair conformation on the left, we have the chlorine in the axial position. This would be the alpha carbon, and the carbons next to the alpha carbon would be the beta carbon. This one on the right is a beta carbon, and the one on the left is a beta carbon. We need a proton that's antiperiplanar, and the only one that fits would be this hydrogen right here. If a base takes that proton, so let me draw in our strong base, taking this proton, these electrons would move into here to form our double bond."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "This would be the alpha carbon, and the carbons next to the alpha carbon would be the beta carbon. This one on the right is a beta carbon, and the one on the left is a beta carbon. We need a proton that's antiperiplanar, and the only one that fits would be this hydrogen right here. If a base takes that proton, so let me draw in our strong base, taking this proton, these electrons would move into here to form our double bond. The electrons come off onto our chlorine. A double bond forms between carbons two and three. When we draw our product, let me put the ring in here, carbon one has a methyl group that's up."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "If a base takes that proton, so let me draw in our strong base, taking this proton, these electrons would move into here to form our double bond. The electrons come off onto our chlorine. A double bond forms between carbons two and three. When we draw our product, let me put the ring in here, carbon one has a methyl group that's up. Let me draw in our CH three, which is up. The double bond formed between carbons two and three. This is carbon one."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "When we draw our product, let me put the ring in here, carbon one has a methyl group that's up. Let me draw in our CH three, which is up. The double bond formed between carbons two and three. This is carbon one. This is carbon two. This is carbon three. We draw in our double bond here."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "This is carbon one. This is carbon two. This is carbon three. We draw in our double bond here. Then we have our isopropyl group going down at carbon four. Let me draw in our isopropyl group going down. This is the only product for the reaction."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "We draw in our double bond here. Then we have our isopropyl group going down at carbon four. Let me draw in our isopropyl group going down. This is the only product for the reaction. We don't have any other protons that are antiperiplanar to our chlorine. When we think about our two chair conformations, the one on the left has all of our groups axial, and the one on the right has all of our groups equatorial. The one on the right is actually much more stable."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "This is the only product for the reaction. We don't have any other protons that are antiperiplanar to our chlorine. When we think about our two chair conformations, the one on the left has all of our groups axial, and the one on the right has all of our groups equatorial. The one on the right is actually much more stable. This is more stable with all of the bulky groups out to the side, which means that this is a relatively slow reaction. The reaction is slow because the equilibrium favors the conformation that has the bulky groups equatorial. But that's not the conformation that has the requirements for our E two mechanism."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "The one on the right is actually much more stable. This is more stable with all of the bulky groups out to the side, which means that this is a relatively slow reaction. The reaction is slow because the equilibrium favors the conformation that has the bulky groups equatorial. But that's not the conformation that has the requirements for our E two mechanism. It's the less stable conformation that has the antiperiplanar hydrogen and chlorine. This reaction is much slower than the one that we just talked about. Let's go back up here so we can compare."}, {"video_title": "E2 mechanism substituted cyclohexane.mp3", "Sentence": "But that's not the conformation that has the requirements for our E two mechanism. It's the less stable conformation that has the antiperiplanar hydrogen and chlorine. This reaction is much slower than the one that we just talked about. Let's go back up here so we can compare. Let's look at this reaction. For this reaction, the more stable conformation is the one with our bulky groups equatorial, so the equilibrium favors this conformation. This conformation is also the one that had our hydrogens antiperiplanar to our halogens."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And for time reasons, I have to assume that you're already familiar with the E2 mechanism and that you understand Newman projections and Sawhorse projections. So the carbon that's bonded to the bromine would be our alpha carbon. The carbon bonded to the alpha carbon would be our beta carbon. And we know in an E2 mechanism, we are going to take this beta proton. Our strong base is gonna take that beta proton. So let's look down that beta alpha bond here. So we're gonna stare down this bond so we can see a Newman projection."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And we know in an E2 mechanism, we are going to take this beta proton. Our strong base is gonna take that beta proton. So let's look down that beta alpha bond here. So we're gonna stare down this bond so we can see a Newman projection. So if you put your eye right here and you look down this bond, you're gonna see a methyl group that's going up and to the right. So down here is our Newman projection. Here's our methyl group."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So we're gonna stare down this bond so we can see a Newman projection. So if you put your eye right here and you look down this bond, you're gonna see a methyl group that's going up and to the right. So down here is our Newman projection. Here's our methyl group. The hydrogen would be going up and to the left. So here's our hydrogen in white. Our phenyl group, which remember, just a benzene ring, so this is going to be going down on our Newman projection."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "Here's our methyl group. The hydrogen would be going up and to the left. So here's our hydrogen in white. Our phenyl group, which remember, just a benzene ring, so this is going to be going down on our Newman projection. And we think about the back carbon, the alpha carbon, we have a bromine, which would be going down and to the right. We know there's also a hydrogen bonded to this alpha carbon. That'd be going down and to the left."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "Our phenyl group, which remember, just a benzene ring, so this is going to be going down on our Newman projection. And we think about the back carbon, the alpha carbon, we have a bromine, which would be going down and to the right. We know there's also a hydrogen bonded to this alpha carbon. That'd be going down and to the left. And then we have another phenyl group, which would be going straight up in this Newman projection. So here is the phenyl group. We know that the E2 mechanism needs to have an antiperiplanar hydrogen and bromine."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "That'd be going down and to the left. And then we have another phenyl group, which would be going straight up in this Newman projection. So here is the phenyl group. We know that the E2 mechanism needs to have an antiperiplanar hydrogen and bromine. So if we look at this Newman projection, we already have that. This hydrogen in white is already antiperiplanar with this bromine. So we already have what we need for the E2 mechanism to occur."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "We know that the E2 mechanism needs to have an antiperiplanar hydrogen and bromine. So if we look at this Newman projection, we already have that. This hydrogen in white is already antiperiplanar with this bromine. So we already have what we need for the E2 mechanism to occur. On the right here, this is really the same Newman projection. I just turned the whole thing a little bit to the right. So that line that I just drew in magenta would be this line."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So we already have what we need for the E2 mechanism to occur. On the right here, this is really the same Newman projection. I just turned the whole thing a little bit to the right. So that line that I just drew in magenta would be this line. And so that would be the hydrogen in white. And that would mean the methyl group would be this one right here in yellow. And the phenyl group would be going down and to the left."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So that line that I just drew in magenta would be this line. And so that would be the hydrogen in white. And that would mean the methyl group would be this one right here in yellow. And the phenyl group would be going down and to the left. So talking about the front carbon. For the back carbon, the bromine would be going down now. The phenyl group would be going up and to the right."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And the phenyl group would be going down and to the left. So talking about the front carbon. For the back carbon, the bromine would be going down now. The phenyl group would be going up and to the right. And the hydrogen would be the one in yellow. So again, this is the same Newman projection, just turned a little bit. So it's a little bit easier to see our antiperiplanar proton and our bromine."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "The phenyl group would be going up and to the right. And the hydrogen would be the one in yellow. So again, this is the same Newman projection, just turned a little bit. So it's a little bit easier to see our antiperiplanar proton and our bromine. You could draw the product at this point from this Newman projection. You think about a strong base coming along and taking your beta proton. These electrons would move in to form your double bond."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So it's a little bit easier to see our antiperiplanar proton and our bromine. You could draw the product at this point from this Newman projection. You think about a strong base coming along and taking your beta proton. These electrons would move in to form your double bond. At the same time, these electrons come off onto the bromine to form the bromide anion. So let's go ahead and draw the product for this reaction. So from the Newman projection, you need to think about this phenyl group and this methyl group being on the same side of the double bond that formed."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "These electrons would move in to form your double bond. At the same time, these electrons come off onto the bromine to form the bromide anion. So let's go ahead and draw the product for this reaction. So from the Newman projection, you need to think about this phenyl group and this methyl group being on the same side of the double bond that formed. So let me draw in our double bond here. So we would have our phenyl group and our methyl group on the same side of the double bond. On the other side of the double bond, we would form, we would have, I should say, this hydrogen and this phenyl group."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So from the Newman projection, you need to think about this phenyl group and this methyl group being on the same side of the double bond that formed. So let me draw in our double bond here. So we would have our phenyl group and our methyl group on the same side of the double bond. On the other side of the double bond, we would form, we would have, I should say, this hydrogen and this phenyl group. So let's draw those in. So on this back carbon here, that would be the hydrogen. And on the front carbon, I would have my phenyl group."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "On the other side of the double bond, we would form, we would have, I should say, this hydrogen and this phenyl group. So let's draw those in. So on this back carbon here, that would be the hydrogen. And on the front carbon, I would have my phenyl group. So that is the product of this reaction. It might be a little bit hard to see from this Newman projection, so here's a Sawhorse projection, another viewpoint, which might be easier to see. So let's highlight everything again."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And on the front carbon, I would have my phenyl group. So that is the product of this reaction. It might be a little bit hard to see from this Newman projection, so here's a Sawhorse projection, another viewpoint, which might be easier to see. So let's highlight everything again. So this would be our alpha carbon and this would be our beta carbon. So our strong base is gonna take this beta proton and these electrons would move into here to form our double bond and these electrons come off onto the bromine. So again, the phenyl group and the methyl group would end up on the same side, so here they are."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So let's highlight everything again. So this would be our alpha carbon and this would be our beta carbon. So our strong base is gonna take this beta proton and these electrons would move into here to form our double bond and these electrons come off onto the bromine. So again, the phenyl group and the methyl group would end up on the same side, so here they are. And on the other side, you would have this phenyl group and this hydrogen, so this phenyl group and this hydrogen. So the two phenyl groups are on opposite sides of the double bond and this would be the E alkene. Let's look at this on a video and I think it might be really helpful to look at the model forming the alkene."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So again, the phenyl group and the methyl group would end up on the same side, so here they are. And on the other side, you would have this phenyl group and this hydrogen, so this phenyl group and this hydrogen. So the two phenyl groups are on opposite sides of the double bond and this would be the E alkene. Let's look at this on a video and I think it might be really helpful to look at the model forming the alkene. Let's look at our alkyl halides. We can see that the purple groups are the phenyl groups, the red is the bromine, and at this carbon, we have a methyl group coming out at us in space and a hydrogen going away from us in space. We're gonna stare down this beta alpha bond and notice that I'm using the stretchy bonds again so we can better visualize the E2 mechanism."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "Let's look at this on a video and I think it might be really helpful to look at the model forming the alkene. Let's look at our alkyl halides. We can see that the purple groups are the phenyl groups, the red is the bromine, and at this carbon, we have a methyl group coming out at us in space and a hydrogen going away from us in space. We're gonna stare down this beta alpha bond and notice that I'm using the stretchy bonds again so we can better visualize the E2 mechanism. So let's stare down that bond and look at the Newman projection. So here in this Newman projection, we already have our beta hydrogen antiperiplanar with our bromine, so if I rotate it a little bit, so there's our Newman projection. I'm also gonna turn it to the side a little bit so we can see it from a sawhorse projection."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "We're gonna stare down this beta alpha bond and notice that I'm using the stretchy bonds again so we can better visualize the E2 mechanism. So let's stare down that bond and look at the Newman projection. So here in this Newman projection, we already have our beta hydrogen antiperiplanar with our bromine, so if I rotate it a little bit, so there's our Newman projection. I'm also gonna turn it to the side a little bit so we can see it from a sawhorse projection. And our strong base, we think about our strong base coming along and taking this proton. So pretend like the proton goes away, pretend like the bromine goes away, and you can see the double bond that formed. And notice how the two phenyl groups are on opposite sides of the double bond."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "I'm also gonna turn it to the side a little bit so we can see it from a sawhorse projection. And our strong base, we think about our strong base coming along and taking this proton. So pretend like the proton goes away, pretend like the bromine goes away, and you can see the double bond that formed. And notice how the two phenyl groups are on opposite sides of the double bond. So that's our product. Let's do another E2 reaction and notice how similar this is to the last one. The only difference is the stereochemistry at this carbon."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And notice how the two phenyl groups are on opposite sides of the double bond. So that's our product. Let's do another E2 reaction and notice how similar this is to the last one. The only difference is the stereochemistry at this carbon. Now the hydrogen is coming out at us in space and the methyl group is going away from us in space. So we'll see how changing the stereochemistry affects the stereochemistry of the product. So we need to stare down our bond again, so this would be our alpha carbon, this would be our beta carbon."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "The only difference is the stereochemistry at this carbon. Now the hydrogen is coming out at us in space and the methyl group is going away from us in space. So we'll see how changing the stereochemistry affects the stereochemistry of the product. So we need to stare down our bond again, so this would be our alpha carbon, this would be our beta carbon. We're gonna stare down this way, just like we did in the previous example. So let's think about the Newman projection that we would have. Now our hydrogen, I'll highlight in white here, would be going up and to the right on our Newman projection."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So we need to stare down our bond again, so this would be our alpha carbon, this would be our beta carbon. We're gonna stare down this way, just like we did in the previous example. So let's think about the Newman projection that we would have. Now our hydrogen, I'll highlight in white here, would be going up and to the right on our Newman projection. Our methyl group would be going up and to the left. And our phenyl group would be going down. So here's the phenyl group, so that's our front carbon."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "Now our hydrogen, I'll highlight in white here, would be going up and to the right on our Newman projection. Our methyl group would be going up and to the left. And our phenyl group would be going down. So here's the phenyl group, so that's our front carbon. Our back carbon would have our bromine going down and to the right. The hydrogen on the alpha carbon would be going down and to the left. And then we have our phenyl group going straight up."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So here's the phenyl group, so that's our front carbon. Our back carbon would have our bromine going down and to the right. The hydrogen on the alpha carbon would be going down and to the left. And then we have our phenyl group going straight up. So there's our Newman projection. Notice that this time our beta hydrogen, so right here, this is not antiperiplanar with our bromine. So we're going to need to rotate about our single bond and we'll do that in a minute."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And then we have our phenyl group going straight up. So there's our Newman projection. Notice that this time our beta hydrogen, so right here, this is not antiperiplanar with our bromine. So we're going to need to rotate about our single bond and we'll do that in a minute. First let's just get to this Newman projection on the right. So I'm just gonna draw a line going through this methyl group and this bromine, just to help us visualize the fact that here is that same methyl group and then here is that bromine. So again, this is the same Newman projection, it's just rotated a little bit to the right."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So we're going to need to rotate about our single bond and we'll do that in a minute. First let's just get to this Newman projection on the right. So I'm just gonna draw a line going through this methyl group and this bromine, just to help us visualize the fact that here is that same methyl group and then here is that bromine. So again, this is the same Newman projection, it's just rotated a little bit to the right. So let me highlight everything. So this would be the hydrogen in white and we have our phenyl group over here. And then on our back carbon, we would have our hydrogen in yellow and then we have our phenyl group."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So again, this is the same Newman projection, it's just rotated a little bit to the right. So let me highlight everything. So this would be the hydrogen in white and we have our phenyl group over here. And then on our back carbon, we would have our hydrogen in yellow and then we have our phenyl group. So our phenyl groups are still anti to each other. So let's go to the video now so we can rotate about our single bond so we can get into a confirmation that has our beta proton antiperiplanar to our bromine. Let's think about the stereochemistry at this carbon."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And then on our back carbon, we would have our hydrogen in yellow and then we have our phenyl group. So our phenyl groups are still anti to each other. So let's go to the video now so we can rotate about our single bond so we can get into a confirmation that has our beta proton antiperiplanar to our bromine. Let's think about the stereochemistry at this carbon. Now the hydrogen's coming out at us and the methyl group is going away from us. So if we stare down this carbon-carbon bond, we will see our Newman projection. And notice how our beta hydrogen is not antiperiplanar with our bromine."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "Let's think about the stereochemistry at this carbon. Now the hydrogen's coming out at us and the methyl group is going away from us. So if we stare down this carbon-carbon bond, we will see our Newman projection. And notice how our beta hydrogen is not antiperiplanar with our bromine. So I'm gonna rotate the whole thing a little bit and then I'm gonna rotate about the single bond so we can get in the proper confirmation. So now our beta proton is antiperiplanar with our bromine. And notice how the two phenyl groups are now gauche to each other."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And notice how our beta hydrogen is not antiperiplanar with our bromine. So I'm gonna rotate the whole thing a little bit and then I'm gonna rotate about the single bond so we can get in the proper confirmation. So now our beta proton is antiperiplanar with our bromine. And notice how the two phenyl groups are now gauche to each other. So if I turn a little bit to see things from a sawhorse projection, we think about taking away this beta proton. So the beta proton goes away and pretend like the bromine goes away as well. And we can see that for our product, our two phenyl groups end up on the same side of the double bond."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And notice how the two phenyl groups are now gauche to each other. So if I turn a little bit to see things from a sawhorse projection, we think about taking away this beta proton. So the beta proton goes away and pretend like the bromine goes away as well. And we can see that for our product, our two phenyl groups end up on the same side of the double bond. Hopefully the video made it clear how to figure out the products. But if you don't have a model set on a test, you have to use your own drawings. So this line in magenta that I drew right here, I forgot to draw it in on this Newman projection so that's this line."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And we can see that for our product, our two phenyl groups end up on the same side of the double bond. Hopefully the video made it clear how to figure out the products. But if you don't have a model set on a test, you have to use your own drawings. So this line in magenta that I drew right here, I forgot to draw it in on this Newman projection so that's this line. And then when you realize that your proton is not antiperiplanar, you'll have to rotate. You'll have to rotate yourself. And I like to think about rotating the front carbon here."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So this line in magenta that I drew right here, I forgot to draw it in on this Newman projection so that's this line. And then when you realize that your proton is not antiperiplanar, you'll have to rotate. You'll have to rotate yourself. And I like to think about rotating the front carbon here. I could take this hydrogen and I'm gonna move it up to here, which would move this methyl group over to this position and it moves the phenyl group over to this position. So the hydrogen in white ends up being vertical. The methyl group ends up going down and to the left and the phenyl group ends up going to the right."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And I like to think about rotating the front carbon here. I could take this hydrogen and I'm gonna move it up to here, which would move this methyl group over to this position and it moves the phenyl group over to this position. So the hydrogen in white ends up being vertical. The methyl group ends up going down and to the left and the phenyl group ends up going to the right. So you don't necessarily need a model set to do this. And now we have our hydrogen antiperiplanar with our bromine so from this Newman projection, you could draw the product. A base takes our beta proton, these electrons move in here and these electrons come off onto our bromine."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "The methyl group ends up going down and to the left and the phenyl group ends up going to the right. So you don't necessarily need a model set to do this. And now we have our hydrogen antiperiplanar with our bromine so from this Newman projection, you could draw the product. A base takes our beta proton, these electrons move in here and these electrons come off onto our bromine. And so our two phenyl groups would be on the same side of the double bond. Notice how they are gauche to each other in this Newman projection. So if I draw in my double bond on the back carbon, I have a phenyl group."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "A base takes our beta proton, these electrons move in here and these electrons come off onto our bromine. And so our two phenyl groups would be on the same side of the double bond. Notice how they are gauche to each other in this Newman projection. So if I draw in my double bond on the back carbon, I have a phenyl group. On the front carbon here, I have a phenyl group. And then the groups on the other side of the double bond, I have a hydrogen on the back carbon and a methyl group on the front. So let me draw those in up here."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So if I draw in my double bond on the back carbon, I have a phenyl group. On the front carbon here, I have a phenyl group. And then the groups on the other side of the double bond, I have a hydrogen on the back carbon and a methyl group on the front. So let me draw those in up here. So I have a methyl group on the front carbon and a hydrogen on the back. If you prefer the saw horse projection point of view, the base would take our beta proton, these electrons move in to form our double bond, these electrons come off onto the bromine and we have our two phenyl groups on the same side. And so here they are, they're on the same side."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So let me draw those in up here. So I have a methyl group on the front carbon and a hydrogen on the back. If you prefer the saw horse projection point of view, the base would take our beta proton, these electrons move in to form our double bond, these electrons come off onto the bromine and we have our two phenyl groups on the same side. And so here they are, they're on the same side. And then we would have our methyl group and our hydrogen on the same side of our double bond. Notice this is the Z alkene. So different stereochemistry for our product as compared to the first reaction."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "And so here they are, they're on the same side. And then we would have our methyl group and our hydrogen on the same side of our double bond. Notice this is the Z alkene. So different stereochemistry for our product as compared to the first reaction. And actually this reaction would be much slower than the first reaction. And that's because of these two phenyl groups here. So this conformation where we have our proton and our bromine antiperiplanar, this is definitely not the most stable conformation."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So different stereochemistry for our product as compared to the first reaction. And actually this reaction would be much slower than the first reaction. And that's because of these two phenyl groups here. So this conformation where we have our proton and our bromine antiperiplanar, this is definitely not the most stable conformation. We have these really bulky phenyl groups gauche to each other. And that means that this would be slower than the first reaction because this is the conformation we must have for this mechanism to occur. But it's not a very stable conformation."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So this conformation where we have our proton and our bromine antiperiplanar, this is definitely not the most stable conformation. We have these really bulky phenyl groups gauche to each other. And that means that this would be slower than the first reaction because this is the conformation we must have for this mechanism to occur. But it's not a very stable conformation. If we go back up here to the first reaction that we did, when we had our proton antiperiplanar to our bromine, our two phenyl groups were anti to each other. So they're very relatively far away from each other. So this would be a stable conformation."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "But it's not a very stable conformation. If we go back up here to the first reaction that we did, when we had our proton antiperiplanar to our bromine, our two phenyl groups were anti to each other. So they're very relatively far away from each other. So this would be a stable conformation. And so this reaction happens a lot faster than our second reaction. Finally, let's summarize the two reactions that we've seen. So on the left, the stereochemistry at this carbon with our methyl group coming out at us and our hydrogen going away from us, that gave us the E alkene as our product with our two phenyl groups on opposite sides of the double bonds."}, {"video_title": "E2 mechanism stereospecificity.mp3", "Sentence": "So this would be a stable conformation. And so this reaction happens a lot faster than our second reaction. Finally, let's summarize the two reactions that we've seen. So on the left, the stereochemistry at this carbon with our methyl group coming out at us and our hydrogen going away from us, that gave us the E alkene as our product with our two phenyl groups on opposite sides of the double bonds. When we change the stereochemistry, so now the hydrogen's coming out at us and the methyl group is going away from us, we got a different product. We got the Z alkene with our two phenyl groups on the same side of our double bond. So that shows you how this is a stereospecific reaction."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "If we look at this blue molecule right here, we always want to start off by finding the longest carbon chain. It looks like it's this chain at the bottom where we have 1, 2, 3, 4, 5, 6, 7, 8 carbons. So it's going to be octane. We don't want to just call it octane because we have these functional groups up here. And maybe one of them is going to take higher priority than just the fact that this is an alkane. And in fact, one of them will. So let's look at them."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "We don't want to just call it octane because we have these functional groups up here. And maybe one of them is going to take higher priority than just the fact that this is an alkane. And in fact, one of them will. So let's look at them. So over here we have an amino group. That would make this molecule an amine. So this is an amino group right here."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So let's look at them. So over here we have an amino group. That would make this molecule an amine. So this is an amino group right here. Then we have this benzene ring. And when benzene is a functional group, we call it a phenyl group. And that's derived from, I think it's either the Greek or the Latin word for light because benzene was first isolated from, I guess, the air around lanterns or something."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So this is an amino group right here. Then we have this benzene ring. And when benzene is a functional group, we call it a phenyl group. And that's derived from, I think it's either the Greek or the Latin word for light because benzene was first isolated from, I guess, the air around lanterns or something. So this is called a phenyl group. And then finally we have this OH, which would make this whole thing an alcohol. Or we could call that a hydroxy group."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "And that's derived from, I think it's either the Greek or the Latin word for light because benzene was first isolated from, I guess, the air around lanterns or something. So this is called a phenyl group. And then finally we have this OH, which would make this whole thing an alcohol. Or we could call that a hydroxy group. So this right here is a hydroxy group. Now this is just something that you need to know. There's no way to kind of deduce it."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "Or we could call that a hydroxy group. So this right here is a hydroxy group. Now this is just something that you need to know. There's no way to kind of deduce it. The people who've decided how to name things just decide that certain groups take higher priority over others. And out of these three groups, the hydroxy group, the thing that's making this whole thing an alcohol, takes the highest priority. So this is what's going to define the suffix."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "There's no way to kind of deduce it. The people who've decided how to name things just decide that certain groups take higher priority over others. And out of these three groups, the hydroxy group, the thing that's making this whole thing an alcohol, takes the highest priority. So this is what's going to define the suffix. So if we start numbering, we want to start numbering this chain closest to the highest priority group. So we'll start numbering at this end. So there's 1, 2, 3, 4, 5, 6, 7, 8."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So this is what's going to define the suffix. So if we start numbering, we want to start numbering this chain closest to the highest priority group. So we'll start numbering at this end. So there's 1, 2, 3, 4, 5, 6, 7, 8. So we have a hydroxy group on the third three carbons. So this is octan 3-ol. If this was not a high priority, we would put a hydroxy in the front."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So there's 1, 2, 3, 4, 5, 6, 7, 8. So we have a hydroxy group on the third three carbons. So this is octan 3-ol. If this was not a high priority, we would put a hydroxy in the front. But since this is the highest priority, it's defining this as an alcohol, octanol. But we put the 3 there to say it's on the three carbon. And then of course, this would be 5-phenyl."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "If this was not a high priority, we would put a hydroxy in the front. But since this is the highest priority, it's defining this as an alcohol, octanol. But we put the 3 there to say it's on the three carbon. And then of course, this would be 5-phenyl. This right here is 5-phenyl. And this right here is 8-amino. And amino comes before phenol in alphabetical order."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "And then of course, this would be 5-phenyl. This right here is 5-phenyl. And this right here is 8-amino. And amino comes before phenol in alphabetical order. So this whole molecule is going to be 8-amino, 5-phenyl, octan 3-ol. And we're done. Octan 3-ol."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "And amino comes before phenol in alphabetical order. So this whole molecule is going to be 8-amino, 5-phenyl, octan 3-ol. And we're done. Octan 3-ol. And we're done. So this is an aromatic compound because of the benzene ring, it's an amine because of the amino group, and it's an alcohol because of the hydroxy group. Now let's look at this molecule here in green."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "Octan 3-ol. And we're done. So this is an aromatic compound because of the benzene ring, it's an amine because of the amino group, and it's an alcohol because of the hydroxy group. Now let's look at this molecule here in green. And actually, I forgot to draw a part of it. This is supposed to be a benzene ring. So let me make clear."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "Now let's look at this molecule here in green. And actually, I forgot to draw a part of it. This is supposed to be a benzene ring. So let me make clear. That is a benzene ring. So let's think about the different functional groups here, and we'll start with the one that you've probably never seen before. This is a nitro group."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So let me make clear. That is a benzene ring. So let's think about the different functional groups here, and we'll start with the one that you've probably never seen before. This is a nitro group. And it's highly explosive. Maybe in a future video I'll show you some nitroglycerin and some TNT, as you know, which are highly explosive. But they have nitro groups on them."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "This is a nitro group. And it's highly explosive. Maybe in a future video I'll show you some nitroglycerin and some TNT, as you know, which are highly explosive. But they have nitro groups on them. So this right here is a nitro group. It takes very low precedence in the whole scheme of things. So this will almost never, actually never, be the cause of naming the suffix."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "But they have nitro groups on them. So this right here is a nitro group. It takes very low precedence in the whole scheme of things. So this will almost never, actually never, be the cause of naming the suffix. We just know that this is a nitro group right here. We, of course, have a bromo group that we've seen many times. That is a bromo group."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So this will almost never, actually never, be the cause of naming the suffix. We just know that this is a nitro group right here. We, of course, have a bromo group that we've seen many times. That is a bromo group. And then we have an amino group. We have an amino group right here. So this is going to be, let me do this in a different color."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "That is a bromo group. And then we have an amino group. We have an amino group right here. So this is going to be, let me do this in a different color. This is definitely an amine. This is an amino group right there. An amino group."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "So this is going to be, let me do this in a different color. This is definitely an amine. This is an amino group right there. An amino group. And out of these three groups, the amino group is definitely at the highest priority. So you might be tempted to name this 1-amino-2-3-bromo-4- nitro-benzene. But when a benzene ring is attached to an amino group, this one also has one of those special names."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "An amino group. And out of these three groups, the amino group is definitely at the highest priority. So you might be tempted to name this 1-amino-2-3-bromo-4- nitro-benzene. But when a benzene ring is attached to an amino group, this one also has one of those special names. And let me just do it on the side right here. So if I just have a benzene ring attached to an amino group, so that's an NH2 right there, we call this molecule right here, and this is just the common name for it. It's something to put in your toolkit."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "But when a benzene ring is attached to an amino group, this one also has one of those special names. And let me just do it on the side right here. So if I just have a benzene ring attached to an amino group, so that's an NH2 right there, we call this molecule right here, and this is just the common name for it. It's something to put in your toolkit. This is aniline. So that will be the base name. This whole thing we call aniline."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "It's something to put in your toolkit. This is aniline. So that will be the base name. This whole thing we call aniline. So let me write it over here. So this is aniline. And then we always want to start numbering at the group that's really defining the molecule."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "This whole thing we call aniline. So let me write it over here. So this is aniline. And then we always want to start numbering at the group that's really defining the molecule. So we start numbering here. And you want to go in the order where you're going to bump into something first. So 1, 2, 3, 4."}, {"video_title": "Amine naming introduction Amines Organic chemistry Khan Academy (2).mp3", "Sentence": "And then we always want to start numbering at the group that's really defining the molecule. So we start numbering here. And you want to go in the order where you're going to bump into something first. So 1, 2, 3, 4. So this is going to be, if we look at them, and bromo comes alphabetically before nitro. So it's going to be 3-bromo, 4-nitro, aniline. 3-bromo, 4-nitro, aniline, and we are done."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "This is the eclipse conformation, and I'm leaving it a little bit off so you can actually see the bonds in the back there. So that's an approximate and eclipse conformation. I can rotate again and get a staggered conformation, and if I keep rotating here, the next one would be an eclipse conformation. So again, I'm leaving the bonds slightly off to the side so you can actually see the ones in the back. I rotate again to get a staggered conformation, and the next one would, of course, be eclipse. So let's look at that. So the eclipse conformation, and then finally, one more time to get back to our staggered conformation of propane."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So again, I'm leaving the bonds slightly off to the side so you can actually see the ones in the back. I rotate again to get a staggered conformation, and the next one would, of course, be eclipse. So let's look at that. So the eclipse conformation, and then finally, one more time to get back to our staggered conformation of propane. So there's your staggered conformation. Here we have the energy diagram for the conformations we saw in the video, and make sure you've seen the conformational analysis of ethane video before you watch this one. On the y-axis, we have potential energy."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So the eclipse conformation, and then finally, one more time to get back to our staggered conformation of propane. So there's your staggered conformation. Here we have the energy diagram for the conformations we saw in the video, and make sure you've seen the conformational analysis of ethane video before you watch this one. On the y-axis, we have potential energy. So as you increase in the y-axis, you're increasing in potential energy, and we started with the staggered conformation of propane, and we rotated it 60 degrees. We held the back carbon stationary, and I rotated the front carbon 60 degrees to give us this conformation, which is the eclipse conformation of propane. Notice the difference in potential energies between these two conformations."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "On the y-axis, we have potential energy. So as you increase in the y-axis, you're increasing in potential energy, and we started with the staggered conformation of propane, and we rotated it 60 degrees. We held the back carbon stationary, and I rotated the front carbon 60 degrees to give us this conformation, which is the eclipse conformation of propane. Notice the difference in potential energies between these two conformations. The staggered conformation has a lower potential energy, and the eclipse conformation has a higher potential energy. Remember, the lower the potential energy, the more stable the conformation. So the staggered conformation is more stable than the eclipse conformation."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "Notice the difference in potential energies between these two conformations. The staggered conformation has a lower potential energy, and the eclipse conformation has a higher potential energy. Remember, the lower the potential energy, the more stable the conformation. So the staggered conformation is more stable than the eclipse conformation. So it takes energy to go from the staggered conformation to the eclipse conformation. And the analogy that I used in the earlier video was a boulder. So if you have a boulder at the bottom of the hill, and you're trying to push the boulder up the hill to the top here, it takes energy to do that."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So the staggered conformation is more stable than the eclipse conformation. So it takes energy to go from the staggered conformation to the eclipse conformation. And the analogy that I used in the earlier video was a boulder. So if you have a boulder at the bottom of the hill, and you're trying to push the boulder up the hill to the top here, it takes energy to do that. And at the top of the hill, the boulder is less stable. So higher the potential energy, less stable. Lower the potential energy, more stable."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So if you have a boulder at the bottom of the hill, and you're trying to push the boulder up the hill to the top here, it takes energy to do that. And at the top of the hill, the boulder is less stable. So higher the potential energy, less stable. Lower the potential energy, more stable. So our staggered conformation is more stable than our eclipse. As we rotate, and we go from this eclipse confirmation to this staggered conformation, that would be a decrease in potential energy. Going from this staggered conformation to this eclipse would be an increase in potential energy."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "Lower the potential energy, more stable. So our staggered conformation is more stable than our eclipse. As we rotate, and we go from this eclipse confirmation to this staggered conformation, that would be a decrease in potential energy. Going from this staggered conformation to this eclipse would be an increase in potential energy. Going from the eclipse to the staggered would be a decrease. And you see the pattern, going from staggered up to this eclipse would take energy, and then going from the eclipse down to this staggered is a decrease in the potential energy. All of our eclipse conformations have the same value for the potential energy."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "Going from this staggered conformation to this eclipse would be an increase in potential energy. Going from the eclipse to the staggered would be a decrease. And you see the pattern, going from staggered up to this eclipse would take energy, and then going from the eclipse down to this staggered is a decrease in the potential energy. All of our eclipse conformations have the same value for the potential energy. They are degenerate in terms of energy. Same thing for the staggered conformations. These all have the same potential energy value."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "All of our eclipse conformations have the same value for the potential energy. They are degenerate in terms of energy. Same thing for the staggered conformations. These all have the same potential energy value. So there's a difference in potential energy between the eclipse conformations and the staggered conformations. And that difference in energy turns out to be 14 kilojoules per mole. So we're talking about the energy difference between the eclipsed and the staggered conformation."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "These all have the same potential energy value. So there's a difference in potential energy between the eclipse conformations and the staggered conformations. And that difference in energy turns out to be 14 kilojoules per mole. So we're talking about the energy difference between the eclipsed and the staggered conformation. We know there's an energy difference of 14 kilojoules per mole between the staggered conformation of propane and the eclipse conformation. And that's called the torsional strain. Let's go ahead and draw a Newman projection for each one of these conformations, so just as practice."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So we're talking about the energy difference between the eclipsed and the staggered conformation. We know there's an energy difference of 14 kilojoules per mole between the staggered conformation of propane and the eclipse conformation. And that's called the torsional strain. Let's go ahead and draw a Newman projection for each one of these conformations, so just as practice. Let's start with the staggered conformation. And we'll start with this carbon in front here, which is represented by a point. So I'll draw in a point here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw a Newman projection for each one of these conformations, so just as practice. Let's start with the staggered conformation. And we'll start with this carbon in front here, which is represented by a point. So I'll draw in a point here. What is bonded to that carbon? Well, there is a CH3 group, a methyl group up here. So let's draw a line straight up and draw in a CH3."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw in a point here. What is bonded to that carbon? Well, there is a CH3 group, a methyl group up here. So let's draw a line straight up and draw in a CH3. And there's a hydrogen going to the right and a hydrogen going to the left. So there's my hydrogen going to the right, and there's my hydrogen going to the left. We know there's a carbon behind this carbon that I marked with a point here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw a line straight up and draw in a CH3. And there's a hydrogen going to the right and a hydrogen going to the left. So there's my hydrogen going to the right, and there's my hydrogen going to the left. We know there's a carbon behind this carbon that I marked with a point here. We just can't see it because the front carbon is eclipsing the back carbon. But we know that these hydrogens in the back here are attached to that back carbon. So we represent the back carbon with a circle when we're doing Newman projections."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "We know there's a carbon behind this carbon that I marked with a point here. We just can't see it because the front carbon is eclipsing the back carbon. But we know that these hydrogens in the back here are attached to that back carbon. So we represent the back carbon with a circle when we're doing Newman projections. So that's supposed to represent the back carbon. And then we would have a hydrogen coming out to the right like that, so that's this hydrogen. A hydrogen coming out to the left, that's this hydrogen."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So we represent the back carbon with a circle when we're doing Newman projections. So that's supposed to represent the back carbon. And then we would have a hydrogen coming out to the right like that, so that's this hydrogen. A hydrogen coming out to the left, that's this hydrogen. And a hydrogen coming straight down, so that would be this hydrogen. So there's your staggered conformation for propane. Next, let's draw the eclipsed conformation as a Newman projection."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "A hydrogen coming out to the left, that's this hydrogen. And a hydrogen coming straight down, so that would be this hydrogen. So there's your staggered conformation for propane. Next, let's draw the eclipsed conformation as a Newman projection. So a little bit harder, but let's start with this carbon again. So this is the one, this is the front carbon. So this is represented with a point right here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's draw the eclipsed conformation as a Newman projection. So a little bit harder, but let's start with this carbon again. So this is the one, this is the front carbon. So this is represented with a point right here. And then we would have a CH3, a methyl group, going off to the right. So let's draw that in. So we have a CH3 going off to the right."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So this is represented with a point right here. And then we would have a CH3, a methyl group, going off to the right. So let's draw that in. So we have a CH3 going off to the right. We have a hydrogen going down. And I'm gonna draw this a little bit off-center. So instead of drawing it straight down, I'm gonna draw it a little bit off to the left, just like I did in the picture here, to make it easier to see the bonds in the back."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So we have a CH3 going off to the right. We have a hydrogen going down. And I'm gonna draw this a little bit off-center. So instead of drawing it straight down, I'm gonna draw it a little bit off to the left, just like I did in the picture here, to make it easier to see the bonds in the back. So there's a hydrogen going down a little bit to the left. And then we have a hydrogen going in this direction. So let me go ahead and draw that in here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So instead of drawing it straight down, I'm gonna draw it a little bit off to the left, just like I did in the picture here, to make it easier to see the bonds in the back. So there's a hydrogen going down a little bit to the left. And then we have a hydrogen going in this direction. So let me go ahead and draw that in here. So here's a hydrogen. Next, let's think about the back carbon. We can't see it, but because this front carbon here is eclipsing the back carbon, but we know that the back carbon has three hydrogens attached to it."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that in here. So here's a hydrogen. Next, let's think about the back carbon. We can't see it, but because this front carbon here is eclipsing the back carbon, but we know that the back carbon has three hydrogens attached to it. This one, this one, and this one, which we can just barely see. So let's add those in on our Newman projection. So the back carbon is represented by a circle here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "We can't see it, but because this front carbon here is eclipsing the back carbon, but we know that the back carbon has three hydrogens attached to it. This one, this one, and this one, which we can just barely see. So let's add those in on our Newman projection. So the back carbon is represented by a circle here. And let's start with this hydrogen right back here. That would be going in this direction. So it's being eclipsed by the methyl group."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So the back carbon is represented by a circle here. And let's start with this hydrogen right back here. That would be going in this direction. So it's being eclipsed by the methyl group. But we draw it a little bit off to the side so we can still see it's there. Next, let's do this hydrogen. So it's going down, pretty much straight down."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So it's being eclipsed by the methyl group. But we draw it a little bit off to the side so we can still see it's there. Next, let's do this hydrogen. So it's going down, pretty much straight down. So we'll draw that in there. And then finally, this hydrogen over here. So this hydrogen, we could represent it like that."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So it's going down, pretty much straight down. So we'll draw that in there. And then finally, this hydrogen over here. So this hydrogen, we could represent it like that. So now we have Newman projections for the staggered conformation and for the eclipsed conformation. Let's go back to that 14 kilojoules per mole, that torsional strain. So let me write that in here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen, we could represent it like that. So now we have Newman projections for the staggered conformation and for the eclipsed conformation. Let's go back to that 14 kilojoules per mole, that torsional strain. So let me write that in here. So 14 kilojoules per mole. In the video on conformations of ethane, we already know that each pair of eclipsed hydrogens has an energy cost of four kilojoules per mole. So this pair of eclipsed hydrogens, that's four kilojoules per mole as an energy cost right here."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So let me write that in here. So 14 kilojoules per mole. In the video on conformations of ethane, we already know that each pair of eclipsed hydrogens has an energy cost of four kilojoules per mole. So this pair of eclipsed hydrogens, that's four kilojoules per mole as an energy cost right here. Same with this one. So this one's four kilojoules per mole. So now we can figure out the energy cost associated with a methyl group eclipsing a hydrogen."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So this pair of eclipsed hydrogens, that's four kilojoules per mole as an energy cost right here. Same with this one. So this one's four kilojoules per mole. So now we can figure out the energy cost associated with a methyl group eclipsing a hydrogen. Because we know the total should add up to equal 14. So four plus four plus what is equal to 14? Obviously the answer is six."}, {"video_title": "Conformational analysis of propane Organic chemistry Khan Academy.mp3", "Sentence": "So now we can figure out the energy cost associated with a methyl group eclipsing a hydrogen. Because we know the total should add up to equal 14. So four plus four plus what is equal to 14? Obviously the answer is six. So this must be six kilojoules per mole. So six plus four plus four gives us our total torsional strain of 14 kilojoules per mole. So now we know that the energy cost of a methyl group eclipsing a hydrogen must be six kilojoules per mole."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "I just think it's really cool looking. And you can see that at each of the corners of the cube, there's a carbon. So there are eight carbons total. And then there's also a hydrogen coming off of each of those carbons for a molecular formula of C8H8. At first, a lot of chemists didn't think this molecule could be made because of the high amount of angle strain that's present in this molecule. But it was made starting in the 1960s. And it's being looked at for a lot of potential uses in medicine and explosives these days."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then there's also a hydrogen coming off of each of those carbons for a molecular formula of C8H8. At first, a lot of chemists didn't think this molecule could be made because of the high amount of angle strain that's present in this molecule. But it was made starting in the 1960s. And it's being looked at for a lot of potential uses in medicine and explosives these days. And because you can nitrate it and make things like octonitrocubane and heptonitrocubane, which are potential explosives for the future. And we're going to look and see if we can name cubane using IUPAC nomenclature in this video. So let's first think about the rules we learned in the video on bicyclic nomenclature."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And it's being looked at for a lot of potential uses in medicine and explosives these days. And because you can nitrate it and make things like octonitrocubane and heptonitrocubane, which are potential explosives for the future. And we're going to look and see if we can name cubane using IUPAC nomenclature in this video. So let's first think about the rules we learned in the video on bicyclic nomenclature. And if you're trying to figure out how many rings are in a system, you have to make cuts and figure out how many cuts does it take to get to an open chain alkane. So if we start with this yellow version of cubane over here on the left, I'm going to start cutting bonds. And let's see how many cuts it takes to get to an open chain alkane."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's first think about the rules we learned in the video on bicyclic nomenclature. And if you're trying to figure out how many rings are in a system, you have to make cuts and figure out how many cuts does it take to get to an open chain alkane. So if we start with this yellow version of cubane over here on the left, I'm going to start cutting bonds. And let's see how many cuts it takes to get to an open chain alkane. For example, I could start by cutting right here. So we'll say that's our first cut. And then our second cut, we could make a cut right back here like that."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And let's see how many cuts it takes to get to an open chain alkane. For example, I could start by cutting right here. So we'll say that's our first cut. And then our second cut, we could make a cut right back here like that. So we could make that my second cut here on my cubane. And then for my third cut, I'm going to go for this one right up here. So we'll take care of that one."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then our second cut, we could make a cut right back here like that. So we could make that my second cut here on my cubane. And then for my third cut, I'm going to go for this one right up here. So we'll take care of that one. So that's three cuts so far. And then if I just go ahead and take out this one, this bond right here, and then this bond right here, so that cuts four and five, I now get an open chain alkane. So it took five cuts for us to do that."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we'll take care of that one. So that's three cuts so far. And then if I just go ahead and take out this one, this bond right here, and then this bond right here, so that cuts four and five, I now get an open chain alkane. So it took five cuts for us to do that. So there are five rings in cubane. So it's pentacyclo. So that's just not immediately obvious to me anyway as to why there are five rings in cubane."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So it took five cuts for us to do that. So there are five rings in cubane. So it's pentacyclo. So that's just not immediately obvious to me anyway as to why there are five rings in cubane. So let's go ahead and write pentacyclo to start the IUPAC name here. So pentacyclo, meaning five rings. And then we start our brackets just like we did in the video on bicyclic nomenclature."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So that's just not immediately obvious to me anyway as to why there are five rings in cubane. So let's go ahead and write pentacyclo to start the IUPAC name here. So pentacyclo, meaning five rings. And then we start our brackets just like we did in the video on bicyclic nomenclature. And to finish naming cubane, we're going to pretend like it is a bicyclic compound. And the first thing we do is identify our bridgehead carbons. So the carbons that are common to both of the two rings here."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then we start our brackets just like we did in the video on bicyclic nomenclature. And to finish naming cubane, we're going to pretend like it is a bicyclic compound. And the first thing we do is identify our bridgehead carbons. So the carbons that are common to both of the two rings here. So hopefully it's obvious those are two rings. And those are the bridgehead carbons that connect those two rings. When you number a bicyclic compound, you start at one of the bridgehead carbons."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So the carbons that are common to both of the two rings here. So hopefully it's obvious those are two rings. And those are the bridgehead carbons that connect those two rings. When you number a bicyclic compound, you start at one of the bridgehead carbons. And then you go the longest path first. So I'm going to start at this carbon. And I'm going to go the longest path, which would be up here."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "When you number a bicyclic compound, you start at one of the bridgehead carbons. And then you go the longest path first. So I'm going to start at this carbon. And I'm going to go the longest path, which would be up here. So this would be number two. Then this would be number three, carbon number four, carbon number five, and carbon number six, which takes me to the other bridgehead carbon. And then you name your next longest path."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to go the longest path, which would be up here. So this would be number two. Then this would be number three, carbon number four, carbon number five, and carbon number six, which takes me to the other bridgehead carbon. And then you name your next longest path. So I'm just going to continue around and make this carbon seven, and then make this one back here carbon eight. So those are my eight carbons of cubane. And so once again, I can continue to pretend like it's a bicyclic molecule."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then you name your next longest path. So I'm just going to continue around and make this carbon seven, and then make this one back here carbon eight. So those are my eight carbons of cubane. And so once again, I can continue to pretend like it's a bicyclic molecule. And the next thing I would do is I would name the number of carbons in my longest path. So the number of carbons in my longest path would be this one. So there would be one, two, three, four carbons."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, I can continue to pretend like it's a bicyclic molecule. And the next thing I would do is I would name the number of carbons in my longest path. So the number of carbons in my longest path would be this one. So there would be one, two, three, four carbons. Remember, you exclude the bridgehead carbons when you're doing this. So we're going to start with a four right here, like that. Next, you do the number of carbons in your second longest path."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So there would be one, two, three, four carbons. Remember, you exclude the bridgehead carbons when you're doing this. So we're going to start with a four right here, like that. Next, you do the number of carbons in your second longest path. So we can see my second longest path would be this one right here. And there are two carbons in my second longest path. So I go ahead and put a two over here, like that."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Next, you do the number of carbons in your second longest path. So we can see my second longest path would be this one right here. And there are two carbons in my second longest path. So I go ahead and put a two over here, like that. And then finally, it's the number of carbons between the bridgehead carbons, which in this example, of course, there are no carbons between my two bridgehead carbons. So I would put a zero here, like that. But of course, cubane is not a bicyclic compound."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I go ahead and put a two over here, like that. And then finally, it's the number of carbons between the bridgehead carbons, which in this example, of course, there are no carbons between my two bridgehead carbons. So I would put a zero here, like that. But of course, cubane is not a bicyclic compound. So we have to keep going. We have to figure out how I can continue naming this molecule. And the way to do it is to next think about how many carbons are there between carbons two and five."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "But of course, cubane is not a bicyclic compound. So we have to keep going. We have to figure out how I can continue naming this molecule. And the way to do it is to next think about how many carbons are there between carbons two and five. So if I draw a dashed line in here, so I can pretend like I'm connecting those two right there. And of course, there are no carbons between two and five. So I can keep going."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And the way to do it is to next think about how many carbons are there between carbons two and five. So if I draw a dashed line in here, so I can pretend like I'm connecting those two right there. And of course, there are no carbons between two and five. So I can keep going. I can make this a zero. And I can put a 2 comma 5, saying there are no carbons between carbons two and five. And I can continue on."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I can keep going. I can make this a zero. And I can put a 2 comma 5, saying there are no carbons between carbons two and five. And I can continue on. I can do that between the three and the eight. So if I were to connect the three and the eight back here, like that, there are no carbons between three and eight. So I can write 0 and then 3 comma 8."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And I can continue on. I can do that between the three and the eight. So if I were to connect the three and the eight back here, like that, there are no carbons between three and eight. So I can write 0 and then 3 comma 8. And of course, I can do the same thing over here on the right. So between four and seven, there are no carbons. So between four and seven, there are no carbons."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I can write 0 and then 3 comma 8. And of course, I can do the same thing over here on the right. So between four and seven, there are no carbons. So between four and seven, there are no carbons. So I can write 0, 4, and 7, like that. So let me just clear up that 7 there. And we're done with our brackets."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So between four and seven, there are no carbons. So I can write 0, 4, and 7, like that. So let me just clear up that 7 there. And we're done with our brackets. So the last thing you need to do when you're naming a polycyclic alkane like this is to figure out how many total carbons are in the molecule. Well, of course, there are eight. So this is octane."}, {"video_title": "Naming cubane Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And we're done with our brackets. So the last thing you need to do when you're naming a polycyclic alkane like this is to figure out how many total carbons are in the molecule. Well, of course, there are eight. So this is octane. So I can go ahead and write octane down here, like that. And I have my IUPAC name for cubane. It is pentacyclo-420-025-038-047-octane."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we were asked to name the molecule on the top left. We would start by numbering our carbons. So this would be carbon one, two, three, and four. Notice we have a double bond starting at carbon two. So the name of this molecule would be 2-butene. Two because we have our double bond starting at carbon two, bute because we have four carbons, and ene because we have a double bond present in the molecule. What about naming the molecule on the right?"}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Notice we have a double bond starting at carbon two. So the name of this molecule would be 2-butene. Two because we have our double bond starting at carbon two, bute because we have four carbons, and ene because we have a double bond present in the molecule. What about naming the molecule on the right? We number our carbons one, two, three, and four. And once again, we have a double bond starting at carbon two. So the name of this molecule would be 2-butene."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What about naming the molecule on the right? We number our carbons one, two, three, and four. And once again, we have a double bond starting at carbon two. So the name of this molecule would be 2-butene. However, these are two different molecules. And the reason why is because there's no free rotation around a double bond. Single bonds have free rotation, but double bonds don't."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the name of this molecule would be 2-butene. However, these are two different molecules. And the reason why is because there's no free rotation around a double bond. Single bonds have free rotation, but double bonds don't. So you couldn't rotate the molecule on the left to look like the molecule on the right. Therefore, they must be isomers of each other. And we need a way to distinguish between our isomers."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Single bonds have free rotation, but double bonds don't. So you couldn't rotate the molecule on the left to look like the molecule on the right. Therefore, they must be isomers of each other. And we need a way to distinguish between our isomers. And so one way to do that is to use cis-trans terminology. So if we look at the molecule on the left, we can see we have two methyl groups. And those two methyl groups are on the same side of our double bonds."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we need a way to distinguish between our isomers. And so one way to do that is to use cis-trans terminology. So if we look at the molecule on the left, we can see we have two methyl groups. And those two methyl groups are on the same side of our double bonds. If I draw a line in here, it's easier to see those two methyl groups are on the same side. And we call that the cis isomer. So we put cis in front of our name here."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And those two methyl groups are on the same side of our double bonds. If I draw a line in here, it's easier to see those two methyl groups are on the same side. And we call that the cis isomer. So we put cis in front of our name here. I'm attempting to write it in italics. So this would be cis-2-butene. On the right, when we look at those methyl groups, these two methyl groups are on opposite sides of the double bonds."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we put cis in front of our name here. I'm attempting to write it in italics. So this would be cis-2-butene. On the right, when we look at those methyl groups, these two methyl groups are on opposite sides of the double bonds. So I draw a line in here to make it easier to see those two methyl groups are on opposite sides. And we call that trans. So this is the trans isomer."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "On the right, when we look at those methyl groups, these two methyl groups are on opposite sides of the double bonds. So I draw a line in here to make it easier to see those two methyl groups are on opposite sides. And we call that trans. So this is the trans isomer. I'm going to write trans here in italics, attempt to anyway. So we have cis-2-butene and trans-2-butene. These are different molecules with different properties."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is the trans isomer. I'm going to write trans here in italics, attempt to anyway. So we have cis-2-butene and trans-2-butene. These are different molecules with different properties. If you want to use cis-trans terminology, you're looking for two identical groups, and you are comparing them. So let's look at these next two examples here and figure out which one is cis and which one is trans. We're looking for identical groups."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "These are different molecules with different properties. If you want to use cis-trans terminology, you're looking for two identical groups, and you are comparing them. So let's look at these next two examples here and figure out which one is cis and which one is trans. We're looking for identical groups. So over here, we have an ethyl group attached to our double bond. And on the right, we have an ethyl group attached to our double bond. Those two ethyl groups are on the same side of our double bond."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're looking for identical groups. So over here, we have an ethyl group attached to our double bond. And on the right, we have an ethyl group attached to our double bond. Those two ethyl groups are on the same side of our double bond. So this must be the cis isomer. On the right, we have this ethyl group and this ethyl group are on opposite sides of our double bond. So that must be the trans isomer."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Those two ethyl groups are on the same side of our double bond. So this must be the cis isomer. On the right, we have this ethyl group and this ethyl group are on opposite sides of our double bond. So that must be the trans isomer. Let's do some more examples. I'll go down to here. On the left, we have the cinnamaldehyde molecule."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that must be the trans isomer. Let's do some more examples. I'll go down to here. On the left, we have the cinnamaldehyde molecule. We're looking for two identical groups. So we can use cis or trans. You can also use hydrogens."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "On the left, we have the cinnamaldehyde molecule. We're looking for two identical groups. So we can use cis or trans. You can also use hydrogens. You don't have to use a methyl group or an ethyl group. So if we look at our double bond, we know there's a hydrogen attached to this carbon, and we know there's a hydrogen attached to this carbon. And those two hydrogens are on opposite sides of our double bond."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You can also use hydrogens. You don't have to use a methyl group or an ethyl group. So if we look at our double bond, we know there's a hydrogen attached to this carbon, and we know there's a hydrogen attached to this carbon. And those two hydrogens are on opposite sides of our double bond. So I'm drawing a line here to make it easier to see. These two hydrogens are on opposite side. So we're talking about trans here."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And those two hydrogens are on opposite sides of our double bond. So I'm drawing a line here to make it easier to see. These two hydrogens are on opposite side. So we're talking about trans here. Those hydrogens are across from each other. What about the tetra substituted alkene on the right? We need two identical groups to use our cis trans."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're talking about trans here. Those hydrogens are across from each other. What about the tetra substituted alkene on the right? We need two identical groups to use our cis trans. And here we have an ethyl group, and here we have an ethyl group. Over here, we have a methyl group and an isopropyl group. But the two methyl groups are on the same side of our double bond."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We need two identical groups to use our cis trans. And here we have an ethyl group, and here we have an ethyl group. Over here, we have a methyl group and an isopropyl group. But the two methyl groups are on the same side of our double bond. So I draw a line in here, and we see that these two groups are on the same side. Therefore, we're talking about cis here. So this double bond has a cis configuration."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But the two methyl groups are on the same side of our double bond. So I draw a line in here, and we see that these two groups are on the same side. Therefore, we're talking about cis here. So this double bond has a cis configuration. Let's compare the drawing on the left to the drawing on the right. The first time you look at these two drawings, you might think these are two isomers, and I could use cis trans terminology to distinguish between them. However, you can't, because these are just two ways to represent the same molecule."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this double bond has a cis configuration. Let's compare the drawing on the left to the drawing on the right. The first time you look at these two drawings, you might think these are two isomers, and I could use cis trans terminology to distinguish between them. However, you can't, because these are just two ways to represent the same molecule. If you picked up this molecule on the left and you flipped it up, you would get the drawing on the right. So they're not isomers of each other. This is the same molecule."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "However, you can't, because these are just two ways to represent the same molecule. If you picked up this molecule on the left and you flipped it up, you would get the drawing on the right. So they're not isomers of each other. This is the same molecule. And a fast way to figure that out is to look at this carbon. And you can see you have two identical groups bonded to that carbon. So you can't use cis trans terminology."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is the same molecule. And a fast way to figure that out is to look at this carbon. And you can see you have two identical groups bonded to that carbon. So you can't use cis trans terminology. That's different from the example we did a minute ago. We had two identical groups, these two ethyl groups here. However, those two ethyl groups weren't bonded to the same carbon."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you can't use cis trans terminology. That's different from the example we did a minute ago. We had two identical groups, these two ethyl groups here. However, those two ethyl groups weren't bonded to the same carbon. Those two ethyl groups are bonded to different carbons. So this ethyl group is bonded to this carbon, and this ethyl group is bonded to this carbon. So we were able to use cis trans terminology."}, {"video_title": "Cis\u2013trans isomerism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "However, those two ethyl groups weren't bonded to the same carbon. Those two ethyl groups are bonded to different carbons. So this ethyl group is bonded to this carbon, and this ethyl group is bonded to this carbon. So we were able to use cis trans terminology. So we looked at our double bond, and we said those two ethyl groups are on the same side of our double bond. So this represents a cis configuration of the double bond. So we can't do that up here, because while we do have two identical groups, those identical groups are bonded to the same carbon."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But that's not what this video is about. I'm just, this is actually just a small little video about different dates. Or maybe a better way to think about it, different ways to specify dates or dating mechanisms. And so if you were to look up Plato's birth, well you might get either 428 or 427, but we'll go with 428. If you were to look up Plato's birth, you might see it written as 428 BC, or you might see it written as 428 BCE. And the natural question is, well what's the difference here, they both have a BC, but this one has an E, it's the same year right now. And the answer is, is that these are referring to the exact same year in history, but the acronyms here do stand for different things."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so if you were to look up Plato's birth, well you might get either 428 or 427, but we'll go with 428. If you were to look up Plato's birth, you might see it written as 428 BC, or you might see it written as 428 BCE. And the natural question is, well what's the difference here, they both have a BC, but this one has an E, it's the same year right now. And the answer is, is that these are referring to the exact same year in history, but the acronyms here do stand for different things. BC literally stands for before Christ. So if the date is written 428 BC, the implication is that this is 428 years before the birth of Christ. We'll see in a second that that's not exactly right, but that's what the implication is."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the answer is, is that these are referring to the exact same year in history, but the acronyms here do stand for different things. BC literally stands for before Christ. So if the date is written 428 BC, the implication is that this is 428 years before the birth of Christ. We'll see in a second that that's not exactly right, but that's what the implication is. If someone writes BCE, they're saying something very different. The B still stands for before, but the C in CE does not stand for Christ anymore. It now stands for common."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We'll see in a second that that's not exactly right, but that's what the implication is. If someone writes BCE, they're saying something very different. The B still stands for before, but the C in CE does not stand for Christ anymore. It now stands for common. And so the CE part is common era. Even though it's not referring to Christ anymore, and this kind of the intention here is so that it's I guess less religious than the term before Christ, it's still kind of putting an importance on Christ's birth. Because it's saying that the common era is the time period after the birth of Christ, which we'll see in a second isn't exactly right."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It now stands for common. And so the CE part is common era. Even though it's not referring to Christ anymore, and this kind of the intention here is so that it's I guess less religious than the term before Christ, it's still kind of putting an importance on Christ's birth. Because it's saying that the common era is the time period after the birth of Christ, which we'll see in a second isn't exactly right. But there's essentially the same exact dating scheme, one not directly referring to Christ, one that is directly referring to Christ. Similarly, this right here is a painting of Christopher Columbus. And if you were to look up in history books, when was his first voyage and when did he first show up in the New World, finding an island in the Bahamas, you would see it written as either 1492 or AD 1492 or 1492 CE."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because it's saying that the common era is the time period after the birth of Christ, which we'll see in a second isn't exactly right. But there's essentially the same exact dating scheme, one not directly referring to Christ, one that is directly referring to Christ. Similarly, this right here is a painting of Christopher Columbus. And if you were to look up in history books, when was his first voyage and when did he first show up in the New World, finding an island in the Bahamas, you would see it written as either 1492 or AD 1492 or 1492 CE. And once again, these are all referring to the same year, just using different acronyms. One of them's a little bit more religious or more directly refers to Christ, and one is a little less religious. So AD, some people think it refers to after death."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if you were to look up in history books, when was his first voyage and when did he first show up in the New World, finding an island in the Bahamas, you would see it written as either 1492 or AD 1492 or 1492 CE. And once again, these are all referring to the same year, just using different acronyms. One of them's a little bit more religious or more directly refers to Christ, and one is a little less religious. So AD, some people think it refers to after death. It does not refer to after death. Because if you think about it, if you have years before the birth of Christ, and if you started numbering after his death, how would you number the years during his life? So AD does not stand for after death."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So AD, some people think it refers to after death. It does not refer to after death. Because if you think about it, if you have years before the birth of Christ, and if you started numbering after his death, how would you number the years during his life? So AD does not stand for after death. It stands for Anno Domini, which literally means year, and domini means Lord or the Lord. So it's the year of the Lord or the year of our Lord. So it's years since."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So AD does not stand for after death. It stands for Anno Domini, which literally means year, and domini means Lord or the Lord. So it's the year of the Lord or the year of our Lord. So it's years since. And one Anno Domini would be the year of Jesus Christ's birth. So not after death. It stands for Anno Domini, but literally year of our Lord, so years since that Jesus was born, with year one being implicitly starting with his birth."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's years since. And one Anno Domini would be the year of Jesus Christ's birth. So not after death. It stands for Anno Domini, but literally year of our Lord, so years since that Jesus was born, with year one being implicitly starting with his birth. And we'll see in a second that's not exactly right. CE stands for Common Era. Once again, 1 CE is the same thing as AD 1."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It stands for Anno Domini, but literally year of our Lord, so years since that Jesus was born, with year one being implicitly starting with his birth. And we'll see in a second that's not exactly right. CE stands for Common Era. Once again, 1 CE is the same thing as AD 1. Sometimes we now write, instead of writing AD 1492, we'll write 1492 AD, all referring to the exact same thing. Now, all of these things refer to, when we say 428 BC, it implies 428 years before the birth of Christ. 1492 AD, that's in the year of our Lord, 1492."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once again, 1 CE is the same thing as AD 1. Sometimes we now write, instead of writing AD 1492, we'll write 1492 AD, all referring to the exact same thing. Now, all of these things refer to, when we say 428 BC, it implies 428 years before the birth of Christ. 1492 AD, that's in the year of our Lord, 1492. It implies 1,492 years since the birth of Christ. But the reality is that we're not really quite sure when Christ was born. And so these aren't exactly."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "1492 AD, that's in the year of our Lord, 1492. It implies 1,492 years since the birth of Christ. But the reality is that we're not really quite sure when Christ was born. And so these aren't exactly. So Columbus didn't sail across the Atlantic exactly 1492 years after the birth of Christ. Most historians put the birth of Christ at 7 to 2 BC or BCE, depending on how you want to view it. Remember, BC is before Christ, which is a little ironic because we're talking about the actual birth of Christ."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so these aren't exactly. So Columbus didn't sail across the Atlantic exactly 1492 years after the birth of Christ. Most historians put the birth of Christ at 7 to 2 BC or BCE, depending on how you want to view it. Remember, BC is before Christ, which is a little ironic because we're talking about the actual birth of Christ. BCE is before the Common Era. And they put his death at 30 to 36 AD, which is 30 to 36 in the year of our Lord. Or, that's what this stands for, Anno Domini, or in the Common Era, CE."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, BC is before Christ, which is a little ironic because we're talking about the actual birth of Christ. BCE is before the Common Era. And they put his death at 30 to 36 AD, which is 30 to 36 in the year of our Lord. Or, that's what this stands for, Anno Domini, or in the Common Era, CE. Now, some people, they obviously don't like BC. They don't like the BC AD naming mechanism because it's explicitly referring to Christ. And every year, it makes Christ the central figure in all of history."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or, that's what this stands for, Anno Domini, or in the Common Era, CE. Now, some people, they obviously don't like BC. They don't like the BC AD naming mechanism because it's explicitly referring to Christ. And every year, it makes Christ the central figure in all of history. So they'll say that this is clearly too Christian. And they would prefer the situation, they would prefer the less Christian naming scheme, where you use BCE and CE. But a lot of people would still say, hey, look, OK, you changed, well, first of all, some Christians wouldn't like this, that you removed the direct references to the birth of Christ or being in the years since Christ's birth."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And every year, it makes Christ the central figure in all of history. So they'll say that this is clearly too Christian. And they would prefer the situation, they would prefer the less Christian naming scheme, where you use BCE and CE. But a lot of people would still say, hey, look, OK, you changed, well, first of all, some Christians wouldn't like this, that you removed the direct references to the birth of Christ or being in the years since Christ's birth. But even here, and they'll say, hey, you've removed it. But even here, some people would complain that although you've made the direct reference, that this is saying common before the Common Era and the Common Era, even though you've removed the direct reference, it still makes Christ's birth the central thing in all of history. But this is the convention, whether people like it or not."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But a lot of people would still say, hey, look, OK, you changed, well, first of all, some Christians wouldn't like this, that you removed the direct references to the birth of Christ or being in the years since Christ's birth. But even here, and they'll say, hey, you've removed it. But even here, some people would complain that although you've made the direct reference, that this is saying common before the Common Era and the Common Era, even though you've removed the direct reference, it still makes Christ's birth the central thing in all of history. But this is the convention, whether people like it or not. In order to have the same reference point, and it would be too logistically difficult to switch at this time, everyone has essentially settled on this. And it's really just a matter of letters of which naming scheme you pick. But the whole point of this video is that you don't get confused between BC and BCE."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is the convention, whether people like it or not. In order to have the same reference point, and it would be too logistically difficult to switch at this time, everyone has essentially settled on this. And it's really just a matter of letters of which naming scheme you pick. But the whole point of this video is that you don't get confused between BC and BCE. You don't think that AD stands for after death. It stands for Anno Domini, the year of the Lord or the year of our Lord. And CE stands for Common Era."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the whole point of this video is that you don't get confused between BC and BCE. You don't think that AD stands for after death. It stands for Anno Domini, the year of the Lord or the year of our Lord. And CE stands for Common Era. But this and this are referring to essentially the same count after the birth of Christ, or this theoretical birth of Christ, which we don't really know when it actually happened. It probably did not happen at the beginning of 1 AD or 1 CE. And these two things both also refer to the same direction in the timeline."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And CE stands for Common Era. But this and this are referring to essentially the same count after the birth of Christ, or this theoretical birth of Christ, which we don't really know when it actually happened. It probably did not happen at the beginning of 1 AD or 1 CE. And these two things both also refer to the same direction in the timeline. One last thing I want to point out is that there is no year zero. So if you take either of these naming schemes, you have, so let's go very close to the year one. So there's just some point, this theoretical birth of Christ, which was probably not the actual birth of Christ, but at that theoretical point, right at that, so New Year's, so you have kind of December 31 of the previous year."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these two things both also refer to the same direction in the timeline. One last thing I want to point out is that there is no year zero. So if you take either of these naming schemes, you have, so let's go very close to the year one. So there's just some point, this theoretical birth of Christ, which was probably not the actual birth of Christ, but at that theoretical point, right at that, so New Year's, so you have kind of December 31 of the previous year. All of a sudden now, you are on January 1 of 1 AD or CE, depending on how you want to refer to it. And the year before that was 1 BC or BCE, depending on how you want to refer to it. So there is no year zero in this scheme."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So there's just some point, this theoretical birth of Christ, which was probably not the actual birth of Christ, but at that theoretical point, right at that, so New Year's, so you have kind of December 31 of the previous year. All of a sudden now, you are on January 1 of 1 AD or CE, depending on how you want to refer to it. And the year before that was 1 BC or BCE, depending on how you want to refer to it. So there is no year zero in this scheme. And then the last thing I want to emphasize, and it might be obvious to you, is the larger a number you have here, the further back you're going in time. Because this is saying how many years before this theoretical birth of Christ. And obviously, larger numbers you have here, this is the further you're going off into the future."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So there is no year zero in this scheme. And then the last thing I want to emphasize, and it might be obvious to you, is the larger a number you have here, the further back you're going in time. Because this is saying how many years before this theoretical birth of Christ. And obviously, larger numbers you have here, this is the further you're going off into the future. And if you wanted to figure out how many years passed between Plato's birth and Columbus sailing across the Atlantic to find the New World, you would say, well, look, it took 428 years to go from Plato's birth to this theoretical birth of Christ. And then you have another 1,492 years to wait until Columbus gets his ship together. So the total number of years would be, I'll do it right over here, 428 years to get to Christ from Plato's birth."}, {"video_title": "Understanding calendar notation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And obviously, larger numbers you have here, this is the further you're going off into the future. And if you wanted to figure out how many years passed between Plato's birth and Columbus sailing across the Atlantic to find the New World, you would say, well, look, it took 428 years to go from Plato's birth to this theoretical birth of Christ. And then you have another 1,492 years to wait until Columbus gets his ship together. So the total number of years would be, I'll do it right over here, 428 years to get to Christ from Plato's birth. And then you have another 1,492 years to wait for Columbus, so let's see, 8 plus 2, that is 10. As you can see, I just wanted to add a little arithmetic in this video. So 1 plus 2 plus 9 is 12."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "I see a lot of mistakes when students draw resonant structures, and so I wanted to make a video on some of the more common mistakes that I've seen. So let's say we wanted to draw a resonant structure for this carbocation. Some students would take these electrons and move them down to here and say, alright, so on the right, now I would have this, and this is my resonant structure. Let me highlight those electrons in blue here. So these electrons here move down to here. But this is incorrect. So let me write no here."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "Let me highlight those electrons in blue here. So these electrons here move down to here. But this is incorrect. So let me write no here. So the resonant structure on the right, this is an incorrect resonant structure. Why is this resonant structure not possible? Well, let's draw in the hydrogens on the carbons, and it'll be much more obvious."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So let me write no here. So the resonant structure on the right, this is an incorrect resonant structure. Why is this resonant structure not possible? Well, let's draw in the hydrogens on the carbons, and it'll be much more obvious. So this carbon right here has one hydrogen on it, same with this carbon, and this carbon right here has two hydrogens on it. And the carbon with a plus one formal charge must have one hydrogen. So let's put in those hydrogens for the resonant structure on the right, and it should be obvious why this resonant structure is incorrect."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "Well, let's draw in the hydrogens on the carbons, and it'll be much more obvious. So this carbon right here has one hydrogen on it, same with this carbon, and this carbon right here has two hydrogens on it. And the carbon with a plus one formal charge must have one hydrogen. So let's put in those hydrogens for the resonant structure on the right, and it should be obvious why this resonant structure is incorrect. Let's focus in on this carbon right here, the one I marked in red. How many bonds are there to that carbon? Well, here's one bond, two, three, four, and five."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in those hydrogens for the resonant structure on the right, and it should be obvious why this resonant structure is incorrect. Let's focus in on this carbon right here, the one I marked in red. How many bonds are there to that carbon? Well, here's one bond, two, three, four, and five. That's five bonds to a carbon. That does not happen. You can't show carbon with five bonds, because that would be 10 electrons around this carbon, and carbon can never exceed an octet of electrons."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "Well, here's one bond, two, three, four, and five. That's five bonds to a carbon. That does not happen. You can't show carbon with five bonds, because that would be 10 electrons around this carbon, and carbon can never exceed an octet of electrons. Because of carbon's position on the periodic table, in the second period, there's four orbitals, and each orbital can hold a maximum of two electrons, which gives us four times two, which is eight. So carbon can never exceed an octet. There's another reason why this is wrong."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "You can't show carbon with five bonds, because that would be 10 electrons around this carbon, and carbon can never exceed an octet of electrons. Because of carbon's position on the periodic table, in the second period, there's four orbitals, and each orbital can hold a maximum of two electrons, which gives us four times two, which is eight. So carbon can never exceed an octet. There's another reason why this is wrong. If we go to this top carbon here, there's only three bonds around that carbon. So that carbon will have a plus one formal charge. So we added another formal charge, and we have carbon with five bonds."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "There's another reason why this is wrong. If we go to this top carbon here, there's only three bonds around that carbon. So that carbon will have a plus one formal charge. So we added another formal charge, and we have carbon with five bonds. So this is incorrect. This is not a correct resonant structure. So what is the proper resonant structure to draw?"}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So we added another formal charge, and we have carbon with five bonds. So this is incorrect. This is not a correct resonant structure. So what is the proper resonant structure to draw? Well, let's show that down here. You take your electrons, and you move them in the direction of the positive charge, of the positive one formal charge. So let's show that."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So what is the proper resonant structure to draw? Well, let's show that down here. You take your electrons, and you move them in the direction of the positive charge, of the positive one formal charge. So let's show that. So the electrons in, let me make them blue again, the electrons in blue, move over to here, like that. And that moves the positive formal charge over to this carbon. If we draw in our hydrogens, it'll be clear why this is correct."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So let's show that. So the electrons in, let me make them blue again, the electrons in blue, move over to here, like that. And that moves the positive formal charge over to this carbon. If we draw in our hydrogens, it'll be clear why this is correct. So we put in a hydrogen here, we put in a hydrogen here, and we put in a hydrogen here. So let me draw in those three hydrogens on the right. Okay, now it's very obvious."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "If we draw in our hydrogens, it'll be clear why this is correct. So we put in a hydrogen here, we put in a hydrogen here, and we put in a hydrogen here. So let me draw in those three hydrogens on the right. Okay, now it's very obvious. Let me point this out in red. It's obvious that this carbon here in red has a plus one formal charge. It has three bonds around it."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "Okay, now it's very obvious. Let me point this out in red. It's obvious that this carbon here in red has a plus one formal charge. It has three bonds around it. So one, two, and three. And this carbon, this carbon over here on the right that had the plus one formal charge, now its formal charge is zero, right? Because there are four bonds around it."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "It has three bonds around it. So one, two, and three. And this carbon, this carbon over here on the right that had the plus one formal charge, now its formal charge is zero, right? Because there are four bonds around it. So one, two, three, and four. So now the formal charge is zero. So this is the correct resonance structure."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "Because there are four bonds around it. So one, two, three, and four. So now the formal charge is zero. So this is the correct resonance structure. Now it looks a little bit confusing when I have those hydrogens drawn in there, which is why we leave them off. Let me go ahead and draw it again on the right just for clarity, right? That's why we leave off those hydrogens when we're drawing our resonance structures, because they get in the way, right?"}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So this is the correct resonance structure. Now it looks a little bit confusing when I have those hydrogens drawn in there, which is why we leave them off. Let me go ahead and draw it again on the right just for clarity, right? That's why we leave off those hydrogens when we're drawing our resonance structures, because they get in the way, right? And once you understand what's going on, it's not necessary to draw in those hydrogens. Let's do another example. And again, I'll start with the wrong way to do it, and then we'll talk about the correct way."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "That's why we leave off those hydrogens when we're drawing our resonance structures, because they get in the way, right? And once you understand what's going on, it's not necessary to draw in those hydrogens. Let's do another example. And again, I'll start with the wrong way to do it, and then we'll talk about the correct way. So a student might say, all right, I have a negative one formal charge on this nitrogen. So I could take this lone pair of electrons and move it into here, which would push these electrons over to here. So let me go ahead and draw what some students might think is a correct resonance structure."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "And again, I'll start with the wrong way to do it, and then we'll talk about the correct way. So a student might say, all right, I have a negative one formal charge on this nitrogen. So I could take this lone pair of electrons and move it into here, which would push these electrons over to here. So let me go ahead and draw what some students might think is a correct resonance structure. So let me put in my lone pair of electrons. Let's follow some electrons along. So electrons in light blue on this nitrogen move into here, and electrons in, let's say, dark blue move down to here."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw what some students might think is a correct resonance structure. So let me put in my lone pair of electrons. Let's follow some electrons along. So electrons in light blue on this nitrogen move into here, and electrons in, let's say, dark blue move down to here. And then finally, electrons in magenta remain behind on the nitrogen. So on the right, why is this not, why is this not a correct resonance structure? So again, this is the wrong way to do it."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So electrons in light blue on this nitrogen move into here, and electrons in, let's say, dark blue move down to here. And then finally, electrons in magenta remain behind on the nitrogen. So on the right, why is this not, why is this not a correct resonance structure? So again, this is the wrong way to do it. Well, think about your hydrogens. So we'll start with this carbon right here. This carbon has one hydrogen."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So again, this is the wrong way to do it. Well, think about your hydrogens. So we'll start with this carbon right here. This carbon has one hydrogen. This carbon has one hydrogen. And this carbon down here has two. So if we put in those hydrogens over here on the right, hopefully it's obvious why this is incorrect."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "This carbon has one hydrogen. This carbon has one hydrogen. And this carbon down here has two. So if we put in those hydrogens over here on the right, hopefully it's obvious why this is incorrect. Let's look at this carbon down at the bottom of the ring. So this carbon right here I just marked in red. How many bonds do we have?"}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So if we put in those hydrogens over here on the right, hopefully it's obvious why this is incorrect. Let's look at this carbon down at the bottom of the ring. So this carbon right here I just marked in red. How many bonds do we have? Well, here's one, two, three, four, and five. So there are five bonds to that carbon. And we know carbon can never have five bonds."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "How many bonds do we have? Well, here's one, two, three, four, and five. So there are five bonds to that carbon. And we know carbon can never have five bonds. Carbon can never exceed an octet of electrons. So immediately we know that this is not a correct resonance structure. All right, let's talk about the right way to do it."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "And we know carbon can never have five bonds. Carbon can never exceed an octet of electrons. So immediately we know that this is not a correct resonance structure. All right, let's talk about the right way to do it. So you take these electrons and you move them into here. And then these pi electrons have to go somewhere, and they move out onto this carbon. So now let's draw the correct other resonance structure here."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's talk about the right way to do it. So you take these electrons and you move them into here. And then these pi electrons have to go somewhere, and they move out onto this carbon. So now let's draw the correct other resonance structure here. So we'll put in our double bond. We'll put our electrons on this carbon. That gives this carbon a negative one formal charge."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So now let's draw the correct other resonance structure here. So we'll put in our double bond. We'll put our electrons on this carbon. That gives this carbon a negative one formal charge. And then we had some electrons, a lone pair of electrons left on the nitrogen. I'll use the same colors as before. So these electrons right here in light blue move in to form our double bond."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "That gives this carbon a negative one formal charge. And then we had some electrons, a lone pair of electrons left on the nitrogen. I'll use the same colors as before. So these electrons right here in light blue move in to form our double bond. The electrons in dark blue move off onto this carbon. So the electrons in dark blue are on this carbon that I just marked in dark blue, which gives that carbon a negative one formal charge. And the electrons in magenta remain behind on the nitrogen."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here in light blue move in to form our double bond. The electrons in dark blue move off onto this carbon. So the electrons in dark blue are on this carbon that I just marked in dark blue, which gives that carbon a negative one formal charge. And the electrons in magenta remain behind on the nitrogen. The reason why the carbon, now I'll go ahead and mark it in a different color. This carbon I just marked in magenta has a negative one formal charge. It's because remember there's one hydrogen on that carbon."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons in magenta remain behind on the nitrogen. The reason why the carbon, now I'll go ahead and mark it in a different color. This carbon I just marked in magenta has a negative one formal charge. It's because remember there's one hydrogen on that carbon. So let me draw in that hydrogen over here. Let me see if I can squeeze it in over here like that. This carbon right here has three bonds to it and a lone pair of electrons, which gives that carbon in magenta a negative one formal charge."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "It's because remember there's one hydrogen on that carbon. So let me draw in that hydrogen over here. Let me see if I can squeeze it in over here like that. This carbon right here has three bonds to it and a lone pair of electrons, which gives that carbon in magenta a negative one formal charge. Now this nitrogen over here would have a formal charge equal to zero. And so this on the right would be the correct resonance structure. And again, drawing in all this, drawing in hydrogens is a waste of time."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here has three bonds to it and a lone pair of electrons, which gives that carbon in magenta a negative one formal charge. Now this nitrogen over here would have a formal charge equal to zero. And so this on the right would be the correct resonance structure. And again, drawing in all this, drawing in hydrogens is a waste of time. It gets in the way. Let me go ahead and draw the resonance structure again. I'll take out that hydrogen so it'll look cleaner."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "And again, drawing in all this, drawing in hydrogens is a waste of time. It gets in the way. Let me go ahead and draw the resonance structure again. I'll take out that hydrogen so it'll look cleaner. It also takes less time when you're not drawing in all of your hydrogens. So you just put a lone pair of electrons and write a negative one formal charge. And you have to know that there's still a hydrogen on this carbon that I marked in magenta."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "I'll take out that hydrogen so it'll look cleaner. It also takes less time when you're not drawing in all of your hydrogens. So you just put a lone pair of electrons and write a negative one formal charge. And you have to know that there's still a hydrogen on this carbon that I marked in magenta. So if you're having trouble drawing resonance structures, usually the problem is not thinking about your hydrogens, right? Forgetting about putting in your hydrogens. And once you put those in, it's a lot easier to see if your dot structure, if your resonance structure I should say, is correct."}, {"video_title": "Common mistakes when drawing resonance structures Organic chemistry Khan Academy.mp3", "Sentence": "And you have to know that there's still a hydrogen on this carbon that I marked in magenta. So if you're having trouble drawing resonance structures, usually the problem is not thinking about your hydrogens, right? Forgetting about putting in your hydrogens. And once you put those in, it's a lot easier to see if your dot structure, if your resonance structure I should say, is correct. So be careful about that. And resonance structures are just practice. So the more you draw, the better you're gonna get."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with physical properties of alcohols. And so we're going to compare, in this case, alcohols to alkanes. And this alkane on the left here, two carbons, so this is, of course, ethane. On the right, if we take off one of those hydrogens and replace it with an OH, we, of course, have ethanol right here. So let's start with boiling point. So the boiling point of ethane is approximately negative 89 degrees Celsius. And since room temperature is somewhere around 20 to 25 degrees Celsius, at room temperature, we are much higher than the boiling point of ethane, which means it's already boiled."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "On the right, if we take off one of those hydrogens and replace it with an OH, we, of course, have ethanol right here. So let's start with boiling point. So the boiling point of ethane is approximately negative 89 degrees Celsius. And since room temperature is somewhere around 20 to 25 degrees Celsius, at room temperature, we are much higher than the boiling point of ethane, which means it's already boiled. It's already turned into a gas. So at room temperature and room pressure, ethane is a gas. Ethanol, however, has a much higher boiling point, somewhere around 78 degrees Celsius."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And since room temperature is somewhere around 20 to 25 degrees Celsius, at room temperature, we are much higher than the boiling point of ethane, which means it's already boiled. It's already turned into a gas. So at room temperature and room pressure, ethane is a gas. Ethanol, however, has a much higher boiling point, somewhere around 78 degrees Celsius. And once again, since room temperature is somewhere around 20 to 25, the boiling point of ethanol is much higher than room temperature. So at room temperature and pressure, ethanol is a liquid. It hasn't boiled yet."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "Ethanol, however, has a much higher boiling point, somewhere around 78 degrees Celsius. And once again, since room temperature is somewhere around 20 to 25, the boiling point of ethanol is much higher than room temperature. So at room temperature and pressure, ethanol is a liquid. It hasn't boiled yet. And these large differences in boiling points between these two molecules can be attributed to the intermolecular forces that are present. So if two molecules of ethane are interacting, really the only intermolecular force that's holding those molecules together would be London dispersion forces, which are the weakest of the intermolecular forces. So it's relatively easy to pull apart two ethane molecules."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "It hasn't boiled yet. And these large differences in boiling points between these two molecules can be attributed to the intermolecular forces that are present. So if two molecules of ethane are interacting, really the only intermolecular force that's holding those molecules together would be London dispersion forces, which are the weakest of the intermolecular forces. So it's relatively easy to pull apart two ethane molecules. And that accounts for the very low boiling point. It doesn't take a lot of energy to pull them apart. So it's easy for it to turn into a gas."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So it's relatively easy to pull apart two ethane molecules. And that accounts for the very low boiling point. It doesn't take a lot of energy to pull them apart. So it's easy for it to turn into a gas. Ethanol, however, is a much higher boiling point, which means it's much harder to pull those molecules apart. It takes more energy. So let's look at why ethanol has such a higher boiling point."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So it's easy for it to turn into a gas. Ethanol, however, is a much higher boiling point, which means it's much harder to pull those molecules apart. It takes more energy. So let's look at why ethanol has such a higher boiling point. So if I show two ethanol molecules interacting, so here is one ethanol molecule. And let's go ahead and draw another ethanol molecule right here. And if I think about the oxygen-hydrogen bond, I know that's a polarized covalent bond."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at why ethanol has such a higher boiling point. So if I show two ethanol molecules interacting, so here is one ethanol molecule. And let's go ahead and draw another ethanol molecule right here. And if I think about the oxygen-hydrogen bond, I know that's a polarized covalent bond. I know that there's a large difference in electronegativity between the oxygen and the hydrogen. Oxygen's much more electronegative, which means the electrons in the bond between oxygen and hydrogen are going to be much closer to the oxygen atom, giving the oxygen atom a partial negative charge. So these electrons in this bond between oxygen and hydrogen are going to move away from the hydrogen."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And if I think about the oxygen-hydrogen bond, I know that's a polarized covalent bond. I know that there's a large difference in electronegativity between the oxygen and the hydrogen. Oxygen's much more electronegative, which means the electrons in the bond between oxygen and hydrogen are going to be much closer to the oxygen atom, giving the oxygen atom a partial negative charge. So these electrons in this bond between oxygen and hydrogen are going to move away from the hydrogen. The hydrogen's going to lose a little bit of electron density, leaving it relatively positive. So we give it a partial positive charge. It's the same thing for the other ethanol molecule, partially negative oxygen, partially positive hydrogen."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in this bond between oxygen and hydrogen are going to move away from the hydrogen. The hydrogen's going to lose a little bit of electron density, leaving it relatively positive. So we give it a partial positive charge. It's the same thing for the other ethanol molecule, partially negative oxygen, partially positive hydrogen. And we know that opposite charges attract. So the partially positive hydrogen is attracted to the partially negative oxygen. And so there's a strong intermolecular force that holds those two molecules together."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "It's the same thing for the other ethanol molecule, partially negative oxygen, partially positive hydrogen. And we know that opposite charges attract. So the partially positive hydrogen is attracted to the partially negative oxygen. And so there's a strong intermolecular force that holds those two molecules together. And that, of course, is hydrogen bonding. So there's hydrogen bonding between alcohol molecules. And since hydrogen bonding is the strongest intermolecular force, it's relatively difficult to pull those molecules apart."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And so there's a strong intermolecular force that holds those two molecules together. And that, of course, is hydrogen bonding. So there's hydrogen bonding between alcohol molecules. And since hydrogen bonding is the strongest intermolecular force, it's relatively difficult to pull those molecules apart. It takes a lot of energy, takes a lot of heat. And that's why the boiling point of ethanol is so much higher than the boiling point of ethane. It also accounts for the state of matter."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And since hydrogen bonding is the strongest intermolecular force, it's relatively difficult to pull those molecules apart. It takes a lot of energy, takes a lot of heat. And that's why the boiling point of ethanol is so much higher than the boiling point of ethane. It also accounts for the state of matter. What about solubility? So is ethanol soluble in water? And of course it is."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "It also accounts for the state of matter. What about solubility? So is ethanol soluble in water? And of course it is. And the reason why is hydrogen bonding, once again. So if we draw a water molecule in here, I know that the water molecule is polarized in the same way that the alcohol molecule is. So the hydrogen is partially positive."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And of course it is. And the reason why is hydrogen bonding, once again. So if we draw a water molecule in here, I know that the water molecule is polarized in the same way that the alcohol molecule is. So the hydrogen is partially positive. And the oxygen is right over here, is partially negative. And so once again, opposite charges attract. The hydrogen is attracted to this oxygen here."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So the hydrogen is partially positive. And the oxygen is right over here, is partially negative. And so once again, opposite charges attract. The hydrogen is attracted to this oxygen here. And so because of hydrogen bonding, there's interaction between the water molecule and between the alcohol molecule. So the water molecule is polar. So if you want to think about it in terms of polarity, because of the difference in electronegativity, water is a polar molecule."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen is attracted to this oxygen here. And so because of hydrogen bonding, there's interaction between the water molecule and between the alcohol molecule. So the water molecule is polar. So if you want to think about it in terms of polarity, because of the difference in electronegativity, water is a polar molecule. Ethanol is a polar molecule, and like dissolves like. So these two molecules will be soluble in each other. So if I look at the structure of ethanol, the reason why it is soluble in water is because of this portion of the molecule, this hydroxyl group, this OH."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So if you want to think about it in terms of polarity, because of the difference in electronegativity, water is a polar molecule. Ethanol is a polar molecule, and like dissolves like. So these two molecules will be soluble in each other. So if I look at the structure of ethanol, the reason why it is soluble in water is because of this portion of the molecule, this hydroxyl group, this OH. It's the differences in electronegativity that allow the hydrogen bonding. So this portion of the molecule is the polar portion of the molecule. And this portion of the molecule is the part that loves water, which is why it is soluble."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at the structure of ethanol, the reason why it is soluble in water is because of this portion of the molecule, this hydroxyl group, this OH. It's the differences in electronegativity that allow the hydrogen bonding. So this portion of the molecule is the polar portion of the molecule. And this portion of the molecule is the part that loves water, which is why it is soluble. So if it loves water, we say it's hydrophilic. Hydro meaning water, phil meaning love, so hydrophilic. Whereas this portion over here on the left, this is more of an alkane type environment, a nonpolar type of environment."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And this portion of the molecule is the part that loves water, which is why it is soluble. So if it loves water, we say it's hydrophilic. Hydro meaning water, phil meaning love, so hydrophilic. Whereas this portion over here on the left, this is more of an alkane type environment, a nonpolar type of environment. So this part of the molecule is scared of water, so it's hydrophobic. So we have the hydrophobic portion of our alcohol molecule, we have the hydrophilic portion of the alcohol molecule. Now, we know that like dissolves like, so nonpolar will not dissolve in polar."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "Whereas this portion over here on the left, this is more of an alkane type environment, a nonpolar type of environment. So this part of the molecule is scared of water, so it's hydrophobic. So we have the hydrophobic portion of our alcohol molecule, we have the hydrophilic portion of the alcohol molecule. Now, we know that like dissolves like, so nonpolar will not dissolve in polar. But as long as we have a relatively small number of carbon atoms in our alkyl group, the OH group is polar enough for the alcohol to be soluble in water. Now, if you have a large number of carbon atoms, your molecule is more nonpolar than polar. And so alcohols will stop being soluble in water if they have a lot of carbon atoms on them."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "Now, we know that like dissolves like, so nonpolar will not dissolve in polar. But as long as we have a relatively small number of carbon atoms in our alkyl group, the OH group is polar enough for the alcohol to be soluble in water. Now, if you have a large number of carbon atoms, your molecule is more nonpolar than polar. And so alcohols will stop being soluble in water if they have a lot of carbon atoms on them. So let's look at now the preparation of alkoxides. So let's look at an alcohol. So here we have our alcohol."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And so alcohols will stop being soluble in water if they have a lot of carbon atoms on them. So let's look at now the preparation of alkoxides. So let's look at an alcohol. So here we have our alcohol. And if we react our alcohol with a strong base, so we'll give it a lone pair of electrons and negative 1 formal charge. So we have a strong base here. And our strong base is going to take this proton and leave these electrons behind on this oxygen."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So here we have our alcohol. And if we react our alcohol with a strong base, so we'll give it a lone pair of electrons and negative 1 formal charge. So we have a strong base here. And our strong base is going to take this proton and leave these electrons behind on this oxygen. So now we have an oxygen that used to have two lone pairs of electrons and now has three lone pairs. That gives it a negative 1 formal charge. And the base is going to form a bond with that proton like that."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And our strong base is going to take this proton and leave these electrons behind on this oxygen. So now we have an oxygen that used to have two lone pairs of electrons and now has three lone pairs. That gives it a negative 1 formal charge. And the base is going to form a bond with that proton like that. So this is an acid-base reaction. So if we react an alcohol with a strong base, so this is a strong base here, we're going to form the conjugate base to an alcohol, which is called an alkoxide. So this is an alkoxide ion right here."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And the base is going to form a bond with that proton like that. So this is an acid-base reaction. So if we react an alcohol with a strong base, so this is a strong base here, we're going to form the conjugate base to an alcohol, which is called an alkoxide. So this is an alkoxide ion right here. So it's a chemical property of alcohols. They are acidic if you use a strong enough base. And the conjugate base to an alcohol is called an alkoxide."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So this is an alkoxide ion right here. So it's a chemical property of alcohols. They are acidic if you use a strong enough base. And the conjugate base to an alcohol is called an alkoxide. Let's look at an example. So let's take ethanol. So here I have my ethanol molecule."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And the conjugate base to an alcohol is called an alkoxide. Let's look at an example. So let's take ethanol. So here I have my ethanol molecule. And we'll react that with a strong base, something like sodium hydride, so NaH, so Na plus, and H with 2, hydrogen with 2 electrons around it, which makes it a negatively charged ion. So that's called the hydride anion. So we have the basic portion, the negatively charged hydrogen."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So here I have my ethanol molecule. And we'll react that with a strong base, something like sodium hydride, so NaH, so Na plus, and H with 2, hydrogen with 2 electrons around it, which makes it a negatively charged ion. So that's called the hydride anion. So we have the basic portion, the negatively charged hydrogen. It's going to function as a base. It's going to take these two electrons. It's going to take that proton right there."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So we have the basic portion, the negatively charged hydrogen. It's going to function as a base. It's going to take these two electrons. It's going to take that proton right there. So the acidic proton on alcohols is the one on the oxygen. And the electrons in here are going to kick off onto our oxygen like that. So we're going to get for our product an alkoxide with three lone pairs of electrons around it, giving it a negative 1 formal charge."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "It's going to take that proton right there. So the acidic proton on alcohols is the one on the oxygen. And the electrons in here are going to kick off onto our oxygen like that. So we're going to get for our product an alkoxide with three lone pairs of electrons around it, giving it a negative 1 formal charge. The sodium is floating around, positively charged. So it's going to electrostatically ionically interact with our alkoxide anion. And the hydride anion picked up a proton."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to get for our product an alkoxide with three lone pairs of electrons around it, giving it a negative 1 formal charge. The sodium is floating around, positively charged. So it's going to electrostatically ionically interact with our alkoxide anion. And the hydride anion picked up a proton. So those two hydrogens combine to form hydrogen gas, which will, of course, bubble out of your solution. So the formation of hydrogen gas will be observed in this reaction. And this is how you form an alkoxide."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And the hydride anion picked up a proton. So those two hydrogens combine to form hydrogen gas, which will, of course, bubble out of your solution. So the formation of hydrogen gas will be observed in this reaction. And this is how you form an alkoxide. This molecule is called sodium ethoxide. So we have sodium ethoxide over here on the right, which is a relatively strong base that is used in a lot of organic chemistry reactions. And let's see, we used a strong base to form it."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And this is how you form an alkoxide. This molecule is called sodium ethoxide. So we have sodium ethoxide over here on the right, which is a relatively strong base that is used in a lot of organic chemistry reactions. And let's see, we used a strong base to form it. We used a sodium hydride over here, sodium hydride, to form that molecule from ethanol. So there's another way to form alkoxides. So let's take a look at a generic way to form alkoxides from group 1 alkali metals."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "And let's see, we used a strong base to form it. We used a sodium hydride over here, sodium hydride, to form that molecule from ethanol. So there's another way to form alkoxides. So let's take a look at a generic way to form alkoxides from group 1 alkali metals. So here we have our alcohol like that. And if we react our alcohol with a group 1 metal, so an alkali metal, those all have one valence electron being in group 1 on the periodic table, so something like lithium or sodium or potassium. We are going to form an alkoxide."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a look at a generic way to form alkoxides from group 1 alkali metals. So here we have our alcohol like that. And if we react our alcohol with a group 1 metal, so an alkali metal, those all have one valence electron being in group 1 on the periodic table, so something like lithium or sodium or potassium. We are going to form an alkoxide. So we're going to form, let's see, three lone pairs of electrons and negative 1 formal charge. And the mechanism, the metal is going to donate its one valence electron, leaving it with a plus 1 charge. So it's going to interact with your alkoxide like that."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "We are going to form an alkoxide. So we're going to form, let's see, three lone pairs of electrons and negative 1 formal charge. And the mechanism, the metal is going to donate its one valence electron, leaving it with a plus 1 charge. So it's going to interact with your alkoxide like that. And this also releases hydrogen gas like that. So that's your general reaction. Let's look at an example where we react."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to interact with your alkoxide like that. And this also releases hydrogen gas like that. So that's your general reaction. Let's look at an example where we react. Let's use cyclohexanol. So we're going to react cyclohexanol with sodium. So let's actually go ahead and redraw that cyclohexanol molecule here, because I want to show a little bit of the mechanism."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an example where we react. Let's use cyclohexanol. So we're going to react cyclohexanol with sodium. So let's actually go ahead and redraw that cyclohexanol molecule here, because I want to show a little bit of the mechanism. So let's go ahead and draw it like that and put our lone pairs of electrons on the oxygen. So sodium metal has one valence electron like that. So if we think about what happens, sodium will donate its one valence electron very easily, because it will then have the stable electronic configuration of a noble gas."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So let's actually go ahead and redraw that cyclohexanol molecule here, because I want to show a little bit of the mechanism. So let's go ahead and draw it like that and put our lone pairs of electrons on the oxygen. So sodium metal has one valence electron like that. So if we think about what happens, sodium will donate its one valence electron very easily, because it will then have the stable electronic configuration of a noble gas. So the first step of the mechanism is donation of this one valence electron. So we're going to show the movement of one electron. So we're going to use a half-headed arrow."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about what happens, sodium will donate its one valence electron very easily, because it will then have the stable electronic configuration of a noble gas. So the first step of the mechanism is donation of this one valence electron. So we're going to show the movement of one electron. So we're going to use a half-headed arrow. And then the two electrons in the bond between oxygen and hydrogen, we're going to use a two-headed arrow to show the movement of those two electrons off onto that oxygen. So let's go ahead and draw what we have now. We have our cyclohexane ring."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to use a half-headed arrow. And then the two electrons in the bond between oxygen and hydrogen, we're going to use a two-headed arrow to show the movement of those two electrons off onto that oxygen. So let's go ahead and draw what we have now. We have our cyclohexane ring. And we now have three lone pairs of electrons around my oxygen, which makes it negatively charged. And the sodium donated its one valence electron. So now it has a plus 1 charge."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "We have our cyclohexane ring. And we now have three lone pairs of electrons around my oxygen, which makes it negatively charged. And the sodium donated its one valence electron. So now it has a plus 1 charge. So it's going to interact with that negatively charged oxygen. And what happened to the hydrogen? That hydrogen there is going to pick up one electron."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "So now it has a plus 1 charge. So it's going to interact with that negatively charged oxygen. And what happened to the hydrogen? That hydrogen there is going to pick up one electron. So now we have hydrogen with one electron around it. That is extremely reactive. Hydrogen prefers to have two electrons around it."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "That hydrogen there is going to pick up one electron. So now we have hydrogen with one electron around it. That is extremely reactive. Hydrogen prefers to have two electrons around it. So when two of those hydrogen atoms get close to each other, they're going to, of course, react and share their electrons to form hydrogen gas. So I could draw it like that. So that's where those two electrons are."}, {"video_title": "Physical properties of alcohols and preparation of alkoxides Organic chemistry Khan Academy.mp3", "Sentence": "Hydrogen prefers to have two electrons around it. So when two of those hydrogen atoms get close to each other, they're going to, of course, react and share their electrons to form hydrogen gas. So I could draw it like that. So that's where those two electrons are. The two electrons are here. So each one from one hydrogen, one from the other hydrogen. So hydrogen gas is going to be produced in this reaction as well."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or why are we even tempted to go through this thought experiment of the number of detectable civilizations in the galaxy? When we don't have a clue of some of these assumptions. We don't know what fraction of planets capable of sustaining life actually do generate life. We don't know of all of the planets that have life, what fraction of those planets go on to have intelligent life. What fractions of those civilizations go on to using electromagnetic radiation as a form of communication. We don't know these answers. In fact, we probably won't know some of these answers for some, some time."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We don't know of all of the planets that have life, what fraction of those planets go on to have intelligent life. What fractions of those civilizations go on to using electromagnetic radiation as a form of communication. We don't know these answers. In fact, we probably won't know some of these answers for some, some time. So what's the point of going through this exercise? And that is a valid point of view. The Drake equation, or even this little equation that we've set up here, it's not an equation in the traditional sense where we can immediately apply it to some engineering problem or some physical problem or anything like that."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, we probably won't know some of these answers for some, some time. So what's the point of going through this exercise? And that is a valid point of view. The Drake equation, or even this little equation that we've set up here, it's not an equation in the traditional sense where we can immediately apply it to some engineering problem or some physical problem or anything like that. I view it more as a bit of a thought experiment. And what's interesting about it is it can structure our thought around the problem. And I think that's where it has the most value."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The Drake equation, or even this little equation that we've set up here, it's not an equation in the traditional sense where we can immediately apply it to some engineering problem or some physical problem or anything like that. I view it more as a bit of a thought experiment. And what's interesting about it is it can structure our thought around the problem. And I think that's where it has the most value. We'll probably not get a solid number on this any time soon. But it does lead us to thinking about these interesting problems of what does it mean, or what do we think has to happen for a planet to start getting life, even if it has all the right ingredients. And then what does it mean for things to eventually get to the point that you have intelligent life?"}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I think that's where it has the most value. We'll probably not get a solid number on this any time soon. But it does lead us to thinking about these interesting problems of what does it mean, or what do we think has to happen for a planet to start getting life, even if it has all the right ingredients. And then what does it mean for things to eventually get to the point that you have intelligent life? And in all fairness to this, is that probably 200 years ago, there would have been no way to even have a decent estimate of the number of stars in the galaxy. Now we're starting to do an okay job on that. 20 or 30 years ago, it would have been viewed impossible to say the fraction of stars that have planets."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then what does it mean for things to eventually get to the point that you have intelligent life? And in all fairness to this, is that probably 200 years ago, there would have been no way to even have a decent estimate of the number of stars in the galaxy. Now we're starting to do an okay job on that. 20 or 30 years ago, it would have been viewed impossible to say the fraction of stars that have planets. But now we're finding exoplanets, we're seeing stars wobble, we're getting more and more accurate instruments so we can start to think about planets that are closer to the size of Earth. So we're making headway there. There's other indirect methods to think about, well, you know, some of these exoplanets look like they're in the right zone, and they look like they have the right chemical signature based on other information that we're getting."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "20 or 30 years ago, it would have been viewed impossible to say the fraction of stars that have planets. But now we're finding exoplanets, we're seeing stars wobble, we're getting more and more accurate instruments so we can start to think about planets that are closer to the size of Earth. So we're making headway there. There's other indirect methods to think about, well, you know, some of these exoplanets look like they're in the right zone, and they look like they have the right chemical signature based on other information that we're getting. Maybe they are capable of sustaining life. So as time goes on and as technology improves, we might be able to get better and better and better at this. But with that said, it's not going to happen any time soon."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's other indirect methods to think about, well, you know, some of these exoplanets look like they're in the right zone, and they look like they have the right chemical signature based on other information that we're getting. Maybe they are capable of sustaining life. So as time goes on and as technology improves, we might be able to get better and better and better at this. But with that said, it's not going to happen any time soon. And the real value of all of this is really to structure our thought about a super, super interesting topic. Now the other thing I want to talk about is a slight clarification of what I talked about in the last video. In the last video for this L, I said it's the civilization's lifespan."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But with that said, it's not going to happen any time soon. And the real value of all of this is really to structure our thought about a super, super interesting topic. Now the other thing I want to talk about is a slight clarification of what I talked about in the last video. In the last video for this L, I said it's the civilization's lifespan. But what's actually relevant is the lifespan of the civilization while it is detectable. So it doesn't matter if the civilization is around 100,000 years, but it's not releasing any type of thing that we can detect. That's not what we care about."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video for this L, I said it's the civilization's lifespan. But what's actually relevant is the lifespan of the civilization while it is detectable. So it doesn't matter if the civilization is around 100,000 years, but it's not releasing any type of thing that we can detect. That's not what we care about. We care about the 5,000 years or the 10,000 years or the 100,000 years when they are actually using some type of communications or some type of electromagnetic radiation that we can eventually detect once those things reach us. Now the other thing I want to make it clear is we're talking about the number of detectable civilizations in the galaxy right now. And I'll write now in quotation marks."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's not what we care about. We care about the 5,000 years or the 10,000 years or the 100,000 years when they are actually using some type of communications or some type of electromagnetic radiation that we can eventually detect once those things reach us. Now the other thing I want to make it clear is we're talking about the number of detectable civilizations in the galaxy right now. And I'll write now in quotation marks. Because we're not talking about a civilization that is maybe even a pure civilization with us that developed radio communication on the order of 100 years ago. Because frankly, they would have to be no more than 100 light years away for us to be able to detect those signals now. If they were on the other side of the galaxy, we wouldn't be able to detect their signals for tens of thousands of years."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'll write now in quotation marks. Because we're not talking about a civilization that is maybe even a pure civilization with us that developed radio communication on the order of 100 years ago. Because frankly, they would have to be no more than 100 light years away for us to be able to detect those signals now. If they were on the other side of the galaxy, we wouldn't be able to detect their signals for tens of thousands of years. So when I talk about now, I'm saying that the signals are getting to us. Signals getting. Signals received."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If they were on the other side of the galaxy, we wouldn't be able to detect their signals for tens of thousands of years. So when I talk about now, I'm saying that the signals are getting to us. Signals getting. Signals received. The signals are being received right now. So you could have a civilization that developed radio 70,000 years ago, but they're 70,000 light years away. And maybe they collapse 10,000 years later, but we're just receiving their first radio signal."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Signals received. The signals are being received right now. So you could have a civilization that developed radio 70,000 years ago, but they're 70,000 light years away. And maybe they collapse 10,000 years later, but we're just receiving their first radio signal. So that would be a civilization that I would count in this equation we're setting up. And so just to make sure we understand it, and then we can play with some numbers, let's remind ourselves. This is the number of stars, our estimate of the number of stars in the galaxy."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And maybe they collapse 10,000 years later, but we're just receiving their first radio signal. So that would be a civilization that I would count in this equation we're setting up. And so just to make sure we understand it, and then we can play with some numbers, let's remind ourselves. This is the number of stars, our estimate of the number of stars in the galaxy. Multiply by this, you now know the number of stars in the galaxy that have planets. You multiply by this N sub p, the average number of planets capable of sustaining life. And these first three terms will give you the average number of planets in the galaxy, or I should say the number, the total number of planets in the galaxy that have been capable of sustaining life at some point in their history."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the number of stars, our estimate of the number of stars in the galaxy. Multiply by this, you now know the number of stars in the galaxy that have planets. You multiply by this N sub p, the average number of planets capable of sustaining life. And these first three terms will give you the average number of planets in the galaxy, or I should say the number, the total number of planets in the galaxy that have been capable of sustaining life at some point in their history. Multiply it by this. This is the number of planets in the galaxy that have sustained actual life, not just capability of it. They actually had life on them at some point in their history."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these first three terms will give you the average number of planets in the galaxy, or I should say the number, the total number of planets in the galaxy that have been capable of sustaining life at some point in their history. Multiply it by this. This is the number of planets in the galaxy that have sustained actual life, not just capability of it. They actually had life on them at some point in their history. Multiply it by this. This is the fraction that have developed intelligent life on these planets, the number of planets with intelligent life at some point in their history. Multiply it by this fraction, all of these terms."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They actually had life on them at some point in their history. Multiply it by this. This is the fraction that have developed intelligent life on these planets, the number of planets with intelligent life at some point in their history. Multiply it by this fraction, all of these terms. You have the number of planets in the galaxy that have developed, that have had intelligent life, that became detectable, that started emitting some type of radio signature. We don't know. Some type of thing like that at some point in their history."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Multiply it by this fraction, all of these terms. You have the number of planets in the galaxy that have developed, that have had intelligent life, that became detectable, that started emitting some type of radio signature. We don't know. Some type of thing like that at some point in their history. So over here, all of these first six terms tell us the number of detectable civilizations that occurred at some point in the history of the stars, the solar systems, the planets that are out there right now. But we care about the ones that are detectable now. We don't care about the ones that came and went, and their radio signature went past us while we were still living in caves, or we were hunter-gatherers."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some type of thing like that at some point in their history. So over here, all of these first six terms tell us the number of detectable civilizations that occurred at some point in the history of the stars, the solar systems, the planets that are out there right now. But we care about the ones that are detectable now. We don't care about the ones that came and went, and their radio signature went past us while we were still living in caves, or we were hunter-gatherers. We care about the ones that their radio signatures are receiving us now. That's why we have this little term right over here. This is the civilization of, I guess you could say, this is the length of the detectable civilization."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We don't care about the ones that came and went, and their radio signature went past us while we were still living in caves, or we were hunter-gatherers. We care about the ones that their radio signatures are receiving us now. That's why we have this little term right over here. This is the civilization of, I guess you could say, this is the length of the detectable civilization. So while they were actually releasing a radio signature, divided by the life of that planet or that solar system or that star. For any given star or planet that meets all of these criterion, what's the probability that it's releasing its... At some point in the history, there was a detectable civilization or more that was releasing some type of a radio signature. But what's the probability that it's doing it right now?"}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the civilization of, I guess you could say, this is the length of the detectable civilization. So while they were actually releasing a radio signature, divided by the life of that planet or that solar system or that star. For any given star or planet that meets all of these criterion, what's the probability that it's releasing its... At some point in the history, there was a detectable civilization or more that was releasing some type of a radio signature. But what's the probability that it's doing it right now? That's the detectable lifespan of that civilization divided by the life of that solar system or of that planet. Because frankly, the star and the solar system and the planet, they're all going to essentially have, give or take, a few hundreds of thousands of years or even a few millions of years, because we're thinking in the billions here. They're going to have roughly the same lifespan."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's the probability that it's doing it right now? That's the detectable lifespan of that civilization divided by the life of that solar system or of that planet. Because frankly, the star and the solar system and the planet, they're all going to essentially have, give or take, a few hundreds of thousands of years or even a few millions of years, because we're thinking in the billions here. They're going to have roughly the same lifespan. And so if you have... Let's say, and just to make this a little bit more tangible, let's say that the sun has a lifespan, and let's say that with the Earth and our solar system, has a lifespan of approximately 10 billion years. And let's say that us as humans, let me be pretty optimistic about it, let's say that we are detectable as a civilization for 1 million years. So we have our best days are ahead of us."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're going to have roughly the same lifespan. And so if you have... Let's say, and just to make this a little bit more tangible, let's say that the sun has a lifespan, and let's say that with the Earth and our solar system, has a lifespan of approximately 10 billion years. And let's say that us as humans, let me be pretty optimistic about it, let's say that we are detectable as a civilization for 1 million years. So we have our best days are ahead of us. So we are detectable for 1 million years. So this term right over here will be 1 million over 10 billion. So this will be 1 over 10,000."}, {"video_title": "Detectable civilizations in our galaxy 2 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we have our best days are ahead of us. So we are detectable for 1 million years. So this term right over here will be 1 million over 10 billion. So this will be 1 over 10,000. So even though we might be around sending out detectable signals for a million years, the odds relative to the entire span of the history of our... And I'm making some simplifying assumptions here. But relative to the entire span of the history of our planet and our sun, if someone is just randomly sampling our solar system at a random time in its history, in a random part of this 10 billion years, there's only a 1 in 10,000 chance that they'll be sampling us at a time that we are releasing signals, assuming that there weren't any other civilizations on Mars or Venus or whatever else, or that there weren't any other civilizations on Earth hundreds of thousands of years ago that were doing this, they'll definitely only have a 1 in 10,000 chance of detecting us, assuming that they're sampling. There could have been a civilization that was around 3 million years ago, and they did this whole search for extraterrestrial life, maybe they're 20 or 100 or 1,000 light years away, and they pointed their radio telescopes at us."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "In the last video, we looked at the mechanism for the SN2 reaction. The hydroxide ion will function as a nucleophile in this case and attack our electrophile, so right here at this carbon. And since the SN2 mechanism is concerted, the nucleophile attacks the electrophile at the same time that our leaving group leaves. So bromine leaves as the bromide anion and the OH, the nucleophile, substitutes for our leaving group. So for our final product, we now have an OH attached to our carbon chain. In this video, we're gonna look at the stereospecificity of the SN2 reaction. And that just means that the stereochemistry of the reactant determines the stereochemistry of the product."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So bromine leaves as the bromide anion and the OH, the nucleophile, substitutes for our leaving group. So for our final product, we now have an OH attached to our carbon chain. In this video, we're gonna look at the stereospecificity of the SN2 reaction. And that just means that the stereochemistry of the reactant determines the stereochemistry of the product. For example, if we look at our substrate, we know that this carbon is a chiral center and our bromine is on a wedge, it's coming out at us in space. So the configuration at this chiral center is R. When we look at our product, this is that chiral center, but now we have our OH going away from us in space and the configuration at this chiral center is S. So the stereochemistry of our product is determined by the stereochemistry of our reactant and that's because of our SN2 mechanism. We observe inversion of configuration."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And that just means that the stereochemistry of the reactant determines the stereochemistry of the product. For example, if we look at our substrate, we know that this carbon is a chiral center and our bromine is on a wedge, it's coming out at us in space. So the configuration at this chiral center is R. When we look at our product, this is that chiral center, but now we have our OH going away from us in space and the configuration at this chiral center is S. So the stereochemistry of our product is determined by the stereochemistry of our reactant and that's because of our SN2 mechanism. We observe inversion of configuration. So let me write that down here. So inversion of configuration. We're going from an R configuration here to an S configuration."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "We observe inversion of configuration. So let me write that down here. So inversion of configuration. We're going from an R configuration here to an S configuration. And that means that the nucleophile can only attack from the side that's opposite of the leaving group and that's consistent with our SN2 mechanism. One way of thinking about this is your bromine is relatively large and it has a large amount of electron density around it with all these lone pairs of electrons. And that would repel your negatively charged nucleophile."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "We're going from an R configuration here to an S configuration. And that means that the nucleophile can only attack from the side that's opposite of the leaving group and that's consistent with our SN2 mechanism. One way of thinking about this is your bromine is relatively large and it has a large amount of electron density around it with all these lone pairs of electrons. And that would repel your negatively charged nucleophile. So your nucleophile has to approach from the side opposite of your leaving group. And that requirement means that you get inversion of configuration. So next we're gonna go to a video where I'm gonna use the model set to show you how R2-bromobutane turns into S2-butanol and hopefully it'll be more obvious."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And that would repel your negatively charged nucleophile. So your nucleophile has to approach from the side opposite of your leaving group. And that requirement means that you get inversion of configuration. So next we're gonna go to a video where I'm gonna use the model set to show you how R2-bromobutane turns into S2-butanol and hopefully it'll be more obvious. Here is R2-bromobutane and I've made bromine yellow and I've left the hydrogens off the alkyl groups just to make it easier to see. So if I turn the model a little bit and we look at our chiral center, so we have tetrahedral geometry around this carbon. There's a methyl group coming out at us in space, a hydrogen going away from us in space, an ethyl group going down, and our bromine is off to the right."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So next we're gonna go to a video where I'm gonna use the model set to show you how R2-bromobutane turns into S2-butanol and hopefully it'll be more obvious. Here is R2-bromobutane and I've made bromine yellow and I've left the hydrogens off the alkyl groups just to make it easier to see. So if I turn the model a little bit and we look at our chiral center, so we have tetrahedral geometry around this carbon. There's a methyl group coming out at us in space, a hydrogen going away from us in space, an ethyl group going down, and our bromine is off to the right. So we get our nucleophile. Our nucleophile is trying to attack this carbon. So the hydroxide ion has to approach our substrate from the side that's opposite of our leaving group."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "There's a methyl group coming out at us in space, a hydrogen going away from us in space, an ethyl group going down, and our bromine is off to the right. So we get our nucleophile. Our nucleophile is trying to attack this carbon. So the hydroxide ion has to approach our substrate from the side that's opposite of our leaving group. So next, let's look at the transition state. Notice that we still have our methyl group coming out at us in space, our hydrogen going away from us in space, our ethyl group going down, and our halogen off to the right. But now there's a partial bond that forms between this oxygen and this carbon."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So the hydroxide ion has to approach our substrate from the side that's opposite of our leaving group. So next, let's look at the transition state. Notice that we still have our methyl group coming out at us in space, our hydrogen going away from us in space, our ethyl group going down, and our halogen off to the right. But now there's a partial bond that forms between this oxygen and this carbon. There's also a partial bond between this halogen and this carbon. And if we look at that central carbon there and we look at the three groups, the methyl, the hydrogen, and the ethyl group, those three groups are all in the same plane. So if I rotate this a little bit, you can see they're all planar there."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "But now there's a partial bond that forms between this oxygen and this carbon. There's also a partial bond between this halogen and this carbon. And if we look at that central carbon there and we look at the three groups, the methyl, the hydrogen, and the ethyl group, those three groups are all in the same plane. So if I rotate this a little bit, you can see they're all planar there. So the transition state has a bond forming at the same time a bond is breaking. And if I keep this model of the transition state up, we can compare it with the final product. So here is the final product, the alcohol."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So if I rotate this a little bit, you can see they're all planar there. So the transition state has a bond forming at the same time a bond is breaking. And if I keep this model of the transition state up, we can compare it with the final product. So here is the final product, the alcohol. We still have the methyl group coming out at us in space with the hydrogen going away from us, our OH on the left, and our ethyl group going down. So this is one way to represent our product, but if I rotate this model set a little bit so we can see our carbon chain. So if I turn it this way, now our OH is going out at us in space."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So here is the final product, the alcohol. We still have the methyl group coming out at us in space with the hydrogen going away from us, our OH on the left, and our ethyl group going down. So this is one way to represent our product, but if I rotate this model set a little bit so we can see our carbon chain. So if I turn it this way, now our OH is going out at us in space. So that's another way to view our final product. I could also turn this model set a little bit. So if I turn it over to view the carbon chain this way, now the OH is going away from us."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So if I turn it this way, now our OH is going out at us in space. So that's another way to view our final product. I could also turn this model set a little bit. So if I turn it over to view the carbon chain this way, now the OH is going away from us. And that's how I drew it in the original reaction. So let's draw out what we saw in the video. So our hydroxide ion, our nucleophile, attacks this carbon."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So if I turn it over to view the carbon chain this way, now the OH is going away from us. And that's how I drew it in the original reaction. So let's draw out what we saw in the video. So our hydroxide ion, our nucleophile, attacks this carbon. At the same time, these electrons come off onto the bromine. Let's draw out the transition state. So we're gonna put some brackets in here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So our hydroxide ion, our nucleophile, attacks this carbon. At the same time, these electrons come off onto the bromine. Let's draw out the transition state. So we're gonna put some brackets in here. And I'm gonna put in my OH. I'll have some lone pairs of electrons here. And we saw that a partial bond forms between this oxygen and this carbon here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So we're gonna put some brackets in here. And I'm gonna put in my OH. I'll have some lone pairs of electrons here. And we saw that a partial bond forms between this oxygen and this carbon here. And in the video, I pointed out how the methyl group is still coming out at us in space. So our methyl group is coming out at us. Our hydrogen is going away from us."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And we saw that a partial bond forms between this oxygen and this carbon here. And in the video, I pointed out how the methyl group is still coming out at us in space. So our methyl group is coming out at us. Our hydrogen is going away from us. And our ethyl group is going down. And those three things, the methyl, the hydrogen, and the ethyl group, are in the same plane. And at the same time, we have a partial bond between our carbon and our leaving group, which is our bromine here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "Our hydrogen is going away from us. And our ethyl group is going down. And those three things, the methyl, the hydrogen, and the ethyl group, are in the same plane. And at the same time, we have a partial bond between our carbon and our leaving group, which is our bromine here. So let me put in these electrons here on the bromine. So this is our transition state. And we designate the transition state with this little symbol up here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And at the same time, we have a partial bond between our carbon and our leaving group, which is our bromine here. So let me put in these electrons here on the bromine. So this is our transition state. And we designate the transition state with this little symbol up here. So let me draw that in. Now the hydroxide ion over here was a full negative charge. But if it's forming a partial bond with this carbon, we would give it a partial negative."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And we designate the transition state with this little symbol up here. So let me draw that in. Now the hydroxide ion over here was a full negative charge. But if it's forming a partial bond with this carbon, we would give it a partial negative. So it's losing some of its electron density as that bond is forming. And as the electrons are coming off onto the bromine to form the bromide anion, we're gonna get a little more negative charge on that bromine. So I'm putting a partial negative here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "But if it's forming a partial bond with this carbon, we would give it a partial negative. So it's losing some of its electron density as that bond is forming. And as the electrons are coming off onto the bromine to form the bromide anion, we're gonna get a little more negative charge on that bromine. So I'm putting a partial negative here. So here's our transition state, where the nucleophile is in the process of forming a bond. At the same time, the bond to the leaving group is breaking. So finally, let's draw our product here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So I'm putting a partial negative here. So here's our transition state, where the nucleophile is in the process of forming a bond. At the same time, the bond to the leaving group is breaking. So finally, let's draw our product here. And just like I did in the video, I'm gonna point out how our methyl group stays coming out at us here. So let me draw that on a wedge. So here's our methyl group."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So finally, let's draw our product here. And just like I did in the video, I'm gonna point out how our methyl group stays coming out at us here. So let me draw that on a wedge. So here's our methyl group. Our hydrogen is still going away from us. And our ethyl group is going down. So let me draw our ethyl group going down."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So here's our methyl group. Our hydrogen is still going away from us. And our ethyl group is going down. So let me draw our ethyl group going down. And now we have a bond to our OH. And we're back to tetrahedral geometry. So this carbon, right, our chiral center, has tetrahedral geometry to start with."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So let me draw our ethyl group going down. And now we have a bond to our OH. And we're back to tetrahedral geometry. So this carbon, right, our chiral center, has tetrahedral geometry to start with. Here, for our three groups, we have planar geometry. But for our final product, we're back to tetrahedral geometry. And this is one way that we drew our product."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So this carbon, right, our chiral center, has tetrahedral geometry to start with. Here, for our three groups, we have planar geometry. But for our final product, we're back to tetrahedral geometry. And this is one way that we drew our product. And in the video, I showed moving the model set around. And one of the viewpoints that we looked at was if your carbon chain looks like this, that OH group is coming out at you in space. And that's the same thing as if you turn the model set around where you would have your carbon chain like this, your OH is going away from you in space."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And this is one way that we drew our product. And in the video, I showed moving the model set around. And one of the viewpoints that we looked at was if your carbon chain looks like this, that OH group is coming out at you in space. And that's the same thing as if you turn the model set around where you would have your carbon chain like this, your OH is going away from you in space. So these are three different ways to represent our product, which is S2-butanol. Let's look at another SN2 reaction with stereochemistry. And let's figure out the product."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And that's the same thing as if you turn the model set around where you would have your carbon chain like this, your OH is going away from you in space. So these are three different ways to represent our product, which is S2-butanol. Let's look at another SN2 reaction with stereochemistry. And let's figure out the product. So first I would say, okay, this ion here is negatively charged, so that must be my nucleophile. And for my substrate, for my alkyl halides, I know that chlorine is withdrawing electron density from this carbon, since chlorine is more electronegative, which means that carbon is partially positive. So my nucleophile is going to attack my electrophile."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And let's figure out the product. So first I would say, okay, this ion here is negatively charged, so that must be my nucleophile. And for my substrate, for my alkyl halides, I know that chlorine is withdrawing electron density from this carbon, since chlorine is more electronegative, which means that carbon is partially positive. So my nucleophile is going to attack my electrophile. At the same time, these electrons are going to come off onto the chlorine to form the chloride anion. So I know that's what happens in my SN2 mechanism. And now that we've looked at stereochemistry, I know that the nucleophile has to attack from a side that's opposite of our leaving group, and we get inversion of configuration."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So my nucleophile is going to attack my electrophile. At the same time, these electrons are going to come off onto the chlorine to form the chloride anion. So I know that's what happens in my SN2 mechanism. And now that we've looked at stereochemistry, I know that the nucleophile has to attack from a side that's opposite of our leaving group, and we get inversion of configuration. So when I draw my product here, if I'm drawing the carbon chain the same way that we started with, since this chlorine is on a dash going away from us in space, we have to show the bond to our nucleophile here coming out as a wedge. And so I'm going to put the SH here. So this would be our final product."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And now that we've looked at stereochemistry, I know that the nucleophile has to attack from a side that's opposite of our leaving group, and we get inversion of configuration. So when I draw my product here, if I'm drawing the carbon chain the same way that we started with, since this chlorine is on a dash going away from us in space, we have to show the bond to our nucleophile here coming out as a wedge. And so I'm going to put the SH here. So this would be our final product. For simple systems, you'll see inversion of configuration. If your starting compound is R, you will get S for your product. And if you start with S, you will get R. However, it doesn't always have to be that way."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So this would be our final product. For simple systems, you'll see inversion of configuration. If your starting compound is R, you will get S for your product. And if you start with S, you will get R. However, it doesn't always have to be that way. If we look at this reaction, on the left is our starting compound. And this is an SN2 reaction. And our nucleophile would be the methoxide anion."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And if you start with S, you will get R. However, it doesn't always have to be that way. If we look at this reaction, on the left is our starting compound. And this is an SN2 reaction. And our nucleophile would be the methoxide anion. So let me go ahead and draw that in. So we have an oxygen, we have a CH3, and we have a negative one formal charge on the oxygen. So our nucleophile is going to attack our partially positive carbon, which is right here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "And our nucleophile would be the methoxide anion. So let me go ahead and draw that in. So we have an oxygen, we have a CH3, and we have a negative one formal charge on the oxygen. So our nucleophile is going to attack our partially positive carbon, which is right here. So our nucleophile attacks our electrophile at the same time that we get loss of a leaving group. And the bromide ion would be the best leaving group here. So these electrons come off onto the bromine to form the bromide ion."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So our nucleophile is going to attack our partially positive carbon, which is right here. So our nucleophile attacks our electrophile at the same time that we get loss of a leaving group. And the bromide ion would be the best leaving group here. So these electrons come off onto the bromine to form the bromide ion. And we form a bond between the oxygen and the carbon. So this lone pair of electrons on the oxygen forms this bond, and we get our product. If we look at the stereochemistry, let's first start on the left here."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So these electrons come off onto the bromine to form the bromide ion. And we form a bond between the oxygen and the carbon. So this lone pair of electrons on the oxygen forms this bond, and we get our product. If we look at the stereochemistry, let's first start on the left here. So we know that this carbon is a chiral center. And if I'm assigning priority to the four groups attached to that chiral center, bromine has the highest atomic number, so that would be priority number one, followed by fluorine, which would be number two, followed by carbon here, which is number three, and then hydrogen is the lowest priority group, number four, which is going away from us. So we're going around in this direction, which is counterclockwise."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "If we look at the stereochemistry, let's first start on the left here. So we know that this carbon is a chiral center. And if I'm assigning priority to the four groups attached to that chiral center, bromine has the highest atomic number, so that would be priority number one, followed by fluorine, which would be number two, followed by carbon here, which is number three, and then hydrogen is the lowest priority group, number four, which is going away from us. So we're going around in this direction, which is counterclockwise. And so that's an S configuration to our starting compound. For our product, again, this carbon here, let me change colors, this carbon is our chiral center. And when we assign priority this time, now the fluorine has the highest atomic number, followed by this oxygen here, followed by this carbon, and then, of course, with the hydrogen."}, {"video_title": "Sn2 mechanism stereospecificity.mp3", "Sentence": "So we're going around in this direction, which is counterclockwise. And so that's an S configuration to our starting compound. For our product, again, this carbon here, let me change colors, this carbon is our chiral center. And when we assign priority this time, now the fluorine has the highest atomic number, followed by this oxygen here, followed by this carbon, and then, of course, with the hydrogen. So when we go around in this direction, again, the configuration is still S. So here's an example of an SN2 mechanism where our nucleophile has to attack from the opposite side of our leaving group. So we still get inversion in terms of the mechanism, but we don't get inversion of configuration because of how we assign priority to our groups. In this case, when our bromine leaves, fluorine becomes the highest priority group, and that's the reason why the configuration stays S."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And that one mistake I did is when I named the ester, all I did is I named kind of the main carbon backbone. I didn't actually tell you how many carbons you have attached to the oxygen over here. Acetate really just refers to this part over here, knowing that it is bonded to, or maybe I should say, it refers to this part over here. But you also have to specify how many carbons you have over here. So this molecule that we drew over here is actually methyl acetate. This is methyl acetate. Or if you want its systematic name, it's methyl ethanoate."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "But you also have to specify how many carbons you have over here. So this molecule that we drew over here is actually methyl acetate. This is methyl acetate. Or if you want its systematic name, it's methyl ethanoate. And this is where we get the methyl from. It is methyl. It is methyl ethanoate."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Or if you want its systematic name, it's methyl ethanoate. And this is where we get the methyl from. It is methyl. It is methyl ethanoate. Now, with that out of the way, let's actually compare these carboxylic acid derivatives. And I'll compare the derivatives of acetic acid. So the first one we saw was the amide acetamide."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "It is methyl ethanoate. Now, with that out of the way, let's actually compare these carboxylic acid derivatives. And I'll compare the derivatives of acetic acid. So the first one we saw was the amide acetamide. Acetamide looks just like this. This is acetamide. Then the next one we looked at was the ether."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So the first one we saw was the amide acetamide. Acetamide looks just like this. This is acetamide. Then the next one we looked at was the ether. And that's the one I just pointed out the mistake on. So this was ethyl acetate. So you have CH3 right here."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Then the next one we looked at was the ether. And that's the one I just pointed out the mistake on. So this was ethyl acetate. So you have CH3 right here. This is where, sorry, this is methyl acetate. This is where the methyl part comes from. So this is methyl acetate."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So you have CH3 right here. This is where, sorry, this is methyl acetate. This is where the methyl part comes from. So this is methyl acetate. And then we will compare that to the anhydride version. So this is acetic anhydride. Let me do this in a new color."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So this is methyl acetate. And then we will compare that to the anhydride version. So this is acetic anhydride. Let me do this in a new color. Acetic anhydride looks like this. Acetic anhydride looks just like that. Acetic anhydride."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Let me do this in a new color. Acetic anhydride looks like this. Acetic anhydride looks just like that. Acetic anhydride. And finally, we had the acetyl chloride. Acetyl chloride looked just like this. And of course, these are all derivatives of acetic acid, which we drew at the top of the last video, which is right over there."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Acetic anhydride. And finally, we had the acetyl chloride. Acetyl chloride looked just like this. And of course, these are all derivatives of acetic acid, which we drew at the top of the last video, which is right over there. Now let's think about which of these may or may not be more stable. And to think about it, we're going to think about any resident structures that these molecules might have. So if we first focus on acetamide, we know that nitrogen forms three bonds and then has two extra electrons that form another lone pair."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And of course, these are all derivatives of acetic acid, which we drew at the top of the last video, which is right over there. Now let's think about which of these may or may not be more stable. And to think about it, we're going to think about any resident structures that these molecules might have. So if we first focus on acetamide, we know that nitrogen forms three bonds and then has two extra electrons that form another lone pair. And it's actually electron rich because it's a good bit more electronegative than the hydrogens here, and actually even a little bit more, reasonably more electronegative than the carbon it's attached to. So you can imagine a situation where a nitrogen could donate an electron to this carbon right over here. And then that carbon can let go of one of the electrons in a double bond with the oxygen."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So if we first focus on acetamide, we know that nitrogen forms three bonds and then has two extra electrons that form another lone pair. And it's actually electron rich because it's a good bit more electronegative than the hydrogens here, and actually even a little bit more, reasonably more electronegative than the carbon it's attached to. So you can imagine a situation where a nitrogen could donate an electron to this carbon right over here. And then that carbon can let go of one of the electrons in a double bond with the oxygen. And so it could go back to the oxygen. And you would have a resident structure that would look like this. Let me actually use the same colors."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And then that carbon can let go of one of the electrons in a double bond with the oxygen. And so it could go back to the oxygen. And you would have a resident structure that would look like this. Let me actually use the same colors. You would have a resident structure that would look like this. This oxygen just gained an electron up there. And actually, let me draw the electron."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Let me actually use the same colors. You would have a resident structure that would look like this. This oxygen just gained an electron up there. And actually, let me draw the electron. Well, actually, I'll just put the negative charge due to that electron. And then we had this bond right over here to the nitrogen. But the nitrogen just gave an electron to this carbon right here, the carbonyl, or what was the carbonyl carbon."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And actually, let me draw the electron. Well, actually, I'll just put the negative charge due to that electron. And then we had this bond right over here to the nitrogen. But the nitrogen just gave an electron to this carbon right here, the carbonyl, or what was the carbonyl carbon. And so now we have a double bond over here. And the nitrogen just gave an electron. So it now has a positive charge."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "But the nitrogen just gave an electron to this carbon right here, the carbonyl, or what was the carbonyl carbon. And so now we have a double bond over here. And the nitrogen just gave an electron. So it now has a positive charge. But this is a resident structure of acetamide. So pretty stable. So this is, maybe I should even say quite stable."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So it now has a positive charge. But this is a resident structure of acetamide. So pretty stable. So this is, maybe I should even say quite stable. So this is quite stable. Now, let's think about an ether here. And we could probably do something fairly similar."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So this is, maybe I should even say quite stable. So this is quite stable. Now, let's think about an ether here. And we could probably do something fairly similar. So if we think about this ether here, oxygen has two extra lone pairs. It is more electronegative than the things that it is bonded to. So it could do a very similar thing to what this nitrogen did."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And we could probably do something fairly similar. So if we think about this ether here, oxygen has two extra lone pairs. It is more electronegative than the things that it is bonded to. So it could do a very similar thing to what this nitrogen did. It could give an electron to this carbon right over here. And then that carbon can give back an electron to the carbonyl oxygen, to that oxygen over there. And so its resident structure would look like this."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So it could do a very similar thing to what this nitrogen did. It could give an electron to this carbon right over here. And then that carbon can give back an electron to the carbonyl oxygen, to that oxygen over there. And so its resident structure would look like this. It would have a resident structure. So it's almost the exact same thing as what we saw with the amide. So it would have a resident structure."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And so its resident structure would look like this. It would have a resident structure. So it's almost the exact same thing as what we saw with the amide. So it would have a resident structure. This oxygen just gained an electron. Now we'll have a negative charge. This bond to the oxygen is still there."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So it would have a resident structure. This oxygen just gained an electron. Now we'll have a negative charge. This bond to the oxygen is still there. But this oxygen just gave another electron, formed another bond with what was the carbonyl carbon. It actually forms a new carbonyl group, if you want to view it that way. And so it will look like this."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "This bond to the oxygen is still there. But this oxygen just gave another electron, formed another bond with what was the carbonyl carbon. It actually forms a new carbonyl group, if you want to view it that way. And so it will look like this. And this oxygen gave away an electron. So now it has a positive charge. So this seems like a pretty good resident structure."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And so it will look like this. And this oxygen gave away an electron. So now it has a positive charge. So this seems like a pretty good resident structure. So the question is, which of these two are going to be more stable? And the answer there really just comes out of the periodic table. If you look at nitrogen and oxygen, they're right there."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So this seems like a pretty good resident structure. So the question is, which of these two are going to be more stable? And the answer there really just comes out of the periodic table. If you look at nitrogen and oxygen, they're right there. They're both near the top right. They're both pretty electronegative. But oxygen is more electronegative."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "If you look at nitrogen and oxygen, they're right there. They're both near the top right. They're both pretty electronegative. But oxygen is more electronegative. It is to the right of nitrogen. So because oxygen is more electronegative, remember, electronegativity is just the tendency to hog electrons. How much do you like to hog electrons?"}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "But oxygen is more electronegative. It is to the right of nitrogen. So because oxygen is more electronegative, remember, electronegativity is just the tendency to hog electrons. How much do you like to hog electrons? Oxygen likes to hog electrons more than nitrogen likes to hog electrons. So it would be less likely, marginally less likely, to give away an electron than nitrogen would be. So this resident structure is a little bit less likely, or when you think of it probabilistically, it's going to happen a little bit less frequently, than this resident structure."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "How much do you like to hog electrons? Oxygen likes to hog electrons more than nitrogen likes to hog electrons. So it would be less likely, marginally less likely, to give away an electron than nitrogen would be. So this resident structure is a little bit less likely, or when you think of it probabilistically, it's going to happen a little bit less frequently, than this resident structure. So this one right here, this is going to stabilize it less. So this is going to be a little bit less stable than the amide. So if we call this quite stable, I'll just say this is just stable over here."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So this resident structure is a little bit less likely, or when you think of it probabilistically, it's going to happen a little bit less frequently, than this resident structure. So this one right here, this is going to stabilize it less. So this is going to be a little bit less stable than the amide. So if we call this quite stable, I'll just say this is just stable over here. Now let's think about what's going to happen with the anhydride. The anhydride, once again, we have an oxygen right here. We have an oxygen right here."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So if we call this quite stable, I'll just say this is just stable over here. Now let's think about what's going to happen with the anhydride. The anhydride, once again, we have an oxygen right here. We have an oxygen right here. It's got its spare electrons. It's more electronegative than the things that it's bonded to. So it seems like you could do something very similar to what we saw with the actual ether."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "We have an oxygen right here. It's got its spare electrons. It's more electronegative than the things that it's bonded to. So it seems like you could do something very similar to what we saw with the actual ether. It could donate an electron to this bond right over here, in which case you would have a resident structure that looks like this. And if it donates an electron to this carbon over here, then an electron can be taken away from the carbon and given back to that oxygen. So that would give us a resident structure that would look like this."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So it seems like you could do something very similar to what we saw with the actual ether. It could donate an electron to this bond right over here, in which case you would have a resident structure that looks like this. And if it donates an electron to this carbon over here, then an electron can be taken away from the carbon and given back to that oxygen. So that would give us a resident structure that would look like this. It would give us a resident structure. That guy now has a negative charge. We now have a double bond to this oxygen."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So that would give us a resident structure that would look like this. It would give us a resident structure. That guy now has a negative charge. We now have a double bond to this oxygen. It still has another bond to the other acyl group, just like that. And now this oxygen gave away an electron, so it has a positive charge. So it seems like a very similar situation to what we saw with an ether."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "We now have a double bond to this oxygen. It still has another bond to the other acyl group, just like that. And now this oxygen gave away an electron, so it has a positive charge. So it seems like a very similar situation to what we saw with an ether. But there's another resident structure here. You could also have a situation where, instead of the electron being donated to this carbon bond and that happening, you could have a situation like this, where the electron gets donated to that carbon bond, and then this electron gets taken back by that oxygen. So you would have this situation, you would have a situation where this acyl group still looks the same, bonded to this oxygen."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So it seems like a very similar situation to what we saw with an ether. But there's another resident structure here. You could also have a situation where, instead of the electron being donated to this carbon bond and that happening, you could have a situation like this, where the electron gets donated to that carbon bond, and then this electron gets taken back by that oxygen. So you would have this situation, you would have a situation where this acyl group still looks the same, bonded to this oxygen. But now we have a double bond over here. This guy took away an electron, so one of the bonds of the double bonds goes away, now has a negative charge, and then you have the rest of the molecule. So you actually have these two resident structures."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So you would have this situation, you would have a situation where this acyl group still looks the same, bonded to this oxygen. But now we have a double bond over here. This guy took away an electron, so one of the bonds of the double bonds goes away, now has a negative charge, and then you have the rest of the molecule. So you actually have these two resident structures. And of course, this is now a positive charge since it gave away an electron. So you might say, hey, more resident structures, more stability. But the key here is to realize is that both resident structures are dependent from the same oxygen."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "So you actually have these two resident structures. And of course, this is now a positive charge since it gave away an electron. So you might say, hey, more resident structures, more stability. But the key here is to realize is that both resident structures are dependent from the same oxygen. They essentially have to share this oxygen's electrons. So each one of these individually is less likely to occur than just what occurs with the ether. You could kind of view it as that both of these bonds have to share what in the ether, this one gets it on its own."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "But the key here is to realize is that both resident structures are dependent from the same oxygen. They essentially have to share this oxygen's electrons. So each one of these individually is less likely to occur than just what occurs with the ether. You could kind of view it as that both of these bonds have to share what in the ether, this one gets it on its own. And we even saw that this is still less likely to occur than this over here. So in this situation, an anhydride, if one guy's getting the double bond, if there's a resident structure on one side of the anhydride, the other side is still reactive. If there's a resident structure on the left side, then the right side is still reactive."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "You could kind of view it as that both of these bonds have to share what in the ether, this one gets it on its own. And we even saw that this is still less likely to occur than this over here. So in this situation, an anhydride, if one guy's getting the double bond, if there's a resident structure on one side of the anhydride, the other side is still reactive. If there's a resident structure on the left side, then the right side is still reactive. So we'll call this less stable. This right here is less stable. And then finally, you have your acetyl chloride."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "If there's a resident structure on the left side, then the right side is still reactive. So we'll call this less stable. This right here is less stable. And then finally, you have your acetyl chloride. In acetyl chloride, there really is no resident structure here. Chlorine is so electronegative that it is unlikely to give away an electron. It is sitting pretty."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "And then finally, you have your acetyl chloride. In acetyl chloride, there really is no resident structure here. Chlorine is so electronegative that it is unlikely to give away an electron. It is sitting pretty. It has no, well, I don't want to, but it is sitting pretty with eight electrons. That's actually true for these other characters right here. But chlorine is very, very unlikely to contribute a resident structure here."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "It is sitting pretty. It has no, well, I don't want to, but it is sitting pretty with eight electrons. That's actually true for these other characters right here. But chlorine is very, very unlikely to contribute a resident structure here. So this gets no stability. So this is the least stable of the carboxylic acid derivatives that we're looking at. Now, why am I even bothering with all of this hierarchy of stability for carboxylic acid derivatives?"}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "But chlorine is very, very unlikely to contribute a resident structure here. So this gets no stability. So this is the least stable of the carboxylic acid derivatives that we're looking at. Now, why am I even bothering with all of this hierarchy of stability for carboxylic acid derivatives? And I'm bothering with it is because we know, or we don't know quite yet, but I've already shown you some mechanisms that can take us from a carboxylic acid to an ether or carboxylic acid to an acyl halide. And in general, you can go between all of these and a carboxylic acid. But since we know their relative stabilities, we know that if you start with an acetyl chloride, you're much more likely to go to an acetamide if you have the proper ingredients there, if you have some amines floating around, than the other way around."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Now, why am I even bothering with all of this hierarchy of stability for carboxylic acid derivatives? And I'm bothering with it is because we know, or we don't know quite yet, but I've already shown you some mechanisms that can take us from a carboxylic acid to an ether or carboxylic acid to an acyl halide. And in general, you can go between all of these and a carboxylic acid. But since we know their relative stabilities, we know that if you start with an acetyl chloride, you're much more likely to go to an acetamide if you have the proper ingredients there, if you have some amines floating around, than the other way around. You're much more likely to go from that to that, because this is much less stable than that, than you are to go from acetamide to acetyl chloride. Likewise, you're much more likely to go from a carboxylic acid to an acetyl chloride. Or let me put it this way."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "But since we know their relative stabilities, we know that if you start with an acetyl chloride, you're much more likely to go to an acetamide if you have the proper ingredients there, if you have some amines floating around, than the other way around. You're much more likely to go from that to that, because this is much less stable than that, than you are to go from acetamide to acetyl chloride. Likewise, you're much more likely to go from a carboxylic acid to an acetyl chloride. Or let me put it this way. You're much more likely to go from a carboxylic acid to an acetamide. And I haven't drawn the carboxylic acid here, but this has better resonance than even its original carboxylic acid than the other way around. Or you're much more likely to go from an anhydride to an ether."}, {"video_title": "Relative stability of amides, esters, anhydrides, and acyl chlorides Khan Academy.mp3", "Sentence": "Or let me put it this way. You're much more likely to go from a carboxylic acid to an acetamide. And I haven't drawn the carboxylic acid here, but this has better resonance than even its original carboxylic acid than the other way around. Or you're much more likely to go from an anhydride to an ether. So that's why I'm doing this hierarchy. You're much more likely to go from something on the right to something on the left. And we'll explore some of those mechanisms in the next few videos."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "On the right is another drawing of the ethane molecule, but this drawing gives us more information. This is a wedge and dash drawing. Remember, a wedge has a bond coming out at you in space, or a bond coming out of the plane of the page. A dash is a bond going away from you in space, or a bond going into the plane of the page. And finally, if you just draw a straight line like this, that means a bond in the plane of the page. This wedge and dash drawing represents one conformation of ethane. Conformations are different arrangements of atoms that result from bond rotation."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "A dash is a bond going away from you in space, or a bond going into the plane of the page. And finally, if you just draw a straight line like this, that means a bond in the plane of the page. This wedge and dash drawing represents one conformation of ethane. Conformations are different arrangements of atoms that result from bond rotation. We know that there's free rotation around this carbon-carbon single bond. And this wedge and dash drawing right now represents what's called the staggered conformation of ethane. But if we rotate about this carbon-carbon bond, we're gonna get different arrangements of the atoms, and therefore we will get different conformations."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Conformations are different arrangements of atoms that result from bond rotation. We know that there's free rotation around this carbon-carbon single bond. And this wedge and dash drawing right now represents what's called the staggered conformation of ethane. But if we rotate about this carbon-carbon bond, we're gonna get different arrangements of the atoms, and therefore we will get different conformations. This is easiest to see with a model set. So up next I have a video where I have a model set of ethane and I rotate around the carbon-carbon bond so we can see different conformations. I'm gonna make these hydrogens green in the video."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "But if we rotate about this carbon-carbon bond, we're gonna get different arrangements of the atoms, and therefore we will get different conformations. This is easiest to see with a model set. So up next I have a video where I have a model set of ethane and I rotate around the carbon-carbon bond so we can see different conformations. I'm gonna make these hydrogens green in the video. So those green hydrogens are attached to this carbon, which will be the front carbon. And then we have hydrogens attached to the back carbon that I will make white. So let's watch the video and look at the different conformations of ethane."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "I'm gonna make these hydrogens green in the video. So those green hydrogens are attached to this carbon, which will be the front carbon. And then we have hydrogens attached to the back carbon that I will make white. So let's watch the video and look at the different conformations of ethane. Here we have the staggered conformation of ethane, looking at it from a wedge and dash perspective. If we rotate this a little bit, we'll see the staggered conformation of ethane from a sawhorse perspective. If we sight down the carbon-carbon bond, we'll see the third way of looking at this molecule, and this is called a Newman projection."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So let's watch the video and look at the different conformations of ethane. Here we have the staggered conformation of ethane, looking at it from a wedge and dash perspective. If we rotate this a little bit, we'll see the staggered conformation of ethane from a sawhorse perspective. If we sight down the carbon-carbon bond, we'll see the third way of looking at this molecule, and this is called a Newman projection. If I take one of these hydrogens here and I rotate it, so I'm keeping the back carbon stationary and I'm rotating the front carbon, every time I rotate this, that represents a different conformation of the ethane molecule. So there are many, many possible conformations in theory. I rotate back to the original position here, so this is the Newman projection where we just were."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "If we sight down the carbon-carbon bond, we'll see the third way of looking at this molecule, and this is called a Newman projection. If I take one of these hydrogens here and I rotate it, so I'm keeping the back carbon stationary and I'm rotating the front carbon, every time I rotate this, that represents a different conformation of the ethane molecule. So there are many, many possible conformations in theory. I rotate back to the original position here, so this is the Newman projection where we just were. The other conformation that we really care about for ethane is what's called the eclipsed conformation. So I just rotated the green hydrogen, so they're in front or eclipsing the white hydrogens in the back. So this is the eclipsed conformation as a Newman projection."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "I rotate back to the original position here, so this is the Newman projection where we just were. The other conformation that we really care about for ethane is what's called the eclipsed conformation. So I just rotated the green hydrogen, so they're in front or eclipsing the white hydrogens in the back. So this is the eclipsed conformation as a Newman projection. I can turn this a little bit so we can see what an eclipsed conformation looks like as a sawhorse, from a sawhorse view. And then finally, I can turn it a little bit so we can see the eclipsed conformation from a wedge and dash perspective. Now that we've seen the video, let's look at three ways to represent the staggered conformation of ethane."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So this is the eclipsed conformation as a Newman projection. I can turn this a little bit so we can see what an eclipsed conformation looks like as a sawhorse, from a sawhorse view. And then finally, I can turn it a little bit so we can see the eclipsed conformation from a wedge and dash perspective. Now that we've seen the video, let's look at three ways to represent the staggered conformation of ethane. And we'll start with the wedge and dash drawing. So you can see that this hydrogen, this carbon, this carbon, and this hydrogen are all in the same plane, so they're all in the same plane here, which is why all these bonds are drawn as straight lines in our wedge and dash drawing. If we look at this carbon on the left, we know that there is a hydrogen in green coming out at us in space, a hydrogen in green going away from us in space, and then this hydrogen, which is in the plane of the page."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Now that we've seen the video, let's look at three ways to represent the staggered conformation of ethane. And we'll start with the wedge and dash drawing. So you can see that this hydrogen, this carbon, this carbon, and this hydrogen are all in the same plane, so they're all in the same plane here, which is why all these bonds are drawn as straight lines in our wedge and dash drawing. If we look at this carbon on the left, we know that there is a hydrogen in green coming out at us in space, a hydrogen in green going away from us in space, and then this hydrogen, which is in the plane of the page. For the hydrogens in white, this hydrogen is in the plane of the page, this one's coming out at us in space, and then we have one going away from us in space. And it's pretty easy to see that in the picture here. So we're just gonna match the drawings to the pictures."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "If we look at this carbon on the left, we know that there is a hydrogen in green coming out at us in space, a hydrogen in green going away from us in space, and then this hydrogen, which is in the plane of the page. For the hydrogens in white, this hydrogen is in the plane of the page, this one's coming out at us in space, and then we have one going away from us in space. And it's pretty easy to see that in the picture here. So we're just gonna match the drawings to the pictures. Our next way to represent the staggered conformation is what's called a sawhorse drawing. So for this sawhorse drawing, this carbon right here is this carbon, and it has three hydrogens in green attached to it. So this is a hydrogen in green, in green, and in green."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So we're just gonna match the drawings to the pictures. Our next way to represent the staggered conformation is what's called a sawhorse drawing. So for this sawhorse drawing, this carbon right here is this carbon, and it has three hydrogens in green attached to it. So this is a hydrogen in green, in green, and in green. So we also have this carbon-carbon bond, so that's this bond here on our sawhorse drawing. And then finally for this carbon in the back, this is the one that has hydrogens in white. So these are the three hydrogens in white."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So this is a hydrogen in green, in green, and in green. So we also have this carbon-carbon bond, so that's this bond here on our sawhorse drawing. And then finally for this carbon in the back, this is the one that has hydrogens in white. So these are the three hydrogens in white. If we look down that carbon-carbon bond, if we look down this axis right here, so if you put your eye, imagine putting your eye right here, so here's the eye, and you look down this axis, you will see the Newman projection, just like we saw in the video. And so this dot, this point here, represents the front carbon, that's this carbon right here. So let me go ahead and write that down."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So these are the three hydrogens in white. If we look down that carbon-carbon bond, if we look down this axis right here, so if you put your eye, imagine putting your eye right here, so here's the eye, and you look down this axis, you will see the Newman projection, just like we saw in the video. And so this dot, this point here, represents the front carbon, that's this carbon right here. So let me go ahead and write that down. So this point represents the front carbon, and the front carbon is bonded to our hydrogens in green. So this would be a hydrogen in green, so is this one, and so is this one. The front carbon blocks the carbon in the back."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write that down. So this point represents the front carbon, and the front carbon is bonded to our hydrogens in green. So this would be a hydrogen in green, so is this one, and so is this one. The front carbon blocks the carbon in the back. You can't see the carbon in the back in this drawing, but we know it's there, and we have to represent it somehow. So in a Newman projection, it's this circle, it's this circle back here that represents the back carbon. So this is the back carbon, the circle, let me write that down, the circle is the back carbon, even though we can't see it when you actually have the model in front of you."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "The front carbon blocks the carbon in the back. You can't see the carbon in the back in this drawing, but we know it's there, and we have to represent it somehow. So in a Newman projection, it's this circle, it's this circle back here that represents the back carbon. So this is the back carbon, the circle, let me write that down, the circle is the back carbon, even though we can't see it when you actually have the model in front of you. And the back carbon is the one that has the hydrogens in white. So this should be a hydrogen in white, so is this one, and so is this one. So here you can see the hydrogens in white back here, and you know they're attached to a carbon, but you can't see it, because the front carbon is blocking it."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So this is the back carbon, the circle, let me write that down, the circle is the back carbon, even though we can't see it when you actually have the model in front of you. And the back carbon is the one that has the hydrogens in white. So this should be a hydrogen in white, so is this one, and so is this one. So here you can see the hydrogens in white back here, and you know they're attached to a carbon, but you can't see it, because the front carbon is blocking it. So let's talk a little bit more about Newman projections. We can talk about the angle between this hydrogen in white and this hydrogen in green. So think about the angle between these."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So here you can see the hydrogens in white back here, and you know they're attached to a carbon, but you can't see it, because the front carbon is blocking it. So let's talk a little bit more about Newman projections. We can talk about the angle between this hydrogen in white and this hydrogen in green. So think about the angle between these. It's 60 degrees, so we have a 60 degree angle between these two hydrogens. So I could write it down here too, so 60 degrees. This angle is called the dihedral angle, or the torsional angle, so I'll write those down."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So think about the angle between these. It's 60 degrees, so we have a 60 degree angle between these two hydrogens. So I could write it down here too, so 60 degrees. This angle is called the dihedral angle, or the torsional angle, so I'll write those down. So you could call this the dihedral angle, or you could also call it the torsional angle. Let me write that down here too, so torsional angle, or you could also call it the torsion angle. And this angle between the hydrogens will be important when you're talking about conformations."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "This angle is called the dihedral angle, or the torsional angle, so I'll write those down. So you could call this the dihedral angle, or you could also call it the torsional angle. Let me write that down here too, so torsional angle, or you could also call it the torsion angle. And this angle between the hydrogens will be important when you're talking about conformations. Here the angle is 60 degrees, which means that these hydrogens are not right on top of each other. There's space between these hydrogens on the Newman projection. The green hydrogens are staggered compared to the white hydrogens."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And this angle between the hydrogens will be important when you're talking about conformations. Here the angle is 60 degrees, which means that these hydrogens are not right on top of each other. There's space between these hydrogens on the Newman projection. The green hydrogens are staggered compared to the white hydrogens. So if you go around, you can see that there's space between all of these, and that's why we call this the staggered conformation of ethane. And finally, let's look at the other important conformation of ethane, which is the eclipsed conformation. And we'll start with the wedge and dash drawing."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "The green hydrogens are staggered compared to the white hydrogens. So if you go around, you can see that there's space between all of these, and that's why we call this the staggered conformation of ethane. And finally, let's look at the other important conformation of ethane, which is the eclipsed conformation. And we'll start with the wedge and dash drawing. So we have this hydrogen, this carbon, this carbon, and this hydrogen. These are all in the same plane, so I draw this line here, and we can see, we can see these bonds are all in the same plane up here on my wedge and dash drawing. For this carbon on the left, we can see that we have a hydrogen in green coming out at us, right?"}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with the wedge and dash drawing. So we have this hydrogen, this carbon, this carbon, and this hydrogen. These are all in the same plane, so I draw this line here, and we can see, we can see these bonds are all in the same plane up here on my wedge and dash drawing. For this carbon on the left, we can see that we have a hydrogen in green coming out at us, right? So here's your hydrogen in green coming out at you, a hydrogen in green going away from you back here, and then this hydrogen in the plane of the page. For the hydrogens in white, this hydrogen is in the plane of the page, right? This hydrogen is coming out at us in space, and this hydrogen is going away from us in space."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "For this carbon on the left, we can see that we have a hydrogen in green coming out at us, right? So here's your hydrogen in green coming out at you, a hydrogen in green going away from you back here, and then this hydrogen in the plane of the page. For the hydrogens in white, this hydrogen is in the plane of the page, right? This hydrogen is coming out at us in space, and this hydrogen is going away from us in space. For the sawhorse drawings, let's move on to here. This carbon is the front carbon, so that's this carbon right here, the one that's bonded to the three hydrogens in green, so let me highlight those. So we have one hydrogen, this one right here is going up, so that's our hydrogen in green going up."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "This hydrogen is coming out at us in space, and this hydrogen is going away from us in space. For the sawhorse drawings, let's move on to here. This carbon is the front carbon, so that's this carbon right here, the one that's bonded to the three hydrogens in green, so let me highlight those. So we have one hydrogen, this one right here is going up, so that's our hydrogen in green going up. This hydrogen over here is going down a little bit to the right, and this hydrogen's going down and to the left. And then we have the carbon-carbon bond, so let me draw that in here. So here's our carbon-carbon bond."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So we have one hydrogen, this one right here is going up, so that's our hydrogen in green going up. This hydrogen over here is going down a little bit to the right, and this hydrogen's going down and to the left. And then we have the carbon-carbon bond, so let me draw that in here. So here's our carbon-carbon bond. I made it much longer up here, just so the atoms wouldn't interfere with each other on our sawhorse drawing. And then we get to the carbon in the back. So here's the carbon in the back, which is bonded to three hydrogens, and I made these the hydrogens in white."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So here's our carbon-carbon bond. I made it much longer up here, just so the atoms wouldn't interfere with each other on our sawhorse drawing. And then we get to the carbon in the back. So here's the carbon in the back, which is bonded to three hydrogens, and I made these the hydrogens in white. So here's one, two, and three. So here are the hydrogens in white on the picture. Finally, we get to the Newman projection for the eclipsed conformation."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So here's the carbon in the back, which is bonded to three hydrogens, and I made these the hydrogens in white. So here's one, two, and three. So here are the hydrogens in white on the picture. Finally, we get to the Newman projection for the eclipsed conformation. And remember, the Newman projection is what we would see if we stare down the carbon-carbon bond. So if we sight down this carbon-carbon bond, if we put our eye right here on our sawhorse projection, and we stare down the carbon-carbon bond, we see this for our Newman projection. First we see this point here representing the front carbon, so that point is the front carbon on our Newman projection."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Finally, we get to the Newman projection for the eclipsed conformation. And remember, the Newman projection is what we would see if we stare down the carbon-carbon bond. So if we sight down this carbon-carbon bond, if we put our eye right here on our sawhorse projection, and we stare down the carbon-carbon bond, we see this for our Newman projection. First we see this point here representing the front carbon, so that point is the front carbon on our Newman projection. And the front carbon is bonded to the hydrogens in green. So here's one, two, and three. So up here, one, two, and three."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "First we see this point here representing the front carbon, so that point is the front carbon on our Newman projection. And the front carbon is bonded to the hydrogens in green. So here's one, two, and three. So up here, one, two, and three. So this would be the bonds. This represents the bonds of that front carbon to those hydrogens. The back carbon is pretty difficult to see, but we know that the circle represents the back carbon."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So up here, one, two, and three. So this would be the bonds. This represents the bonds of that front carbon to those hydrogens. The back carbon is pretty difficult to see, but we know that the circle represents the back carbon. The front carbon is in the way. I'm not sure if you might be able to see a tiny bit of the back carbon here. But the back carbon should be blocked by the front carbon."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "The back carbon is pretty difficult to see, but we know that the circle represents the back carbon. The front carbon is in the way. I'm not sure if you might be able to see a tiny bit of the back carbon here. But the back carbon should be blocked by the front carbon. The back carbon, we know, has our hydrogens in white. So here are the three hydrogens in white. So back here, you can just barely see them."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "But the back carbon should be blocked by the front carbon. The back carbon, we know, has our hydrogens in white. So here are the three hydrogens in white. So back here, you can just barely see them. You can just barely see them poking out. This time, when you think about the angle between your hydrogens, this time your dihedral or your torsional angle is zero degrees. Think about this hydrogen being perfectly upright and the hydrogen in the back being perfectly upright."}, {"video_title": "Conformations of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So back here, you can just barely see them. You can just barely see them poking out. This time, when you think about the angle between your hydrogens, this time your dihedral or your torsional angle is zero degrees. Think about this hydrogen being perfectly upright and the hydrogen in the back being perfectly upright. So the angle between those is zero. So your dihedral angle is zero degrees here for the eclipsed conformation. The green hydrogens are eclipsing the white hydrogen."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So first we're gonna draw out the mechanism, figure out the products, and then we'll talk about why it's a stereoselective reaction. On the left we have our alcohol, the carbon that's bonded to the OH is our alpha carbon, and the carbon next to that would be our beta carbon. So this beta carbon has two beta protons, because we know in our mechanism, in our E1 mechanism, we're gonna lose a beta proton. On the left here, this carbon's next to the alpha carbon, but there are no hydrogens on that carbon, so we only have to worry about the carbon on the right. Sulfuric acid is a strong acid, and it will protonate our alcohol. So I'll just draw an H plus in here to save some time. So a lone pair of electrons on the oxygen picks up a proton from sulfuric acid, and let's draw what we would have now."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "On the left here, this carbon's next to the alpha carbon, but there are no hydrogens on that carbon, so we only have to worry about the carbon on the right. Sulfuric acid is a strong acid, and it will protonate our alcohol. So I'll just draw an H plus in here to save some time. So a lone pair of electrons on the oxygen picks up a proton from sulfuric acid, and let's draw what we would have now. So I'll put in my benzene ring right in here. So let's put in those electrons, and I'll put in my carbon chain, and if we're protonating our OH, let me draw this wedge in here, now oxygen would have two bonds to hydrogen, one lone pair of electrons on the oxygen, and the oxygen would have a plus one formal charge. So let me show those electrons here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So a lone pair of electrons on the oxygen picks up a proton from sulfuric acid, and let's draw what we would have now. So I'll put in my benzene ring right in here. So let's put in those electrons, and I'll put in my carbon chain, and if we're protonating our OH, let me draw this wedge in here, now oxygen would have two bonds to hydrogen, one lone pair of electrons on the oxygen, and the oxygen would have a plus one formal charge. So let me show those electrons here. So these electrons in magenta pick up a proton from sulfuric acid, and we could say that it formed this bond right in here. And then we also know there's a hydrogen on this carbon going away from us in space, so I'll draw in that hydrogen. And then for our beta carbon here, we know we have two beta protons, I'll put those in, one would be on a wedge, and one would be on a dash like that."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So let me show those electrons here. So these electrons in magenta pick up a proton from sulfuric acid, and we could say that it formed this bond right in here. And then we also know there's a hydrogen on this carbon going away from us in space, so I'll draw in that hydrogen. And then for our beta carbon here, we know we have two beta protons, I'll put those in, one would be on a wedge, and one would be on a dash like that. And then I'll go ahead and draw in the CH3 right here. The next step in our E1 mechanism is loss of our leaving group. So now we have water as a leaving group, and we know water is a good leaving group."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And then for our beta carbon here, we know we have two beta protons, I'll put those in, one would be on a wedge, and one would be on a dash like that. And then I'll go ahead and draw in the CH3 right here. The next step in our E1 mechanism is loss of our leaving group. So now we have water as a leaving group, and we know water is a good leaving group. These electrons in here can come off onto the oxygen, and we would lose water. We're taking a bond away from this carbon, so we're going to form a carbocation. So actually, let me just go down here real fast and get some more space, and let me draw in our carbocation."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So now we have water as a leaving group, and we know water is a good leaving group. These electrons in here can come off onto the oxygen, and we would lose water. We're taking a bond away from this carbon, so we're going to form a carbocation. So actually, let me just go down here real fast and get some more space, and let me draw in our carbocation. So we would have our benzene ring right here, so let me put in these electrons. And we would have, let me draw in the chain like that, we would have a plus one formal charge on this carbon. So after the water leaves, we get a plus one formal charge right here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So actually, let me just go down here real fast and get some more space, and let me draw in our carbocation. So we would have our benzene ring right here, so let me put in these electrons. And we would have, let me draw in the chain like that, we would have a plus one formal charge on this carbon. So after the water leaves, we get a plus one formal charge right here. We know in an E1 mechanism, we need a stable carbocation, and this is a benzylic carbocation, so it's very stable. Because we can draw several resonance structures. I won't draw them all in to save time, but just to give you an idea, I could take these electrons and move them out to here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So after the water leaves, we get a plus one formal charge right here. We know in an E1 mechanism, we need a stable carbocation, and this is a benzylic carbocation, so it's very stable. Because we can draw several resonance structures. I won't draw them all in to save time, but just to give you an idea, I could take these electrons and move them out to here. And let me go ahead and draw that now. So if I move those electrons out, I would have a plus one charge. I took a bond away from, let me use red for this, I took a bond away from this carbon, so now that's where I have a plus one charge."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "I won't draw them all in to save time, but just to give you an idea, I could take these electrons and move them out to here. And let me go ahead and draw that now. So if I move those electrons out, I would have a plus one charge. I took a bond away from, let me use red for this, I took a bond away from this carbon, so now that's where I have a plus one charge. So a plus one charge right here, and then I would have these other pi electrons in the ring. And you could keep going and draw more resonance structures. For example, you could move these electrons into here, but I won't do that."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "I took a bond away from, let me use red for this, I took a bond away from this carbon, so now that's where I have a plus one charge. So a plus one charge right here, and then I would have these other pi electrons in the ring. And you could keep going and draw more resonance structures. For example, you could move these electrons into here, but I won't do that. I just wanted to show you that this is a benzylic carbocation that is resonance stabilized. So let's go back up to here, so we can figure out our products for this reaction. And I also want you to think about the possibility of free rotation around this sigma bond here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "For example, you could move these electrons into here, but I won't do that. I just wanted to show you that this is a benzylic carbocation that is resonance stabilized. So let's go back up to here, so we can figure out our products for this reaction. And I also want you to think about the possibility of free rotation around this sigma bond here. So there's free rotation around this sigma bond. And let me go ahead and draw in one way to view our carbocation here. So I'm going to call this benzene ring here a phenyl group, so I'm gonna write pH for a phenyl group."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And I also want you to think about the possibility of free rotation around this sigma bond here. So there's free rotation around this sigma bond. And let me go ahead and draw in one way to view our carbocation here. So I'm going to call this benzene ring here a phenyl group, so I'm gonna write pH for a phenyl group. And then this will be our carbocation, so we need to try to show planar geometry around our carbocation. And this carbon right here, the one in magenta, is an sp2 hybridized carbon in our carbocation, so there is an unhybridized p orbital. Let me draw in that p orbital right here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So I'm going to call this benzene ring here a phenyl group, so I'm gonna write pH for a phenyl group. And then this will be our carbocation, so we need to try to show planar geometry around our carbocation. And this carbon right here, the one in magenta, is an sp2 hybridized carbon in our carbocation, so there is an unhybridized p orbital. Let me draw in that p orbital right here. And there's a plus one formal charge on this carbon. There's a plus one formal charge, so that's our carbocation. And then we would have a carbon right here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "Let me draw in that p orbital right here. And there's a plus one formal charge on this carbon. There's a plus one formal charge, so that's our carbocation. And then we would have a carbon right here. And since we know there's free rotation around this sigma bond, so over here, so this arrow that I drew, let me just highlight it in red, I'm gonna pick a particular conformation. I'm gonna have one of the hydrogens parallel with the p orbital, and another hydrogen over here, and then that would mean a methyl group right here. So this is one possible conformation."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And then we would have a carbon right here. And since we know there's free rotation around this sigma bond, so over here, so this arrow that I drew, let me just highlight it in red, I'm gonna pick a particular conformation. I'm gonna have one of the hydrogens parallel with the p orbital, and another hydrogen over here, and then that would mean a methyl group right here. So this is one possible conformation. And to show you how you get this conformation, let's go and look at a video. And in the video, I have the methyl group as being red, so it's easier to see. And the phenyl group over here, the phenyl group is gonna be purple in the video."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So this is one possible conformation. And to show you how you get this conformation, let's go and look at a video. And in the video, I have the methyl group as being red, so it's easier to see. And the phenyl group over here, the phenyl group is gonna be purple in the video. On the left, we have our carbocation. So you can see I put in these paddles here for the p orbital. And the geometry around this carbocation is planar, so hopefully you can see that with the bonds here."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And the phenyl group over here, the phenyl group is gonna be purple in the video. On the left, we have our carbocation. So you can see I put in these paddles here for the p orbital. And the geometry around this carbocation is planar, so hopefully you can see that with the bonds here. And we have our purple for our phenyl group. We know that we have free rotation about this single bond, so I'm going to rotate to get to the conformation that we saw earlier here. So this conformation has a carbon-hydrogen bond that's parallel with the p orbital and can donate some electron density into that p orbital."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And the geometry around this carbocation is planar, so hopefully you can see that with the bonds here. And we have our purple for our phenyl group. We know that we have free rotation about this single bond, so I'm going to rotate to get to the conformation that we saw earlier here. So this conformation has a carbon-hydrogen bond that's parallel with the p orbital and can donate some electron density into that p orbital. In E1 mechanism, we take a proton, so I'm gonna take this proton here, so pretend like I'm taking it away. But now we can see the alkene that forms. This would be the trans alkene with the bulky phenyl group on an opposite side of the double bond from our methyl group."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So this conformation has a carbon-hydrogen bond that's parallel with the p orbital and can donate some electron density into that p orbital. In E1 mechanism, we take a proton, so I'm gonna take this proton here, so pretend like I'm taking it away. But now we can see the alkene that forms. This would be the trans alkene with the bulky phenyl group on an opposite side of the double bond from our methyl group. So we know that there's another possible conformation that has another carbon-hydrogen bond parallel with our p orbital, and so it can donate some electron density into the p orbital there. If I took away this proton, you can see we would get the cis alkene for the product. So in this case, the bulky phenyl group and methyl group would be on the same side of the double bond."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "This would be the trans alkene with the bulky phenyl group on an opposite side of the double bond from our methyl group. So we know that there's another possible conformation that has another carbon-hydrogen bond parallel with our p orbital, and so it can donate some electron density into the p orbital there. If I took away this proton, you can see we would get the cis alkene for the product. So in this case, the bulky phenyl group and methyl group would be on the same side of the double bond. As we saw in the video, the p orbital of the carbocation aligns parallel with the breaking carbon-hydrogen bond. So our p orbital would be in this direction, and our carbon-hydrogen bond would be parallel with that. So the electron density from this bond can be donated into the p orbital."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So in this case, the bulky phenyl group and methyl group would be on the same side of the double bond. As we saw in the video, the p orbital of the carbocation aligns parallel with the breaking carbon-hydrogen bond. So our p orbital would be in this direction, and our carbon-hydrogen bond would be parallel with that. So the electron density from this bond can be donated into the p orbital. So we know in an E1 mechanism, a weak base comes along at this point and takes that proton. So we're gonna take this proton here, and those electrons in light blue would move in to form our double bond. And this conformation gives us the trans product."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So the electron density from this bond can be donated into the p orbital. So we know in an E1 mechanism, a weak base comes along at this point and takes that proton. So we're gonna take this proton here, and those electrons in light blue would move in to form our double bond. And this conformation gives us the trans product. So let me go ahead and draw in our trans product here. So we have our phenyl group, and then we have our double bond, and then we would have our methyl group, so CH3. So this would be the trans product."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And this conformation gives us the trans product. So let me go ahead and draw in our trans product here. So we have our phenyl group, and then we have our double bond, and then we would have our methyl group, so CH3. So this would be the trans product. And we know that a different conformation gives us the cis product. So let's draw in the other conformation or the conformation that gives us the cis product here. So here's our phenyl, and then we would have our carbon, our carbocation."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So this would be the trans product. And we know that a different conformation gives us the cis product. So let's draw in the other conformation or the conformation that gives us the cis product here. So here's our phenyl, and then we would have our carbon, our carbocation. There's a hydrogen going away. And then this would be, let me show the planar geometry around that carbocation. So we'd have our p orbital in here like that, and a plus one formal charge on that carbon."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So here's our phenyl, and then we would have our carbon, our carbocation. There's a hydrogen going away. And then this would be, let me show the planar geometry around that carbocation. So we'd have our p orbital in here like that, and a plus one formal charge on that carbon. And then we would have our other hydrogen is now, the carbon-hydrogen bond is parallel with our p orbital. And so this was after we rotated it, so it's a different conformation than the one on the left. And so this conformation has the methyl group over here like that."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So we'd have our p orbital in here like that, and a plus one formal charge on that carbon. And then we would have our other hydrogen is now, the carbon-hydrogen bond is parallel with our p orbital. And so this was after we rotated it, so it's a different conformation than the one on the left. And so this conformation has the methyl group over here like that. And so now, again, we have this electron density that can be donated into this p orbital. And so our base comes along and takes this proton. And if that happens, then our electrons move into here, and we would form the cis product."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And so this conformation has the methyl group over here like that. And so now, again, we have this electron density that can be donated into this p orbital. And so our base comes along and takes this proton. And if that happens, then our electrons move into here, and we would form the cis product. So let me draw in the cis products. We would have our phenyl group, and then our double bond, and then our methyl group, CH3. So this is the cis product."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "And if that happens, then our electrons move into here, and we would form the cis product. So let me draw in the cis products. We would have our phenyl group, and then our double bond, and then our methyl group, CH3. So this is the cis product. So now that we've figured out our products, let's talk about why this reaction is said to be stereoselective. So we formed two stereoisomers as our product. We have a trans isomer, which is actually 95% of our product, and a cis isomer, which is only about 5% of our product."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "So this is the cis product. So now that we've figured out our products, let's talk about why this reaction is said to be stereoselective. So we formed two stereoisomers as our product. We have a trans isomer, which is actually 95% of our product, and a cis isomer, which is only about 5% of our product. And we can explain that by looking at our two conformations down here. So the conformation on the left has the bulky phenyl and methyl group relatively far away from each other in space, so decreased steric hindrance. But the conformation on the right has these two bulky groups pretty close together in space."}, {"video_title": "E1 mechanism stereoselectivity.mp3", "Sentence": "We have a trans isomer, which is actually 95% of our product, and a cis isomer, which is only about 5% of our product. And we can explain that by looking at our two conformations down here. So the conformation on the left has the bulky phenyl and methyl group relatively far away from each other in space, so decreased steric hindrance. But the conformation on the right has these two bulky groups pretty close together in space. And so that destabilizes this conformation. And that's why we don't get as much of the cis products. So the trans product forms because of decreased steric hindrance, and our trans product is more stable."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about what will happen if we have this molecule. Let's name it. We have 1, 2, 3, 4, 5 carbons, no double bond. So 5 tells us pentane. And it has two groups on the number 3 carbon. 1, 2, 3, doesn't matter which side we start counting from. We have a bromo group and we have an ethyl group, two carbons right there."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So 5 tells us pentane. And it has two groups on the number 3 carbon. 1, 2, 3, doesn't matter which side we start counting from. We have a bromo group and we have an ethyl group, two carbons right there. So on the 3 carbon, we have 3 bromo, 3 ethyl pentane right here, so we have 3 bromo, 3 ethyl pentane dissolved in a solvent of, in a solvent, and this right here, it's an alcohol and it has two carbons right there. So meth, eth, so it is ethanol. So this right there is ethanol."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We have a bromo group and we have an ethyl group, two carbons right there. So on the 3 carbon, we have 3 bromo, 3 ethyl pentane right here, so we have 3 bromo, 3 ethyl pentane dissolved in a solvent of, in a solvent, and this right here, it's an alcohol and it has two carbons right there. So meth, eth, so it is ethanol. So this right there is ethanol. So let's think about what might happen if we have 3 bromo, 3 ethyl pentane dissolved in some ethanol. Now, ethanol already has a hydrogen. It's not super eager to get another proton, although it does have a partial negative charge."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this right there is ethanol. So let's think about what might happen if we have 3 bromo, 3 ethyl pentane dissolved in some ethanol. Now, ethanol already has a hydrogen. It's not super eager to get another proton, although it does have a partial negative charge. It is polar. Oxygen is very electronegative, so it has a partial negative charge. So maybe it might be willing to take on another proton, but it doesn't want to do so very badly."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It's not super eager to get another proton, although it does have a partial negative charge. It is polar. Oxygen is very electronegative, so it has a partial negative charge. So maybe it might be willing to take on another proton, but it doesn't want to do so very badly. So it's actually a weak base. So ethanol right here is a weak base. So it's not strong enough to just go nabbing hydrogens off of carbons like we saw in an E2 reaction."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So maybe it might be willing to take on another proton, but it doesn't want to do so very badly. So it's actually a weak base. So ethanol right here is a weak base. So it's not strong enough to just go nabbing hydrogens off of carbons like we saw in an E2 reaction. So it's just going to sit passively here and maybe wait for something to happen. Now, what might happen? Well, we have this bromo group right here."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it's not strong enough to just go nabbing hydrogens off of carbons like we saw in an E2 reaction. So it's just going to sit passively here and maybe wait for something to happen. Now, what might happen? Well, we have this bromo group right here. We have this bromine, and the bromide anion is actually a pretty good leaving group. It's a fairly large molecule. It's able to keep that charge because it's spread out over a large electron cloud."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have this bromo group right here. We have this bromine, and the bromide anion is actually a pretty good leaving group. It's a fairly large molecule. It's able to keep that charge because it's spread out over a large electron cloud. And it's connected to a tertiary carbon. This carbon right here is connected to one, two, three carbons. So if it were to lose its electron, if it were to lose that electron right there, it might not like to do it, but it would be reasonably stable."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It's able to keep that charge because it's spread out over a large electron cloud. And it's connected to a tertiary carbon. This carbon right here is connected to one, two, three carbons. So if it were to lose its electron, if it were to lose that electron right there, it might not like to do it, but it would be reasonably stable. It's within the realm of possibilities. It could occur. So maybe in this first step, since bromine is a good leaving group and this carbon can be stable as a carbocation, maybe, and the bromine is already more electronegative, so it was already hogging this electron, maybe it takes it all together."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So if it were to lose its electron, if it were to lose that electron right there, it might not like to do it, but it would be reasonably stable. It's within the realm of possibilities. It could occur. So maybe in this first step, since bromine is a good leaving group and this carbon can be stable as a carbocation, maybe, and the bromine is already more electronegative, so it was already hogging this electron, maybe it takes it all together. So let me draw. So neutral bromine has one, two, three, four, five, six, seven valence electrons. Maybe it swipes this electron from the carbon, and now it'll have eight valence electrons and become bromide."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So maybe in this first step, since bromine is a good leaving group and this carbon can be stable as a carbocation, maybe, and the bromine is already more electronegative, so it was already hogging this electron, maybe it takes it all together. So let me draw. So neutral bromine has one, two, three, four, five, six, seven valence electrons. Maybe it swipes this electron from the carbon, and now it'll have eight valence electrons and become bromide. So what happens now? And of course, the ethanol did nothing. It's a weak base."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Maybe it swipes this electron from the carbon, and now it'll have eight valence electrons and become bromide. So what happens now? And of course, the ethanol did nothing. It's a weak base. It wasn't strong enough to react with this just yet. So what is happening now? So this is going to be the slow reaction."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It's a weak base. It wasn't strong enough to react with this just yet. So what is happening now? So this is going to be the slow reaction. So right here, what I said, this isn't going to happen super fast, but it could happen. This is actually the rate determining step. So what happens after that?"}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be the slow reaction. So right here, what I said, this isn't going to happen super fast, but it could happen. This is actually the rate determining step. So what happens after that? So let me just paste everything again. So this is our setup to begin with. But now that this little reaction occurred, what will it look like?"}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So what happens after that? So let me just paste everything again. So this is our setup to begin with. But now that this little reaction occurred, what will it look like? Well, the bromine has left. So let me clear that out. Let me clear out the bromine."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But now that this little reaction occurred, what will it look like? Well, the bromine has left. So let me clear that out. Let me clear out the bromine. Edit, clear. And I actually took an electron with it, so it's bromide. So let me draw it like this."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let me clear out the bromine. Edit, clear. And I actually took an electron with it, so it's bromide. So let me draw it like this. So I'll do it in blue. So this is the bromine. So the bromine is right over here."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw it like this. So I'll do it in blue. So this is the bromine. So the bromine is right over here. It had one, two, three, four, five, six, seven valence electrons. It swiped this magenta electron from the carbon. Now it has eight valence electrons."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So the bromine is right over here. It had one, two, three, four, five, six, seven valence electrons. It swiped this magenta electron from the carbon. Now it has eight valence electrons. It has a negative charge. The carbon lost an electron, so it has a positive charge. And it's somewhat stable because it's a tertiary carbocation."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now it has eight valence electrons. It has a negative charge. The carbon lost an electron, so it has a positive charge. And it's somewhat stable because it's a tertiary carbocation. So now let's think about what's happening. And I want to point out one thing. In this first step of a reaction, only one of the reactants was involved."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And it's somewhat stable because it's a tertiary carbocation. So now let's think about what's happening. And I want to point out one thing. In this first step of a reaction, only one of the reactants was involved. This rate determining, the slow step of a reaction. If this doesn't occur, nothing else will. But now that this does occur, everything else will happen quickly."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "In this first step of a reaction, only one of the reactants was involved. This rate determining, the slow step of a reaction. If this doesn't occur, nothing else will. But now that this does occur, everything else will happen quickly. But in our rate determining step, we only had one of the reactants involved. So it's analogous to the SN1 reaction. What we're going to see here is that we're actually eliminating."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But now that this does occur, everything else will happen quickly. But in our rate determining step, we only had one of the reactants involved. So it's analogous to the SN1 reaction. What we're going to see here is that we're actually eliminating. And we're going to call this an E1 reaction. We're going to see that in a second. Actually, the elimination has already occurred."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "What we're going to see here is that we're actually eliminating. And we're going to call this an E1 reaction. We're going to see that in a second. Actually, the elimination has already occurred. The bromide has already left. So hopefully you see why this is called an E1 reaction. It's elimination, E for elimination."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Actually, the elimination has already occurred. The bromide has already left. So hopefully you see why this is called an E1 reaction. It's elimination, E for elimination. And the rate determining step only involves one of the reactants right here. It didn't involve, in this case, the weak base. But now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It's elimination, E for elimination. And the rate determining step only involves one of the reactants right here. It didn't involve, in this case, the weak base. But now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge. And on these ends, it has partial positive charges."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But now that the bromide has left, let's think about whether this weak base, this ethanol, can actually do anything. It does have a partial negative charge over here. It does have a partial negative charge. And on these ends, it has partial positive charges. So it is somewhat attracted to hydrogen, or to protons, I should say, to positive charges. But not so much that it can swipe it off of things that aren't reasonably acidic. Now that this guy is a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And on these ends, it has partial positive charges. So it is somewhat attracted to hydrogen, or to protons, I should say, to positive charges. But not so much that it can swipe it off of things that aren't reasonably acidic. Now that this guy is a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Or I guess another way you could view it is it wants to hog, it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid or the Lewis definition. Either way, it wants to give away a proton. It could be this one, it could be that one."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now that this guy is a carbocation, this entire molecule actually now becomes pretty acidic, which means it wants to give away protons. Or I guess another way you could view it is it wants to hog, it wants to take electrons, depending on whether you want to use the Bronsted-Lowry definition of acid or the Lewis definition. Either way, it wants to give away a proton. It could be this one, it could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It could be this one, it could be that one. It has excess positive charge. It wants to get rid of its excess positive charge. So it's reasonably acidic, enough so that it can react with this weak base. And what you have now is a situation where on this partial negative charge of this oxygen, let me pick a nice color here, let's say this purple electron right here. It can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken by the larger molecule."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it's reasonably acidic, enough so that it can react with this weak base. And what you have now is a situation where on this partial negative charge of this oxygen, let me pick a nice color here, let's say this purple electron right here. It can be donated, or it will swipe the hydrogen proton. Then hydrogen's electron will be taken by the larger molecule. In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs?"}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Then hydrogen's electron will be taken by the larger molecule. In fact, it'll be attracted to the carbocation. So it will go to the carbocation just like that. Now in that situation, what occurs? What's our final product? So let me draw it here. So this part of the reaction is going to happen fast."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now in that situation, what occurs? What's our final product? So let me draw it here. So this part of the reaction is going to happen fast. The rate determining step happens slow. The leaving group had to leave, the carbocation had to form, that's not going to happen super fast. But once that forms, it's not that stable, and then this thing will happen."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this part of the reaction is going to happen fast. The rate determining step happens slow. The leaving group had to leave, the carbocation had to form, that's not going to happen super fast. But once that forms, it's not that stable, and then this thing will happen. This is fast. So let me just paste everything. Actually, let me just paste everything again."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "But once that forms, it's not that stable, and then this thing will happen. This is fast. So let me just paste everything. Actually, let me just paste everything again. Aid the drawing. So now we already had the bromide it had left. And now the hydrogen is gone."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me just paste everything again. Aid the drawing. So now we already had the bromide it had left. And now the hydrogen is gone. Let me clear this. The hydrogen from that carbon right there is gone. Now this electron, that electron is still on this carbon, but the electron that was with this hydrogen is now on what was the carbocation."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And now the hydrogen is gone. Let me clear this. The hydrogen from that carbon right there is gone. Now this electron, that electron is still on this carbon, but the electron that was with this hydrogen is now on what was the carbocation. So that electron right here is now over here, and now this bond right over here is this bond. We formed an alkene. And now what was an ethanol now took a hydrogen proton and now becomes a positive cation."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now this electron, that electron is still on this carbon, but the electron that was with this hydrogen is now on what was the carbocation. So that electron right here is now over here, and now this bond right over here is this bond. We formed an alkene. And now what was an ethanol now took a hydrogen proton and now becomes a positive cation. So let me draw that. So this right here, this electron ends up being given. So it's no longer with the ethanol."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And now what was an ethanol now took a hydrogen proton and now becomes a positive cation. So let me draw that. So this right here, this electron ends up being given. So it's no longer with the ethanol. It gets given to this hydrogen right here. And now they have formed a new bond. And since this oxygen gave away an electron, it now has a positive charge."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it's no longer with the ethanol. It gets given to this hydrogen right here. And now they have formed a new bond. And since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in the previous step. So the bromide anion is floating around with its 8 valence electrons. 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And since this oxygen gave away an electron, it now has a positive charge. And all along, the bromide anion had left in the previous step. So the bromide anion is floating around with its 8 valence electrons. 1, 2, 3, 4, 5, 6, 7. And then it has this one right over here that makes it negative. And then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7. And then it has this one right over here that makes it negative. And then our reaction is done. We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We had a weak base and a good leaving group, a tertiary carbon, and the leaving group left. We only had one of the reactants involved. It was eliminated. And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule. And that allowed this molecule to become an alkene, formed a double bond. This is called an E. And I already told you, this is an E 1 reaction. E for elimination, in this case, of the halide."}, {"video_title": "E1 reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then once it was eliminated, then the weak base was then able to take a hydrogen off of this molecule. And that allowed this molecule to become an alkene, formed a double bond. This is called an E. And I already told you, this is an E 1 reaction. E for elimination, in this case, of the halide. 1 because the rate determining step only involved one of the molecules. It did not involve the weak base. We'll talk more about this, and especially different circumstances where you might have, well, one, the different types of E1 reactions."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, we named some fairly simple ethers. In this video, we're going to think about slightly more complicated ones. And in particular, what happens if in the process of having an ether, we actually have a ring as opposed to just a long chain? So you could imagine a molecule that looks something like this. You have your oxygen. On this side of the oxygen, you have this carbon chain right here. You have a carbon chain like this."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So you could imagine a molecule that looks something like this. You have your oxygen. On this side of the oxygen, you have this carbon chain right here. You have a carbon chain like this. But then that chain bonds back to the oxygen. So we have a ring here. And so it's not obvious how to name this."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "You have a carbon chain like this. But then that chain bonds back to the oxygen. So we have a ring here. And so it's not obvious how to name this. You can't just look at this side and call it methyl, and then that side and call it methyl as well. It's the same side. It connects back to itself."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "And so it's not obvious how to name this. You can't just look at this side and call it methyl, and then that side and call it methyl as well. It's the same side. It connects back to itself. So how do you name this type of ether? So what you do is you just number it. You number the longest carbon chain, like we've always done in the case of an alkane."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "It connects back to itself. So how do you name this type of ether? So what you do is you just number it. You number the longest carbon chain, like we've always done in the case of an alkane. So we could start numbering here, 1, 2, 3, 4. And so if we just think about the carbon chain by itself, we know if it's one carbon, the prefix is met. Meth-2, it's eth-3, it's prop-4, it's but-."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "You number the longest carbon chain, like we've always done in the case of an alkane. So we could start numbering here, 1, 2, 3, 4. And so if we just think about the carbon chain by itself, we know if it's one carbon, the prefix is met. Meth-2, it's eth-3, it's prop-4, it's but-. So if this was just a carbon chain, we would call this butane. So if we only looked at this carbon chain right here, you would call this butane. But obviously, this isn't butane."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "Meth-2, it's eth-3, it's prop-4, it's but-. So if this was just a carbon chain, we would call this butane. So if we only looked at this carbon chain right here, you would call this butane. But obviously, this isn't butane. We have this oxygen that's bonding to the 1 and 4 carbons of the butane. So to make that clear, we call this, and let me color code this part right here, this oxygen right there, it's bonded to the 1 and the 4 carbons. So we call this 1, 4, and this is our new word that we're going to learn in this video."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "But obviously, this isn't butane. We have this oxygen that's bonding to the 1 and 4 carbons of the butane. So to make that clear, we call this, and let me color code this part right here, this oxygen right there, it's bonded to the 1 and the 4 carbons. So we call this 1, 4, and this is our new word that we're going to learn in this video. 1, 4-epoxybutane. And it doesn't just apply when the ether forms a large ring. It can actually form a little subset ring on a regular chain."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So we call this 1, 4, and this is our new word that we're going to learn in this video. 1, 4-epoxybutane. And it doesn't just apply when the ether forms a large ring. It can actually form a little subset ring on a regular chain. So you could imagine something like this. Let me draw a chain of carbons. Let's say we have five carbons."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "It can actually form a little subset ring on a regular chain. So you could imagine something like this. Let me draw a chain of carbons. Let's say we have five carbons. 1, 2, 3, 4, 5, just like that. And then let's say that between this carbon and this carbon, instead of having a double bond, this carbon actually bonds to an oxygen, which then bonds to this carbon over here. And obviously, every carbon has four bonds."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we have five carbons. 1, 2, 3, 4, 5, just like that. And then let's say that between this carbon and this carbon, instead of having a double bond, this carbon actually bonds to an oxygen, which then bonds to this carbon over here. And obviously, every carbon has four bonds. The ones that we're not drawing, those are hydrogens. How do we name this? Well, same exact process."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "And obviously, every carbon has four bonds. The ones that we're not drawing, those are hydrogens. How do we name this? Well, same exact process. We actually start numbering the chain closer to where the oxygen is bonded. So we start numbering at this end over here, 1, 2, 3, 4, 5. So this is pentane."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "Well, same exact process. We actually start numbering the chain closer to where the oxygen is bonded. So we start numbering at this end over here, 1, 2, 3, 4, 5. So this is pentane. And then the oxygen is bonded to the 1 and the 2 carbons. So we call this 1, 2-epoxypentane. Now, in the last video, I told you that in general, ethers are fairly nonreactive."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So this is pentane. And then the oxygen is bonded to the 1 and the 2 carbons. So we call this 1, 2-epoxypentane. Now, in the last video, I told you that in general, ethers are fairly nonreactive. They actually make for good solvents. But what I've just drawn here is a special case of ethers called epoxides. So when you just have this kind of three-atom chain right here, where it's two carbons and an oxygen, this is a special case of an ether called an epoxide."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "Now, in the last video, I told you that in general, ethers are fairly nonreactive. They actually make for good solvents. But what I've just drawn here is a special case of ethers called epoxides. So when you just have this kind of three-atom chain right here, where it's two carbons and an oxygen, this is a special case of an ether called an epoxide. And this, unlike most ethers, is very reactive. Or another way you could think about it, it's very unstable. This is very reactive."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So when you just have this kind of three-atom chain right here, where it's two carbons and an oxygen, this is a special case of an ether called an epoxide. And this, unlike most ethers, is very reactive. Or another way you could think about it, it's very unstable. This is very reactive. So sometimes people kind of consider these separate from ethers. And the reason why they're very reactive is this three-member ring right here. There's a lot of strain on these bonds."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "This is very reactive. So sometimes people kind of consider these separate from ethers. And the reason why they're very reactive is this three-member ring right here. There's a lot of strain on these bonds. These electrons, these bonds don't like to be that close to each other. If you actually tried to make it with an actual model set with molecules, you would have trouble making it bend enough to actually make this bond. So this is highly, highly, highly unstable."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "There's a lot of strain on these bonds. These electrons, these bonds don't like to be that close to each other. If you actually tried to make it with an actual model set with molecules, you would have trouble making it bend enough to actually make this bond. So this is highly, highly, highly unstable. And there's actually an alternate way to name epoxides. This is a completely legitimate way. You could name it just like an ether with a ring."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So this is highly, highly, highly unstable. And there's actually an alternate way to name epoxides. This is a completely legitimate way. You could name it just like an ether with a ring. This is 1,2-epoxypentane. But the alternate way is to pretend like you had a double bond here. Instead of this oxygen here, you had a double bond."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "You could name it just like an ether with a ring. This is 1,2-epoxypentane. But the alternate way is to pretend like you had a double bond here. Instead of this oxygen here, you had a double bond. If you had a double bond here, this thing would be called, depending how you want to name it, you could be called 1-pentene. That's if there was not this oxygen here, but if there was a double bond here. 1-pentene would look like this."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "Instead of this oxygen here, you had a double bond. If you had a double bond here, this thing would be called, depending how you want to name it, you could be called 1-pentene. That's if there was not this oxygen here, but if there was a double bond here. 1-pentene would look like this. 1, 2, 3, 4, 5. This is the 1 carbon. So 1, 2, 3, 4, 5."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "1-pentene would look like this. 1, 2, 3, 4, 5. This is the 1 carbon. So 1, 2, 3, 4, 5. So this is what 1-pentene looks like. We've learned that many, many, many videos ago. Sometimes it's called pent-1-ene, depending on which convention."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5. So this is what 1-pentene looks like. We've learned that many, many, many videos ago. Sometimes it's called pent-1-ene, depending on which convention. This is kind of the more common one. But we have this oxygen here instead of this double bond. So instead of calling it just 1-pentene, we call it 1-pentene oxide."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes it's called pent-1-ene, depending on which convention. This is kind of the more common one. But we have this oxygen here instead of this double bond. So instead of calling it just 1-pentene, we call it 1-pentene oxide. Just like that. So both of these are the names for the same exact molecule. This makes it clear that it's an epoxide."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So instead of calling it just 1-pentene, we call it 1-pentene oxide. Just like that. So both of these are the names for the same exact molecule. This makes it clear that it's an epoxide. That's kind of the special ether that is more reactive. This is just a general way that we name any type of cyclic ether. So let's just do one more just to make the point clear."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "This makes it clear that it's an epoxide. That's kind of the special ether that is more reactive. This is just a general way that we name any type of cyclic ether. So let's just do one more just to make the point clear. Let's have a cycle branching off of a cycle. Let's have an epoxide off of another ring, just to make the point clear. These aren't too hard to name, but the first time you see them a little daunting."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So let's just do one more just to make the point clear. Let's have a cycle branching off of a cycle. Let's have an epoxide off of another ring, just to make the point clear. These aren't too hard to name, but the first time you see them a little daunting. So let's say we have a cyclohexane ring right here. So this is cyclohexane. But let's say we have a little epoxy branching off of it, just like this."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "These aren't too hard to name, but the first time you see them a little daunting. So let's say we have a cyclohexane ring right here. So this is cyclohexane. But let's say we have a little epoxy branching off of it, just like this. We have that going on. So if we wanted to make it clear that this is an epoxide, we would essentially pretend, or first pretend, that this is just a double bond. If this was just a double bond, this would be cyclohexene."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "But let's say we have a little epoxy branching off of it, just like this. We have that going on. So if we wanted to make it clear that this is an epoxide, we would essentially pretend, or first pretend, that this is just a double bond. If this was just a double bond, this would be cyclohexene. If this oxygen wasn't there and instead we just had a double bond here, and you actually don't have to specify the number for just when you only have one double bond in cyclohexane, because it could have been anywhere and it would have essentially been the same molecule. But since we have this oxygen here, instead of a double bond that's bonding to both of these carbons, we call this cyclohexene oxide. So this part right here makes us name this cyclohexene oxide."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "If this was just a double bond, this would be cyclohexene. If this oxygen wasn't there and instead we just had a double bond here, and you actually don't have to specify the number for just when you only have one double bond in cyclohexane, because it could have been anywhere and it would have essentially been the same molecule. But since we have this oxygen here, instead of a double bond that's bonding to both of these carbons, we call this cyclohexene oxide. So this part right here makes us name this cyclohexene oxide. Or if we wanted to just name this as a traditional ether, the way we would do it, we would just name this cyclohexane and put the epoxy in front of it. Either of these are valid. And once again, you don't have to number it, because you could call it 1, 2 epoxy cyclohexane if you made this the 1 or the 2 carbon."}, {"video_title": "Cyclic ethers and epoxide naming Organic chemistry Khan Academy.mp3", "Sentence": "So this part right here makes us name this cyclohexene oxide. Or if we wanted to just name this as a traditional ether, the way we would do it, we would just name this cyclohexane and put the epoxy in front of it. Either of these are valid. And once again, you don't have to number it, because you could call it 1, 2 epoxy cyclohexane if you made this the 1 or the 2 carbon. But you know it's going to be on adjacent carbons, and it could have really been on any of these two. It could have been on the 3 and the 4, and it would essentially have been the same molecule. So this actually makes it clear exactly what the molecular structure of the molecule is."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What I want to do in this video is talk a little bit about seismic waves. One, because they're interesting by themselves, but they're also really useful for figuring out what the actual composition of the Earth is. You've seen my video on the actual layers of the Earth. And seismic waves are crucial to actually realizing how people figure it out what the different layers of the Earth are. And just to be clear, seismic waves, they're normally associated with earthquakes. But there are any waves that travel through the Earth that could be due to an earthquake, or just really any kind of large explosion, an explosion or anything that really essentially starts sending energy through the rock on Earth, or really through Earth itself. Now, there's two fundamentally different types of seismic waves."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And seismic waves are crucial to actually realizing how people figure it out what the different layers of the Earth are. And just to be clear, seismic waves, they're normally associated with earthquakes. But there are any waves that travel through the Earth that could be due to an earthquake, or just really any kind of large explosion, an explosion or anything that really essentially starts sending energy through the rock on Earth, or really through Earth itself. Now, there's two fundamentally different types of seismic waves. And we're going to focus on one more than the other. One is surface waves, and the other is body waves. Now, surface waves are ones that literally travel across the surface of something."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, there's two fundamentally different types of seismic waves. And we're going to focus on one more than the other. One is surface waves, and the other is body waves. Now, surface waves are ones that literally travel across the surface of something. In this case, we're talking about the surface of the ground. And this right here is a depiction of surface waves. And these really are more analogous to the type of waves we normally associate with the surface of water."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, surface waves are ones that literally travel across the surface of something. In this case, we're talking about the surface of the ground. And this right here is a depiction of surface waves. And these really are more analogous to the type of waves we normally associate with the surface of water. And there's two types of surface waves, rally waves and love waves. We won't go into a lot of details. But you can see that rally waves are kind of the ground moving up and down."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these really are more analogous to the type of waves we normally associate with the surface of water. And there's two types of surface waves, rally waves and love waves. We won't go into a lot of details. But you can see that rally waves are kind of the ground moving up and down. Here the ground is moving up. Here it's moving down. Here it's moving up."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But you can see that rally waves are kind of the ground moving up and down. Here the ground is moving up. Here it's moving down. Here it's moving up. Here it's moving down. So you can kind of view it as kind of a ground roll. The love waves are essentially the ground shifting left and right."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Here it's moving up. Here it's moving down. So you can kind of view it as kind of a ground roll. The love waves are essentially the ground shifting left and right. So here it's not moving up and down, but here it's moving. If you're facing the direction of the wave movement, it's moving to the left here. Here it's moving to the right."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The love waves are essentially the ground shifting left and right. So here it's not moving up and down, but here it's moving. If you're facing the direction of the wave movement, it's moving to the left here. Here it's moving to the right. Here it's moving to the left. Here it's moving to the right. In both cases, the movement of the surface wave is perpendicular to the direction of motion."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Here it's moving to the right. Here it's moving to the left. Here it's moving to the right. In both cases, the movement of the surface wave is perpendicular to the direction of motion. So we sometimes call these transverse waves. And these are essentially analogous to, as I said, kind of what we see in water waves. Now the more interesting thing are the body waves."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In both cases, the movement of the surface wave is perpendicular to the direction of motion. So we sometimes call these transverse waves. And these are essentially analogous to, as I said, kind of what we see in water waves. Now the more interesting thing are the body waves. Because the body waves, first of all, they're the fastest moving waves. And these are also the waves that are used to figure out the structure of the Earth. So the body waves come in two varieties."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now the more interesting thing are the body waves. Because the body waves, first of all, they're the fastest moving waves. And these are also the waves that are used to figure out the structure of the Earth. So the body waves come in two varieties. You have your P waves or primary waves. You have your P waves, and you have your S waves, or secondary waves. And they're depicted right over here."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the body waves come in two varieties. You have your P waves or primary waves. You have your P waves, and you have your S waves, or secondary waves. And they're depicted right over here. And this is actually energy that's being transferred through a body. So it's not just moving along the surface of one. And so here in this diagram that I got from Wikipedia, which I think Wikipedia got from the US Geological Survey, we have a hammer being hit on some rock or whatever."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they're depicted right over here. And this is actually energy that's being transferred through a body. So it's not just moving along the surface of one. And so here in this diagram that I got from Wikipedia, which I think Wikipedia got from the US Geological Survey, we have a hammer being hit on some rock or whatever. And what you see is right when the hammer gets hit at this end of the rock, and I can zoom in a little bit. So let's say I have this rock over here. And I hit it right over here with a hammer or something."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so here in this diagram that I got from Wikipedia, which I think Wikipedia got from the US Geological Survey, we have a hammer being hit on some rock or whatever. And what you see is right when the hammer gets hit at this end of the rock, and I can zoom in a little bit. So let's say I have this rock over here. And I hit it right over here with a hammer or something. What that's immediately going to do is it's going to compress the rock that the hammer comes in touch with. It's going to compress that rock. But then that energy, essentially the molecules, are going to bump into the adjacent molecules."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I hit it right over here with a hammer or something. What that's immediately going to do is it's going to compress the rock that the hammer comes in touch with. It's going to compress that rock. But then that energy, essentially the molecules, are going to bump into the adjacent molecules. And then those adjacent molecules are then going to bump into the molecules right next to it. And then they're going to bump into the molecules right next to it. So you're going to have this kind of compressed part of rock moving through the wave."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But then that energy, essentially the molecules, are going to bump into the adjacent molecules. And then those adjacent molecules are then going to bump into the molecules right next to it. And then they're going to bump into the molecules right next to it. So you're going to have this kind of compressed part of rock moving through the wave. So these are compressed. And those molecules are going to bump into the adjacent molecules. So immediately after that, the rock will be denser right over here."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you're going to have this kind of compressed part of rock moving through the wave. So these are compressed. And those molecules are going to bump into the adjacent molecules. So immediately after that, the rock will be denser right over here. The first things that were bumped, those will essentially bump into the ones right above them. And then they will move back to where they were. And so now the compression will have moved."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So immediately after that, the rock will be denser right over here. The first things that were bumped, those will essentially bump into the ones right above them. And then they will move back to where they were. And so now the compression will have moved. And if you fast forward, it will have moved a little bit forward. So you essentially have this compression wave. You hit the hammer here."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so now the compression will have moved. And if you fast forward, it will have moved a little bit forward. So you essentially have this compression wave. You hit the hammer here. And you essentially have a changing density that is moving in the same direction of the wave. In this situation, that is the direction of the wave. And you see that the molecules are kind of going back and forth along that same axis."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You hit the hammer here. And you essentially have a changing density that is moving in the same direction of the wave. In this situation, that is the direction of the wave. And you see that the molecules are kind of going back and forth along that same axis. They're going along the same direction as the wave. So those are P waves. And P waves can travel through air."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you see that the molecules are kind of going back and forth along that same axis. They're going along the same direction as the wave. So those are P waves. And P waves can travel through air. That's what essentially sound waves are, compression waves. They can travel through liquid. And they can obviously travel through solids."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And P waves can travel through air. That's what essentially sound waves are, compression waves. They can travel through liquid. And they can obviously travel through solids. And depending on the air, they'll travel the slowest. They'll essentially move at the speed of sound, 330 meters per second, which isn't really slow by everyday human standards. In a liquid, they'll move about 1,500 meters per second."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they can obviously travel through solids. And depending on the air, they'll travel the slowest. They'll essentially move at the speed of sound, 330 meters per second, which isn't really slow by everyday human standards. In a liquid, they'll move about 1,500 meters per second. And then in granite, which is most of the crustal material of the Earth, they'll move at around 5,000 meters per second. Let me write that down. So 5,000 meters per second, or essentially 5 kilometers per second if they're moving through granite."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In a liquid, they'll move about 1,500 meters per second. And then in granite, which is most of the crustal material of the Earth, they'll move at around 5,000 meters per second. Let me write that down. So 5,000 meters per second, or essentially 5 kilometers per second if they're moving through granite. Now, S waves are essentially if you were to hit a hammer on the side of this rock. So let me draw another diagram since this is pretty small. If you were to hit a hammer right over here, what it would do is it would temporarily kind of push all the rock over here."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So 5,000 meters per second, or essentially 5 kilometers per second if they're moving through granite. Now, S waves are essentially if you were to hit a hammer on the side of this rock. So let me draw another diagram since this is pretty small. If you were to hit a hammer right over here, what it would do is it would temporarily kind of push all the rock over here. It would deform it a little bit. And that would pull a little bit of the rock back with it. And then this rock that's right above it would slowly be pulled down while this rock that was initially hit will be moved back up."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to hit a hammer right over here, what it would do is it would temporarily kind of push all the rock over here. It would deform it a little bit. And that would pull a little bit of the rock back with it. And then this rock that's right above it would slowly be pulled down while this rock that was initially hit will be moved back up. So you fast forward maybe a millisecond. And now the next layer of rock right above that will be kind of deformed to the right. And if you keep fast forwarding it, the deformation will move upwards."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then this rock that's right above it would slowly be pulled down while this rock that was initially hit will be moved back up. So you fast forward maybe a millisecond. And now the next layer of rock right above that will be kind of deformed to the right. And if you keep fast forwarding it, the deformation will move upwards. And notice over here, once again, the movement of the wave is upwards. But now the movement of the material is not going along the same axis that we saw with the P waves or the compression waves. It's now moving perpendicular."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if you keep fast forwarding it, the deformation will move upwards. And notice over here, once again, the movement of the wave is upwards. But now the movement of the material is not going along the same axis that we saw with the P waves or the compression waves. It's now moving perpendicular. It's now moving along a perpendicular axis. Or you could call this a transverse wave. The movement of the particles is now on a perpendicular axis to the actual movement of the waves."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's now moving perpendicular. It's now moving along a perpendicular axis. Or you could call this a transverse wave. The movement of the particles is now on a perpendicular axis to the actual movement of the waves. And so that's what an S wave is. And they move a little bit slower than the P waves. So if an earthquake were to happen, you would see the P waves first."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The movement of the particles is now on a perpendicular axis to the actual movement of the waves. And so that's what an S wave is. And they move a little bit slower than the P waves. So if an earthquake were to happen, you would see the P waves first. And then at about 60% of the speed of the P waves, you would see the S waves. Now the most important thing to think about, especially from the point of view of figuring out the composition of the Earth, is that the S waves can only travel through solid. And you might say, wait, I've seen transverse waves on water that look like this."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if an earthquake were to happen, you would see the P waves first. And then at about 60% of the speed of the P waves, you would see the S waves. Now the most important thing to think about, especially from the point of view of figuring out the composition of the Earth, is that the S waves can only travel through solid. And you might say, wait, I've seen transverse waves on water that look like this. Remember, that is a surface wave. We're talking about body waves. We're talking about things that are actually going through the body of water."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you might say, wait, I've seen transverse waves on water that look like this. Remember, that is a surface wave. We're talking about body waves. We're talking about things that are actually going through the body of water. And one way to think about this is if I had some water over here. So let's say that this is a pool. I'll draw a cross section of water."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking about things that are actually going through the body of water. And one way to think about this is if I had some water over here. So let's say that this is a pool. I'll draw a cross section of water. If I have a cross section of water right over here, let's think about it. And hopefully it'll make intuitive sense to you. If I were to compress some of the water, if I were to kind of slam some part of the water here with like a big, I don't know, some type of, I would just compress it really fast, it would do, a P wave could transmit."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll draw a cross section of water. If I have a cross section of water right over here, let's think about it. And hopefully it'll make intuitive sense to you. If I were to compress some of the water, if I were to kind of slam some part of the water here with like a big, I don't know, some type of, I would just compress it really fast, it would do, a P wave could transmit. Because those water molecules would bump into the water molecules next to it, which would bump into the water molecules next to that. And so you would have a compression wave or a P wave moving in the direction of my bump. So P waves, it makes sense."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If I were to compress some of the water, if I were to kind of slam some part of the water here with like a big, I don't know, some type of, I would just compress it really fast, it would do, a P wave could transmit. Because those water molecules would bump into the water molecules next to it, which would bump into the water molecules next to that. And so you would have a compression wave or a P wave moving in the direction of my bump. So P waves, it makes sense. And the same thing is true with air or sound waves. That it makes sense that it could travel through a liquid. But let's say that you, let's say that if, and remember, we're under the water."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So P waves, it makes sense. And the same thing is true with air or sound waves. That it makes sense that it could travel through a liquid. But let's say that you, let's say that if, and remember, we're under the water. We're not thinking about the surface. We're thinking about moving through the body of the water. Let's say that you were to kind of take that hammer, if you were to take a hammer and kind of slap the side of this little volume of water here."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's say that you, let's say that if, and remember, we're under the water. We're not thinking about the surface. We're thinking about moving through the body of the water. Let's say that you were to kind of take that hammer, if you were to take a hammer and kind of slap the side of this little volume of water here. Well, essentially all that would do, it would send a compression wave in that direction. It really wouldn't do anything. It wouldn't allow a transverse wave to go that way because it's not going to, the water doesn't allow it to kind of, it doesn't have this elastic property where if something bounces that way, it's going to immediately bounce back that way."}, {"video_title": "Seismic waves Earth geological and climatic history Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say that you were to kind of take that hammer, if you were to take a hammer and kind of slap the side of this little volume of water here. Well, essentially all that would do, it would send a compression wave in that direction. It really wouldn't do anything. It wouldn't allow a transverse wave to go that way because it's not going to, the water doesn't allow it to kind of, it doesn't have this elastic property where if something bounces that way, it's going to immediately bounce back that way. It's not being pulled back like a solid would. So S waves only travel through solids. So we're going to use, essentially, our understanding of P waves, which travel through air, liquid, or solid, and our understanding of S waves to essentially figure out what the composition of Earth is."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And let's say they're separated by a distance of d here. We know that a proton and an electron have the same magnitude of charge. So both have a magnitude of charge, q, equal to 1.6 times 10 to the negative 19. So of course, a proton would have a positively charged q. So let's go ahead and make this a positively charged q. And an electron would have a negatively charged q, like that. If we were to calculate the dipole moment, the definition of a dipole moment, symbolized by the Greek letter mu, dipole moment is equal to the magnitude of that charge, q, times the distance between those charges, d. So mu is equal to q times d. And we're not really going to get into math in this video."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So of course, a proton would have a positively charged q. So let's go ahead and make this a positively charged q. And an electron would have a negatively charged q, like that. If we were to calculate the dipole moment, the definition of a dipole moment, symbolized by the Greek letter mu, dipole moment is equal to the magnitude of that charge, q, times the distance between those charges, d. So mu is equal to q times d. And we're not really going to get into math in this video. But if you were to go ahead and do that calculation, you would end up with units of Debye's. So you'd get a number, and that number would be in Debye's here. So we're more concerned with analyzing dipole moments in terms of the molecular structure."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If we were to calculate the dipole moment, the definition of a dipole moment, symbolized by the Greek letter mu, dipole moment is equal to the magnitude of that charge, q, times the distance between those charges, d. So mu is equal to q times d. And we're not really going to get into math in this video. But if you were to go ahead and do that calculation, you would end up with units of Debye's. So you'd get a number, and that number would be in Debye's here. So we're more concerned with analyzing dipole moments in terms of the molecular structure. So let's go ahead and look at the dot structure for HCl. So if I look at this covalent bond between the hydrogen and the chlorine, I know that that covalent bond consists of two electrons. And chlorine is more electronegative than hydrogen, which means that those two electrons are going to be pulled closer to the chlorine."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we're more concerned with analyzing dipole moments in terms of the molecular structure. So let's go ahead and look at the dot structure for HCl. So if I look at this covalent bond between the hydrogen and the chlorine, I know that that covalent bond consists of two electrons. And chlorine is more electronegative than hydrogen, which means that those two electrons are going to be pulled closer to the chlorine. So I'm going to go ahead and show that here with this arrow. The arrow is pointing in the direction of movement of electrons, if you will. So those electrons in yellow are going to move closer towards the chlorine."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And chlorine is more electronegative than hydrogen, which means that those two electrons are going to be pulled closer to the chlorine. So I'm going to go ahead and show that here with this arrow. The arrow is pointing in the direction of movement of electrons, if you will. So those electrons in yellow are going to move closer towards the chlorine. So chlorine is going to get a little bit more electron density around it. And so we represent that with a partial negative charge. So I'm going to do a lowercase Greek delta here."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So those electrons in yellow are going to move closer towards the chlorine. So chlorine is going to get a little bit more electron density around it. And so we represent that with a partial negative charge. So I'm going to do a lowercase Greek delta here. And it's partially negative, since it has increased in electron density. That's one way of thinking about it. And since hydrogen is losing a little bit of electron density, it's losing a little bit of negative charge."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to do a lowercase Greek delta here. And it's partially negative, since it has increased in electron density. That's one way of thinking about it. And since hydrogen is losing a little bit of electron density, it's losing a little bit of negative charge. And so it is partially positive. So we can go ahead and draw a partial positive sign here. And so we're setting up a situation where we are polarizing the molecule."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And since hydrogen is losing a little bit of electron density, it's losing a little bit of negative charge. And so it is partially positive. So we can go ahead and draw a partial positive sign here. And so we're setting up a situation where we are polarizing the molecule. So this part of the molecule over here on the right is increasing in electron density. And so that is our partial negative side. That's one pole."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so we're setting up a situation where we are polarizing the molecule. So this part of the molecule over here on the right is increasing in electron density. And so that is our partial negative side. That's one pole. And then this other side here is losing some electron density. And so it's partially positive. So we have it like that."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That's one pole. And then this other side here is losing some electron density. And so it's partially positive. So we have it like that. So that's where the positive sign comes in. You can think about on this arrow here, this little positive sign giving you the distribution of charge in this molecule. And so you have these two poles, a positive pole and a negative pole."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we have it like that. So that's where the positive sign comes in. You can think about on this arrow here, this little positive sign giving you the distribution of charge in this molecule. And so you have these two poles, a positive pole and a negative pole. And if you think about those two poles as having a center of mass, you could have a distance between them. And you could calculate the dipole moment for this molecule. And so when you calculate the dipole moment for HCl, mu turns out to be equal to approximately 1.11 debyes."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so you have these two poles, a positive pole and a negative pole. And if you think about those two poles as having a center of mass, you could have a distance between them. And you could calculate the dipole moment for this molecule. And so when you calculate the dipole moment for HCl, mu turns out to be equal to approximately 1.11 debyes. And so we have a polarized bond. We have a polarized molecule. And so therefore, we can say that HCl is relatively polar."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so when you calculate the dipole moment for HCl, mu turns out to be equal to approximately 1.11 debyes. And so we have a polarized bond. We have a polarized molecule. And so therefore, we can say that HCl is relatively polar. It has a dipole moment. So that's how to think about analyzing these molecules. Let's do another one here."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, we can say that HCl is relatively polar. It has a dipole moment. So that's how to think about analyzing these molecules. Let's do another one here. Let's do carbon dioxide. So I know that the CO2 molecule is linear. So after you draw the dot structure, you're going to get a linear shape, which is going to be important when we're trying to predict the dipole moment."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one here. Let's do carbon dioxide. So I know that the CO2 molecule is linear. So after you draw the dot structure, you're going to get a linear shape, which is going to be important when we're trying to predict the dipole moment. If I analyze the electrons in this carbon-oxygen bond, so we have a double bond between carbon and oxygen. Oxygen is more electronegative than carbon. So oxygen is going to try to pull those electrons closer to itself."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So after you draw the dot structure, you're going to get a linear shape, which is going to be important when we're trying to predict the dipole moment. If I analyze the electrons in this carbon-oxygen bond, so we have a double bond between carbon and oxygen. Oxygen is more electronegative than carbon. So oxygen is going to try to pull those electrons closer to itself. And so we can go ahead and draw our arrow or our vector pointing towards the right here. And so we have a bond-dipole situation here. On the left, we have the exact same situation."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen is going to try to pull those electrons closer to itself. And so we can go ahead and draw our arrow or our vector pointing towards the right here. And so we have a bond-dipole situation here. On the left, we have the exact same situation. Oxygen is more electronegative than carbon. And so these electrons are going to be pulled closer to this oxygen. So we draw another arrow or another vector in this case."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "On the left, we have the exact same situation. Oxygen is more electronegative than carbon. And so these electrons are going to be pulled closer to this oxygen. So we draw another arrow or another vector in this case. So even though we have these individual bond dipoles, if you think about this molecule as being linear, and you can see we have these two vectors that are equal in magnitude but opposite in direction, those two vectors are going to cancel out. And therefore, we would not expect to have a dipole moment for the molecule. There's no molecular dipole here."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we draw another arrow or another vector in this case. So even though we have these individual bond dipoles, if you think about this molecule as being linear, and you can see we have these two vectors that are equal in magnitude but opposite in direction, those two vectors are going to cancel out. And therefore, we would not expect to have a dipole moment for the molecule. There's no molecular dipole here. So mu turns out to be equal to 0. A simplistic way of thinking about this would be like a tug of war. You have these really strong atoms, if you will, these oxygens."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There's no molecular dipole here. So mu turns out to be equal to 0. A simplistic way of thinking about this would be like a tug of war. You have these really strong atoms, if you will, these oxygens. But they're equally strong. And if they're pulling with equal force in opposite directions, it's going to cancel out. So the individual bond dipoles cancel out."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "You have these really strong atoms, if you will, these oxygens. But they're equally strong. And if they're pulling with equal force in opposite directions, it's going to cancel out. So the individual bond dipoles cancel out. And so there's no overall dipole moment for this molecule. And carbon dioxide is considered to be nonpolar. Let's go ahead and analyze a water molecule over here on the right."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the individual bond dipoles cancel out. And so there's no overall dipole moment for this molecule. And carbon dioxide is considered to be nonpolar. Let's go ahead and analyze a water molecule over here on the right. So the electrons in this covalent bond between the hydrogen and oxygen, oxygen is more electronegative than hydrogen. So those electrons are going to be pulled closer to the oxygen. Same thing for this bond over here."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and analyze a water molecule over here on the right. So the electrons in this covalent bond between the hydrogen and oxygen, oxygen is more electronegative than hydrogen. So those electrons are going to be pulled closer to the oxygen. Same thing for this bond over here. And we also have lone pairs of electrons on our central atom to think about. And that's, of course, going to increase the electron density going in this direction for that lone pair and in this direction for that lone pair. And so even though we know the geometry of the water molecule is bent, and it's hard to represent that on this two-dimensional surface here, if you use a molymod set, you will kind of see that your net dipole moment would be directed upward in this case."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Same thing for this bond over here. And we also have lone pairs of electrons on our central atom to think about. And that's, of course, going to increase the electron density going in this direction for that lone pair and in this direction for that lone pair. And so even though we know the geometry of the water molecule is bent, and it's hard to represent that on this two-dimensional surface here, if you use a molymod set, you will kind of see that your net dipole moment would be directed upward in this case. And so the individual bond dipoles you're going to add to give you a molecular dipole, in this case, pointed up. And so therefore, you're going to have a dipole moment associated with your water molecule. So mu turns out to be approximately 1.85."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so even though we know the geometry of the water molecule is bent, and it's hard to represent that on this two-dimensional surface here, if you use a molymod set, you will kind of see that your net dipole moment would be directed upward in this case. And so the individual bond dipoles you're going to add to give you a molecular dipole, in this case, pointed up. And so therefore, you're going to have a dipole moment associated with your water molecule. So mu turns out to be approximately 1.85. And we can consider water to be a polar molecule. Let's do two more examples. So on the left is CCl4, or carbon tetrachloride."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So mu turns out to be approximately 1.85. And we can consider water to be a polar molecule. Let's do two more examples. So on the left is CCl4, or carbon tetrachloride. And so you can see that we have carbon bonded to chlorine here. And since this is a straight line, this means in the plane of the page, if you will. And so we know the geometry is tetrahedral around this carbon."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So on the left is CCl4, or carbon tetrachloride. And so you can see that we have carbon bonded to chlorine here. And since this is a straight line, this means in the plane of the page, if you will. And so we know the geometry is tetrahedral around this carbon. So let's go ahead and analyze that as well. So I have a wedge drawn here, which means this chlorine is coming out at you in space. And then I have a dash back here, meaning this chlorine back here is going away from you in space."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so we know the geometry is tetrahedral around this carbon. So let's go ahead and analyze that as well. So I have a wedge drawn here, which means this chlorine is coming out at you in space. And then I have a dash back here, meaning this chlorine back here is going away from you in space. So that's how to think about it. But it's really much easier to go ahead and make this using a molymod set. And you can see that however you rotate this molecule, it's going to look the same in all directions."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then I have a dash back here, meaning this chlorine back here is going away from you in space. So that's how to think about it. But it's really much easier to go ahead and make this using a molymod set. And you can see that however you rotate this molecule, it's going to look the same in all directions. So a tetrahedral arrangement of four of the same atoms around a central atom, you can turn the molecule over. It's always going to look the same in three dimensions. And that's really important when you're analyzing the dipole moment for this molecule."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And you can see that however you rotate this molecule, it's going to look the same in all directions. So a tetrahedral arrangement of four of the same atoms around a central atom, you can turn the molecule over. It's always going to look the same in three dimensions. And that's really important when you're analyzing the dipole moment for this molecule. So let's go ahead and do that. We'll start with our electronegativity differences. So if I look at this top carbon chlorine bond, these two electrons in this top carbon chlorine bond, chlorine is more electronegative than carbon."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And that's really important when you're analyzing the dipole moment for this molecule. So let's go ahead and do that. We'll start with our electronegativity differences. So if I look at this top carbon chlorine bond, these two electrons in this top carbon chlorine bond, chlorine is more electronegative than carbon. And so we can think about those electrons being pulled closer to the chlorines. Let me go ahead and use green for that. So those two electrons are going in this direction."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at this top carbon chlorine bond, these two electrons in this top carbon chlorine bond, chlorine is more electronegative than carbon. And so we can think about those electrons being pulled closer to the chlorines. Let me go ahead and use green for that. So those two electrons are going in this direction. And it's the same thing for all of these chlorines. Chlorine is more electronegative than carbon. So we can draw these individual bond dipoles."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So those two electrons are going in this direction. And it's the same thing for all of these chlorines. Chlorine is more electronegative than carbon. So we can draw these individual bond dipoles. We can draw four of them here. And in this case, we have four dipoles. But they're going to cancel out in three dimensions."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we can draw these individual bond dipoles. We can draw four of them here. And in this case, we have four dipoles. But they're going to cancel out in three dimensions. So again, this is a tough one to visualize on a two-dimensional surface. But if you have the molecule in front of you, it's a little bit easier to see that if you keep rotating the molecule, it looks the same. And so these individual bond dipoles cancel."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But they're going to cancel out in three dimensions. So again, this is a tough one to visualize on a two-dimensional surface. But if you have the molecule in front of you, it's a little bit easier to see that if you keep rotating the molecule, it looks the same. And so these individual bond dipoles cancel. There's no dipole moment for this molecule. And so mu is equal to 0. And we would expect the carbon tetrachloride molecule to be nonpolar."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so these individual bond dipoles cancel. There's no dipole moment for this molecule. And so mu is equal to 0. And we would expect the carbon tetrachloride molecule to be nonpolar. Let's look at the example on the right, where we have substituted in a hydrogen for one of the chlorines. And so now we have CHCl3 or chloroform. So now if we analyze the molecule, so let's think about this bond in here."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we would expect the carbon tetrachloride molecule to be nonpolar. Let's look at the example on the right, where we have substituted in a hydrogen for one of the chlorines. And so now we have CHCl3 or chloroform. So now if we analyze the molecule, so let's think about this bond in here. Carbon is actually a little bit more electronegative than hydrogen. So we can show the electrons in that bond in red moving towards the carbon this time. And once again, carbon versus chlorine."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So now if we analyze the molecule, so let's think about this bond in here. Carbon is actually a little bit more electronegative than hydrogen. So we can show the electrons in that bond in red moving towards the carbon this time. And once again, carbon versus chlorine. Chlorine is more electronegative. So we're going to have a bond dipole in that direction, which we can do for all of our chlorines here. And so hopefully, it's a little bit easier to see."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And once again, carbon versus chlorine. Chlorine is more electronegative. So we're going to have a bond dipole in that direction, which we can do for all of our chlorines here. And so hopefully, it's a little bit easier to see. In this case, the individual bond dipoles are going to combine to give you a net dipole located in the downward direction for this molecule. So I'm attempting to draw the molecular dipole, the dipole for the entire molecule, going a little bit down in terms of how I've drawn this molecule. And so since we have a hydrogen here, there's no upward pole in this case to balance out the downward pole."}, {"video_title": "Electronegativity and bonding Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so hopefully, it's a little bit easier to see. In this case, the individual bond dipoles are going to combine to give you a net dipole located in the downward direction for this molecule. So I'm attempting to draw the molecular dipole, the dipole for the entire molecule, going a little bit down in terms of how I've drawn this molecule. And so since we have a hydrogen here, there's no upward pole in this case to balance out the downward pole. And so we would expect this molecule to have a dipole moment. And so mu turns out to be approximately 1.01 for chloroform. So it is certainly more polar than our carbon tetrachloride example."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here's a lone pair of electrons. I'm going to highlight it in magenta. That lone pair of electrons is located on this carbon. Let me go ahead and put this carbon in green here. I'm saying there's a negative one formal charge in that carbon in the green. That carbon in green is also bonded to a hydrogen. Once again, you need to be very familiar with assigning formal charges."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and put this carbon in green here. I'm saying there's a negative one formal charge in that carbon in the green. That carbon in green is also bonded to a hydrogen. Once again, you need to be very familiar with assigning formal charges. We have a lone pair of electrons next to a pi bond because over here we have a double bond between the carbon and the oxygen. One of those bonds is a sigma bond and one of those bonds is a pi bond. I'm just going to say that these are the pi electrons."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Once again, you need to be very familiar with assigning formal charges. We have a lone pair of electrons next to a pi bond because over here we have a double bond between the carbon and the oxygen. One of those bonds is a sigma bond and one of those bonds is a pi bond. I'm just going to say that these are the pi electrons. Our goal when drawing a resonance structure is to delocalize that negative one formal charge, so spread out some electron density. We could take the electrons in magenta and move them into here to form a double bond between the carbon in green and this carbon right here. That'd be too many bonds to the carbon in yellow."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm just going to say that these are the pi electrons. Our goal when drawing a resonance structure is to delocalize that negative one formal charge, so spread out some electron density. We could take the electrons in magenta and move them into here to form a double bond between the carbon in green and this carbon right here. That'd be too many bonds to the carbon in yellow. The electrons in blue have to come off onto this top oxygen here. We go ahead and draw in our brackets and we put our double-headed resonance arrow and we draw the other resonance structure. We have our ring like that and then we have now a double bond between those two carbons."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That'd be too many bonds to the carbon in yellow. The electrons in blue have to come off onto this top oxygen here. We go ahead and draw in our brackets and we put our double-headed resonance arrow and we draw the other resonance structure. We have our ring like that and then we have now a double bond between those two carbons. Then this top oxygen here now has only one bond to it. The oxygen used to have two lone pairs of electrons. Now it has three because it just picked up a pair of electrons from that pi bond."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have our ring like that and then we have now a double bond between those two carbons. Then this top oxygen here now has only one bond to it. The oxygen used to have two lone pairs of electrons. Now it has three because it just picked up a pair of electrons from that pi bond. Let's go ahead and follow the electrons. The electrons in magenta moved in here to form our pi bond like that and the electrons in the pi bond in blue moved off onto this oxygen. I'm saying that they are those electrons."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now it has three because it just picked up a pair of electrons from that pi bond. Let's go ahead and follow the electrons. The electrons in magenta moved in here to form our pi bond like that and the electrons in the pi bond in blue moved off onto this oxygen. I'm saying that they are those electrons. That gives the top oxygen a negative one formal charge. We have our two resonance structures for the enolate anion. We know that both resonance structures contribute to the overall hybrid."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm saying that they are those electrons. That gives the top oxygen a negative one formal charge. We have our two resonance structures for the enolate anion. We know that both resonance structures contribute to the overall hybrid. If you think about which one contributes more, for the example on the left, we had a negative one formal charge on the carbon in green. That's a carb anion. For the resonance structure on the right, we had a negative one formal charge on the oxygen."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that both resonance structures contribute to the overall hybrid. If you think about which one contributes more, for the example on the left, we had a negative one formal charge on the carbon in green. That's a carb anion. For the resonance structure on the right, we had a negative one formal charge on the oxygen. That's an oxyanion. Oxygen is more electronegative than carbon which means it's more likely to support a negative one formal charge. The resonance structure on the right contributes more to the overall hybrid for an enolate anion."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "For the resonance structure on the right, we had a negative one formal charge on the oxygen. That's an oxyanion. Oxygen is more electronegative than carbon which means it's more likely to support a negative one formal charge. The resonance structure on the right contributes more to the overall hybrid for an enolate anion. Let's do another pattern. A lone pair of electrons next to a positive charge this time. Let's look at nitromethane."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The resonance structure on the right contributes more to the overall hybrid for an enolate anion. Let's do another pattern. A lone pair of electrons next to a positive charge this time. Let's look at nitromethane. We can look at this lone pair of electrons here on this oxygen. That lone pair of electrons is next to a positive charge. This nitrogen actually has a plus one formal charge on it."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at nitromethane. We can look at this lone pair of electrons here on this oxygen. That lone pair of electrons is next to a positive charge. This nitrogen actually has a plus one formal charge on it. Let's think about drawing the resonance structure. Our goal is to delocalize charge, to spread charge out. We could take the electrons in magenta and move them into here to form a double bond between the nitrogen and the oxygen."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This nitrogen actually has a plus one formal charge on it. Let's think about drawing the resonance structure. Our goal is to delocalize charge, to spread charge out. We could take the electrons in magenta and move them into here to form a double bond between the nitrogen and the oxygen. That's too many bonds to this nitrogen. That would give us five bonds to that nitrogen which we know doesn't happen because of nitrogen's position on the periodic table. That means that the electrons in this pi bond here are going to come off onto the oxygen."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could take the electrons in magenta and move them into here to form a double bond between the nitrogen and the oxygen. That's too many bonds to this nitrogen. That would give us five bonds to that nitrogen which we know doesn't happen because of nitrogen's position on the periodic table. That means that the electrons in this pi bond here are going to come off onto the oxygen. These electrons in blue come off onto this oxygen. We draw our other resonance structure for nitromethane. We have a CH3."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That means that the electrons in this pi bond here are going to come off onto the oxygen. These electrons in blue come off onto this oxygen. We draw our other resonance structure for nitromethane. We have a CH3. We now have a double bond between nitrogen and this oxygen. This oxygen used to have three lone pairs of electrons. The electrons in magenta moved in here to form this bond."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have a CH3. We now have a double bond between nitrogen and this oxygen. This oxygen used to have three lone pairs of electrons. The electrons in magenta moved in here to form this bond. That means we have only two lone pairs left on this oxygen. For the oxygen on the bottom right, there's only one bond now between nitrogen and the oxygen because the electrons in blue moved off onto this oxygen. That means this oxygen has two more lone pairs of electrons."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in magenta moved in here to form this bond. That means we have only two lone pairs left on this oxygen. For the oxygen on the bottom right, there's only one bond now between nitrogen and the oxygen because the electrons in blue moved off onto this oxygen. That means this oxygen has two more lone pairs of electrons. When we go ahead and put in our resonance bracket here, you always need to think about assigning formal charge. What happened to the charge? This oxygen now has a negative one formal charge and this nitrogen still has a plus one formal charge."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That means this oxygen has two more lone pairs of electrons. When we go ahead and put in our resonance bracket here, you always need to think about assigning formal charge. What happened to the charge? This oxygen now has a negative one formal charge and this nitrogen still has a plus one formal charge. We've delocalized that negative charge. It's actually over both of those oxygens. Notice that the overall charge for nitromethane is zero for both resonance structures."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen now has a negative one formal charge and this nitrogen still has a plus one formal charge. We've delocalized that negative charge. It's actually over both of those oxygens. Notice that the overall charge for nitromethane is zero for both resonance structures. We have one positive charge and one negative charge on the left, so that gives us zero. We have one positive charge and one negative charge on the right, so that gives us zero, so conservation of charge. Let's do another example for a pattern that we might see."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Notice that the overall charge for nitromethane is zero for both resonance structures. We have one positive charge and one negative charge on the left, so that gives us zero. We have one positive charge and one negative charge on the right, so that gives us zero, so conservation of charge. Let's do another example for a pattern that we might see. For this one, we have a positive charge next to a pi bond. Let's look at this carbon. I'm saying it has a plus one formal charge."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another example for a pattern that we might see. For this one, we have a positive charge next to a pi bond. Let's look at this carbon. I'm saying it has a plus one formal charge. If it has a plus one formal charge, it must have only three bonds. Since it's already bonded to another carbon, it's already bonded to, let me go ahead and label these. The carbon in yellow there is bonded to this carbon in green."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm saying it has a plus one formal charge. If it has a plus one formal charge, it must have only three bonds. Since it's already bonded to another carbon, it's already bonded to, let me go ahead and label these. The carbon in yellow there is bonded to this carbon in green. Because it has a plus one formal charge, it must have only two other bonds. Those must be to hydrogen, so I draw in those hydrogens. Now it makes a little bit more sense why it's a plus one formal charge."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The carbon in yellow there is bonded to this carbon in green. Because it has a plus one formal charge, it must have only two other bonds. Those must be to hydrogen, so I draw in those hydrogens. Now it makes a little bit more sense why it's a plus one formal charge. It'd be four minus three, giving us plus one. The carbon in green has a formal charge of zero, so it already has three bonds, so it needs one more to hydrogen. Let's go ahead and make this carbon over here."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now it makes a little bit more sense why it's a plus one formal charge. It'd be four minus three, giving us plus one. The carbon in green has a formal charge of zero, so it already has three bonds, so it needs one more to hydrogen. Let's go ahead and make this carbon over here. Let's make it red. This carbon over here in red already has two bonds. It has a formal charge of zero, so it needs two more hydrogens."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and make this carbon over here. Let's make it red. This carbon over here in red already has two bonds. It has a formal charge of zero, so it needs two more hydrogens. Once again, our pattern is a positive charge next to a pi bond. Let me go ahead and highlight these things here. We have a positive charge next to a pi bond."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It has a formal charge of zero, so it needs two more hydrogens. Once again, our pattern is a positive charge next to a pi bond. Let me go ahead and highlight these things here. We have a positive charge next to a pi bond. Here, let's say this one is our pi bond, like that. When you're drawing resonance structures, again, your goal is to delocalize that charge. We could spread out that positive charge by taking the electrons in blue, the pi electrons, and moving them into here."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have a positive charge next to a pi bond. Here, let's say this one is our pi bond, like that. When you're drawing resonance structures, again, your goal is to delocalize that charge. We could spread out that positive charge by taking the electrons in blue, the pi electrons, and moving them into here. Let's draw the resonance structure. We now have a double bond between the two carbons on the right. The hydrogens haven't moved, so I'm going to leave those hydrogens in there."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could spread out that positive charge by taking the electrons in blue, the pi electrons, and moving them into here. Let's draw the resonance structure. We now have a double bond between the two carbons on the right. The hydrogens haven't moved, so I'm going to leave those hydrogens in there. There's still one hydrogen in the carbon in the middle, two hydrogens in the carbon on the right, and two hydrogens on the carbon on the left. The electrons in blue moved to here, like that. Let me go ahead and highlight those carbons."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogens haven't moved, so I'm going to leave those hydrogens in there. There's still one hydrogen in the carbon in the middle, two hydrogens in the carbon on the right, and two hydrogens on the carbon on the left. The electrons in blue moved to here, like that. Let me go ahead and highlight those carbons. The carbon in green right here, and the carbon in red. What happened to the plus one formal charge? You can see that it's actually moved to the carbon in the red."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and highlight those carbons. The carbon in green right here, and the carbon in red. What happened to the plus one formal charge? You can see that it's actually moved to the carbon in the red. The carbon in red right here has only three bonds, so four minus three gives us a plus one formal charge. Let's go ahead and finish our resonance bracket here. I put that in."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can see that it's actually moved to the carbon in the red. The carbon in red right here has only three bonds, so four minus three gives us a plus one formal charge. Let's go ahead and finish our resonance bracket here. I put that in. When you're doing this for cations, you're not going to move a positive charge. When you're drawing your arrows, you're showing the movement of electrons. The arrow that I drew over here, let me go ahead and mark it in magenta."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I put that in. When you're doing this for cations, you're not going to move a positive charge. When you're drawing your arrows, you're showing the movement of electrons. The arrow that I drew over here, let me go ahead and mark it in magenta. This arrow in magenta is showing the movement of those electrons in blue. When those electrons in blue move, that creates a plus one formal charge on this carbon. Don't try to move positive charges."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The arrow that I drew over here, let me go ahead and mark it in magenta. This arrow in magenta is showing the movement of those electrons in blue. When those electrons in blue move, that creates a plus one formal charge on this carbon. Don't try to move positive charges. Remember, you're always pushing electrons around. Finally, let's do one more. For this situation, we have a, this is for acetone, we have a carbon right here double bonded to an oxygen."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Don't try to move positive charges. Remember, you're always pushing electrons around. Finally, let's do one more. For this situation, we have a, this is for acetone, we have a carbon right here double bonded to an oxygen. We know that there are differences in electronegativity between carbon and oxygen. Oxygen is more electronegative. What would happen if we took those pi electrons, let me go ahead and highlight those."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "For this situation, we have a, this is for acetone, we have a carbon right here double bonded to an oxygen. We know that there are differences in electronegativity between carbon and oxygen. Oxygen is more electronegative. What would happen if we took those pi electrons, let me go ahead and highlight those. I'm using blue for pi electrons. These pi electrons right here. We move those pi electrons off onto the more electronegative atom like that."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "What would happen if we took those pi electrons, let me go ahead and highlight those. I'm using blue for pi electrons. These pi electrons right here. We move those pi electrons off onto the more electronegative atom like that. Let's go ahead and draw our resonance structure. We would now have this top oxygen would have three lone pairs of electrons. One of those lone pairs are the ones in blue, those pi electrons."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We move those pi electrons off onto the more electronegative atom like that. Let's go ahead and draw our resonance structure. We would now have this top oxygen would have three lone pairs of electrons. One of those lone pairs are the ones in blue, those pi electrons. That's going to give the oxygen a negative one formal charge. We took a bond away from this carbon. We took a bond away from this carbon."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "One of those lone pairs are the ones in blue, those pi electrons. That's going to give the oxygen a negative one formal charge. We took a bond away from this carbon. We took a bond away from this carbon. That's going to give that carbon a plus one formal charge. When you think about your resonance structures, first of all, I should point out that one negative charge and one positive charge give you an overall charge of zero. Charge is conserved."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We took a bond away from this carbon. That's going to give that carbon a plus one formal charge. When you think about your resonance structures, first of all, I should point out that one negative charge and one positive charge give you an overall charge of zero. Charge is conserved. Over here, of course, the charge is zero. You're thinking about the resonance hybrid. We know that both structures contribute to the overall hybrid."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Charge is conserved. Over here, of course, the charge is zero. You're thinking about the resonance hybrid. We know that both structures contribute to the overall hybrid. This one on the right isn't going to contribute as much. This one on the right is pretty minor. That's because you have a positive and a negative charge."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that both structures contribute to the overall hybrid. This one on the right isn't going to contribute as much. This one on the right is pretty minor. That's because you have a positive and a negative charge. The goal, of course, is to get to overall neutral. What's nice about drawing this resonance structure and thinking about this resonance structure is it's emphasizing the difference in electronegativity. For this one, you could just say oxygen gets a partial negative and this carbon right here gets a partial positive."}, {"video_title": "Resonance structure patterns Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's because you have a positive and a negative charge. The goal, of course, is to get to overall neutral. What's nice about drawing this resonance structure and thinking about this resonance structure is it's emphasizing the difference in electronegativity. For this one, you could just say oxygen gets a partial negative and this carbon right here gets a partial positive. That's one way of thinking about it which is very helpful for reactions. Drawing this resonance structure is just another way of thinking about emphasizing the fact that when you're thinking about the hybrid, you're thinking about a little bit more electron density on that oxygen. Once again, do lots of practice."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about what might happen if we had a solution of this carboxylic acid here. We might as well name it just to get some practice. We have 1, 2, 3, 4, 5, 6, 7 carbons. So this is heptan. And then we don't write heptane because this is a carboxylic acid. It is heptanoic acid. Heptanoic acid."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this is heptan. And then we don't write heptane because this is a carboxylic acid. It is heptanoic acid. Heptanoic acid. So let's see what happens if we have heptanoic acid reacting with this is 1, 2 carbons and then it has an OH group. So this is ethanol. That's what the OH group does."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Heptanoic acid. So let's see what happens if we have heptanoic acid reacting with this is 1, 2 carbons and then it has an OH group. So this is ethanol. That's what the OH group does. It mixes in alcohol. And it's in the presence of a sulfuric acid catalyst. This is right here is sulfuric acid."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "That's what the OH group does. It mixes in alcohol. And it's in the presence of a sulfuric acid catalyst. This is right here is sulfuric acid. One of the stronger acids. Sulfuric acid. And I actually draw it structured because I always find it frustrating when people just write the formula here without the actual structure because the structure actually shows you why it's so acidic."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This is right here is sulfuric acid. One of the stronger acids. Sulfuric acid. And I actually draw it structured because I always find it frustrating when people just write the formula here without the actual structure because the structure actually shows you why it's so acidic. So sulfuric acid. Sulfur has 6 valence electrons just like oxygen. So it has a double bond to an oxygen."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And I actually draw it structured because I always find it frustrating when people just write the formula here without the actual structure because the structure actually shows you why it's so acidic. So sulfuric acid. Sulfur has 6 valence electrons just like oxygen. So it has a double bond to an oxygen. Another double bond to an oxygen. And then it has a single bond to an OH group. And then it has another single bond to an OH group."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it has a double bond to an oxygen. Another double bond to an oxygen. And then it has a single bond to an OH group. And then it has another single bond to an OH group. And notice it has 1, 2, 3, 4, 5, 6 valence electrons. Now the reason why this is such a strong acid is that if either of these oxygens take the electron from this proton and it's actually give away the proton to the solution, there's a ton of resonance structures here. And maybe I'll make a whole video on sulfuric acid just to show you all of the resonance structures."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then it has another single bond to an OH group. And notice it has 1, 2, 3, 4, 5, 6 valence electrons. Now the reason why this is such a strong acid is that if either of these oxygens take the electron from this proton and it's actually give away the proton to the solution, there's a ton of resonance structures here. And maybe I'll make a whole video on sulfuric acid just to show you all of the resonance structures. But in general, whenever you see a reaction, when they say it's catalyzed by an acid, all you have to do is realize that it's just going to make the surrounding solution a lot more acidic. Just a ton more acidic. And maybe we're in a solution of ethanol."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And maybe I'll make a whole video on sulfuric acid just to show you all of the resonance structures. But in general, whenever you see a reaction, when they say it's catalyzed by an acid, all you have to do is realize that it's just going to make the surrounding solution a lot more acidic. Just a ton more acidic. And maybe we're in a solution of ethanol. And if we are in a solution of ethanol, it'll just add protons to the ethanol itself. So you can imagine this guy right over here. Let me draw the sulfuric acid a little bit differently."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And maybe we're in a solution of ethanol. And if we are in a solution of ethanol, it'll just add protons to the ethanol itself. So you can imagine this guy right over here. Let me draw the sulfuric acid a little bit differently. Let me draw these oxygen-hydrogen bonds. So you have this oxygen and then it is bonded to a hydrogen there. This is floating around the solution."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw the sulfuric acid a little bit differently. Let me draw these oxygen-hydrogen bonds. So you have this oxygen and then it is bonded to a hydrogen there. This is floating around the solution. You have your ethanol. You have your ethanol that looks like this. So two carbons and then bonded to an oxygen."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This is floating around the solution. You have your ethanol. You have your ethanol that looks like this. So two carbons and then bonded to an oxygen. And then that oxygen is bonded to a hydrogen. The oxygen has two lone pairs just like that. And so this guy really is good at getting rid of the protons."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So two carbons and then bonded to an oxygen. And then that oxygen is bonded to a hydrogen. The oxygen has two lone pairs just like that. And so this guy really is good at getting rid of the protons. So you have the situation where this electron can be taken back by this oxygen and then it can actually be given here and there's all these resonance structures. But it's just very good at taking that electron. And that can happen at the exact same time that one of the ethanols capture the hydrogen proton."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so this guy really is good at getting rid of the protons. So you have the situation where this electron can be taken back by this oxygen and then it can actually be given here and there's all these resonance structures. But it's just very good at taking that electron. And that can happen at the exact same time that one of the ethanols capture the hydrogen proton. At the exact same time that one of these electrons or this oxygen captures that hydrogen proton. And if you just look at this part right here, this will just result in a ton of having these protonated ethanols flying around. Actually, let me draw it over here."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And that can happen at the exact same time that one of the ethanols capture the hydrogen proton. At the exact same time that one of these electrons or this oxygen captures that hydrogen proton. And if you just look at this part right here, this will just result in a ton of having these protonated ethanols flying around. Actually, let me draw it over here. I want to make sure I'm using my screen real estate properly. So this will result in a ton of these protonated ethanols. So one, two carbons."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me draw it over here. I want to make sure I'm using my screen real estate properly. So this will result in a ton of these protonated ethanols. So one, two carbons. It has this original hydrogen over here. But now it has one electron in that original pair. And then it gave the other electron to this hydrogen."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So one, two carbons. It has this original hydrogen over here. But now it has one electron in that original pair. And then it gave the other electron to this hydrogen. So it gave the other electron to that hydrogen nucleus. Let me do it in that same color. To that hydrogen nucleus right there."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then it gave the other electron to this hydrogen. So it gave the other electron to that hydrogen nucleus. Let me do it in that same color. To that hydrogen nucleus right there. So it took a proton. It gave away an electron. It now has a positive charge."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "To that hydrogen nucleus right there. So it took a proton. It gave away an electron. It now has a positive charge. And now what was a sulfuric acid now has a negative charge over here. So if I were to draw it, it would now look like this. Plus sulfur, two double bonds, two oxygens."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It now has a positive charge. And now what was a sulfuric acid now has a negative charge over here. So if I were to draw it, it would now look like this. Plus sulfur, two double bonds, two oxygens. You still have this OH group. And actually, you could still donate this. It's still acidic."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Plus sulfur, two double bonds, two oxygens. You still have this OH group. And actually, you could still donate this. It's still acidic. But now this oxygen right now gained an electron. It now has a negative charge. Now the whole reason I did this is to give you a tangible sense of what sulfuric acid looks like and why it's acidic."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's still acidic. But now this oxygen right now gained an electron. It now has a negative charge. Now the whole reason I did this is to give you a tangible sense of what sulfuric acid looks like and why it's acidic. But really, you just have to kind of internalize that you're just going to have a bunch of hydrogen protons floating around. They could be attached to an ethanol. If this was a water solution, they could create hydronium."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now the whole reason I did this is to give you a tangible sense of what sulfuric acid looks like and why it's acidic. But really, you just have to kind of internalize that you're just going to have a bunch of hydrogen protons floating around. They could be attached to an ethanol. If this was a water solution, they could create hydronium. So you have a bunch of hydrogen protons floating around that will catalyze this reaction. They will be used to facilitate the reaction we're going to explore. And then they will be let go."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "If this was a water solution, they could create hydronium. So you have a bunch of hydrogen protons floating around that will catalyze this reaction. They will be used to facilitate the reaction we're going to explore. And then they will be let go. So hopefully, this right here, if I start involving some of these protonated ethanols in our reaction, you won't view that as a huge stretch of the imagination because they would have gotten protonated by the sulfuric acid. Or if I just actually just grab protons in our reaction, that actually might make it a little bit simpler. So let's start with the actual reaction."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then they will be let go. So hopefully, this right here, if I start involving some of these protonated ethanols in our reaction, you won't view that as a huge stretch of the imagination because they would have gotten protonated by the sulfuric acid. Or if I just actually just grab protons in our reaction, that actually might make it a little bit simpler. So let's start with the actual reaction. So let me redraw my heptanoic acid. We have 1, 2, 3, 4, 5, 6, 7 carbons double bond to an oxygen. And then we have an OH group right over here."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with the actual reaction. So let me redraw my heptanoic acid. We have 1, 2, 3, 4, 5, 6, 7 carbons double bond to an oxygen. And then we have an OH group right over here. Now, the first step of this reaction, this oxygen right here, we have all of these protons floating around. Very acidic environment. We have sulfuric acid there just giving protons away to the ethanol or to other things."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then we have an OH group right over here. Now, the first step of this reaction, this oxygen right here, we have all of these protons floating around. Very acidic environment. We have sulfuric acid there just giving protons away to the ethanol or to other things. So this guy can grab a proton either directly from sulfuric acid or maybe from one of the protonated ethanols. Either one. So we could just draw it like this."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We have sulfuric acid there just giving protons away to the ethanol or to other things. So this guy can grab a proton either directly from sulfuric acid or maybe from one of the protonated ethanols. Either one. So we could just draw it like this. He just grabs a hydrogen proton. The hydrogen proton might have had an electron associated with it. That would then go back to a sulfuric acid."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we could just draw it like this. He just grabs a hydrogen proton. The hydrogen proton might have had an electron associated with it. That would then go back to a sulfuric acid. But just to make things simple, I'll just say grabs a proton from something else. So once he grabs that proton, then it looks like this. So I'll draw my 2, 3, 4, 5, 6, 7 carbons double bond to an oxygen, single bond to an OH."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "That would then go back to a sulfuric acid. But just to make things simple, I'll just say grabs a proton from something else. So once he grabs that proton, then it looks like this. So I'll draw my 2, 3, 4, 5, 6, 7 carbons double bond to an oxygen, single bond to an OH. This oxygen had 2 lone pairs, but now one of the lone pairs is broken up because it gave an electron to this hydrogen proton right over there. Hydrogen proton, it was positively charged. It gained an electron."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw my 2, 3, 4, 5, 6, 7 carbons double bond to an oxygen, single bond to an OH. This oxygen had 2 lone pairs, but now one of the lone pairs is broken up because it gave an electron to this hydrogen proton right over there. Hydrogen proton, it was positively charged. It gained an electron. Now it is neutral. But this oxygen right over here gave away an electron. So it is now positive."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It gained an electron. Now it is neutral. But this oxygen right over here gave away an electron. So it is now positive. Now the next step, we have all of this ethanol floating around. We have all of this ethanol floating around. So let me introduce some ethanol."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it is now positive. Now the next step, we have all of this ethanol floating around. We have all of this ethanol floating around. So let me introduce some ethanol. I'll do this in a different color. So we have this ethanol floating around. This is 1, 2 carbons."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let me introduce some ethanol. I'll do this in a different color. So we have this ethanol floating around. This is 1, 2 carbons. This is our ethanol. You have 2 electron pairs on that oxygen. Then you could imagine, especially because this is now a good leaving group, you could imagine that this acts as a nucleophile."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This is 1, 2 carbons. This is our ethanol. You have 2 electron pairs on that oxygen. Then you could imagine, especially because this is now a good leaving group, you could imagine that this acts as a nucleophile. It would do a nucleophile attack on this carbonyl carbon right here. So what you could imagine is that this electron attacks or gets given to this carbon right here on the carbonyl. At the exact same time that this happens, this oxygen is positively charged."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Then you could imagine, especially because this is now a good leaving group, you could imagine that this acts as a nucleophile. It would do a nucleophile attack on this carbonyl carbon right here. So what you could imagine is that this electron attacks or gets given to this carbon right here on the carbonyl. At the exact same time that this happens, this oxygen is positively charged. It wants an electron. It wants to get neutral again. It's already hogging these electrons."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "At the exact same time that this happens, this oxygen is positively charged. It wants an electron. It wants to get neutral again. It's already hogging these electrons. That's why you have a partial positive charge on this carbon. That's why this guy might be attracted to this. We've seen this in SN2 reactions many, many, many, many times already."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's already hogging these electrons. That's why you have a partial positive charge on this carbon. That's why this guy might be attracted to this. We've seen this in SN2 reactions many, many, many, many times already. So this electron will be taken up by this top oxygen over there. So after that happens, what does our newly formed molecule look like? Let me just scroll down and have some clear space."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We've seen this in SN2 reactions many, many, many, many times already. So this electron will be taken up by this top oxygen over there. So after that happens, what does our newly formed molecule look like? Let me just scroll down and have some clear space. So after that, our newly formed molecule will look like this. It will have 2, 3, 4, 5, 6, 7 carbons to get to the carbonyl carbon. Now it will have a single bond to that oxygen up there."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let me just scroll down and have some clear space. So after that, our newly formed molecule will look like this. It will have 2, 3, 4, 5, 6, 7 carbons to get to the carbonyl carbon. Now it will have a single bond to that oxygen up there. That oxygen had one lone pair already. So it had one lone pair already. It had this bond to this hydrogen that it nabbed from the solution, that proton that it nabbed from the solution."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now it will have a single bond to that oxygen up there. That oxygen had one lone pair already. So it had one lone pair already. It had this bond to this hydrogen that it nabbed from the solution, that proton that it nabbed from the solution. Now it has another pair. It had this electron that was participating in a double bond with this carbonyl carbon. It took the other side of that bond back."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It had this bond to this hydrogen that it nabbed from the solution, that proton that it nabbed from the solution. Now it has another pair. It had this electron that was participating in a double bond with this carbonyl carbon. It took the other side of that bond back. So now it has that electron and the other one that it took from the carbon. So it has 2 lone pairs again. It had a positive charge, but now it took back an electron."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It took the other side of that bond back. So now it has that electron and the other one that it took from the carbon. So it has 2 lone pairs again. It had a positive charge, but now it took back an electron. Now it is neutral. Now the rest of the molecule, we have this OH group right over here. We have that OH group right over there."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It had a positive charge, but now it took back an electron. Now it is neutral. Now the rest of the molecule, we have this OH group right over here. We have that OH group right over there. Now we have the actual ethanol that has attached itself. So it's no longer ethanol. It's attached itself to this larger molecule."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We have that OH group right over there. Now we have the actual ethanol that has attached itself. So it's no longer ethanol. It's attached itself to this larger molecule. This oxygen right over here, it still has one lone pair, but the other lone pair has been broken up and it has given an electron to this carbon. It has given an electron to that carbon. It is now bonded to the larger molecule."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's attached itself to this larger molecule. This oxygen right over here, it still has one lone pair, but the other lone pair has been broken up and it has given an electron to this carbon. It has given an electron to that carbon. It is now bonded to the larger molecule. And then the rest of it, you have these 1, 2 carbons right there. And then you have this hydrogen right over there. Now, the next step of the reaction, remember, we have a bunch of protons floating around."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It is now bonded to the larger molecule. And then the rest of it, you have these 1, 2 carbons right there. And then you have this hydrogen right over there. Now, the next step of the reaction, remember, we have a bunch of protons floating around. This oxygen right here, and actually I should make it clear, these are all kind of reversible reactions, so I actually shouldn't even draw one-way arrows here. A better thing to do, instead of drawing these one-way arrows, is to show that the reaction can actually go in either direction. It's just as likely to go from here to here as it is from there to there."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now, the next step of the reaction, remember, we have a bunch of protons floating around. This oxygen right here, and actually I should make it clear, these are all kind of reversible reactions, so I actually shouldn't even draw one-way arrows here. A better thing to do, instead of drawing these one-way arrows, is to show that the reaction can actually go in either direction. It's just as likely to go from here to here as it is from there to there. So let me draw that. These are kind of in equilibrium. These are in equilibrium with each other."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's just as likely to go from here to here as it is from there to there. So let me draw that. These are kind of in equilibrium. These are in equilibrium with each other. So you can imagine the next thing that could happen is another oxygen could grab a proton from the medium. And actually, before I do that, let me make sure this guy can lose a proton to the medium. And actually, this guy could take the proton from that guy, but I won't do it that way."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "These are in equilibrium with each other. So you can imagine the next thing that could happen is another oxygen could grab a proton from the medium. And actually, before I do that, let me make sure this guy can lose a proton to the medium. And actually, this guy could take the proton from that guy, but I won't do it that way. So you could imagine a situation where you have this proton just jumps off. This guy right here has a, oh, and I should have pointed it out. This guy gave away an electron to this carbonyl carbon."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And actually, this guy could take the proton from that guy, but I won't do it that way. So you could imagine a situation where you have this proton just jumps off. This guy right here has a, oh, and I should have pointed it out. This guy gave away an electron to this carbonyl carbon. So this right here has a positive charge. And in general, the oxygen, very electronegative, it's going to be hogging the electrons of this proton. So it can take them away."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This guy gave away an electron to this carbonyl carbon. So this right here has a positive charge. And in general, the oxygen, very electronegative, it's going to be hogging the electrons of this proton. So it can take them away. So, so far we gained a proton, and now we can give back a proton. So this electron right here can go back to this oxygen, making it neutral. And the proton can be picked up by anything."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it can take them away. So, so far we gained a proton, and now we can give back a proton. So this electron right here can go back to this oxygen, making it neutral. And the proton can be picked up by anything. It could be picked up by another molecule of heptanoic acid to do this first protonation we saw. It could be picked up by another molecule of ethanol. Let me draw it like that."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And the proton can be picked up by anything. It could be picked up by another molecule of heptanoic acid to do this first protonation we saw. It could be picked up by another molecule of ethanol. Let me draw it like that. Let me draw it as getting picked up by another molecule of ethanol. It just gets thrown back in the solution. So this guy gives an electron to the proton, but the end result is the hydrogen leaves, the electron goes back to the oxygen."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw it like that. Let me draw it as getting picked up by another molecule of ethanol. It just gets thrown back in the solution. So this guy gives an electron to the proton, but the end result is the hydrogen leaves, the electron goes back to the oxygen. It now has a neutral charge. So, and once again, these are all reversible reactions, can go in either direction. But we're now over here, and let me redraw my heptanoic acid."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this guy gives an electron to the proton, but the end result is the hydrogen leaves, the electron goes back to the oxygen. It now has a neutral charge. So, and once again, these are all reversible reactions, can go in either direction. But we're now over here, and let me redraw my heptanoic acid. 2, 3, 4, 5, 6, 7, double, actually single bond now to the oxygen, single bond to this OH. And now we have this bond right over here to this now deprotonated, what was ethanol. So we have O, and then we have a 1, 2 carbons just like that."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But we're now over here, and let me redraw my heptanoic acid. 2, 3, 4, 5, 6, 7, double, actually single bond now to the oxygen, single bond to this OH. And now we have this bond right over here to this now deprotonated, what was ethanol. So we have O, and then we have a 1, 2 carbons just like that. And this guy has a hydrogen attached to it. This guy right over here has a hydrogen, and I won't even bother to draw this protonated ethanol. The protons are flying around everywhere."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we have O, and then we have a 1, 2 carbons just like that. And this guy has a hydrogen attached to it. This guy right over here has a hydrogen, and I won't even bother to draw this protonated ethanol. The protons are flying around everywhere. Now the next step, this guy, this OH group, especially the oxygen in it, he could grab a proton from the surrounding solution. So he's got these two lone pairs. He can grab a proton from one of these protonated ethanol, from the sulfuric acid, or maybe from one of these other kind of intermediate molecules, from anywhere."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "The protons are flying around everywhere. Now the next step, this guy, this OH group, especially the oxygen in it, he could grab a proton from the surrounding solution. So he's got these two lone pairs. He can grab a proton from one of these protonated ethanol, from the sulfuric acid, or maybe from one of these other kind of intermediate molecules, from anywhere. So he could grab, that's the whole point of having this acid catalyst. So he could donate an electron to a proton and then form a bond with it. And so if that, let me do that in a different, I'll keep it in that color."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "He can grab a proton from one of these protonated ethanol, from the sulfuric acid, or maybe from one of these other kind of intermediate molecules, from anywhere. So he could grab, that's the whole point of having this acid catalyst. So he could donate an electron to a proton and then form a bond with it. And so if that, let me do that in a different, I'll keep it in that color. And so if that happened, then the next step in our reaction, and remember these can all go in either direction, the next step in our reaction will look like this. You would have your 2, 3, 4, 5, 6, 7 carbons, single bond to an oxygen, single bond to an OH, and then you have your bond to this oxygen, which is bond bound to 2 carbons. One, two carbons, just like that."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so if that, let me do that in a different, I'll keep it in that color. And so if that happened, then the next step in our reaction, and remember these can all go in either direction, the next step in our reaction will look like this. You would have your 2, 3, 4, 5, 6, 7 carbons, single bond to an oxygen, single bond to an OH, and then you have your bond to this oxygen, which is bond bound to 2 carbons. One, two carbons, just like that. And this guy on top is bonded to an oxygen. He's got two lone pairs. And this guy over here grabbed a proton."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "One, two carbons, just like that. And this guy on top is bonded to an oxygen. He's got two lone pairs. And this guy over here grabbed a proton. He gave an electron to a hydrogen proton. So now he gave an electron, he had two, but now he gave one of them to this proton. And so he gave it to that hydrogen."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this guy over here grabbed a proton. He gave an electron to a hydrogen proton. So now he gave an electron, he had two, but now he gave one of them to this proton. And so he gave it to that hydrogen. That hydrogen is now neutral, gained an electron. This oxygen is now positive because it gave away an electron. It still has this other lone pair over here."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so he gave it to that hydrogen. That hydrogen is now neutral, gained an electron. This oxygen is now positive because it gave away an electron. It still has this other lone pair over here. It still has this other lone pair. It is now positive. And frankly, it is now a good leaving group."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It still has this other lone pair over here. It still has this other lone pair. It is now positive. And frankly, it is now a good leaving group. So in the next step, you could have someone else. Remember, other people need protons earlier in this reaction. So this proton might get lost maybe by one of the other ethanol molecules."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And frankly, it is now a good leaving group. So in the next step, you could have someone else. Remember, other people need protons earlier in this reaction. So this proton might get lost maybe by one of the other ethanol molecules. So let me draw that. Let me do a color I haven't used yet. I'll use orange."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this proton might get lost maybe by one of the other ethanol molecules. So let me draw that. Let me do a color I haven't used yet. I'll use orange. So maybe one of the other ethanol molecules or one of the other intermediaries in this whole reaction. So ethanol, I'll just do ethanol because it's easier to draw. It might give an electron to just the nucleus."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "I'll use orange. So maybe one of the other ethanol molecules or one of the other intermediaries in this whole reaction. So ethanol, I'll just do ethanol because it's easier to draw. It might give an electron to just the nucleus. And then this guy can take the electron back. This guy can take that hydrogen's electron back, give it to this carbon that was the carbonyl carbon several steps ago. And then since he's got that electron, he can then give back this electron to this what was an OH group, but now it got another hydrogen there, so he can then give it back to him."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It might give an electron to just the nucleus. And then this guy can take the electron back. This guy can take that hydrogen's electron back, give it to this carbon that was the carbonyl carbon several steps ago. And then since he's got that electron, he can then give back this electron to this what was an OH group, but now it got another hydrogen there, so he can then give it back to him. And then the resulting products will look like this. And this is all in equilibrium. We now have 2, 3, 4, 5, 6, 7 carbons."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then since he's got that electron, he can then give back this electron to this what was an OH group, but now it got another hydrogen there, so he can then give it back to him. And then the resulting products will look like this. And this is all in equilibrium. We now have 2, 3, 4, 5, 6, 7 carbons. Now this guy has a double bond again. So you have an oxygen right there. It is now a double bond."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We now have 2, 3, 4, 5, 6, 7 carbons. Now this guy has a double bond again. So you have an oxygen right there. It is now a double bond. I'll draw this newly formed double bond in magenta. This guy has left as water, so I can just draw that. So now you have this other OH group that is bonded to this hydrogen over here."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It is now a double bond. I'll draw this newly formed double bond in magenta. This guy has left as water, so I can just draw that. So now you have this other OH group that is bonded to this hydrogen over here. He is now left as water. And now you have this other thing over here. You have what was that ethanol group has now attached itself."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So now you have this other OH group that is bonded to this hydrogen over here. He is now left as water. And now you have this other thing over here. You have what was that ethanol group has now attached itself. That ethanol has lost its hydrogen, it's attached itself to what was a carboxylic acid, so now it looks like this. It now is bonded to an oxygen and is bonded to one, two carbons just like that. And this whole reaction that I've showed you is called the esterification, one of those words that I have trouble saying."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "You have what was that ethanol group has now attached itself. That ethanol has lost its hydrogen, it's attached itself to what was a carboxylic acid, so now it looks like this. It now is bonded to an oxygen and is bonded to one, two carbons just like that. And this whole reaction that I've showed you is called the esterification, one of those words that I have trouble saying. This one in particular is called the Fischer esterification. So this is the Fischer, and he won the Nobel Prize in 1902, actually generally for his work in organic chemistry, but this is the Fischer esterification. And the reason why it's called that is we started with the carboxylic acid, we started with the heptanoic acid, and now we have an ester."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this whole reaction that I've showed you is called the esterification, one of those words that I have trouble saying. This one in particular is called the Fischer esterification. So this is the Fischer, and he won the Nobel Prize in 1902, actually generally for his work in organic chemistry, but this is the Fischer esterification. And the reason why it's called that is we started with the carboxylic acid, we started with the heptanoic acid, and now we have an ester. An ester is something that instead of an OH group, like you have in a carboxylic acid, you have an OR, you have an oxygen with an actual alkyl group attached to it. And this ester right here, and we'll probably talk more about esters in future videos, this ester right here, just to give you a hint of how to name it, we first start with the group that's attached to this oxygen right here. There are two carbons over there, so we'll use ethyl to name that over there."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why it's called that is we started with the carboxylic acid, we started with the heptanoic acid, and now we have an ester. An ester is something that instead of an OH group, like you have in a carboxylic acid, you have an OR, you have an oxygen with an actual alkyl group attached to it. And this ester right here, and we'll probably talk more about esters in future videos, this ester right here, just to give you a hint of how to name it, we first start with the group that's attached to this oxygen right here. There are two carbons over there, so we'll use ethyl to name that over there. And then we have the rest of this, and we already know that that is one, two, three, four, five, six, seven carbons, including the carbon in the carbonyl group. And so that is heptan, and you would be tempted to call it heptanoic acid, but it is no longer a carboxylic acid. It is now an ester."}, {"video_title": "Fischer esterification Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "There are two carbons over there, so we'll use ethyl to name that over there. And then we have the rest of this, and we already know that that is one, two, three, four, five, six, seven carbons, including the carbon in the carbonyl group. And so that is heptan, and you would be tempted to call it heptanoic acid, but it is no longer a carboxylic acid. It is now an ester. So you call it heptanoate. And this is what tells you that you are dealing with an ester, and this tells you what's on the other side of the oxygen in the ester. This is telling you how many carbons you have attached to, I guess you could kind of view it as the carbonyl chain of the actual ester."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "A meso compound is a compound that has chirality centers but is achiral. We're gonna come back to this definition in a few minutes. Right now, let's focus in on this drawing. And our goal is to draw all possible stereoisomers for this dot structure. So we know from earlier videos that this carbon is a chirality center and so is this one. And we would expect two to the n stereoisomers where n is the number of chirality centers. And since we have two chiral centers here, we would expect two to the second power or four stereoisomers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And our goal is to draw all possible stereoisomers for this dot structure. So we know from earlier videos that this carbon is a chirality center and so is this one. And we would expect two to the n stereoisomers where n is the number of chirality centers. And since we have two chiral centers here, we would expect two to the second power or four stereoisomers. So this is really just a maximum number. So I'm gonna put a question mark right here. So do we get four stereoisomers?"}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And since we have two chiral centers here, we would expect two to the second power or four stereoisomers. So this is really just a maximum number. So I'm gonna put a question mark right here. So do we get four stereoisomers? Well, let's draw out the four possibilities. Our first stereoisomer could have both bromines coming out at us in space. So let me go ahead and draw that in."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So do we get four stereoisomers? Well, let's draw out the four possibilities. Our first stereoisomer could have both bromines coming out at us in space. So let me go ahead and draw that in. So we could have both bromines coming out at us. For the second possibility, we might have both bromines going away from us in space. So I'll draw that in there."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that in. So we could have both bromines coming out at us. For the second possibility, we might have both bromines going away from us in space. So I'll draw that in there. For the third possibility, we could have one bromine up and one bromine down. So I'll put those in. And for the fourth, just reverse them."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw that in there. For the third possibility, we could have one bromine up and one bromine down. So I'll put those in. And for the fourth, just reverse them. Have the top bromine down and the bottom bromine up. Like that. Alright, let's examine the relationships between our stereoisomers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And for the fourth, just reverse them. Have the top bromine down and the bottom bromine up. Like that. Alright, let's examine the relationships between our stereoisomers. And let's start with stereoisomer possibility three and four. So let's compare these and let's figure out the relationship. On the left is stereoisomer three."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Alright, let's examine the relationships between our stereoisomers. And let's start with stereoisomer possibility three and four. So let's compare these and let's figure out the relationship. On the left is stereoisomer three. We can see there's a bromine coming out at us and a bromine going away from us. On the right is stereoisomer four. Now we have a bromine going away from us at the top carbon and a bromine coming out at us here."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left is stereoisomer three. We can see there's a bromine coming out at us and a bromine going away from us. On the right is stereoisomer four. Now we have a bromine going away from us at the top carbon and a bromine coming out at us here. So let's compare our two stereoisomers. If I rotate the one on the right, we can see these are mirror images of each other. And if I try to superimpose one on top of the other, here we get one pair of bromines to line up but the other pair doesn't match."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now we have a bromine going away from us at the top carbon and a bromine coming out at us here. So let's compare our two stereoisomers. If I rotate the one on the right, we can see these are mirror images of each other. And if I try to superimpose one on top of the other, here we get one pair of bromines to line up but the other pair doesn't match. If we try to get the other pair of bromines to line up, now the first pair doesn't match. So these are non-superimposable mirror images of each other. These are enantiomers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if I try to superimpose one on top of the other, here we get one pair of bromines to line up but the other pair doesn't match. If we try to get the other pair of bromines to line up, now the first pair doesn't match. So these are non-superimposable mirror images of each other. These are enantiomers. So three and four are enantiomers. They are non-superimposable mirror images. And we could have guessed that by looking at the drawing because at this carbon, we have bromine coming out at us in space."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are enantiomers. So three and four are enantiomers. They are non-superimposable mirror images. And we could have guessed that by looking at the drawing because at this carbon, we have bromine coming out at us in space. And then now we have bromine going away from us in space for the other stereoisomer. So that's an opposite configuration at that carbon. And at this carbon, we go from a dash to a wedge."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we could have guessed that by looking at the drawing because at this carbon, we have bromine coming out at us in space. And then now we have bromine going away from us in space for the other stereoisomer. So that's an opposite configuration at that carbon. And at this carbon, we go from a dash to a wedge. So that's an opposite configuration at this one too. So since we have opposite configurations at all chirality centers, we would expect these two to be enantiomers of each other. What about the relationship between one and two?"}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And at this carbon, we go from a dash to a wedge. So that's an opposite configuration at this one too. So since we have opposite configurations at all chirality centers, we would expect these two to be enantiomers of each other. What about the relationship between one and two? Well, at first we might say, oh, those are enantiomers because here we have a wedge and then over here we have a dash and here we have a wedge and here we have a dash. So that should be the opposite configuration at both chirality centers. So those might be enantiomers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "What about the relationship between one and two? Well, at first we might say, oh, those are enantiomers because here we have a wedge and then over here we have a dash and here we have a wedge and here we have a dash. So that should be the opposite configuration at both chirality centers. So those might be enantiomers. But let's go to the video to see if that's true. On the left is a model of drawing one with the two bromines coming out at us in space. On the right is a model of drawing two with the two bromines going away from us in space."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So those might be enantiomers. But let's go to the video to see if that's true. On the left is a model of drawing one with the two bromines coming out at us in space. On the right is a model of drawing two with the two bromines going away from us in space. And if I rotate the model on the right, we can see that these are mirror images of each other. But they are superimposable mirror images. If I put that one on top of the other, you'll see that they are superimposable."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the right is a model of drawing two with the two bromines going away from us in space. And if I rotate the model on the right, we can see that these are mirror images of each other. But they are superimposable mirror images. If I put that one on top of the other, you'll see that they are superimposable. So these actually are two models of the same molecule. This is a meso compound. It's a compound that has chirality centers, but it is achiral."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If I put that one on top of the other, you'll see that they are superimposable. So these actually are two models of the same molecule. This is a meso compound. It's a compound that has chirality centers, but it is achiral. The mirror image is superimposable. So one and two really represent the same molecule. This is a meso compound, a compound that has chirality centers, but is achiral."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's a compound that has chirality centers, but it is achiral. The mirror image is superimposable. So one and two really represent the same molecule. This is a meso compound, a compound that has chirality centers, but is achiral. The mirror image is superimposable on itself. So we thought we would have four stereoisomers, but really we only have three. We have a pair of enantiomers and we have one meso compound."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a meso compound, a compound that has chirality centers, but is achiral. The mirror image is superimposable on itself. So we thought we would have four stereoisomers, but really we only have three. We have a pair of enantiomers and we have one meso compound. So to look for a meso compound, one thing you could do is what we did in the video. We had the mirror image and we were able to superimpose the mirror image on itself. Another way to look for a meso compound is to look for a plane of symmetry."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have a pair of enantiomers and we have one meso compound. So to look for a meso compound, one thing you could do is what we did in the video. We had the mirror image and we were able to superimpose the mirror image on itself. Another way to look for a meso compound is to look for a plane of symmetry. So if I draw a line here, think about this as being a plane and look for symmetry on either side. So you can see it's symmetrical. I drew in the plane of symmetry with a dashed line here, but it's hard to visualize it."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Another way to look for a meso compound is to look for a plane of symmetry. So if I draw a line here, think about this as being a plane and look for symmetry on either side. So you can see it's symmetrical. I drew in the plane of symmetry with a dashed line here, but it's hard to visualize it. So up here is a better picture. Here you can see the plane dividing the molecule in half. And on the left side we have our bonds here and then we have our bromine going up and our hydrogen going down."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I drew in the plane of symmetry with a dashed line here, but it's hard to visualize it. So up here is a better picture. Here you can see the plane dividing the molecule in half. And on the left side we have our bonds here and then we have our bromine going up and our hydrogen going down. The right side is symmetrical with the left side. So look for symmetry on both sides of the plane. Let's do another example."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And on the left side we have our bonds here and then we have our bromine going up and our hydrogen going down. The right side is symmetrical with the left side. So look for symmetry on both sides of the plane. Let's do another example. This one's a little bit harder than the last one. We know that we have two chiral centers. So that's a chiral center and so is this one."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another example. This one's a little bit harder than the last one. We know that we have two chiral centers. So that's a chiral center and so is this one. So we would expect two to the second stereoisomers. So that's of course four. So I'll put a question mark here again because we're not sure if we actually will get four stereoisomers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's a chiral center and so is this one. So we would expect two to the second stereoisomers. So that's of course four. So I'll put a question mark here again because we're not sure if we actually will get four stereoisomers. That's a maximum number. Down here I have the four possibilities. So I've drawn them out just to save some time."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'll put a question mark here again because we're not sure if we actually will get four stereoisomers. That's a maximum number. Down here I have the four possibilities. So I've drawn them out just to save some time. So we have one, two, three, and four. And let's examine the relationship between one and two first. On the left is stereoisomer one."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I've drawn them out just to save some time. So we have one, two, three, and four. And let's examine the relationship between one and two first. On the left is stereoisomer one. And I've left the hydrogens off the methyl groups and the OH just so we can see the models better. So here's our carbon chain and we have both OHs coming out at us in space. And then for stereoisomer two, here's the carbon chain and we have both OHs going away from us."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left is stereoisomer one. And I've left the hydrogens off the methyl groups and the OH just so we can see the models better. So here's our carbon chain and we have both OHs coming out at us in space. And then for stereoisomer two, here's the carbon chain and we have both OHs going away from us. So I'm gonna hold them in the way they are in the drawing and I'm gonna rotate the one on the right. And when I do that, we can see that these are mirror images of each other. But if I try to superimpose one on the other, so I'll just rotate this one back here and flip it over, you can see they don't match up."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then for stereoisomer two, here's the carbon chain and we have both OHs going away from us. So I'm gonna hold them in the way they are in the drawing and I'm gonna rotate the one on the right. And when I do that, we can see that these are mirror images of each other. But if I try to superimpose one on the other, so I'll just rotate this one back here and flip it over, you can see they don't match up. So the atoms don't line up here. So they're non-superimposable. Doesn't matter how you do it."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But if I try to superimpose one on the other, so I'll just rotate this one back here and flip it over, you can see they don't match up. So the atoms don't line up here. So they're non-superimposable. Doesn't matter how you do it. I'll rotate it again and we can see we can't superimpose our atoms. So these are non-superimposable mirror images of each other. These are enantiomers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Doesn't matter how you do it. I'll rotate it again and we can see we can't superimpose our atoms. So these are non-superimposable mirror images of each other. These are enantiomers. And we could check for a plane of symmetry. So I could take one of these and I could rotate it so we have our hydrogens going away from us in space. And I look for a plane of symmetry but I don't see one."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are enantiomers. And we could check for a plane of symmetry. So I could take one of these and I could rotate it so we have our hydrogens going away from us in space. And I look for a plane of symmetry but I don't see one. So this is not a meso compound. So one and two are enantiomers. They're non-superimposable mirror images of each other."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I look for a plane of symmetry but I don't see one. So this is not a meso compound. So one and two are enantiomers. They're non-superimposable mirror images of each other. And we could have guessed that because if we look at our chiral centers here, so this one has an OH coming out at us, and then that one has it going away from us. This one has an OH coming out at us and this one has it going away from us. We have opposite configurations at both chirality centers."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "They're non-superimposable mirror images of each other. And we could have guessed that because if we look at our chiral centers here, so this one has an OH coming out at us, and then that one has it going away from us. This one has an OH coming out at us and this one has it going away from us. We have opposite configurations at both chirality centers. Let's look at three and four next. So what is the relationship between these two? Well, at first we might think these could be enantiomers because at this carbon we have OH on a wedge and then here we have OH on a dash."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have opposite configurations at both chirality centers. Let's look at three and four next. So what is the relationship between these two? Well, at first we might think these could be enantiomers because at this carbon we have OH on a wedge and then here we have OH on a dash. And then here we have OH on a dash and here we have it on a wedge. So that might be your first guess. But let's look at the video and let's look at the model sets to help us out."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, at first we might think these could be enantiomers because at this carbon we have OH on a wedge and then here we have OH on a dash. And then here we have OH on a dash and here we have it on a wedge. So that might be your first guess. But let's look at the video and let's look at the model sets to help us out. Remember, I'm leaving the hydrogens off the methyl groups and the hydrogens off the oxygens in the video just to help us see the molecule more clearly. On the left we have a model of drawing three. So here's our carbon chain with an OH going away from us in space and an OH coming out at us in space."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But let's look at the video and let's look at the model sets to help us out. Remember, I'm leaving the hydrogens off the methyl groups and the hydrogens off the oxygens in the video just to help us see the molecule more clearly. On the left we have a model of drawing three. So here's our carbon chain with an OH going away from us in space and an OH coming out at us in space. On the right is a model of drawing four. Here's our carbon chain with an OH coming out at us in space and an OH going away from us in space. So I'll hold the two models and we'll compare them."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's our carbon chain with an OH going away from us in space and an OH coming out at us in space. On the right is a model of drawing four. Here's our carbon chain with an OH coming out at us in space and an OH going away from us in space. So I'll hold the two models and we'll compare them. First let's see if they are mirror images of each other. So I'll take the one on the right and I'll rotate it and I'll hold it up next to the one on the left. And now we can see that these two are mirror images of each other."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'll hold the two models and we'll compare them. First let's see if they are mirror images of each other. So I'll take the one on the right and I'll rotate it and I'll hold it up next to the one on the left. And now we can see that these two are mirror images of each other. So next let's see if one is superimposable on the other. So I'll go back to the starting point and I'll rotate it around like that. And let's see if we can superimpose the one on the right, the mirror image, on the molecule on the left."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And now we can see that these two are mirror images of each other. So next let's see if one is superimposable on the other. So I'll go back to the starting point and I'll rotate it around like that. And let's see if we can superimpose the one on the right, the mirror image, on the molecule on the left. And notice that we can. All of the atoms line up. So all of the hydrogens, carbons, and oxygens are in the same place."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's see if we can superimpose the one on the right, the mirror image, on the molecule on the left. And notice that we can. All of the atoms line up. So all of the hydrogens, carbons, and oxygens are in the same place. So this is a compound that has chirality centers but it is achiral. The mirror image is superimposable on itself. So we should be able to find a plane of symmetry."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So all of the hydrogens, carbons, and oxygens are in the same place. So this is a compound that has chirality centers but it is achiral. The mirror image is superimposable on itself. So we should be able to find a plane of symmetry. So I'll just pick one of these models. It doesn't matter which one because they represent the same compound. And I'll rotate."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we should be able to find a plane of symmetry. So I'll just pick one of these models. It doesn't matter which one because they represent the same compound. And I'll rotate. I'll rotate around so we can see a plane of symmetry. So right there is our plane of symmetry. This is a meso compound."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I'll rotate. I'll rotate around so we can see a plane of symmetry. So right there is our plane of symmetry. This is a meso compound. So three and four actually represent the same compound. So this is one meso compound. So these two are the same and we have one meso compound."}, {"video_title": "Meso compounds Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a meso compound. So three and four actually represent the same compound. So this is one meso compound. So these two are the same and we have one meso compound. It's not really obvious looking at these bond line structures that these represent the same molecule. So definitely get a model set and try this out for yourself. So we thought there might be four stereoisomers but actually there are only three."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That these electrons in these pi orbitals that form these double bonds, they're actually just not in this double bond, they can keep swapping, this one can go here, this one can go there, that one can go there. Actually, they don't go back and forth, they actually just completely go around the entire ring. When a molecule is aromatic, it stabilizes it. But we've seen examples of aromatic, or actually in particular, we've seen examples of benzene rings that have other things bumping off of them, whether they're halides or whether they're OH groups. What we want to do in this video is think about how that might happen. How do things get added on to a benzene ring? We're going to learn about electrophilic aromatic substitution."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But we've seen examples of aromatic, or actually in particular, we've seen examples of benzene rings that have other things bumping off of them, whether they're halides or whether they're OH groups. What we want to do in this video is think about how that might happen. How do things get added on to a benzene ring? We're going to learn about electrophilic aromatic substitution. Let me write that down. Electrophilic aromatic substitution. You might say, well Sal, you just said you're adding things to the ring, but the reality is that there's six hydrogens here."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We're going to learn about electrophilic aromatic substitution. Let me write that down. Electrophilic aromatic substitution. You might say, well Sal, you just said you're adding things to the ring, but the reality is that there's six hydrogens here. There's one hydrogen, two hydrogens, three hydrogens, four hydrogens, five hydrogens, and six hydrogens. They're always there. If you don't draw them, they are implicitly there."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You might say, well Sal, you just said you're adding things to the ring, but the reality is that there's six hydrogens here. There's one hydrogen, two hydrogens, three hydrogens, four hydrogens, five hydrogens, and six hydrogens. They're always there. If you don't draw them, they are implicitly there. What we're actually doing, when you add a chlorine or a bromine or an OH group, it's actually replacing one of these hydrogens. That's why it is substitution. It's aromatic because we're dealing with a benzene ring."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If you don't draw them, they are implicitly there. What we're actually doing, when you add a chlorine or a bromine or an OH group, it's actually replacing one of these hydrogens. That's why it is substitution. It's aromatic because we're dealing with a benzene ring. We're dealing with an aromatic molecule. We're going to see that we need a really strong electrophile in order to do this. Let's think about how this will happen."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's aromatic because we're dealing with a benzene ring. We're dealing with an aromatic molecule. We're going to see that we need a really strong electrophile in order to do this. Let's think about how this will happen. Before I do that, let me just copy and paste this because I don't want to have to redraw this. Let me just copy it just like that. Let's say we have a really strong electrophile."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about how this will happen. Before I do that, let me just copy and paste this because I don't want to have to redraw this. Let me just copy it just like that. Let's say we have a really strong electrophile. I'll give you particular cases in the next few videos so you can better visualize what a really strong electrophile is. Just from the word itself, electrophile, you could imagine it's something that loves electrons. It wants electrons really, really, really, really, really badly."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we have a really strong electrophile. I'll give you particular cases in the next few videos so you can better visualize what a really strong electrophile is. Just from the word itself, electrophile, you could imagine it's something that loves electrons. It wants electrons really, really, really, really, really badly. Usually, it has a positive charge. It wants electrons badly. Actually, let me make it very clear."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It wants electrons really, really, really, really, really badly. Usually, it has a positive charge. It wants electrons badly. Actually, let me make it very clear. Instead of saying wants electrons badly, because when you're talking about electrophiles or nucleophiles, you're actually talking about how good something is reacting. You're not actually talking about the actual energies involved. Let me put it a different way."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me make it very clear. Instead of saying wants electrons badly, because when you're talking about electrophiles or nucleophiles, you're actually talking about how good something is reacting. You're not actually talking about the actual energies involved. Let me put it a different way. Good at getting electrons. Really, really, really, really, really good at getting electrons. What would happen?"}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me put it a different way. Good at getting electrons. Really, really, really, really, really good at getting electrons. What would happen? We already said this is already pretty stable. These electrons, these pi electrons can circulate all around. If it bumps just in the right way to something that's really good at getting electrons, what might happen, let's say we have this electron right here."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "What would happen? We already said this is already pretty stable. These electrons, these pi electrons can circulate all around. If it bumps just in the right way to something that's really good at getting electrons, what might happen, let's say we have this electron right here. The way we've drawn it, it's on this carbon right here. Obviously, the carbon is just at the intersection. I never drew the carbon."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If it bumps just in the right way to something that's really good at getting electrons, what might happen, let's say we have this electron right here. The way we've drawn it, it's on this carbon right here. Obviously, the carbon is just at the intersection. I never drew the carbon. But if this electrophile, which is really good at getting electrons, bumps in just the right way, this electron can go to that electrophile. Then what would be left with? Let me copy and paste our original molecule."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I never drew the carbon. But if this electrophile, which is really good at getting electrons, bumps in just the right way, this electron can go to that electrophile. Then what would be left with? Let me copy and paste our original molecule. Then what would we be left with? We no longer have this bond right here. It has now been bonded to the electrophile."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me copy and paste our original molecule. Then what would we be left with? We no longer have this bond right here. It has now been bonded to the electrophile. Let me make it clear. We had this electron right here. That electron is still on this carbon right over here at this intersection."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It has now been bonded to the electrophile. Let me make it clear. We had this electron right here. That electron is still on this carbon right over here at this intersection. But the other end of it, the other electron has now been given to the electrophile, the thing that's good at getting them. The other side has been given to this electrophile. This electrophile now gained an electron, so it had a positive charge."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That electron is still on this carbon right over here at this intersection. But the other end of it, the other electron has now been given to the electrophile, the thing that's good at getting them. The other side has been given to this electrophile. This electrophile now gained an electron, so it had a positive charge. Now it will be neutral. Once again, I'll show you several particular cases of this in the next few videos. Let me just make it clear."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This electrophile now gained an electron, so it had a positive charge. Now it will be neutral. Once again, I'll show you several particular cases of this in the next few videos. Let me just make it clear. This bond, you can now view it as being this bond. This carbon right over here, this lost an electron. If it lost an electron, it will now have a positive charge."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me just make it clear. This bond, you can now view it as being this bond. This carbon right over here, this lost an electron. If it lost an electron, it will now have a positive charge. This is hard to do to a resident stabilized molecule, to a benzene ring. Once again, I said I'm being a little bit repetitive. This has to be a very good electrophile to do it."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If it lost an electron, it will now have a positive charge. This is hard to do to a resident stabilized molecule, to a benzene ring. Once again, I said I'm being a little bit repetitive. This has to be a very good electrophile to do it. Once this is there, this is actually a relatively stable carbocation. The reason why it is, it's only a secondary carbocation, but it's actually a resident stabilized carbocation. Because this guy can go, this electron right here can be given to that."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This has to be a very good electrophile to do it. Once this is there, this is actually a relatively stable carbocation. The reason why it is, it's only a secondary carbocation, but it's actually a resident stabilized carbocation. Because this guy can go, this electron right here can be given to that. If this electron goes there, then it would look like this. Let me redraw it. I'll draw the resident structures quickly."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Because this guy can go, this electron right here can be given to that. If this electron goes there, then it would look like this. Let me redraw it. I'll draw the resident structures quickly. You have your hydrogen, you have your electrophile. It's not an electrophile anymore, but you have that E that's now been added. You have that hydrogen, you have a double bond here."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'll draw the resident structures quickly. You have your hydrogen, you have your electrophile. It's not an electrophile anymore, but you have that E that's now been added. You have that hydrogen, you have a double bond here. Let me make that, draw a little bit neater. You have this hydrogen, you have this hydrogen, this hydrogen, and this hydrogen. What I said is, this is stabilized."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You have that hydrogen, you have a double bond here. Let me make that, draw a little bit neater. You have this hydrogen, you have this hydrogen, this hydrogen, and this hydrogen. What I said is, this is stabilized. An electron here can actually jump over here. If this electron jumps over here, the double bond is now over there. If that goes over there like that, the double bond is now over here."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "What I said is, this is stabilized. An electron here can actually jump over here. If this electron jumps over here, the double bond is now over there. If that goes over there like that, the double bond is now over here. Now this guy lost his electron and it would have a positive charge. Then that is resident stabilized. It can either go back to this guy, or this electron over here can jump over there."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If that goes over there like that, the double bond is now over here. Now this guy lost his electron and it would have a positive charge. Then that is resident stabilized. It can either go back to this guy, or this electron over here can jump over there. Let me redraw the whole thing over again. Let me draw all the hydrogens. This right here, you have the E and the hydrogen."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It can either go back to this guy, or this electron over here can jump over there. Let me redraw the whole thing over again. Let me draw all the hydrogens. This right here, you have the E and the hydrogen. You have a hydrogen here, hydrogen here, hydrogen here, hydrogen here. Normally you don't worry about the hydrogens, but one of the hydrogens is going to be nabbed later on in this mechanism. I want to draw the hydrogens just so you know that they are there."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This right here, you have the E and the hydrogen. You have a hydrogen here, hydrogen here, hydrogen here, hydrogen here. Normally you don't worry about the hydrogens, but one of the hydrogens is going to be nabbed later on in this mechanism. I want to draw the hydrogens just so you know that they are there. As I said, this is resident stabilized. If this electron right here jumps over there, then this double bond is now this double bond. Now this guy over here lost an electron, so it would have a positive charge."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I want to draw the hydrogens just so you know that they are there. As I said, this is resident stabilized. If this electron right here jumps over there, then this double bond is now this double bond. Now this guy over here lost an electron, so it would have a positive charge. Again, once you had this double bond up here, this double bond up there is that double bond. We can go back and forth between these. The electrons are just swishing around the ring."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now this guy over here lost an electron, so it would have a positive charge. Again, once you had this double bond up here, this double bond up there is that double bond. We can go back and forth between these. The electrons are just swishing around the ring. It's not going to be maybe as great as the situation that we had when we had a nice benzene ring that was completely aromatic. The electrons could just go around the p orbitals, round and round the ring, stabilize the structure. But this is still a relatively stable carbocation because the electrons can move around."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The electrons are just swishing around the ring. It's not going to be maybe as great as the situation that we had when we had a nice benzene ring that was completely aromatic. The electrons could just go around the p orbitals, round and round the ring, stabilize the structure. But this is still a relatively stable carbocation because the electrons can move around. You can kind of view it as a positive charge that gets dispersed between this carbon, this carbon, and that carbon over there. Now, as I said, this is still not a great situation. The molecule wants to go back to being aromatic, wants to go to that really stable state."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But this is still a relatively stable carbocation because the electrons can move around. You can kind of view it as a positive charge that gets dispersed between this carbon, this carbon, and that carbon over there. Now, as I said, this is still not a great situation. The molecule wants to go back to being aromatic, wants to go to that really stable state. The way it can go back to that really stable state is somehow an electron can be added to this thing. The way that an electron can be added to this thing is if we have some base flying around, and that base nabs this proton, this proton right here that's on the same carbon as where the electrophile is attached. If this base nabs a proton, so it just nabs the hydrogen nucleus, then that electron that the hydrogen had that electron, let me do that in a different color I reused, that electron that the hydrogen had right over there could then be returned to this carbon up there."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The molecule wants to go back to being aromatic, wants to go to that really stable state. The way it can go back to that really stable state is somehow an electron can be added to this thing. The way that an electron can be added to this thing is if we have some base flying around, and that base nabs this proton, this proton right here that's on the same carbon as where the electrophile is attached. If this base nabs a proton, so it just nabs the hydrogen nucleus, then that electron that the hydrogen had that electron, let me do that in a different color I reused, that electron that the hydrogen had right over there could then be returned to this carbon up there. Maybe that makes it a little confusing when I cross lines. It can be returned to that carbon right there. What would it look like after that?"}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If this base nabs a proton, so it just nabs the hydrogen nucleus, then that electron that the hydrogen had that electron, let me do that in a different color I reused, that electron that the hydrogen had right over there could then be returned to this carbon up there. Maybe that makes it a little confusing when I cross lines. It can be returned to that carbon right there. What would it look like after that? After that, it would look like this. Let me draw my... If that happened, we do it in yellow, we have our 6-carbon ring."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "What would it look like after that? After that, it would look like this. Let me draw my... If that happened, we do it in yellow, we have our 6-carbon ring. Let me draw all the hydrogens. What did I do that in? That looks like a slightly green color I did that in."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If that happened, we do it in yellow, we have our 6-carbon ring. Let me draw all the hydrogens. What did I do that in? That looks like a slightly green color I did that in. I have all the hydrogens on that ring, all of the hydrogens. Now I have to be careful. This hydrogen right there, just the nucleus of it got nabbed by this base."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That looks like a slightly green color I did that in. I have all the hydrogens on that ring, all of the hydrogens. Now I have to be careful. This hydrogen right there, just the nucleus of it got nabbed by this base. That hydrogen has now been nabbed by the base. If you view this electron right here, it has now been given to this hydrogen. That electron has now been given to this hydrogen."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This hydrogen right there, just the nucleus of it got nabbed by this base. That hydrogen has now been nabbed by the base. If you view this electron right here, it has now been given to this hydrogen. That electron has now been given to this hydrogen. Then the other electron in the pair is still with the base. Now this is the conjugate acid of the base. It has gained a proton."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That electron has now been given to this hydrogen. Then the other electron in the pair is still with the base. Now this is the conjugate acid of the base. It has gained a proton. On this carbon right here, we just have what was the electrophile. We just have what was the electrophile. I'll do the same colors just to make it clear."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It has gained a proton. On this carbon right here, we just have what was the electrophile. We just have what was the electrophile. I'll do the same colors just to make it clear. What was the electrophile right over there? This bond is this bond. Then finally, we had..."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'll do the same colors just to make it clear. What was the electrophile right over there? This bond is this bond. Then finally, we had... I'll color-code it here just to make it clear. We had this double bond here, which is this double bond right over here. We had this double bond, which is that double bond there."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Then finally, we had... I'll color-code it here just to make it clear. We had this double bond here, which is this double bond right over here. We had this double bond, which is that double bond there. This electron gets returned to this top carbon right here. That electron... Let me make it very, very clear. The bond and that electron are returned to that top carbon."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We had this double bond, which is that double bond there. This electron gets returned to this top carbon right here. That electron... Let me make it very, very clear. The bond and that electron are returned to that top carbon. Then we have the bond and that electron returned to that top carbon. That top carbon is now going to be neutral. Once again, we are resonance stabilized."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The bond and that electron are returned to that top carbon. Then we have the bond and that electron returned to that top carbon. That top carbon is now going to be neutral. Once again, we are resonance stabilized. One thing I forgot, just to make the charge stabilize, maybe this base had a negative charge to begin with. It didn't have to. If this base did have a negative charge to begin with, it now gave an electron to the hydrogen, so it is now neutral."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we are resonance stabilized. One thing I forgot, just to make the charge stabilize, maybe this base had a negative charge to begin with. It didn't have to. If this base did have a negative charge to begin with, it now gave an electron to the hydrogen, so it is now neutral. This should make sense, because before we had a plus charge and a negative charge. Then when everything reacted, everything is neutral again. The total net charge is zero."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If this base did have a negative charge to begin with, it now gave an electron to the hydrogen, so it is now neutral. This should make sense, because before we had a plus charge and a negative charge. Then when everything reacted, everything is neutral again. The total net charge is zero. This is the electrophilic aromatic substitution. We substituted one of the hydrogens. We substituted this hydrogen right here with this electrophile, or what was previously an electrophile."}, {"video_title": "Electrophilic aromatic substitution Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The total net charge is zero. This is the electrophilic aromatic substitution. We substituted one of the hydrogens. We substituted this hydrogen right here with this electrophile, or what was previously an electrophile. Then once it got an electron, it's just kind of a group that is now on the benzene ring. By going through this little convoluted process, we finally got to another aromatic molecule that now has this E group on it. In the next video, I'll show you this with particular examples of electrophiles and bases."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "However, they're different from typical amides. And first, let's look at a typical amide. We know that the lone pair of electrons on nitrogen is not localized to the nitrogen. It's delocalized. It participates in resonance. So when we drew the resonance structure for an amide, this top oxygen here gets a negative 1 formal charge. And there'd be a double bond between the carbon and the nitrogen."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's delocalized. It participates in resonance. So when we drew the resonance structure for an amide, this top oxygen here gets a negative 1 formal charge. And there'd be a double bond between the carbon and the nitrogen. So let me go ahead and draw in our groups. That would give the nitrogen a plus 1 formal charge. And if we look at the resonance structure on the right, and we think about the hybridization state of nitrogen in this resonance structure, it's obviously sp2 hybridized here, indicating that the nitrogen is planar."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And there'd be a double bond between the carbon and the nitrogen. So let me go ahead and draw in our groups. That would give the nitrogen a plus 1 formal charge. And if we look at the resonance structure on the right, and we think about the hybridization state of nitrogen in this resonance structure, it's obviously sp2 hybridized here, indicating that the nitrogen is planar. And in an ideal amide, the planar nitrogen gives the best overlap of orbitals. And that allows this lone pair of electrons to be delocalized, which increases the electron density around our carbonyl carbon. And so that makes our carbonyl carbon less electrophilic and therefore less reactive."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And if we look at the resonance structure on the right, and we think about the hybridization state of nitrogen in this resonance structure, it's obviously sp2 hybridized here, indicating that the nitrogen is planar. And in an ideal amide, the planar nitrogen gives the best overlap of orbitals. And that allows this lone pair of electrons to be delocalized, which increases the electron density around our carbonyl carbon. And so that makes our carbonyl carbon less electrophilic and therefore less reactive. And so we said this is why amides are generally unreactive here. So that's an ideal amide. There's a special one in penicillin, so an amide in a ring, which we call a lactam."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so that makes our carbonyl carbon less electrophilic and therefore less reactive. And so we said this is why amides are generally unreactive here. So that's an ideal amide. There's a special one in penicillin, so an amide in a ring, which we call a lactam. And so let's look at the general structure for penicillin here, or a penicillin derivative, because you could change the derivative by changing the R group. You could change it into a amoxicillin or ampicillin or anything like that. So looking for our lactam ring, so it's an amide in a ring."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "There's a special one in penicillin, so an amide in a ring, which we call a lactam. And so let's look at the general structure for penicillin here, or a penicillin derivative, because you could change the derivative by changing the R group. You could change it into a amoxicillin or ampicillin or anything like that. So looking for our lactam ring, so it's an amide in a ring. And we can see that here is our lactam. If we wanted to classify this lactam, so the carbon next to the carbonyl is the alpha carbon. The carbon next to that is the beta carbon."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So looking for our lactam ring, so it's an amide in a ring. And we can see that here is our lactam. If we wanted to classify this lactam, so the carbon next to the carbonyl is the alpha carbon. The carbon next to that is the beta carbon. And then we hit the nitrogen. And so that's why we call this a beta lactam ring. During World War II, there was a huge effort to synthesize penicillin."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "The carbon next to that is the beta carbon. And then we hit the nitrogen. And so that's why we call this a beta lactam ring. During World War II, there was a huge effort to synthesize penicillin. So chemists didn't know the exact structure. But obviously, if you could make it, it would be a huge help in the war effort. It was known that penicillin was easily hydrolyzed in acid or in base."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "During World War II, there was a huge effort to synthesize penicillin. So chemists didn't know the exact structure. But obviously, if you could make it, it would be a huge help in the war effort. It was known that penicillin was easily hydrolyzed in acid or in base. And so some chemists thought that a lactam ring could not be present, because there's such strong resonance in amides, that should decrease the reactivity. And it shouldn't be so easy to hydrolyze it. However, other chemists, like R.B."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It was known that penicillin was easily hydrolyzed in acid or in base. And so some chemists thought that a lactam ring could not be present, because there's such strong resonance in amides, that should decrease the reactivity. And it shouldn't be so easy to hydrolyze it. However, other chemists, like R.B. Woodward, favored the beta lactam structure. And of course, those chemists proved to be correct. And Woodward thought that this interesting arrangement in penicillin of these two rings, let me go ahead and show you these two rings here."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "However, other chemists, like R.B. Woodward, favored the beta lactam structure. And of course, those chemists proved to be correct. And Woodward thought that this interesting arrangement in penicillin of these two rings, let me go ahead and show you these two rings here. So we have a four-membered ring, which is our beta lactam. And then we have, if you think about it, this ring is separate over here on the right, a five-membered ring. So it's a fused four-five ring system."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And Woodward thought that this interesting arrangement in penicillin of these two rings, let me go ahead and show you these two rings here. So we have a four-membered ring, which is our beta lactam. And then we have, if you think about it, this ring is separate over here on the right, a five-membered ring. So it's a fused four-five ring system. And if you look at the model structure, I should say, that I took a picture of over here on the left, you can see that this fused four-five ring system prevents the nitrogen from being planar. So let me go ahead and highlight some of these atoms here. So this blue atom is the nitrogen."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it's a fused four-five ring system. And if you look at the model structure, I should say, that I took a picture of over here on the left, you can see that this fused four-five ring system prevents the nitrogen from being planar. So let me go ahead and highlight some of these atoms here. So this blue atom is the nitrogen. And then we can see that our carbonyl's over here on the left. And then there are four-five, our fused four-five ring system. And thinking about this nitrogen, looking at this geometry here, you can see this bond is up."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this blue atom is the nitrogen. And then we can see that our carbonyl's over here on the left. And then there are four-five, our fused four-five ring system. And thinking about this nitrogen, looking at this geometry here, you can see this bond is up. This bond is up a little bit. And so this is definitely not a planar nitrogen here. And because it's not planar, you're not going to get the same kind of resonance stabilization that we had and we talked about over here."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And thinking about this nitrogen, looking at this geometry here, you can see this bond is up. This bond is up a little bit. And so this is definitely not a planar nitrogen here. And because it's not planar, you're not going to get the same kind of resonance stabilization that we had and we talked about over here. So the nitrogen can't donate as much electron density to our carbonyl carbon because of this fused four-five ring system. So the orbitals don't overlap well enough. And so because there isn't as much donation of electron density to our carbonyl carbon, that's going to make this carbonyl carbon more partially positive, more electrophilic, and therefore more reactive."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And because it's not planar, you're not going to get the same kind of resonance stabilization that we had and we talked about over here. So the nitrogen can't donate as much electron density to our carbonyl carbon because of this fused four-five ring system. So the orbitals don't overlap well enough. And so because there isn't as much donation of electron density to our carbonyl carbon, that's going to make this carbonyl carbon more partially positive, more electrophilic, and therefore more reactive. And so that's one of the reasons why this beta-lactam turned out to be easily hydrolyzed. So another reason why this beta-lactam can break is due to ring strain or angle strain. So let's take a look at this fused four-five ring system once again."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so because there isn't as much donation of electron density to our carbonyl carbon, that's going to make this carbonyl carbon more partially positive, more electrophilic, and therefore more reactive. And so that's one of the reasons why this beta-lactam turned out to be easily hydrolyzed. So another reason why this beta-lactam can break is due to ring strain or angle strain. So let's take a look at this fused four-five ring system once again. Let me go ahead and use black so we can see what we're talking about here. So here's our beta-lactam. I'm just going to draw it in here so you can see the nitrogen right there in blue, our four-membered ring."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a look at this fused four-five ring system once again. Let me go ahead and use black so we can see what we're talking about here. So here's our beta-lactam. I'm just going to draw it in here so you can see the nitrogen right there in blue, our four-membered ring. And so if we think about the hybridization state of, let's say, this carbon right here. So this carbon is bonded to four atoms. It's sp3 hybridized."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "I'm just going to draw it in here so you can see the nitrogen right there in blue, our four-membered ring. And so if we think about the hybridization state of, let's say, this carbon right here. So this carbon is bonded to four atoms. It's sp3 hybridized. The ideal bond angle for sp3 hybridized carbon is 109.5 degrees. So that's the ideal. But we can see that we're far from the ideal in this situation."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's sp3 hybridized. The ideal bond angle for sp3 hybridized carbon is 109.5 degrees. So that's the ideal. But we can see that we're far from the ideal in this situation. So I'm not sure exactly what it is, but it's definitely less than 109.5. So if you think about this as being a square, it might be closer to 90 degrees, so somewhere in there. So a bond angle of somewhere around 90 degrees or somewhere close to it, I'm sure it's not exactly 90 degrees, is a deviation from this ideal bond angle of 109.5."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But we can see that we're far from the ideal in this situation. So I'm not sure exactly what it is, but it's definitely less than 109.5. So if you think about this as being a square, it might be closer to 90 degrees, so somewhere in there. So a bond angle of somewhere around 90 degrees or somewhere close to it, I'm sure it's not exactly 90 degrees, is a deviation from this ideal bond angle of 109.5. And the more you deviate from 109.5, the more strain there is. So you call that ring strain or angle strain. And when you're making a model set, you can actually feel these bonds bend."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So a bond angle of somewhere around 90 degrees or somewhere close to it, I'm sure it's not exactly 90 degrees, is a deviation from this ideal bond angle of 109.5. And the more you deviate from 109.5, the more strain there is. So you call that ring strain or angle strain. And when you're making a model set, you can actually feel these bonds bend. And this just gives you an idea about the strain that's present. So making this model set, you can actually feel this angle strain. And the best way to alleviate that angle strain would be to break the ring."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And when you're making a model set, you can actually feel these bonds bend. And this just gives you an idea about the strain that's present. So making this model set, you can actually feel this angle strain. And the best way to alleviate that angle strain would be to break the ring. So you could hydrolyze your amide. So you could break the ring right here. And you could see that's what I've drawn over here on the right."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And the best way to alleviate that angle strain would be to break the ring. So you could hydrolyze your amide. So you could break the ring right here. And you could see that's what I've drawn over here on the right. So we've hydrolyzed our amide. We have relieved this angle strain. So looking at the angle now, so this angle has increased."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And you could see that's what I've drawn over here on the right. So we've hydrolyzed our amide. We have relieved this angle strain. So looking at the angle now, so this angle has increased. It's no longer somewhere around 90. It's definitely increased. It's gotten closer to our ideal bond angle of 109.5."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So looking at the angle now, so this angle has increased. It's no longer somewhere around 90. It's definitely increased. It's gotten closer to our ideal bond angle of 109.5. So that's the idea of angle strain or ring strain. Opening the ring alleviates that strain and gets your bond angle closer to the ideal value, if we're thinking about just this carbon, for example, being sp3 hybridized. So we have these two factors that make a beta-lactam ring very reactive."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's gotten closer to our ideal bond angle of 109.5. So that's the idea of angle strain or ring strain. Opening the ring alleviates that strain and gets your bond angle closer to the ideal value, if we're thinking about just this carbon, for example, being sp3 hybridized. So we have these two factors that make a beta-lactam ring very reactive. One is there's not as much resonance stabilization. And the other one is ring strain. And so the two of these things combined to make this extremely reactive."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we have these two factors that make a beta-lactam ring very reactive. One is there's not as much resonance stabilization. And the other one is ring strain. And so the two of these things combined to make this extremely reactive. So let's take a look at the mechanism of action of penicillin. And so here we have our penicillin derivative. And over here we have the transpeptidase enzyme, which is an enzyme in bacteria that's used to build the cell walls of that bacteria."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so the two of these things combined to make this extremely reactive. So let's take a look at the mechanism of action of penicillin. And so here we have our penicillin derivative. And over here we have the transpeptidase enzyme, which is an enzyme in bacteria that's used to build the cell walls of that bacteria. And so this is the active enzyme right here. And we can see the active enzyme has an OH on it. And this OH is going to function as a nucleophile."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And over here we have the transpeptidase enzyme, which is an enzyme in bacteria that's used to build the cell walls of that bacteria. And so this is the active enzyme right here. And we can see the active enzyme has an OH on it. And this OH is going to function as a nucleophile. And it's going to attack our carbonyl carbon right here on our beta-lactam ring. So we know that this carbonyl carbon is more electrophilic than for most amides. And we also know that there is significant ring or angle strain here."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this OH is going to function as a nucleophile. And it's going to attack our carbonyl carbon right here on our beta-lactam ring. So we know that this carbonyl carbon is more electrophilic than for most amides. And we also know that there is significant ring or angle strain here. So that's actually the reactive portion of our penicillin derivative molecule. And so the nucleophile attacks the electrophile. So nucleophile attacks the electrophile."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we also know that there is significant ring or angle strain here. So that's actually the reactive portion of our penicillin derivative molecule. And so the nucleophile attacks the electrophile. So nucleophile attacks the electrophile. And these electrons would kick off onto your oxygen. And then when you reform the carbonyl, so those electrons would move back in to reform the carbonyl, which would kick these electrons off onto the nitrogen. And so this is pretty much just a nucleophilic acyl substitution mechanism."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So nucleophile attacks the electrophile. And these electrons would kick off onto your oxygen. And then when you reform the carbonyl, so those electrons would move back in to reform the carbonyl, which would kick these electrons off onto the nitrogen. And so this is pretty much just a nucleophilic acyl substitution mechanism. So we're going to break the amides. And let's go ahead and show the product here. So what would happen?"}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so this is pretty much just a nucleophilic acyl substitution mechanism. So we're going to break the amides. And let's go ahead and show the product here. So what would happen? Well, let me go ahead and highlight some of these atoms so we can follow along. This oxygen right here is this oxygen. And this carbon right here is this carbon right here."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So what would happen? Well, let me go ahead and highlight some of these atoms so we can follow along. This oxygen right here is this oxygen. And this carbon right here is this carbon right here. And so we broke the bond. We broke the bond between the carbon and the nitrogen. And so that would be this nitrogen over here."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon right here is this carbon right here. And so we broke the bond. We broke the bond between the carbon and the nitrogen. And so that would be this nitrogen over here. So let me go ahead and circle it. And this nitrogen picked up a proton in the process. And so the point is we've now disabled the enzyme."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so that would be this nitrogen over here. So let me go ahead and circle it. And this nitrogen picked up a proton in the process. And so the point is we've now disabled the enzyme. So this is now a disabled enzyme. There's no longer this free OH here. So if it's disabled, it can't build cell walls for the bacteria."}, {"video_title": "Beta-lactam antibiotics Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so the point is we've now disabled the enzyme. So this is now a disabled enzyme. There's no longer this free OH here. So if it's disabled, it can't build cell walls for the bacteria. And if the bacteria can't build cell walls, that means that our immune system can fight off any kind of a bacterial infection without a cell wall. And so this is the idea behind how beta-lactam antibiotics like penicillin work. So they prevent the bacteria from building cell walls."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The first realization that there were actually distinct layers of the Earth came from this guy right over here, Andrija Mohorovicic. And I apologize ahead of time to any Croatians for butchering any of the pronunciation. And he was a meteorologist and seismologist, and he was the first one to notice in 1909 when there was an earthquake in Croatia, a little bit southeast of Zagreb. So the earthquake was roughly over here, and lucky for him and lucky for us, before that earthquake there was actually a bunch of seismographic stations already in the area. And all these seismographic stations are, they essentially, instruments were installed so that if there was any essentially seismic waves passing, they would be able to measure it when the waves got there. And what was interesting about this, Andrija realized that if the entire Earth was just kind of a uniform material, so let's draw that scenario, it would get denser as you go down, and so you would have kind of this refraction, this continuous refraction or these curved paths happening. But he realized that, let's say we had an earthquake right over here, so this is the uniform case, uniform layer, only one layer, although it does get denser."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the earthquake was roughly over here, and lucky for him and lucky for us, before that earthquake there was actually a bunch of seismographic stations already in the area. And all these seismographic stations are, they essentially, instruments were installed so that if there was any essentially seismic waves passing, they would be able to measure it when the waves got there. And what was interesting about this, Andrija realized that if the entire Earth was just kind of a uniform material, so let's draw that scenario, it would get denser as you go down, and so you would have kind of this refraction, this continuous refraction or these curved paths happening. But he realized that, let's say we had an earthquake right over here, so this is the uniform case, uniform layer, only one layer, although it does get denser. Then the closer you are to the earthquake, so waves would get there first, then waves would get over there, then waves would get over there, and these are the body waves, these are the ones that are traveling through the Earth's crust. But in general, the further you are away from the earthquake, or the time it takes for the waves to get to a point, is going to be proportional to the distance that point is away from the earthquake. So you would expect to see something like this."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But he realized that, let's say we had an earthquake right over here, so this is the uniform case, uniform layer, only one layer, although it does get denser. Then the closer you are to the earthquake, so waves would get there first, then waves would get over there, then waves would get over there, and these are the body waves, these are the ones that are traveling through the Earth's crust. But in general, the further you are away from the earthquake, or the time it takes for the waves to get to a point, is going to be proportional to the distance that point is away from the earthquake. So you would expect to see something like this. So if you were to plot on the horizontal axis, if you were to plot distance, and on the vertical axis, if you were to plot time, you should see something like this. You should see a straight line. And that's just because it's traveling roughly the same velocity along any of these arcs."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you would expect to see something like this. So if you were to plot on the horizontal axis, if you were to plot distance, and on the vertical axis, if you were to plot time, you should see something like this. You should see a straight line. And that's just because it's traveling roughly the same velocity along any of these arcs. It's maybe getting a little bit faster as it's getting deeper, but roughly the same velocity is traveling along these arcs, and the distance of these arcs are proportional to the distance along the surface. So essentially, they're all traveling roughly at the same velocity, and they're just traveling different distances, so the time it takes is just going to be proportional to the distance. But he noticed something interesting."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's just because it's traveling roughly the same velocity along any of these arcs. It's maybe getting a little bit faster as it's getting deeper, but roughly the same velocity is traveling along these arcs, and the distance of these arcs are proportional to the distance along the surface. So essentially, they're all traveling roughly at the same velocity, and they're just traveling different distances, so the time it takes is just going to be proportional to the distance. But he noticed something interesting. When he actually measured when the waves from that earthquake reached different seismographic stations, he saw something interesting. So this is the theoretical, if we had this uniform one-layered Earth. But he saw something interesting."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But he noticed something interesting. When he actually measured when the waves from that earthquake reached different seismographic stations, he saw something interesting. So this is the theoretical, if we had this uniform one-layered Earth. But he saw something interesting. So once again, this is distance, and this right over here is time. And at 200 kilometers away from the earthquake, so until 200 kilometers, he saw exactly what you would expect from a uniform Earth. It was just the time was proportional to the distance."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But he saw something interesting. So once again, this is distance, and this right over here is time. And at 200 kilometers away from the earthquake, so until 200 kilometers, he saw exactly what you would expect from a uniform Earth. It was just the time was proportional to the distance. But at 200 kilometers, he saw something interesting. All of a sudden, the waves were reaching there faster. The slope of this line changed."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It was just the time was proportional to the distance. But at 200 kilometers, he saw something interesting. All of a sudden, the waves were reaching there faster. The slope of this line changed. It took less time for each incremental distance. So for some reason, the waves that were going at these farther stations, the stations that were more than 200 kilometers away, the stations, somehow they were accelerated. Somehow they were able to move faster."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The slope of this line changed. It took less time for each incremental distance. So for some reason, the waves that were going at these farther stations, the stations that were more than 200 kilometers away, the stations, somehow they were accelerated. Somehow they were able to move faster. And he's the one that realized that this was because the waves that were getting to these further stations must have traveled through a more dense layer of the Earth. So let's just think about it. So if we have a more dense layer, it will fit this information right over here."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Somehow they were able to move faster. And he's the one that realized that this was because the waves that were getting to these further stations must have traveled through a more dense layer of the Earth. So let's just think about it. So if we have a more dense layer, it will fit this information right over here. So if we have a layer like this, which we now know to be the crust, and then you have a denser layer, which we now know to be the mantle, then what you would have is, so you have your earthquake right over here. Closer by, while you're still within the crust, it would be proportional. It would be proportional."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if we have a more dense layer, it will fit this information right over here. So if we have a layer like this, which we now know to be the crust, and then you have a denser layer, which we now know to be the mantle, then what you would have is, so you have your earthquake right over here. Closer by, while you're still within the crust, it would be proportional. It would be proportional. And then let's say that this is exactly, this right here is 200 kilometers away. But then if you go any further, the waves would have to travel. So they would travel, so they would go like this."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would be proportional. And then let's say that this is exactly, this right here is 200 kilometers away. But then if you go any further, the waves would have to travel. So they would travel, so they would go like this. And then they would get refracted even harder. So they would get refracted even, so they would be a little bit curved ahead of time. But then they're going into a much denser material, or it's not gradually dense."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they would travel, so they would go like this. And then they would get refracted even harder. So they would get refracted even, so they would be a little bit curved ahead of time. But then they're going into a much denser material, or it's not gradually dense. It's actually kind of a, all of a sudden, a considerably more dense material. So it would get refracted even more. And then it'll go over here."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But then they're going into a much denser material, or it's not gradually dense. It's actually kind of a, all of a sudden, a considerably more dense material. So it would get refracted even more. And then it'll go over here. And since it was able to travel all of this distance in a denser material, it would have traveled faster along this path. And so it would get to this distance on the surface that's more than 200 kilometers away. It would get there faster."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then it'll go over here. And since it was able to travel all of this distance in a denser material, it would have traveled faster along this path. And so it would get to this distance on the surface that's more than 200 kilometers away. It would get there faster. And so he said that there must be a denser layer that those waves are traveling through, which we now know to be the mantle. And the boundary between what we now know to be the crust and this denser layer, which we now know to be the mantle, is actually named after him. It's called the Mohorovicic discontinuity."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would get there faster. And so he said that there must be a denser layer that those waves are traveling through, which we now know to be the mantle. And the boundary between what we now know to be the crust and this denser layer, which we now know to be the mantle, is actually named after him. It's called the Mohorovicic discontinuity. And sometimes it's just called the Moho for short. So that boundary between the crust and the mantle is now named for him. But this was a huge discovery, because not only was he able to tell us, based on the data, based on kind of indirect data, just based on earthquakes happening and measuring when the earthquakes reach different points of the Earth, that there probably is a denser layer."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's called the Mohorovicic discontinuity. And sometimes it's just called the Moho for short. So that boundary between the crust and the mantle is now named for him. But this was a huge discovery, because not only was he able to tell us, based on the data, based on kind of indirect data, just based on earthquakes happening and measuring when the earthquakes reach different points of the Earth, that there probably is a denser layer. And if you do the math, under continental crust, that denser layer is about 35 kilometers down. He was able to tell us that there is that layer. But even more importantly, he was able to give the clue that just using information from earthquakes, we could essentially figure out the actual composition of the Earth, because no one has ever dug that deep."}, {"video_title": "The mohorovicic seismic discontinuity Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this was a huge discovery, because not only was he able to tell us, based on the data, based on kind of indirect data, just based on earthquakes happening and measuring when the earthquakes reach different points of the Earth, that there probably is a denser layer. And if you do the math, under continental crust, that denser layer is about 35 kilometers down. He was able to tell us that there is that layer. But even more importantly, he was able to give the clue that just using information from earthquakes, we could essentially figure out the actual composition of the Earth, because no one has ever dug that deep. No one has ever dug into the mantle, much less the outer core or the inner core. In the next few videos, we're going to kind of take this insight, that we can use information from earthquakes, to actually think about how we know that there is an outer liquid core and that there's an inner core as well. And then obviously you could keep going and think about all of the different densities within the mantle and all of that."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And it's easily one of the most important mechanisms and reactions in all of organic chemistry, because it's a powerful way to actually create carbon-carbon bonds. And it'll actually be a little bit of a review of what we saw with enol and the enolate ions and the keto-enol tautorism. I have trouble saying that. Anyway, let's start with a couple of aldehydes. And just for convenience, I'll make them identical. So let's say that this is one of the aldehydes. It just has a carbon chain there."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Anyway, let's start with a couple of aldehydes. And just for convenience, I'll make them identical. So let's say that this is one of the aldehydes. It just has a carbon chain there. That's what the R is. It could be of any length. Who knows what's there?"}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It just has a carbon chain there. That's what the R is. It could be of any length. Who knows what's there? And then I have another carbon here. And then this is bound to the carbonyl group. And we're going to make it an aldehyde."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Who knows what's there? And then I have another carbon here. And then this is bound to the carbonyl group. And we're going to make it an aldehyde. Although you could do this reaction with a ketone as well, and just to make things clear, this carbon right here, which is going to be involved in a lot of the business here. Let me draw its hydrogens. Normally we don't have to draw its hydrogens."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to make it an aldehyde. Although you could do this reaction with a ketone as well, and just to make things clear, this carbon right here, which is going to be involved in a lot of the business here. Let me draw its hydrogens. Normally we don't have to draw its hydrogens. And just as a bit of review, the carbon that is next to the carbonyl carbon is called an alpha carbon. If this was a ketone, this would have also been an alpha carbon if this was a carbon. And we're going to see in this reaction, besides just exploring the reaction, is that these hydrogens are actually much more acidic than traditional hydrogens attached to carbons on the rest of the chain."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Normally we don't have to draw its hydrogens. And just as a bit of review, the carbon that is next to the carbonyl carbon is called an alpha carbon. If this was a ketone, this would have also been an alpha carbon if this was a carbon. And we're going to see in this reaction, besides just exploring the reaction, is that these hydrogens are actually much more acidic than traditional hydrogens attached to carbons on the rest of the chain. And it comes from the fact that this proton can be given to something else. The electron can go to that carbon, and then it'll be resonance stabilized. And we're going to see that in a second."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to see in this reaction, besides just exploring the reaction, is that these hydrogens are actually much more acidic than traditional hydrogens attached to carbons on the rest of the chain. And it comes from the fact that this proton can be given to something else. The electron can go to that carbon, and then it'll be resonance stabilized. And we're going to see that in a second. Now I said I would draw two molecules of that, because we need two molecules. We're actually going to be, to some degree, joining the two molecules. So let me draw another aldehyde right over here."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to see that in a second. Now I said I would draw two molecules of that, because we need two molecules. We're actually going to be, to some degree, joining the two molecules. So let me draw another aldehyde right over here. And I'm going to draw it symmetric to this, because it'll make it, I think, a little bit easier to visualize the two molecules. Actually, let me just draw it the same way. But I'll draw it in a different color."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw another aldehyde right over here. And I'm going to draw it symmetric to this, because it'll make it, I think, a little bit easier to visualize the two molecules. Actually, let me just draw it the same way. But I'll draw it in a different color. So you have the R group, and then you have the oxygen right there. And I won't draw all of the hydrogens on this guy. But this and this are the exact same molecules."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But I'll draw it in a different color. So you have the R group, and then you have the oxygen right there. And I won't draw all of the hydrogens on this guy. But this and this are the exact same molecules. Just the hydrogens are implicit here. Now, the aldol reaction I'll show you will be in a basic environment. So you can imagine that it'll be catalyzed by a base."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But this and this are the exact same molecules. Just the hydrogens are implicit here. Now, the aldol reaction I'll show you will be in a basic environment. So you can imagine that it'll be catalyzed by a base. And so imagine we have some hydroxide laying around, some of the hydroxide anion. Let me do that in a different color. So let's say we have some hydroxide anion floating around, negative charge, just like that."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So you can imagine that it'll be catalyzed by a base. And so imagine we have some hydroxide laying around, some of the hydroxide anion. Let me do that in a different color. So let's say we have some hydroxide anion floating around, negative charge, just like that. I just told you that these hydrogens are much more acidic than hydrogens anywhere else on a carbon chain, these alpha hydrogens. So you could imagine a situation where an electron from the hydroxide is given to one of these hydrogen protons. And then the electron that was associated with that hydrogen is now given back to this alpha carbon."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's say we have some hydroxide anion floating around, negative charge, just like that. I just told you that these hydrogens are much more acidic than hydrogens anywhere else on a carbon chain, these alpha hydrogens. So you could imagine a situation where an electron from the hydroxide is given to one of these hydrogen protons. And then the electron that was associated with that hydrogen is now given back to this alpha carbon. And so if that were to happen, the next step in our reaction would look like this. And I'll draw it in equilibrium. Actually, let me draw it this way."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then the electron that was associated with that hydrogen is now given back to this alpha carbon. And so if that were to happen, the next step in our reaction would look like this. And I'll draw it in equilibrium. Actually, let me draw it this way. So the next step in our reaction, or the products of that step, would be in equilibrium with you have your carbon chain or the rest of your molecule right there. And that's just to show that it could be anything. It's attached here to the alpha carbon, which is now going to be negative."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Actually, let me draw it this way. So the next step in our reaction, or the products of that step, would be in equilibrium with you have your carbon chain or the rest of your molecule right there. And that's just to show that it could be anything. It's attached here to the alpha carbon, which is now going to be negative. I'll show that in a second, which is attached to the carbonyl group, which is attached to a hydrogen. And actually, I'll stop drawing that hydrogen for now too, just we know it's there. But I'll keep drawing this hydrogen right over here."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's attached here to the alpha carbon, which is now going to be negative. I'll show that in a second, which is attached to the carbonyl group, which is attached to a hydrogen. And actually, I'll stop drawing that hydrogen for now too, just we know it's there. But I'll keep drawing this hydrogen right over here. The other hydrogen was taken away, and this alpha carbon now has a negative charge because it got the electron from that proton. And of course, we have the hydroxide. It grabbed this hydrogen, and it is now water."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But I'll keep drawing this hydrogen right over here. The other hydrogen was taken away, and this alpha carbon now has a negative charge because it got the electron from that proton. And of course, we have the hydroxide. It grabbed this hydrogen, and it is now water. Now, the reason why this was acidic to begin with is because this is resonance stabilized. And I'll show you that it's resonance stabilized right now. This alpha carbon right here can give its electron to the carbonyl carbon."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It grabbed this hydrogen, and it is now water. Now, the reason why this was acidic to begin with is because this is resonance stabilized. And I'll show you that it's resonance stabilized right now. This alpha carbon right here can give its electron to the carbonyl carbon. And if the carbonyl carbon gets an electron, it can give an electron to this oxygen up here. It'll break the double bond. So this configuration is resonance stabilized with this, so I could draw it like this."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This alpha carbon right here can give its electron to the carbonyl carbon. And if the carbonyl carbon gets an electron, it can give an electron to this oxygen up here. It'll break the double bond. So this configuration is resonance stabilized with this, so I could draw it like this. You have your R, and then you have a single bond to this oxygen. It now gained an electron. It is now negative."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this configuration is resonance stabilized with this, so I could draw it like this. You have your R, and then you have a single bond to this oxygen. It now gained an electron. It is now negative. And you now have a double bond just like that. And I could draw this hydrogen if I like, or I don't have to. It's implicitly over there."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It is now negative. And you now have a double bond just like that. And I could draw this hydrogen if I like, or I don't have to. It's implicitly over there. Now, and you might be familiar with this. This is the enolate anion. This right here is the enolate ion."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's implicitly over there. Now, and you might be familiar with this. This is the enolate anion. This right here is the enolate ion. If we had a hydrogen right here, it would be an enol. And we would say, hey, this is the keto form. This is the enol form."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This right here is the enolate ion. If we had a hydrogen right here, it would be an enol. And we would say, hey, this is the keto form. This is the enol form. We've seen this before. Now, what's interesting about the enolate ion is it can act as a nucleophile. It can do a nucleophilic attack on the other aldehyde's carbonyl group."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is the enol form. We've seen this before. Now, what's interesting about the enolate ion is it can act as a nucleophile. It can do a nucleophilic attack on the other aldehyde's carbonyl group. But it does it in kind of a nonconventional way. And I'll show you how it does it right now. So it does the attack like this."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It can do a nucleophilic attack on the other aldehyde's carbonyl group. But it does it in kind of a nonconventional way. And I'll show you how it does it right now. So it does the attack like this. So let me draw this guy over here. So you have the carbonyl group, and then you have its alpha carbon, and then you have an R group right over there. There's actually a hydrogen right over here as well."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it does the attack like this. So let me draw this guy over here. So you have the carbonyl group, and then you have its alpha carbon, and then you have an R group right over there. There's actually a hydrogen right over here as well. I just flipped it over. And this are the same molecule. But let me make it clear."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's actually a hydrogen right over here as well. I just flipped it over. And this are the same molecule. But let me make it clear. These two guys right here are resonance forms. And this is, once again, the reason why it's easier to take this hydrogen than other hydrogens on a traditional carbon chain. Easier to take an alpha hydrogen to a carbonyl group because you have this resonance structure."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But let me make it clear. These two guys right here are resonance forms. And this is, once again, the reason why it's easier to take this hydrogen than other hydrogens on a traditional carbon chain. Easier to take an alpha hydrogen to a carbonyl group because you have this resonance structure. But this enolate ion, especially this configuration of it, you can imagine it doing something like this. You can imagine this oxygen giving back the electron to the carbonyl carbon, to this carbon right here. And when that happens, then this guy is going to be giving up an electron."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Easier to take an alpha hydrogen to a carbonyl group because you have this resonance structure. But this enolate ion, especially this configuration of it, you can imagine it doing something like this. You can imagine this oxygen giving back the electron to the carbonyl carbon, to this carbon right here. And when that happens, then this guy is going to be giving up an electron. And that electron that he gives up, let me do it in a new color, this electron that he gives up could go and do a nucleophilic attack on this carbonyl group. And so if that carbonyl carbon gets, let me do this in a new color, if this carbonyl carbon gets an electron, then it could give away an electron to that oxygen right up there. So the next step after this, we would have something like this."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And when that happens, then this guy is going to be giving up an electron. And that electron that he gives up, let me do it in a new color, this electron that he gives up could go and do a nucleophilic attack on this carbonyl group. And so if that carbonyl carbon gets, let me do this in a new color, if this carbonyl carbon gets an electron, then it could give away an electron to that oxygen right up there. So the next step after this, we would have something like this. And once again, I'll show it as happening in equilibrium. So from here, we go right over there. And what we have is a situation where, let me draw this guy on the left first."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the next step after this, we would have something like this. And once again, I'll show it as happening in equilibrium. So from here, we go right over there. And what we have is a situation where, let me draw this guy on the left first. So we have a double bond to this oxygen. Now actually, let me draw the second. So this is this oxygen."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And what we have is a situation where, let me draw this guy on the left first. So we have a double bond to this oxygen. Now actually, let me draw the second. So this is this oxygen. We now have a double bond. Let me do it in the same purple color right over here. And then we have the rest of what was an aldehyde."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is this oxygen. We now have a double bond. Let me do it in the same purple color right over here. And then we have the rest of what was an aldehyde. Where you have, let me do it in that same color. And then you have your R group right over there. But now this electron get an attack on this other aldehyde."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we have the rest of what was an aldehyde. Where you have, let me do it in that same color. And then you have your R group right over there. But now this electron get an attack on this other aldehyde. So this guy right here, this alpha carbon, it's that same alpha carbon we've been dealing with. The same alpha carbon is now bonded to this carbonyl carbon. So it is now bound to this carbonyl carbon right over here, and so it will look like this."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But now this electron get an attack on this other aldehyde. So this guy right here, this alpha carbon, it's that same alpha carbon we've been dealing with. The same alpha carbon is now bonded to this carbonyl carbon. So it is now bound to this carbonyl carbon right over here, and so it will look like this. Let me draw it with the right colors. Get the orange out. So that carbonyl carbon, it now has a single bond to this oxygen."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it is now bound to this carbonyl carbon right over here, and so it will look like this. Let me draw it with the right colors. Get the orange out. So that carbonyl carbon, it now has a single bond to this oxygen. This electron was taken back by it, so this oxygen now has a negative charge. And it is bound to its alpha carbon, and then that is bonded to another group, probably a carbon chain or something that contains a carbon chain, another functional group, whatever you want to call it. And then the final step, this anion can get rid of its negative charge by essentially grabbing a hydrogen from maybe from this water that was formed before."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that carbonyl carbon, it now has a single bond to this oxygen. This electron was taken back by it, so this oxygen now has a negative charge. And it is bound to its alpha carbon, and then that is bonded to another group, probably a carbon chain or something that contains a carbon chain, another functional group, whatever you want to call it. And then the final step, this anion can get rid of its negative charge by essentially grabbing a hydrogen from maybe from this water that was formed before. Obviously not going to be the same molecule, but it could grab it from this in a previous step, this water molecule that was formed in a previous step. And of course, this is all in a basic environment. So it can give an electron to this hydrogen, and then the hydrogen proton would lose an electron to the hydroxide, and the hydroxide will become negative again."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then the final step, this anion can get rid of its negative charge by essentially grabbing a hydrogen from maybe from this water that was formed before. Obviously not going to be the same molecule, but it could grab it from this in a previous step, this water molecule that was formed in a previous step. And of course, this is all in a basic environment. So it can give an electron to this hydrogen, and then the hydrogen proton would lose an electron to the hydroxide, and the hydroxide will become negative again. And so what will be the final product? The final product will be, and I'm just going to try my best to redraw this thing right over here, you have this part of the molecule. So you have this carbonyl group right over here."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it can give an electron to this hydrogen, and then the hydrogen proton would lose an electron to the hydroxide, and the hydroxide will become negative again. And so what will be the final product? The final product will be, and I'm just going to try my best to redraw this thing right over here, you have this part of the molecule. So you have this carbonyl group right over here. It is attached to this radical group right over there. So that is this part, and I can even do the same colors. This bond right over here is this bond right over here."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So you have this carbonyl group right over here. It is attached to this radical group right over there. So that is this part, and I can even do the same colors. This bond right over here is this bond right over here. And then this carbon is attached to a carbon that's attached to a hydroxyl group now. So it'll look like this. And let me draw it."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This bond right over here is this bond right over here. And then this carbon is attached to a carbon that's attached to a hydroxyl group now. So it'll look like this. And let me draw it. So this oxygen is now this oxygen, and it just captured this hydrogen. So it is now a hydroxyl group. It's now an OH."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let me draw it. So this oxygen is now this oxygen, and it just captured this hydrogen. So it is now a hydroxyl group. It's now an OH. It is now an OH group. And then finally, this guy is bound to what was an alpha carbon, it's not anymore, which is then bound to a radical group. And if we want, we can remember that there was always from the get-go, there was always a hydrogen over here."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's now an OH. It is now an OH group. And then finally, this guy is bound to what was an alpha carbon, it's not anymore, which is then bound to a radical group. And if we want, we can remember that there was always from the get-go, there was always a hydrogen over here. So why is this called the aldol reaction? Why does it matter? Well, it's called the aldol reaction because what we formed is both an aldehyde."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if we want, we can remember that there was always from the get-go, there was always a hydrogen over here. So why is this called the aldol reaction? Why does it matter? Well, it's called the aldol reaction because what we formed is both an aldehyde. Notice this is an aldehyde, and it's an alcohol. So that's where the word aldol comes from. But the more important thing about this, and I don't want to mislead you, you could have also have done this with a ketone."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, it's called the aldol reaction because what we formed is both an aldehyde. Notice this is an aldehyde, and it's an alcohol. So that's where the word aldol comes from. But the more important thing about this, and I don't want to mislead you, you could have also have done this with a ketone. You could have had a methyl group or an ethyl group, or you could have had a big carbon chain here. It still would have worked. So the aldol reaction doesn't only form things that are aldehydes and alcohols."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But the more important thing about this, and I don't want to mislead you, you could have also have done this with a ketone. You could have had a methyl group or an ethyl group, or you could have had a big carbon chain here. It still would have worked. So the aldol reaction doesn't only form things that are aldehydes and alcohols. It could have formed something that's both a ketone and an alcohol. But that's why it's called the aldol reaction. But the more important thing about the aldol reaction is, one, it shows you how the enolate ion can be a nucleophile."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the aldol reaction doesn't only form things that are aldehydes and alcohols. It could have formed something that's both a ketone and an alcohol. But that's why it's called the aldol reaction. But the more important thing about the aldol reaction is, one, it shows you how the enolate ion can be a nucleophile. It shows you why the alpha hydrogens are more acidic than hydrogens on other parts of carbon chains. But the most useful aspect of it is it's a useful way to actually join two carbon chains together. Notice, we were able to join this alpha carbon right here to this carbonyl carbon over here to form this aldol."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But the more important thing about the aldol reaction is, one, it shows you how the enolate ion can be a nucleophile. It shows you why the alpha hydrogens are more acidic than hydrogens on other parts of carbon chains. But the most useful aspect of it is it's a useful way to actually join two carbon chains together. Notice, we were able to join this alpha carbon right here to this carbonyl carbon over here to form this aldol. Or sometimes this will be called, because this is still an alpha carbon right here. This is an alpha carbon. This is a beta carbon."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Notice, we were able to join this alpha carbon right here to this carbonyl carbon over here to form this aldol. Or sometimes this will be called, because this is still an alpha carbon right here. This is an alpha carbon. This is a beta carbon. And so sometimes this will be referred to as a beta hydroxy, and we've probably used things from the pharmacy that has this word in it. This is also called a beta hydroxy. This is alpha."}, {"video_title": "Aldol reaction Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a beta carbon. And so sometimes this will be referred to as a beta hydroxy, and we've probably used things from the pharmacy that has this word in it. This is also called a beta hydroxy. This is alpha. This is beta. It has a hydroxyl group on the beta carbon. Beta hydroxy aldehyde."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "Does your reagent function as a nucleophile or does it function as a base? So first let's look at nucleophile. Let's consider the idea of charge. We know that water can function as a nucleophile, so we have a region of high electron density around the oxygen. The oxygen is partially negative because oxygen is more electronegative than hydrogen, so these hydrogens are partially positive. So water can function as a nucleophile, however, it is a weak nucleophile because we do not have as high a region of electron density as we would with the hydroxide ion. So the hydroxide ion is a strong nucleophile."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "We know that water can function as a nucleophile, so we have a region of high electron density around the oxygen. The oxygen is partially negative because oxygen is more electronegative than hydrogen, so these hydrogens are partially positive. So water can function as a nucleophile, however, it is a weak nucleophile because we do not have as high a region of electron density as we would with the hydroxide ion. So the hydroxide ion is a strong nucleophile. It has a full negative charge on the oxygen instead of only a partial negative charge. Next, let's consider the polarizability of the nucleophile. Let's compare the hydroxide ion with hydrogen sulfide."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So the hydroxide ion is a strong nucleophile. It has a full negative charge on the oxygen instead of only a partial negative charge. Next, let's consider the polarizability of the nucleophile. Let's compare the hydroxide ion with hydrogen sulfide. So we've already seen the hydroxide ion is a strong nucleophile with its negative charge on the oxygen, but hydrogen sulfide also turns out to be a strong nucleophile, even though it doesn't have a negative charge on the sulfur. And that's because of the concept of polarizability, which is related to the size of the atom and the distance of electrons from the nucleus. Sulfur is a larger atom than oxygen, and we have several electrons that are far away from the nucleus, and since those electrons are far away from the nucleus, the nucleus doesn't have as much of a pull on them, and it's easier to polarize those electrons."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "Let's compare the hydroxide ion with hydrogen sulfide. So we've already seen the hydroxide ion is a strong nucleophile with its negative charge on the oxygen, but hydrogen sulfide also turns out to be a strong nucleophile, even though it doesn't have a negative charge on the sulfur. And that's because of the concept of polarizability, which is related to the size of the atom and the distance of electrons from the nucleus. Sulfur is a larger atom than oxygen, and we have several electrons that are far away from the nucleus, and since those electrons are far away from the nucleus, the nucleus doesn't have as much of a pull on them, and it's easier to polarize those electrons. So it's easier for, let's say, a pair of these electrons to function as a nucleophile, to get closer to an electrophile. And so that's the reason why hydrogen sulfide is a strong nucleophile. Hydroxide, we've already seen is a strong nucleophile."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "Sulfur is a larger atom than oxygen, and we have several electrons that are far away from the nucleus, and since those electrons are far away from the nucleus, the nucleus doesn't have as much of a pull on them, and it's easier to polarize those electrons. So it's easier for, let's say, a pair of these electrons to function as a nucleophile, to get closer to an electrophile. And so that's the reason why hydrogen sulfide is a strong nucleophile. Hydroxide, we've already seen is a strong nucleophile. The electrons are closer to the nucleus, so the nucleus has a stronger pull on them, so it's not as polarizable, but it still turns out to be a strong nucleophile because of the negative charge. When you get to something like the hydride ion, this is very small. We know that hydrogen is the smallest atom, so these two electrons are pretty close to the nucleus, so the nucleus has a very strong pull on them."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "Hydroxide, we've already seen is a strong nucleophile. The electrons are closer to the nucleus, so the nucleus has a stronger pull on them, so it's not as polarizable, but it still turns out to be a strong nucleophile because of the negative charge. When you get to something like the hydride ion, this is very small. We know that hydrogen is the smallest atom, so these two electrons are pretty close to the nucleus, so the nucleus has a very strong pull on them. So because this electron cloud is so close to the nucleus, it's not polarizable, even though it has a negative charge on it. So this does not function as a nucleophile, so the hydride ion can't function as a nucleophile because it's not polarizable. When you're trying to figure out the nature of the reagent, there are four different categories, and you want to assign your reagent to one of those four categories."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "We know that hydrogen is the smallest atom, so these two electrons are pretty close to the nucleus, so the nucleus has a very strong pull on them. So because this electron cloud is so close to the nucleus, it's not polarizable, even though it has a negative charge on it. So this does not function as a nucleophile, so the hydride ion can't function as a nucleophile because it's not polarizable. When you're trying to figure out the nature of the reagent, there are four different categories, and you want to assign your reagent to one of those four categories. The first one is where your reagent acts only as a nucleophile and not as a base. And a good example of that would be the chloride anion. So it can act as a nucleophile because we have a negative one charge here."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "When you're trying to figure out the nature of the reagent, there are four different categories, and you want to assign your reagent to one of those four categories. The first one is where your reagent acts only as a nucleophile and not as a base. And a good example of that would be the chloride anion. So it can act as a nucleophile because we have a negative one charge here. We have a region of high electron density. But it can't act as a base, and think about why. The conjugate acid to the chloride anion would be HCl."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So it can act as a nucleophile because we have a negative one charge here. We have a region of high electron density. But it can't act as a base, and think about why. The conjugate acid to the chloride anion would be HCl. Just add an H plus to Cl minus, and you get HCl. And we know that HCl is a strong acid, and we also know the stronger the acid, the weaker the conjugate base. So the chloride anion is a very weak base, and that's why it's only gonna function as a nucleophile in our reactions."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "The conjugate acid to the chloride anion would be HCl. Just add an H plus to Cl minus, and you get HCl. And we know that HCl is a strong acid, and we also know the stronger the acid, the weaker the conjugate base. So the chloride anion is a very weak base, and that's why it's only gonna function as a nucleophile in our reactions. So the same idea for the bromide anion and the iodide anion. And then we also have our sulfur nucleophiles, which we just saw, this hydrogen sulfide here is a strong nucleophile because of the polarizability of sulfur, but these are also only gonna function as a nucleophile in our reactions, and that's because the conjugate acids are fairly acidic. So the same reason we talked about over here."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So the chloride anion is a very weak base, and that's why it's only gonna function as a nucleophile in our reactions. So the same idea for the bromide anion and the iodide anion. And then we also have our sulfur nucleophiles, which we just saw, this hydrogen sulfide here is a strong nucleophile because of the polarizability of sulfur, but these are also only gonna function as a nucleophile in our reactions, and that's because the conjugate acids are fairly acidic. So the same reason we talked about over here. The second category is when your reagent only functions as a base and not a nucleophile. An example of that would be the hydride ion. We've already seen why the hydride ion does not function as a nucleophile, but now let's talk about why it's a strong base."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So the same reason we talked about over here. The second category is when your reagent only functions as a base and not a nucleophile. An example of that would be the hydride ion. We've already seen why the hydride ion does not function as a nucleophile, but now let's talk about why it's a strong base. If you think about the conjugate acid to H minus, just add an H plus, and you of course get H two. We know that H two is a very stable molecule, which makes it a very weak acid, and the weaker the acid, the stronger the conjugate base, which makes the hydride anion a very strong base. And if you see it in a reaction, think base only as the reagent."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "We've already seen why the hydride ion does not function as a nucleophile, but now let's talk about why it's a strong base. If you think about the conjugate acid to H minus, just add an H plus, and you of course get H two. We know that H two is a very stable molecule, which makes it a very weak acid, and the weaker the acid, the stronger the conjugate base, which makes the hydride anion a very strong base. And if you see it in a reaction, think base only as the reagent. You would get the hydride ion from something like sodium hydride, so NaH would be your source. Another example is this molecule, which has the abbreviation DBN. So DBN functions as a base only and not a nucleophile."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "And if you see it in a reaction, think base only as the reagent. You would get the hydride ion from something like sodium hydride, so NaH would be your source. Another example is this molecule, which has the abbreviation DBN. So DBN functions as a base only and not a nucleophile. You might think a lone pair of electrons on a nitrogen could function as a nucleophile, but not when you have this fused ring system here. That would be too bulky and prevents this from acting as a nucleophile. It does act as a base, though, and let's figure out which nitrogen gets protonated."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So DBN functions as a base only and not a nucleophile. You might think a lone pair of electrons on a nitrogen could function as a nucleophile, but not when you have this fused ring system here. That would be too bulky and prevents this from acting as a nucleophile. It does act as a base, though, and let's figure out which nitrogen gets protonated. Is it this sp three hybridized nitrogen or this sp two hybridized nitrogen? It turns out to be the sp two hybridized nitrogen, and let's look at why. So I'm gonna make this lone pair of electrons magenta, and that lone pair of electrons is gonna pick up a proton and form a bond."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "It does act as a base, though, and let's figure out which nitrogen gets protonated. Is it this sp three hybridized nitrogen or this sp two hybridized nitrogen? It turns out to be the sp two hybridized nitrogen, and let's look at why. So I'm gonna make this lone pair of electrons magenta, and that lone pair of electrons is gonna pick up a proton and form a bond. So that lone pair turns into this bond here, and now this nitrogen would have a plus one formal charge. So plus one formal charge on this nitrogen. This is the conjugate acid to DBN, and this conjugate acid is resonance stabilized."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So I'm gonna make this lone pair of electrons magenta, and that lone pair of electrons is gonna pick up a proton and form a bond. So that lone pair turns into this bond here, and now this nitrogen would have a plus one formal charge. So plus one formal charge on this nitrogen. This is the conjugate acid to DBN, and this conjugate acid is resonance stabilized. I could push these electrons into here and then push these electrons off onto the nitrogen, and let's follow those electrons. Let's make these electrons red. So this lone pair moves into here to form a double bond, and then let's make these electrons in here blue."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "This is the conjugate acid to DBN, and this conjugate acid is resonance stabilized. I could push these electrons into here and then push these electrons off onto the nitrogen, and let's follow those electrons. Let's make these electrons red. So this lone pair moves into here to form a double bond, and then let's make these electrons in here blue. The blue electrons come off onto this nitrogen, and we still have a bond to our hydrogen in here, which moves the formal charge to this other nitrogen. So this nitrogen now has a plus one formal charge. So our conjugate acid is resonance stabilized."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So this lone pair moves into here to form a double bond, and then let's make these electrons in here blue. The blue electrons come off onto this nitrogen, and we still have a bond to our hydrogen in here, which moves the formal charge to this other nitrogen. So this nitrogen now has a plus one formal charge. So our conjugate acid is resonance stabilized. The positive charge is delocalized over two nitrogens, and because our conjugate acid is resonance stabilized, it's not very acidic, which means that the conjugate base is strong, and that's why this is a strong conjugate base even though it's a neutral molecule. So protonation occurs at the sp2 hybridized nitrogen and not the sp3. If you tried protonating the sp3, you wouldn't be able to delocalize the positive charge over both nitrogens."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So our conjugate acid is resonance stabilized. The positive charge is delocalized over two nitrogens, and because our conjugate acid is resonance stabilized, it's not very acidic, which means that the conjugate base is strong, and that's why this is a strong conjugate base even though it's a neutral molecule. So protonation occurs at the sp2 hybridized nitrogen and not the sp3. If you tried protonating the sp3, you wouldn't be able to delocalize the positive charge over both nitrogens. So there's a similar molecule to DBN, which is abbreviated DBU, and it acts in the same way, only as a base in a reaction. Our third category is where our reagent is a strong nucleophile and a strong base, and a good example of that is the hydroxide ion. We've already talked about why the hydroxide ion is a strong nucleophile, and we know just from experience that hydroxide is a strong base."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "If you tried protonating the sp3, you wouldn't be able to delocalize the positive charge over both nitrogens. So there's a similar molecule to DBN, which is abbreviated DBU, and it acts in the same way, only as a base in a reaction. Our third category is where our reagent is a strong nucleophile and a strong base, and a good example of that is the hydroxide ion. We've already talked about why the hydroxide ion is a strong nucleophile, and we know just from experience that hydroxide is a strong base. So something like sodium hydroxide is used all the time in general chemistry. If we replace the hydrogen with an alkyl group, we form an alkoxide ion, which functions in a similar way to the hydroxide ion. So they're both examples of strong nucleophiles, strong bases."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "We've already talked about why the hydroxide ion is a strong nucleophile, and we know just from experience that hydroxide is a strong base. So something like sodium hydroxide is used all the time in general chemistry. If we replace the hydrogen with an alkyl group, we form an alkoxide ion, which functions in a similar way to the hydroxide ion. So they're both examples of strong nucleophiles, strong bases. Our fourth and last category is weak nucleophile, weak base. And the water molecule we know is a weak nucleophile. It does not have a negative one formal charge on the oxygen."}, {"video_title": "Elimination vs substitution reagent.mp3", "Sentence": "So they're both examples of strong nucleophiles, strong bases. Our fourth and last category is weak nucleophile, weak base. And the water molecule we know is a weak nucleophile. It does not have a negative one formal charge on the oxygen. And water, of course, is a weak base. So the conjugate acid would be H3O+, right? So just add a H plus to H2O, and you get H3O+."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I want to make a quick correction to the last video. It doesn't really affect the learning of the last video, but I just want to make sure that you understand that I got the math a little bit wrong in the last video. I said that you had this state 300,000 years. So we talk about the Big Bang happening 13.7 billion years ago. And then I talk about this state of affairs where we're maybe 30 million light years away from the edge of the observable universe, the current observable universe. And I said that this was about 300,000 years after the Big Bang. That's what I talked about in the last video."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we talk about the Big Bang happening 13.7 billion years ago. And then I talk about this state of affairs where we're maybe 30 million light years away from the edge of the observable universe, the current observable universe. And I said that this was about 300,000 years after the Big Bang. That's what I talked about in the last video. That was our starting point when the photon started leaving that point, and obviously the universe expanding. The photon, it kind of traversed more and more, but still had more and more to travel as the universe expanded, as all of space expanded. But this is 300,000 years after the Big Bang."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's what I talked about in the last video. That was our starting point when the photon started leaving that point, and obviously the universe expanding. The photon, it kind of traversed more and more, but still had more and more to travel as the universe expanded, as all of space expanded. But this is 300,000 years after the Big Bang. Now my brain, because I was kind of not thinking hard enough about it, I said, hey, this was 13.4 billion years ago. That's what I incorrectly said in the last video. I said that this is 13.4 billion years ago."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is 300,000 years after the Big Bang. Now my brain, because I was kind of not thinking hard enough about it, I said, hey, this was 13.4 billion years ago. That's what I incorrectly said in the last video. I said that this is 13.4 billion years ago. That's what I said in the last video, and that is wrong. Because if this was 13.4 billion years ago, this would have been 300 million years after the Big Bang. We were talking about only 300,000 years after the Big Bang, so it wouldn't have taken that many decimal places off of something in the billions."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I said that this is 13.4 billion years ago. That's what I said in the last video, and that is wrong. Because if this was 13.4 billion years ago, this would have been 300 million years after the Big Bang. We were talking about only 300,000 years after the Big Bang, so it wouldn't have taken that many decimal places off of something in the billions. The correct answer is this would have been only a little less than 13.7 billion years. It actually wouldn't have even made the significant digits. This is still approximately 13.7 billion years ago."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We were talking about only 300,000 years after the Big Bang, so it wouldn't have taken that many decimal places off of something in the billions. The correct answer is this would have been only a little less than 13.7 billion years. It actually wouldn't have even made the significant digits. This is still approximately 13.7 billion years ago. So I wanted to just make that correction. It was a slight error. I shouldn't have viewed this as 0.3 billion years."}, {"video_title": "Radius of observable universe (correction) Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is still approximately 13.7 billion years ago. So I wanted to just make that correction. It was a slight error. I shouldn't have viewed this as 0.3 billion years. This is only 0.3 million years. It doesn't even basically change the precision on this number right over here. So I just wanted to clear that up, but hopefully it doesn't affect your understanding too much."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's dependent upon K, which is the spring constant, or the force constant, and the reduced mass. Where the reduced mass is equal to the mass, M1 times M2, over M1 plus M2. And so here's your bond as a spring, with M1 and M2 on either end. And so if you increase the force constant, increase the spring constant, so that's like increasing the strength of the bond, obviously you would increase the frequency, just looking at the math here. So you increase the frequency of bond vibration. And so a stronger bond, a stronger bond vibrates faster than a weaker bond. And also if you decrease the reduced mass, so if you decrease the reduced mass, mathematically that's also going to increase the frequency of vibration here like that."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if you increase the force constant, increase the spring constant, so that's like increasing the strength of the bond, obviously you would increase the frequency, just looking at the math here. So you increase the frequency of bond vibration. And so a stronger bond, a stronger bond vibrates faster than a weaker bond. And also if you decrease the reduced mass, so if you decrease the reduced mass, mathematically that's also going to increase the frequency of vibration here like that. In the first video, we talked about relating the frequency to wave numbers, right? So a wave number, you take the frequency, divide by the speed of light in centimeters per second. And so if you take this over here, and just divide by the speed of light, which is C, you're going to get wave numbers."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And also if you decrease the reduced mass, so if you decrease the reduced mass, mathematically that's also going to increase the frequency of vibration here like that. In the first video, we talked about relating the frequency to wave numbers, right? So a wave number, you take the frequency, divide by the speed of light in centimeters per second. And so if you take this over here, and just divide by the speed of light, which is C, you're going to get wave numbers. And so this is good to use to approximate the wave number, where you would find your signals in IR spectroscopy. There's one more thing we have to do though, to get the proper units for wave number. And that's because here the reduced mass is in grams, and so we need to make that atomic mass units, or AMU."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if you take this over here, and just divide by the speed of light, which is C, you're going to get wave numbers. And so this is good to use to approximate the wave number, where you would find your signals in IR spectroscopy. There's one more thing we have to do though, to get the proper units for wave number. And that's because here the reduced mass is in grams, and so we need to make that atomic mass units, or AMU. And so we can divide by Avogadro's number to get that. So I'm not too concerned with units here. I'm just going to show you real quickly where this equation comes from."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that's because here the reduced mass is in grams, and so we need to make that atomic mass units, or AMU. And so we can divide by Avogadro's number to get that. So I'm not too concerned with units here. I'm just going to show you real quickly where this equation comes from. So wave number is equal to one over two pi C, times the square root of K over the reduced mass. And to get our proper units, we have to divide by Avogadro's number, which is 6.02 times 10 to the 23rd. All right, so let's continue here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I'm just going to show you real quickly where this equation comes from. So wave number is equal to one over two pi C, times the square root of K over the reduced mass. And to get our proper units, we have to divide by Avogadro's number, which is 6.02 times 10 to the 23rd. All right, so let's continue here. That'd be one over two pi C, times the square root. I'm going to make a little bit of a long square root sign here, because we still have the force constant over the reduced mass. And now we could just move Avogadro's number up to the numerator here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's continue here. That'd be one over two pi C, times the square root. I'm going to make a little bit of a long square root sign here, because we still have the force constant over the reduced mass. And now we could just move Avogadro's number up to the numerator here. So that's 6.02 times 10 to the 23rd. Let's go ahead and find the square root of Avogadro's number. All right, so the square root of Avogadro's number is equal to, so we have the square root of 6.02 times 10 to the 23rd."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And now we could just move Avogadro's number up to the numerator here. So that's 6.02 times 10 to the 23rd. Let's go ahead and find the square root of Avogadro's number. All right, so the square root of Avogadro's number is equal to, so we have the square root of 6.02 times 10 to the 23rd. And we get that number is 7.76 times 10 to the 11th. So let's go ahead and write that down here. So now we have wave number is equal to 7.76 times 10 to the 11th, divided by two pi, C is the speed of light in centimeters per second, that's three times 10 to the 10th."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so the square root of Avogadro's number is equal to, so we have the square root of 6.02 times 10 to the 23rd. And we get that number is 7.76 times 10 to the 11th. So let's go ahead and write that down here. So now we have wave number is equal to 7.76 times 10 to the 11th, divided by two pi, C is the speed of light in centimeters per second, that's three times 10 to the 10th. And then we still have the square root of the force constant over the reduced mass. So let's go ahead and solve it even further here. We need to divide that number by two."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So now we have wave number is equal to 7.76 times 10 to the 11th, divided by two pi, C is the speed of light in centimeters per second, that's three times 10 to the 10th. And then we still have the square root of the force constant over the reduced mass. So let's go ahead and solve it even further here. We need to divide that number by two. We need to divide that number by pi. And we need to divide that number by the speed of light in centimeters per second. So divide by three times 10 to the 10th."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We need to divide that number by two. We need to divide that number by pi. And we need to divide that number by the speed of light in centimeters per second. So divide by three times 10 to the 10th. And we get 4.12. And so we arrive at this wave number is equal to 4.12 times the square root of K over the reduced mass. And so here we have a nice little equation where we can approximate the wave number for different bonds."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So divide by three times 10 to the 10th. And we get 4.12. And so we arrive at this wave number is equal to 4.12 times the square root of K over the reduced mass. And so here we have a nice little equation where we can approximate the wave number for different bonds. So where we would expect to see the signal for different bonds stretching. And let me use a different color to point these things out here. So K, call this the spring constant or the force constant."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so here we have a nice little equation where we can approximate the wave number for different bonds. So where we would expect to see the signal for different bonds stretching. And let me use a different color to point these things out here. So K, call this the spring constant or the force constant. The units are dynes per centimeter. And then we have the reduced mass here in AMUs. And so let's do some calculations."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So K, call this the spring constant or the force constant. The units are dynes per centimeter. And then we have the reduced mass here in AMUs. And so let's do some calculations. And so let's see if we can approximate where we would find the signals for some bonds. And so let's get some more room down here. And let's start with a carbon-hydrogen bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so let's do some calculations. And so let's see if we can approximate where we would find the signals for some bonds. And so let's get some more room down here. And let's start with a carbon-hydrogen bond. So we have a carbon-hydrogen bond here. First thing we could do is calculate the reduced mass. So the reduced mass is equal to the atomic mass of carbon is 12."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's start with a carbon-hydrogen bond. So we have a carbon-hydrogen bond here. First thing we could do is calculate the reduced mass. So the reduced mass is equal to the atomic mass of carbon is 12. Mass of hydrogen is one. So 12 times one over 12 plus one. And we did this calculation in the previous video and we got.923."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the reduced mass is equal to the atomic mass of carbon is 12. Mass of hydrogen is one. So 12 times one over 12 plus one. And we did this calculation in the previous video and we got.923. So we're gonna plug this into our equation for wave number. So wave number is equal to 4.12 times the square root of the force constant. For a single bond, right, we're talking about a carbon-hydrogen bond here, it's a single bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we did this calculation in the previous video and we got.923. So we're gonna plug this into our equation for wave number. So wave number is equal to 4.12 times the square root of the force constant. For a single bond, right, we're talking about a carbon-hydrogen bond here, it's a single bond. You can use a force constant of five times 10 to the fifth dynes per centimeter. And we're gonna divide that by 0.923, the reduced mass. And so let's get out the calculator and do that calculation."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For a single bond, right, we're talking about a carbon-hydrogen bond here, it's a single bond. You can use a force constant of five times 10 to the fifth dynes per centimeter. And we're gonna divide that by 0.923, the reduced mass. And so let's get out the calculator and do that calculation. Alright, so we have five times 10 to the fifth. We're going to divide that number by.923. We're going to take the square root of our answer here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so let's get out the calculator and do that calculation. Alright, so we have five times 10 to the fifth. We're going to divide that number by.923. We're going to take the square root of our answer here. And then we're going to multiply that by 4.12. And this gives us 3,032. So 3,032 wave numbers."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're going to take the square root of our answer here. And then we're going to multiply that by 4.12. And this gives us 3,032. So 3,032 wave numbers. So 3,032 units would be one over centimeters. And so this is what we calculated, right? So this is the approximate wave number."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 3,032 wave numbers. So 3,032 units would be one over centimeters. And so this is what we calculated, right? So this is the approximate wave number. So this is approximately where you would find the signal for a carbon-hydrogen bond stretching. So if you're looking at an IR spectrum. And the actual signal is pretty close to this number."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is the approximate wave number. So this is approximately where you would find the signal for a carbon-hydrogen bond stretching. So if you're looking at an IR spectrum. And the actual signal is pretty close to this number. So this is a good approximation. And that's why it's kind of useful to use this equation here. Let's do it again for carbon-oxygen."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the actual signal is pretty close to this number. So this is a good approximation. And that's why it's kind of useful to use this equation here. Let's do it again for carbon-oxygen. So it's still a single bond, but this is carbon-oxygen this time. And so the reduced mass is going to change. But the force constant is going to stay the same because in both cases, we're talking about a single bond here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's do it again for carbon-oxygen. So it's still a single bond, but this is carbon-oxygen this time. And so the reduced mass is going to change. But the force constant is going to stay the same because in both cases, we're talking about a single bond here. So we're going to pretend like they're exactly the same strength, which, just to help us for our calculations to approximate things. All right, so the reduced mass would be equal to, carbon is 12, oxygen is 16. So we have 12 times 16 over 12 plus 16."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But the force constant is going to stay the same because in both cases, we're talking about a single bond here. So we're going to pretend like they're exactly the same strength, which, just to help us for our calculations to approximate things. All right, so the reduced mass would be equal to, carbon is 12, oxygen is 16. So we have 12 times 16 over 12 plus 16. And let's do that math really quickly. So we have 12 times 16, which is 192, divided by 28, gives us about 6.9. So we're going to say the reduced mass is approximately 6.9."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have 12 times 16 over 12 plus 16. And let's do that math really quickly. So we have 12 times 16, which is 192, divided by 28, gives us about 6.9. So we're going to say the reduced mass is approximately 6.9. And let's calculate the approximate wave number. So where would we expect to find this signal? So 4.12 times the square root."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to say the reduced mass is approximately 6.9. And let's calculate the approximate wave number. So where would we expect to find this signal? So 4.12 times the square root. We're going to use the same number, right? We're going to use five times 10 to the fifth. So we're saying we have a single bond here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 4.12 times the square root. We're going to use the same number, right? We're going to use five times 10 to the fifth. So we're saying we have a single bond here. So again, these are just approximations. So this is five times 10 to the fifth. We're going to divide that by the reduced mass, which is 6.9."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're saying we have a single bond here. So again, these are just approximations. So this is five times 10 to the fifth. We're going to divide that by the reduced mass, which is 6.9. And let's see what we get for the wave number. So let's do that math. All right, so we have five times 10 to the fifth."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're going to divide that by the reduced mass, which is 6.9. And let's see what we get for the wave number. So let's do that math. All right, so we have five times 10 to the fifth. We're going to divide that by 6.9. We're going to take the square root of our answer. And then we're going to multiply by 4.12 to get the approximate wave number, which is 1109."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we have five times 10 to the fifth. We're going to divide that by 6.9. We're going to take the square root of our answer. And then we're going to multiply by 4.12 to get the approximate wave number, which is 1109. All right, so we have 1109, one over centimeters for our wave number here. And let's compare these two wave numbers. So what did we do in our calculations?"}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And then we're going to multiply by 4.12 to get the approximate wave number, which is 1109. All right, so we have 1109, one over centimeters for our wave number here. And let's compare these two wave numbers. So what did we do in our calculations? So let me use blue here. So what did we do? We switched out a hydrogen for an oxygen."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So what did we do in our calculations? So let me use blue here. So what did we do? We switched out a hydrogen for an oxygen. So we increased the mass of the second atom, or the nucleus if you want to think about it that way. So you increase the mass of the second atom. And what happened to the wave number?"}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We switched out a hydrogen for an oxygen. So we increased the mass of the second atom, or the nucleus if you want to think about it that way. So you increase the mass of the second atom. And what happened to the wave number? The wave number decreased. We went from 3032 to 1109. And notice what we did, right?"}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And what happened to the wave number? The wave number decreased. We went from 3032 to 1109. And notice what we did, right? The reduced mass was.923 and it went up to 6.9. So we increased the reduced mass and we decreased the frequency, or we decreased the wave number. And so again, this is approximately where you would find the signal for a carbon-oxygen single bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And notice what we did, right? The reduced mass was.923 and it went up to 6.9. So we increased the reduced mass and we decreased the frequency, or we decreased the wave number. And so again, this is approximately where you would find the signal for a carbon-oxygen single bond. Not exact, because obviously we're just using really simple numbers here, but it's pretty close. All right, let's do a carbon-carbon double bond now. So carbon-carbon double bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so again, this is approximately where you would find the signal for a carbon-oxygen single bond. Not exact, because obviously we're just using really simple numbers here, but it's pretty close. All right, let's do a carbon-carbon double bond now. So carbon-carbon double bond. Let's calculate the reduced mass. The reduced mass would be 12 times 12 over 12 plus 12. And when you do that math, you're gonna get six."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So carbon-carbon double bond. Let's calculate the reduced mass. The reduced mass would be 12 times 12 over 12 plus 12. And when you do that math, you're gonna get six. So I think we did that in the previous video too. So let's plug this into our equation. So the approximate wave number would be equal to 4.12 times the square root."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And when you do that math, you're gonna get six. So I think we did that in the previous video too. So let's plug this into our equation. So the approximate wave number would be equal to 4.12 times the square root. All right, so what are we gonna plug in now for the force constant? Well, we have a double bond here, right? We have a double bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the approximate wave number would be equal to 4.12 times the square root. All right, so what are we gonna plug in now for the force constant? Well, we have a double bond here, right? We have a double bond. So a carbon-carbon double bond. And if we just approximate and say that a double bond is twice as strong as a single bond, we could just take this number and multiply it by two. And so that would be 10 times 10 to the fifth."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have a double bond. So a carbon-carbon double bond. And if we just approximate and say that a double bond is twice as strong as a single bond, we could just take this number and multiply it by two. And so that would be 10 times 10 to the fifth. So we're just saying that a double bond is close to being twice as strong, and again, not a perfect number, but it's a good approximation, and it just gives us an idea about where these wave numbers come from. So now we solve for the wave number, the approximate wave number here. So we have 10 times 10 to the fifth."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that would be 10 times 10 to the fifth. So we're just saying that a double bond is close to being twice as strong, and again, not a perfect number, but it's a good approximation, and it just gives us an idea about where these wave numbers come from. So now we solve for the wave number, the approximate wave number here. So we have 10 times 10 to the fifth. We're going to divide that by six. We're going to take the square root of our answer, and then we're going to multiply by 4.12. And this is going to give us somewhere around 1682."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have 10 times 10 to the fifth. We're going to divide that by six. We're going to take the square root of our answer, and then we're going to multiply by 4.12. And this is going to give us somewhere around 1682. All right, so our calculation gives us 1,682, where we would find the signal. Again, not perfect, but a decent approximation. Not a perfect number, but pretty close to where you would find the signal for a carbon-carbon double bond on an IR spectrum."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this is going to give us somewhere around 1682. All right, so our calculation gives us 1,682, where we would find the signal. Again, not perfect, but a decent approximation. Not a perfect number, but pretty close to where you would find the signal for a carbon-carbon double bond on an IR spectrum. Let's do one more. One more, let's get some more room down here. So let's see if we can squeeze in one more calculation."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Not a perfect number, but pretty close to where you would find the signal for a carbon-carbon double bond on an IR spectrum. Let's do one more. One more, let's get some more room down here. So let's see if we can squeeze in one more calculation. Let's do a carbon-carbon triple bond. So a carbon-carbon triple bond, the reduced mass obviously would be the same as above. It's six."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's see if we can squeeze in one more calculation. Let's do a carbon-carbon triple bond. So a carbon-carbon triple bond, the reduced mass obviously would be the same as above. It's six. So we're going to be changing the force constant here. So the wave number would be equal to 4.12 times. So what would we plug in for the force constant now?"}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's six. So we're going to be changing the force constant here. So the wave number would be equal to 4.12 times. So what would we plug in for the force constant now? Well, this would be, this was double a single bond, so we need to go for triple a single bond. We have a triple bond here. So three times five times 10 to the fifth."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So what would we plug in for the force constant now? Well, this would be, this was double a single bond, so we need to go for triple a single bond. We have a triple bond here. So three times five times 10 to the fifth. So that's 15 times 10 to the fifth. So this would be three, we're approximating it. This thing is three times as strong as a single bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So three times five times 10 to the fifth. So that's 15 times 10 to the fifth. So this would be three, we're approximating it. This thing is three times as strong as a single bond. 15 times 10 to the fifth. We're going to divide that by six. And let's see what happens now."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This thing is three times as strong as a single bond. 15 times 10 to the fifth. We're going to divide that by six. And let's see what happens now. So we have 15 times 10 to the fifth. We're going to divide that by six. We're going to take the square root of our answer once again."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's see what happens now. So we have 15 times 10 to the fifth. We're going to divide that by six. We're going to take the square root of our answer once again. So square root of our answer. And we get 500. We multiply that by 4.12 and we get 2,060."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're going to take the square root of our answer once again. So square root of our answer. And we get 500. We multiply that by 4.12 and we get 2,060. So we get 2,060 here for the wave number. So however you want to write this. So what happened?"}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We multiply that by 4.12 and we get 2,060. So we get 2,060 here for the wave number. So however you want to write this. So what happened? What happened between these last two examples here? We increased k. That's the only thing that changed between these two calculations. We increased k, we increased the force constant."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So what happened? What happened between these last two examples here? We increased k. That's the only thing that changed between these two calculations. We increased k, we increased the force constant. We're saying a triple bond is stronger than a double bond. And what happened to the frequency, or the wave number? The wave number increased."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We increased k, we increased the force constant. We're saying a triple bond is stronger than a double bond. And what happened to the frequency, or the wave number? The wave number increased. So increase k, you increase the frequency, you increase the wave number here. Finally, let's plot these on an IR spectrum. So let's first start with the carbon-hydrogen single bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The wave number increased. So increase k, you increase the frequency, you increase the wave number here. Finally, let's plot these on an IR spectrum. So let's first start with the carbon-hydrogen single bond. So somewhere in the 3,032. So let's go down here for our IR spectrum. So 3,032."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's first start with the carbon-hydrogen single bond. So somewhere in the 3,032. So let's go down here for our IR spectrum. So 3,032. So that would be somewhere around in here. So this would be approximately where we would find the signal for a carbon-hydrogen bond. And this just allows us to think about regions of our IR spectra."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 3,032. So that would be somewhere around in here. So this would be approximately where we would find the signal for a carbon-hydrogen bond. And this just allows us to think about regions of our IR spectra. So somewhere in this region, right in here, is where you would find a bond to hydrogen. So we just said this was carbon to hydrogen here, but you could generalize this by saying it's any bond to hydrogen, because it's the hydrogen, it's the smaller mass that causes the increased frequency, the increased value for the wave number. Let's look at our next one here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this just allows us to think about regions of our IR spectra. So somewhere in this region, right in here, is where you would find a bond to hydrogen. So we just said this was carbon to hydrogen here, but you could generalize this by saying it's any bond to hydrogen, because it's the hydrogen, it's the smaller mass that causes the increased frequency, the increased value for the wave number. Let's look at our next one here. Let's look at carbon-oxygen. So for carbon-oxygen, we have 1,109. So that's a big difference."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at our next one here. Let's look at carbon-oxygen. So for carbon-oxygen, we have 1,109. So that's a big difference. It's a single bond. 1,109, that's way down here. So 1,109 would be somewhere around in here for this carbon-oxygen bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's a big difference. It's a single bond. 1,109, that's way down here. So 1,109 would be somewhere around in here for this carbon-oxygen bond. And so it turns out that this region right in here is where you find, it's a single bond region. So this is the single bond region when you're not talking about hydrogen. So this would be the single bond region."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 1,109 would be somewhere around in here for this carbon-oxygen bond. And so it turns out that this region right in here is where you find, it's a single bond region. So this is the single bond region when you're not talking about hydrogen. So this would be the single bond region. This over here would be the bond to hydrogen region. And then we have two more regions to talk about. We have our double bond region."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this would be the single bond region. This over here would be the bond to hydrogen region. And then we have two more regions to talk about. We have our double bond region. Let's use green for that. So carbon-carbon double bond, we calculated it's approximately 1,682 wave numbers. So let's find that on our spectrum here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have our double bond region. Let's use green for that. So carbon-carbon double bond, we calculated it's approximately 1,682 wave numbers. So let's find that on our spectrum here. So 1,682. So that would be approximately in here. So this would be like somewhere in there, in that range."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's find that on our spectrum here. So 1,682. So that would be approximately in here. So this would be like somewhere in there, in that range. And so we could say that's approximately where we find the signal for a carbon-carbon double bond. And your double bond region, so let me write double bond region, double bond region is right around in that area. So double bond region, expect to see a signal for a double bond in this area here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this would be like somewhere in there, in that range. And so we could say that's approximately where we find the signal for a carbon-carbon double bond. And your double bond region, so let me write double bond region, double bond region is right around in that area. So double bond region, expect to see a signal for a double bond in this area here. So somewhere in here. So this is approximately your double bond region on IR spectrum. And then finally, our last thing that we're gonna think about is our triple bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So double bond region, expect to see a signal for a double bond in this area here. So somewhere in here. So this is approximately your double bond region on IR spectrum. And then finally, our last thing that we're gonna think about is our triple bond. So our triple bond, we calculated an approximate value for a carbon-carbon triple bond. We got 2,060. So 2,060 would be somewhere in here."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, our last thing that we're gonna think about is our triple bond. So our triple bond, we calculated an approximate value for a carbon-carbon triple bond. We got 2,060. So 2,060 would be somewhere in here. So that's approximately where we'd find our triple bond region. And that usually does go about 2,100 to 2,300. So in here somewhere."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So 2,060 would be somewhere in here. So that's approximately where we'd find our triple bond region. And that usually does go about 2,100 to 2,300. So in here somewhere. So pretty small, but just to give you an idea of these different regions. So here's the triple bond region. So triple bond."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in here somewhere. So pretty small, but just to give you an idea of these different regions. So here's the triple bond region. So triple bond. And so hopefully that allows you to think about why you get these different wave numbers. It has to do with two factors. It has to do with the force constant and it has to do with the reduced mass."}, {"video_title": "Signal characteristics - wavenumber Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So triple bond. And so hopefully that allows you to think about why you get these different wave numbers. It has to do with two factors. It has to do with the force constant and it has to do with the reduced mass. And if you think about those two factors, you can think about where the signal should appear for these bonds, for these bonds stretching. So you can figure out the approximate wave number. And you should know these wave numbers because it's gonna help you when you read your IR spectrum."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So to do that, let's think about the different layers. And actually, I want to make one quick correction on the last video. Over here, I had drawn these arrows going in that direction, and based on how I defined them, they should be going into the page, and so they should have had these Xs there. Now, with that out of the way, let's draw a little diagram of what happens at the early stages of these divergent plate boundaries. So you might have your crust, and maybe it's continental crust. So this right here is the Earth's crust. And then you have the solid part of the mantle, and the combination of them is the lithosphere."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Now, with that out of the way, let's draw a little diagram of what happens at the early stages of these divergent plate boundaries. So you might have your crust, and maybe it's continental crust. So this right here is the Earth's crust. And then you have the solid part of the mantle, and the combination of them is the lithosphere. And then you have the liquid part, or the super-hot part of the mantle. So this down here is magma. It hasn't solidified."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And then you have the solid part of the mantle, and the combination of them is the lithosphere. And then you have the liquid part, or the super-hot part of the mantle. So this down here is magma. It hasn't solidified. It's hot enough to be in the liquid state. And all of this combined, so this right here, we consider the mantle. Now, there's some debate, and we'll talk about this in the next video, of how hot spots actually form."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "It hasn't solidified. It's hot enough to be in the liquid state. And all of this combined, so this right here, we consider the mantle. Now, there's some debate, and we'll talk about this in the next video, of how hot spots actually form. It could be these mantle plumes that start at the border between the mantle and the core. It could be some type of convection currents in the actual mantle. We'll talk more about that in the next video, or maybe a few videos from now."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Now, there's some debate, and we'll talk about this in the next video, of how hot spots actually form. It could be these mantle plumes that start at the border between the mantle and the core. It could be some type of convection currents in the actual mantle. We'll talk more about that in the next video, or maybe a few videos from now. But let's take it for granted that hot spots form in the mantle. So let's say we have an area of magma right here that is particularly hot. Let me draw it."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "We'll talk more about that in the next video, or maybe a few videos from now. But let's take it for granted that hot spots form in the mantle. So let's say we have an area of magma right here that is particularly hot. Let me draw it. Let me do this in another color, blue and pink. So this is particularly hot magma here. And we know, or maybe we don't know, or you learn known right now, if you take the same material and you make it hotter, it's going to become less dense, because the particles essentially are going to bump into each other with more kinetic energy and have more space in between them."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Let me draw it. Let me do this in another color, blue and pink. So this is particularly hot magma here. And we know, or maybe we don't know, or you learn known right now, if you take the same material and you make it hotter, it's going to become less dense, because the particles essentially are going to bump into each other with more kinetic energy and have more space in between them. And so this really hot part of the magma, or this really hot part of the mantle, it is going to move upwards because it is less dense. It will have buoyancy. And as it moves upwards, it will heat up the things around it, and it will eventually make its way into the lithosphere."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And we know, or maybe we don't know, or you learn known right now, if you take the same material and you make it hotter, it's going to become less dense, because the particles essentially are going to bump into each other with more kinetic energy and have more space in between them. And so this really hot part of the magma, or this really hot part of the mantle, it is going to move upwards because it is less dense. It will have buoyancy. And as it moves upwards, it will heat up the things around it, and it will eventually make its way into the lithosphere. And it will kind of be able to break through the lithosphere, because it's so hot, it can kind of melt its way through. So let's fast forward this a little bit. So this is step one up here."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And as it moves upwards, it will heat up the things around it, and it will eventually make its way into the lithosphere. And it will kind of be able to break through the lithosphere, because it's so hot, it can kind of melt its way through. So let's fast forward this a little bit. So this is step one up here. Now step two, this hot magma is rising now through the lithosphere, and so it's going to create a hot spot. It's going to create a dome in the lithosphere and actually on the crust. And so it might look like this."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So this is step one up here. Now step two, this hot magma is rising now through the lithosphere, and so it's going to create a hot spot. It's going to create a dome in the lithosphere and actually on the crust. And so it might look like this. So the crust is now going to have a dome in it. And this is the original lithosphere. And it's now kind of been broken in two by this hot spot."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And so it might look like this. So the crust is now going to have a dome in it. And this is the original lithosphere. And it's now kind of been broken in two by this hot spot. So the lithosphere is now broken in two, or it's about to be broken in two, by this hot spot. So all of this is still the lithosphere. I'll just write litho for short."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And it's now kind of been broken in two by this hot spot. So the lithosphere is now broken in two, or it's about to be broken in two, by this hot spot. So all of this is still the lithosphere. I'll just write litho for short. This up here is the crust. And if you take any rigid material, and the crust and the lithosphere, for that matter, they're rigid, and you push outward on it, it won't stretch nicely like a nice elastic balloon. It'll start to crack and have to be pulled apart in order to kind of take the pushing from below."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "I'll just write litho for short. This up here is the crust. And if you take any rigid material, and the crust and the lithosphere, for that matter, they're rigid, and you push outward on it, it won't stretch nicely like a nice elastic balloon. It'll start to crack and have to be pulled apart in order to kind of take the pushing from below. So this crust is going to start to crack. And actually the best example where you see this is actually in like sourdough bread that has really hard shells around it. You see sourdough bread."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "It'll start to crack and have to be pulled apart in order to kind of take the pushing from below. So this crust is going to start to crack. And actually the best example where you see this is actually in like sourdough bread that has really hard shells around it. You see sourdough bread. Let me see if I can draw a roll of sourdough bread. It has all of these cracks in the surface, and that's because the outer layer, the outer shell of the bread is really rigid. And so when the inside heats up and the surface area has to expand, these kind of rifts form in the bread to allow that kind of rigid shell to actually expand."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "You see sourdough bread. Let me see if I can draw a roll of sourdough bread. It has all of these cracks in the surface, and that's because the outer layer, the outer shell of the bread is really rigid. And so when the inside heats up and the surface area has to expand, these kind of rifts form in the bread to allow that kind of rigid shell to actually expand. And that exact same thing would happen to the crust, or actually the entire lithosphere. So let me draw this hot spot again. So now the hot spot, I want to do it in that pink color."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And so when the inside heats up and the surface area has to expand, these kind of rifts form in the bread to allow that kind of rigid shell to actually expand. And that exact same thing would happen to the crust, or actually the entire lithosphere. So let me draw this hot spot again. So now the hot spot, I want to do it in that pink color. Now the hot spot has gotten this far. This is the hot magma right over here. And if we fast forward a little bit, if we fast forward even a little bit more, then you could actually have the crust start to be fully pulled apart."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So now the hot spot, I want to do it in that pink color. Now the hot spot has gotten this far. This is the hot magma right over here. And if we fast forward a little bit, if we fast forward even a little bit more, then you could actually have the crust start to be fully pulled apart. So you fast forward a little bit more. The bottom boundary of the lithosphere maybe now starts to look something like this. The magma has kind of broken through the hardened part, the rigid part of the mantle."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And if we fast forward a little bit, if we fast forward even a little bit more, then you could actually have the crust start to be fully pulled apart. So you fast forward a little bit more. The bottom boundary of the lithosphere maybe now starts to look something like this. The magma has kind of broken through the hardened part, the rigid part of the mantle. So maybe it looks like this. Maybe it looks like this right now. You have the hot spot right over here."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "The magma has kind of broken through the hardened part, the rigid part of the mantle. So maybe it looks like this. Maybe it looks like this right now. You have the hot spot right over here. It's gotten that far now. And the crust on top, what we normally see, has now been pulled apart to kind of have to cover this new surface area. So now it kind of looks something like this."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "You have the hot spot right over here. It's gotten that far now. And the crust on top, what we normally see, has now been pulled apart to kind of have to cover this new surface area. So now it kind of looks something like this. So it's been pushed apart. Let me see how well I can draw this. So now it's been pushed apart."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So now it kind of looks something like this. So it's been pushed apart. Let me see how well I can draw this. So now it's been pushed apart. And as it gets pushed apart, it kind of thins out a little bit, as you can imagine it doing. It's almost exactly as the bread analogy. When you look at bread like this, the rift, the parts, the depressions where it was expanding most vigorously, those parts of the bread are actually thinner."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So now it's been pushed apart. And as it gets pushed apart, it kind of thins out a little bit, as you can imagine it doing. It's almost exactly as the bread analogy. When you look at bread like this, the rift, the parts, the depressions where it was expanding most vigorously, those parts of the bread are actually thinner. Like these parts of the bread are actually thinner, and they're not as hard as the parts that moved away. And you see that exact same thing happening with the land. And all of this stuff is going to start getting pushed."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "When you look at bread like this, the rift, the parts, the depressions where it was expanding most vigorously, those parts of the bread are actually thinner. Like these parts of the bread are actually thinner, and they're not as hard as the parts that moved away. And you see that exact same thing happening with the land. And all of this stuff is going to start getting pushed. All of this stuff is continuously getting pushed outward, essentially to kind of make space for this hot spot. Now this step right over here, you might have a volcano or two, but more important, you're going to have what's called a rift valley. Right now we're assuming that we're not at a below sea level yet, or we're assuming that this kind of depression in the land that you see here hasn't come in contact with another body of water, and so it'll just kind of become a little valley in between higher land."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And all of this stuff is going to start getting pushed. All of this stuff is continuously getting pushed outward, essentially to kind of make space for this hot spot. Now this step right over here, you might have a volcano or two, but more important, you're going to have what's called a rift valley. Right now we're assuming that we're not at a below sea level yet, or we're assuming that this kind of depression in the land that you see here hasn't come in contact with another body of water, and so it'll just kind of become a little valley in between higher land. And you actually see that on Earth. And the most famous is the African rift valley that's right about this region here. I actually have a better diagram that depicts the African rift valley right over here."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Right now we're assuming that we're not at a below sea level yet, or we're assuming that this kind of depression in the land that you see here hasn't come in contact with another body of water, and so it'll just kind of become a little valley in between higher land. And you actually see that on Earth. And the most famous is the African rift valley that's right about this region here. I actually have a better diagram that depicts the African rift valley right over here. It's this whole region of Africa is actually kind of a big valley created by a hot spot right over there. Now as the hot spot kind of keeps maturing, eventually some of the rift will become so depressed that it'll actually be below sea level. Remember, all of this land up here is being stretched apart."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "I actually have a better diagram that depicts the African rift valley right over here. It's this whole region of Africa is actually kind of a big valley created by a hot spot right over there. Now as the hot spot kind of keeps maturing, eventually some of the rift will become so depressed that it'll actually be below sea level. Remember, all of this land up here is being stretched apart. So let me go to the next step. The next step will be right over here. The land on top is now maybe below sea level, this next step, and it comes in contact with maybe an ocean or a sea."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Remember, all of this land up here is being stretched apart. So let me go to the next step. The next step will be right over here. The land on top is now maybe below sea level, this next step, and it comes in contact with maybe an ocean or a sea. And so now it might look like this. Now the land is super thin on top. I'll do my best to draw it."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "The land on top is now maybe below sea level, this next step, and it comes in contact with maybe an ocean or a sea. And so now it might look like this. Now the land is super thin on top. I'll do my best to draw it. So it's super thin on top, and remember, it kind of keeps getting pulled apart. It keeps getting pulled apart from this bubble of hot magma that's essentially coming up from below. Let me draw it like this."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "I'll do my best to draw it. So it's super thin on top, and remember, it kind of keeps getting pulled apart. It keeps getting pulled apart from this bubble of hot magma that's essentially coming up from below. Let me draw it like this. This is all solid rock here. What I drew in orange is the crust. This is kind of the rocky part."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Let me draw it like this. This is all solid rock here. What I drew in orange is the crust. This is kind of the rocky part. Actually, I shouldn't draw it. This is the rocky part of the mantle, so the combination is the lithosphere. And now you have the hot magma coming up like this."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "This is kind of the rocky part. Actually, I shouldn't draw it. This is the rocky part of the mantle, so the combination is the lithosphere. And now you have the hot magma coming up like this. And it might peek through every now and then and create a volcano there. Maybe it'll peek through and create a volcano there. But in general, it's going to keep pushing the land up and outwards."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And now you have the hot magma coming up like this. And it might peek through every now and then and create a volcano there. Maybe it'll peek through and create a volcano there. But in general, it's going to keep pushing the land up and outwards. And so this land, even though you're saying, hey, it's being pushed up, because of the outward motion, this land over here is going to be lower than the land around it, like the loaf of bread. If it gets low enough and below sea level, actually, and it comes in contact with the body of water, or even if it doesn't, actually, water will start to gather over there. And once again, we actually see that in the rift forming between the African and the Arabian plates."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "But in general, it's going to keep pushing the land up and outwards. And so this land, even though you're saying, hey, it's being pushed up, because of the outward motion, this land over here is going to be lower than the land around it, like the loaf of bread. If it gets low enough and below sea level, actually, and it comes in contact with the body of water, or even if it doesn't, actually, water will start to gather over there. And once again, we actually see that in the rift forming between the African and the Arabian plates. The Red Sea is actually an example of exactly that. The Arabian plate is moving away from the African plate because of this hot spot. This is pushing all of the land up and out right over here."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And once again, we actually see that in the rift forming between the African and the Arabian plates. The Red Sea is actually an example of exactly that. The Arabian plate is moving away from the African plate because of this hot spot. This is pushing all of the land up and out right over here. And so this is going out, that is going out. It's moving outward in every direction. And so it creates these depressions where water can flow inwards."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "This is pushing all of the land up and out right over here. And so this is going out, that is going out. It's moving outward in every direction. And so it creates these depressions where water can flow inwards. The rift valley hasn't had water flow into it the way the Red Sea has just yet. But if it kept happening, eventually, it's going to get low enough so that the water will flow into it. So the Red Sea is exactly that."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And so it creates these depressions where water can flow inwards. The rift valley hasn't had water flow into it the way the Red Sea has just yet. But if it kept happening, eventually, it's going to get low enough so that the water will flow into it. So the Red Sea is exactly that. You essentially have the Indian Ocean flowing into this rift that formed from this hot spot. And then if you fast forward a bunch, so that finally the magma can kind of surface. So let's fast forward from even this point even more."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So the Red Sea is exactly that. You essentially have the Indian Ocean flowing into this rift that formed from this hot spot. And then if you fast forward a bunch, so that finally the magma can kind of surface. So let's fast forward from even this point even more. So let's fast forward even more. And let's say now the land has been pushed a good bit apart. Now the hot spot has actually surfaced."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So let's fast forward from even this point even more. So let's fast forward even more. And let's say now the land has been pushed a good bit apart. Now the hot spot has actually surfaced. Now the crust might look something like this. The crust might look something like this. So it's been pushed apart a good bit at this point."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "Now the hot spot has actually surfaced. Now the crust might look something like this. The crust might look something like this. So it's been pushed apart a good bit at this point. Now we're talking about in the order of hundreds of thousands of years or tens of thousands of years. So the land, for example, the land that was here, this part of the land might now be out here. And this part of the land might now be out here."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "So it's been pushed apart a good bit at this point. Now we're talking about in the order of hundreds of thousands of years or tens of thousands of years. So the land, for example, the land that was here, this part of the land might now be out here. And this part of the land might now be out here. What's going to happen is that this hot spot is going to continue to fuel, and we're assuming everything's underwater at this point, since this depression that was created is now so low. The crust was stretched thin. We can assume that all of this is underwater."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And this part of the land might now be out here. What's going to happen is that this hot spot is going to continue to fuel, and we're assuming everything's underwater at this point, since this depression that was created is now so low. The crust was stretched thin. We can assume that all of this is underwater. The hot spot is essentially going to come out of underwater volcanoes and start creating what's now, this body of water's gotten large enough that we can call it a mid-oceanic ridge. And so it'll actually start creating, let me do this in a different color, it'll start creating an actual ridge. It'll actually create a ridge with volcanoes in the center."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "We can assume that all of this is underwater. The hot spot is essentially going to come out of underwater volcanoes and start creating what's now, this body of water's gotten large enough that we can call it a mid-oceanic ridge. And so it'll actually start creating, let me do this in a different color, it'll start creating an actual ridge. It'll actually create a ridge with volcanoes in the center. And so that's why, one, we see things like the Rift Valley in Africa. We see things like the Red Sea. And maybe even more importantly, that's why we see something like the mid-Atlantic Rift in the middle of the Atlantic Ocean, where you have all of this depressed land that was essentially analogous to that Rift Valley, but it's at a much later stage, and that's why it's able to collect water, because when the land was pushed out and stretched thin, water was able to flow into it."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "It'll actually create a ridge with volcanoes in the center. And so that's why, one, we see things like the Rift Valley in Africa. We see things like the Red Sea. And maybe even more importantly, that's why we see something like the mid-Atlantic Rift in the middle of the Atlantic Ocean, where you have all of this depressed land that was essentially analogous to that Rift Valley, but it's at a much later stage, and that's why it's able to collect water, because when the land was pushed out and stretched thin, water was able to flow into it. Going back to the bread analogy, it's essentially when this bread was baking and this part of the crust pushed outwards, you had this rift form, and then if there was some water on the bread, if it was raining or if it was connected to a body of water, water would have eventually flowed in here. And if that bread kept growing, this rift would have kept growing, eventually to the size of the Atlantic Ocean in our theoretical bread. And so that's why you have this huge depressed area where an ocean can form, but in the middle of it you have this actual submerged mountain chain, this submerged chain of volcanoes, this submerged ridge where the land actually does go up a little bit because of all that magma flowing directly out of it."}, {"video_title": "Plate tectonics Geological features of divergent plate boundaries Khan Academy.mp3", "Sentence": "And maybe even more importantly, that's why we see something like the mid-Atlantic Rift in the middle of the Atlantic Ocean, where you have all of this depressed land that was essentially analogous to that Rift Valley, but it's at a much later stage, and that's why it's able to collect water, because when the land was pushed out and stretched thin, water was able to flow into it. Going back to the bread analogy, it's essentially when this bread was baking and this part of the crust pushed outwards, you had this rift form, and then if there was some water on the bread, if it was raining or if it was connected to a body of water, water would have eventually flowed in here. And if that bread kept growing, this rift would have kept growing, eventually to the size of the Atlantic Ocean in our theoretical bread. And so that's why you have this huge depressed area where an ocean can form, but in the middle of it you have this actual submerged mountain chain, this submerged chain of volcanoes, this submerged ridge where the land actually does go up a little bit because of all that magma flowing directly out of it. So hopefully that clears up a little bit. That was always confusing to me, why you see uplifted land, but then everything around the uplifted land is much lower, and why the whole thing is submerged as it's moving away. So hopefully that cleared things up a little bit."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I number my carbons, this would be carbon one, two, three, four, five. You can see there's an OH at carbon two, and there are five carbons, so that's where we get the two-pentanol. And there's a methyl group at carbon three, so that's three-methyl, two-pentanol. But we need to put in stereochemistry, because we know from an earlier video that carbon two is a chiral center, and carbon three is a chiral center. So we need to use the RS system to finish our name. So let's focus in at carbon two, and this drawing on the right will determine the configuration at this chiral center. So here's our carbon."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But we need to put in stereochemistry, because we know from an earlier video that carbon two is a chiral center, and carbon three is a chiral center. So we need to use the RS system to finish our name. So let's focus in at carbon two, and this drawing on the right will determine the configuration at this chiral center. So here's our carbon. If there's an OH coming out at us in space, we know there's a hydrogen going away from us in space. And our chiral center is directly connected to a carbon on the left, and this carbon is bonded to three hydrogens, so I'll draw those in there. And our chiral center is directly bonded to a carbon on the right, and this carbon on the right is directly bonded to another carbon, another carbon, and there must be a hydrogen coming out at us in space here."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's our carbon. If there's an OH coming out at us in space, we know there's a hydrogen going away from us in space. And our chiral center is directly connected to a carbon on the left, and this carbon is bonded to three hydrogens, so I'll draw those in there. And our chiral center is directly bonded to a carbon on the right, and this carbon on the right is directly bonded to another carbon, another carbon, and there must be a hydrogen coming out at us in space here. So let's assign priority to our four groups. Remember, that was our first step when we're trying to determine the configuration of a chirality center. So here's our chiral center."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And our chiral center is directly bonded to a carbon on the right, and this carbon on the right is directly bonded to another carbon, another carbon, and there must be a hydrogen coming out at us in space here. So let's assign priority to our four groups. Remember, that was our first step when we're trying to determine the configuration of a chirality center. So here's our chiral center. We look at the atoms directly bonded to that carbon. There's an oxygen directly bonded to that carbon. There's a hydrogen directly bonded to it."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's our chiral center. We look at the atoms directly bonded to that carbon. There's an oxygen directly bonded to that carbon. There's a hydrogen directly bonded to it. There's a carbon on the left, and there's a carbon on the right. So we assign priority based on atomic number. Out of those atoms, oxygen has the highest atomic number, so oxygen gets highest priority."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's a hydrogen directly bonded to it. There's a carbon on the left, and there's a carbon on the right. So we assign priority based on atomic number. Out of those atoms, oxygen has the highest atomic number, so oxygen gets highest priority. So we call that group a number one. So the OH group gets a number one. Hydrogen has the lowest atomic number, so hydrogen gets lowest priority, and we call that group four here."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Out of those atoms, oxygen has the highest atomic number, so oxygen gets highest priority. So we call that group a number one. So the OH group gets a number one. Hydrogen has the lowest atomic number, so hydrogen gets lowest priority, and we call that group four here. So hydrogen is a four. So now we have a tie. We have these two carbons."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Hydrogen has the lowest atomic number, so hydrogen gets lowest priority, and we call that group four here. So hydrogen is a four. So now we have a tie. We have these two carbons. Let me go ahead and circle them. So carbon obviously has the same atomic number. So to break this tie, we look at the atoms directly bonded to these carbons."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have these two carbons. Let me go ahead and circle them. So carbon obviously has the same atomic number. So to break this tie, we look at the atoms directly bonded to these carbons. The carbon on the left is directly bonded to three hydrogens. So we write down here three hydrogens, and the carbon on the right is directly bonded to two carbons and one hydrogen. So we write carbon, carbon, hydrogen."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So to break this tie, we look at the atoms directly bonded to these carbons. The carbon on the left is directly bonded to three hydrogens. So we write down here three hydrogens, and the carbon on the right is directly bonded to two carbons and one hydrogen. So we write carbon, carbon, hydrogen. Next, we look for the first point of difference, and that's the first atom here. So carbon versus hydrogen, carbon has the higher atomic number, so this group wins. So this group on the right is higher priority, which means this must be number two for our groups, and the methyl group gets a number three."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we write carbon, carbon, hydrogen. Next, we look for the first point of difference, and that's the first atom here. So carbon versus hydrogen, carbon has the higher atomic number, so this group wins. So this group on the right is higher priority, which means this must be number two for our groups, and the methyl group gets a number three. Now that we've assigned priority to our four groups, the next step was to orient the molecule so that the lowest priority group is projecting away from us, and that's what we have here because the hydrogen is the lowest priority group, so we can ignore this, and we can focus in on one, two, and three. So let me label this on the right. The OH was the highest priority, and this group to the right was the second highest priority, and the methyl group was the third highest priority."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this group on the right is higher priority, which means this must be number two for our groups, and the methyl group gets a number three. Now that we've assigned priority to our four groups, the next step was to orient the molecule so that the lowest priority group is projecting away from us, and that's what we have here because the hydrogen is the lowest priority group, so we can ignore this, and we can focus in on one, two, and three. So let me label this on the right. The OH was the highest priority, and this group to the right was the second highest priority, and the methyl group was the third highest priority. Next, we draw a circle and determine whether we're going around clockwise or counterclockwise. So we draw a circle from one to two to three, and obviously we're going around in a clockwise direction, and clockwise is R. So we are R at carbon two, so I write here two R. Now let's focus on carbon three. So we know that carbon three is a chiral center, so what is directly bonded to this carbon?"}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The OH was the highest priority, and this group to the right was the second highest priority, and the methyl group was the third highest priority. Next, we draw a circle and determine whether we're going around clockwise or counterclockwise. So we draw a circle from one to two to three, and obviously we're going around in a clockwise direction, and clockwise is R. So we are R at carbon two, so I write here two R. Now let's focus on carbon three. So we know that carbon three is a chiral center, so what is directly bonded to this carbon? Well, there's a carbon attached to three hydrogens, so there's a methyl group. There's a hydrogen that must be coming out at us in space, and on the right, there's a carbon bonded to two hydrogens and directly bonded to another carbon. On the left, there's a carbon directly bonded to an oxygen."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we know that carbon three is a chiral center, so what is directly bonded to this carbon? Well, there's a carbon attached to three hydrogens, so there's a methyl group. There's a hydrogen that must be coming out at us in space, and on the right, there's a carbon bonded to two hydrogens and directly bonded to another carbon. On the left, there's a carbon directly bonded to an oxygen. There must be a hydrogen going away from us, and this carbon is directly bonded to another carbon. So let's assign priority to our four groups. So here is our chiral center."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left, there's a carbon directly bonded to an oxygen. There must be a hydrogen going away from us, and this carbon is directly bonded to another carbon. So let's assign priority to our four groups. So here is our chiral center. We look at the atoms directly bonded to that carbon. There are three carbons, one, two, three, and a hydrogen. We know that hydrogen has the lowest atomic number, so hydrogen is the lowest priority, and we call that group four."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here is our chiral center. We look at the atoms directly bonded to that carbon. There are three carbons, one, two, three, and a hydrogen. We know that hydrogen has the lowest atomic number, so hydrogen is the lowest priority, and we call that group four. Now we have three carbons. We have a tie because carbon, of course, has the same atomic number. To break a tie, we look at the atoms that are bonded to those carbons."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that hydrogen has the lowest atomic number, so hydrogen is the lowest priority, and we call that group four. Now we have three carbons. We have a tie because carbon, of course, has the same atomic number. To break a tie, we look at the atoms that are bonded to those carbons. So let's start with the carbon on the right here. So the carbon on the right is directly bonded to one carbon and then two hydrogens. So we write carbon, hydrogen, hydrogen, so in decreasing atomic number, in order of decreasing atomic number."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "To break a tie, we look at the atoms that are bonded to those carbons. So let's start with the carbon on the right here. So the carbon on the right is directly bonded to one carbon and then two hydrogens. So we write carbon, hydrogen, hydrogen, so in decreasing atomic number, in order of decreasing atomic number. This carbon down here is directly bonded to three hydrogens, so one, two, and three. The carbon on the left is directly bonded to an oxygen, a carbon, and a hydrogen. So we put these in order of decreasing atomic number, so we put oxygen first, then carbon, then hydrogen."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we write carbon, hydrogen, hydrogen, so in decreasing atomic number, in order of decreasing atomic number. This carbon down here is directly bonded to three hydrogens, so one, two, and three. The carbon on the left is directly bonded to an oxygen, a carbon, and a hydrogen. So we put these in order of decreasing atomic number, so we put oxygen first, then carbon, then hydrogen. Let's look at the first point of difference here. So the first atom, we're comparing a carbon to a hydrogen to an oxygen. Out of those atoms, oxygen has the highest atomic number, which means that this group gets highest priority."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we put these in order of decreasing atomic number, so we put oxygen first, then carbon, then hydrogen. Let's look at the first point of difference here. So the first atom, we're comparing a carbon to a hydrogen to an oxygen. Out of those atoms, oxygen has the highest atomic number, which means that this group gets highest priority. So the group with the OH has the highest priority. This gets a number one. Next is carbon, because carbon has a higher atomic number than hydrogen, so that wins it for this group."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Out of those atoms, oxygen has the highest atomic number, which means that this group gets highest priority. So the group with the OH has the highest priority. This gets a number one. Next is carbon, because carbon has a higher atomic number than hydrogen, so that wins it for this group. So this group on the right, this ethyl group, gets second highest priority. And finally, the methyl group would get third highest priority. So now that we've assigned priority, step two is to orient the molecule so that the lowest priority group is pointing away from you in space."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next is carbon, because carbon has a higher atomic number than hydrogen, so that wins it for this group. So this group on the right, this ethyl group, gets second highest priority. And finally, the methyl group would get third highest priority. So now that we've assigned priority, step two is to orient the molecule so that the lowest priority group is pointing away from you in space. And here, our hydrogen is coming out at us in space, so that's not what we want. We want the hydrogen to point away from us in space. So one thing we could do is to think about an axis through our chiral center, and we could rotate the molecule about that axis, and that would put the hydrogen going away from us in space."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So now that we've assigned priority, step two is to orient the molecule so that the lowest priority group is pointing away from you in space. And here, our hydrogen is coming out at us in space, so that's not what we want. We want the hydrogen to point away from us in space. So one thing we could do is to think about an axis through our chiral center, and we could rotate the molecule about that axis, and that would put the hydrogen going away from us in space. So let's go to the video so we can visualize what happens. So here's our compound. You can see at carbon two, there's an OH coming out at us in space, and at carbon three, there's a methyl group going away from us in space."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So one thing we could do is to think about an axis through our chiral center, and we could rotate the molecule about that axis, and that would put the hydrogen going away from us in space. So let's go to the video so we can visualize what happens. So here's our compound. You can see at carbon two, there's an OH coming out at us in space, and at carbon three, there's a methyl group going away from us in space. If we rotate about an axis through this carbon, let's go ahead and do that, we can see what happens to those two groups. Now, at carbon two, the OH is going away from us in space, and at carbon three, the methyl group is coming out at us in space. Let's draw our compound with the hydrogen going away from us."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can see at carbon two, there's an OH coming out at us in space, and at carbon three, there's a methyl group going away from us in space. If we rotate about an axis through this carbon, let's go ahead and do that, we can see what happens to those two groups. Now, at carbon two, the OH is going away from us in space, and at carbon three, the methyl group is coming out at us in space. Let's draw our compound with the hydrogen going away from us. So first we put in our carbon chain, and we know at carbon two, our OH is going away from us, so here is our OH, and then at carbon three, our methyl group is coming out at us, so let me go ahead and put that in, so there's a CH three coming out, and there's a hydrogen going away from us in space. We can't really see it very well in the picture, but we know it's there. So we know that the group with the OH had the highest priority, so this was number one."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw our compound with the hydrogen going away from us. So first we put in our carbon chain, and we know at carbon two, our OH is going away from us, so here is our OH, and then at carbon three, our methyl group is coming out at us, so let me go ahead and put that in, so there's a CH three coming out, and there's a hydrogen going away from us in space. We can't really see it very well in the picture, but we know it's there. So we know that the group with the OH had the highest priority, so this was number one. We know the ethyl group had the second highest priority, so we say that's a number two. The methyl group was a number three, and the hydrogen was a number four. So with our lowest priority group going away from us, we can now ignore it, and see that we're going around in this direction, so we're going around counterclockwise, and that would be S. So we are S at carbon three, so I could finish off the name by writing in here 3S."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we know that the group with the OH had the highest priority, so this was number one. We know the ethyl group had the second highest priority, so we say that's a number two. The methyl group was a number three, and the hydrogen was a number four. So with our lowest priority group going away from us, we can now ignore it, and see that we're going around in this direction, so we're going around counterclockwise, and that would be S. So we are S at carbon three, so I could finish off the name by writing in here 3S. So it's 2R, 3S, 3-methyl, 2-pensinol. Now, you didn't have to do this whole trick with rotating about an axis. I showed you another way to do it in an earlier video, so let's go back to our original drawing over here on the left."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So with our lowest priority group going away from us, we can now ignore it, and see that we're going around in this direction, so we're going around counterclockwise, and that would be S. So we are S at carbon three, so I could finish off the name by writing in here 3S. So it's 2R, 3S, 3-methyl, 2-pensinol. Now, you didn't have to do this whole trick with rotating about an axis. I showed you another way to do it in an earlier video, so let's go back to our original drawing over here on the left. Let me use dark blue. There is a trick that you can use. So if your hydrogen's coming out at you in space, you can just ignore it for the time being, and look at one, two, and three, and one, two, and three are going around this way."}, {"video_title": "More R,S practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I showed you another way to do it in an earlier video, so let's go back to our original drawing over here on the left. Let me use dark blue. There is a trick that you can use. So if your hydrogen's coming out at you in space, you can just ignore it for the time being, and look at one, two, and three, and one, two, and three are going around this way. They're going around clockwise, so it looks R. It looks like it's R, but since the hydrogen is coming out at you in space, you can just reverse it. You can take the opposite. If it looks R with the hydrogen coming out at you, it must be S. So that's a nice trick, and it means you don't have to rotate the molecule in your head."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "And then everything else will hopefully not be too difficult. So just to start off, and this is really a little bit of review of regular chemistry, if I just have a chain of carbons, and organic chemistry is dealing with chains of carbons. Let me just draw a one-carbon chain. So it's really kind of ridiculous to call it a chain. But if we have one carbon over here, and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry for all molecules. That's the stable valence structure, I guess you could say."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "So it's really kind of ridiculous to call it a chain. But if we have one carbon over here, and it has four valence electrons, it wants to get to eight. That's the magic number we learned in just regular chemistry for all molecules. That's the stable valence structure, I guess you could say. So if a good partner to bond with is hydrogen. So it has four valence electrons, and then hydrogen has one valence electron. So they can each share an electron with each other."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "That's the stable valence structure, I guess you could say. So if a good partner to bond with is hydrogen. So it has four valence electrons, and then hydrogen has one valence electron. So they can each share an electron with each other. And then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they are only trying to fill their 1s orbitals."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "So they can each share an electron with each other. And then they both look pretty happy. I said eight's the magic number for everybody except for hydrogen and helium. Both of them are happy because they are only trying to fill their 1s orbitals. So the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Both of them are happy because they are only trying to fill their 1s orbitals. So the magic number for those two guys is two. So all of the hydrogens now feel like they have two electrons. The carbon feels like it has eight. Now there are several ways to write this. You could write it just like this, and you can see the electrons explicitly. Or you can draw little lines here."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "The carbon feels like it has eight. Now there are several ways to write this. You could write it just like this, and you can see the electrons explicitly. Or you can draw little lines here. So I could also write this exact molecule, which is methane. And we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Or you can draw little lines here. So I could also write this exact molecule, which is methane. And we'll talk a little bit more about why it's called methane later in this video. I can write this exact structure like this. A carbon bonded to four hydrogens. And the way that I've written these bonds right here, you could imagine that each of these bonds consists of two electrons. One from the carbon and one from the hydrogen."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "I can write this exact structure like this. A carbon bonded to four hydrogens. And the way that I've written these bonds right here, you could imagine that each of these bonds consists of two electrons. One from the carbon and one from the hydrogen. Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Let me do a three-carbon chain so it really looks like a chain."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "One from the carbon and one from the hydrogen. Now let's explore slightly larger chains. So let's say I have a two-carbon chain. Let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly, it might look like this. So I have a carbon. It has one, two, three, four electrons."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Let me do a three-carbon chain so it really looks like a chain. So if I were to draw everything explicitly, it might look like this. So I have a carbon. It has one, two, three, four electrons. Maybe I have another carbon here that has Let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "It has one, two, three, four electrons. Maybe I have another carbon here that has Let me do the carbons in slightly different shades of yellow. I have another carbon here that has one, two, three, four electrons. And then let me do the other carbon in that first yellow. And then I have another carbon. So we have a three-carbon chain that has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "And then let me do the other carbon in that first yellow. And then I have another carbon. So we have a three-carbon chain that has one, two, three, four valence electrons. Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen. So let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a hydrogen over there, a hydrogen over here. Almost done."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Now, these other guys are unpaired, and if you don't specify it, it's normally going to be hydrogen. So let me draw some hydrogens over here. So you're going to have a hydrogen there, a hydrogen over there, a hydrogen over here, a hydrogen over here, a hydrogen over there, a hydrogen over here. Almost done. Hydrogen there, and then a hydrogen there. Now notice, in this molecular structure that I've drawn, I have three carbons. They were each able to form four bonds."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Almost done. Hydrogen there, and then a hydrogen there. Now notice, in this molecular structure that I've drawn, I have three carbons. They were each able to form four bonds. This guy has bonds with three hydrogens and another carbon. This guy has a bond with two hydrogens and two carbons. This guy has a bond with three hydrogens and then this carbon right here."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "They were each able to form four bonds. This guy has bonds with three hydrogens and another carbon. This guy has a bond with two hydrogens and two carbons. This guy has a bond with three hydrogens and then this carbon right here. And so this is a completely valid molecular structure, but it was kind of a pain to draw all of these valence electrons here. So what we typically would want to do is at least this structure, and we're going to see later in this video there's even simpler ways to write it. So if we want to at least do it with these lines, we can draw it like this."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "This guy has a bond with three hydrogens and then this carbon right here. And so this is a completely valid molecular structure, but it was kind of a pain to draw all of these valence electrons here. So what we typically would want to do is at least this structure, and we're going to see later in this video there's even simpler ways to write it. So if we want to at least do it with these lines, we can draw it like this. We have a carbon, carbon, carbon, and then they are bonded to the hydrogens. So you'll almost never see it written like this because this is just kind of crazy. Hydrogen, hydrogen, at least crazy to write."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "So if we want to at least do it with these lines, we can draw it like this. We have a carbon, carbon, carbon, and then they are bonded to the hydrogens. So you'll almost never see it written like this because this is just kind of crazy. Hydrogen, hydrogen, at least crazy to write. It takes forever. And it might be messy. Like it might not be clear where these electrons belong."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Hydrogen, hydrogen, at least crazy to write. It takes forever. And it might be messy. Like it might not be clear where these electrons belong. I didn't write it as clearly as I could, so they have two electrons there. They're shared with these two guys. Hopefully that was reasonably clear."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Like it might not be clear where these electrons belong. I didn't write it as clearly as I could, so they have two electrons there. They're shared with these two guys. Hopefully that was reasonably clear. But if we were to draw it with the lines, it looks just like that. So it's a little bit neater, faster to draw. Same exact idea here and here."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Hopefully that was reasonably clear. But if we were to draw it with the lines, it looks just like that. So it's a little bit neater, faster to draw. Same exact idea here and here. And in general, and we'll go in more detail on it, this three-carbon chain where everything is a single bond is propane. Let me write these words down because it's helpful to get. This is methane."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Same exact idea here and here. And in general, and we'll go in more detail on it, this three-carbon chain where everything is a single bond is propane. Let me write these words down because it's helpful to get. This is methane. And you're going to see the rhyme, you're going to see the reason to this naming soon enough. This is methane. This is propane."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "This is methane. And you're going to see the rhyme, you're going to see the reason to this naming soon enough. This is methane. This is propane. And there's an even simpler way to write propane. You could write it like this. Instead of explicitly drawing these bonds, you could say that this part right here, you could write that that part right there, that is CH3."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "This is propane. And there's an even simpler way to write propane. You could write it like this. Instead of explicitly drawing these bonds, you could say that this part right here, you could write that that part right there, that is CH3. So you have a CH3 connected to a CH2, which is then connected to another CH3. And the important thing is no matter what the notation, as long as you can figure out the exact molecular structure, so there's this last CH3, whether you have this, this, or this, you know what the molecular structure is. You could draw any one of these given any of the others."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Instead of explicitly drawing these bonds, you could say that this part right here, you could write that that part right there, that is CH3. So you have a CH3 connected to a CH2, which is then connected to another CH3. And the important thing is no matter what the notation, as long as you can figure out the exact molecular structure, so there's this last CH3, whether you have this, this, or this, you know what the molecular structure is. You could draw any one of these given any of the others. Now there's an even simpler way to write this. You could write it just like this. Let me do it in a different color."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "You could draw any one of these given any of the others. Now there's an even simpler way to write this. You could write it just like this. Let me do it in a different color. You literally could write it, so we have three carbons, so one, two, three. Now this seems ridiculously simple, and you're like, how can this thing right here give you the same information as all of these more complicated ways to draw it? Well, in chemistry, in organic chemistry in particular, any of these, this is called a line diagram or a line angle diagram."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Let me do it in a different color. You literally could write it, so we have three carbons, so one, two, three. Now this seems ridiculously simple, and you're like, how can this thing right here give you the same information as all of these more complicated ways to draw it? Well, in chemistry, in organic chemistry in particular, any of these, this is called a line diagram or a line angle diagram. It's the simplest way, and it's actually probably the most useful way to show chains of carbons or to show organic molecules once they start to get really, really complicated because then it's a pain to draw all of the H's. But when you see something like this, you assume that the endpoints of any lines have a carbon on it. If you see something like that, you assume that there's a carbon at that endpoint, a carbon at that endpoint, and a carbon at that endpoint."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "Well, in chemistry, in organic chemistry in particular, any of these, this is called a line diagram or a line angle diagram. It's the simplest way, and it's actually probably the most useful way to show chains of carbons or to show organic molecules once they start to get really, really complicated because then it's a pain to draw all of the H's. But when you see something like this, you assume that the endpoints of any lines have a carbon on it. If you see something like that, you assume that there's a carbon at that endpoint, a carbon at that endpoint, and a carbon at that endpoint. Then you know that carbon makes four bonds. There are no kind of charges here. All of the carbons are going to make four bonds."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "If you see something like that, you assume that there's a carbon at that endpoint, a carbon at that endpoint, and a carbon at that endpoint. Then you know that carbon makes four bonds. There are no kind of charges here. All of the carbons are going to make four bonds. Each of the carbons here, this carbon has two bonds, so the other two bonds are implicitly going to be with hydrogens. If they don't draw them, you assume that they're going to be with hydrogens. This guy has one bond, so the other three must be with hydrogen."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "All of the carbons are going to make four bonds. Each of the carbons here, this carbon has two bonds, so the other two bonds are implicitly going to be with hydrogens. If they don't draw them, you assume that they're going to be with hydrogens. This guy has one bond, so the other three must be with hydrogen. This guy has one bond, so the other three must be hydrogens. Just drawing that little line angle thing right there, I actually did convey the exact same information as this depiction, this depiction, or this depiction. You're going to see a lot of this."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "This guy has one bond, so the other three must be with hydrogen. This guy has one bond, so the other three must be hydrogens. Just drawing that little line angle thing right there, I actually did convey the exact same information as this depiction, this depiction, or this depiction. You're going to see a lot of this. This really simplifies things. Sometimes you see things that are in between. You might see someone draw it like this, where they'll write CH3, and then they'll draw it like that."}, {"video_title": "Representing structures of organic molecules Biology Khan Academy.mp3", "Sentence": "You're going to see a lot of this. This really simplifies things. Sometimes you see things that are in between. You might see someone draw it like this, where they'll write CH3, and then they'll draw it like that. That's kind of combining this way of writing the molecule, where you write the CH3s for the endpoints, but then you implicitly have the CH2 on the inside. You assume that this endpoint right here is a C, and it's bonded to two hydrogens. These are all completely valid ways of drawing the molecular structures of these carbon chains or of these organic compounds."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's the dot structure for natural cholesterol. It's an optically active compound. And we dissolve our cholesterol in 15.0 milliliters of chloroform. And we put that solution in a 10.0 centimeter polarimeter tube. The observed rotation at 20 degrees C using the D line of sodium turns out to be negative.630 degrees. And our goal is to calculate the specific rotation of cholesterol. We saw how to do this in the last video."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we put that solution in a 10.0 centimeter polarimeter tube. The observed rotation at 20 degrees C using the D line of sodium turns out to be negative.630 degrees. And our goal is to calculate the specific rotation of cholesterol. We saw how to do this in the last video. The specific rotation is equal to the observed rotation divided by the concentration times the path length. So let's plug in some numbers here. The specific rotation is equal to the observed rotation, which is negative.630 degrees."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We saw how to do this in the last video. The specific rotation is equal to the observed rotation divided by the concentration times the path length. So let's plug in some numbers here. The specific rotation is equal to the observed rotation, which is negative.630 degrees. So we put that in, negative.630 degrees. We divide by the concentration, which is in grams per ml. So that's.300 grams divided by 15.0 mls."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The specific rotation is equal to the observed rotation, which is negative.630 degrees. So we put that in, negative.630 degrees. We divide by the concentration, which is in grams per ml. So that's.300 grams divided by 15.0 mls. So.300 grams divided by 15.0 mls. We multiply that by the path length. And the path length needs to be in decimeters."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's.300 grams divided by 15.0 mls. So.300 grams divided by 15.0 mls. We multiply that by the path length. And the path length needs to be in decimeters. So we have a 10.0 centimeter tube. 10.0 centimeters is one decimeter. So that makes our math easy here."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the path length needs to be in decimeters. So we have a 10.0 centimeter tube. 10.0 centimeters is one decimeter. So that makes our math easy here. So this would be 1.00 decimeter. Alright, let's do the math. So let's get out the calculator and let's solve for the specific rotation."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that makes our math easy here. So this would be 1.00 decimeter. Alright, let's do the math. So let's get out the calculator and let's solve for the specific rotation. That would be negative.630 divided by, we have.300 divided by 15.0. And then we multiply that by one. I don't really need to do that, but I'll go ahead and do it anyway."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's get out the calculator and let's solve for the specific rotation. That would be negative.630 divided by, we have.300 divided by 15.0. And then we multiply that by one. I don't really need to do that, but I'll go ahead and do it anyway. So that's multiplied by 1.00 here. And we get negative 31.5. So that is our specific rotation."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I don't really need to do that, but I'll go ahead and do it anyway. So that's multiplied by 1.00 here. And we get negative 31.5. So that is our specific rotation. So let's write that down here. So we have our specific rotation at 20 degrees C. So we put a 20 here. Using the D line of sodium, so we put a D here."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that is our specific rotation. So let's write that down here. So we have our specific rotation at 20 degrees C. So we put a 20 here. Using the D line of sodium, so we put a D here. And this is equal to negative 31.5. Now sometimes you see this with a degree sign. So sometimes you see it written like that."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Using the D line of sodium, so we put a D here. And this is equal to negative 31.5. Now sometimes you see this with a degree sign. So sometimes you see it written like that. But I'm gonna take that out because normally we don't have any units for our specific rotation. So it just depends on what book you're looking in. For our next problem, problem two, let's talk about percent enantiomeric excess, or optical purity."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So sometimes you see it written like that. But I'm gonna take that out because normally we don't have any units for our specific rotation. So it just depends on what book you're looking in. For our next problem, problem two, let's talk about percent enantiomeric excess, or optical purity. This is where you take the percentage of one enantiomer and from that you subtract the percentage of the other enantiomer. So for part A, let's calculate the percent enantiomeric excess for a solution that contains a single enantiomer. So if we have only one enantiomer, this is like the first problem that we did with natural cholesterol."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "For our next problem, problem two, let's talk about percent enantiomeric excess, or optical purity. This is where you take the percentage of one enantiomer and from that you subtract the percentage of the other enantiomer. So for part A, let's calculate the percent enantiomeric excess for a solution that contains a single enantiomer. So if we have only one enantiomer, this is like the first problem that we did with natural cholesterol. That means you have 100% of this enantiomer and obviously 0% of the other one. So the percent enantiomeric excess would just be 100 minus zero, or 100%. So we have 100% optical purity."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we have only one enantiomer, this is like the first problem that we did with natural cholesterol. That means you have 100% of this enantiomer and obviously 0% of the other one. So the percent enantiomeric excess would just be 100 minus zero, or 100%. So we have 100% optical purity. So this is an optically pure solution. For part B, let's do this for a solution that contains equal amounts of both enantiomers. So when that happens, it's called a racemic mixture."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have 100% optical purity. So this is an optically pure solution. For part B, let's do this for a solution that contains equal amounts of both enantiomers. So when that happens, it's called a racemic mixture. So if we have equal amounts of both, that must mean we have 50% of one enantiomer and 50% of the other. So the percent enantiomeric excess would be equal to 50 minus 50, which of course is equal to zero. So this has an optical purity of 0% and a racemic mixture is not optically active."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So when that happens, it's called a racemic mixture. So if we have equal amounts of both, that must mean we have 50% of one enantiomer and 50% of the other. So the percent enantiomeric excess would be equal to 50 minus 50, which of course is equal to zero. So this has an optical purity of 0% and a racemic mixture is not optically active. You get a net rotation of zero if you have equal amounts of both enantiomers. For part C, we have a solution that contains 75% of one enantiomer and 25% of the other. So the percent enantiomeric excess is equal to, this would be 75%, minus 25%, which of course is equal to 50%."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this has an optical purity of 0% and a racemic mixture is not optically active. You get a net rotation of zero if you have equal amounts of both enantiomers. For part C, we have a solution that contains 75% of one enantiomer and 25% of the other. So the percent enantiomeric excess is equal to, this would be 75%, minus 25%, which of course is equal to 50%. So we have 50% excess of this enantiomer and we have a 50% optically pure solution. For our last problem, we have a mixture of natural cholesterol and its enantiomer. And our mixture has a specific rotation of negative 27."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the percent enantiomeric excess is equal to, this would be 75%, minus 25%, which of course is equal to 50%. So we have 50% excess of this enantiomer and we have a 50% optically pure solution. For our last problem, we have a mixture of natural cholesterol and its enantiomer. And our mixture has a specific rotation of negative 27. Our goal is to calculate the percent enantiomeric excess of this mixture. And we can do that using this equation up here. So the percentage enantiomeric excess is equal to the observed specific rotation divided by the specific rotation of the pure enantiomer."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And our mixture has a specific rotation of negative 27. Our goal is to calculate the percent enantiomeric excess of this mixture. And we can do that using this equation up here. So the percentage enantiomeric excess is equal to the observed specific rotation divided by the specific rotation of the pure enantiomer. And to get a percentage, we multiply by 100. So the percent enantiomeric excess is equal to the observed specific rotation, which is negative 27, so we write that in here. So negative 27."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the percentage enantiomeric excess is equal to the observed specific rotation divided by the specific rotation of the pure enantiomer. And to get a percentage, we multiply by 100. So the percent enantiomeric excess is equal to the observed specific rotation, which is negative 27, so we write that in here. So negative 27. We divide that by the specific rotation of the pure enantiomer. And for natural cholesterol, we saw what the specific rotation of the pure enantiomer was in the first problem. We got negative 31.5."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So negative 27. We divide that by the specific rotation of the pure enantiomer. And for natural cholesterol, we saw what the specific rotation of the pure enantiomer was in the first problem. We got negative 31.5. So I'll write it here, negative 31.5. And we multiply by 100. So that gives us our percent enantiomeric excess."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We got negative 31.5. So I'll write it here, negative 31.5. And we multiply by 100. So that gives us our percent enantiomeric excess. So let's get out the calculator here. We don't need to worry about negative signs. So we can just take 27 and divide that by 31.5."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that gives us our percent enantiomeric excess. So let's get out the calculator here. We don't need to worry about negative signs. So we can just take 27 and divide that by 31.5. And multiply by 100, and we get 85.7. And let's round that to 86%. So our percent enantiomeric excess is 86%."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can just take 27 and divide that by 31.5. And multiply by 100, and we get 85.7. And let's round that to 86%. So our percent enantiomeric excess is 86%. So we're done with our calculation here. Our next question is, what percentage of the mixture is natural cholesterol? Well, 86%, this was our enantiomeric excess."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So our percent enantiomeric excess is 86%. So we're done with our calculation here. Our next question is, what percentage of the mixture is natural cholesterol? Well, 86%, this was our enantiomeric excess. So if we think about this as being 86% of natural cholesterol, so let me write this down here, 86% of natural cholesterol, and the remaining 14% must be a racemic mixture. So if the remaining 14% is a racemic mixture, that means half of it is natural cholesterol, and half of it is the enantiomer. So that means that 7% is our natural cholesterol, and 7% is the enantiomer."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, 86%, this was our enantiomeric excess. So if we think about this as being 86% of natural cholesterol, so let me write this down here, 86% of natural cholesterol, and the remaining 14% must be a racemic mixture. So if the remaining 14% is a racemic mixture, that means half of it is natural cholesterol, and half of it is the enantiomer. So that means that 7% is our natural cholesterol, and 7% is the enantiomer. So seven plus seven is, of course, equal to 14. So what's the total percentage of natural cholesterol in our mixture? That would be 86 plus seven, which, of course, is 93%."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that means that 7% is our natural cholesterol, and 7% is the enantiomer. So seven plus seven is, of course, equal to 14. So what's the total percentage of natural cholesterol in our mixture? That would be 86 plus seven, which, of course, is 93%. So that's our answer. So 93% of our mixture is natural cholesterol. This can get a little bit confusing sometimes, so you can check this answer to make sure it's correct."}, {"video_title": "Optical activity calculations Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That would be 86 plus seven, which, of course, is 93%. So that's our answer. So 93% of our mixture is natural cholesterol. This can get a little bit confusing sometimes, so you can check this answer to make sure it's correct. You know that the total of natural cholesterol and its enantiomer should be 100%. So if natural cholesterol is 93%, and its enantiomer is 7%, obviously 93% plus 7% is 100%. Also, we know from the previous problem that the percentage enantiomeric excess is equal to the percent of one enantiomer minus the percent of the other enantiomer."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I've talked a bunch about the Drake equation, or our own version of the Drake equation, that starts with the number of stars in the galaxy, but I haven't given it a shot yet. I haven't tried my own attempt at thinking about how many detectable civilizations there are. So let's actually do that here. So let's just assume that there are 100 billion stars. So that's my first term right over there. Let's say that one-fourth will develop planets. And let's say of the solar systems that develop planets, on average, let's say that they develop an average of 0.1 planets capable of sustaining life."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's just assume that there are 100 billion stars. So that's my first term right over there. Let's say that one-fourth will develop planets. And let's say of the solar systems that develop planets, on average, let's say that they develop an average of 0.1 planets capable of sustaining life. Or really, that you'll have one planet for every 10 of these solar systems with planets. That's just my assumption there. I don't know if that's right."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say of the solar systems that develop planets, on average, let's say that they develop an average of 0.1 planets capable of sustaining life. Or really, that you'll have one planet for every 10 of these solar systems with planets. That's just my assumption there. I don't know if that's right. Now let's multiply that times the fraction of these planets capable of sustaining life that actually will get life. And I don't know what that is, but I hinted in previous videos that life is one of those things that it seems like if you have all the right ingredients, it's so robust that you have life at these underwater volcanoes. You have bacteria that can process all sorts of weird things."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't know if that's right. Now let's multiply that times the fraction of these planets capable of sustaining life that actually will get life. And I don't know what that is, but I hinted in previous videos that life is one of those things that it seems like if you have all the right ingredients, it's so robust that you have life at these underwater volcanoes. You have bacteria that can process all sorts of weird things. So let's say that that probability is pretty high. Let's say that that is 50% or half of the planets that are capable of getting life actually do have life. I would guess that that might even be higher."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have bacteria that can process all sorts of weird things. So let's say that that probability is pretty high. Let's say that that is 50% or half of the planets that are capable of getting life actually do have life. I would guess that that might even be higher. But once again, just a guess. Now we have to think about of the life, what fraction becomes intelligent? What becomes intelligent over some point in the history?"}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I would guess that that might even be higher. But once again, just a guess. Now we have to think about of the life, what fraction becomes intelligent? What becomes intelligent over some point in the history? Well, I'll say it's a tenth. Maybe if the asteroids didn't kill the dinosaurs, it wouldn't have happened on Earth. Who knows?"}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What becomes intelligent over some point in the history? Well, I'll say it's a tenth. Maybe if the asteroids didn't kill the dinosaurs, it wouldn't have happened on Earth. Who knows? Or maybe we just have some very intelligent dinosaurs around. We don't know. And maybe there's other forms of intelligent life even on our own planet that we haven't fully appreciated."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Who knows? Or maybe we just have some very intelligent dinosaurs around. We don't know. And maybe there's other forms of intelligent life even on our own planet that we haven't fully appreciated. Dolphins are a good candidate. Some people believe that octopuses, because they have such flexible arms, there's a theory that they could develop eventually the ability to kind of one day, if their brains mature and all of the rest make tools the same way primitive primates eventually were able to have larger brain sizes and actually manipulate things to make tools. So who knows?"}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And maybe there's other forms of intelligent life even on our own planet that we haven't fully appreciated. Dolphins are a good candidate. Some people believe that octopuses, because they have such flexible arms, there's a theory that they could develop eventually the ability to kind of one day, if their brains mature and all of the rest make tools the same way primitive primates eventually were able to have larger brain sizes and actually manipulate things to make tools. So who knows? I don't want to get into all of that. So there's a 1 in 10 chance that you get intelligent life. And then assuming that intelligent life shows up, what fraction is going to become detectable?"}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So who knows? I don't want to get into all of that. So there's a 1 in 10 chance that you get intelligent life. And then assuming that intelligent life shows up, what fraction is going to become detectable? I don't know. I don't know whether dolphins will ever communicate via radio or not. So let's just say that that is, I don't know, let's say that that is another 1 in 10 chance or I'll say 0.1."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then assuming that intelligent life shows up, what fraction is going to become detectable? I don't know. I don't know whether dolphins will ever communicate via radio or not. So let's just say that that is, I don't know, let's say that that is another 1 in 10 chance or I'll say 0.1. And then we have to multiply it times the detectable life of the civilization on average. Once again, huge assumptions being here. But the detectable life of a civilization, let me just put it at 10,000 years."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's just say that that is, I don't know, let's say that that is another 1 in 10 chance or I'll say 0.1. And then we have to multiply it times the detectable life of the civilization on average. Once again, huge assumptions being here. But the detectable life of a civilization, let me just put it at 10,000 years. Either they destroy themselves or they get beyond that type of radio-type communication, electromagnetic-type communication. Maybe they start doing all sorts of weird, wacky things. Probably won't take you 10,000 years to even progress it."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the detectable life of a civilization, let me just put it at 10,000 years. Either they destroy themselves or they get beyond that type of radio-type communication, electromagnetic-type communication. Maybe they start doing all sorts of weird, wacky things. Probably won't take you 10,000 years to even progress it. That might take you less time. But let's just do this just for the sake of fun. And then the lifespan of your average star, that's probably one of the things that we have the best sense of."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Probably won't take you 10,000 years to even progress it. That might take you less time. But let's just do this just for the sake of fun. And then the lifespan of your average star, that's probably one of the things that we have the best sense of. So on average, let's put it at 10 billion years. So let's calculate all of this. Let's get my handy TI-85 out."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then the lifespan of your average star, that's probably one of the things that we have the best sense of. So on average, let's put it at 10 billion years. So let's calculate all of this. Let's get my handy TI-85 out. And so we're going to have 100 billion. That's 1 times 10 to the 9th. Sorry, that's 100 times 10 to the 9th."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's get my handy TI-85 out. And so we're going to have 100 billion. That's 1 times 10 to the 9th. Sorry, that's 100 times 10 to the 9th. So let me clear it. Or you could have 1 times 10 to the 11th. That is 100 billion times 0.25 times 0.1 times 0.5 times 0.5 times 0.1 again times 0.1 times 0.1 again times 0.1 times 10,000 divided by 10 billion."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Sorry, that's 100 times 10 to the 9th. So let me clear it. Or you could have 1 times 10 to the 11th. That is 100 billion times 0.25 times 0.1 times 0.5 times 0.5 times 0.1 again times 0.1 times 0.1 again times 0.1 times 10,000 divided by 10 billion. So times 10,000 divided by 10 billion. So that's 1E10. 1 times 10 to the 10th power."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is 100 billion times 0.25 times 0.1 times 0.5 times 0.5 times 0.1 again times 0.1 times 0.1 again times 0.1 times 10,000 divided by 10 billion. So times 10,000 divided by 10 billion. So that's 1E10. 1 times 10 to the 10th power. 1 with 10 zeros. So let's see what we get. We get 12.5, which is kind of a neat number."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "1 times 10 to the 10th power. 1 with 10 zeros. So let's see what we get. We get 12.5, which is kind of a neat number. But these are heavily dependent on this. So we're saying, given these assumptions, there should be 12.5 detectable civilizations in our galaxy right now. So the question is, why aren't we detecting it?"}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We get 12.5, which is kind of a neat number. But these are heavily dependent on this. So we're saying, given these assumptions, there should be 12.5 detectable civilizations in our galaxy right now. So the question is, why aren't we detecting it? Maybe their radio signals, maybe their electromagnetic waves are getting to us. But we can't differentiate it from noise right now. And that's what the whole SETI project is all about, trying to keep track of all of this information, all of these radio waves and electromagnetic waves that are coming from outer space towards Earth."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the question is, why aren't we detecting it? Maybe their radio signals, maybe their electromagnetic waves are getting to us. But we can't differentiate it from noise right now. And that's what the whole SETI project is all about, trying to keep track of all of this information, all of these radio waves and electromagnetic waves that are coming from outer space towards Earth. And seeing if any of them actually have any non-noise signal that actually look like they're being generated by some type of intelligent civilization. So maybe we're getting them and we're just not detecting them. Or maybe something else is at play."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's what the whole SETI project is all about, trying to keep track of all of this information, all of these radio waves and electromagnetic waves that are coming from outer space towards Earth. And seeing if any of them actually have any non-noise signal that actually look like they're being generated by some type of intelligent civilization. So maybe we're getting them and we're just not detecting them. Or maybe something else is at play. Maybe we've overestimated one of these. Maybe there is a lot of life, but maybe they're not using electromagnetic waves to communicate. Maybe that's some type of primitive way of communicating."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or maybe something else is at play. Maybe we've overestimated one of these. Maybe there is a lot of life, but maybe they're not using electromagnetic waves to communicate. Maybe that's some type of primitive way of communicating. Maybe they start doing telepathy or something crazy. Or they start using some type of quantum thing that allows them to communicate more directly without having to wait for the speed of light. Maybe that is a very slow way to communicate."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe that's some type of primitive way of communicating. Maybe they start doing telepathy or something crazy. Or they start using some type of quantum thing that allows them to communicate more directly without having to wait for the speed of light. Maybe that is a very slow way to communicate. And it is a slow way, frankly, if you're trying to communicate across solar systems and stars and planets or even across galaxies, one could imagine. So maybe we're just kind of in a transition state of communication. That electromagnetic waves, radio, and all the rest is just a transition state."}, {"video_title": "Detectable civilizations in our galaxy 4 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe that is a very slow way to communicate. And it is a slow way, frankly, if you're trying to communicate across solar systems and stars and planets or even across galaxies, one could imagine. So maybe we're just kind of in a transition state of communication. That electromagnetic waves, radio, and all the rest is just a transition state. Maybe in 100 years we'll figure out another better way that's not detectable in our traditional ways. Maybe we're being bombarded with another type of communication mechanism that we're just not ready to perceive yet. Who knows?"}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at the molecular formula, CH4, which is methane, and I want to draw a dot structure for the methane molecule, I would go over here to my organic periodic table and find carbon. And I can see carbon is in group 4. Therefore, carbon will have four valence electrons. So I can draw carbon with its four valence electrons around it like that. Remember from general chemistry, valence electrons are the electrons in the outermost energy level. So carbon has four valence electrons in its outermost energy level. Next, I have to think about hydrogen."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I can draw carbon with its four valence electrons around it like that. Remember from general chemistry, valence electrons are the electrons in the outermost energy level. So carbon has four valence electrons in its outermost energy level. Next, I have to think about hydrogen. And hydrogen is in group 1 on the periodic table. Therefore, hydrogen will have one valence electron. And so I can go ahead and put a hydrogen in there with one valence electron."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, I have to think about hydrogen. And hydrogen is in group 1 on the periodic table. Therefore, hydrogen will have one valence electron. And so I can go ahead and put a hydrogen in there with one valence electron. And I know I have to do that three more times. So I keep putting in hydrogens, each with one valence electron, so a total of four hydrogens. And now I can start connecting my dots."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so I can go ahead and put a hydrogen in there with one valence electron. And I know I have to do that three more times. So I keep putting in hydrogens, each with one valence electron, so a total of four hydrogens. And now I can start connecting my dots. I know that two valence electrons equals one single covalent bond. So there is a single covalent bond. There is a single covalent bond."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And now I can start connecting my dots. I know that two valence electrons equals one single covalent bond. So there is a single covalent bond. There is a single covalent bond. And then I have two more here. So this is my complete dot structure for methane. Now, I can see that carbon is surrounded by eight electrons here."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There is a single covalent bond. And then I have two more here. So this is my complete dot structure for methane. Now, I can see that carbon is surrounded by eight electrons here. So we can go ahead and highlight those. So if I'm counting the electrons around carbon, it would be 2, 4, 6, and 8, like that. And eight electrons around carbon makes carbon very stable."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now, I can see that carbon is surrounded by eight electrons here. So we can go ahead and highlight those. So if I'm counting the electrons around carbon, it would be 2, 4, 6, and 8, like that. And eight electrons around carbon makes carbon very stable. And if we look at the periodic table, we can see why. So if I look at the second period, I can see that the valence electrons for carbon would be 1, 2, 3, and 4. And to get to eight electrons, we would go 5, 6, 7, 8."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And eight electrons around carbon makes carbon very stable. And if we look at the periodic table, we can see why. So if I look at the second period, I can see that the valence electrons for carbon would be 1, 2, 3, and 4. And to get to eight electrons, we would go 5, 6, 7, 8. So if carbon is surrounded by eight electrons, it's like it has the electron configuration of a noble gas, which makes it very stable, because all of the orbitals in that energy level are now full. So an octet of electrons is the maximum number of electrons for carbon. If we look at hydrogen, we can see that each hydrogen is surrounded by two electrons."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And to get to eight electrons, we would go 5, 6, 7, 8. So if carbon is surrounded by eight electrons, it's like it has the electron configuration of a noble gas, which makes it very stable, because all of the orbitals in that energy level are now full. So an octet of electrons is the maximum number of electrons for carbon. If we look at hydrogen, we can see that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron, and here's two electrons. So in the first energy level, there's only an s orbital."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If we look at hydrogen, we can see that each hydrogen is surrounded by two electrons. And so if I find hydrogen here, hydrogen is in the first energy level. And so here's one electron, and here's two electrons. So in the first energy level, there's only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So in the first energy level, there's only an s orbital. And so that s orbital holds a maximum of two electrons. And we get to the electron configuration of a noble gas. And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3NH2, I'm going to, once again, start with carbon in the center with its four valence electrons around it, like that."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so hydrogen is stable with having only two electrons around it. Let's look at another dot structure. And let's do one that has nitrogen in it. So if I look at the molecular formula CH3NH2, I'm going to, once again, start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at the molecular formula CH3NH2, I'm going to, once again, start with carbon in the center with its four valence electrons around it, like that. And I know that there are three hydrogens on that carbon. So I can go ahead and put in those three hydrogens. Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group 5."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Each hydrogen has one valence electron, like that. And then on the right side, I'm going to think about nitrogen. So I need to find nitrogen on my periodic table. Nitrogen is in group 5. Therefore, nitrogen has five valence electrons. I can represent those valence electrons as 1, 2, 3, 4, and 5, like that. And I still have two hydrogens to worry about."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Nitrogen is in group 5. Therefore, nitrogen has five valence electrons. I can represent those valence electrons as 1, 2, 3, 4, and 5, like that. And I still have two hydrogens to worry about. So I still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here, and a hydrogen in here, and connect the dots."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And I still have two hydrogens to worry about. So I still have these two hydrogens here. And I can see there's a place for them on the nitrogen. So I can go ahead and put a hydrogen in here, and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and put a hydrogen in here, and a hydrogen in here, and connect the dots. And I have my dot structure. And I can also check on my octet rule. So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, 4, 6, and 8."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So carbon has an octet. And nitrogen has an octet as well. So let's go ahead and verify that. So there's two electrons here, 4, 6, and 8. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So there's two electrons here, 4, 6, and 8. So nitrogen is in the second period. And so nitrogen is also going to follow the octet rule when you're drawing your dot structures. Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3OH. And so once again, I start with carbon with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one with oxygen next. So if I wanted to draw the dot structure for methanol, methanol is CH3OH. And so once again, I start with carbon with the four valence electrons. And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next, I have oxygen. So I need to find oxygen on my organic periodic table."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And I have three hydrogens, each one with one valence electron, like that. And so I can go ahead and put in those three hydrogens. Next, I have oxygen. So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group 6 right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I need to find oxygen on my organic periodic table. And I can see that oxygen is in group 6 right here. So oxygen is going to have six valence electrons around it. So I can go ahead and draw in oxygen. And I can put its six valence electrons in, 1, 2, 3, 4, 5, and 6, like that. And then I'm going to put in the hydrogen. So now I have a hydrogen to worry about."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and draw in oxygen. And I can put its six valence electrons in, 1, 2, 3, 4, 5, and 6, like that. And then I'm going to put in the hydrogen. So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So now I have a hydrogen to worry about. And I know that hydrogen has one valence electron. So I can see there's a place for it over here. And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond. And then I can see the carbon is bonded to the oxygen."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And once again, I can connect the dots and see all of the single covalent bonds in this molecule. So that's one bond. That's another bond. And then I can see the carbon is bonded to the oxygen. And the oxygen is bonded to this hydrogen as well. Again, we can check our octet rule. So the carbon has eight electrons around it."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then I can see the carbon is bonded to the oxygen. And the oxygen is bonded to this hydrogen as well. Again, we can check our octet rule. So the carbon has eight electrons around it. And so does the oxygen. So this would be 2 right here, and then 4, and then 6, and then 8. So oxygen is going to follow the octet rule."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the carbon has eight electrons around it. And so does the oxygen. So this would be 2 right here, and then 4, and then 6, and then 8. So oxygen is going to follow the octet rule. Now, when you're drawing dot structures, you don't always have to do this step where you're drawing each individual atom and summing all of your valence electrons that way. You can just start drawing it. So for an example, if I gave you C2H6, which is ethane, another way to do it would just be starting to draw some bonds here."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen is going to follow the octet rule. Now, when you're drawing dot structures, you don't always have to do this step where you're drawing each individual atom and summing all of your valence electrons that way. You can just start drawing it. So for an example, if I gave you C2H6, which is ethane, another way to do it would just be starting to draw some bonds here. And so I have two carbons. And it's a pretty good bet those two carbons are going to be connected to each other. And then I have six hydrogens."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So for an example, if I gave you C2H6, which is ethane, another way to do it would just be starting to draw some bonds here. And so I have two carbons. And it's a pretty good bet those two carbons are going to be connected to each other. And then I have six hydrogens. And if I look at what's possible around those carbons, I could put those six hydrogens around those two carbon atoms like that. And if I do that, I'll have an octet around each carbon atom. So this would be my dot structure for ethane."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then I have six hydrogens. And if I look at what's possible around those carbons, I could put those six hydrogens around those two carbon atoms like that. And if I do that, I'll have an octet around each carbon atom. So this would be my dot structure for ethane. To double check yourself, you could make sure that your dot structure has the correct number of valence electrons. So if I'm thinking about each carbon having four valence electrons, and I have two of them, I'm going to get eight valence electrons from those two carbons that I have to represent in my dot structure. Each hydrogen has one valence electron."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this would be my dot structure for ethane. To double check yourself, you could make sure that your dot structure has the correct number of valence electrons. So if I'm thinking about each carbon having four valence electrons, and I have two of them, I'm going to get eight valence electrons from those two carbons that I have to represent in my dot structure. Each hydrogen has one valence electron. And I have six of them. So I need to worry about six valence electrons from the hydrogen, so for a total of 14. So when I look at my dot structure, I can check to make sure I have the correct number of valence electrons."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Each hydrogen has one valence electron. And I have six of them. So I need to worry about six valence electrons from the hydrogen, so for a total of 14. So when I look at my dot structure, I can check to make sure I have the correct number of valence electrons. I need 14. So let's go ahead and count them. So this would be 2 here, 4, 6, 8, 10, 12, and 14."}, {"video_title": "Dot structures I Single bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So when I look at my dot structure, I can check to make sure I have the correct number of valence electrons. I need 14. So let's go ahead and count them. So this would be 2 here, 4, 6, 8, 10, 12, and 14. So I have the correct number of valence electrons represented in my dot structure. I also have an octet of electrons around my carbons. And so this would be the dot structure for ethane."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the stretching vibration of a bond is like the oscillation of a spring. So you can think about a bond as being like a spring. So let's think about this bond right here. So the bond between the carbon and the hydrogen. We're going to model that bond as a spring. So I'm going to attempt to draw a spring in here. So here's the spring, and then let's put in the carbon on one side."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the bond between the carbon and the hydrogen. We're going to model that bond as a spring. So I'm going to attempt to draw a spring in here. So here's the spring, and then let's put in the carbon on one side. So we have the carbon on one side, and the hydrogen on the other side. And so the stretching vibration of the bond is like the oscillation of the spring. So if you had a spring, and you had two masses on either end of the spring, and if you put some energy in, you can stretch that spring."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's the spring, and then let's put in the carbon on one side. So we have the carbon on one side, and the hydrogen on the other side. And so the stretching vibration of the bond is like the oscillation of the spring. So if you had a spring, and you had two masses on either end of the spring, and if you put some energy in, you can stretch that spring. So you could pull the carbon this way, you could pull the hydrogen that way. So that's just like the stretching of this bond here. And we know that springs also contract."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you had a spring, and you had two masses on either end of the spring, and if you put some energy in, you can stretch that spring. So you could pull the carbon this way, you could pull the hydrogen that way. So that's just like the stretching of this bond here. And we know that springs also contract. So then the spring could pull back in this direction, and you get an oscillation of a spring, which is, once again, how we model the stretching vibration of a bond. So let's look at the IR spectrum for this molecule. So we're talking about one octane here."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we know that springs also contract. So then the spring could pull back in this direction, and you get an oscillation of a spring, which is, once again, how we model the stretching vibration of a bond. So let's look at the IR spectrum for this molecule. So we're talking about one octane here. And if you shine a range of infrared frequencies through a sample of this compound, some of the frequencies are absorbed by the compound. And you can tell which frequencies are absorbed by looking at your infrared spectrum here. So for right now, let's think about these numbers, like 3,000 or 4,000."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're talking about one octane here. And if you shine a range of infrared frequencies through a sample of this compound, some of the frequencies are absorbed by the compound. And you can tell which frequencies are absorbed by looking at your infrared spectrum here. So for right now, let's think about these numbers, like 3,000 or 4,000. Let's think about those as representing frequencies of light. And so over here we have percent transmittance. And so if you had 100% transmittance, let me go ahead and draw a line up here."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So for right now, let's think about these numbers, like 3,000 or 4,000. Let's think about those as representing frequencies of light. And so over here we have percent transmittance. And so if you had 100% transmittance, let me go ahead and draw a line up here. So 100% transmittance. So let's say we're talking about this frequency of light. So I look at this frequency of light."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if you had 100% transmittance, let me go ahead and draw a line up here. So 100% transmittance. So let's say we're talking about this frequency of light. So I look at this frequency of light. I go up to here, and I can see I have 100% transmittance. 100% transmittance means all of that light was transmitted through your sample. If all the light went through your sample, nothing was absorbed."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I look at this frequency of light. I go up to here, and I can see I have 100% transmittance. 100% transmittance means all of that light was transmitted through your sample. If all the light went through your sample, nothing was absorbed. And so this particular frequency was not absorbed by your compound. So if you're talking about less than 100% transmittance, so let's say for this frequency right here. So for this frequency, we have this signal here appearing at this frequency."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If all the light went through your sample, nothing was absorbed. And so this particular frequency was not absorbed by your compound. So if you're talking about less than 100% transmittance, so let's say for this frequency right here. So for this frequency, we have this signal here appearing at this frequency. We don't have 100% transmittance. So that means not all of the light went through the compound. Some of it was absorbed."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So for this frequency, we have this signal here appearing at this frequency. We don't have 100% transmittance. So that means not all of the light went through the compound. Some of it was absorbed. So this specific frequency was absorbed by the molecule. And that energy can cause a bond to stretch, and we get a stretching vibration. And actually, this signal corresponds to the bond that we've been talking about."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Some of it was absorbed. So this specific frequency was absorbed by the molecule. And that energy can cause a bond to stretch, and we get a stretching vibration. And actually, this signal corresponds to the bond that we've been talking about. So this signal indicates the stretching of this bond right here. All right, let's think about wave number next. So we just talked about percent transmittance."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And actually, this signal corresponds to the bond that we've been talking about. So this signal indicates the stretching of this bond right here. All right, let's think about wave number next. So we just talked about percent transmittance. Let's think about wave number. So I've been calling all these things frequencies, different frequencies of light. And let's see how wave number relates to the frequency of light and also the wavelength of light."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we just talked about percent transmittance. Let's think about wave number. So I've been calling all these things frequencies, different frequencies of light. And let's see how wave number relates to the frequency of light and also the wavelength of light. So the definition of wave number, so a wave number, here's the symbol for wave number. The definition of wave number is it's equal to one over the wavelength in centimeters. So if we had a wavelength of light of.002 centimeters, so let's go ahead and plug that in."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's see how wave number relates to the frequency of light and also the wavelength of light. So the definition of wave number, so a wave number, here's the symbol for wave number. The definition of wave number is it's equal to one over the wavelength in centimeters. So if we had a wavelength of light of.002 centimeters, so let's go ahead and plug that in. So a wavelength of light of.002 centimeters, what would be the wave number? So let's get out the calculator here, and let's do that math. One divided by.002 is equal to 500."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we had a wavelength of light of.002 centimeters, so let's go ahead and plug that in. So a wavelength of light of.002 centimeters, what would be the wave number? So let's get out the calculator here, and let's do that math. One divided by.002 is equal to 500. So that's equal to 500. Your units would be one over centimeters, or you could write that, meaning the same thing. So that's the wave number."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One divided by.002 is equal to 500. So that's equal to 500. Your units would be one over centimeters, or you could write that, meaning the same thing. So that's the wave number. That's the wave number. So if we go over here, a wave number of 500, you could think about this corresponding to a particular wavelength of light. And of course, this also relates to frequency because we know that wavelength and frequency are related to each other."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's the wave number. That's the wave number. So if we go over here, a wave number of 500, you could think about this corresponding to a particular wavelength of light. And of course, this also relates to frequency because we know that wavelength and frequency are related to each other. So let's get some more room down here. And we know that the wavelength times the frequency, so lambda times nu, is equal to the speed of light. That's equal to c. So if I wanted to solve for the frequency, so solve for nu, the frequency is equal to the speed of light divided by lambda, divided by the wavelength."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And of course, this also relates to frequency because we know that wavelength and frequency are related to each other. So let's get some more room down here. And we know that the wavelength times the frequency, so lambda times nu, is equal to the speed of light. That's equal to c. So if I wanted to solve for the frequency, so solve for nu, the frequency is equal to the speed of light divided by lambda, divided by the wavelength. And that's the same thing as one over lambda times the speed of light. One over lambda was our definition for wave number. So you could say that the frequency of light is equal to the wave number times the speed of light."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's equal to c. So if I wanted to solve for the frequency, so solve for nu, the frequency is equal to the speed of light divided by lambda, divided by the wavelength. And that's the same thing as one over lambda times the speed of light. One over lambda was our definition for wave number. So you could say that the frequency of light is equal to the wave number times the speed of light. So let's go ahead and do that calculation here. So we talked about this as being a particular wavelength. We found the wave number."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So you could say that the frequency of light is equal to the wave number times the speed of light. So let's go ahead and do that calculation here. So we talked about this as being a particular wavelength. We found the wave number. Let's plug that wave number into here and let's see what we get. So the wave number was 500, units were one over centimeters. Multiply that by the speed of light."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We found the wave number. Let's plug that wave number into here and let's see what we get. So the wave number was 500, units were one over centimeters. Multiply that by the speed of light. We need to have the speed of light in centimeters. So that's three times, approximately, three times 10 to the 10th centimeters per second is the speed of light. Notice what happens to the units."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Multiply that by the speed of light. We need to have the speed of light in centimeters. So that's three times, approximately, three times 10 to the 10th centimeters per second is the speed of light. Notice what happens to the units. The centimeters cancel. And let's do the math. So let's get some more over here."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Notice what happens to the units. The centimeters cancel. And let's do the math. So let's get some more over here. So we take the wave number and we multiply that by the speed of light in centimeters. So three times 10 to the 10th. And we get 1.5 times 10 to the 13th."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's get some more over here. So we take the wave number and we multiply that by the speed of light in centimeters. So three times 10 to the 10th. And we get 1.5 times 10 to the 13th. So the frequency, the frequency would be 1.5 times 10 to the 13th. The units would be, this would be one over seconds or you could say, or use hertz for that. So a wave number, right, corresponds to the wavelength and you can also get the frequency from that."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we get 1.5 times 10 to the 13th. So the frequency, the frequency would be 1.5 times 10 to the 13th. The units would be, this would be one over seconds or you could say, or use hertz for that. So a wave number, right, corresponds to the wavelength and you can also get the frequency from that. So we have, let me just rewrite this really quickly. So frequency is equal to wave number times the speed of light. And so wave number is equal to the frequency divided by the speed of light."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a wave number, right, corresponds to the wavelength and you can also get the frequency from that. So we have, let me just rewrite this really quickly. So frequency is equal to wave number times the speed of light. And so wave number is equal to the frequency divided by the speed of light. And we'll use this in a later video. So we'll come back to this idea. But for right now, the frequency of light is directly proportional to the wave number."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so wave number is equal to the frequency divided by the speed of light. And we'll use this in a later video. So we'll come back to this idea. But for right now, the frequency of light is directly proportional to the wave number. And so you could look at an infrared spectrum, let me go back up here. You could look at an infrared spectrum and call this down here, you could call this a wave number, you could refer to it as a frequency, you could call it whatever you want as long as you understand what's going on here. Let's look more in detail at this infrared spectrum."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But for right now, the frequency of light is directly proportional to the wave number. And so you could look at an infrared spectrum, let me go back up here. You could look at an infrared spectrum and call this down here, you could call this a wave number, you could refer to it as a frequency, you could call it whatever you want as long as you understand what's going on here. Let's look more in detail at this infrared spectrum. And let's draw a line at approximately 1,500 wave numbers right here. And the left side, the left side of that line, so we've divided our spectrum into two regions. The region on the left is called the diagnostic region."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's look more in detail at this infrared spectrum. And let's draw a line at approximately 1,500 wave numbers right here. And the left side, the left side of that line, so we've divided our spectrum into two regions. The region on the left is called the diagnostic region. So this is called the diagnostic region of our spectrum. And it's because a signal in this region can be diagnostic for a certain functional group. For example, this signal right here, if we go down here to the wave number, that signal's at approximately 2,100 for this wave number here."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The region on the left is called the diagnostic region. So this is called the diagnostic region of our spectrum. And it's because a signal in this region can be diagnostic for a certain functional group. For example, this signal right here, if we go down here to the wave number, that signal's at approximately 2,100 for this wave number here. And that's corresponding to the triple bond here. So this tells us a functional group. This tells us that a functional group is present, this triple bond is present."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For example, this signal right here, if we go down here to the wave number, that signal's at approximately 2,100 for this wave number here. And that's corresponding to the triple bond here. So this tells us a functional group. This tells us that a functional group is present, this triple bond is present. And so it's diagnostic. It helps you figure out the structure of the molecule. So you can figure out different functional groups present in molecules using IR spectra."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This tells us that a functional group is present, this triple bond is present. And so it's diagnostic. It helps you figure out the structure of the molecule. So you can figure out different functional groups present in molecules using IR spectra. So the right side, the right side of this line is called the fingerprint region. So this is the fingerprint region. And it's harder to interpret the fingerprint region."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So you can figure out different functional groups present in molecules using IR spectra. So the right side, the right side of this line is called the fingerprint region. So this is the fingerprint region. And it's harder to interpret the fingerprint region. It's much more complicated. It's not as easy to see different signals. It's extremely complicated, but it is unique to each molecule."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it's harder to interpret the fingerprint region. It's much more complicated. It's not as easy to see different signals. It's extremely complicated, but it is unique to each molecule. And so it's like a fingerprint for the molecule. And so you can match up IR spectra. If you have an unknown, you can look at the fingerprint region."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's extremely complicated, but it is unique to each molecule. And so it's like a fingerprint for the molecule. And so you can match up IR spectra. If you have an unknown, you can look at the fingerprint region. And again, it's unique. All these different lines are unique to that molecule. So we have the diagnostic region and the fingerprint region."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you have an unknown, you can look at the fingerprint region. And again, it's unique. All these different lines are unique to that molecule. So we have the diagnostic region and the fingerprint region. We're not gonna deal much with the fingerprint region, maybe a little bit. We're gonna focus in on the diagnostic region. We're gonna focus where the signals appear."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have the diagnostic region and the fingerprint region. We're not gonna deal much with the fingerprint region, maybe a little bit. We're gonna focus in on the diagnostic region. We're gonna focus where the signals appear. So I look at the signal, I go down to here, and I get a specific wave number. So the location of the signal is pretty important. I did wanna point out that if you look at what I use for the wave number here, I changed, I kind of changed how I did everything."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're gonna focus where the signals appear. So I look at the signal, I go down to here, and I get a specific wave number. So the location of the signal is pretty important. I did wanna point out that if you look at what I use for the wave number here, I changed, I kind of changed how I did everything. So the spacing is different. It doesn't really matter. I only did this to fit this video and to make it work for this video."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I did wanna point out that if you look at what I use for the wave number here, I changed, I kind of changed how I did everything. So the spacing is different. It doesn't really matter. I only did this to fit this video and to make it work for this video. And I'm also gonna hand draw all of my IR spectra. And so it's certainly not gonna be perfect. The idea is not worry too much about what they give you here for your scaling for the wave number, but think about where the signals appear."}, {"video_title": "Introduction to infrared spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I only did this to fit this video and to make it work for this video. And I'm also gonna hand draw all of my IR spectra. And so it's certainly not gonna be perfect. The idea is not worry too much about what they give you here for your scaling for the wave number, but think about where the signals appear. So the location of this signal, approximately 2,100 wave numbers. You also wanna think about the intensity of the signal and the shape of the signal. We'll talk much more about those in later videos."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's see how to determine the number of expected signals in an NMR spectrum. And the best way to do this is just to do a lot of practice problems. And we'll start with methane. And methane has four protons in the same environment. Therefore, we say those four protons are chemically equivalent, and we would expect to see only one signal on our NMR spectrum. So for methane, we would expect to see one signal because all four protons are chemically equivalent. If we move on to propane, so this carbon right here has two protons on it, and these two protons are in the same environment."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And methane has four protons in the same environment. Therefore, we say those four protons are chemically equivalent, and we would expect to see only one signal on our NMR spectrum. So for methane, we would expect to see one signal because all four protons are chemically equivalent. If we move on to propane, so this carbon right here has two protons on it, and these two protons are in the same environment. Therefore, they are chemically equivalent, and we would expect to see only one signal for these two protons. If we look at these protons over here, so let me go ahead and draw them in, so these methyl protons, these methyl protons are in their own environment here. So we'd expect one signal for these methyl protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we move on to propane, so this carbon right here has two protons on it, and these two protons are in the same environment. Therefore, they are chemically equivalent, and we would expect to see only one signal for these two protons. If we look at these protons over here, so let me go ahead and draw them in, so these methyl protons, these methyl protons are in their own environment here. So we'd expect one signal for these methyl protons. But for a molecule like this, you need to think about symmetry. So if I draw a line right here, it's easy to see that these three protons are in the same environment as these. So really, really, we have six protons in the same environment."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we'd expect one signal for these methyl protons. But for a molecule like this, you need to think about symmetry. So if I draw a line right here, it's easy to see that these three protons are in the same environment as these. So really, really, we have six protons in the same environment. So these six protons are chemically equivalent, and we would expect to see one signal for all six. So for propane, we would expect to see a total of two signals. All right, let's move on to this one."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So really, really, we have six protons in the same environment. So these six protons are chemically equivalent, and we would expect to see one signal for all six. So for propane, we would expect to see a total of two signals. All right, let's move on to this one. So let's draw in our protons here. So let's just go ahead and draw in all of them first. So we're drawing in all of the protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's move on to this one. So let's draw in our protons here. So let's just go ahead and draw in all of them first. So we're drawing in all of the protons. So we have these methyl protons here, and then right here, we have a proton on this carbon, and then we have all of these protons. So a lot of them. Let's focus in on the methyl protons first."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're drawing in all of the protons. So we have these methyl protons here, and then right here, we have a proton on this carbon, and then we have all of these protons. So a lot of them. Let's focus in on the methyl protons first. So these methyl protons here are in their own environment, and that environment is the same as all of these methyl protons. They're all in the exact same environment. They're all right next to a carbon with a hydrogen on it."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's focus in on the methyl protons first. So these methyl protons here are in their own environment, and that environment is the same as all of these methyl protons. They're all in the exact same environment. They're all right next to a carbon with a hydrogen on it. And so we would expect to see only one signal for all these protons. So one signal for the magenta protons. All right, so again, think about symmetry."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "They're all right next to a carbon with a hydrogen on it. And so we would expect to see only one signal for all these protons. So one signal for the magenta protons. All right, so again, think about symmetry. Think about the symmetry of this. And then if I look at this proton right here, this is in a different environment from the magenta protons, but it's in the same environment as this proton. So we'd expect one signal for these red protons here."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so again, think about symmetry. Think about the symmetry of this. And then if I look at this proton right here, this is in a different environment from the magenta protons, but it's in the same environment as this proton. So we'd expect one signal for these red protons here. So one signal for the red protons. So a total of two signals for the molecule. All right, let's do some more examples."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we'd expect one signal for these red protons here. So one signal for the red protons. So a total of two signals for the molecule. All right, let's do some more examples. So let's look at this one next. All right, so we have a carbon right here with two protons on it, and these two protons are in the same environment. They're chemically equivalent."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's do some more examples. So let's look at this one next. All right, so we have a carbon right here with two protons on it, and these two protons are in the same environment. They're chemically equivalent. These two protons are next to this oxygen here. We know oxygen is more electronegative than carbon, so oxygen is going to withdraw some electron density. And so these two protons are in a different environment from these two protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "They're chemically equivalent. These two protons are next to this oxygen here. We know oxygen is more electronegative than carbon, so oxygen is going to withdraw some electron density. And so these two protons are in a different environment from these two protons. These two protons right here, let me go ahead and change colors. These protons in red are further away from the oxygen, so they're in a different environment. So we'd expect one signal from the protons in red, and we would expect one signal from the protons in yellow."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so these two protons are in a different environment from these two protons. These two protons right here, let me go ahead and change colors. These protons in red are further away from the oxygen, so they're in a different environment. So we'd expect one signal from the protons in red, and we would expect one signal from the protons in yellow. All right, we still have our methyl protons. Let me go ahead and draw those in. So we have three protons over here on this carbon."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we'd expect one signal from the protons in red, and we would expect one signal from the protons in yellow. All right, we still have our methyl protons. Let me go ahead and draw those in. So we have three protons over here on this carbon. They're further away from the oxygen. They're in their own environment. They're chemically equivalent, so one signal expected for these protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have three protons over here on this carbon. They're further away from the oxygen. They're in their own environment. They're chemically equivalent, so one signal expected for these protons. Finally, don't forget about the proton right here, the proton on the oxygen. So that's in its own environment, so we would expect one signal for that proton. So really, a total, if you count them all up, a total of four signals."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "They're chemically equivalent, so one signal expected for these protons. Finally, don't forget about the proton right here, the proton on the oxygen. So that's in its own environment, so we would expect one signal for that proton. So really, a total, if you count them all up, a total of four signals. So we would expect four signals. Not sure why I put an S on here, only one signal. So a total of four for this alcohol."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So really, a total, if you count them all up, a total of four signals. So we would expect four signals. Not sure why I put an S on here, only one signal. So a total of four for this alcohol. Let's look at this alcohol down here. So we have these two protons, and these protons over here, and these protons over here. So let's think about that."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a total of four for this alcohol. Let's look at this alcohol down here. So we have these two protons, and these protons over here, and these protons over here. So let's think about that. So symmetry helps for an example like this. These two protons are in the same environment. They're between two CH2s, and so we'd expect one signal for these protons, so they're chemically equivalent."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about that. So symmetry helps for an example like this. These two protons are in the same environment. They're between two CH2s, and so we'd expect one signal for these protons, so they're chemically equivalent. Let's look at these protons next. So these right here are next to a CH2. They're also next to an OH."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "They're between two CH2s, and so we'd expect one signal for these protons, so they're chemically equivalent. Let's look at these protons next. So these right here are next to a CH2. They're also next to an OH. That's the exact same environment as these protons. They're next to a CH2 and next to an OH. So if you think about symmetry right here, we'd expect only one signal, only one signal for these four protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "They're also next to an OH. That's the exact same environment as these protons. They're next to a CH2 and next to an OH. So if you think about symmetry right here, we'd expect only one signal, only one signal for these four protons. And then finally, we still have the, we had the proton on the oxygen, and once again, symmetry helps us think about the fact that this proton is in the same environment as this proton, so we would expect, we would expect one signal there. So a total of three expected signals for this molecule. All right, let's move on to this compound over here on the right, and we can see that there's a chiral center right here on this carbon, and therefore must be a hydrogen, a proton on that carbon."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if you think about symmetry right here, we'd expect only one signal, only one signal for these four protons. And then finally, we still have the, we had the proton on the oxygen, and once again, symmetry helps us think about the fact that this proton is in the same environment as this proton, so we would expect, we would expect one signal there. So a total of three expected signals for this molecule. All right, let's move on to this compound over here on the right, and we can see that there's a chiral center right here on this carbon, and therefore must be a hydrogen, a proton on that carbon. So let's go ahead and draw in all of our protons, and then let's analyze them here. So we have three protons on this carbon. On this carbon right here, we have two protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's move on to this compound over here on the right, and we can see that there's a chiral center right here on this carbon, and therefore must be a hydrogen, a proton on that carbon. So let's go ahead and draw in all of our protons, and then let's analyze them here. So we have three protons on this carbon. On this carbon right here, we have two protons. And I'm drawing them in with a wedge and a dash because we're gonna need to think about those two protons. And then over here, we have three like that. All right, so let's think about how many different signals we're going to get."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "On this carbon right here, we have two protons. And I'm drawing them in with a wedge and a dash because we're gonna need to think about those two protons. And then over here, we have three like that. All right, so let's think about how many different signals we're going to get. All right, so in the past, let me go back and use a different color here. When we talked about a CH2 group, and we didn't have any chiral centers like these two protons right here, we didn't have any chiral centers in this molecule. These two protons are generally equivalent."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's think about how many different signals we're going to get. All right, so in the past, let me go back and use a different color here. When we talked about a CH2 group, and we didn't have any chiral centers like these two protons right here, we didn't have any chiral centers in this molecule. These two protons are generally equivalent. All right, but if I think about over here, I have a chiral center at this carbon, and that's going to affect these two protons. These two protons are no longer chemically equivalent, right, because a chiral center is present. So in general, if a chiral center is present, the two protons on a methylene group, on a CH2 group, are generally not equivalent."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These two protons are generally equivalent. All right, but if I think about over here, I have a chiral center at this carbon, and that's going to affect these two protons. These two protons are no longer chemically equivalent, right, because a chiral center is present. So in general, if a chiral center is present, the two protons on a methylene group, on a CH2 group, are generally not equivalent. So we would expect two different signals. So this one might give us one signal. I'll put that in magenta."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So in general, if a chiral center is present, the two protons on a methylene group, on a CH2 group, are generally not equivalent. So we would expect two different signals. So this one might give us one signal. I'll put that in magenta. So one signal here. And this one, we would expect a different signal. They're in slightly different environments because of this chiral center that is present."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I'll put that in magenta. So one signal here. And this one, we would expect a different signal. They're in slightly different environments because of this chiral center that is present. All right, so let's keep on going, and let's figure out how many more signals we would expect. So over here, right, we have, these three protons are chemically equivalent, and they're in their own unique environment. So we would expect one signal for those three protons."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "They're in slightly different environments because of this chiral center that is present. All right, so let's keep on going, and let's figure out how many more signals we would expect. So over here, right, we have, these three protons are chemically equivalent, and they're in their own unique environment. So we would expect one signal for those three protons. All right, and those are in a different environment from these over here. So we would expect one signal for these. So for a methyl group, so methyl protons are always in the same environment, so they're chemically equivalent."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we would expect one signal for those three protons. All right, and those are in a different environment from these over here. So we would expect one signal for these. So for a methyl group, so methyl protons are always in the same environment, so they're chemically equivalent. So you don't have to worry about them. All right, and then let's see, what else do we have here? We have the proton on the oxygen, so we would expect one signal for that."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So for a methyl group, so methyl protons are always in the same environment, so they're chemically equivalent. So you don't have to worry about them. All right, and then let's see, what else do we have here? We have the proton on the oxygen, so we would expect one signal for that. And then finally, we're going to have, let's see, what color should I choose here? Let's go with orange. We still have a proton right here in its own unique environment, so we would expect one signal for that proton."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have the proton on the oxygen, so we would expect one signal for that. And then finally, we're going to have, let's see, what color should I choose here? Let's go with orange. We still have a proton right here in its own unique environment, so we would expect one signal for that proton. So how many total signals do we expect? So let's count them up. So one, two, three, four, five, and six."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We still have a proton right here in its own unique environment, so we would expect one signal for that proton. So how many total signals do we expect? So let's count them up. So one, two, three, four, five, and six. So a total of six expected signals for this alcohol. All right, let's move on to a benzene ring here. So let's look at benzene."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So one, two, three, four, five, and six. So a total of six expected signals for this alcohol. All right, let's move on to a benzene ring here. So let's look at benzene. And we know that benzene has six protons, right? So I could draw in the six protons here. And those six protons are all in the exact same environment."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at benzene. And we know that benzene has six protons, right? So I could draw in the six protons here. And those six protons are all in the exact same environment. They're all chemically equivalent. And sometimes it's easier to just go ahead and draw your circle in here, and that just helps you to see that these protons are equivalent. It allows you to see the symmetry a little bit better."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And those six protons are all in the exact same environment. They're all chemically equivalent. And sometimes it's easier to just go ahead and draw your circle in here, and that just helps you to see that these protons are equivalent. It allows you to see the symmetry a little bit better. So again, thinking about symmetry, all six protons are in the same environment, therefore we would expect only one signal for benzene on an NMR spectrum. So let me go ahead and write that here. So one, only one signal for benzene because all six protons are chemically equivalent."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It allows you to see the symmetry a little bit better. So again, thinking about symmetry, all six protons are in the same environment, therefore we would expect only one signal for benzene on an NMR spectrum. So let me go ahead and write that here. So one, only one signal for benzene because all six protons are chemically equivalent. Move over here to this molecule down here. All right, so let's think about what we have. So we have a methyl group here, a methyl group here, so that's three protons, right?"}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So one, only one signal for benzene because all six protons are chemically equivalent. Move over here to this molecule down here. All right, so let's think about what we have. So we have a methyl group here, a methyl group here, so that's three protons, right? So three protons for each methyl group. And those protons are in the same environment, right? So these three protons in this methyl group are in the same environment as these three protons, right?"}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have a methyl group here, a methyl group here, so that's three protons, right? So three protons for each methyl group. And those protons are in the same environment, right? So these three protons in this methyl group are in the same environment as these three protons, right? They're right next to an oxygen, which is next to this benzene ring. So next to an oxygen, next to a benzene ring. And so we'd expect only one signal, right?"}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So these three protons in this methyl group are in the same environment as these three protons, right? They're right next to an oxygen, which is next to this benzene ring. So next to an oxygen, next to a benzene ring. And so we'd expect only one signal, right? Only one signal for these six protons right here. If we look at our ring, if we look at our ring, we know that there's a proton here, we know there's a proton here, a proton here, and a proton here. And if we think about those protons, they're all in the exact same environment, right?"}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so we'd expect only one signal, right? Only one signal for these six protons right here. If we look at our ring, if we look at our ring, we know that there's a proton here, we know there's a proton here, a proton here, and a proton here. And if we think about those protons, they're all in the exact same environment, right? So if I think about, let's say, this proton, this proton is next to a CH, right, with a double bond. It's next to a carbon right here with an oxygen. This is in the same environment as this one, right?"}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if we think about those protons, they're all in the exact same environment, right? So if I think about, let's say, this proton, this proton is next to a CH, right, with a double bond. It's next to a carbon right here with an oxygen. This is in the same environment as this one, right? This one's in the exact same environment. This one is, this one is. So again, symmetry helps you realize that all four protons are in the same environment, they're chemically equivalent, therefore one signal."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is in the same environment as this one, right? This one's in the exact same environment. This one is, this one is. So again, symmetry helps you realize that all four protons are in the same environment, they're chemically equivalent, therefore one signal. So we would expect to see two signals. We would expect to see two signals for this molecule on an NMR spectrum. Finally, let's just, for fun, let's just look at cubane here, right?"}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So again, symmetry helps you realize that all four protons are in the same environment, they're chemically equivalent, therefore one signal. So we would expect to see two signals. We would expect to see two signals for this molecule on an NMR spectrum. Finally, let's just, for fun, let's just look at cubane here, right? So one of the most interesting molecules, in my opinion. There are eight protons on cubane. Cubane is just this cube here."}, {"video_title": "Chemical equivalence Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Finally, let's just, for fun, let's just look at cubane here, right? So one of the most interesting molecules, in my opinion. There are eight protons on cubane. Cubane is just this cube here. And those eight protons are all equivalent, right? So if you think about, if you just rotated this cube, right, you wouldn't be able to tell a difference here. So we have eight protons, all equivalent, therefore only one signal."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Here's an example of a nucleophilic addition reaction to an aldehyde or a ketone. So over here on the left, right, it could be an aldehyde or you could change that to form a ketone. If you add water to an aldehyde or a ketone, you form this product over here on the right, which is called a hydrate, or also called a gem diol, or geminal diol, because these two OHs here are on the same carbon, so they're like twins. And this reaction is at equilibrium. So let's think about the aldehyde or the ketone, right? We know the carbonyl in aldehyde or ketone is polarized, so we know that the oxygen is more electronegative than carbon, so it withdraws some electron density. So this oxygen here is partially negative, and this carbonyl carbon is partially positive, like that."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And this reaction is at equilibrium. So let's think about the aldehyde or the ketone, right? We know the carbonyl in aldehyde or ketone is polarized, so we know that the oxygen is more electronegative than carbon, so it withdraws some electron density. So this oxygen here is partially negative, and this carbonyl carbon is partially positive, like that. And therefore, the carbonyl carbon, since it's partially positive, is electrophilic, so it wants electrons, and it can get electrons from water. So let's go ahead and draw the water molecule right here. Water can function as a nucleophile, right?"}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen here is partially negative, and this carbonyl carbon is partially positive, like that. And therefore, the carbonyl carbon, since it's partially positive, is electrophilic, so it wants electrons, and it can get electrons from water. So let's go ahead and draw the water molecule right here. Water can function as a nucleophile, right? It has two lone pairs of electrons. So this oxygen here is partially negative. And so we're going to get a nucleophile attacking our electrophile."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Water can function as a nucleophile, right? It has two lone pairs of electrons. So this oxygen here is partially negative. And so we're going to get a nucleophile attacking our electrophile. So a lone pair of electrons on the oxygen is going to attack our carbonyl carbon, like that. So the nucleophile attacks the electrophilic portion of the molecule, and these pi electrons here kick off onto the oxygen. So let's go ahead and draw the result of our nucleophilic attack here."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to get a nucleophile attacking our electrophile. So a lone pair of electrons on the oxygen is going to attack our carbonyl carbon, like that. So the nucleophile attacks the electrophilic portion of the molecule, and these pi electrons here kick off onto the oxygen. So let's go ahead and draw the result of our nucleophilic attack here. And so we now have our oxygen, right, bonded to this carbon. And this oxygen still has two hydrogens bonded to it. So I'm going to go ahead and draw in those two hydrogens."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the result of our nucleophilic attack here. And so we now have our oxygen, right, bonded to this carbon. And this oxygen still has two hydrogens bonded to it. So I'm going to go ahead and draw in those two hydrogens. There's still a lone pair of electrons on that oxygen, which gives that oxygen a plus 1 formal charge. And then this carbon here is bonded to another oxygen, which had two lone pairs of electrons around it, and now it picked up another one, so a negative 1 formal charge on this oxygen. And there's still an R group bonded to it, and a hydrogen over here, like that."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and draw in those two hydrogens. There's still a lone pair of electrons on that oxygen, which gives that oxygen a plus 1 formal charge. And then this carbon here is bonded to another oxygen, which had two lone pairs of electrons around it, and now it picked up another one, so a negative 1 formal charge on this oxygen. And there's still an R group bonded to it, and a hydrogen over here, like that. And so let's try to follow some electrons here. So one of the lone pairs of electrons on the oxygen formed a bond with our carbon. So I'm saying that these electrons are right here, like that."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And there's still an R group bonded to it, and a hydrogen over here, like that. And so let's try to follow some electrons here. So one of the lone pairs of electrons on the oxygen formed a bond with our carbon. So I'm saying that these electrons are right here, like that. And then we can think about our pi electrons. So our pi electrons in here is kicking off onto our oxygen. So it doesn't really matter which one of these three it is."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So I'm saying that these electrons are right here, like that. And then we can think about our pi electrons. So our pi electrons in here is kicking off onto our oxygen. So it doesn't really matter which one of these three it is. Let's just say it's that one. And we get this as our intermediate. And so next, we can think about an acid-base reaction."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it doesn't really matter which one of these three it is. Let's just say it's that one. And we get this as our intermediate. And so next, we can think about an acid-base reaction. So another water molecule comes along right here. And so we know water can function as an acid or a base. And so this lone pair of electrons could take, let's say it takes this proton right here and leaves these electrons behind on our oxygen."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so next, we can think about an acid-base reaction. So another water molecule comes along right here. And so we know water can function as an acid or a base. And so this lone pair of electrons could take, let's say it takes this proton right here and leaves these electrons behind on our oxygen. So let's go ahead and draw the result of that acid-base reaction. And so we would have our oxygen here would now be bonded to only one hydrogen. And let's see, we still have our negatively charged oxygen over here on the right."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so this lone pair of electrons could take, let's say it takes this proton right here and leaves these electrons behind on our oxygen. So let's go ahead and draw the result of that acid-base reaction. And so we would have our oxygen here would now be bonded to only one hydrogen. And let's see, we still have our negatively charged oxygen over here on the right. And then we have our R group and our hydrogen like that. And we still have this lone pair of electrons. And we just picked up another lone pair of electrons."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And let's see, we still have our negatively charged oxygen over here on the right. And then we have our R group and our hydrogen like that. And we still have this lone pair of electrons. And we just picked up another lone pair of electrons. So let me go ahead and show those. So let me go ahead and use blue for those electrons. So these electrons in here ended up on this oxygen."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And we just picked up another lone pair of electrons. So let me go ahead and show those. So let me go ahead and use blue for those electrons. So these electrons in here ended up on this oxygen. And so we're almost to our final product. We've almost formed our hydrate. We just need to do one more acid-base reaction."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here ended up on this oxygen. And so we're almost to our final product. We've almost formed our hydrate. We just need to do one more acid-base reaction. So of course, another water molecule could come along. So water can function as an acid or a base. And a lone pair of electrons on this oxygen could take this proton and leave these electrons behind."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We just need to do one more acid-base reaction. So of course, another water molecule could come along. So water can function as an acid or a base. And a lone pair of electrons on this oxygen could take this proton and leave these electrons behind. And then that, of course, would give us our final product. So that would give us our hydrate over here. And you could, of course, change this hydrogen to an R prime, change this hydrogen to an R prime if you wanted to."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And a lone pair of electrons on this oxygen could take this proton and leave these electrons behind. And then that, of course, would give us our final product. So that would give us our hydrate over here. And you could, of course, change this hydrogen to an R prime, change this hydrogen to an R prime if you wanted to. And you would get this over here. So your hydrate or your gem diol. And so that's the general mechanism for uncatalyzed reaction here."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And you could, of course, change this hydrogen to an R prime, change this hydrogen to an R prime if you wanted to. And you would get this over here. So your hydrate or your gem diol. And so that's the general mechanism for uncatalyzed reaction here. Let's look at some examples. And it's important to remember that this reaction is at equilibrium. And so let's look at three examples."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so that's the general mechanism for uncatalyzed reaction here. Let's look at some examples. And it's important to remember that this reaction is at equilibrium. And so let's look at three examples. Our first example is formaldehyde. So if you react formaldehyde with water, you form your hydrate over here on the right. And in this case, the equilibrium is to the right."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so let's look at three examples. Our first example is formaldehyde. So if you react formaldehyde with water, you form your hydrate over here on the right. And in this case, the equilibrium is to the right. It favors the formation of the hydrate. And that's because aldehydes are very reactive. And we talked about why in the video on the reactivity of aldehydes and ketones."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And in this case, the equilibrium is to the right. It favors the formation of the hydrate. And that's because aldehydes are very reactive. And we talked about why in the video on the reactivity of aldehydes and ketones. So we have a very polarized carbonyl situation here. So this carbon right here is partially positive. And we have these hydrogens here, which don't take up a whole lot of space."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And we talked about why in the video on the reactivity of aldehydes and ketones. So we have a very polarized carbonyl situation here. So this carbon right here is partially positive. And we have these hydrogens here, which don't take up a whole lot of space. And so we have greater polarization than ketones and also decreased steric hindrance. And so because of the high reactivity of aldehydes, then this product is favored. If we look at the next reaction, we no longer have an aldehyde."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And we have these hydrogens here, which don't take up a whole lot of space. And so we have greater polarization than ketones and also decreased steric hindrance. And so because of the high reactivity of aldehydes, then this product is favored. If we look at the next reaction, we no longer have an aldehyde. This is a ketone. This is acetone. And so adding water to acetone gives us this as our product for our hydrate."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If we look at the next reaction, we no longer have an aldehyde. This is a ketone. This is acetone. And so adding water to acetone gives us this as our product for our hydrate. Except we know that ketones are not as reactive as aldehydes. And so this time, the equilibrium is to the left. So it favors the formation of the ketone."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so adding water to acetone gives us this as our product for our hydrate. Except we know that ketones are not as reactive as aldehydes. And so this time, the equilibrium is to the left. So it favors the formation of the ketone. Let's look at another one. So this is acetaldehyde. And so if we add water to acetaldehyde over here, we form this as our hydrate product."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So it favors the formation of the ketone. Let's look at another one. So this is acetaldehyde. And so if we add water to acetaldehyde over here, we form this as our hydrate product. And once again, we know that these reactions occur because of our carbonyl carbon right here being partially positive. So the oxygen withdraws some electron density like that. So we could make aldehydes or ketones more reactive by adding something else that withdraws electron density from that carbonyl carbon."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so if we add water to acetaldehyde over here, we form this as our hydrate product. And once again, we know that these reactions occur because of our carbonyl carbon right here being partially positive. So the oxygen withdraws some electron density like that. So we could make aldehydes or ketones more reactive by adding something else that withdraws electron density from that carbonyl carbon. And one thing you could do is add an electronegative atom like a halogen. So let's go ahead and do that. Let's add three halogens here."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we could make aldehydes or ketones more reactive by adding something else that withdraws electron density from that carbonyl carbon. And one thing you could do is add an electronegative atom like a halogen. So let's go ahead and do that. Let's add three halogens here. Let's add three chlorines to this carbon, the one adjacent to our carbonyl carbon. And those electron withdrawing groups, those very electronegative atoms, withdraw some electron density. So they're going to withdraw electron density this way, once again, away from our carbonyl carbon."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's add three halogens here. Let's add three chlorines to this carbon, the one adjacent to our carbonyl carbon. And those electron withdrawing groups, those very electronegative atoms, withdraw some electron density. So they're going to withdraw electron density this way, once again, away from our carbonyl carbon. And so this carbonyl carbon gets even more partially positive by the addition of these electronegative atoms. And the more positive you make that carbonyl carbon, the more electrophilic you make it. And therefore, the more the nucleophile, which is water, is going to attack."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So they're going to withdraw electron density this way, once again, away from our carbonyl carbon. And so this carbonyl carbon gets even more partially positive by the addition of these electronegative atoms. And the more positive you make that carbonyl carbon, the more electrophilic you make it. And therefore, the more the nucleophile, which is water, is going to attack. And so you make it even more reactive by adding these. And so you can push the equilibrium even more. To the right, you can form more of your product."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, the more the nucleophile, which is water, is going to attack. And so you make it even more reactive by adding these. And so you can push the equilibrium even more. To the right, you can form more of your product. So let's go ahead and draw the product of that reaction. So we would put three chlorines on here like that. And so you could also do this with ketones."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "To the right, you can form more of your product. So let's go ahead and draw the product of that reaction. So we would put three chlorines on here like that. And so you could also do this with ketones. And you could make ketones much more reactive by doing that. And so this particular reaction is a little bit famous. Over here on the left is trichloroacetaldehyde."}, {"video_title": "Formation of hydrates Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so you could also do this with ketones. And you could make ketones much more reactive by doing that. And so this particular reaction is a little bit famous. Over here on the left is trichloroacetaldehyde. And then once you form the hydrate of that, you form chloral hydrate. And this is famous for being knockout drops. And so some of the old references to it are slip someone a Mickey Finn."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Here we have two compounds, so this one and this one, that would both be called 1,2-dimethylcyclohexane. But the top one here has two methyl groups going up in space, or coming out at us. So those two methyl groups must be on the same side of the ring, and we call that cis. For this one, we have one methyl group with a wedge and one methyl group with a dash. So those two methyl groups are on opposite sides of the ring, and we call that trans. So first we're gonna look at the cis-1,2-dimethylcyclohexane compound, and we're gonna draw both chair conformations. And then we'll look at the trans."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "For this one, we have one methyl group with a wedge and one methyl group with a dash. So those two methyl groups are on opposite sides of the ring, and we call that trans. So first we're gonna look at the cis-1,2-dimethylcyclohexane compound, and we're gonna draw both chair conformations. And then we'll look at the trans. Here we have our chair conformation. At carbon one, we have a methyl group that is up axial. We also have a hydrogen that's down equatorial, which is a little bit hard to see here."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll look at the trans. Here we have our chair conformation. At carbon one, we have a methyl group that is up axial. We also have a hydrogen that's down equatorial, which is a little bit hard to see here. The hydrogen's green so we can see it better. At carbon two, we have a methyl group that is up equatorial, and then we have a hydrogen that's down axial. So again, it's green to see it better."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We also have a hydrogen that's down equatorial, which is a little bit hard to see here. The hydrogen's green so we can see it better. At carbon two, we have a methyl group that is up equatorial, and then we have a hydrogen that's down axial. So again, it's green to see it better. When this chair conformation undergoes a ring flip, I rotate this carbon up and I rotate this carbon down. And then we turn it a little bit so we can see our other chair conformation. And then we analyze what happened to our groups."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So again, it's green to see it better. When this chair conformation undergoes a ring flip, I rotate this carbon up and I rotate this carbon down. And then we turn it a little bit so we can see our other chair conformation. And then we analyze what happened to our groups. Now at carbon one, we have a methyl group that's up equatorial, and our hydrogen is now down axial. And at carbon two, our methyl group is up axial, and our hydrogen is down equatorial. Now let's draw our chair conformations."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we analyze what happened to our groups. Now at carbon one, we have a methyl group that's up equatorial, and our hydrogen is now down axial. And at carbon two, our methyl group is up axial, and our hydrogen is down equatorial. Now let's draw our chair conformations. So we'll start with the one on the left. So you've seen how to draw chair conformations in earlier videos. We start with our two parallel lines that are offset from each other."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now let's draw our chair conformations. So we'll start with the one on the left. So you've seen how to draw chair conformations in earlier videos. We start with our two parallel lines that are offset from each other. So there's one line and there's our other line. Next we draw a dotted line that just touches the top part of the top line. So there's our first dotted line, so it's supposed to come close to this point."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We start with our two parallel lines that are offset from each other. So there's one line and there's our other line. Next we draw a dotted line that just touches the top part of the top line. So there's our first dotted line, so it's supposed to come close to this point. And then our other dotted line is gonna go right here and touch the bottom point of the bottom line. Next we think about another set of parallel lines. So if we draw a line from the top one down to here, and then we draw one parallel to that over here, and then finally we put in our last set of parallel lines."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So there's our first dotted line, so it's supposed to come close to this point. And then our other dotted line is gonna go right here and touch the bottom point of the bottom line. Next we think about another set of parallel lines. So if we draw a line from the top one down to here, and then we draw one parallel to that over here, and then finally we put in our last set of parallel lines. So from this point to here, and then from this point to here. If we call this carbon one, we start up axial at carbon one, as we've seen in earlier videos, and then carbon two would be down axial since we alternate. I'll stop there, I won't draw in the rest of the hydrogens just to save time."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So if we draw a line from the top one down to here, and then we draw one parallel to that over here, and then finally we put in our last set of parallel lines. So from this point to here, and then from this point to here. If we call this carbon one, we start up axial at carbon one, as we've seen in earlier videos, and then carbon two would be down axial since we alternate. I'll stop there, I won't draw in the rest of the hydrogens just to save time. If we go back to carbon one, we know that next we go down, down equatorial, so I put that one in, and then at carbon two this would be up. So let's look at our picture here, and we can see that our methyl group is up axial at carbon one. So we put in a CH three up axial."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "I'll stop there, I won't draw in the rest of the hydrogens just to save time. If we go back to carbon one, we know that next we go down, down equatorial, so I put that one in, and then at carbon two this would be up. So let's look at our picture here, and we can see that our methyl group is up axial at carbon one. So we put in a CH three up axial. Then we have a hydrogen down equatorial right here at carbon one. At carbon two, our methyl group is up equatorial, so we put that in. And then we have a hydrogen that's down axial."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we put in a CH three up axial. Then we have a hydrogen down equatorial right here at carbon one. At carbon two, our methyl group is up equatorial, so we put that in. And then we have a hydrogen that's down axial. So this is the cis compound. Both these methyl groups are on the same side. They're both going up relative to a flat plane of the ring."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a hydrogen that's down axial. So this is the cis compound. Both these methyl groups are on the same side. They're both going up relative to a flat plane of the ring. We know that this chair conformation is in equilibrium with our other chair conformation, so let's go ahead and draw the one on the right now. So we start with our two parallel lines that are a little offset. So here's one of the lines, and then here is the other one."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "They're both going up relative to a flat plane of the ring. We know that this chair conformation is in equilibrium with our other chair conformation, so let's go ahead and draw the one on the right now. So we start with our two parallel lines that are a little offset. So here's one of the lines, and then here is the other one. Next we draw our dotted line, so intersecting that top point and intersecting the bottom point here. Finally, we think about our next set of parallel lines, so this one and then this one, and finally our last set of parallel lines, so from here to here and then from here to here. Now we can see that this is carbon one, so here is carbon one after we undergo a ring flip."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So here's one of the lines, and then here is the other one. Next we draw our dotted line, so intersecting that top point and intersecting the bottom point here. Finally, we think about our next set of parallel lines, so this one and then this one, and finally our last set of parallel lines, so from here to here and then from here to here. Now we can see that this is carbon one, so here is carbon one after we undergo a ring flip. And now we start down axial at carbon one, so I'll draw down axial at carbon one. At carbon two, now it's up axial, so up axial at carbon two. So let's put in our methyl groups, and to do that we need to put in the other bonds, so this would be up and then down."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now we can see that this is carbon one, so here is carbon one after we undergo a ring flip. And now we start down axial at carbon one, so I'll draw down axial at carbon one. At carbon two, now it's up axial, so up axial at carbon two. So let's put in our methyl groups, and to do that we need to put in the other bonds, so this would be up and then down. So at carbon one, we can see our methyl group is up equatorial, so we put in our CH three here, and our hydrogen is down axial. And then at carbon two, our methyl group is up axial, and then our hydrogen is down equatorial. So when this compound underwent a ring flip, we know this is carbon one, where our methyl group was up axial, and then this is carbon one."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in our methyl groups, and to do that we need to put in the other bonds, so this would be up and then down. So at carbon one, we can see our methyl group is up equatorial, so we put in our CH three here, and our hydrogen is down axial. And then at carbon two, our methyl group is up axial, and then our hydrogen is down equatorial. So when this compound underwent a ring flip, we know this is carbon one, where our methyl group was up axial, and then this is carbon one. Our methyl group is still up relative to the plane of the ring, but now it is equatorial, so that's what happens when you do a ring flip. And then carbon two, let me change colors here. So carbon two, we had a methyl group that was up equatorial, and now our methyl group is up axial."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So when this compound underwent a ring flip, we know this is carbon one, where our methyl group was up axial, and then this is carbon one. Our methyl group is still up relative to the plane of the ring, but now it is equatorial, so that's what happens when you do a ring flip. And then carbon two, let me change colors here. So carbon two, we had a methyl group that was up equatorial, and now our methyl group is up axial. So for both chair conformations, the methyl groups are on the same side, and that's why we say cis. In terms of which one is the more stable conformation, for both of these, we have one methyl group axial and one methyl group equatorial, so they are equivalent, so they're the same in terms of energy. Next, let's look at trans, one, two dimethylcyclohexane."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So carbon two, we had a methyl group that was up equatorial, and now our methyl group is up axial. So for both chair conformations, the methyl groups are on the same side, and that's why we say cis. In terms of which one is the more stable conformation, for both of these, we have one methyl group axial and one methyl group equatorial, so they are equivalent, so they're the same in terms of energy. Next, let's look at trans, one, two dimethylcyclohexane. Here we have trans, one, two dimethylcyclohexane, and you can see at carbon one, we have a methyl group that's up axial, and a hydrogen that's down equatorial. And at carbon two, we have a hydrogen that's up equatorial and a methyl group that's down axial. If this undergoes a ring flip, so I lift this carbon up and pull this carbon down, and then turn it so we can see our other chair conformation, now let's analyze what happened to our methyl groups."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at trans, one, two dimethylcyclohexane. Here we have trans, one, two dimethylcyclohexane, and you can see at carbon one, we have a methyl group that's up axial, and a hydrogen that's down equatorial. And at carbon two, we have a hydrogen that's up equatorial and a methyl group that's down axial. If this undergoes a ring flip, so I lift this carbon up and pull this carbon down, and then turn it so we can see our other chair conformation, now let's analyze what happened to our methyl groups. At carbon one, now our methyl group is up equatorial, and at carbon two, our methyl group is down equatorial. Now let's draw the chair conformations for the trans compound. The cis and trans compounds are isomers of each other."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "If this undergoes a ring flip, so I lift this carbon up and pull this carbon down, and then turn it so we can see our other chair conformation, now let's analyze what happened to our methyl groups. At carbon one, now our methyl group is up equatorial, and at carbon two, our methyl group is down equatorial. Now let's draw the chair conformations for the trans compound. The cis and trans compounds are isomers of each other. They're different molecules. So let's draw this chair conformation on the left, and we start with our parallel lines that are offset from each other, so there's one line and then there's the other one. Next, we put in our dotted lines, our dashed lines here, so that one hits the top part, this one hits the bottom part, like that."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "The cis and trans compounds are isomers of each other. They're different molecules. So let's draw this chair conformation on the left, and we start with our parallel lines that are offset from each other, so there's one line and then there's the other one. Next, we put in our dotted lines, our dashed lines here, so that one hits the top part, this one hits the bottom part, like that. And then we draw a line from here to here and then a parallel to that, and then we put in our last set of parallel lines. So that gives us our carbon skeleton. And we start at carbon one, we start axial up, and then at carbon two is axial down, and then we have down and then up."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Next, we put in our dotted lines, our dashed lines here, so that one hits the top part, this one hits the bottom part, like that. And then we draw a line from here to here and then a parallel to that, and then we put in our last set of parallel lines. So that gives us our carbon skeleton. And we start at carbon one, we start axial up, and then at carbon two is axial down, and then we have down and then up. So now, since this is carbon one, we see our methyl group is up axial, so we put our methyl group up axial here, and then we have our hydrogen down equatorial, so we put that in. At carbon two, the hydrogen is up equatorial and the methyl group is down axial, so let's put that in. Alright, you can see those two methyl groups are on opposite sides of the ring."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And we start at carbon one, we start axial up, and then at carbon two is axial down, and then we have down and then up. So now, since this is carbon one, we see our methyl group is up axial, so we put our methyl group up axial here, and then we have our hydrogen down equatorial, so we put that in. At carbon two, the hydrogen is up equatorial and the methyl group is down axial, so let's put that in. Alright, you can see those two methyl groups are on opposite sides of the ring. So one is going up, right, this one is going up, and then this one is going down, so they're trans to each other. This chair conformation is in equilibrium with our other one. Let's draw the one on the right."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Alright, you can see those two methyl groups are on opposite sides of the ring. So one is going up, right, this one is going up, and then this one is going down, so they're trans to each other. This chair conformation is in equilibrium with our other one. Let's draw the one on the right. So we start with our parallel lines offset, then we draw in our dashed or dotted line here. So next, we put in our next set of parallel lines, so here to here, here to here, and then connect these points, so that's just a really rough chair. And then now this is carbon one."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw the one on the right. So we start with our parallel lines offset, then we draw in our dashed or dotted line here. So next, we put in our next set of parallel lines, so here to here, here to here, and then connect these points, so that's just a really rough chair. And then now this is carbon one. So we start axial down at carbon one, carbon two would be axial up, and then let's put in our other ones here like that. Alright, for this one, this is carbon one, and we see our methyl group is now up equatorial, so now our methyl group of carbon one is here, and we have a hydrogen going down at carbon one. At carbon two, our hydrogen is up axial, and our methyl group is down equatorial, so here's our CH three."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then now this is carbon one. So we start axial down at carbon one, carbon two would be axial up, and then let's put in our other ones here like that. Alright, for this one, this is carbon one, and we see our methyl group is now up equatorial, so now our methyl group of carbon one is here, and we have a hydrogen going down at carbon one. At carbon two, our hydrogen is up axial, and our methyl group is down equatorial, so here's our CH three. It's hard sometimes to see these equatorial bonds, so a good hint is to look at the axial ones. For example, if you look at this bond here, you think, is this up or down relative to the plane of the ring? It's pretty hard to see, but this one is clearly up, so that means that this one must be down, so there's a little trick."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "At carbon two, our hydrogen is up axial, and our methyl group is down equatorial, so here's our CH three. It's hard sometimes to see these equatorial bonds, so a good hint is to look at the axial ones. For example, if you look at this bond here, you think, is this up or down relative to the plane of the ring? It's pretty hard to see, but this one is clearly up, so that means that this one must be down, so there's a little trick. Alright, let's look at those two groups and what happened in our ring flip. So let's start with carbon one, so here's carbon one with our methyl group that's up axial. For the ring flip, that methyl group stayed up relative to the plane of the ring, but now it is equatorial, so there's that same methyl group."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It's pretty hard to see, but this one is clearly up, so that means that this one must be down, so there's a little trick. Alright, let's look at those two groups and what happened in our ring flip. So let's start with carbon one, so here's carbon one with our methyl group that's up axial. For the ring flip, that methyl group stayed up relative to the plane of the ring, but now it is equatorial, so there's that same methyl group. For carbon two, this is carbon two. Our methyl group was down axial, and here it is down equatorial, so for both of our chair conformations, the methyl groups are on opposite sides of the ring. If we analyze these two chair conformations in terms of stability, well, for this one on the left, we have two axial substituents."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "For the ring flip, that methyl group stayed up relative to the plane of the ring, but now it is equatorial, so there's that same methyl group. For carbon two, this is carbon two. Our methyl group was down axial, and here it is down equatorial, so for both of our chair conformations, the methyl groups are on opposite sides of the ring. If we analyze these two chair conformations in terms of stability, well, for this one on the left, we have two axial substituents. That's not as stable as the one on the right. The one on the right has our two relatively bulky methyl groups equatorial, and we know that's the best place for them. That decreases the steric hindrance, so the one on the right is the more stable conformation."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "If we analyze these two chair conformations in terms of stability, well, for this one on the left, we have two axial substituents. That's not as stable as the one on the right. The one on the right has our two relatively bulky methyl groups equatorial, and we know that's the best place for them. That decreases the steric hindrance, so the one on the right is the more stable conformation. Finally, let's do a problem. So our goal is to draw the most stable conformation of trans-1-terbutyl-3-methyl-cyclohexane. We know the most stable conformation is going to be a chair conformation, so let's start by drawing a chair."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "That decreases the steric hindrance, so the one on the right is the more stable conformation. Finally, let's do a problem. So our goal is to draw the most stable conformation of trans-1-terbutyl-3-methyl-cyclohexane. We know the most stable conformation is going to be a chair conformation, so let's start by drawing a chair. So we have one parallel line, so here's one of them, and then here is the other one, so they're a little offset. Next, we put in our dotted lines just as guidelines to help us with the drawing here. And our next set of parallel lines goes in, so this point to this point."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We know the most stable conformation is going to be a chair conformation, so let's start by drawing a chair. So we have one parallel line, so here's one of them, and then here is the other one, so they're a little offset. Next, we put in our dotted lines just as guidelines to help us with the drawing here. And our next set of parallel lines goes in, so this point to this point. We try to draw a line parallel to that over here, and finally, our last set, so from here to here and from here to here. All right, so trans, which means that our two groups are on opposite sides of our ring. What are the two groups?"}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And our next set of parallel lines goes in, so this point to this point. We try to draw a line parallel to that over here, and finally, our last set, so from here to here and from here to here. All right, so trans, which means that our two groups are on opposite sides of our ring. What are the two groups? We have a tert-butyl group at carbon one, and we have a methyl group at carbon three. So we know this is carbon one, so let me go ahead and start up axial. So here's up axial, and then we alternate, so carbon two must be down axial, carbon three must be up axial."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "What are the two groups? We have a tert-butyl group at carbon one, and we have a methyl group at carbon three. So we know this is carbon one, so let me go ahead and start up axial. So here's up axial, and then we alternate, so carbon two must be down axial, carbon three must be up axial. Let's go back and put in equatorial, so carbon one must be down, carbon two must be up, and carbon three must be down. So let's put in our tert-butyl group at carbon one. So now we have two choices."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So here's up axial, and then we alternate, so carbon two must be down axial, carbon three must be up axial. Let's go back and put in equatorial, so carbon one must be down, carbon two must be up, and carbon three must be down. So let's put in our tert-butyl group at carbon one. So now we have two choices. Where do we put our tert-butyl group? Do we put it axial, or do we put it equatorial? Well, we know this is a very bulky group, and we put bulky groups equatorial, so we have only one choice at carbon one."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So now we have two choices. Where do we put our tert-butyl group? Do we put it axial, or do we put it equatorial? Well, we know this is a very bulky group, and we put bulky groups equatorial, so we have only one choice at carbon one. We must put the tert-butyl group right here. All right, so now we chose to put the tert-butyl group down at carbon one, so this is down, and since this says trans, that means we must put the methyl group up at carbon three. So we go to carbon three, and then here's carbon three."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Well, we know this is a very bulky group, and we put bulky groups equatorial, so we have only one choice at carbon one. We must put the tert-butyl group right here. All right, so now we chose to put the tert-butyl group down at carbon one, so this is down, and since this says trans, that means we must put the methyl group up at carbon three. So we go to carbon three, and then here's carbon three. We have only one choice. We must put the methyl group axial. That's the only one that's going up, so here is our CH three like that."}, {"video_title": "Disubstituted cyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we go to carbon three, and then here's carbon three. We have only one choice. We must put the methyl group axial. That's the only one that's going up, so here is our CH three like that. So let's go ahead and erase these other bonds here just to make this look a little prettier, so I'll just erase this stuff, and I'll just erase this stuff. You could put in all of your hydrogens. You could leave this stuff out."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, we saw how to draw dot structures for molecules with single covalent bonds. In this video, we'll talk about multiple covalent bonds. And so we start the same way we did in the last video. If I wanted to draw the dot structure for C2H4, I would find carbon over here. And once again, carbon is in group four, so it has four valence electrons. So I'm going to go ahead and put in one carbon with four valence electrons. And I now have another carbon in my dot structure."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If I wanted to draw the dot structure for C2H4, I would find carbon over here. And once again, carbon is in group four, so it has four valence electrons. So I'm going to go ahead and put in one carbon with four valence electrons. And I now have another carbon in my dot structure. It also has four valence electrons like that. And so immediately, I can see there's going to be a single covalent bond between my two carbons like that. Four hydrogens."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And I now have another carbon in my dot structure. It also has four valence electrons like that. And so immediately, I can see there's going to be a single covalent bond between my two carbons like that. Four hydrogens. And I know that since hydrogen is in group one on our periodic table, hydrogen has one valence electron. Now it makes sense to go ahead and put two hydrogens on each carbon. So if I put one hydrogen over here and then another hydrogen on this carbon, I have two more hydrogens."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Four hydrogens. And I know that since hydrogen is in group one on our periodic table, hydrogen has one valence electron. Now it makes sense to go ahead and put two hydrogens on each carbon. So if I put one hydrogen over here and then another hydrogen on this carbon, I have two more hydrogens. And so I can go ahead and put in those two hydrogens and the carbon on the right. When I connect my dots, I can see that I have a bond between carbon and hydrogen here. And this is not the correct dot structure."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if I put one hydrogen over here and then another hydrogen on this carbon, I have two more hydrogens. And so I can go ahead and put in those two hydrogens and the carbon on the right. When I connect my dots, I can see that I have a bond between carbon and hydrogen here. And this is not the correct dot structure. Because if you count up the number of electrons around carbon, so let's go ahead and do it. Let's do the carbon on the left. I'm going to get two, four, six, and then seven electrons."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And this is not the correct dot structure. Because if you count up the number of electrons around carbon, so let's go ahead and do it. Let's do the carbon on the left. I'm going to get two, four, six, and then seven electrons. So with only seven electrons around each carbon, carbon does not satisfy the octet rule. So the only way for carbon to get an octet of electrons around it would be if this magenta electron moved in here and this electron moves in here to form a double covalent bond between those two carbons. And so now, instead of only having one bond between those carbons, now there are going to be two bonds around it like that."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to get two, four, six, and then seven electrons. So with only seven electrons around each carbon, carbon does not satisfy the octet rule. So the only way for carbon to get an octet of electrons around it would be if this magenta electron moved in here and this electron moves in here to form a double covalent bond between those two carbons. And so now, instead of only having one bond between those carbons, now there are going to be two bonds around it like that. So I can go ahead and put in my hydrogen, so two on each carbon. And that is the correct dot structure for ethene or ethylene. We can double check by checking the octet rule here."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so now, instead of only having one bond between those carbons, now there are going to be two bonds around it like that. So I can go ahead and put in my hydrogen, so two on each carbon. And that is the correct dot structure for ethene or ethylene. We can double check by checking the octet rule here. So if I look at each carbon, there would be two electrons, four, six, and then eight. So each carbon is following the octet rule for my dot structure. Let's do one for CH2O."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We can double check by checking the octet rule here. So if I look at each carbon, there would be two electrons, four, six, and then eight. So each carbon is following the octet rule for my dot structure. Let's do one for CH2O. So if I want to draw the dot structure for the molecular formula, CH2O, once again, I start with carbon in the center, so four valence electrons like that, and two hydrogens. So I'll just put one hydrogen over here on the left and another hydrogen over here on the right. Oxygen, let's go back to our organic periodic table to refresh our memory about how many valence electrons oxygen will have."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one for CH2O. So if I want to draw the dot structure for the molecular formula, CH2O, once again, I start with carbon in the center, so four valence electrons like that, and two hydrogens. So I'll just put one hydrogen over here on the left and another hydrogen over here on the right. Oxygen, let's go back to our organic periodic table to refresh our memory about how many valence electrons oxygen will have. And we can see it's in group six over here, so six valence electrons. So I can go ahead and put in my six valence electrons for oxygen. And so we'll go one, and then two, and then three, and then four, and then five, and then six, like that."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen, let's go back to our organic periodic table to refresh our memory about how many valence electrons oxygen will have. And we can see it's in group six over here, so six valence electrons. So I can go ahead and put in my six valence electrons for oxygen. And so we'll go one, and then two, and then three, and then four, and then five, and then six, like that. And when I start connecting my bonds here, I know there's a bond between carbon and hydrogen on the left. I know there's a bond between carbon and hydrogen on the right. I know there's a bond between this carbon and oxygen."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so we'll go one, and then two, and then three, and then four, and then five, and then six, like that. And when I start connecting my bonds here, I know there's a bond between carbon and hydrogen on the left. I know there's a bond between carbon and hydrogen on the right. I know there's a bond between this carbon and oxygen. And unfortunately, I still don't have an octet of electrons around my carbon or around my oxygen. So if I go ahead and highlight the electrons around carbon, once again, I have two, four, six, and then seven. So I still don't have an octet of electrons around carbon."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I know there's a bond between this carbon and oxygen. And unfortunately, I still don't have an octet of electrons around my carbon or around my oxygen. So if I go ahead and highlight the electrons around carbon, once again, I have two, four, six, and then seven. So I still don't have an octet of electrons around carbon. And so I need to share an electron with oxygen here. So carbon is going to contribute to an electron, and oxygen is going to contribute to an electron like that. And so instead of a single bond between carbon and oxygen, there's actually going to be a double bond like that."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I still don't have an octet of electrons around carbon. And so I need to share an electron with oxygen here. So carbon is going to contribute to an electron, and oxygen is going to contribute to an electron like that. And so instead of a single bond between carbon and oxygen, there's actually going to be a double bond like that. There's still two lone pairs of electrons around that oxygen, and then I have my two hydrogens coming off of my carbon like that. So that is the correct dot structure for formaldehyde. We can go ahead and double check an octet."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And so instead of a single bond between carbon and oxygen, there's actually going to be a double bond like that. There's still two lone pairs of electrons around that oxygen, and then I have my two hydrogens coming off of my carbon like that. So that is the correct dot structure for formaldehyde. We can go ahead and double check an octet. So if I look at this carbon here, this would be two, four, six, and eight. And obviously, there's also an octet around that oxygen as well. So this is the dot structure for formaldehyde."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We can go ahead and double check an octet. So if I look at this carbon here, this would be two, four, six, and eight. And obviously, there's also an octet around that oxygen as well. So this is the dot structure for formaldehyde. Let's do one more example of a molecule with a multiple covalent bond, so C2H2. So once again, carbon with four valence electrons. And I have two carbons this time, so I go ahead and put in my second carbon with four valence electrons like that."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this is the dot structure for formaldehyde. Let's do one more example of a molecule with a multiple covalent bond, so C2H2. So once again, carbon with four valence electrons. And I have two carbons this time, so I go ahead and put in my second carbon with four valence electrons like that. I can see immediately a single covalent bond between my two carbons. Now I have two hydrogens. So once again, I'm going to put one hydrogen on the carbon on the left and connect that for a single covalent bond."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And I have two carbons this time, so I go ahead and put in my second carbon with four valence electrons like that. I can see immediately a single covalent bond between my two carbons. Now I have two hydrogens. So once again, I'm going to put one hydrogen on the carbon on the left and connect that for a single covalent bond. And it would make sense to put the other hydrogen over here on the right and connect that for another bond. Now once again, if I look at the valence electrons for carbon, I have a total of two, four, five, and six. And carbon wants to get to eight."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So once again, I'm going to put one hydrogen on the carbon on the left and connect that for a single covalent bond. And it would make sense to put the other hydrogen over here on the right and connect that for another bond. Now once again, if I look at the valence electrons for carbon, I have a total of two, four, five, and six. And carbon wants to get to eight. It wants an octet. So it needs to share two more electrons. So each carbon needs to share two more electrons so each carbon can get to an octet."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And carbon wants to get to eight. It wants an octet. So it needs to share two more electrons. So each carbon needs to share two more electrons so each carbon can get to an octet. So we're going to go ahead and move this electron in here and this electron in here. And then we'd also have to move these electrons in here as well. So we're going to end up with a triple bond between my two carbon atoms there."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So each carbon needs to share two more electrons so each carbon can get to an octet. So we're going to go ahead and move this electron in here and this electron in here. And then we'd also have to move these electrons in here as well. So we're going to end up with a triple bond between my two carbon atoms there. So if I go ahead and show now three bonds between my two carbons and then a hydrogen on either side, we can double check the octets on the carbon here. So I'll get out the magenta again. So two, four, six, and eight."}, {"video_title": "Dot structures II Multiple bonds Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to end up with a triple bond between my two carbon atoms there. So if I go ahead and show now three bonds between my two carbons and then a hydrogen on either side, we can double check the octets on the carbon here. So I'll get out the magenta again. So two, four, six, and eight. So there is now an octet. And so this is the correct dot structure for ethine or acetylene. And so we've seen how to draw molecules with single and multiple covalent bonds."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Obviously, the sun's gravity isn't so strong that it keeps light from escaping. So why would something, or even a star that's two or three solar masses, its gravity isn't so strong that it keeps light from escaping. Why would a black hole that has the same mass, why would that keep light from escaping? And to understand that, let's just think a little bit about, and I'll just do Newtonian classical physics right here. I won't get into the whole general relativity of things. And this really will just give us the intuition of why a smaller, denser thing of the same mass can exert a stronger gravitational pull. So let's imagine, so let's take two examples."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And to understand that, let's just think a little bit about, and I'll just do Newtonian classical physics right here. I won't get into the whole general relativity of things. And this really will just give us the intuition of why a smaller, denser thing of the same mass can exert a stronger gravitational pull. So let's imagine, so let's take two examples. Let's say I have some star here. Let's just call that mass m1. And let's say that its radius, let's just call this r. And let's say that I have some other mass right at the surface of this star."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So let's imagine, so let's take two examples. Let's say I have some star here. Let's just call that mass m1. And let's say that its radius, let's just call this r. And let's say that I have some other mass right at the surface of this star. It's somehow able to survive those surface temperatures. And this mass over here has a mass of m1. This mass over here has a mass of m2."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And let's say that its radius, let's just call this r. And let's say that I have some other mass right at the surface of this star. It's somehow able to survive those surface temperatures. And this mass over here has a mass of m1. This mass over here has a mass of m2. The universal law of gravitation tells us that the force between these two masses is going to be equal to the gravitational constant times the product of the masses. So m1 times m2, all of that over the square of the distance. Now let me be very clear."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This mass over here has a mass of m2. The universal law of gravitation tells us that the force between these two masses is going to be equal to the gravitational constant times the product of the masses. So m1 times m2, all of that over the square of the distance. Now let me be very clear. You might say, wait, this magenta mass right here is touching this larger mass. Isn't the distance 0? And you have to be very careful."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Now let me be very clear. You might say, wait, this magenta mass right here is touching this larger mass. Isn't the distance 0? And you have to be very careful. This is the distance between their center of masses. So the center of mass of this large mass over here is r away from this mass that's on the surface. Now with that said, let's take another example."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And you have to be very careful. This is the distance between their center of masses. So the center of mass of this large mass over here is r away from this mass that's on the surface. Now with that said, let's take another example. Let's say that this large, massive star, or whatever it might be, eventually condenses into something 1,000 times smaller. So let me draw it like this. And obviously I'm not drawing it to scale."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Now with that said, let's take another example. Let's say that this large, massive star, or whatever it might be, eventually condenses into something 1,000 times smaller. So let me draw it like this. And obviously I'm not drawing it to scale. So let's say we have another case like this. And I'm not drawing it to scale. So this object, maybe it's the same object, or maybe it's a different object, that has the exact same mass as this larger object, but now it has a much smaller radius."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And obviously I'm not drawing it to scale. So let's say we have another case like this. And I'm not drawing it to scale. So this object, maybe it's the same object, or maybe it's a different object, that has the exact same mass as this larger object, but now it has a much smaller radius. It now has a much smaller radius. So that radius now, the radius is 1 over, let's just say it's 1,000th of this radius over here. So it's 1, maybe I'll just call it r over 1,000."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this object, maybe it's the same object, or maybe it's a different object, that has the exact same mass as this larger object, but now it has a much smaller radius. It now has a much smaller radius. So that radius now, the radius is 1 over, let's just say it's 1,000th of this radius over here. So it's 1, maybe I'll just call it r over 1,000. So if this had a million kilometer radius, so that would make it roughly about twice the radius of the sun. If this was a million kilometer radius right over here, this would be 1,000 kilometer radius. So maybe we're talking about something that's approaching a neutron star."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So it's 1, maybe I'll just call it r over 1,000. So if this had a million kilometer radius, so that would make it roughly about twice the radius of the sun. If this was a million kilometer radius right over here, this would be 1,000 kilometer radius. So maybe we're talking about something that's approaching a neutron star. But we don't have to think about what it actually is. Let's just think about the thought experiment here. So let's say I have this thing over here."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So maybe we're talking about something that's approaching a neutron star. But we don't have to think about what it actually is. Let's just think about the thought experiment here. So let's say I have this thing over here. And let's say I have something on the surface of this. So let's say I have that same mass, it's on the surface of this thing. So this is m2 right over here."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So let's say I have this thing over here. And let's say I have something on the surface of this. So let's say I have that same mass, it's on the surface of this thing. So this is m2 right over here. So what is going to be the force between these two masses? How strong are they going to want to, what's the force pulling them together? So let's just do the universal law of gravitation again."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this is m2 right over here. So what is going to be the force between these two masses? How strong are they going to want to, what's the force pulling them together? So let's just do the universal law of gravitation again. The force, let's just call this force 1, and let's call this force 2. Once again, it's going to be the gravitational constant times the product of their masses. So the big m1 times the smaller mass, m2, all of that over this distance squared, this radius squared."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So let's just do the universal law of gravitation again. The force, let's just call this force 1, and let's call this force 2. Once again, it's going to be the gravitational constant times the product of their masses. So the big m1 times the smaller mass, m2, all of that over this distance squared, this radius squared. Remember, it's the distance to the center of masses. This center of mass here, we're considering m2 to kind of be just a point mass right over there. So what's the radius squared?"}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So the big m1 times the smaller mass, m2, all of that over this distance squared, this radius squared. Remember, it's the distance to the center of masses. This center of mass here, we're considering m2 to kind of be just a point mass right over there. So what's the radius squared? It's going to be r over 1,000 squared. Or if we simplify this, what will this be? This is the same thing, and I'll just write it in one color just because it takes less time."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So what's the radius squared? It's going to be r over 1,000 squared. Or if we simplify this, what will this be? This is the same thing, and I'll just write it in one color just because it takes less time. Gravitational constant m1, m2 over r squared over 1,000 squared, or over 1 million. Over 1 million. That's just 1,000 squared."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This is the same thing, and I'll just write it in one color just because it takes less time. Gravitational constant m1, m2 over r squared over 1,000 squared, or over 1 million. Over 1 million. That's just 1,000 squared. Or we can multiply the numerator and the denominator by a million, and this is going to be equal to 1 million, I'm going to write it out, 1 million, let me scroll to the right a little bit, times the gravitational constant times m1, m2, all of that over r squared. Now what is this thing right over here? That's the same thing as this F1."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "That's just 1,000 squared. Or we can multiply the numerator and the denominator by a million, and this is going to be equal to 1 million, I'm going to write it out, 1 million, let me scroll to the right a little bit, times the gravitational constant times m1, m2, all of that over r squared. Now what is this thing right over here? That's the same thing as this F1. So this is going to be 1 million times F1. So even though the masses involved are the same, this yellow object right here is the same mass as this larger object over here. It's able to exert a million times the gravitational force on this point mass, and actually vice versa."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "That's the same thing as this F1. So this is going to be 1 million times F1. So even though the masses involved are the same, this yellow object right here is the same mass as this larger object over here. It's able to exert a million times the gravitational force on this point mass, and actually vice versa. They're both being attracted. They're both exerting this on each other. And the reality is, is because this thing is smaller, because this m1 on the right here, this one I'm coloring in, because this one is smaller and denser, this particle is able to get closer to its center of mass."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's able to exert a million times the gravitational force on this point mass, and actually vice versa. They're both being attracted. They're both exerting this on each other. And the reality is, is because this thing is smaller, because this m1 on the right here, this one I'm coloring in, because this one is smaller and denser, this particle is able to get closer to its center of mass. Now you might be saying, OK, well I can buy that. This just comes straight from the universal law of gravitation. But wouldn't something closer to this center of mass experience that same thing?"}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And the reality is, is because this thing is smaller, because this m1 on the right here, this one I'm coloring in, because this one is smaller and denser, this particle is able to get closer to its center of mass. Now you might be saying, OK, well I can buy that. This just comes straight from the universal law of gravitation. But wouldn't something closer to this center of mass experience that same thing? If this was a star, wouldn't photons that are over here, wouldn't this experience the same force? If this distance right here is r over 1,000, wouldn't some photon here, or atom here, or molecule, or whatever it's over here, wouldn't that experience the same force, this million times the force as this thing? And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening?"}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But wouldn't something closer to this center of mass experience that same thing? If this was a star, wouldn't photons that are over here, wouldn't this experience the same force? If this distance right here is r over 1,000, wouldn't some photon here, or atom here, or molecule, or whatever it's over here, wouldn't that experience the same force, this million times the force as this thing? And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening? It no longer has the entire mass is no longer pulling on it in that direction. It's no longer pulling it in that inward direction. You now have all of this mass over here."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And you've got to remember, all of a sudden when this thing is inside of this larger mass, what's happening? It no longer has the entire mass is no longer pulling on it in that direction. It's no longer pulling it in that inward direction. You now have all of this mass over here. Let me think of the best way of doing it. So you could think of it all of this mass over here is pulling it in an outward direction. It's not telling you what that mass out there is doing, since that mass itself is being pulled inward."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "You now have all of this mass over here. Let me think of the best way of doing it. So you could think of it all of this mass over here is pulling it in an outward direction. It's not telling you what that mass out there is doing, since that mass itself is being pulled inward. It is pushing down on this. It is exerting pressure on that point. But the actual gravitational force that that point is experiencing is actually going to be less."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's not telling you what that mass out there is doing, since that mass itself is being pulled inward. It is pushing down on this. It is exerting pressure on that point. But the actual gravitational force that that point is experiencing is actually going to be less. It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction. And so you can imagine if you were in the center of a really massive object, there would be no net gravitational force being pulled on you, because you're at its center of mass. The rest of the mass is outward."}, {"video_title": "Why gravity gets so strong near dense objects Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But the actual gravitational force that that point is experiencing is actually going to be less. It's actually going to be mitigated by the fact that there's so much mass over here pulling in the other direction. And so you can imagine if you were in the center of a really massive object, there would be no net gravitational force being pulled on you, because you're at its center of mass. The rest of the mass is outward. So at every point, it will be pulling you outward. And so that's why if you were to enter the core of a star, if you were to get a lot closer to its center of mass, it's not going to be pulling on you with this type of force. And the only way you can get these type of forces is if the entire mass is contained in a very dense region, in a very small region."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "The SN1 and SN2 reactions involve leaving groups. So let's look at this pKa table to study leaving groups in more detail. On the left we have the acid, for example, hydroiodic acid, HI, with an approximate pKa of negative 11. Remember, the lower the pKa value, the stronger the acid. So on this table, with a pKa value of negative 11, hydroiodic acid is the strongest acid. And the stronger the acid, the more stable the conjugate base. So the conjugate base to HI is I minus the iodide anion."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "Remember, the lower the pKa value, the stronger the acid. So on this table, with a pKa value of negative 11, hydroiodic acid is the strongest acid. And the stronger the acid, the more stable the conjugate base. So the conjugate base to HI is I minus the iodide anion. And since this is the conjugate base to the strongest acid, this is the most stable base. So let me write that down here. So this is the most stable base on the table."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So the conjugate base to HI is I minus the iodide anion. And since this is the conjugate base to the strongest acid, this is the most stable base. So let me write that down here. So this is the most stable base on the table. Which means that the iodide anion is an excellent leaving group because it is very stable. Next we have hydrobromic acid, approximate pKa of negative nine. So the conjugate base would be the bromide anion."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So this is the most stable base on the table. Which means that the iodide anion is an excellent leaving group because it is very stable. Next we have hydrobromic acid, approximate pKa of negative nine. So the conjugate base would be the bromide anion. So also a stable conjugate base, so therefore a good leaving group. For HCl, it's the chloride anion, also a good leaving group. So you see these halide anions as leaving groups all the time in organic mechanisms."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So the conjugate base would be the bromide anion. So also a stable conjugate base, so therefore a good leaving group. For HCl, it's the chloride anion, also a good leaving group. So you see these halide anions as leaving groups all the time in organic mechanisms. Let me write this down here. So these are all examples of good leaving groups. Next, let's look at this acid on the left here."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So you see these halide anions as leaving groups all the time in organic mechanisms. Let me write this down here. So these are all examples of good leaving groups. Next, let's look at this acid on the left here. This is p-toluene sulfonic acid with a pKa value of negative three, so it's still pretty acidic. The conjugate base to this is on the right here. And we call this anion a tosylate group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "Next, let's look at this acid on the left here. This is p-toluene sulfonic acid with a pKa value of negative three, so it's still pretty acidic. The conjugate base to this is on the right here. And we call this anion a tosylate group. So let me write this down. This is called a tosylate group. And since it's kind of a bulky group, instead of drawing this out all the time, you often see OTS written."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "And we call this anion a tosylate group. So let me write this down. This is called a tosylate group. And since it's kind of a bulky group, instead of drawing this out all the time, you often see OTS written. So O-T-S, like that. And you could put a negative charge on the oxygen here if you wanted to. So you'll see the tosylate group function as a leaving group in many reactions."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "And since it's kind of a bulky group, instead of drawing this out all the time, you often see OTS written. So O-T-S, like that. And you could put a negative charge on the oxygen here if you wanted to. So you'll see the tosylate group function as a leaving group in many reactions. Let's look at an example of another acid. So if I move down here to H3O+, the hydronium ion with a pKa value of negative two. The conjugate base to H3O+, is H2O."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So you'll see the tosylate group function as a leaving group in many reactions. Let's look at an example of another acid. So if I move down here to H3O+, the hydronium ion with a pKa value of negative two. The conjugate base to H3O+, is H2O. And water is also a good leaving group. So let's go back up here to the top again, and we can see that all the acids that we talked about have negative pKa values. So negative 11, negative nine, negative seven, negative three, and negative two."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "The conjugate base to H3O+, is H2O. And water is also a good leaving group. So let's go back up here to the top again, and we can see that all the acids that we talked about have negative pKa values. So negative 11, negative nine, negative seven, negative three, and negative two. And notice all of the conjugate bases are good leaving groups. So you can say that if an acid has a negative value for the pKa, the conjugate base will be a good leaving group. Let's look at another example of an acid."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So negative 11, negative nine, negative seven, negative three, and negative two. And notice all of the conjugate bases are good leaving groups. So you can say that if an acid has a negative value for the pKa, the conjugate base will be a good leaving group. Let's look at another example of an acid. So water. Water's pKa value is positive 15.7. So it's not a very strong acid."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "Let's look at another example of an acid. So water. Water's pKa value is positive 15.7. So it's not a very strong acid. The conjugate base to water is the hydroxide anion, OH-. And this is a bad leaving group. So hydroxide ion is a bad leaving group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So it's not a very strong acid. The conjugate base to water is the hydroxide anion, OH-. And this is a bad leaving group. So hydroxide ion is a bad leaving group. And that's because water is not a strong acid. Look at this value for the pKa. Positive 15.7."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So hydroxide ion is a bad leaving group. And that's because water is not a strong acid. Look at this value for the pKa. Positive 15.7. So if we look at ethanol, similar story here. So ethanol's pKa value of positive 16. So the ethoxide anion is not a good leaving group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "Positive 15.7. So if we look at ethanol, similar story here. So ethanol's pKa value of positive 16. So the ethoxide anion is not a good leaving group. So these pKa values are in the positive. And these conjugate bases must not be very stable, which means they are bad leaving groups. Let me write that down here."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So the ethoxide anion is not a good leaving group. So these pKa values are in the positive. And these conjugate bases must not be very stable, which means they are bad leaving groups. Let me write that down here. So these are examples of bad leaving groups. Both SN1 and SN2 reactions need good leaving groups. However, the SN1 reaction is even more sensitive."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "Let me write that down here. So these are examples of bad leaving groups. Both SN1 and SN2 reactions need good leaving groups. However, the SN1 reaction is even more sensitive. So let's look at tert-butyl chloride. And let's say it's reacting via an SN1 mechanism. The first step should be loss of a leaving group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "However, the SN1 reaction is even more sensitive. So let's look at tert-butyl chloride. And let's say it's reacting via an SN1 mechanism. The first step should be loss of a leaving group. So if these electrons come off onto the chlorine, we would form the chloride anion, which has a negative one formal charge. We just saw on our pKa table that the chloride anion is a stable conjugate base. So therefore, this is a good leaving group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "The first step should be loss of a leaving group. So if these electrons come off onto the chlorine, we would form the chloride anion, which has a negative one formal charge. We just saw on our pKa table that the chloride anion is a stable conjugate base. So therefore, this is a good leaving group. We're taking a bond away from the carbon in red. So the carbon in red gets a plus one formal charge. And we form a tertiary carbocation as well."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So therefore, this is a good leaving group. We're taking a bond away from the carbon in red. So the carbon in red gets a plus one formal charge. And we form a tertiary carbocation as well. Since this is the rate-determining step of our SN1 mechanism, the formation of our stable anion, the formation of a good leaving group, helps the SN1 mechanism occur. Next, let's look at this alcohol here. If we approach it the same way as we did in the previous problem, and we said, okay, first step is loss of a leaving group, and these electrons come off onto the oxygen, think about what leaving group that is."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "And we form a tertiary carbocation as well. Since this is the rate-determining step of our SN1 mechanism, the formation of our stable anion, the formation of a good leaving group, helps the SN1 mechanism occur. Next, let's look at this alcohol here. If we approach it the same way as we did in the previous problem, and we said, okay, first step is loss of a leaving group, and these electrons come off onto the oxygen, think about what leaving group that is. That would be the hydroxide ion, which we know from our pKa table is not a good leaving group. So the hydroxide ion is not as stable of an anion as the chloride anion. So the chloride anion is a good leaving group, the hydroxide anion is a bad leaving group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "If we approach it the same way as we did in the previous problem, and we said, okay, first step is loss of a leaving group, and these electrons come off onto the oxygen, think about what leaving group that is. That would be the hydroxide ion, which we know from our pKa table is not a good leaving group. So the hydroxide ion is not as stable of an anion as the chloride anion. So the chloride anion is a good leaving group, the hydroxide anion is a bad leaving group. So that's not the first step of this mechanism. We need to make a better leaving group. And you can do that by having a proton source."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So the chloride anion is a good leaving group, the hydroxide anion is a bad leaving group. So that's not the first step of this mechanism. We need to make a better leaving group. And you can do that by having a proton source. So let's say we have a source of protons, an acid in solution, so let's say there's an H plus here. The first step would be to protonate our alcohol, so our alcohol's gonna act as a base and pick up a proton. So let's draw the result of that."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "And you can do that by having a proton source. So let's say we have a source of protons, an acid in solution, so let's say there's an H plus here. The first step would be to protonate our alcohol, so our alcohol's gonna act as a base and pick up a proton. So let's draw the result of that. So we have our ring. Let's put in that methyl group. And now our oxygen is bonded to two hydrogens."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "So let's draw the result of that. So we have our ring. Let's put in that methyl group. And now our oxygen is bonded to two hydrogens. There's still a lone pair of electrons on this oxygen, which gives the oxygen a plus one formal charge. So the electrons here in magenta, let's say, pick up this proton to form this bond. Now we're ready for loss of a leaving group, because if these electrons come off onto the oxygen now, we form water as a leaving group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "And now our oxygen is bonded to two hydrogens. There's still a lone pair of electrons on this oxygen, which gives the oxygen a plus one formal charge. So the electrons here in magenta, let's say, pick up this proton to form this bond. Now we're ready for loss of a leaving group, because if these electrons come off onto the oxygen now, we form water as a leaving group. So let me draw that in here. So here is the water molecule. And let me highlight those electrons in blue."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "Now we're ready for loss of a leaving group, because if these electrons come off onto the oxygen now, we form water as a leaving group. So let me draw that in here. So here is the water molecule. And let me highlight those electrons in blue. So if these electrons come off onto the oxygen, then we form water. And we know from our pKa table that water is a good leaving group. We're taking a bond away from this carbon in red, so we're also going to form a tertiary carbocation."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "And let me highlight those electrons in blue. So if these electrons come off onto the oxygen, then we form water. And we know from our pKa table that water is a good leaving group. We're taking a bond away from this carbon in red, so we're also going to form a tertiary carbocation. So let me draw that in here. So here's our ring. Here's our methyl group."}, {"video_title": "Sn1 and Sn2 leaving group.mp3", "Sentence": "We're taking a bond away from this carbon in red, so we're also going to form a tertiary carbocation. So let me draw that in here. So here's our ring. Here's our methyl group. A plus one formal charge on the carbon in red. So by thinking about your pKa values, you can determine the stability of the conjugate base, and therefore, if a leaving group is a good leaving group or a bad leaving group. And that helps you out when you're drawing mechanisms."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I said, maybe it's 5,730 years since this bone was part of a living animal, or it's roughly that old. Now, when I did that, I made a pretty big assumption. And some of you all have touched on this in the comments on YouTube on the last video. It's how do I know that this estimate I made is based on the assumption that the amount of carbon-14 in the atmosphere would have been roughly constant from when this bone was living to now. And so the question is, is the amount of carbon-14 in the atmosphere and in the water and in living plants and animals, is it constant? And if it isn't constant, how do you calibrate your measurement so you can actually figure out how much carbon-14 there is relative to living plants and animals at that time? And the way that you can make that calibration, because it turns out it isn't perfectly constant, the way that you can make that calibration, there's two ways."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's how do I know that this estimate I made is based on the assumption that the amount of carbon-14 in the atmosphere would have been roughly constant from when this bone was living to now. And so the question is, is the amount of carbon-14 in the atmosphere and in the water and in living plants and animals, is it constant? And if it isn't constant, how do you calibrate your measurement so you can actually figure out how much carbon-14 there is relative to living plants and animals at that time? And the way that you can make that calibration, because it turns out it isn't perfectly constant, the way that you can make that calibration, there's two ways. And I have pictures here of both of them. One is to look at tree rings. And I'm told that this will work up to about 10,000 years old."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the way that you can make that calibration, because it turns out it isn't perfectly constant, the way that you can make that calibration, there's two ways. And I have pictures here of both of them. One is to look at tree rings. And I'm told that this will work up to about 10,000 years old. I don't know of any 10,000-year-old trees. I don't think anyone does. But maybe there are some remains of old trees."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm told that this will work up to about 10,000 years old. I don't know of any 10,000-year-old trees. I don't think anyone does. But maybe there are some remains of old trees. And you can look at their tree rings. And I think most of us are familiar with this idea that every year that a tree grows, it forms another layer of bark. And so you can look back to that layer of bark, adjust for the half-life of carbon-14, and then figure out how much carbon-14 was there in the atmosphere at that period in time."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But maybe there are some remains of old trees. And you can look at their tree rings. And I think most of us are familiar with this idea that every year that a tree grows, it forms another layer of bark. And so you can look back to that layer of bark, adjust for the half-life of carbon-14, and then figure out how much carbon-14 was there in the atmosphere at that period in time. And so it's kind of a record of the atmosphere up to 10,000 years. If you want to go even further back, you can look at cave deposits. And the fancy word for these cave deposits are speleothems."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you can look back to that layer of bark, adjust for the half-life of carbon-14, and then figure out how much carbon-14 was there in the atmosphere at that period in time. And so it's kind of a record of the atmosphere up to 10,000 years. If you want to go even further back, you can look at cave deposits. And the fancy word for these cave deposits are speleothems. You might be familiar with stalagmites. Those are those speleothems that are kind of coming out of the bottom of the cave. Or stalactites, those are the speleothems that are coming from the top of the cave."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the fancy word for these cave deposits are speleothems. You might be familiar with stalagmites. Those are those speleothems that are kind of coming out of the bottom of the cave. Or stalactites, those are the speleothems that are coming from the top of the cave. But the reason why these are useful is these are formed by calcium carbonate. So they have carbon in them. And slowly, over really tens of thousands of years, the water in the cave deposits that calcium carbonate."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or stalactites, those are the speleothems that are coming from the top of the cave. But the reason why these are useful is these are formed by calcium carbonate. So they have carbon in them. And slowly, over really tens of thousands of years, the water in the cave deposits that calcium carbonate. So it's a record of the fraction of carbon-14 in some of those years. And you can go down to resolutions of as small as 10 years. And so this will give us pretty good estimates over tens of thousands of years, up to 50,000 years."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And slowly, over really tens of thousands of years, the water in the cave deposits that calcium carbonate. So it's a record of the fraction of carbon-14 in some of those years. And you can go down to resolutions of as small as 10 years. And so this will give us pretty good estimates over tens of thousands of years, up to 50,000 years. And frankly, carbon-14 isn't even useful beyond really 50,000 or 60,000 years. So this gives us a pretty good record of carbon-14 in the atmosphere, assuming that it's fairly uniform throughout the atmosphere. And all evidence suggests that."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this will give us pretty good estimates over tens of thousands of years, up to 50,000 years. And frankly, carbon-14 isn't even useful beyond really 50,000 or 60,000 years. So this gives us a pretty good record of carbon-14 in the atmosphere, assuming that it's fairly uniform throughout the atmosphere. And all evidence suggests that. And that uniformity through the atmosphere also goes into the water supply and into living plants and animals. Now the other thing, and I looked into this a little bit, it actually turns out because we are spewing so much fossil fuel right now, we are changing the amount or the proportion of carbon-14 much, much faster than has happened in other time periods. So just to answer the question, it's actually probably in the last 50 years where the fossil fuel use has really exploded that we've really been changing the proportion of carbon-14 relative to the other isotopes of carbon."}, {"video_title": "Carbon 14 dating 2 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And all evidence suggests that. And that uniformity through the atmosphere also goes into the water supply and into living plants and animals. Now the other thing, and I looked into this a little bit, it actually turns out because we are spewing so much fossil fuel right now, we are changing the amount or the proportion of carbon-14 much, much faster than has happened in other time periods. So just to answer the question, it's actually probably in the last 50 years where the fossil fuel use has really exploded that we've really been changing the proportion of carbon-14 relative to the other isotopes of carbon. But anyway, hopefully that rests some of your worries about the assumption that I made in the last video about carbon-14 being relatively constant. There are ways to look back at specific years and figure out the relative amounts of carbon-14. So it is a pretty good way of estimating how old living things are, especially things that are less than 50,000 years old."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "This carbon, let's call beta one. This carbon, let's say this is the beta two carbon. And finally, this would be the beta three carbon. If you react a tertiary alcohol with sulfuric acid and you heat up your reaction mixture, this is gonna be an E1 mechanism. And we'll talk about the regiochemistry for this reaction and why this is a regioselective reaction in a few minutes. First, let's go through the mechanism. We know that with an alcohol, the alcohol will be protonated by the sulfuric acid."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "If you react a tertiary alcohol with sulfuric acid and you heat up your reaction mixture, this is gonna be an E1 mechanism. And we'll talk about the regiochemistry for this reaction and why this is a regioselective reaction in a few minutes. First, let's go through the mechanism. We know that with an alcohol, the alcohol will be protonated by the sulfuric acid. So instead of drawing out the dot structure for sulfuric acid, I'll just write H plus here. So sulfuric acid is a source of protons. And one of the lone pairs of electrons on oxygen picks up that proton."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "We know that with an alcohol, the alcohol will be protonated by the sulfuric acid. So instead of drawing out the dot structure for sulfuric acid, I'll just write H plus here. So sulfuric acid is a source of protons. And one of the lone pairs of electrons on oxygen picks up that proton. So our first step is a proton transfer. And let me draw that in here. So now this oxygen would be bonded to two hydrogens with one lone pair of electrons and a plus one formal charge on the oxygen."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "And one of the lone pairs of electrons on oxygen picks up that proton. So our first step is a proton transfer. And let me draw that in here. So now this oxygen would be bonded to two hydrogens with one lone pair of electrons and a plus one formal charge on the oxygen. So the lone pair, let's say this lone pair here in magenta, picks up a proton from sulfuric acid to form this bond. And that gives us water as a leaving group. And we know water is a good leaving group."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So now this oxygen would be bonded to two hydrogens with one lone pair of electrons and a plus one formal charge on the oxygen. So the lone pair, let's say this lone pair here in magenta, picks up a proton from sulfuric acid to form this bond. And that gives us water as a leaving group. And we know water is a good leaving group. The electrons in this bond can come off onto the oxygen to form H2O. And when that happens, we take a bond away from this carbon in red. So we're gonna form a carbocation."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "And we know water is a good leaving group. The electrons in this bond can come off onto the oxygen to form H2O. And when that happens, we take a bond away from this carbon in red. So we're gonna form a carbocation. So we take away a bond from the carbon in red. Let me go ahead and draw in our carbocation here. So the carbon in red would be this carbon."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So we're gonna form a carbocation. So we take away a bond from the carbon in red. Let me go ahead and draw in our carbocation here. So the carbon in red would be this carbon. And that carbon would have a plus one formal charge. This is a tertiary carbocation because the carbon in red is directly bonded to three other carbons. So this one, this one, and this one."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So the carbon in red would be this carbon. And that carbon would have a plus one formal charge. This is a tertiary carbocation because the carbon in red is directly bonded to three other carbons. So this one, this one, and this one. So this is a stable carbocation. Next, let's think about the next step of an E1 mechanism. A base is gonna come along and take a proton from one of the beta carbons."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So this one, this one, and this one. So this is a stable carbocation. Next, let's think about the next step of an E1 mechanism. A base is gonna come along and take a proton from one of the beta carbons. And let's start with beta two. So let's think about a proton on that carbon. So let me draw one in here."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "A base is gonna come along and take a proton from one of the beta carbons. And let's start with beta two. So let's think about a proton on that carbon. So let me draw one in here. And our weak base comes along and takes this proton, which would leave these electrons to move into here to form a double bond. So let's draw that product. So we would have our double bond forming right here."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So let me draw one in here. And our weak base comes along and takes this proton, which would leave these electrons to move into here to form a double bond. So let's draw that product. So we would have our double bond forming right here. Let me draw in the rest of the molecule. So our electrons in, let me make these blue here. So the electrons in light blue are going to move in to form our double bond."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So we would have our double bond forming right here. Let me draw in the rest of the molecule. So our electrons in, let me make these blue here. So the electrons in light blue are going to move in to form our double bond. And so we would get this alkene. So that's beta two. Let's think about what would happen if we took a proton away from beta one."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So the electrons in light blue are going to move in to form our double bond. And so we would get this alkene. So that's beta two. Let's think about what would happen if we took a proton away from beta one. So let's draw that one in next. So I'm gonna draw the carbocation again. So let me draw that in here."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "Let's think about what would happen if we took a proton away from beta one. So let's draw that one in next. So I'm gonna draw the carbocation again. So let me draw that in here. And beta one would be up here. So let me put in a proton on beta one. We think about a weak base coming along and taking this proton."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So let me draw that in here. And beta one would be up here. So let me put in a proton on beta one. We think about a weak base coming along and taking this proton. So our base is probably water. And these electrons would move into here this time. So if that happens, let's draw that product."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "We think about a weak base coming along and taking this proton. So our base is probably water. And these electrons would move into here this time. So if that happens, let's draw that product. Our double bond would form up here. And let me draw in the rest of this molecule. So let me use red for those electrons."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So if that happens, let's draw that product. Our double bond would form up here. And let me draw in the rest of this molecule. So let me use red for those electrons. So the electrons in red are going to move into here to form this alkene. Notice that the two alkenes that we just drew, these are really the same molecule. This is the same compound."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So let me use red for those electrons. So the electrons in red are going to move into here to form this alkene. Notice that the two alkenes that we just drew, these are really the same molecule. This is the same compound. So we haven't formed two different products. If you take a proton away from the beta one or the beta two carbon, you're gonna make the same alkene. But what about the beta three carbon?"}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "This is the same compound. So we haven't formed two different products. If you take a proton away from the beta one or the beta two carbon, you're gonna make the same alkene. But what about the beta three carbon? So that's our last example. And let me go ahead, I forgot to put in a plus one formal charge on our carbocation. Let me draw one more carbocation."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "But what about the beta three carbon? So that's our last example. And let me go ahead, I forgot to put in a plus one formal charge on our carbocation. Let me draw one more carbocation. The same one, a tertiary carbocation. The difference is this time we're gonna take a proton away from our beta three carbon. And so let me draw in a proton there."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "Let me draw one more carbocation. The same one, a tertiary carbocation. The difference is this time we're gonna take a proton away from our beta three carbon. And so let me draw in a proton there. And we think about a weak base coming along and taking that proton. So I'll draw in my weak base here. So it takes this proton, and these electrons move into here."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "And so let me draw in a proton there. And we think about a weak base coming along and taking that proton. So I'll draw in my weak base here. So it takes this proton, and these electrons move into here. So let's draw this product. So we would have our alkene that looks like this. So let's follow those electrons."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So it takes this proton, and these electrons move into here. So let's draw this product. So we would have our alkene that looks like this. So let's follow those electrons. I'll make them dark blue. The electrons in this bond move into here to form our double bond. And so now we've gone through the complete mechanism, and we have two products."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So let's follow those electrons. I'll make them dark blue. The electrons in this bond move into here to form our double bond. And so now we've gone through the complete mechanism, and we have two products. So let me circle our two products. So this is really just one product. And then this would be our second product."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "And so now we've gone through the complete mechanism, and we have two products. So let me circle our two products. So this is really just one product. And then this would be our second product. For this reaction, we actually get 90% of the alkene on the right, and 10% of the alkene on the left. And so let's look at the degree of substitution of our two products. And let's start with the one on the right."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "And then this would be our second product. For this reaction, we actually get 90% of the alkene on the right, and 10% of the alkene on the left. And so let's look at the degree of substitution of our two products. And let's start with the one on the right. So let me use red for this. If we think about the degree of substitution for the alkene on the right, if I draw in my hydrogen right here, it makes it a little bit easier to see. We have three alkyl groups."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "And let's start with the one on the right. So let me use red for this. If we think about the degree of substitution for the alkene on the right, if I draw in my hydrogen right here, it makes it a little bit easier to see. We have three alkyl groups. So this one, this one, and this one. So this would be a tri-substituted alkene. So the one on the right is a tri-substituted alkene."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "We have three alkyl groups. So this one, this one, and this one. So this would be a tri-substituted alkene. So the one on the right is a tri-substituted alkene. And the one on the left, so this one right here, would be a di-substituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "So the one on the right is a tri-substituted alkene. And the one on the left, so this one right here, would be a di-substituted alkene. These are the two carbons across our double bond. We have two hydrogens on this carbon. And the carbon on the right has two alkyl groups bonded to it. So this one is a di-substituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a di-substituted product and a tri-substituted product."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "We have two hydrogens on this carbon. And the carbon on the right has two alkyl groups bonded to it. So this one is a di-substituted alkene. Now we've gone through the whole E1 mechanism, and we've seen that we get a di-substituted product and a tri-substituted product. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. For the di-substituted product, the double bond formed in this region of the molecule."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "Now we've gone through the whole E1 mechanism, and we've seen that we get a di-substituted product and a tri-substituted product. Now let's think about regiochemistry. For this reaction, it's the region of the molecule where the double bond forms. For the di-substituted product, the double bond formed in this region of the molecule. And for the tri-substituted product, the double bond formed in this region. The tri-substituted product is the major product. And it's also the more stable alkene."}, {"video_title": "E1 mechanism regioselectivity.mp3", "Sentence": "For the di-substituted product, the double bond formed in this region of the molecule. And for the tri-substituted product, the double bond formed in this region. The tri-substituted product is the major product. And it's also the more stable alkene. So remember from the video on alkene stability, the more substituted your alkene is, the more stable it is. So this product is more stable, and that's why we form more of it. And the more stable product, or the more substituted product, is called the Zaitsev product."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "So A minus is the conjugate base to HA. If A minus is stable, then HA is more likely to donate this proton. Therefore, if you want to determine the acidity of a compound, you can look at the stability of the conjugate base. The more stable the conjugate base, the more likely the acid is to donate a proton. Therefore, the more stable the conjugate base, the stronger the acid. Let's use that concept and let's look at these four compounds down here. So we'll start with methane."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "The more stable the conjugate base, the more likely the acid is to donate a proton. Therefore, the more stable the conjugate base, the stronger the acid. Let's use that concept and let's look at these four compounds down here. So we'll start with methane. The pKa for this proton on methane is approximately 48. For ammonia, the pKa for this proton is about 36. And if we look at water, the pKa for this proton is about 16."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start with methane. The pKa for this proton on methane is approximately 48. For ammonia, the pKa for this proton is about 36. And if we look at water, the pKa for this proton is about 16. And finally, for HF, this proton has a pKa of about three. We know that the lower the pKa value, the stronger the acid. So as we move to the right, we see a decrease in pKa values from 48 to 36 to 16 to three."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "And if we look at water, the pKa for this proton is about 16. And finally, for HF, this proton has a pKa of about three. We know that the lower the pKa value, the stronger the acid. So as we move to the right, we see a decrease in pKa values from 48 to 36 to 16 to three. Therefore, as we go to the right, we see an increase in acid strength. We see an increase in acid strength. So HF is the strongest acid out of these four."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "So as we move to the right, we see a decrease in pKa values from 48 to 36 to 16 to three. Therefore, as we go to the right, we see an increase in acid strength. We see an increase in acid strength. So HF is the strongest acid out of these four. And if HF is the strongest acid out of these four, then HF must have the most stable conjugate base. So now let's think about the conjugate bases for all four of these compounds. So let me go down here and we'll get some more room."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "So HF is the strongest acid out of these four. And if HF is the strongest acid out of these four, then HF must have the most stable conjugate base. So now let's think about the conjugate bases for all four of these compounds. So let me go down here and we'll get some more room. If we take this proton from methane, then these electrons are left behind on the carbon. So the carbon gets a negative one formal charge. For ammonia, if we took this proton, then these electrons are left on the nitrogen."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "So let me go down here and we'll get some more room. If we take this proton from methane, then these electrons are left behind on the carbon. So the carbon gets a negative one formal charge. For ammonia, if we took this proton, then these electrons are left on the nitrogen. So the nitrogen has a negative one formal charge. For water, if we took this proton, these electrons are left on the oxygen to form the hydroxide anion as our conjugate base with a negative one formal charge on the oxygen. And finally, if we took this proton, then these electrons would be left behind on the fluorine to form the fluoride anion."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "For ammonia, if we took this proton, then these electrons are left on the nitrogen. So the nitrogen has a negative one formal charge. For water, if we took this proton, these electrons are left on the oxygen to form the hydroxide anion as our conjugate base with a negative one formal charge on the oxygen. And finally, if we took this proton, then these electrons would be left behind on the fluorine to form the fluoride anion. We already know that HF is the strongest acid out of these four, and the strongest acid must have the most stable conjugate base. So the fluoride anion must be the most stable conjugate base. So as we move to the right, we're increasing in the stability, we're increasing in the stability of the conjugate base."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "And finally, if we took this proton, then these electrons would be left behind on the fluorine to form the fluoride anion. We already know that HF is the strongest acid out of these four, and the strongest acid must have the most stable conjugate base. So the fluoride anion must be the most stable conjugate base. So as we move to the right, we're increasing in the stability, we're increasing in the stability of the conjugate base. And we can explain that trend by looking at the element that has the negative charge. This has a negative charge on carbon, and this was our least stable conjugate base. Then we go to nitrogen with a negative charge, we get a little bit more stable."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "So as we move to the right, we're increasing in the stability, we're increasing in the stability of the conjugate base. And we can explain that trend by looking at the element that has the negative charge. This has a negative charge on carbon, and this was our least stable conjugate base. Then we go to nitrogen with a negative charge, we get a little bit more stable. We go to oxygen with a negative one charge, we get a little bit more stable. And finally, we get to fluorine with a negative charge, and we have the most stable conjugate base. That's the same trend as electronegativity."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "Then we go to nitrogen with a negative charge, we get a little bit more stable. We go to oxygen with a negative one charge, we get a little bit more stable. And finally, we get to fluorine with a negative charge, and we have the most stable conjugate base. That's the same trend as electronegativity. So if you look at carbon, nitrogen, oxygen, and fluorine, as you move to the right on the periodic table, you know you increase in electronegativity, with fluorine being the most electronegative element. And the most electronegative element attracts electrons the most. It likes to have electrons around it."}, {"video_title": "Stabilization of a conjugate base electronegativity Organic chemistry Khan Academy.mp3", "Sentence": "That's the same trend as electronegativity. So if you look at carbon, nitrogen, oxygen, and fluorine, as you move to the right on the periodic table, you know you increase in electronegativity, with fluorine being the most electronegative element. And the most electronegative element attracts electrons the most. It likes to have electrons around it. And therefore, it makes sense that fluorine is the best at stabilizing a negative charge. And that makes this the most stable conjugate base. And if this is the most stable conjugate base, the fluoride anion is the most stable conjugate base, that means that HF must be the strongest acid."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We've already seen that formal charge is equal to the number of valence electrons in the free atom minus the number of valence electrons in the bonded atom. Another way of saying that is the formal charge is equal to the number of valence electrons the atom is supposed to have minus the number of valence electrons the atom actually has in the drawing. Let's assign a formal charge to oxygen in this molecule. Remember that each bond is made up of two electrons. This bond right here is made up of two electrons. This bond over here is made up of two electrons. Our goal is to find the formal charge on oxygen."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Remember that each bond is made up of two electrons. This bond right here is made up of two electrons. This bond over here is made up of two electrons. Our goal is to find the formal charge on oxygen. The formal charge on oxygen is equal to the number of valence electrons in the free atom, so the number of valence electrons that oxygen is supposed to have. We know that's six. Oxygen is supposed to have six valence electrons minus the number of electrons that oxygen actually has in our drawing."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Our goal is to find the formal charge on oxygen. The formal charge on oxygen is equal to the number of valence electrons in the free atom, so the number of valence electrons that oxygen is supposed to have. We know that's six. Oxygen is supposed to have six valence electrons minus the number of electrons that oxygen actually has in our drawing. Remember, when you have a bond with two electrons, we give one electron to one atom and the other electron to the other atom. From these two electrons, oxygen gets one of those electrons. Same thing for this other bond here."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen is supposed to have six valence electrons minus the number of electrons that oxygen actually has in our drawing. Remember, when you have a bond with two electrons, we give one electron to one atom and the other electron to the other atom. From these two electrons, oxygen gets one of those electrons. Same thing for this other bond here. Oxygen gets one of those electrons and the other electron goes to hydrogen. Now we have a total of six electrons around our oxygen. I'll highlight those."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Same thing for this other bond here. Oxygen gets one of those electrons and the other electron goes to hydrogen. Now we have a total of six electrons around our oxygen. I'll highlight those. We have one, two, three, four, five, six. There are six valence electrons around the oxygen in our drawing. Six minus six is equal to zero."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll highlight those. We have one, two, three, four, five, six. There are six valence electrons around the oxygen in our drawing. Six minus six is equal to zero. The formal charge on oxygen is equal to zero. Let me go ahead and write down this pattern that we've just seen. When oxygen has two bonds and two lone pairs of electrons, so when oxygen has two bonds and two lone pairs of electrons, the formal charge is equal to zero."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Six minus six is equal to zero. The formal charge on oxygen is equal to zero. Let me go ahead and write down this pattern that we've just seen. When oxygen has two bonds and two lone pairs of electrons, so when oxygen has two bonds and two lone pairs of electrons, the formal charge is equal to zero. Sometimes the lone pairs are just left off for convenience reasons. You could draw this with your oxygen and your hydrogen like that or you could even go like this. All of those are just different ways of representing the same molecule, so leaving off lone pairs of electrons."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "When oxygen has two bonds and two lone pairs of electrons, so when oxygen has two bonds and two lone pairs of electrons, the formal charge is equal to zero. Sometimes the lone pairs are just left off for convenience reasons. You could draw this with your oxygen and your hydrogen like that or you could even go like this. All of those are just different ways of representing the same molecule, so leaving off lone pairs of electrons. Let's look at some other examples where the formal charge on oxygen is equal to zero. We'll look at the one on the left first. The formal charge is equal to zero on this oxygen."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All of those are just different ways of representing the same molecule, so leaving off lone pairs of electrons. Let's look at some other examples where the formal charge on oxygen is equal to zero. We'll look at the one on the left first. The formal charge is equal to zero on this oxygen. We can see we have two bonds here to oxygen. Here's one of the bonds to oxygen. Here's the other bond to oxygen."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The formal charge is equal to zero on this oxygen. We can see we have two bonds here to oxygen. Here's one of the bonds to oxygen. Here's the other bond to oxygen. The lone pairs of electrons have been left off this dot structure, but we know since the formal charge is zero, we already have our two bonds here. There should be two lone pairs of electrons on that oxygen. You could put them in there or you could leave them off."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here's the other bond to oxygen. The lone pairs of electrons have been left off this dot structure, but we know since the formal charge is zero, we already have our two bonds here. There should be two lone pairs of electrons on that oxygen. You could put them in there or you could leave them off. I'll go ahead and put them in. We have a total of eight electrons around our oxygen. Oxygen's following the octet rule here."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You could put them in there or you could leave them off. I'll go ahead and put them in. We have a total of eight electrons around our oxygen. Oxygen's following the octet rule here. I'll highlight them, two, four, six, and eight. On the right, we have another example where oxygen has a formal charge of zero. This oxygen has two bonds to it."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen's following the octet rule here. I'll highlight them, two, four, six, and eight. On the right, we have another example where oxygen has a formal charge of zero. This oxygen has two bonds to it. Here's one of the bonds and here's the other bond. This oxygen would also have to have two lone pairs of electrons on it. Again, I didn't draw them in here."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen has two bonds to it. Here's one of the bonds and here's the other bond. This oxygen would also have to have two lone pairs of electrons on it. Again, I didn't draw them in here. Sometimes you don't draw them in here for convenience, but I could go ahead and add them so it's easier to see that that oxygen has a formal charge of zero and a total of an octet of electrons around it. Let me highlight those, two, four, six, and eight. These patterns are important."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Again, I didn't draw them in here. Sometimes you don't draw them in here for convenience, but I could go ahead and add them so it's easier to see that that oxygen has a formal charge of zero and a total of an octet of electrons around it. Let me highlight those, two, four, six, and eight. These patterns are important. For oxygen, two bonds and two lone pairs of electrons give us a formal charge of zero. Let's move on to another formal charge situation for oxygen. Let's find the formal charge on oxygen here."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These patterns are important. For oxygen, two bonds and two lone pairs of electrons give us a formal charge of zero. Let's move on to another formal charge situation for oxygen. Let's find the formal charge on oxygen here. We start by drawing in the electrons in our bonds. Each bond consists of two electrons. I'm going to draw those in."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's find the formal charge on oxygen here. We start by drawing in the electrons in our bonds. Each bond consists of two electrons. I'm going to draw those in. What is the formal charge on oxygen this time? The formal charge is equal to the number of valence electrons oxygen is supposed to have, which is six, minus the number of valence electrons oxygen actually has in our drawing. We divide up our electrons again."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to draw those in. What is the formal charge on oxygen this time? The formal charge is equal to the number of valence electrons oxygen is supposed to have, which is six, minus the number of valence electrons oxygen actually has in our drawing. We divide up our electrons again. Oxygen gets one from this bond and one from this bond and one from this bond. How many electrons are around oxygen now? It would be one, two, three, four, and five."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We divide up our electrons again. Oxygen gets one from this bond and one from this bond and one from this bond. How many electrons are around oxygen now? It would be one, two, three, four, and five. Six minus five is equal to plus one. It's like oxygen has lost an electron here. Oxygen has a formal charge of plus one."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It would be one, two, three, four, and five. Six minus five is equal to plus one. It's like oxygen has lost an electron here. Oxygen has a formal charge of plus one. I could redraw that. Let me go ahead and do that over here. I could redraw that over here."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Oxygen has a formal charge of plus one. I could redraw that. Let me go ahead and do that over here. I could redraw that over here. We have oxygen with our bond to hydrogens. Oxygen has a lone pair of electrons on it. This oxygen has a plus one formal charge."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I could redraw that over here. We have oxygen with our bond to hydrogens. Oxygen has a lone pair of electrons on it. This oxygen has a plus one formal charge. We can come up with another pattern here. Here oxygen has three bonds. Let me highlight those bonds."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen has a plus one formal charge. We can come up with another pattern here. Here oxygen has three bonds. Let me highlight those bonds. Let me use red this time. Here's one bond, two bonds, and then three bonds, and then one lone pair of electrons. The pattern of three bonds plus one lone pair of electrons for oxygen will give you a formal charge of plus one."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me highlight those bonds. Let me use red this time. Here's one bond, two bonds, and then three bonds, and then one lone pair of electrons. The pattern of three bonds plus one lone pair of electrons for oxygen will give you a formal charge of plus one. Again, it's good to recognize these patterns. You should be able to do the calculation, and then after you do enough of these problems, you can just look at it and figure out what the formal charge is. Let's look at some more examples where oxygen has a formal charge of plus one."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The pattern of three bonds plus one lone pair of electrons for oxygen will give you a formal charge of plus one. Again, it's good to recognize these patterns. You should be able to do the calculation, and then after you do enough of these problems, you can just look at it and figure out what the formal charge is. Let's look at some more examples where oxygen has a formal charge of plus one. The lone pairs were left off of this one, again, for convenience reasons. We'll start with this example on the left. We can see that the oxygen with a plus one formal charge has three bonds to it."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at some more examples where oxygen has a formal charge of plus one. The lone pairs were left off of this one, again, for convenience reasons. We'll start with this example on the left. We can see that the oxygen with a plus one formal charge has three bonds to it. Here's one, here's two, and then here's three. Three bonds. In order for that oxygen to have a plus one formal charge, it must also have one lone pair of electrons on it."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can see that the oxygen with a plus one formal charge has three bonds to it. Here's one, here's two, and then here's three. Three bonds. In order for that oxygen to have a plus one formal charge, it must also have one lone pair of electrons on it. You could just leave them off and know that they're there, or you could go ahead and draw them in. I'll draw them in on that oxygen. Now to our other example over here on the right."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "In order for that oxygen to have a plus one formal charge, it must also have one lone pair of electrons on it. You could just leave them off and know that they're there, or you could go ahead and draw them in. I'll draw them in on that oxygen. Now to our other example over here on the right. This is the oxygen with a plus one formal charge. That oxygen must have three bonds and one lone pair. Here are the three bonds."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now to our other example over here on the right. This is the oxygen with a plus one formal charge. That oxygen must have three bonds and one lone pair. Here are the three bonds. Here's one, two, and three. Again, I didn't draw in the lone pair of electrons on the oxygen, but the lone pair is there. I'll go ahead and put it in like that."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here are the three bonds. Here's one, two, and three. Again, I didn't draw in the lone pair of electrons on the oxygen, but the lone pair is there. I'll go ahead and put it in like that. Recognize this pattern. Three bonds plus one lone pair for oxygen gives us a formal charge of plus one. You could have also figured out how many electrons are necessary."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll go ahead and put it in like that. Recognize this pattern. Three bonds plus one lone pair for oxygen gives us a formal charge of plus one. You could have also figured out how many electrons are necessary. Let's use this example by, let me go ahead and redraw it here. How else could we figure out how many electrons are on that oxygen if that oxygen has a plus one formal charge? You could say that the calculation for a formal charge would be six minus, six minus x is what we don't know, but we do know the formal charge is plus one."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You could have also figured out how many electrons are necessary. Let's use this example by, let me go ahead and redraw it here. How else could we figure out how many electrons are on that oxygen if that oxygen has a plus one formal charge? You could say that the calculation for a formal charge would be six minus, six minus x is what we don't know, but we do know the formal charge is plus one. Let me just put this in a little box over here. Six minus x is equal to plus one. Obviously, x would have to be equal to five, meaning that oxygen would have five electrons around it."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You could say that the calculation for a formal charge would be six minus, six minus x is what we don't know, but we do know the formal charge is plus one. Let me just put this in a little box over here. Six minus x is equal to plus one. Obviously, x would have to be equal to five, meaning that oxygen would have five electrons around it. Let's think about how many electrons that we see right now around oxygen. Let me draw in some electrons here. I'll use red."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Obviously, x would have to be equal to five, meaning that oxygen would have five electrons around it. Let's think about how many electrons that we see right now around oxygen. Let me draw in some electrons here. I'll use red. This bond is two electrons. Then we have a bunch of electrons in here. How many electrons around oxygen do we have so far?"}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll use red. This bond is two electrons. Then we have a bunch of electrons in here. How many electrons around oxygen do we have so far? We would have three, just these three. We need five, which means we need two more, which means we need a lone pair of electrons on the oxygen. That's a little bit too complicated, I think, for figuring it out, but you could use that method, or you could just learn this pattern."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "How many electrons around oxygen do we have so far? We would have three, just these three. We need five, which means we need two more, which means we need a lone pair of electrons on the oxygen. That's a little bit too complicated, I think, for figuring it out, but you could use that method, or you could just learn this pattern. Eventually, you'll have to have this pattern down pretty well. Let's look at another example for assigning formal charge to oxygen. Our goal is to find the formal charge on oxygen in this example."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's a little bit too complicated, I think, for figuring it out, but you could use that method, or you could just learn this pattern. Eventually, you'll have to have this pattern down pretty well. Let's look at another example for assigning formal charge to oxygen. Our goal is to find the formal charge on oxygen in this example. We put in our electrons in this bond. Each bond represents two electrons. The formal charge on oxygen is equal to the number of valence electrons that oxygen is supposed to have, which is six, minus the number of valence electrons that oxygen actually has."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Our goal is to find the formal charge on oxygen in this example. We put in our electrons in this bond. Each bond represents two electrons. The formal charge on oxygen is equal to the number of valence electrons that oxygen is supposed to have, which is six, minus the number of valence electrons that oxygen actually has. In this example, we would take one of these electrons from this bond, and how many electrons is that total around oxygen? This would be one, two, three, four, five, six, seven. Six minus seven gives us a formal charge of negative one."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The formal charge on oxygen is equal to the number of valence electrons that oxygen is supposed to have, which is six, minus the number of valence electrons that oxygen actually has. In this example, we would take one of these electrons from this bond, and how many electrons is that total around oxygen? This would be one, two, three, four, five, six, seven. Six minus seven gives us a formal charge of negative one. I could redraw that over here. I could say oxygen has three lone pairs of electrons and a negative one formal charge. Our pattern, this time our pattern is one bond, here's the one bond, and then three lone pairs of electrons."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Six minus seven gives us a formal charge of negative one. I could redraw that over here. I could say oxygen has three lone pairs of electrons and a negative one formal charge. Our pattern, this time our pattern is one bond, here's the one bond, and then three lone pairs of electrons. Let me write that down. The pattern is one bond, when oxygen has one bond and three lone pairs of electrons, the formal charge is negative one, just like we saw up here with the calculation. We could leave those electrons off."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Our pattern, this time our pattern is one bond, here's the one bond, and then three lone pairs of electrons. Let me write that down. The pattern is one bond, when oxygen has one bond and three lone pairs of electrons, the formal charge is negative one, just like we saw up here with the calculation. We could leave those electrons off. If we wanted to save some time, we could just say, oh, this is oxygen with a negative one formal charge, and we should know that there must be three lone pairs of electrons on that oxygen. Let's look at one more example where formal charge is negative one. Right here, this oxygen has a negative one formal charge, and we can see it already has one bond to it."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could leave those electrons off. If we wanted to save some time, we could just say, oh, this is oxygen with a negative one formal charge, and we should know that there must be three lone pairs of electrons on that oxygen. Let's look at one more example where formal charge is negative one. Right here, this oxygen has a negative one formal charge, and we can see it already has one bond to it. The pattern, of course, is one bond plus three lone pairs of electrons. We already have the one bond. In order for that oxygen to have a negative one formal charge, we need three lone pairs of electrons."}, {"video_title": "Formal charge on oxygen Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Right here, this oxygen has a negative one formal charge, and we can see it already has one bond to it. The pattern, of course, is one bond plus three lone pairs of electrons. We already have the one bond. In order for that oxygen to have a negative one formal charge, we need three lone pairs of electrons. We could redraw this. That is one way to represent that ion. We could also represent it like this with putting three lone pairs of electrons on that oxygen with a negative one formal charge."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "However, if you react them with strong acids, you get acidic cleavage of the ether. So if we start with our ether over here on the left, and we add excess hydrogen halide, and we heat things up, the ether gets cleaved to form an alcohol and an alkyl halide. Normally, the reaction conditions will then convert that alcohol into yet another alkyl halide. So you will usually end up with two alkyl halides as your product, however, not always. In terms of what hydrogen halide do you use, HBr and HI work the best. And that's because when you have the bromide or the iodide anion, the electrons are further away from the nucleus, which increases the nucleophilic strength of those anions, compared to, say, something like chlorine, where the electrons are closer to the nucleus, since it's a smaller size. Let's look at the mechanism for the acidic cleavage of ethers."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So you will usually end up with two alkyl halides as your product, however, not always. In terms of what hydrogen halide do you use, HBr and HI work the best. And that's because when you have the bromide or the iodide anion, the electrons are further away from the nucleus, which increases the nucleophilic strength of those anions, compared to, say, something like chlorine, where the electrons are closer to the nucleus, since it's a smaller size. Let's look at the mechanism for the acidic cleavage of ethers. And we start with our ether like this. So this is our ether, which we are going to react with our hydrogen halide. So we start with an acid-base reaction."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the mechanism for the acidic cleavage of ethers. And we start with our ether like this. So this is our ether, which we are going to react with our hydrogen halide. So we start with an acid-base reaction. So a strong acid like hydrobromic acid will donate protons. The ether is going to act as a base. Lone pair of electrons are going to pick up the proton from our hydrogen halide."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we start with an acid-base reaction. So a strong acid like hydrobromic acid will donate protons. The ether is going to act as a base. Lone pair of electrons are going to pick up the proton from our hydrogen halide. The electrons will kick off onto the halogen like that. So we will end up with, we're going to protonate our ether. So we have a hydrogen there."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Lone pair of electrons are going to pick up the proton from our hydrogen halide. The electrons will kick off onto the halogen like that. So we will end up with, we're going to protonate our ether. So we have a hydrogen there. We have a lone pair of electrons on this oxygen, which gives this oxygen a plus 1 formal charge. So we also have our halogen. It used to have three lone pairs of electrons."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we have a hydrogen there. We have a lone pair of electrons on this oxygen, which gives this oxygen a plus 1 formal charge. So we also have our halogen. It used to have three lone pairs of electrons. It just picked up one more lone pair of electrons to form a halide anion, which is going to function as our nucleophile. So our nucleophile is going to attack the carbon bonded to the oxygen. And then the electrons between the carbon and the oxygen are going to kick off onto the oxygen."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It used to have three lone pairs of electrons. It just picked up one more lone pair of electrons to form a halide anion, which is going to function as our nucleophile. So our nucleophile is going to attack the carbon bonded to the oxygen. And then the electrons between the carbon and the oxygen are going to kick off onto the oxygen. So this is an SN2 type mechanism, where your halide anion is going to function as your nucleophile. So this is why bromide anions and iodide anions work better than chloride anions. So the result of that nucleophilic attack, we now have our alcohol."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then the electrons between the carbon and the oxygen are going to kick off onto the oxygen. So this is an SN2 type mechanism, where your halide anion is going to function as your nucleophile. So this is why bromide anions and iodide anions work better than chloride anions. So the result of that nucleophilic attack, we now have our alcohol. So we got an extra lone pair of electrons on our oxygen there. And our R prime group is now bonded to our halogen. So we've made our first alkyl halide like that."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the result of that nucleophilic attack, we now have our alcohol. So we got an extra lone pair of electrons on our oxygen there. And our R prime group is now bonded to our halogen. So we've made our first alkyl halide like that. And sometimes the reaction will stop here. And sometimes the reaction will keep going. So if the reaction keeps going, we can get another hydrogen halide molecule in here."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we've made our first alkyl halide like that. And sometimes the reaction will stop here. And sometimes the reaction will keep going. So if the reaction keeps going, we can get another hydrogen halide molecule in here. And so the lone pair of electrons on our alcohol are now going to function as a base and pick up a proton, which kicks these electrons off onto our halogen. So if we draw the result of that acid-base reaction, we're going to protonate our alcohol. So our alcohol now has two hydrogens, lone pair of electrons on our oxygen plus one formal charge."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if the reaction keeps going, we can get another hydrogen halide molecule in here. And so the lone pair of electrons on our alcohol are now going to function as a base and pick up a proton, which kicks these electrons off onto our halogen. So if we draw the result of that acid-base reaction, we're going to protonate our alcohol. So our alcohol now has two hydrogens, lone pair of electrons on our oxygen plus one formal charge. And our halogen now has an extra lone pair of electrons, giving it a negative charge, making it a halide anion. So in the next step of the mechanism, the halide anion will function as our nucleophile and attack the carbon bonded to the oxygen, kicks these electrons off onto the oxygen. And that's how we make our second alkyl halide."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So our alcohol now has two hydrogens, lone pair of electrons on our oxygen plus one formal charge. And our halogen now has an extra lone pair of electrons, giving it a negative charge, making it a halide anion. So in the next step of the mechanism, the halide anion will function as our nucleophile and attack the carbon bonded to the oxygen, kicks these electrons off onto the oxygen. And that's how we make our second alkyl halide. So we'd form RX as our other product. And also water would be produced in this as well, so H2O like that. Now I drew this second part of the mechanism like it's an SN2 mechanism."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And that's how we make our second alkyl halide. So we'd form RX as our other product. And also water would be produced in this as well, so H2O like that. Now I drew this second part of the mechanism like it's an SN2 mechanism. And it would be an SN2 mechanism if we were starting with a primary alcohol. So if this guy over here is a primary alcohol and after it gets protonated, a primary alcohol would work the best for an SN2 mechanism because that would be decreased steric hindrance. However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now I drew this second part of the mechanism like it's an SN2 mechanism. And it would be an SN2 mechanism if we were starting with a primary alcohol. So if this guy over here is a primary alcohol and after it gets protonated, a primary alcohol would work the best for an SN2 mechanism because that would be decreased steric hindrance. However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism. So it's important to look at the structure of the alkyl halide. Let's do an example of the acidic cleavage of ethers. And we'll start with an ether that looks like this."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "However, if we were dealing with something like a tertiary alcohol at this point, things would likely proceed via an SN1 type mechanism. So it's important to look at the structure of the alkyl halide. Let's do an example of the acidic cleavage of ethers. And we'll start with an ether that looks like this. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether is going to go away."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with an ether that looks like this. So we're going to react this ether with excess hydrobromic acid. And we're going to heat things up. And when we think about our products, we know that the ether is going to go away. And we know that we're going to get two alkyl halides out of this. So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And when we think about our products, we know that the ether is going to go away. And we know that we're going to get two alkyl halides out of this. So we just need to find our alkyl groups. So if I look over here, here's one of my alkyl groups. And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I look over here, here's one of my alkyl groups. And if I look over here, here's my other alkyl group. So all I have to do is turn those alkyl groups into alkyl halides. And they're going to be alkyl bromides, since we're using hydrobromic acid here. So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide would be this methyl group over here."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And they're going to be alkyl bromides, since we're using hydrobromic acid here. So I'm going to draw one of my alkyl halides like that. So it would be bromine attached to the ring. And then my other alkyl halide would be this methyl group over here. So I take a methyl group, and I attach it to bromine. So methyl bromide would be my other product. And we'd also produce water as well."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And then my other alkyl halide would be this methyl group over here. So I take a methyl group, and I attach it to bromine. So methyl bromide would be my other product. And we'd also produce water as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make a little bit tricky."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we'd also produce water as well. But your major organic products would be these two alkyl halides. Let's do another one. This one will make a little bit tricky. So in three dimensions. So we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This one will make a little bit tricky. So in three dimensions. So we're going to have to think about acidic cleavage of ethers in three dimensions. So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether. And when I'm trying to figure out the products, I know that oxygen is going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it makes it much harder. So if I look at an ether, which looks like this, and I add, once again, excess hydrobromic acid, and I heat things up, I'm going to get acidic cleavage of that ether. And when I'm trying to figure out the products, I know that oxygen is going to go away. And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides. So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen. And if I look at this carbon on the other side of my ether, that carbon is also going to be bonded to a halogen."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And I know that the carbons that are bonded to that oxygen are the ones that are going to form my alkyl halides. So I look at this carbon that's bonded to my oxygen. That's going to be bonded to a halogen. And if I look at this carbon on the other side of my ether, that carbon is also going to be bonded to a halogen. So when I'm trying to draw the products, so this ring here is going to stay the same. So I'm just going to go ahead and draw my ring. Like that."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And if I look at this carbon on the other side of my ether, that carbon is also going to be bonded to a halogen. So when I'm trying to draw the products, so this ring here is going to stay the same. So I'm just going to go ahead and draw my ring. Like that. And let's first look at this carbon right here. This one in red, which is this one over here. So it was already bonded to a methyl group right here."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Like that. And let's first look at this carbon right here. This one in red, which is this one over here. So it was already bonded to a methyl group right here. And the bond between the carbon that oxygen is going to break, and we're going to form a new bond with our halogen, which happens to be bromine. So we're going to put bromine there like that. And the opposite side, when we look at this carbon over here."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it was already bonded to a methyl group right here. And the bond between the carbon that oxygen is going to break, and we're going to form a new bond with our halogen, which happens to be bromine. So we're going to put bromine there like that. And the opposite side, when we look at this carbon over here. So let's go ahead and draw that carbon in. So we're going to get like that. So it's the carbon in blue."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And the opposite side, when we look at this carbon over here. So let's go ahead and draw that carbon in. So we're going to get like that. So it's the carbon in blue. The bond between the carbon in blue and the oxygen is going to break, and there's going to be a new bond formed to our bromine, which is our halogen. So we can go ahead and draw our bromine in there like that. And so that will be our product."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it's the carbon in blue. The bond between the carbon in blue and the oxygen is going to break, and there's going to be a new bond formed to our bromine, which is our halogen. So we can go ahead and draw our bromine in there like that. And so that will be our product. So we get two alkyl halides in the same molecule this time. We could draw the product like that, or we could kind of flatten it out a little bit. And let's think about that ring system there."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so that will be our product. So we get two alkyl halides in the same molecule this time. We could draw the product like that, or we could kind of flatten it out a little bit. And let's think about that ring system there. So what kind of a ring system do we have? We have 1, 2, 3, 4, 5, 6 carbons present. So if I'm going to draw this molecule in a different way, I would have a 6-carbon ring, so cyclohexane."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And let's think about that ring system there. So what kind of a ring system do we have? We have 1, 2, 3, 4, 5, 6 carbons present. So if I'm going to draw this molecule in a different way, I would have a 6-carbon ring, so cyclohexane. And if I take a look at, let's do this carbon right here. So that's this carbon. What is attached to that carbon?"}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I'm going to draw this molecule in a different way, I would have a 6-carbon ring, so cyclohexane. And if I take a look at, let's do this carbon right here. So that's this carbon. What is attached to that carbon? Well, we have this carbon is attached to another carbon. There are two methyl groups and then a bromine like that. Doesn't really matter the order that you put your methyl groups and your bromine."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "What is attached to that carbon? Well, we have this carbon is attached to another carbon. There are two methyl groups and then a bromine like that. Doesn't really matter the order that you put your methyl groups and your bromine. This is because this carbon right here is not a chirality center. So you don't have to worry too much about that. Let's look at this carbon down here."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Doesn't really matter the order that you put your methyl groups and your bromine. This is because this carbon right here is not a chirality center. So you don't have to worry too much about that. Let's look at this carbon down here. So in green, so I'll make this one down here in green. There's a methyl group attached to that carbon. So if I make a methyl group attached to that carbon right here, and there's also a bromine attached to that carbon."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at this carbon down here. So in green, so I'll make this one down here in green. There's a methyl group attached to that carbon. So if I make a methyl group attached to that carbon right here, and there's also a bromine attached to that carbon. So these are just two different ways to represent the exact same molecule for your product after the acidic cleavage of the original ether. So kind of a tough one, because you're thinking in three dimensions here. Let's do one more example for the acidic cleavage of ethers."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I make a methyl group attached to that carbon right here, and there's also a bromine attached to that carbon. So these are just two different ways to represent the exact same molecule for your product after the acidic cleavage of the original ether. So kind of a tough one, because you're thinking in three dimensions here. Let's do one more example for the acidic cleavage of ethers. And we'll start with ethyl. Let's do ethyl phenyl ether here. So we'll draw a phenyl group."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more example for the acidic cleavage of ethers. And we'll start with ethyl. Let's do ethyl phenyl ether here. So we'll draw a phenyl group. So here is our phenyl group. And then we'll have an ethyl group on this side. So what will be the product of ethyl phenyl ether reacted with excess hydrobromic acid, and everything is heated up?"}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we'll draw a phenyl group. So here is our phenyl group. And then we'll have an ethyl group on this side. So what will be the product of ethyl phenyl ether reacted with excess hydrobromic acid, and everything is heated up? All right, so let's think about the mechanism. We know that in our mechanism, the first step is to protonate the ether. So the ether is going to function as a base, pick up a proton from hydrobromic acids."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So what will be the product of ethyl phenyl ether reacted with excess hydrobromic acid, and everything is heated up? All right, so let's think about the mechanism. We know that in our mechanism, the first step is to protonate the ether. So the ether is going to function as a base, pick up a proton from hydrobromic acids. So if we go ahead and draw the first step, the acid-base reaction there. So we're going to protonate our ether. And that means there's going to be a hydrogen attached to our oxygen now."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the ether is going to function as a base, pick up a proton from hydrobromic acids. So if we go ahead and draw the first step, the acid-base reaction there. So we're going to protonate our ether. And that means there's going to be a hydrogen attached to our oxygen now. Lone pair of electrons still left on our oxygen, plus one formal charge on our oxygen like that. Well, we have the bromide anion left over, right? After HBr donates a proton, Br negative is left."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And that means there's going to be a hydrogen attached to our oxygen now. Lone pair of electrons still left on our oxygen, plus one formal charge on our oxygen like that. Well, we have the bromide anion left over, right? After HBr donates a proton, Br negative is left. So Br negative. The bromide anion is going to function as my nucleophile. And it's going to proceed via an SN2 mechanism."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "After HBr donates a proton, Br negative is left. So Br negative. The bromide anion is going to function as my nucleophile. And it's going to proceed via an SN2 mechanism. Therefore, it's going to attack the least sterically hindered alkyl group, which is, of course, the alkyl group on the right. So a lone pair of electrons is going to attack this carbon right in here. And these electrons would then kick off onto my oxygen."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And it's going to proceed via an SN2 mechanism. Therefore, it's going to attack the least sterically hindered alkyl group, which is, of course, the alkyl group on the right. So a lone pair of electrons is going to attack this carbon right in here. And these electrons would then kick off onto my oxygen. So if I draw the result of that nucleophilic attack, I now have my ring over here. And I just turned everything into an OH group, right? Because now I have an oxygen with two lone pairs of electrons on it like that."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And these electrons would then kick off onto my oxygen. So if I draw the result of that nucleophilic attack, I now have my ring over here. And I just turned everything into an OH group, right? Because now I have an oxygen with two lone pairs of electrons on it like that. So I make phenol as one of my products here. And the bromine just added on the bromide anion, just added onto an ethyl group. So we're going to form ethyl bromide as our other product."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Because now I have an oxygen with two lone pairs of electrons on it like that. So I make phenol as one of my products here. And the bromine just added on the bromide anion, just added onto an ethyl group. So we're going to form ethyl bromide as our other product. So ethyl bromide would be the other product here. And let me just highlight those carbons, right? So these two carbons are these two carbons right there for my product."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to form ethyl bromide as our other product. So ethyl bromide would be the other product here. And let me just highlight those carbons, right? So these two carbons are these two carbons right there for my product. And it turns out that this reaction will stop at this point. So these are your two products, phenol and ethyl bromide. And the reaction doesn't really continue on."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So these two carbons are these two carbons right there for my product. And it turns out that this reaction will stop at this point. So these are your two products, phenol and ethyl bromide. And the reaction doesn't really continue on. And let's pretend like the reaction does continue on. And we'll see why it actually stops here. So if we were to continue on in our mechanism, we would next protonate the phenol, right?"}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And the reaction doesn't really continue on. And let's pretend like the reaction does continue on. And we'll see why it actually stops here. So if we were to continue on in our mechanism, we would next protonate the phenol, right? So we have HBr here again. So we have HBr, the addition of hydrobromic acid. We would protonate the phenol."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if we were to continue on in our mechanism, we would next protonate the phenol, right? So we have HBr here again. So we have HBr, the addition of hydrobromic acid. We would protonate the phenol. So if we protonated that phenol right there, we would get something that looks like this, with a plus 1 formal charge on our oxygen. And we would also form the bromide anion. And if this mechanism were to continue like usual, the bromide anion will function as our nucleophile."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We would protonate the phenol. So if we protonated that phenol right there, we would get something that looks like this, with a plus 1 formal charge on our oxygen. And we would also form the bromide anion. And if this mechanism were to continue like usual, the bromide anion will function as our nucleophile. And it would attack this carbon right here on our ring. But that's going to be a problem. That's going to be a problem if you're talking about an SN2 mechanism."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And if this mechanism were to continue like usual, the bromide anion will function as our nucleophile. And it would attack this carbon right here on our ring. But that's going to be a problem. That's going to be a problem if you're talking about an SN2 mechanism. Because the benzene ring has all this stuff that gets in the way to sterically hinder that bromide anion as a nucleophile. So because of that steric hindrance, the bromide anion cannot attack. So there's no SN2 mechanism here to form another alkyl halide."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That's going to be a problem if you're talking about an SN2 mechanism. Because the benzene ring has all this stuff that gets in the way to sterically hinder that bromide anion as a nucleophile. So because of that steric hindrance, the bromide anion cannot attack. So there's no SN2 mechanism here to form another alkyl halide. So that won't work. What about an SN1 type mechanism? So if this was an SN1 type mechanism, we would need to form a carbocation."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So there's no SN2 mechanism here to form another alkyl halide. So that won't work. What about an SN1 type mechanism? So if this was an SN1 type mechanism, we would need to form a carbocation. And so these electrons in here would kick off onto the oxygen. And we could form a carbocation that way. So it would be plus 1 charge on that carbon."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if this was an SN1 type mechanism, we would need to form a carbocation. And so these electrons in here would kick off onto the oxygen. And we could form a carbocation that way. So it would be plus 1 charge on that carbon. The problem with this carbocation is there's no resonance stabilization. We can't draw any resonance structures to help stabilize this. And so therefore, an SN1 mechanism won't work either."}, {"video_title": "Acidic cleavage of ethers Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So it would be plus 1 charge on that carbon. The problem with this carbocation is there's no resonance stabilization. We can't draw any resonance structures to help stabilize this. And so therefore, an SN1 mechanism won't work either. So no SN1 mechanism to turn everything into a second alkyl halide. No SN2 mechanism because of steric hindrance. And for those reasons, we are stuck with our phenol as our product and ethyl bromide as our other product."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as we see as we start getting into the galaxy and the universe, it just becomes almost impossible to imagine. But we'll at least give our best shot. So I think most of us watching this video know that this right here is Earth. And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half and go about 100 miles. And on the Earth, that would be about this far. It would be a speck that would look something like that."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just to get a sense of scale here, I think probably the largest distance that we can somehow relate to is about 100 miles. You can get into a car for an hour, hour and a half and go about 100 miles. And on the Earth, that would be about this far. It would be a speck that would look something like that. That is 100 miles. And also to get a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be maybe the speed of a bullet."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would be a speck that would look something like that. That is 100 miles. And also to get a bit of scale, let's think about a speed that at least we can kind of comprehend. And that would be maybe the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we can maybe kind of comprehend. It goes about, and there are different types of bullets depending on the type of gun and all of that, about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that would be maybe the speed of a bullet. Maybe we can't comprehend it, but I'll say this is the fastest thing that we can maybe kind of comprehend. It goes about, and there are different types of bullets depending on the type of gun and all of that, about 280 meters per second, which is about 1,000 kilometers per hour. And this is also roughly the speed of a jet. So just to give a sense of scale here, the Earth's circumference, so if you were to go around the planet, the Earth's circumference, just like that, is about 40,000 kilometers. It is 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the Earth."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is also roughly the speed of a jet. So just to give a sense of scale here, the Earth's circumference, so if you were to go around the planet, the Earth's circumference, just like that, is about 40,000 kilometers. It is 40,000 kilometers. So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the Earth. 40 hours to go around the Earth. And I think none of this information is too surprising. You might have taken 12 or 15 hour flights that get you not all the way around the Earth, but get you pretty far, San Francisco to Australia or something like that."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you were to travel at the speed of a bullet or the speed of a jetliner, at 1,000 kilometers an hour, it would take you 40 hours to circumnavigate the Earth. 40 hours to go around the Earth. And I think none of this information is too surprising. You might have taken 12 or 15 hour flights that get you not all the way around the Earth, but get you pretty far, San Francisco to Australia or something like that. So right now these aren't at scales that are too crazy, although, you know, even for me, even the Earth itself is a pretty mind-blowingly large object. Now, with that out of the way, let's think about the Sun, because the Sun starts to approach something far huger. So this obviously here is the Sun."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You might have taken 12 or 15 hour flights that get you not all the way around the Earth, but get you pretty far, San Francisco to Australia or something like that. So right now these aren't at scales that are too crazy, although, you know, even for me, even the Earth itself is a pretty mind-blowingly large object. Now, with that out of the way, let's think about the Sun, because the Sun starts to approach something far huger. So this obviously here is the Sun. And I think most people appreciate that the Sun is larger, that it's much larger than the Earth, and that it's pretty far away from the Earth. But I don't think most people, including myself, fully appreciate how large the Sun is or how far it is away from the Earth. So just to give you a sense, the Sun is 109 times the circumference of the Earth."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this obviously here is the Sun. And I think most people appreciate that the Sun is larger, that it's much larger than the Earth, and that it's pretty far away from the Earth. But I don't think most people, including myself, fully appreciate how large the Sun is or how far it is away from the Earth. So just to give you a sense, the Sun is 109 times the circumference of the Earth. So if we do that same thought exercise there, if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the Earth. Well, how long would it take to go around the Sun? So if you were to get on a jet plane and try to go around the Sun, or if you were to somehow ride a bullet and try to go around the Sun, do a complete circumnavigation of the Sun, it's going to take you 109 times as long as it would have taken you to do the Earth."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So just to give you a sense, the Sun is 109 times the circumference of the Earth. So if we do that same thought exercise there, if we said, OK, if I'm traveling at the speed of a bullet or the speed of a jetliner, it would take me 40 hours to go around the Earth. Well, how long would it take to go around the Sun? So if you were to get on a jet plane and try to go around the Sun, or if you were to somehow ride a bullet and try to go around the Sun, do a complete circumnavigation of the Sun, it's going to take you 109 times as long as it would have taken you to do the Earth. So it would be 100 times, I could do 109, but just for approximate. It's roughly 100 times the circumference of the Earth. So 109 times 40 is equal to 4,000 hours."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you were to get on a jet plane and try to go around the Sun, or if you were to somehow ride a bullet and try to go around the Sun, do a complete circumnavigation of the Sun, it's going to take you 109 times as long as it would have taken you to do the Earth. So it would be 100 times, I could do 109, but just for approximate. It's roughly 100 times the circumference of the Earth. So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is, actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the Earth times 40 hours. That's what it would take to do the circumference of the Earth."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So 109 times 40 is equal to 4,000 hours. And just to get a sense of what 4,000 is, actually, since I have the calculator out, let's do the exact calculation. It's 109 times the circumference of the Earth times 40 hours. That's what it would take to do the circumference of the Earth. So it's 4,360 hours to circumnavigate the Sun going at the speed of a bullet or a jetliner. And so that is 24 hours of the day. That is 181 days."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's what it would take to do the circumference of the Earth. So it's 4,360 hours to circumnavigate the Sun going at the speed of a bullet or a jetliner. And so that is 24 hours of the day. That is 181 days. It would take you roughly half a year. It would take half a year to go around the Sun at the speed of a jetliner. Let me write this down."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That is 181 days. It would take you roughly half a year. It would take half a year to go around the Sun at the speed of a jetliner. Let me write this down. Half a year. The Sun is huge. That by itself may or may not be... Actually, let me give you a sense of scale here because I have this other diagram of a Sun."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me write this down. Half a year. The Sun is huge. That by itself may or may not be... Actually, let me give you a sense of scale here because I have this other diagram of a Sun. We'll talk more about the rest of the solar system in the next video. But over here, at this scale, at the Sun, at least on my screen, if I were to complete it, it would probably be about 20 inches in diameter. The Earth is just this little thing over here, smaller than a raindrop."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That by itself may or may not be... Actually, let me give you a sense of scale here because I have this other diagram of a Sun. We'll talk more about the rest of the solar system in the next video. But over here, at this scale, at the Sun, at least on my screen, if I were to complete it, it would probably be about 20 inches in diameter. The Earth is just this little thing over here, smaller than a raindrop. It's just a small little thing over here. If I were to draw it on this scale where the Sun is even smaller, the Earth would be right about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade or we always see these diagrams of the solar system that look something like this, is that these planets are way further away."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The Earth is just this little thing over here, smaller than a raindrop. It's just a small little thing over here. If I were to draw it on this scale where the Sun is even smaller, the Earth would be right about that big. Now, what isn't obvious, because we've all done our science projects in third and fourth grade or we always see these diagrams of the solar system that look something like this, is that these planets are way further away. Even though these are depicted to scale, they're way further away from the Sun than this makes it look. The Earth is 150 million kilometers from the Sun. If this is the Sun right here, at this scale, you wouldn't even be able to see the Earth."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, what isn't obvious, because we've all done our science projects in third and fourth grade or we always see these diagrams of the solar system that look something like this, is that these planets are way further away. Even though these are depicted to scale, they're way further away from the Sun than this makes it look. The Earth is 150 million kilometers from the Sun. If this is the Sun right here, at this scale, you wouldn't even be able to see the Earth. It wouldn't even be a pixel, but it would be 150 million kilometers from the Earth. This distance right here is called an astronomical unit. We'll be using that term in the next few videos, just because it's an easier way to think about distance."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If this is the Sun right here, at this scale, you wouldn't even be able to see the Earth. It wouldn't even be a pixel, but it would be 150 million kilometers from the Earth. This distance right here is called an astronomical unit. We'll be using that term in the next few videos, just because it's an easier way to think about distance. Sometimes abbreviated AU, astronomical unit. Just to give a sense of how far this is, light, which is something that we think is almost infinitely fast, something that looks instantaneous, that takes eight minutes to travel from the Sun to the Earth. If the Sun were to disappear, it would take eight minutes for that light for us to know that it disappeared on Earth."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We'll be using that term in the next few videos, just because it's an easier way to think about distance. Sometimes abbreviated AU, astronomical unit. Just to give a sense of how far this is, light, which is something that we think is almost infinitely fast, something that looks instantaneous, that takes eight minutes to travel from the Sun to the Earth. If the Sun were to disappear, it would take eight minutes for that light for us to know that it disappeared on Earth. Or another way, just to put it in the sense of this jet airplane, let's get the calculator back out. We're talking about 150 million kilometers. If we're going at 1,000 kilometers an hour, it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the Sun."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If the Sun were to disappear, it would take eight minutes for that light for us to know that it disappeared on Earth. Or another way, just to put it in the sense of this jet airplane, let's get the calculator back out. We're talking about 150 million kilometers. If we're going at 1,000 kilometers an hour, it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the Sun. Just to put that in perspective, if we wanted it in days, there's 24 hours per day. This would be 6,250 days, or if we divide by 365, roughly 17 years. If you were to shoot a bullet straight at the Sun, it would take 17 years to get there, if it could maintain its velocity somehow."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we're going at 1,000 kilometers an hour, it would take us 150,000 hours at the speed of a bullet or at the speed of a jet plane to get to the Sun. Just to put that in perspective, if we wanted it in days, there's 24 hours per day. This would be 6,250 days, or if we divide by 365, roughly 17 years. If you were to shoot a bullet straight at the Sun, it would take 17 years to get there, if it could maintain its velocity somehow. This would take a bullet or a jet plane 17 years to get to the Sun. 17 years. Or another way to visualize it, this Sun right over here, it looks on my screen, it has about a 5 or 6 inch diameter."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to shoot a bullet straight at the Sun, it would take 17 years to get there, if it could maintain its velocity somehow. This would take a bullet or a jet plane 17 years to get to the Sun. 17 years. Or another way to visualize it, this Sun right over here, it looks on my screen, it has about a 5 or 6 inch diameter. If I were to actually do it as scale, this little dot right here, this little dot which is the Earth, this speck, if I actually wanted to draw this distance at scale, I would have to put this speck about 50 feet away from the Sun. 50 or 60 feet away from the Sun. If you were to look at the solar system, and obviously there's other things in the solar system, we'll talk more about them in the next video, you wouldn't even notice this speck."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or another way to visualize it, this Sun right over here, it looks on my screen, it has about a 5 or 6 inch diameter. If I were to actually do it as scale, this little dot right here, this little dot which is the Earth, this speck, if I actually wanted to draw this distance at scale, I would have to put this speck about 50 feet away from the Sun. 50 or 60 feet away from the Sun. If you were to look at the solar system, and obviously there's other things in the solar system, we'll talk more about them in the next video, you wouldn't even notice this speck. This is a little dust thing flying around this Sun. As we go further and further out of this solar system, you're going to see even this distance starts to become ridiculously small. Or another way to think about it, if the Sun was about this size, then this speck, then the Earth at this scale would be about 200 feet away from it."}, {"video_title": "Scale of earth and sun Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to look at the solar system, and obviously there's other things in the solar system, we'll talk more about them in the next video, you wouldn't even notice this speck. This is a little dust thing flying around this Sun. As we go further and further out of this solar system, you're going to see even this distance starts to become ridiculously small. Or another way to think about it, if the Sun was about this size, then this speck, then the Earth at this scale would be about 200 feet away from it. So you can imagine, if you had a football field, these are the end zones, one end zone, another end zone, and if you were to stick something maybe the size of a medicine ball, a little bit bigger than a basketball at one end zone, this little speck would be about 60 yards away, or roughly 60 meters away. So it's a little speck, you wouldn't even notice it on the scale of a football field, something this size. Anyway, I'm going to leave you there."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We've already seen how to form enolates from ketones, but what happens if you don't start with a symmetrical ketone? So right here, this ketone is not symmetrical. If we look at the right side, we have a methyl group. On the left side of our carbonyl, there's a CH2 and then an R group, an alkyl group. So this is different, right? And if we look at our alpha carbons, an alpha carbon is the one next to our carbonyl. So on the left side, this is an alpha carbon."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left side of our carbonyl, there's a CH2 and then an R group, an alkyl group. So this is different, right? And if we look at our alpha carbons, an alpha carbon is the one next to our carbonyl. So on the left side, this is an alpha carbon. On the right side, this is an alpha carbon. And so when we look at alpha protons, on this alpha carbon on the right, there would be three alpha protons. And on the alpha carbon on the left, there would be two alpha protons."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So on the left side, this is an alpha carbon. On the right side, this is an alpha carbon. And so when we look at alpha protons, on this alpha carbon on the right, there would be three alpha protons. And on the alpha carbon on the left, there would be two alpha protons. So the question is, which one of those alpha protons are we going to take with our base? And the answer lies in what kind of base we use and also reaction conditions. So if we use a base like LDA, lithium diisopropyl amide, we talked about this base in a previous video."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And on the alpha carbon on the left, there would be two alpha protons. So the question is, which one of those alpha protons are we going to take with our base? And the answer lies in what kind of base we use and also reaction conditions. So if we use a base like LDA, lithium diisopropyl amide, we talked about this base in a previous video. It's a very strong base, but it's also very sterically hindered. We have these big isopropyl groups, which means that LDA wants to approach our ketone from the least sterically hindered side. This left side here has our alkyl group, which is much bulkier than the hydrogen."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we use a base like LDA, lithium diisopropyl amide, we talked about this base in a previous video. It's a very strong base, but it's also very sterically hindered. We have these big isopropyl groups, which means that LDA wants to approach our ketone from the least sterically hindered side. This left side here has our alkyl group, which is much bulkier than the hydrogen. So this could interfere if it approached from the left side. So it's much more likely to approach from the right side and therefore take one of the alpha protons on the alpha carbon on the right. So let's say that this lone pair of electrons here takes this proton, which moves these electrons in here to form a double bond, kicks these electrons off onto our oxygen."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This left side here has our alkyl group, which is much bulkier than the hydrogen. So this could interfere if it approached from the left side. So it's much more likely to approach from the right side and therefore take one of the alpha protons on the alpha carbon on the right. So let's say that this lone pair of electrons here takes this proton, which moves these electrons in here to form a double bond, kicks these electrons off onto our oxygen. So we can go ahead and draw the enolate anion that would result. So we would have our oxygen up here with three lone pairs of electrons, negative 1 formal charge, a double bond over here on the right, and then we still have two hydrogens attached to that carbon. And so this is our enolate anion."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that this lone pair of electrons here takes this proton, which moves these electrons in here to form a double bond, kicks these electrons off onto our oxygen. So we can go ahead and draw the enolate anion that would result. So we would have our oxygen up here with three lone pairs of electrons, negative 1 formal charge, a double bond over here on the right, and then we still have two hydrogens attached to that carbon. And so this is our enolate anion. Let's follow those electrons. The electrons in here and magenta moved in to form our double bond. And then we could say that these electrons in here moved out onto our oxygen to form our enolate."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this is our enolate anion. Let's follow those electrons. The electrons in here and magenta moved in to form our double bond. And then we could say that these electrons in here moved out onto our oxygen to form our enolate. And since lithium is present, if you wanted to, you could put lithium here, Li plus, like that. So this is the enolate anion that we would get. And we call this the kinetic enolate."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we could say that these electrons in here moved out onto our oxygen to form our enolate. And since lithium is present, if you wanted to, you could put lithium here, Li plus, like that. So this is the enolate anion that we would get. And we call this the kinetic enolate. So let me go ahead and write that down here. This is the kinetic enolate. And it's called the kinetic enolate because this is the one that forms the fastest."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we call this the kinetic enolate. So let me go ahead and write that down here. This is the kinetic enolate. And it's called the kinetic enolate because this is the one that forms the fastest. So think about kinetic and speed. So this is the one that would form the fastest because we used LDA as our base, because of the choice of our base, and also because of probability. On the alpha carbon on the right, we had three alpha protons."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And it's called the kinetic enolate because this is the one that forms the fastest. So think about kinetic and speed. So this is the one that would form the fastest because we used LDA as our base, because of the choice of our base, and also because of probability. On the alpha carbon on the right, we had three alpha protons. So a greater chance of taking one of these, as opposed to these two over here on the left. And so the kinetic enolate is the one that forms the fastest. Let's look at forming another kind of enolate."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the alpha carbon on the right, we had three alpha protons. So a greater chance of taking one of these, as opposed to these two over here on the left. And so the kinetic enolate is the one that forms the fastest. Let's look at forming another kind of enolate. So let's think about deprotonating the alpha carbon on the left this time. So this is our alpha carbon over here with two alpha protons. Let's use a different base this time."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at forming another kind of enolate. So let's think about deprotonating the alpha carbon on the left this time. So this is our alpha carbon over here with two alpha protons. Let's use a different base this time. Let's use sodium hydride, so Na plus H minus. Or we could use potassium hydride. These are sources of hydride anions, which we know can act as a base."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's use a different base this time. Let's use sodium hydride, so Na plus H minus. Or we could use potassium hydride. These are sources of hydride anions, which we know can act as a base. So the hydride anion could take this proton, leaving these electrons in here, pushing these electrons off onto our oxygen. And we can draw a different enolate. We can show the double bond now is between those two carbons."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are sources of hydride anions, which we know can act as a base. So the hydride anion could take this proton, leaving these electrons in here, pushing these electrons off onto our oxygen. And we can draw a different enolate. We can show the double bond now is between those two carbons. And then we have our oxygen up here with a negative 1 formal charge. And then we have our methyl group over here. And then we left one hydrogen behind."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can show the double bond now is between those two carbons. And then we have our oxygen up here with a negative 1 formal charge. And then we have our methyl group over here. And then we left one hydrogen behind. And so we have a different enolate. Once again, following some electrons, electrons in magenta moved in here to form our double bond. And then the electrons in blue here moved out onto our oxygen."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we left one hydrogen behind. And so we have a different enolate. Once again, following some electrons, electrons in magenta moved in here to form our double bond. And then the electrons in blue here moved out onto our oxygen. So this is a different enolate. And we call this the thermodynamic enolate. So let me go ahead and write that."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then the electrons in blue here moved out onto our oxygen. So this is a different enolate. And we call this the thermodynamic enolate. So let me go ahead and write that. So the thermodynamic enolate. Let's analyze these two enolates that we've formed in terms of stability. And to do that, we need to look at the substitution at the double bond."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write that. So the thermodynamic enolate. Let's analyze these two enolates that we've formed in terms of stability. And to do that, we need to look at the substitution at the double bond. So going back up here to the kinetic enolate, I look at my double bond. I think about how substituted it is. We have these two hydrogens over here on this side."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And to do that, we need to look at the substitution at the double bond. So going back up here to the kinetic enolate, I look at my double bond. I think about how substituted it is. We have these two hydrogens over here on this side. And so that's actually not as substituted as the thermodynamic enolate. If we look at the thermodynamic enolate in the double bond, we only have one hydrogen here. We have an R group over here."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have these two hydrogens over here on this side. And so that's actually not as substituted as the thermodynamic enolate. If we look at the thermodynamic enolate in the double bond, we only have one hydrogen here. We have an R group over here. So we also have this alkyl group, if you're thinking about it. So this is actually the more stable enolate that forms. Because we know the more substituted the double bond, the more stable it is."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have an R group over here. So we also have this alkyl group, if you're thinking about it. So this is actually the more stable enolate that forms. Because we know the more substituted the double bond, the more stable it is. So the thermodynamic enolate is the more stable enolate. It's more substituted. The kinetic enolate is not as stable."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Because we know the more substituted the double bond, the more stable it is. So the thermodynamic enolate is the more stable enolate. It's more substituted. The kinetic enolate is not as stable. But it is the one that forms the fastest. And so once again, you can control which one of these enolates you form, depending on the base that you use. Let's look at a problem where we have a ketone."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The kinetic enolate is not as stable. But it is the one that forms the fastest. And so once again, you can control which one of these enolates you form, depending on the base that you use. Let's look at a problem where we have a ketone. Then we're going to add our two different bases to our ketone. So here's our ketone. And first, let's add some sodium hydride."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at a problem where we have a ketone. Then we're going to add our two different bases to our ketone. So here's our ketone. And first, let's add some sodium hydride. We know that sodium hydride puts our ketone under thermodynamic control. And so when we identify our alpha carbons, let's go ahead and do that. Alpha carbons are the ones next to our carbonyl."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And first, let's add some sodium hydride. We know that sodium hydride puts our ketone under thermodynamic control. And so when we identify our alpha carbons, let's go ahead and do that. Alpha carbons are the ones next to our carbonyl. So this would be an alpha carbon over here on the right. And then this would be an alpha carbon over here on the left. We think about how many alpha protons we have."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Alpha carbons are the ones next to our carbonyl. So this would be an alpha carbon over here on the right. And then this would be an alpha carbon over here on the left. We think about how many alpha protons we have. For the alpha carbon on the right, there's only one. And for the alpha carbon on the left, there would be two alpha protons. So if I think about thermodynamic control, I know that's going to form the more stable enolate, the thermodynamic enolate."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We think about how many alpha protons we have. For the alpha carbon on the right, there's only one. And for the alpha carbon on the left, there would be two alpha protons. So if I think about thermodynamic control, I know that's going to form the more stable enolate, the thermodynamic enolate. And so I know that hydride is going to take the proton on the right, leaving these electrons in here and pushing those electrons off onto your oxygen. So let's go ahead and draw this enolate. So we would push the electrons in here."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I think about thermodynamic control, I know that's going to form the more stable enolate, the thermodynamic enolate. And so I know that hydride is going to take the proton on the right, leaving these electrons in here and pushing those electrons off onto your oxygen. So let's go ahead and draw this enolate. So we would push the electrons in here. And then we would have our oxygen up here with three lone pairs of electrons, so negative 1 formal charge. And then we still have our methyl group right here, so CH3. So showing those electrons in magenta, moving in here to form our double bond."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we would push the electrons in here. And then we would have our oxygen up here with three lone pairs of electrons, so negative 1 formal charge. And then we still have our methyl group right here, so CH3. So showing those electrons in magenta, moving in here to form our double bond. And so this is the thermodynamic enolate. It's more substituted. So because of this methyl group here, this double bond is more substituted than if we showed an enolate forming from the other side."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So showing those electrons in magenta, moving in here to form our double bond. And so this is the thermodynamic enolate. It's more substituted. So because of this methyl group here, this double bond is more substituted than if we showed an enolate forming from the other side. So this is our thermodynamic enolate. What if we use something like LDA and we make our temperatures very cold, so negative 78 degrees Celsius? So this is going to take a proton from the least sterically hindered side, which we know is going to be the left side."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So because of this methyl group here, this double bond is more substituted than if we showed an enolate forming from the other side. So this is our thermodynamic enolate. What if we use something like LDA and we make our temperatures very cold, so negative 78 degrees Celsius? So this is going to take a proton from the least sterically hindered side, which we know is going to be the left side. Over here on the right side, we have this methyl group. So the LDA is going to approach from the left side. So let me go ahead and draw that in."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to take a proton from the least sterically hindered side, which we know is going to be the left side. Over here on the right side, we have this methyl group. So the LDA is going to approach from the left side. So let me go ahead and draw that in. So here we have our LDA, negative 1 formal charge on the nitrogen. So we could show a lone pair of electrons taking this proton, least sterically hindered one. These electrons move in here."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that in. So here we have our LDA, negative 1 formal charge on the nitrogen. So we could show a lone pair of electrons taking this proton, least sterically hindered one. These electrons move in here. These electrons kick off onto your oxygen. So let's go ahead and show the product of that. We would have our ring."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These electrons move in here. These electrons kick off onto your oxygen. So let's go ahead and show the product of that. We would have our ring. We would have our methyl group over here. We would now have a double bond here. And then we would have our oxygen with three lone pairs of electrons and a negative 1 formal charge."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We would have our ring. We would have our methyl group over here. We would now have a double bond here. And then we would have our oxygen with three lone pairs of electrons and a negative 1 formal charge. And again, if you want to draw your lithium in there, this would be the enolate anion that would form. So showing those electrons, let's go ahead and make them in blue this time. So these electrons in here moved in to form our double bond."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have our oxygen with three lone pairs of electrons and a negative 1 formal charge. And again, if you want to draw your lithium in there, this would be the enolate anion that would form. So showing those electrons, let's go ahead and make them in blue this time. So these electrons in here moved in to form our double bond. And so this is our kinetic enolate. So once again, we have our kinetic versus our thermodynamic. The kinetic enolate is not as substituted because we have a hydrogen here."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here moved in to form our double bond. And so this is our kinetic enolate. So once again, we have our kinetic versus our thermodynamic. The kinetic enolate is not as substituted because we have a hydrogen here. So it's not as substituted. It's not as stable. But it's favored by low temperatures and strong sterically hindered base like LDA."}, {"video_title": "Kinetic and thermodynamic enolates Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The kinetic enolate is not as substituted because we have a hydrogen here. So it's not as substituted. It's not as stable. But it's favored by low temperatures and strong sterically hindered base like LDA. So we're going to form this as our enolate using this base and these reaction conditions. At a little bit of a higher temperature and use of a non-sterically hindered base like sodium hydride, we're going to form the more stable enolate, the thermodynamic enolate. And so you can control which enolate you form, once again, based on the type of base that you use and also your reaction conditions."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the mechanism for the formation of this. So if we start with our acetaldehydes, we'll go ahead and draw it out here, and we look for our alpha carbon, right? It's the one right next to our carbonyl. So here is our alpha carbon. There are three alpha protons on that alpha carbon. I'm just gonna show one here, and then we're gonna have sodium hydroxide come along as act as a base. So we have the hydroxide anion, so negative one formal charge on the oxygen."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here is our alpha carbon. There are three alpha protons on that alpha carbon. I'm just gonna show one here, and then we're gonna have sodium hydroxide come along as act as a base. So we have the hydroxide anion, so negative one formal charge on the oxygen. It's gonna take this proton, right, leaving these electrons behind on our alpha carbon. So let's go ahead and show the carb anion that would result. All right, so we have our carbonyl, and then we have our hydrogen on the right side."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have the hydroxide anion, so negative one formal charge on the oxygen. It's gonna take this proton, right, leaving these electrons behind on our alpha carbon. So let's go ahead and show the carb anion that would result. All right, so we have our carbonyl, and then we have our hydrogen on the right side. On the left side, we have a lone pair of electrons on this carbon, so we have a carb anion. So the electrons in magenta here moved off onto this carbon. We can draw a resonance structure, right?"}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we have our carbonyl, and then we have our hydrogen on the right side. On the left side, we have a lone pair of electrons on this carbon, so we have a carb anion. So the electrons in magenta here moved off onto this carbon. We can draw a resonance structure, right? These electrons move into here. These electrons go off onto our oxygen, and we would form the oxyanion, right? So we have a double bond here, and then our oxygen up here would have three lone pairs of electrons, so negative one formal charge, and then a hydrogen."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can draw a resonance structure, right? These electrons move into here. These electrons go off onto our oxygen, and we would form the oxyanion, right? So we have a double bond here, and then our oxygen up here would have three lone pairs of electrons, so negative one formal charge, and then a hydrogen. So the electrons in magenta moved in here to form our double bond. And so we know that the negative charge is best supported on oxygen, so the oxyanion is probably the best way to show an enolate anion. However, for the aldol condensations, I'm always gonna use the carb anion as our enolate anion."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have a double bond here, and then our oxygen up here would have three lone pairs of electrons, so negative one formal charge, and then a hydrogen. So the electrons in magenta moved in here to form our double bond. And so we know that the negative charge is best supported on oxygen, so the oxyanion is probably the best way to show an enolate anion. However, for the aldol condensations, I'm always gonna use the carb anion as our enolate anion. It's actually a little bit easier for students to look at the mechanism that way. So we're gonna start with our carb anion. So I'm just gonna redraw it down here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "However, for the aldol condensations, I'm always gonna use the carb anion as our enolate anion. It's actually a little bit easier for students to look at the mechanism that way. So we're gonna start with our carb anion. So I'm just gonna redraw it down here. So here's our carb anion. And once you form your enolate anion, your enolate anion can act as a nucleophile, because this carbon here is negatively charged. And because we use something like sodium hydroxide, sodium hydroxide is not basic enough to completely form enolate anions, so only a small amount of your enolate anion is formed."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just gonna redraw it down here. So here's our carb anion. And once you form your enolate anion, your enolate anion can act as a nucleophile, because this carbon here is negatively charged. And because we use something like sodium hydroxide, sodium hydroxide is not basic enough to completely form enolate anions, so only a small amount of your enolate anion is formed. And so this can act as a nucleophile and attack another molecule of acetaldehyde. So I'm gonna go ahead and draw in another molecule of acetaldehyde. And I'm gonna draw it in a particular way."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And because we use something like sodium hydroxide, sodium hydroxide is not basic enough to completely form enolate anions, so only a small amount of your enolate anion is formed. And so this can act as a nucleophile and attack another molecule of acetaldehyde. So I'm gonna go ahead and draw in another molecule of acetaldehyde. And I'm gonna draw it in a particular way. I'm going to leave off the hydrogen, and I'm going to put it right next to this. It just helps me when I'm doing an aldol reaction here. All right, so if I know the carbonyl of my other molecule of acetaldehyde is polarized, right, the oxygen is partially negative, and the carbonyl carbon here is partially positive."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I'm gonna draw it in a particular way. I'm going to leave off the hydrogen, and I'm going to put it right next to this. It just helps me when I'm doing an aldol reaction here. All right, so if I know the carbonyl of my other molecule of acetaldehyde is polarized, right, the oxygen is partially negative, and the carbonyl carbon here is partially positive. So the carbonyl carbon is electrophilic. It wants electrons, and it's going to get electrons from our nucleophilic enolate anion. So these electrons here are going to attack our carbon, right, push these electrons off onto our oxygen."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, so if I know the carbonyl of my other molecule of acetaldehyde is polarized, right, the oxygen is partially negative, and the carbonyl carbon here is partially positive. So the carbonyl carbon is electrophilic. It wants electrons, and it's going to get electrons from our nucleophilic enolate anion. So these electrons here are going to attack our carbon, right, push these electrons off onto our oxygen. So let's go ahead and show the result of that. So on the left side, right, we would have our oxygen with three lone pairs of electrons, and then we form a carbon-carbon bond here. So that's extremely important."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons here are going to attack our carbon, right, push these electrons off onto our oxygen. So let's go ahead and show the result of that. So on the left side, right, we would have our oxygen with three lone pairs of electrons, and then we form a carbon-carbon bond here. So that's extremely important. And then we also have over here on the right our aldehydes. So let's show the movement of electrons, right. So these electrons right here, right, this is our carb anion, attack our carbonyl carbon, form this bond, right, a carbon-carbon bond-forming reaction to form an alkoxide intermediate."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's extremely important. And then we also have over here on the right our aldehydes. So let's show the movement of electrons, right. So these electrons right here, right, this is our carb anion, attack our carbonyl carbon, form this bond, right, a carbon-carbon bond-forming reaction to form an alkoxide intermediate. And so we could protonate our alkoxide with water, right. So water is right here. So we could show a lone pair of electrons picking up this proton."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here, right, this is our carb anion, attack our carbonyl carbon, form this bond, right, a carbon-carbon bond-forming reaction to form an alkoxide intermediate. And so we could protonate our alkoxide with water, right. So water is right here. So we could show a lone pair of electrons picking up this proton. And we can go ahead and draw our aldol product now. So we draw our aldol right here like that. So the formation of a carbon-carbon bond makes an aldol addition, and then connects an aldol condensation, a very important reaction in organic chemistry here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we could show a lone pair of electrons picking up this proton. And we can go ahead and draw our aldol product now. So we draw our aldol right here like that. So the formation of a carbon-carbon bond makes an aldol addition, and then connects an aldol condensation, a very important reaction in organic chemistry here. So there's our aldol product. All right, let's look at an aldol condensation, which is kind of a continuation of what we were just talking about. And you can form different products depending on reaction conditions."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the formation of a carbon-carbon bond makes an aldol addition, and then connects an aldol condensation, a very important reaction in organic chemistry here. So there's our aldol product. All right, let's look at an aldol condensation, which is kind of a continuation of what we were just talking about. And you can form different products depending on reaction conditions. So for example, here we're starting off with acetaldehyde again and sodium hydroxide and water. We're adding heat, so we're changing the reaction conditions. And so we're going to get a different product."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And you can form different products depending on reaction conditions. So for example, here we're starting off with acetaldehyde again and sodium hydroxide and water. We're adding heat, so we're changing the reaction conditions. And so we're going to get a different product. And so we call this an aldol condensation. So you can see the product that we get is an enal. So we get an enal."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to get a different product. And so we call this an aldol condensation. So you can see the product that we get is an enal. So we get an enal. We get a double bond here for the ene. And then we have an aldehyde. So also here referred to as alpha-beta unsaturated because this carbon right here is the alpha carbon."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we get an enal. We get a double bond here for the ene. And then we have an aldehyde. So also here referred to as alpha-beta unsaturated because this carbon right here is the alpha carbon. And this carbon is the beta carbon. And it's unsaturated because we formed a double bond here. So this is an aldol condensation."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So also here referred to as alpha-beta unsaturated because this carbon right here is the alpha carbon. And this carbon is the beta carbon. And it's unsaturated because we formed a double bond here. So this is an aldol condensation. And to form this enal product, we're going to start with the aldol that we just made in the previous reaction. So acetaldehyde with sodium hydroxide can form an aldol as we just saw. So let's go ahead and draw in the aldol product."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is an aldol condensation. And to form this enal product, we're going to start with the aldol that we just made in the previous reaction. So acetaldehyde with sodium hydroxide can form an aldol as we just saw. So let's go ahead and draw in the aldol product. So the same mechanism that we just discussed. And I should point out that we're doing a base-catalyzed aldol condensations here. So we have a hydrogen here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in the aldol product. So the same mechanism that we just discussed. And I should point out that we're doing a base-catalyzed aldol condensations here. So we have a hydrogen here. And then our aldol had an oxygen and then a hydrogen right here. So there's our aldol. And if we look at our aldol, the carbon right next to the carbonyl is our alpha carbon."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have a hydrogen here. And then our aldol had an oxygen and then a hydrogen right here. So there's our aldol. And if we look at our aldol, the carbon right next to the carbonyl is our alpha carbon. And there's still two acidic protons on that alpha carbon. So I'm just going to draw one in there. And then we can think about sodium hydroxide acting as a base and forming another enolate anion."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if we look at our aldol, the carbon right next to the carbonyl is our alpha carbon. And there's still two acidic protons on that alpha carbon. So I'm just going to draw one in there. And then we can think about sodium hydroxide acting as a base and forming another enolate anion. So I'm going to have sodium hydroxide come along. So OH minus. It's going to function as a base."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we can think about sodium hydroxide acting as a base and forming another enolate anion. So I'm going to have sodium hydroxide come along. So OH minus. It's going to function as a base. It's going to take this proton, leave these electrons behind on this carbon. So let's go ahead and draw that. So we would have our oxygen right here with two lone pairs of electrons."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's going to function as a base. It's going to take this proton, leave these electrons behind on this carbon. So let's go ahead and draw that. So we would have our oxygen right here with two lone pairs of electrons. We would have our carbon now with a lone pair of electrons making a carbanion. And then we have our carbonyl over here like that. So let's once again show those electrons."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our oxygen right here with two lone pairs of electrons. We would have our carbon now with a lone pair of electrons making a carbanion. And then we have our carbonyl over here like that. So let's once again show those electrons. These electrons are right in here forming our carbanion. And then to form our enal product, we can just think about these electrons moving in here to form a double bond and kicking these electrons off onto the oxygen to form hydroxide as a leaving group. So we can say that hydroxide leaves, and then we form our double bond."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's once again show those electrons. These electrons are right in here forming our carbanion. And then to form our enal product, we can just think about these electrons moving in here to form a double bond and kicking these electrons off onto the oxygen to form hydroxide as a leaving group. So we can say that hydroxide leaves, and then we form our double bond. So I can go ahead and show that. Let's say these electrons right here were the magenta ones to form our enal product. And so once again, I'm using the carbanion form of our resonance structure instead of the oxyanion."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can say that hydroxide leaves, and then we form our double bond. So I can go ahead and show that. Let's say these electrons right here were the magenta ones to form our enal product. And so once again, I'm using the carbanion form of our resonance structure instead of the oxyanion. So again, it's probably more appropriate to use the oxyanion. But it's just easier from a student perspective to use a carbanion to go ahead and form your enal product that way. So in terms of another product, this is the major product that we would form here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, I'm using the carbanion form of our resonance structure instead of the oxyanion. So again, it's probably more appropriate to use the oxyanion. But it's just easier from a student perspective to use a carbanion to go ahead and form your enal product that way. So in terms of another product, this is the major product that we would form here. This is the trans product. So we're thinking about this double bond. These hydrogens are on opposite sides."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in terms of another product, this is the major product that we would form here. This is the trans product. So we're thinking about this double bond. These hydrogens are on opposite sides. So this is the trans product. So that's the major one. You would form a small amount of the cis product as well."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "These hydrogens are on opposite sides. So this is the trans product. So that's the major one. You would form a small amount of the cis product as well. So let me go ahead and draw that in. But the cis product would be the minor product because it has more steric hindrance than the trans product. So this is the major product."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You would form a small amount of the cis product as well. So let me go ahead and draw that in. But the cis product would be the minor product because it has more steric hindrance than the trans product. So this is the major product. And I'm sure the one that your professor wants you to put on an exam question. So the trans product has decreased steric hindrance. All right, let's do one more of these aldol condensations."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is the major product. And I'm sure the one that your professor wants you to put on an exam question. So the trans product has decreased steric hindrance. All right, let's do one more of these aldol condensations. And so here I've drawn out the reaction. And so the question would be, write a mechanism, write a plausible mechanism for the formation of this enone. So we have an enone this time."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's do one more of these aldol condensations. And so here I've drawn out the reaction. And so the question would be, write a mechanism, write a plausible mechanism for the formation of this enone. So we have an enone this time. We're starting with a ketone. And so we have a double bond and then the ketone. So an enone product."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have an enone this time. We're starting with a ketone. And so we have a double bond and then the ketone. So an enone product. So we're going to use sodium hydroxide. And we're going to heat things up here. So let's think about what would happen."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So an enone product. So we're going to use sodium hydroxide. And we're going to heat things up here. So let's think about what would happen. So we have our ketone. And we have two alpha carbons, which are symmetrical. So it doesn't really matter which one we take a proton from."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about what would happen. So we have our ketone. And we have two alpha carbons, which are symmetrical. So it doesn't really matter which one we take a proton from. So I'm just going to say take a proton from this one. So we take a proton from here to form our enolate anion. And once again, I'm going to form the carb anion just to make things simpler."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it doesn't really matter which one we take a proton from. So I'm just going to say take a proton from this one. So we take a proton from here to form our enolate anion. And once again, I'm going to form the carb anion just to make things simpler. So we have our carb anion now. And then we have another molecule. Cyclohexanone is going to come along."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And once again, I'm going to form the carb anion just to make things simpler. So we have our carb anion now. And then we have another molecule. Cyclohexanone is going to come along. So let me go ahead and draw that in. And we know we have a polarized carbonyl situation, partial negative, partial positive. So our nucleophilic enolate anion is going to attack right here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Cyclohexanone is going to come along. So let me go ahead and draw that in. And we know we have a polarized carbonyl situation, partial negative, partial positive. So our nucleophilic enolate anion is going to attack right here. So we can show these electrons attacking right here, pushing those electrons off onto your oxygen. So I'm just going a little bit fast through this mechanism here. And so we have now a bond that formed."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So our nucleophilic enolate anion is going to attack right here. So we can show these electrons attacking right here, pushing those electrons off onto your oxygen. So I'm just going a little bit fast through this mechanism here. And so we have now a bond that formed. And now this oxygen right here is going to have three lone pairs of electrons around it, negative 1 formal charge. And I can go ahead and draw in the rest of my ring like that. So let's follow some of those electrons here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we have now a bond that formed. And now this oxygen right here is going to have three lone pairs of electrons around it, negative 1 formal charge. And I can go ahead and draw in the rest of my ring like that. So let's follow some of those electrons here. So I'll make these electrons in blue this time, forming the all-important carbon-carbon bonds right here. And then these electrons came off onto our oxygen, forming our alkoxide. So we could think about water protonating our alkoxide intermediate."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow some of those electrons here. So I'll make these electrons in blue this time, forming the all-important carbon-carbon bonds right here. And then these electrons came off onto our oxygen, forming our alkoxide. So we could think about water protonating our alkoxide intermediate. So lone pair takes that proton, leaves those electrons behind. And hopefully we have enough room to draw the aldol. So over here we would have our aldol intermediate."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we could think about water protonating our alkoxide intermediate. So lone pair takes that proton, leaves those electrons behind. And hopefully we have enough room to draw the aldol. So over here we would have our aldol intermediate. We have our OH. We have our ring like that. Thinking about what happens next, the carbon next to the carbonyl is the alpha carbon."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So over here we would have our aldol intermediate. We have our OH. We have our ring like that. Thinking about what happens next, the carbon next to the carbonyl is the alpha carbon. So that's our alpha carbon. And so there's still a proton on that alpha carbon. And I don't really have a lot of room here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Thinking about what happens next, the carbon next to the carbonyl is the alpha carbon. So that's our alpha carbon. And so there's still a proton on that alpha carbon. And I don't really have a lot of room here. But let me see if I can draw in an alpha proton like that. And so we could have hydroxide, once again, functioning as a base. So picking up this proton and then leaving these electrons behind on that carbon in red."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I don't really have a lot of room here. But let me see if I can draw in an alpha proton like that. And so we could have hydroxide, once again, functioning as a base. So picking up this proton and then leaving these electrons behind on that carbon in red. So let's get some more room here. So now what we would have is our ring. We have our carbonyl there."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So picking up this proton and then leaving these electrons behind on that carbon in red. So let's get some more room here. So now what we would have is our ring. We have our carbonyl there. And then we have a lone pair of electrons on this carbon, a negative 1 formal charge. Over here we have our OH like that. And then we have our ring."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have our carbonyl there. And then we have a lone pair of electrons on this carbon, a negative 1 formal charge. Over here we have our OH like that. And then we have our ring. So in the next step, once again, I'm not drawing an oxyanion. So I'm not going to draw the oxyanion form of this enolate anion here. But we could show these electrons moving into here and pushing these electrons off onto the oxygen."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our ring. So in the next step, once again, I'm not drawing an oxyanion. So I'm not going to draw the oxyanion form of this enolate anion here. But we could show these electrons moving into here and pushing these electrons off onto the oxygen. And so let's go ahead and draw our product once we do all of that. So we would have our ring. We would have our carbonyl here."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "But we could show these electrons moving into here and pushing these electrons off onto the oxygen. And so let's go ahead and draw our product once we do all of that. So we would have our ring. We would have our carbonyl here. And then we would form a double bond. And then now we can see that we have our product. So let's follow those electrons."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We would have our carbonyl here. And then we would form a double bond. And then now we can see that we have our product. So let's follow those electrons. So electrons in blue here, these electrons move in here to form our double bond. And then let's use red to show these electrons in here coming off. And so we get hydroxide as our leaving group."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons. So electrons in blue here, these electrons move in here to form our double bond. And then let's use red to show these electrons in here coming off. And so we get hydroxide as our leaving group. And we form our conjugated enone as our product. And so this is one way to represent the mechanism. And once again, the reaction conditions will determine what kind of product you get."}, {"video_title": "Aldol condensation Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we get hydroxide as our leaving group. And we form our conjugated enone as our product. And so this is one way to represent the mechanism. And once again, the reaction conditions will determine what kind of product you get. Our enone here is conjugated. We have an alternating single double bond situation. And so that favors this product here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we'll see how to synthesize alcohols using Grignard reagents. So first, we have to learn how to make a Grignard reagent. So you start with an alkyl halide, so over here on the left. And you add magnesium metal. And you need to add something like diethyl ether as your solvent. You can't have any water present, because water will react with the Grignard reagent. And so this is what you make over here on the right."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And you add magnesium metal. And you need to add something like diethyl ether as your solvent. You can't have any water present, because water will react with the Grignard reagent. And so this is what you make over here on the right. You end up with a carbon atom bonded to a metal. So carbon is bonded to magnesium. This is called an organometallic bond."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And so this is what you make over here on the right. You end up with a carbon atom bonded to a metal. So carbon is bonded to magnesium. This is called an organometallic bond. And you can do this with other metals. You can do this with lithium, for example. But Grignard reagents are one of those things that's always talked about in undergraduate organic chemistry classes."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "This is called an organometallic bond. And you can do this with other metals. You can do this with lithium, for example. But Grignard reagents are one of those things that's always talked about in undergraduate organic chemistry classes. And you can see that these two electrons here, these red ones, the ones in red, I've drawn it like a covalent bond, the bond between carbon and magnesium. But in reality, it's more ionic than covalent. So it's equivalent to the second structure down here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "But Grignard reagents are one of those things that's always talked about in undergraduate organic chemistry classes. And you can see that these two electrons here, these red ones, the ones in red, I've drawn it like a covalent bond, the bond between carbon and magnesium. But in reality, it's more ionic than covalent. So it's equivalent to the second structure down here. Now in terms of electronegativities, carbon is actually more electronegative than magnesium. So the two electrons in red are actually going to be closer to the carbon atom itself, giving the carbon a negative charge and forming a carbanion. So this is a carbanion that is formed."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So it's equivalent to the second structure down here. Now in terms of electronegativities, carbon is actually more electronegative than magnesium. So the two electrons in red are actually going to be closer to the carbon atom itself, giving the carbon a negative charge and forming a carbanion. So this is a carbanion that is formed. And this is unique, because this carbanion can now act as a nucleophile in your mechanism to make alcohols. So this is the preparation of a Grignard reagent. It's proved to be a very, very useful thing in organic synthesis, so much so that Victor Grignard won the Nobel Prize for his research into this chemistry."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this is a carbanion that is formed. And this is unique, because this carbanion can now act as a nucleophile in your mechanism to make alcohols. So this is the preparation of a Grignard reagent. It's proved to be a very, very useful thing in organic synthesis, so much so that Victor Grignard won the Nobel Prize for his research into this chemistry. Let's take a look at the mechanism to form a Grignard reagent. So I'm going to start with my alkyl halide. And this time, I'll draw in all of my lone pairs on my halogen like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "It's proved to be a very, very useful thing in organic synthesis, so much so that Victor Grignard won the Nobel Prize for his research into this chemistry. Let's take a look at the mechanism to form a Grignard reagent. So I'm going to start with my alkyl halide. And this time, I'll draw in all of my lone pairs on my halogen like that. And we're going to add magnesium, which we know, being in group two, magnesium has two valence electrons. I'm going to draw magnesium's two valence electrons like that. In the first step of the mechanism, magnesium is going to donate one of its electrons."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And this time, I'll draw in all of my lone pairs on my halogen like that. And we're going to add magnesium, which we know, being in group two, magnesium has two valence electrons. I'm going to draw magnesium's two valence electrons like that. In the first step of the mechanism, magnesium is going to donate one of its electrons. So we're going to show the movement of one of its electrons over here to this carbon. We're going to use a half-headed arrow like that. So we're going to make a new anion here, because this carbon actually picks up an electron."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "In the first step of the mechanism, magnesium is going to donate one of its electrons. So we're going to show the movement of one of its electrons over here to this carbon. We're going to use a half-headed arrow like that. So we're going to make a new anion here, because this carbon actually picks up an electron. So it actually picks up a negative charge. And we call this an anion radical. So this intermediate here is called an anion radical."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to make a new anion here, because this carbon actually picks up an electron. So it actually picks up a negative charge. And we call this an anion radical. So this intermediate here is called an anion radical. So I'll go ahead and write that. So it's an anion radical. It's an anion because it picked up an electron, giving it a negative charge."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this intermediate here is called an anion radical. So I'll go ahead and write that. So it's an anion radical. It's an anion because it picked up an electron, giving it a negative charge. And it's a radical because that electron is unpaired. So this anion radical is unstable, and it's going to fragment. So these two electrons right here are going to come off onto the halogen."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "It's an anion because it picked up an electron, giving it a negative charge. And it's a radical because that electron is unpaired. So this anion radical is unstable, and it's going to fragment. So these two electrons right here are going to come off onto the halogen. So let's go ahead and draw what that would give us. We now have this carbon with one electron around it on the right side like that. And now our halogen over here had three lone pairs."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So these two electrons right here are going to come off onto the halogen. So let's go ahead and draw what that would give us. We now have this carbon with one electron around it on the right side like that. And now our halogen over here had three lone pairs. It just picked up another. So it is now negatively charged like that. And the magnesium that we started with donated an electron."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And now our halogen over here had three lone pairs. It just picked up another. So it is now negatively charged like that. And the magnesium that we started with donated an electron. So this magnesium has one electron left around it, one valence electron. And it donated an electron, which gives it a plus 1 charge. So this magnesium is now positively charged, because it donated an electron in the first step."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And the magnesium that we started with donated an electron. So this magnesium has one electron left around it, one valence electron. And it donated an electron, which gives it a plus 1 charge. So this magnesium is now positively charged, because it donated an electron in the first step. And in the next step of the mechanism, magnesium can donate its second valence electron. And it's stable for it to do so, because then it'll have an electron configuration like a noble gas. So magnesium is going to go ahead and donate its second electron over here to the carbon like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this magnesium is now positively charged, because it donated an electron in the first step. And in the next step of the mechanism, magnesium can donate its second valence electron. And it's stable for it to do so, because then it'll have an electron configuration like a noble gas. So magnesium is going to go ahead and donate its second electron over here to the carbon like that. And let's go ahead and draw what would result. So we're going to now have our carbon right here with three bonds with now two electrons around it. And that's what gives it its negative 1 formal charge to make a carbanion."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So magnesium is going to go ahead and donate its second electron over here to the carbon like that. And let's go ahead and draw what would result. So we're going to now have our carbon right here with three bonds with now two electrons around it. And that's what gives it its negative 1 formal charge to make a carbanion. Magnesium has donated both of its electrons. So this magnesium is now a plus 2 charged cation, so Mg2 plus like that. And then the halogen is going to form an ionic bond with the magnesium on the right side here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And that's what gives it its negative 1 formal charge to make a carbanion. Magnesium has donated both of its electrons. So this magnesium is now a plus 2 charged cation, so Mg2 plus like that. And then the halogen is going to form an ionic bond with the magnesium on the right side here. So we have our halogen, which is negatively charged. So we have a magnesium with two positive charges. And then we have two things with negative charges around it forming ionic bonds."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And then the halogen is going to form an ionic bond with the magnesium on the right side here. So we have our halogen, which is negatively charged. So we have a magnesium with two positive charges. And then we have two things with negative charges around it forming ionic bonds. So this is our carbanion. And so I could redraw the stuff on the left with this carbon here. I can make it an R group with a lone pair of electrons, a negative 1 formal charge."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And then we have two things with negative charges around it forming ionic bonds. So this is our carbanion. And so I could redraw the stuff on the left with this carbon here. I can make it an R group with a lone pair of electrons, a negative 1 formal charge. And all the stuff on the right, I could just write it like this. There are several ways that you'll see this written. I could just say this is Mgx with a plus 1 charge."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "I can make it an R group with a lone pair of electrons, a negative 1 formal charge. And all the stuff on the right, I could just write it like this. There are several ways that you'll see this written. I could just say this is Mgx with a plus 1 charge. So this just allows us to focus in on this carbanion here with a negative 1 charge. Or I could just pretend like everything's covalent and just save myself some time. I could go like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "I could just say this is Mgx with a plus 1 charge. So this just allows us to focus in on this carbanion here with a negative 1 charge. Or I could just pretend like everything's covalent and just save myself some time. I could go like that. And organic chemists understand what this organometallic compound means, that the R group is negatively charged as a carbanion. So you'll see several different ways to write a Grignard reagent. Just as long as you understand what's going on, that's the most important thing."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "I could go like that. And organic chemists understand what this organometallic compound means, that the R group is negatively charged as a carbanion. So you'll see several different ways to write a Grignard reagent. Just as long as you understand what's going on, that's the most important thing. So let's now take the Grignard reagent we just formed and let's make an alcohol with it. So I'll go ahead and write it the last way I did. It doesn't really matter how you do it."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "Just as long as you understand what's going on, that's the most important thing. So let's now take the Grignard reagent we just formed and let's make an alcohol with it. So I'll go ahead and write it the last way I did. It doesn't really matter how you do it. So this is our generic reaction. We're going to introduce a carbonyl compound. So for this generic reaction, I'm just going to say it's some generic carbonyl."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't really matter how you do it. So this is our generic reaction. We're going to introduce a carbonyl compound. So for this generic reaction, I'm just going to say it's some generic carbonyl. So I'm not going to show what's attached to either side of my carbonyl carbon here. And diethyl ether, once again, is our solvent. You have to exclude water from this reaction, again, because the Grignard reagent will react with it."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So for this generic reaction, I'm just going to say it's some generic carbonyl. So I'm not going to show what's attached to either side of my carbonyl carbon here. And diethyl ether, once again, is our solvent. You have to exclude water from this reaction, again, because the Grignard reagent will react with it. So in the first step, you want it to react with your carbonyl. And the second step, once it's reacted with its carbonyl, it's OK to add water in the form of H3O+. And this is going to form our alcohol."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "You have to exclude water from this reaction, again, because the Grignard reagent will react with it. So in the first step, you want it to react with your carbonyl. And the second step, once it's reacted with its carbonyl, it's OK to add water in the form of H3O+. And this is going to form our alcohol. So we're going to form an alcohol as our product. And this carbon here came from our carbonyl. And this R group is going to attach to that carbonyl carbon."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And this is going to form our alcohol. So we're going to form an alcohol as our product. And this carbon here came from our carbonyl. And this R group is going to attach to that carbonyl carbon. That comes from our Grignard reagent. So it's a very useful reaction because it's a carbon-carbon bond forming reaction. So this R group had a carbon on the end, and then this carbon right here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And this R group is going to attach to that carbonyl carbon. That comes from our Grignard reagent. So it's a very useful reaction because it's a carbon-carbon bond forming reaction. So this R group had a carbon on the end, and then this carbon right here. So forming carbon-carbon bonds is very important when you're trying to build large organic molecules using synthesis. So this is a very useful way to form either a primary, secondary, or a tertiary alcohol. It all depends on what sort of carbonyl compound that you're starting with."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this R group had a carbon on the end, and then this carbon right here. So forming carbon-carbon bonds is very important when you're trying to build large organic molecules using synthesis. So this is a very useful way to form either a primary, secondary, or a tertiary alcohol. It all depends on what sort of carbonyl compound that you're starting with. So let's show the mechanism for what's happening. So I'm going to say the Grignard reagent is a source of carbanions. So I'm just going to draw my carbanion here like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "It all depends on what sort of carbonyl compound that you're starting with. So let's show the mechanism for what's happening. So I'm going to say the Grignard reagent is a source of carbanions. So I'm just going to draw my carbanion here like that. And then my carbonyl, I have carbon double-bonded to an oxygen. And I'm going to make this just anything could be attached to the carbon for right now. Once again, when you're looking at carbonyl chemistry, all you have to do is think about electronegativity differences between carbon and oxygen."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So I'm just going to draw my carbanion here like that. And then my carbonyl, I have carbon double-bonded to an oxygen. And I'm going to make this just anything could be attached to the carbon for right now. Once again, when you're looking at carbonyl chemistry, all you have to do is think about electronegativity differences between carbon and oxygen. So go back and watch the electronegativity video. We know that oxygen being more electronegative will draw these electrons in this double bond closer to it, giving it a partial negative charge, leaving our carbon partially positive. So carbon being partially positive, carbon wants electrons."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "Once again, when you're looking at carbonyl chemistry, all you have to do is think about electronegativity differences between carbon and oxygen. So go back and watch the electronegativity video. We know that oxygen being more electronegative will draw these electrons in this double bond closer to it, giving it a partial negative charge, leaving our carbon partially positive. So carbon being partially positive, carbon wants electrons. Carbon is an electrophile. And from our Grignard reagent, we have a nucleophile. The carbanion is going to act as a nucleophile."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So carbon being partially positive, carbon wants electrons. Carbon is an electrophile. And from our Grignard reagent, we have a nucleophile. The carbanion is going to act as a nucleophile. The negative charge is attracted to the positive charge. So the carbanion attacks the carbonyl carbon like that, which would kick these electrons off onto our oxygen. And we'll go ahead and draw the intermediate here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "The carbanion is going to act as a nucleophile. The negative charge is attracted to the positive charge. So the carbanion attacks the carbonyl carbon like that, which would kick these electrons off onto our oxygen. And we'll go ahead and draw the intermediate here. So we now have our R group directly attached to what used to be our carbonyl carbon. And now our oxygen has three lone pairs of electrons around it, which give our oxygen a negative 1 formal charge like that. So that's the first step of our reaction."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And we'll go ahead and draw the intermediate here. So we now have our R group directly attached to what used to be our carbonyl carbon. And now our oxygen has three lone pairs of electrons around it, which give our oxygen a negative 1 formal charge like that. So that's the first step of our reaction. And the second step, we have hydronium ions floating around. So H3O plus. So go ahead and draw H3O plus here like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So that's the first step of our reaction. And the second step, we have hydronium ions floating around. So H3O plus. So go ahead and draw H3O plus here like that. And the second step, of course, will be acid-base chemistry. So a lone pair of electrons on our oxygen takes a proton from H3O plus, leaving these electrons behind to form water. And that is how we get our alcohol."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So go ahead and draw H3O plus here like that. And the second step, of course, will be acid-base chemistry. So a lone pair of electrons on our oxygen takes a proton from H3O plus, leaving these electrons behind to form water. And that is how we get our alcohol. So that's going to give us our alcohol as our product after our acid-base reaction. Let's look at three different examples of synthesis of alcohols. And let's show the different types of alcohols that can be produced."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And that is how we get our alcohol. So that's going to give us our alcohol as our product after our acid-base reaction. Let's look at three different examples of synthesis of alcohols. And let's show the different types of alcohols that can be produced. So we'll start with formaldehyde here. So we'll start with a very simple molecule like that. And we'll go ahead and already make our Grignard reagent."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And let's show the different types of alcohols that can be produced. So we'll start with formaldehyde here. So we'll start with a very simple molecule like that. And we'll go ahead and already make our Grignard reagent. And we're going to make methyl magnesium bromide. So methyl magnesium bromide we're going to add in our first step. We're going to use ether as our solvent."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And we'll go ahead and already make our Grignard reagent. And we're going to make methyl magnesium bromide. So methyl magnesium bromide we're going to add in our first step. We're going to use ether as our solvent. And in our second step, we're going to add H3O plus. Now when you're analyzing a Grignard reagent, you pretty much have to think, where is my carbanion? So this carbon right here is negatively charged."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "We're going to use ether as our solvent. And in our second step, we're going to add H3O plus. Now when you're analyzing a Grignard reagent, you pretty much have to think, where is my carbanion? So this carbon right here is negatively charged. That's the carbon that's going to attack my carbonyl. So if this carbon attacks my carbonyl and the electrons kick off onto here as our intermediate, we would have hydrogens on either side of our carbon, a negatively charged oxygen up here. And the R group this time is a methyl group like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here is negatively charged. That's the carbon that's going to attack my carbonyl. So if this carbon attacks my carbonyl and the electrons kick off onto here as our intermediate, we would have hydrogens on either side of our carbon, a negatively charged oxygen up here. And the R group this time is a methyl group like that. So after you protonate it in the second step, so after these lone pair of electrons are going to pick up a proton from H3O plus, we would form this carbon with two hydrogens. We're going to protonate our alkoxide to form an alcohol up here for our product. And if you look at that molecule closely, you'll notice it is ethanol."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And the R group this time is a methyl group like that. So after you protonate it in the second step, so after these lone pair of electrons are going to pick up a proton from H3O plus, we would form this carbon with two hydrogens. We're going to protonate our alkoxide to form an alcohol up here for our product. And if you look at that molecule closely, you'll notice it is ethanol. So you make primary alcohols if you use formaldehyde. So this is a primary alcohol. It's primary because the carbon attached to the OH is attached to one other carbon."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And if you look at that molecule closely, you'll notice it is ethanol. So you make primary alcohols if you use formaldehyde. So this is a primary alcohol. It's primary because the carbon attached to the OH is attached to one other carbon. Let's see how we can make secondary alcohols. So this time, you need to start with an aldehyde. So instead of two hydrogens on either side of your carbon as we did before, this time you have to have an R group on one side."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "It's primary because the carbon attached to the OH is attached to one other carbon. Let's see how we can make secondary alcohols. So this time, you need to start with an aldehyde. So instead of two hydrogens on either side of your carbon as we did before, this time you have to have an R group on one side. So this will be our aldehyde like that. And let's go ahead and use the same Grignard reagent. We'll use methyl magnesium bromide again like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So instead of two hydrogens on either side of your carbon as we did before, this time you have to have an R group on one side. So this will be our aldehyde like that. And let's go ahead and use the same Grignard reagent. We'll use methyl magnesium bromide again like that. And second step, H3O plus. So once again, think about what is your nucleophile. So the negatively charged carbon is going to be my nucleophile."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "We'll use methyl magnesium bromide again like that. And second step, H3O plus. So once again, think about what is your nucleophile. So the negatively charged carbon is going to be my nucleophile. It's going to attack my carbonyl, kick these electrons off. So once again, when we draw the intermediate, up at the top here we have our alkoxide anion, negatively charged. The hydrogen is still there."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So the negatively charged carbon is going to be my nucleophile. It's going to attack my carbonyl, kick these electrons off. So once again, when we draw the intermediate, up at the top here we have our alkoxide anion, negatively charged. The hydrogen is still there. And what we did was we added a methyl group on. So this CH3 at the bottom of our intermediate came from our Grignard reagent. And once again, acid-base chemistry to protonate the alkoxide will form our secondary alcohol like that."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen is still there. And what we did was we added a methyl group on. So this CH3 at the bottom of our intermediate came from our Grignard reagent. And once again, acid-base chemistry to protonate the alkoxide will form our secondary alcohol like that. So that would be the secondary alcohol that is produced from this reaction. Once again, this carbon is attached to two other carbons, making this a secondary alcohol. So one more example here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And once again, acid-base chemistry to protonate the alkoxide will form our secondary alcohol like that. So that would be the secondary alcohol that is produced from this reaction. Once again, this carbon is attached to two other carbons, making this a secondary alcohol. So one more example here. This time we will react our Grignard reagent with a ketone. So we'll start with our ketone over here on the left. So here's our ketone."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So one more example here. This time we will react our Grignard reagent with a ketone. So we'll start with our ketone over here on the left. So here's our ketone. So it's cyclohexanone. And once again, let's stick with methyl magnesium bromide. So we have methyl magnesium bromide that we add."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So here's our ketone. So it's cyclohexanone. And once again, let's stick with methyl magnesium bromide. So we have methyl magnesium bromide that we add. And again, our solvent is ether, excluding water. And second step, we're going to add a source of protons, so H3O+. So once again, exact same mechanism, exact same thinking involved."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So we have methyl magnesium bromide that we add. And again, our solvent is ether, excluding water. And second step, we're going to add a source of protons, so H3O+. So once again, exact same mechanism, exact same thinking involved. Our negatively charged carbanion attacks our carbonyl carbon, kicking these electrons off onto our oxygen. So we form our intermediate. So this top oxygen here now has three lone pairs of electrons, negatively charged."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So once again, exact same mechanism, exact same thinking involved. Our negatively charged carbanion attacks our carbonyl carbon, kicking these electrons off onto our oxygen. So we form our intermediate. So this top oxygen here now has three lone pairs of electrons, negatively charged. And actually, let me go ahead and take that off there so we can better show the atoms attached to that carbonyl carbon. So if I'm showing that methyl group attacking that carbonyl, I'm going to push the alkoxide over here a little bit to the left. And that's going to get my negative charge."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this top oxygen here now has three lone pairs of electrons, negatively charged. And actually, let me go ahead and take that off there so we can better show the atoms attached to that carbonyl carbon. So if I'm showing that methyl group attacking that carbonyl, I'm going to push the alkoxide over here a little bit to the left. And that's going to get my negative charge. And that way, that just gives me some space to put my methyl group right here, like that. And that just looks a little bit closer to what my final product will look like. So this lone pair of electrons, as usual, this lone pair, one of these lone pairs is going to pick up a proton right here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "And that's going to get my negative charge. And that way, that just gives me some space to put my methyl group right here, like that. And that just looks a little bit closer to what my final product will look like. So this lone pair of electrons, as usual, this lone pair, one of these lone pairs is going to pick up a proton right here. And that's going to form our product. So we now protonate our alkoxide to form our alcohol, like that. So our alcohol is going to form right here."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So this lone pair of electrons, as usual, this lone pair, one of these lone pairs is going to pick up a proton right here. And that's going to form our product. So we now protonate our alkoxide to form our alcohol, like that. So our alcohol is going to form right here. And then our methyl group adds on right here. And if you look closely, you can see this is a tertiary alcohol that we just made. So this carbon is connected to one, two, three other carbons."}, {"video_title": "Synthesis of alcohols using Grignard reagents I Organic chemistry Khan Academy.mp3", "Sentence": "So our alcohol is going to form right here. And then our methyl group adds on right here. And if you look closely, you can see this is a tertiary alcohol that we just made. So this carbon is connected to one, two, three other carbons. So Grignard reagents are very useful for making alcohols. And you can make either primary, secondary, or tertiary alcohols from them. So it's a very versatile reagent to use."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Now that we understand hybridization states, let's do a couple of examples. We're going to identify the hybridization states and predict the geometries for all the atoms in this molecule, except for hydrogen. Let's start with this carbon right here. The fast way of identifying a hybridization state is to say, okay, that carbon has a double bond to it, therefore it must be sp2 hybridized. If it's sp2 hybridized, we know the geometry around that carbon must be trigonal planar, with bond angles approximately 120 degrees. This carbon over here also has a double bond to it, so it's also sp2 hybridized with trigonal planar geometry. Let's move to this carbon right here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The fast way of identifying a hybridization state is to say, okay, that carbon has a double bond to it, therefore it must be sp2 hybridized. If it's sp2 hybridized, we know the geometry around that carbon must be trigonal planar, with bond angles approximately 120 degrees. This carbon over here also has a double bond to it, so it's also sp2 hybridized with trigonal planar geometry. Let's move to this carbon right here. That carbon has only single bonds around it. The fast way of doing it is if you see all single bonds, it must be sp3 hybridized. If that carbon is sp3 hybridized, we know the geometry is tetrahedral, so tetrahedral geometry, with an ideal bond angles of 109.5 degrees around that carbon."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Let's move to this carbon right here. That carbon has only single bonds around it. The fast way of doing it is if you see all single bonds, it must be sp3 hybridized. If that carbon is sp3 hybridized, we know the geometry is tetrahedral, so tetrahedral geometry, with an ideal bond angles of 109.5 degrees around that carbon. Let's move over to this carbon right here. This carbon has a triple bond on the right side of it. The fast way of doing this is if it has a triple bond, it must be sp hybridized here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "If that carbon is sp3 hybridized, we know the geometry is tetrahedral, so tetrahedral geometry, with an ideal bond angles of 109.5 degrees around that carbon. Let's move over to this carbon right here. This carbon has a triple bond on the right side of it. The fast way of doing this is if it has a triple bond, it must be sp hybridized here. Therefore, the geometry would be linear, with a bond angle of 180 degrees. Same with this carbon. This carbon has a triple bond to it, so it also must be sp hybridized with linear geometry."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The fast way of doing this is if it has a triple bond, it must be sp hybridized here. Therefore, the geometry would be linear, with a bond angle of 180 degrees. Same with this carbon. This carbon has a triple bond to it, so it also must be sp hybridized with linear geometry. That's why I drew it this way, so it's linear around those two carbons here. Let's go ahead and count up the total number of sigma and pi bonds for this. That's also something we talked about in the previous videos here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "This carbon has a triple bond to it, so it also must be sp hybridized with linear geometry. That's why I drew it this way, so it's linear around those two carbons here. Let's go ahead and count up the total number of sigma and pi bonds for this. That's also something we talked about in the previous videos here. First, let's count the number of sigma bonds. Let's go back over to start with this carbon here. Here's a sigma bond to that carbon."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "That's also something we talked about in the previous videos here. First, let's count the number of sigma bonds. Let's go back over to start with this carbon here. Here's a sigma bond to that carbon. Here's a sigma bond to that carbon. We know that a double bond, one of those bonds is a sigma bond, and one of those bonds is a pi bond. Let me go ahead and also draw in our pi bonds in red."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Here's a sigma bond to that carbon. Here's a sigma bond to that carbon. We know that a double bond, one of those bonds is a sigma bond, and one of those bonds is a pi bond. Let me go ahead and also draw in our pi bonds in red. I already colored the sigma bond blue, so let's say this one is the pi bond. Let's continue assigning all of our bonds here. I know this single bond is a sigma bond."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Let me go ahead and also draw in our pi bonds in red. I already colored the sigma bond blue, so let's say this one is the pi bond. Let's continue assigning all of our bonds here. I know this single bond is a sigma bond. I know this single bond is a sigma bond. All of these single bonds here are sigma. When I get to the triple bond, I know one of those is a sigma bond, and two of those are pi bonds."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "I know this single bond is a sigma bond. I know this single bond is a sigma bond. All of these single bonds here are sigma. When I get to the triple bond, I know one of those is a sigma bond, and two of those are pi bonds. Two of those are pi bonds here. Then finally, I have one more bond. It's a single bond, so I know that it is a sigma bond here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "When I get to the triple bond, I know one of those is a sigma bond, and two of those are pi bonds. Two of those are pi bonds here. Then finally, I have one more bond. It's a single bond, so I know that it is a sigma bond here. If you count up all of those sigma bonds, you should get 10. Let's do that really quickly. Let me go ahead and change colors here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "It's a single bond, so I know that it is a sigma bond here. If you count up all of those sigma bonds, you should get 10. Let's do that really quickly. Let me go ahead and change colors here. You get one, two, three, four, five, six, seven, eight, nine, and 10. We have 10 sigma bonds total. In terms of pi bonds, we had three pi bonds, so three pi bonds for this molecule."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Let me go ahead and change colors here. You get one, two, three, four, five, six, seven, eight, nine, and 10. We have 10 sigma bonds total. In terms of pi bonds, we had three pi bonds, so three pi bonds for this molecule. You can also find hybridization states using a steric number. Let's go ahead and do that really quickly. Let's go back to this carbon, and let's find the hybridization state of that carbon using steric number."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "In terms of pi bonds, we had three pi bonds, so three pi bonds for this molecule. You can also find hybridization states using a steric number. Let's go ahead and do that really quickly. Let's go back to this carbon, and let's find the hybridization state of that carbon using steric number. Let's use green for this. Steric number is equal to the number of sigma bonds plus lone pairs of electrons. One, two, three sigma bonds around that carbon, so three plus zero gives me a steric number of three."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Let's go back to this carbon, and let's find the hybridization state of that carbon using steric number. Let's use green for this. Steric number is equal to the number of sigma bonds plus lone pairs of electrons. One, two, three sigma bonds around that carbon, so three plus zero gives me a steric number of three. Therefore, I need three hybrid orbitals, and sp2 hybridization gives me three hybrid orbitals. If I wanted to do for this carbon, I would have one, two, three, four. The steric number would be equal to four sigma bonds and zero lone pairs of electrons, giving me a total of four for my steric number."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "One, two, three sigma bonds around that carbon, so three plus zero gives me a steric number of three. Therefore, I need three hybrid orbitals, and sp2 hybridization gives me three hybrid orbitals. If I wanted to do for this carbon, I would have one, two, three, four. The steric number would be equal to four sigma bonds and zero lone pairs of electrons, giving me a total of four for my steric number. I need four hybrid orbitals. I have four sp3 hybridized orbitals at that carbon. Then finally, let's do it for this carbon right here, so using steric number."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The steric number would be equal to four sigma bonds and zero lone pairs of electrons, giving me a total of four for my steric number. I need four hybrid orbitals. I have four sp3 hybridized orbitals at that carbon. Then finally, let's do it for this carbon right here, so using steric number. Steric number is equal to number of sigma bonds plus numbers of lone pairs of electrons. There are two sigma bonds around that carbon, zero lone pairs of electrons, steric number of two. It means I need two hybridized orbitals, and in sp hybridization, that's what you get."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Then finally, let's do it for this carbon right here, so using steric number. Steric number is equal to number of sigma bonds plus numbers of lone pairs of electrons. There are two sigma bonds around that carbon, zero lone pairs of electrons, steric number of two. It means I need two hybridized orbitals, and in sp hybridization, that's what you get. You get two sp hybridized orbitals like that. All right, let's move on to another example. Let's do a similar analysis."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "It means I need two hybridized orbitals, and in sp hybridization, that's what you get. You get two sp hybridized orbitals like that. All right, let's move on to another example. Let's do a similar analysis. Before we do, notice I excluded hydrogen here, and that's because hydrogen is only bonded to one other atom, so there's no real geometry to talk about. All right, let's move on to this example. This molecule is diethyl ether."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Let's do a similar analysis. Before we do, notice I excluded hydrogen here, and that's because hydrogen is only bonded to one other atom, so there's no real geometry to talk about. All right, let's move on to this example. This molecule is diethyl ether. Let's start with this carbon right here, so the hybridization state. The fast way of doing it is to notice that there are only single bonds around that carbon, only sigma bonds. Therefore, we know that carbon is sp3 hybridized with tetrahedral geometry."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "This molecule is diethyl ether. Let's start with this carbon right here, so the hybridization state. The fast way of doing it is to notice that there are only single bonds around that carbon, only sigma bonds. Therefore, we know that carbon is sp3 hybridized with tetrahedral geometry. So sp3 hybridized tetrahedral geometry. All right, let's look at this carbon right here. It's the exact same situation, right?"}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Therefore, we know that carbon is sp3 hybridized with tetrahedral geometry. So sp3 hybridized tetrahedral geometry. All right, let's look at this carbon right here. It's the exact same situation, right? Only sigma or single bonds around it, so this carbon is also sp3 hybridized, and so therefore tetrahedral geometry. Let's next look at the oxygen here. If I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "It's the exact same situation, right? Only sigma or single bonds around it, so this carbon is also sp3 hybridized, and so therefore tetrahedral geometry. Let's next look at the oxygen here. If I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here. Let's go ahead and calculate the steric number of this oxygen, so that's number of sigma bonds. Here's a single bond, so that's a sigma bond. Then here's another one, so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "If I wanted to figure out the hybridization and the geometry of this oxygen, steric number is useful here. Let's go ahead and calculate the steric number of this oxygen, so that's number of sigma bonds. Here's a single bond, so that's a sigma bond. Then here's another one, so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom. Here's a lone pair of electrons, and here's a lone pair of electrons. I have two lone pairs of electrons, so two plus two gives me a steric number of four. I need four hybridized orbitals for this oxygen."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Then here's another one, so I have two sigma bonds, so two plus number of lone pairs of electrons around the atom. Here's a lone pair of electrons, and here's a lone pair of electrons. I have two lone pairs of electrons, so two plus two gives me a steric number of four. I need four hybridized orbitals for this oxygen. We know that occurs when you have sp3 hybridization. Therefore, this oxygen is sp3 hybridized. There are four sp3 hybrid orbitals around that oxygen."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "I need four hybridized orbitals for this oxygen. We know that occurs when you have sp3 hybridization. Therefore, this oxygen is sp3 hybridized. There are four sp3 hybrid orbitals around that oxygen. Let's do geometry of this oxygen. The electron groups, there are four electron groups around that oxygen, so each electron group is in an sp3 hybridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "There are four sp3 hybrid orbitals around that oxygen. Let's do geometry of this oxygen. The electron groups, there are four electron groups around that oxygen, so each electron group is in an sp3 hybridized orbital. The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here. The geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is bent. Even though that oxygen is sp3 hybridized, its geometry is not tetrahedral. The geometry of that oxygen there is bent or angular."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The geometry of those electron groups might be tetrahedral, but not the geometry around the oxygen here. The geometry around the oxygen, if you ignore the lone pairs of electrons, you can see that it is bent. Even though that oxygen is sp3 hybridized, its geometry is not tetrahedral. The geometry of that oxygen there is bent or angular. Because of symmetry, this carbon right here is the same as this carbon, so it's also sp3 hybridized. Then this carbon over here is the same as this carbon, so it's also sp3 hybridized. Symmetry made our lives easy on this one."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The geometry of that oxygen there is bent or angular. Because of symmetry, this carbon right here is the same as this carbon, so it's also sp3 hybridized. Then this carbon over here is the same as this carbon, so it's also sp3 hybridized. Symmetry made our lives easy on this one. Let's do one more example. Once again, our goal is to find the hybridization states and the geometries for all the atoms except for hydrogen. Once again, let's start with carbon."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Symmetry made our lives easy on this one. Let's do one more example. Once again, our goal is to find the hybridization states and the geometries for all the atoms except for hydrogen. Once again, let's start with carbon. Let's start with this carbon right here. Once again, our goal is to find the hybridization states. The fast way of doing it is to notice that there's one double bond to that carbon, so it must be sp2 hybridized, and therefore the geometry is trigonal planar."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Once again, let's start with carbon. Let's start with this carbon right here. Once again, our goal is to find the hybridization states. The fast way of doing it is to notice that there's one double bond to that carbon, so it must be sp2 hybridized, and therefore the geometry is trigonal planar. So trigonal planar geometry. Let's do the steric number way. If I were to calculate the steric number, steric number is equal to number of sigma bonds."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The fast way of doing it is to notice that there's one double bond to that carbon, so it must be sp2 hybridized, and therefore the geometry is trigonal planar. So trigonal planar geometry. Let's do the steric number way. If I were to calculate the steric number, steric number is equal to number of sigma bonds. Here's a sigma bond. Here's a sigma bond. I have a double bond between the carbon and the oxygen, so one of those is a sigma bond, and one of those is a pi bond, which I'll draw in red here."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "If I were to calculate the steric number, steric number is equal to number of sigma bonds. Here's a sigma bond. Here's a sigma bond. I have a double bond between the carbon and the oxygen, so one of those is a sigma bond, and one of those is a pi bond, which I'll draw in red here. I have three sigma bonds around that carbon, so three plus zero lone pairs of electrons. Gives me a steric number of three, so I need three hybridized orbitals, and so once again, sp2 hybridization. All right, let's do the next carbon."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "I have a double bond between the carbon and the oxygen, so one of those is a sigma bond, and one of those is a pi bond, which I'll draw in red here. I have three sigma bonds around that carbon, so three plus zero lone pairs of electrons. Gives me a steric number of three, so I need three hybridized orbitals, and so once again, sp2 hybridization. All right, let's do the next carbon. Let's move on to this one. I see only single bonds around that carbon. Therefore, it must be sp3 hybridized with tetrahedral geometry."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "All right, let's do the next carbon. Let's move on to this one. I see only single bonds around that carbon. Therefore, it must be sp3 hybridized with tetrahedral geometry. So sp3 hybridized tetrahedral geometry. Same thing for this carbon. Only single bonds around it, only sigma bonds, so it's sp3 hybridized with tetrahedral geometry."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Therefore, it must be sp3 hybridized with tetrahedral geometry. So sp3 hybridized tetrahedral geometry. Same thing for this carbon. Only single bonds around it, only sigma bonds, so it's sp3 hybridized with tetrahedral geometry. Let's finally look at this nitrogen here. If I want to find the hybridization state of this nitrogen, I could use steric number. The steric number is equal to number of sigma bonds."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Only single bonds around it, only sigma bonds, so it's sp3 hybridized with tetrahedral geometry. Let's finally look at this nitrogen here. If I want to find the hybridization state of this nitrogen, I could use steric number. The steric number is equal to number of sigma bonds. Around this nitrogen, here's a sigma bond. It's a single bond. Here's another one, and here's another one, so I have three sigma bonds."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "The steric number is equal to number of sigma bonds. Around this nitrogen, here's a sigma bond. It's a single bond. Here's another one, and here's another one, so I have three sigma bonds. I have one lone pair of electrons. So three plus one gives me four. A steric number of four means I need four hybridized orbitals and that's our situation with sp3 hybridization."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "Here's another one, and here's another one, so I have three sigma bonds. I have one lone pair of electrons. So three plus one gives me four. A steric number of four means I need four hybridized orbitals and that's our situation with sp3 hybridization. And so this nitrogen is sp3 hybridized, but its geometry is not tetrahedral. So the geometry for that nitrogen, as we discussed in an earlier video, it has these three sigma bonds like this and a lone pair of electrons. And that lone pair of electrons is an sp3 hybridized orbital."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "A steric number of four means I need four hybridized orbitals and that's our situation with sp3 hybridization. And so this nitrogen is sp3 hybridized, but its geometry is not tetrahedral. So the geometry for that nitrogen, as we discussed in an earlier video, it has these three sigma bonds like this and a lone pair of electrons. And that lone pair of electrons is an sp3 hybridized orbital. And if we look at that geometry and ignore the lone pair of electrons, because you always ignore the lone pairs of electrons when you're looking at geometry, we can see we have this sort of shape here. So the nitrogen's bonded to three atoms, the carbon, hydrogen, and hydrogen. And then we have this sort of a shape like that."}, {"video_title": "Worked examples Finding the hybridization of atoms in organic molecules Khan Academy.mp3", "Sentence": "And that lone pair of electrons is an sp3 hybridized orbital. And if we look at that geometry and ignore the lone pair of electrons, because you always ignore the lone pairs of electrons when you're looking at geometry, we can see we have this sort of shape here. So the nitrogen's bonded to three atoms, the carbon, hydrogen, and hydrogen. And then we have this sort of a shape like that. So in the back there, and you can see, we call this trigonal pyramidal. So the geometry around that nitrogen is trigonal pyramidal. All right, so that does it for three examples of organic hybridization."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are stereoisomers that are non-superimposable mirror images of each other. And they have opposite configurations at all chirality centers. Diastereomers are also stereoisomers, but these are stereoisomers that are non-superimposable, non-mirror images of each other. So these are stereoisomers that are not enantiomers. And diastereomers have opposite configurations at some chirality centers. If we look at this compound up here, we have a cyclopropane ring with a bromine coming off and a chlorine coming off. We know from earlier videos that there are two chirality centers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these are stereoisomers that are not enantiomers. And diastereomers have opposite configurations at some chirality centers. If we look at this compound up here, we have a cyclopropane ring with a bromine coming off and a chlorine coming off. We know from earlier videos that there are two chirality centers. So this carbon is a chiral center and so is this one. The total number of stereoisomers is two to the n, where n is equal to the number of chiral centers. And since n is equal to two for this drawing, we would expect to be able to draw two to the second power, or four stereoisomers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know from earlier videos that there are two chirality centers. So this carbon is a chiral center and so is this one. The total number of stereoisomers is two to the n, where n is equal to the number of chiral centers. And since n is equal to two for this drawing, we would expect to be able to draw two to the second power, or four stereoisomers. So there should be a total of four stereoisomers. Actually, two to the n is just a maximum, and we'll talk about that in later videos. So let's draw all four stereoisomers and let's look at the relationship between them."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And since n is equal to two for this drawing, we would expect to be able to draw two to the second power, or four stereoisomers. So there should be a total of four stereoisomers. Actually, two to the n is just a maximum, and we'll talk about that in later videos. So let's draw all four stereoisomers and let's look at the relationship between them. Let's think about how to draw our four stereoisomers. For the first one, we could have both halogens coming out at us in space. So I put the bromine on a wedge and I put the chlorine on a wedge."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw all four stereoisomers and let's look at the relationship between them. Let's think about how to draw our four stereoisomers. For the first one, we could have both halogens coming out at us in space. So I put the bromine on a wedge and I put the chlorine on a wedge. Next, we could have both halogens going away from us in space. So I put the bromine on a dash and same with the chlorine. Next, for our third stereoisomer, we could have one halogen coming out at us, so I'll make that the bromine, and one going away from us."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I put the bromine on a wedge and I put the chlorine on a wedge. Next, we could have both halogens going away from us in space. So I put the bromine on a dash and same with the chlorine. Next, for our third stereoisomer, we could have one halogen coming out at us, so I'll make that the bromine, and one going away from us. And for the last one, we could just reverse it. We could have the bromine going away from us and the chlorine coming out at us. Next, let's look at the relationships between our stereoisomers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next, for our third stereoisomer, we could have one halogen coming out at us, so I'll make that the bromine, and one going away from us. And for the last one, we could just reverse it. We could have the bromine going away from us and the chlorine coming out at us. Next, let's look at the relationships between our stereoisomers. Let's start with the relationship between stereoisomer one and stereoisomer two. Model sets really help in stereochemistry, so we're gonna look at videos for a lot of these. Let's look at the video comparing stereoisomer one and stereoisomer two."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at the relationships between our stereoisomers. Let's start with the relationship between stereoisomer one and stereoisomer two. Model sets really help in stereochemistry, so we're gonna look at videos for a lot of these. Let's look at the video comparing stereoisomer one and stereoisomer two. On the left, we have stereoisomer one. Both halogens are coming out at us in space. On the right is stereoisomer two, where both halogens are going away from us in space."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the video comparing stereoisomer one and stereoisomer two. On the left, we have stereoisomer one. Both halogens are coming out at us in space. On the right is stereoisomer two, where both halogens are going away from us in space. If I hold these two stereoisomers next to each other and I rotate the one on the right, we can see they're actually mirror images of each other, and they're non-superimposable mirror images. If I line up the chlorines, then the bromines are not in the right position. And if I try to line up the bromines, now the chlorines are not in the right position."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the right is stereoisomer two, where both halogens are going away from us in space. If I hold these two stereoisomers next to each other and I rotate the one on the right, we can see they're actually mirror images of each other, and they're non-superimposable mirror images. If I line up the chlorines, then the bromines are not in the right position. And if I try to line up the bromines, now the chlorines are not in the right position. So these are non-superimposable mirror images of each other. These are enantiomers. So we saw in the video that one and two are enantiomers of each other."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if I try to line up the bromines, now the chlorines are not in the right position. So these are non-superimposable mirror images of each other. These are enantiomers. So we saw in the video that one and two are enantiomers of each other. They are non-superimposable mirror images, and they have opposite configurations at all chirality centers. And that's easy to see if you look at the drawings here. So at this carbon, we have bromine on a wedge, and if we change it to a dash, we see we have this one on the right."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we saw in the video that one and two are enantiomers of each other. They are non-superimposable mirror images, and they have opposite configurations at all chirality centers. And that's easy to see if you look at the drawings here. So at this carbon, we have bromine on a wedge, and if we change it to a dash, we see we have this one on the right. If we look at this chiral center, we have chlorine on a wedge, and here it's changed to a dash. So that's an opposite configuration at both chiral centers. And so that's how we know, that's one way of knowing that this one on the right is the mirror image of the one on the left."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So at this carbon, we have bromine on a wedge, and if we change it to a dash, we see we have this one on the right. If we look at this chiral center, we have chlorine on a wedge, and here it's changed to a dash. So that's an opposite configuration at both chiral centers. And so that's how we know, that's one way of knowing that this one on the right is the mirror image of the one on the left. They are enantiomers. Let's look at the relationship between stereoisomers three and four. On the left is stereoisomer three, with bromine up and chlorine down."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so that's how we know, that's one way of knowing that this one on the right is the mirror image of the one on the left. They are enantiomers. Let's look at the relationship between stereoisomers three and four. On the left is stereoisomer three, with bromine up and chlorine down. On the right is four, with bromine down and chlorine up. If we hold them together and I rotate the one on the right, it's easy to see that these are mirror images of each other, and they are non-superimposable. If I put the chlorines on top of each other, now the bromines don't line up."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left is stereoisomer three, with bromine up and chlorine down. On the right is four, with bromine down and chlorine up. If we hold them together and I rotate the one on the right, it's easy to see that these are mirror images of each other, and they are non-superimposable. If I put the chlorines on top of each other, now the bromines don't line up. And if I try to line up the bromines, then the chlorines don't. So these are non-superimposable mirror images of each other. These are enantiomers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If I put the chlorines on top of each other, now the bromines don't line up. And if I try to line up the bromines, then the chlorines don't. So these are non-superimposable mirror images of each other. These are enantiomers. So three and four are enantiomers of each other. They are non-superimposable mirror images. And they have opposite configurations at all chirality centers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are enantiomers. So three and four are enantiomers of each other. They are non-superimposable mirror images. And they have opposite configurations at all chirality centers. So at this chiral center, we have bromine on a wedge, and over here we have bromine on a dash. At this chiral center, we have chlorine on a dash, and over here we have it on a wedge. So we have opposite configurations at both chirality centers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And they have opposite configurations at all chirality centers. So at this chiral center, we have bromine on a wedge, and over here we have bromine on a dash. At this chiral center, we have chlorine on a dash, and over here we have it on a wedge. So we have opposite configurations at both chirality centers. Next, let's compare stereoisomers two and three. So what's the relationship between two and three? On the left is stereoisomer two, with the bromine and the chlorine going away from us in space."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have opposite configurations at both chirality centers. Next, let's compare stereoisomers two and three. So what's the relationship between two and three? On the left is stereoisomer two, with the bromine and the chlorine going away from us in space. On the right is three, with bromine up and chlorine down. If I hold the two stereoisomers next to each other, and I rotate the one on the right, we can see these are not mirror images of each other. The bromines look right, but the chlorines don't."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left is stereoisomer two, with the bromine and the chlorine going away from us in space. On the right is three, with bromine up and chlorine down. If I hold the two stereoisomers next to each other, and I rotate the one on the right, we can see these are not mirror images of each other. The bromines look right, but the chlorines don't. One chlorine is up, and one chlorine is down. If I try to superimpose these, I can get the bromines to match, but not the chlorines. And if I try to make the chlorines line up, then the bromines won't."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The bromines look right, but the chlorines don't. One chlorine is up, and one chlorine is down. If I try to superimpose these, I can get the bromines to match, but not the chlorines. And if I try to make the chlorines line up, then the bromines won't. So these are non-superimposable, non-mirror images. These are diastereomers. We saw in the video that two and three are non-superimposable, and they're also non-mirror images."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if I try to make the chlorines line up, then the bromines won't. So these are non-superimposable, non-mirror images. These are diastereomers. We saw in the video that two and three are non-superimposable, and they're also non-mirror images. Therefore, they are diastereomers. Let me write that down here. Two and three represent a pair of diastereomers."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We saw in the video that two and three are non-superimposable, and they're also non-mirror images. Therefore, they are diastereomers. Let me write that down here. Two and three represent a pair of diastereomers. Diastereomers have opposite configurations at some chirality centers. If we look at this carbon, we have bromine on a dash, and over here we have bromine on a wedge. That's opposite."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Two and three represent a pair of diastereomers. Diastereomers have opposite configurations at some chirality centers. If we look at this carbon, we have bromine on a dash, and over here we have bromine on a wedge. That's opposite. But if we look at this one, we have chlorine on a dash, and over here we have chlorine on a dash. That's the same. We only have an opposite configuration at one chiral center."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's opposite. But if we look at this one, we have chlorine on a dash, and over here we have chlorine on a dash. That's the same. We only have an opposite configuration at one chiral center. These are diastereomers. What about comparing stereoisomer two with stereoisomer four? Let's look at the video for that."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We only have an opposite configuration at one chiral center. These are diastereomers. What about comparing stereoisomer two with stereoisomer four? Let's look at the video for that. On the left is stereoisomer two with the bromine and chlorine going away from us. On the right is four with bromine down and chlorine up. We hold the two stereoisomers next to each other, and we rotate the one on the right."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the video for that. On the left is stereoisomer two with the bromine and chlorine going away from us. On the right is four with bromine down and chlorine up. We hold the two stereoisomers next to each other, and we rotate the one on the right. We can see these are not mirror images. The chlorines look right, but the bromines don't. One bromine's down and one bromine is up."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We hold the two stereoisomers next to each other, and we rotate the one on the right. We can see these are not mirror images. The chlorines look right, but the bromines don't. One bromine's down and one bromine is up. If we try to superimpose one on top of the other, the chlorines line up, but not the bromines. If I try matching the bromines, now the chlorines don't. These are non-superimposable, non-mirror images."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "One bromine's down and one bromine is up. If we try to superimpose one on top of the other, the chlorines line up, but not the bromines. If I try matching the bromines, now the chlorines don't. These are non-superimposable, non-mirror images. These are diastereomers. Two and four are diastereomers. They are non-superimposable, non-mirror images of each other."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are non-superimposable, non-mirror images. These are diastereomers. Two and four are diastereomers. They are non-superimposable, non-mirror images of each other. They only have opposite configurations at some chirality centers, in this case, one. If we look at this chiral center, the bromine's on a dash. If we look at this one, the bromine's also on a dash."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "They are non-superimposable, non-mirror images of each other. They only have opposite configurations at some chirality centers, in this case, one. If we look at this chiral center, the bromine's on a dash. If we look at this one, the bromine's also on a dash. That's the same. If we look at this one, the chlorine is on a dash. Over here, the chlorine's on a wedge."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we look at this one, the bromine's also on a dash. That's the same. If we look at this one, the chlorine is on a dash. Over here, the chlorine's on a wedge. That's an opposite configuration at only one chiral center. Two and four are diastereomers. What about comparing one and three?"}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Over here, the chlorine's on a wedge. That's an opposite configuration at only one chiral center. Two and four are diastereomers. What about comparing one and three? Thinking about one and three. We don't need a video for this anymore. I think we've got the hang of it."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "What about comparing one and three? Thinking about one and three. We don't need a video for this anymore. I think we've got the hang of it. If we look at this carbon, we have bromine on a wedge. At this carbon, we have bromine on a wedge. That's the same."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I think we've got the hang of it. If we look at this carbon, we have bromine on a wedge. At this carbon, we have bromine on a wedge. That's the same. At this carbon, we have chlorine on a wedge. At this one, we have chlorine on a dash. That's different."}, {"video_title": "Enantiomers and diastereomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's the same. At this carbon, we have chlorine on a wedge. At this one, we have chlorine on a dash. That's different. We only have an opposite configuration at one chirality center. One and three are diastereomers of each other. The same thing if you're thinking about one and four."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, I have my alkene. And I'm going to add a halogen to it, so something like bromine or chlorine. And I can see that those two halogen atoms are going to add anti to each other. So they'll add on opposite sides of where the double bond used to be. Let's take a look at the mechanism so we can figure out why we get an anti addition. So I start with my alkene down here. And I'm going to show the halogen approaching that alkene."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So they'll add on opposite sides of where the double bond used to be. Let's take a look at the mechanism so we can figure out why we get an anti addition. So I start with my alkene down here. And I'm going to show the halogen approaching that alkene. So it's going to approach this way. And I put in my lone pairs of electrons like that. If I think about that halogen molecule, I know that it's nonpolar."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to show the halogen approaching that alkene. So it's going to approach this way. And I put in my lone pairs of electrons like that. If I think about that halogen molecule, I know that it's nonpolar. Because if I think about the electrons and the bond between my two halogens here, both halogen atoms, of course, have the exact same electronegativity. So neither one is pulling more strongly. And so overall, the molecule is nonpolar."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If I think about that halogen molecule, I know that it's nonpolar. Because if I think about the electrons and the bond between my two halogens here, both halogen atoms, of course, have the exact same electronegativity. So neither one is pulling more strongly. And so overall, the molecule is nonpolar. However, if the pi electrons in my alkenes, I'm going to say that these electrons right here are pi electrons. If those electrons get too close to the electrons in blue, they would, of course, repel them since electrons repel each other since they're like charge. And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so overall, the molecule is nonpolar. However, if the pi electrons in my alkenes, I'm going to say that these electrons right here are pi electrons. If those electrons get too close to the electrons in blue, they would, of course, repel them since electrons repel each other since they're like charge. And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge. It's losing a little bit of electron density. Now I can think about that bottom halogen as acting like an electrophile because it wants electrons. And so in this mechanism, the pi electrons are going to function as a nucleophile."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And if the electrons in magenta repel the electrons in blue, the electrons in blue would be forced closer to the top halogen like that, giving the top halogen a partial negative charge and leaving the bottom halogen with a partial positive charge. It's losing a little bit of electron density. Now I can think about that bottom halogen as acting like an electrophile because it wants electrons. And so in this mechanism, the pi electrons are going to function as a nucleophile. And the pi electrons are going to attack my electrophile like that. At the same time, these electrons over here, this electron pair on the left side of the halogen, is going to attack this carbon. And the electrons in blue are also going to kick off onto the top halogen like that."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so in this mechanism, the pi electrons are going to function as a nucleophile. And the pi electrons are going to attack my electrophile like that. At the same time, these electrons over here, this electron pair on the left side of the halogen, is going to attack this carbon. And the electrons in blue are also going to kick off onto the top halogen like that. So let's go ahead and draw the result of all those electrons moving around. So now I have carbon singly bonded to another carbon like that. The electrons in magenta formed a new bond between the carbon on the right and my halogen like that."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons in blue are also going to kick off onto the top halogen like that. So let's go ahead and draw the result of all those electrons moving around. So now I have carbon singly bonded to another carbon like that. The electrons in magenta formed a new bond between the carbon on the right and my halogen like that. And these electrons over here I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left like that. And that halogen still has two lone pairs of electrons on it."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in magenta formed a new bond between the carbon on the right and my halogen like that. And these electrons over here I'm going to mark in red. So this lone pair of electrons on my halogen are going to form a bond with the carbon on the left like that. And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that halogen still has two lone pairs of electrons on it. So I'm going to put in those two lone pairs of electrons. And that halogen has a plus 1 formal charge. This is called a cyclic halonium ion. And it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here, losing a little bit of electron density, giving it a partial positive charge."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is called a cyclic halonium ion. And it's been proven to occur in this mechanism. If I think about that positively charged halogen, halogens are very electronegative and they want electrons. So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here, losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. And our nucleophile is going to be the halide anion created in the previous step. So we had a halogen that had three lone pairs of electrons around it."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta, let's say, are going to be pulled a little bit closer to that halogen, which would leave this carbon down here, losing a little bit of electron density, giving it a partial positive charge. And so that's going to function as our electrophile in the next step. And our nucleophile is going to be the halide anion created in the previous step. So we had a halogen that had three lone pairs of electrons around it. It picked up the electrons in blue. So now it has four lone pairs of electrons, eight total electrons, giving a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we had a halogen that had three lone pairs of electrons around it. It picked up the electrons in blue. So now it has four lone pairs of electrons, eight total electrons, giving a negative 1 formal charge, meaning it can now function as a nucleophile. So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here, this carbon. And that's going to kick these electrons in magenta off onto this halogen here."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I think about this cyclic halonium ion here, the halogen on top is going to prevent the nucleophile from attacking from the top. It's going to have to attack from below here. So this negatively charged halide anion is going to nucleophilic attack this electrophile here, this carbon. And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. So now I'm going to have my two carbons still bonded to each other like that. And the top halogen has swung over here to the carbon on the left."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that's going to kick these electrons in magenta off onto this halogen here. So let's go ahead and draw the results of that nucleophilic attack. So now I'm going to have my two carbons still bonded to each other like that. And the top halogen has swung over here to the carbon on the left. It used to have two lone pairs of electrons. It picked up the electrons in magenta. So that's what the carbon on the left will look like."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the top halogen has swung over here to the carbon on the left. It used to have two lone pairs of electrons. It picked up the electrons in magenta. So that's what the carbon on the left will look like. The carbon on the right is still bonded to two other things. And the halide anion had to add from below. So now we're going to have this halogen down here like that."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's what the carbon on the left will look like. The carbon on the right is still bonded to two other things. And the halide anion had to add from below. So now we're going to have this halogen down here like that. And so now we understand why it's an anti-addition of my two halogen atoms. Let's go ahead and do a reaction. So we're going to start with cyclohexene as our reactant here."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now we're going to have this halogen down here like that. And so now we understand why it's an anti-addition of my two halogen atoms. Let's go ahead and do a reaction. So we're going to start with cyclohexene as our reactant here. And we're going to react cyclohexene with bromine, so Br2. Now if I think about the first step of the mechanism, I know I'm going to form a cyclic halonium ion. So I'm going to draw that ring."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to start with cyclohexene as our reactant here. And we're going to react cyclohexene with bromine, so Br2. Now if I think about the first step of the mechanism, I know I'm going to form a cyclic halonium ion. So I'm going to draw that ring. And I'm going to show the formation of my cyclic halonium ion. It's called a bromonium ion. So I'm going to form a ring like this."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to draw that ring. And I'm going to show the formation of my cyclic halonium ion. It's called a bromonium ion. So I'm going to form a ring like this. And the bromine's going to have two lone pairs of electrons. It's going to have a plus 1 formal charge like that. And also, I know for my mechanism, I'm going to form a bromide anion at the same time."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to form a ring like this. And the bromine's going to have two lone pairs of electrons. It's going to have a plus 1 formal charge like that. And also, I know for my mechanism, I'm going to form a bromide anion at the same time. So I'm going to have a negatively charged bromide anion like that. When I think about where that bromide anion is going to attack, I know that it's going to attack one of these two carbons here. So it could attack the one on the left."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And also, I know for my mechanism, I'm going to form a bromide anion at the same time. So I'm going to have a negatively charged bromide anion like that. When I think about where that bromide anion is going to attack, I know that it's going to attack one of these two carbons here. So it could attack the one on the left. It could attack the one on the right. Let's go ahead and start with the carbon on the right and draw the product. So if a lone pair of electrons in the bromide anion attack this carbon right here, that would kick these electrons off onto the bromine."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it could attack the one on the left. It could attack the one on the right. Let's go ahead and start with the carbon on the right and draw the product. So if a lone pair of electrons in the bromide anion attack this carbon right here, that would kick these electrons off onto the bromine. And we could go ahead and draw the results of that. So I would have my ring. And the bromine on top is going to swing over to the carbon on the left here."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if a lone pair of electrons in the bromide anion attack this carbon right here, that would kick these electrons off onto the bromine. And we could go ahead and draw the results of that. So I would have my ring. And the bromine on top is going to swing over to the carbon on the left here. So now this bromine is going to go look like that. And the bromide anion has added from below the plane of the ring like that. So that's one possible product."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the bromine on top is going to swing over to the carbon on the left here. So now this bromine is going to go look like that. And the bromide anion has added from below the plane of the ring like that. So that's one possible product. The bromide anion could also attack the bromonium ion from the left side. So this lone pair of electrons could attack this carbon, which would kick these electrons off onto the bromine. And so we could go ahead and draw the result of that nucleophilic attack."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's one possible product. The bromide anion could also attack the bromonium ion from the left side. So this lone pair of electrons could attack this carbon, which would kick these electrons off onto the bromine. And so we could go ahead and draw the result of that nucleophilic attack. So in this case, the top bromine would swing over to the carbon on the right. And it would pick up an extra lone pair of electrons. And the bromide anion, again, added from below the plane of the ring like that."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we could go ahead and draw the result of that nucleophilic attack. So in this case, the top bromine would swing over to the carbon on the right. And it would pick up an extra lone pair of electrons. And the bromide anion, again, added from below the plane of the ring like that. Now these two molecules are actually different molecules. Let's go ahead and redraw them so it's a little bit easier to see. And we're going to stare down this way at the molecule."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the bromide anion, again, added from below the plane of the ring like that. Now these two molecules are actually different molecules. Let's go ahead and redraw them so it's a little bit easier to see. And we're going to stare down this way at the molecule. This is the top of your head. So let's go ahead and redraw that molecule. So this is the same molecule as if we're looking down on it."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to stare down this way at the molecule. This is the top of your head. So let's go ahead and redraw that molecule. So this is the same molecule as if we're looking down on it. If we're looking down on it, this carbon right here would be this carbon. And I can see there's a bromine coming out at me in space. So I'm going to put a bromine coming out at me in space at that carbon."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is the same molecule as if we're looking down on it. If we're looking down on it, this carbon right here would be this carbon. And I can see there's a bromine coming out at me in space. So I'm going to put a bromine coming out at me in space at that carbon. I move to this carbon over here. That's this carbon. So there must be a bromine going away from me in space at that carbon."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to put a bromine coming out at me in space at that carbon. I move to this carbon over here. That's this carbon. So there must be a bromine going away from me in space at that carbon. So that's one possible product. On the right, we do the same thing. So we're going to put our eye right here."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So there must be a bromine going away from me in space at that carbon. So that's one possible product. On the right, we do the same thing. So we're going to put our eye right here. We're going to look down. This is the top. So I can look at this carbon first."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to put our eye right here. We're going to look down. This is the top. So I can look at this carbon first. I can see there's a bromine down at that carbon. So I go ahead and draw my cyclohexane ring. And at this carbon, there's now a bromine down."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can look at this carbon first. I can see there's a bromine down at that carbon. So I go ahead and draw my cyclohexane ring. And at this carbon, there's now a bromine down. And of course, at this carbon over here, there's a bromine coming out at me. So I represent that with a wedge. And these are my products."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And at this carbon, there's now a bromine down. And of course, at this carbon over here, there's a bromine coming out at me. So I represent that with a wedge. And these are my products. And if you've already had stereochemistry, you know that these two products are enantiomers to each other. They're actually different molecules. They're non-superimposable mirror images."}, {"video_title": "Halogenation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And these are my products. And if you've already had stereochemistry, you know that these two products are enantiomers to each other. They're actually different molecules. They're non-superimposable mirror images. So we can see that the absolute configurations have been reversed. So if I think about this carbon right here, bromine coming out at me, bromine going away from me. This one down here, bromine going away from me."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Here's a dot structure for propane dioic acid, or malonic acid, and if you heat it up, it's going to undergo a decarboxylation reaction. So if we show free rotation about this bond, it's a sigma bond, so we can show a different conformation. Let me go ahead and draw in this carboxylic acid on the left, and then we're gonna have a carboxylic acid on the right, too. This time the carbonyl is going to be going to the right. Let me go ahead and put in those electrons. And the OH would be going to the left. So there we have it."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This time the carbonyl is going to be going to the right. Let me go ahead and put in those electrons. And the OH would be going to the left. So there we have it. Alright, so in this mechanism, we're actually gonna form a bond between this oxygen and this proton, and it's a cyclic mechanism. So if these electrons in here move into here, that's gonna push these electrons into here, and then these electrons are gonna form the bond between the oxygen and the hydrogen. So let's go ahead and show the result of our cyclic mechanism."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So there we have it. Alright, so in this mechanism, we're actually gonna form a bond between this oxygen and this proton, and it's a cyclic mechanism. So if these electrons in here move into here, that's gonna push these electrons into here, and then these electrons are gonna form the bond between the oxygen and the hydrogen. So let's go ahead and show the result of our cyclic mechanism. Alright, we'd have a carbon bonded to an oxygen, right, bonded to hydrogen, and then we have an OH over here. And then we'd have a double bond between this carbon and another carbon. On the right, we would actually form CO2."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the result of our cyclic mechanism. Alright, we'd have a carbon bonded to an oxygen, right, bonded to hydrogen, and then we have an OH over here. And then we'd have a double bond between this carbon and another carbon. On the right, we would actually form CO2. So let me go ahead and put in lone pairs of electrons on that oxygen, alright, so we can see that we would form our carbon dioxide molecule here. So let me go ahead and draw those in. And let's follow some of those electrons, right?"}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "On the right, we would actually form CO2. So let me go ahead and put in lone pairs of electrons on that oxygen, alright, so we can see that we would form our carbon dioxide molecule here. So let me go ahead and draw those in. And let's follow some of those electrons, right? So the electrons in magenta right in here are going to move in, right, to form this bond, right? To form our double bond for CO2. And then these electrons in here in blue, so between this carbon and this carbon, right, are going to move over here to form this double bond, right, between this carbon, and then there's a carbon right here, and then there's also two hydrogens bonded to this carbon."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And let's follow some of those electrons, right? So the electrons in magenta right in here are going to move in, right, to form this bond, right? To form our double bond for CO2. And then these electrons in here in blue, so between this carbon and this carbon, right, are going to move over here to form this double bond, right, between this carbon, and then there's a carbon right here, and then there's also two hydrogens bonded to this carbon. Let me go ahead and draw those in so we can see it a little bit better. And then finally, let's make these electrons in here red. So these electrons are the ones that are gonna form this bond between the oxygen and the hydrogen."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons in here in blue, so between this carbon and this carbon, right, are going to move over here to form this double bond, right, between this carbon, and then there's a carbon right here, and then there's also two hydrogens bonded to this carbon. Let me go ahead and draw those in so we can see it a little bit better. And then finally, let's make these electrons in here red. So these electrons are the ones that are gonna form this bond between the oxygen and the hydrogen. So we've formed our CO2, and we've also formed an acid enol, right? So this right here is called an acid enol. And we saw in earlier videos how the enol is in equilibrium with the keto form, right, with keto enol tautomerization."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons are the ones that are gonna form this bond between the oxygen and the hydrogen. So we've formed our CO2, and we've also formed an acid enol, right? So this right here is called an acid enol. And we saw in earlier videos how the enol is in equilibrium with the keto form, right, with keto enol tautomerization. And so this is actually the enol form of acetic acid. And so that's actually gonna be our product. So let me go ahead and draw acetic acid up here."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we saw in earlier videos how the enol is in equilibrium with the keto form, right, with keto enol tautomerization. And so this is actually the enol form of acetic acid. And so that's actually gonna be our product. So let me go ahead and draw acetic acid up here. And here we have the OH on the left side. And then we would have a carbon over here with three hydrogens bonded to it, right? So this would be, if we're thinking about it, the keto type form, right?"}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw acetic acid up here. And here we have the OH on the left side. And then we would have a carbon over here with three hydrogens bonded to it, right? So this would be, if we're thinking about it, the keto type form, right? So the difference between the enol and the keto form are the movement of one proton, right? So there's a proton here in the oxygen, and here it's one of these on the carbon, and then the double bond. Here we have the double bond between the two carbons, and here we've moved the double bond between the carbon and the oxygen."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this would be, if we're thinking about it, the keto type form, right? So the difference between the enol and the keto form are the movement of one proton, right? So there's a proton here in the oxygen, and here it's one of these on the carbon, and then the double bond. Here we have the double bond between the two carbons, and here we've moved the double bond between the carbon and the oxygen. So once again, we've seen how to do that in earlier videos. And then we also produce CO2, so carbon dioxide as the other product for this reaction. So the key to a decarboxylation reaction is having a carbonyl beta to a carboxylic acid."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Here we have the double bond between the two carbons, and here we've moved the double bond between the carbon and the oxygen. So once again, we've seen how to do that in earlier videos. And then we also produce CO2, so carbon dioxide as the other product for this reaction. So the key to a decarboxylation reaction is having a carbonyl beta to a carboxylic acid. So for example, here's our carboxylic acid, and we know the carbon next to a carboxylic acid is the alpha carbon, and the carbon next to that is the beta carbon. And we saw how this carbonyl was necessary in the mechanism. And so the fact that there's an OH here isn't really necessary."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the key to a decarboxylation reaction is having a carbonyl beta to a carboxylic acid. So for example, here's our carboxylic acid, and we know the carbon next to a carboxylic acid is the alpha carbon, and the carbon next to that is the beta carbon. And we saw how this carbonyl was necessary in the mechanism. And so the fact that there's an OH here isn't really necessary. And what we really need is a carbonyl that's beta to our carboxylic acid in order for a decarboxylation reaction to occur. So let's look at another example, where we don't have a dioic acid anymore. We have a carboxylic acid on the right, and then over here on the left we have a ketone."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so the fact that there's an OH here isn't really necessary. And what we really need is a carbonyl that's beta to our carboxylic acid in order for a decarboxylation reaction to occur. So let's look at another example, where we don't have a dioic acid anymore. We have a carboxylic acid on the right, and then over here on the left we have a ketone. But again, the key point is, here's the alpha carbon and here's the beta carbon. We have a carbonyl that's beta to our carboxylic acid, and so therefore a decarboxylation reaction can take place. So if we heat up this molecule, once again thinking about the mechanism, rotating about that sigma bond, let's go ahead and redraw this."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We have a carboxylic acid on the right, and then over here on the left we have a ketone. But again, the key point is, here's the alpha carbon and here's the beta carbon. We have a carbonyl that's beta to our carboxylic acid, and so therefore a decarboxylation reaction can take place. So if we heat up this molecule, once again thinking about the mechanism, rotating about that sigma bond, let's go ahead and redraw this. So we have our benzene ring, and then we have our carbonyl right here. And then we would have, once again, our carbonyl going off to the right this time, and then our OH over here on the left. So thinking about our mechanism, once again we know it's a cyclic mechanism."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So if we heat up this molecule, once again thinking about the mechanism, rotating about that sigma bond, let's go ahead and redraw this. So we have our benzene ring, and then we have our carbonyl right here. And then we would have, once again, our carbonyl going off to the right this time, and then our OH over here on the left. So thinking about our mechanism, once again we know it's a cyclic mechanism. We know this oxygen is going to bond to this proton, and so these electrons are going to move into here, and these electrons move into here, and these electrons move into here. So that's our cyclic mechanism. When we draw what happens, moving all those electrons around, we have our benzene ring."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So thinking about our mechanism, once again we know it's a cyclic mechanism. We know this oxygen is going to bond to this proton, and so these electrons are going to move into here, and these electrons move into here, and these electrons move into here. So that's our cyclic mechanism. When we draw what happens, moving all those electrons around, we have our benzene ring. We would have it bonded to an oxygen. This oxygen was bonded to this proton now, and then we have a double bond right in here. And then we also form CO2."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "When we draw what happens, moving all those electrons around, we have our benzene ring. We would have it bonded to an oxygen. This oxygen was bonded to this proton now, and then we have a double bond right in here. And then we also form CO2. So once again, let's run through those electrons and try to follow some of those electrons here. So let me draw on these lone pairs of electrons on our oxygens, so we can see where the CO2 comes from. So once again, these electrons in here, and magenta are going to move in here to form the double bond on CO2."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then we also form CO2. So once again, let's run through those electrons and try to follow some of those electrons here. So let me draw on these lone pairs of electrons on our oxygens, so we can see where the CO2 comes from. So once again, these electrons in here, and magenta are going to move in here to form the double bond on CO2. At the same time, these electrons in here are going to move into here to form this double bond, and then finally, these electrons in here are going to move out to form the bond between the oxygen and the hydrogen. So we've made our CO2. So the oxygen-carbon-oxygen comes from this oxygen-carbon-oxygen right here on the molecule on the left."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So once again, these electrons in here, and magenta are going to move in here to form the double bond on CO2. At the same time, these electrons in here are going to move into here to form this double bond, and then finally, these electrons in here are going to move out to form the bond between the oxygen and the hydrogen. So we've made our CO2. So the oxygen-carbon-oxygen comes from this oxygen-carbon-oxygen right here on the molecule on the left. So hopefully that's a little bit easier to see now. And once again, we have an enol. So we form our CO2."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the oxygen-carbon-oxygen comes from this oxygen-carbon-oxygen right here on the molecule on the left. So hopefully that's a little bit easier to see now. And once again, we have an enol. So we form our CO2. We also create our enol. And so we could think about keto-enol tautomerization for our product. So the enol is going to be in equilibrium with the keto form."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we form our CO2. We also create our enol. And so we could think about keto-enol tautomerization for our product. So the enol is going to be in equilibrium with the keto form. So let's go ahead and draw the keto form, which we know is moving one proton and moving the double bond. So we move the double bond between the carbon and the oxygen, and we move that proton to this carbon right here. So this carbon right here picks up a proton."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So the enol is going to be in equilibrium with the keto form. So let's go ahead and draw the keto form, which we know is moving one proton and moving the double bond. So we move the double bond between the carbon and the oxygen, and we move that proton to this carbon right here. So this carbon right here picks up a proton. So down here, that carbon had two hydrogens. And it doesn't have to be this one right here, but we are going to add a hydrogen to that carbon, and that gives us our final product, our ketone. So this decarboxylation reaction produces a ketone."}, {"video_title": "Decarboxylation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here picks up a proton. So down here, that carbon had two hydrogens. And it doesn't have to be this one right here, but we are going to add a hydrogen to that carbon, and that gives us our final product, our ketone. So this decarboxylation reaction produces a ketone. And once again, it doesn't really matter what this R group is here. Here we have a benzene ring instead of the OH in the previous example. And then of course, we're also going to make CO2."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "We've learned that axial precession, it's not a change in the tilt or the obliquity of our rotational axis, it's a change in the direction and over a long period of time, 26,000 years, it kind of traces out a circle. And the main effect of that is that if we wait long enough, our rotational axis, or you could say almost the North Pole, will be pointed in a different direction. And so if our rotational axis is pointed in a different direction after a long enough time, then the absolute point in our orbit, if we use the sun as our frame of reference, the point in our orbit when we are most pointed away from the sun, or when the Northern Hemisphere is most pointed away from the sun, will be earlier in the orbit. Now I emphasize that that won't necessarily mean earlier in our calendar, because our calendar, by definition, takes into consideration, I guess, or it's more based on when we are furthest away from, furthest tilted away from the sun, or furthest tilted towards the sun. So, even though, if we wait 1,800 years, like the example I gave, we will be most tilted away from the sun, we will have our, or the Northern Hemisphere will have its winter, will have its winter equinox at an earlier point in the orbit, according to our calendar, it will still be December 22nd. If our calendar instead was based, and it's not based on this, but if our calendar was based on the exact point in orbit, if our calendar was based on the exact point in orbit, then our year would be about 20-25 minutes longer every year, and then the dates actually would, then the date for the start of winter actually would go back, it would, you know, 1,800 years later, the date of the start of winter would be November 22nd. But that's not how we measure our calendar."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "Now I emphasize that that won't necessarily mean earlier in our calendar, because our calendar, by definition, takes into consideration, I guess, or it's more based on when we are furthest away from, furthest tilted away from the sun, or furthest tilted towards the sun. So, even though, if we wait 1,800 years, like the example I gave, we will be most tilted away from the sun, we will have our, or the Northern Hemisphere will have its winter, will have its winter equinox at an earlier point in the orbit, according to our calendar, it will still be December 22nd. If our calendar instead was based, and it's not based on this, but if our calendar was based on the exact point in orbit, if our calendar was based on the exact point in orbit, then our year would be about 20-25 minutes longer every year, and then the dates actually would, then the date for the start of winter actually would go back, it would, you know, 1,800 years later, the date of the start of winter would be November 22nd. But that's not how we measure our calendar. Our calendar is actually measured from equinox to equinox, from December 22nd or 21st, there's slight fluctuations depending on the calendar, but that will always be the date that we are most pointed away from the sun. That will not be necessarily the date that we are at this exact position, at this exact position relative to the sun itself. And that's why the actual perihelion does change, because if this is always December 22nd, and if we at first assume that the perihelion is always at the same fixed point in space relative to the sun, although that's not exactly the case, but if we make that assumption, then it will be further and further after that December 22nd, further and further after that time that we are most pointed away from the sun."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "But that's not how we measure our calendar. Our calendar is actually measured from equinox to equinox, from December 22nd or 21st, there's slight fluctuations depending on the calendar, but that will always be the date that we are most pointed away from the sun. That will not be necessarily the date that we are at this exact position, at this exact position relative to the sun itself. And that's why the actual perihelion does change, because if this is always December 22nd, and if we at first assume that the perihelion is always at the same fixed point in space relative to the sun, although that's not exactly the case, but if we make that assumption, then it will be further and further after that December 22nd, further and further after that time that we are most pointed away from the sun. And that's why you have this kind of pushing back of the perihelion. Now, what I want to add to this video is that the perihelion itself is also changing. So if I draw the sun again, and right now our orbit looks something like this, and I'm going to exaggerate the eccentricity of it, I'm going to exaggerate the eccentricity of it, just so that the perihelion and the aphelion are a little bit clearer."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "And that's why the actual perihelion does change, because if this is always December 22nd, and if we at first assume that the perihelion is always at the same fixed point in space relative to the sun, although that's not exactly the case, but if we make that assumption, then it will be further and further after that December 22nd, further and further after that time that we are most pointed away from the sun. And that's why you have this kind of pushing back of the perihelion. Now, what I want to add to this video is that the perihelion itself is also changing. So if I draw the sun again, and right now our orbit looks something like this, and I'm going to exaggerate the eccentricity of it, I'm going to exaggerate the eccentricity of it, just so that the perihelion and the aphelion are a little bit clearer. So right now this is the perihelion, this is the aphelion, based on the way I drew it right over there. We drew it in different colors. I'm not going to show that that's necessarily where Earth is."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "So if I draw the sun again, and right now our orbit looks something like this, and I'm going to exaggerate the eccentricity of it, I'm going to exaggerate the eccentricity of it, just so that the perihelion and the aphelion are a little bit clearer. So right now this is the perihelion, this is the aphelion, based on the way I drew it right over there. We drew it in different colors. I'm not going to show that that's necessarily where Earth is. Perihelion and aphelion. There is also a rotation of this, of the perihelion, and sometimes this is called the precession of the perihelion, or perihelion precession, or apsidal precession. These are all very hard to say."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "I'm not going to show that that's necessarily where Earth is. Perihelion and aphelion. There is also a rotation of this, of the perihelion, and sometimes this is called the precession of the perihelion, or perihelion precession, or apsidal precession. These are all very hard to say. And so if we wait several thousands of years, our orbit might look a little bit like this. Our orbit will look like this. The actual perihelion will have rotated."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "These are all very hard to say. And so if we wait several thousands of years, our orbit might look a little bit like this. Our orbit will look like this. The actual perihelion will have rotated. So our orbit will look like this. The actual ellipse would have rotated a little bit. You wait a little bit longer, it will go, it will look like, it might look like this."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "The actual perihelion will have rotated. So our orbit will look like this. The actual ellipse would have rotated a little bit. You wait a little bit longer, it will go, it will look like, it might look like this. And obviously I'm once again talking about over thousands and thousands of years. From a year-to-year basis, you really wouldn't notice the difference. But what that does is, is we talked about the axial precession, that this change in direction of our rotational axis, it takes 26,000 years to complete one period."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "You wait a little bit longer, it will go, it will look like, it might look like this. And obviously I'm once again talking about over thousands and thousands of years. From a year-to-year basis, you really wouldn't notice the difference. But what that does is, is we talked about the axial precession, that this change in direction of our rotational axis, it takes 26,000 years to complete one period. 26,000 years. So 26,000 years from today, our polar axis, if we don't think about our rotational axis, if we aren't too concerned about the actual change in tilt, which there will be some small change in tilt, but 26,000 years from now, our pole will roughly point in the same direction again. We would have completed one whole period of axial precession."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "But what that does is, is we talked about the axial precession, that this change in direction of our rotational axis, it takes 26,000 years to complete one period. 26,000 years. So 26,000 years from today, our polar axis, if we don't think about our rotational axis, if we aren't too concerned about the actual change in tilt, which there will be some small change in tilt, but 26,000 years from now, our pole will roughly point in the same direction again. We would have completed one whole period of axial precession. However, it does not take 26,000 years for whatever our date of perihelion is today, so it's in January, I actually don't know the exact date, you could look that up, but whatever that date is in January, it will not take 26,000 years for it to be that date again. And it would have taken 26,000 years if the perihelion itself were not changing, if it always stayed fixed over here. If we did not have this abscidal precession."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "We would have completed one whole period of axial precession. However, it does not take 26,000 years for whatever our date of perihelion is today, so it's in January, I actually don't know the exact date, you could look that up, but whatever that date is in January, it will not take 26,000 years for it to be that date again. And it would have taken 26,000 years if the perihelion itself were not changing, if it always stayed fixed over here. If we did not have this abscidal precession. But since it is also changing, you can kind of say it is over thousands of years moving in that direction, while our date, our January date, is moving in that direction, they will actually meet sooner, so that the precession will be back on whatever date it is on January, less than 26,000 years from now, and actually the exact time, and I haven't done the calculation, but this is what I've read, is that it will be 21,000 years from now. And then on top of that, if that's not enough for you, that not only is the direction of Earth's rotational axis changing, and the tilt is changing, and that the perihelion and the aphelion are also rotating around, it's also the case that the eccentricity of the orbit itself is changing. So over long periods of time, Earth's orbit becomes more or less eccentric, and we've learned that almost circular has, well if you're circular you have no eccentricity, and then you can become more and more eccentric, which means you're more and more of kind of this flattened out ellipse."}, {"video_title": "Apsidal precession (perihelion precession) and Milankovitch cycles Khan Academy.mp3", "Sentence": "If we did not have this abscidal precession. But since it is also changing, you can kind of say it is over thousands of years moving in that direction, while our date, our January date, is moving in that direction, they will actually meet sooner, so that the precession will be back on whatever date it is on January, less than 26,000 years from now, and actually the exact time, and I haven't done the calculation, but this is what I've read, is that it will be 21,000 years from now. And then on top of that, if that's not enough for you, that not only is the direction of Earth's rotational axis changing, and the tilt is changing, and that the perihelion and the aphelion are also rotating around, it's also the case that the eccentricity of the orbit itself is changing. So over long periods of time, Earth's orbit becomes more or less eccentric, and we've learned that almost circular has, well if you're circular you have no eccentricity, and then you can become more and more eccentric, which means you're more and more of kind of this flattened out ellipse. And these cycles occur, so these eccentricity cycles occur over approximately 100,000 years. And so to revisit the Milankovitch cycles, and once again this is a theory, we're not sure whether this is necessarily causing our ice ages, or whether this is necessarily a major influence over long-term climate change, but the Milankovitch cycle, or the theory of Milankovitch cycles, is that over long periods of time, if the eccentricity changes enough, and if it coincides with when the perihelion and the seasons also coincide, maybe that's enough to start an ice age, or maybe that's enough to take us out of an ice age. And actually if you want to throw something even more on that, the actual plane of our orbit also changes over time, mainly because of interactions with the outer planets."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "A functional group is a group of atoms that has a predictable chemical behavior. And there are many functional groups in organic chemistry. And I'm gonna cover some of the more common ones, the ones you would have to know for your class. So we start with an alkene. An alkene has a carbon-carbon double bond in it. So here is a carbon-carbon double bond. R refers to the rest of the compound, or the remainder of the compound."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we start with an alkene. An alkene has a carbon-carbon double bond in it. So here is a carbon-carbon double bond. R refers to the rest of the compound, or the remainder of the compound. And normally we're talking about carbons and hydrogens. Let's look at an example of an alkene. So over here on the right, we can see that this molecule contains a carbon-carbon double bond."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "R refers to the rest of the compound, or the remainder of the compound. And normally we're talking about carbons and hydrogens. Let's look at an example of an alkene. So over here on the right, we can see that this molecule contains a carbon-carbon double bond. So this is an an alkene. So how do we name this? Well, if we had a four-carbon alkane, let me go ahead and write out a four-carbon alkane here."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the right, we can see that this molecule contains a carbon-carbon double bond. So this is an an alkene. So how do we name this? Well, if we had a four-carbon alkane, let me go ahead and write out a four-carbon alkane here. So one, two, three, four. We already know we should call that butane. So this one should be called butane."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Well, if we had a four-carbon alkane, let me go ahead and write out a four-carbon alkane here. So one, two, three, four. We already know we should call that butane. So this one should be called butane. And since we have a four-carbon alkene, we're gonna lose our ending here, A-N-E, and we're gonna add on E-N-E, the ending for an alkene. So the name of this molecule is butene. Let me go ahead and write that out here."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this one should be called butane. And since we have a four-carbon alkene, we're gonna lose our ending here, A-N-E, and we're gonna add on E-N-E, the ending for an alkene. So the name of this molecule is butene. Let me go ahead and write that out here. So this is butene. And if we number this molecule, we start with a double bond, and we give this carbon right here a number one. And then we have a number two right here, and a number three for this carbon, and a number four for this carbon."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and write that out here. So this is butene. And if we number this molecule, we start with a double bond, and we give this carbon right here a number one. And then we have a number two right here, and a number three for this carbon, and a number four for this carbon. Our double bond starts at carbon one. So we put a one in front here, and we call this one butene. Let's look at another example of an alkene."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a number two right here, and a number three for this carbon, and a number four for this carbon. Our double bond starts at carbon one. So we put a one in front here, and we call this one butene. Let's look at another example of an alkene. If we drew out a ring like this, we already know this is called cyclohexane. If we put in a double bond, now it's cyclohexene. And you can see this has a carbon-carbon double bond, and so does this molecule."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at another example of an alkene. If we drew out a ring like this, we already know this is called cyclohexane. If we put in a double bond, now it's cyclohexene. And you can see this has a carbon-carbon double bond, and so does this molecule. So they both contain an alkene functional group, and therefore, both these molecules will undergo the same types of reactions. They have predictable chemical behaviors. And that's the usefulness of grouping functional groups, of identifying functional groups on molecules."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And you can see this has a carbon-carbon double bond, and so does this molecule. So they both contain an alkene functional group, and therefore, both these molecules will undergo the same types of reactions. They have predictable chemical behaviors. And that's the usefulness of grouping functional groups, of identifying functional groups on molecules. Our next example of a functional group is an alkyne. So this time, it's a carbon-carbon triple bond. So here you can see the carbon-carbon triple bond in an alkyne."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And that's the usefulness of grouping functional groups, of identifying functional groups on molecules. Our next example of a functional group is an alkyne. So this time, it's a carbon-carbon triple bond. So here you can see the carbon-carbon triple bond in an alkyne. So if we look at the right, here's our example. And we number our carbons. We call this carbon one."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So here you can see the carbon-carbon triple bond in an alkyne. So if we look at the right, here's our example. And we number our carbons. We call this carbon one. This is carbon two, carbon three, and carbon four. So when you're drawing an alkyne, we know that this is linear at this portion of the molecule. This bond angle is linear."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "We call this carbon one. This is carbon two, carbon three, and carbon four. So when you're drawing an alkyne, we know that this is linear at this portion of the molecule. This bond angle is linear. So that's why it's drawn in a straight line. So four carbons. So we would use but, and we have a YNE ending for an alkyne."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "This bond angle is linear. So that's why it's drawn in a straight line. So four carbons. So we would use but, and we have a YNE ending for an alkyne. So this is butyne. So let me write out butyne here. And our triple bond starts at carbon two."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we would use but, and we have a YNE ending for an alkyne. So this is butyne. So let me write out butyne here. And our triple bond starts at carbon two. So here is where our triple bond starts. So we could write 2-butyne for the IUPAC name. Next, we're looking at an arene, also called an aromatic ring."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And our triple bond starts at carbon two. So here is where our triple bond starts. So we could write 2-butyne for the IUPAC name. Next, we're looking at an arene, also called an aromatic ring. And you're looking for this. You're looking for a six-carbon ring. And you have alternating single-double bonds, so a total of three double bonds."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Next, we're looking at an arene, also called an aromatic ring. And you're looking for this. You're looking for a six-carbon ring. And you have alternating single-double bonds, so a total of three double bonds. Now, just because this has double bonds, this doesn't mean that it's going to be an alkene. Actually, arenes react in different ways from alkenes. And that's why this is a separate functional group."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And you have alternating single-double bonds, so a total of three double bonds. Now, just because this has double bonds, this doesn't mean that it's going to be an alkene. Actually, arenes react in different ways from alkenes. And that's why this is a separate functional group. So on the right, here's an example of an arene or an aromatic ring. And this right here, this portion of the molecule, would be a benzene. So let's write out that."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And that's why this is a separate functional group. So on the right, here's an example of an arene or an aromatic ring. And this right here, this portion of the molecule, would be a benzene. So let's write out that. So benzene is a very famous organic chemistry molecule. And then we have a methyl group coming off of our benzene ring. So one name for this would be methylbenzene."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let's write out that. So benzene is a very famous organic chemistry molecule. And then we have a methyl group coming off of our benzene ring. So one name for this would be methylbenzene. That's not the name you would usually see. Normally, you'll hear this molecule called toluene. So it's so common that toluene is used most often."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So one name for this would be methylbenzene. That's not the name you would usually see. Normally, you'll hear this molecule called toluene. So it's so common that toluene is used most often. So toluene is an example of an arene. It contains an arene functional group. And so toluene would react in similar ways to benzene."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So it's so common that toluene is used most often. So toluene is an example of an arene. It contains an arene functional group. And so toluene would react in similar ways to benzene. All right, let's look at our next functional group here. This is an alkyl halide. So halide refers to a halogen."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And so toluene would react in similar ways to benzene. All right, let's look at our next functional group here. This is an alkyl halide. So halide refers to a halogen. So over here on the right, we have our halogen. So X could be a halogen like chlorine or bromine. And R will be our alkyl group."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So halide refers to a halogen. So over here on the right, we have our halogen. So X could be a halogen like chlorine or bromine. And R will be our alkyl group. For the example on the right, let me get a little more room down here. Our halogen is chlorine. And our alkyl group here would be an ethyl group."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And R will be our alkyl group. For the example on the right, let me get a little more room down here. Our halogen is chlorine. And our alkyl group here would be an ethyl group. We have a CH2 and a CH3. So one possible name for this molecule would be ethyl chloride. You could also call this something like chloroethane."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And our alkyl group here would be an ethyl group. We have a CH2 and a CH3. So one possible name for this molecule would be ethyl chloride. You could also call this something like chloroethane. All right, next functional group is an alcohol. So an alcohol has an OH, an OH right here, and an R for the rest of the molecule. As an example, here's an OH or a hydroxyl group."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "You could also call this something like chloroethane. All right, next functional group is an alcohol. So an alcohol has an OH, an OH right here, and an R for the rest of the molecule. As an example, here's an OH or a hydroxyl group. And then we have a CH2 and a CH3. So we have two carbons in this molecule. And we know two carbons is eth."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "As an example, here's an OH or a hydroxyl group. And then we have a CH2 and a CH3. So we have two carbons in this molecule. And we know two carbons is eth. And for the full name, this would be ethanol. So a very famous molecule, obviously. So this is ethanol."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "And we know two carbons is eth. And for the full name, this would be ethanol. So a very famous molecule, obviously. So this is ethanol. Next we have an ether. So an ether has an oxygen with an R group on either side. So I'm sorry for the bad joke, but if it helps you remember that you have an R group on either side of your oxygen, it's worth putting in a bad joke into this video."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is ethanol. Next we have an ether. So an ether has an oxygen with an R group on either side. So I'm sorry for the bad joke, but if it helps you remember that you have an R group on either side of your oxygen, it's worth putting in a bad joke into this video. So now these R groups could be the same R groups or they could be different R groups. For the example on the right, we have R groups that happen to be the same. So here's our oxygen, and here's an R group, and here's an R group."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'm sorry for the bad joke, but if it helps you remember that you have an R group on either side of your oxygen, it's worth putting in a bad joke into this video. So now these R groups could be the same R groups or they could be different R groups. For the example on the right, we have R groups that happen to be the same. So here's our oxygen, and here's an R group, and here's an R group. Both of those R groups are ethyl groups. So we call this diethyl ether. So diethyl ether is the most famous ether."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So here's our oxygen, and here's an R group, and here's an R group. Both of those R groups are ethyl groups. So we call this diethyl ether. So diethyl ether is the most famous ether. So famous that usually it's just referred to as ether. But diethyl ether would be one way to name this molecule. Next we're gonna look at a thiol."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So diethyl ether is the most famous ether. So famous that usually it's just referred to as ether. But diethyl ether would be one way to name this molecule. Next we're gonna look at a thiol. So a thiol is similar to an alcohol. Instead of an OH though, we have an SH. So we have sulfur instead of an oxygen."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "Next we're gonna look at a thiol. So a thiol is similar to an alcohol. Instead of an OH though, we have an SH. So we have sulfur instead of an oxygen. So for the example of a thiol over here on the right, we have an SH here, and then we have an ethyl group. So one name for this would be ethane thiol. So I'm gonna write this out here."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So we have sulfur instead of an oxygen. So for the example of a thiol over here on the right, we have an SH here, and then we have an ethyl group. So one name for this would be ethane thiol. So I'm gonna write this out here. So ethane thiol is one possible name for this molecule. Next we have a sulfide. So a sulfide is similar to an ether."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna write this out here. So ethane thiol is one possible name for this molecule. Next we have a sulfide. So a sulfide is similar to an ether. Remember for an ether we had ROR. For a sulfide we have RSR. So it's completely analogous to an ether."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So a sulfide is similar to an ether. Remember for an ether we had ROR. For a sulfide we have RSR. So it's completely analogous to an ether. So on the right here's our sulfur, and again we have two ethyl groups. So you could call this diethyl sulfide. So let me write that out here."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So it's completely analogous to an ether. So on the right here's our sulfur, and again we have two ethyl groups. So you could call this diethyl sulfide. So let me write that out here. So this is diethyl sulfide. And finally for this video, one more functional group. So this is an amine."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So let me write that out here. So this is diethyl sulfide. And finally for this video, one more functional group. So this is an amine. So for this amine we have nitrogen, and then we have three R groups attached to nitrogen. Or you could be talking about hydrogens here. So any of these could be hydrogens as well."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So this is an amine. So for this amine we have nitrogen, and then we have three R groups attached to nitrogen. Or you could be talking about hydrogens here. So any of these could be hydrogens as well. And so on the right is an example of an amine. We have a nitrogen with a lone pair of electrons. This one has two ethyl groups, and one of our groups here is a hydrogen."}, {"video_title": "Functional groups Alkanes, cycloalkanes, and functional groups Organic chemistry Khan Academy.mp3", "Sentence": "So any of these could be hydrogens as well. And so on the right is an example of an amine. We have a nitrogen with a lone pair of electrons. This one has two ethyl groups, and one of our groups here is a hydrogen. So one name for this is diethylamine. So let me write out that one. So you could call this diethylamine."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "We're going to use a definition that Linus Pauling gives in his book, The Nature of the Chemical Bond. So Linus Pauling says that electronegativity refers to the power of an atom in a molecule to attract electrons to itself. So if I look at a molecule, I'm going to compare two atoms in that molecule. I'm going to compare carbon to oxygen in terms of the electronegativity. And to do that, I need to look over here on the right at the organic periodic table, which shows the elements most commonly used in organic chemistry. And then in blue, it gives us the Pauling scale for electronegativity. So Linus Pauling actually calculated electronegativity values for the elements and put them into the table."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "I'm going to compare carbon to oxygen in terms of the electronegativity. And to do that, I need to look over here on the right at the organic periodic table, which shows the elements most commonly used in organic chemistry. And then in blue, it gives us the Pauling scale for electronegativity. So Linus Pauling actually calculated electronegativity values for the elements and put them into the table. And that allows us to compare different elements in terms of their electronegativities. For example, we are concerned with carbon, which has an electronegativity value of 2.5. And we're going to compare that to oxygen, which has an electronegativity value of 3.5."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So Linus Pauling actually calculated electronegativity values for the elements and put them into the table. And that allows us to compare different elements in terms of their electronegativities. For example, we are concerned with carbon, which has an electronegativity value of 2.5. And we're going to compare that to oxygen, which has an electronegativity value of 3.5. So oxygen is more electronegative than carbon. And the definition tells us that if oxygen is more electronegative, oxygen has a greater power to attract electrons to itself than carbon does. And so if you think about the electrons in the covalent bond between carbon and oxygen that are shared, they're shared unequally because oxygen is more electronegative."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And we're going to compare that to oxygen, which has an electronegativity value of 3.5. So oxygen is more electronegative than carbon. And the definition tells us that if oxygen is more electronegative, oxygen has a greater power to attract electrons to itself than carbon does. And so if you think about the electrons in the covalent bond between carbon and oxygen that are shared, they're shared unequally because oxygen is more electronegative. Oxygen is going to pull those electrons in red closer to itself. And since electrons are negatively charged, the oxygen is going to get a little bit more negative charge. And so it's going to have what we call a partial negative charge on it."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And so if you think about the electrons in the covalent bond between carbon and oxygen that are shared, they're shared unequally because oxygen is more electronegative. Oxygen is going to pull those electrons in red closer to itself. And since electrons are negatively charged, the oxygen is going to get a little bit more negative charge. And so it's going to have what we call a partial negative charge on it. So partial negative. Its partial sign is a lowercase Greek letter delta. And so the oxygen is partially negative."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And so it's going to have what we call a partial negative charge on it. So partial negative. Its partial sign is a lowercase Greek letter delta. And so the oxygen is partially negative. It's pulling the electrons in red closer to itself. Another way to show the movement of those electrons in red closer to the oxygen would be this funny arrow here. So the arrow points in the direction of the movement of the electrons in red."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And so the oxygen is partially negative. It's pulling the electrons in red closer to itself. Another way to show the movement of those electrons in red closer to the oxygen would be this funny arrow here. So the arrow points in the direction of the movement of the electrons in red. So carbon is losing some of those electrons in red. Carbon is losing a little bit of electron density. Carbon is losing a little bit of negative charge."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So the arrow points in the direction of the movement of the electrons in red. So carbon is losing some of those electrons in red. Carbon is losing a little bit of electron density. Carbon is losing a little bit of negative charge. So carbon used to be neutral. But since it's losing a little bit of negative charge, this carbon will end up being partially positive like that. So the carbon is partially positive."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Carbon is losing a little bit of negative charge. So carbon used to be neutral. But since it's losing a little bit of negative charge, this carbon will end up being partially positive like that. So the carbon is partially positive. And the oxygen is partially negative. That's a polarized situation. You have a little bit of negative charge on one side, a little bit of positive charge on the other side."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So the carbon is partially positive. And the oxygen is partially negative. That's a polarized situation. You have a little bit of negative charge on one side, a little bit of positive charge on the other side. So it's still a covalent bond. But it's a polarized covalent bond due to the differences in electronegativities between those two atoms. Let's do a few more examples here where we show the differences in electronegativity."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "You have a little bit of negative charge on one side, a little bit of positive charge on the other side. So it's still a covalent bond. But it's a polarized covalent bond due to the differences in electronegativities between those two atoms. Let's do a few more examples here where we show the differences in electronegativity. So if we're thinking about a molecule that has two carbons in it, and I'm thinking about what happens to the electrons in red. Well, for this example, each carbon has the same value for electronegativity. So the carbon on the left has a value of 2.5."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Let's do a few more examples here where we show the differences in electronegativity. So if we're thinking about a molecule that has two carbons in it, and I'm thinking about what happens to the electrons in red. Well, for this example, each carbon has the same value for electronegativity. So the carbon on the left has a value of 2.5. The carbon on the right has a value of 2.5. That's a difference in electronegativity of 0, which means that the electrons in red aren't going to move towards one carbon or towards the other carbon. They're going to stay in the middle."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So the carbon on the left has a value of 2.5. The carbon on the right has a value of 2.5. That's a difference in electronegativity of 0, which means that the electrons in red aren't going to move towards one carbon or towards the other carbon. They're going to stay in the middle. They're going to be shared between those two atoms. So this is a covalent bond. And there's no polarity situation created here since there's no difference in electronegativity."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "They're going to stay in the middle. They're going to be shared between those two atoms. So this is a covalent bond. And there's no polarity situation created here since there's no difference in electronegativity. So we call this a nonpolar covalent bond. So this is a nonpolar covalent bond like that. Let's do another example."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And there's no polarity situation created here since there's no difference in electronegativity. So we call this a nonpolar covalent bond. So this is a nonpolar covalent bond like that. Let's do another example. Let's compare carbon to hydrogen. So if I had a molecule, and I have a bond between carbon and hydrogen, and I want to know what happens to the electrons in red between the carbon and the hydrogen, we've seen that carbon has an electronegativity value of 2.5. And we go up here to hydrogen, which has a value of 2.1."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Let's do another example. Let's compare carbon to hydrogen. So if I had a molecule, and I have a bond between carbon and hydrogen, and I want to know what happens to the electrons in red between the carbon and the hydrogen, we've seen that carbon has an electronegativity value of 2.5. And we go up here to hydrogen, which has a value of 2.1. So that's a difference of 0.4. So there is a difference in electronegativity between those two atoms. But it's a very small difference."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And we go up here to hydrogen, which has a value of 2.1. So that's a difference of 0.4. So there is a difference in electronegativity between those two atoms. But it's a very small difference. And so most textbooks would consider the bond between carbon and hydrogen to still be a nonpolar covalent bond. Let's go ahead and put in the example we did above, where we compared the electronegativities of carbon and oxygen like that. When we looked up the values, we saw that carbon had an electronegativity value of 2.5, and oxygen had a value of 3.5 for a difference of 1."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "But it's a very small difference. And so most textbooks would consider the bond between carbon and hydrogen to still be a nonpolar covalent bond. Let's go ahead and put in the example we did above, where we compared the electronegativities of carbon and oxygen like that. When we looked up the values, we saw that carbon had an electronegativity value of 2.5, and oxygen had a value of 3.5 for a difference of 1. And that's enough to have a polar covalent bond. This is a polar covalent bond between the carbon and the oxygen. So when we think about the electrons in red, the electrons in red are pulled closer to the oxygen, giving the oxygen a partial negative charge."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "When we looked up the values, we saw that carbon had an electronegativity value of 2.5, and oxygen had a value of 3.5 for a difference of 1. And that's enough to have a polar covalent bond. This is a polar covalent bond between the carbon and the oxygen. So when we think about the electrons in red, the electrons in red are pulled closer to the oxygen, giving the oxygen a partial negative charge. And since electron density is moving away from the carbon, the carbon gets a partial positive charge. And so we can see that if your difference in electronegativity is 1, it's considered to be a polar covalent bond. And if your difference in electronegativity is 0.4, that's considered to be a nonpolar covalent bond."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So when we think about the electrons in red, the electrons in red are pulled closer to the oxygen, giving the oxygen a partial negative charge. And since electron density is moving away from the carbon, the carbon gets a partial positive charge. And so we can see that if your difference in electronegativity is 1, it's considered to be a polar covalent bond. And if your difference in electronegativity is 0.4, that's considered to be a nonpolar covalent bond. So somewhere in between there must be the difference between nonpolar covalent bond and a polar covalent bond. And most textbooks will tell you approximately somewhere in the 0.5 range. So if the difference in electronegativity is greater than 0.5, you can go ahead and consider it to be mostly a polar covalent bond."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And if your difference in electronegativity is 0.4, that's considered to be a nonpolar covalent bond. So somewhere in between there must be the difference between nonpolar covalent bond and a polar covalent bond. And most textbooks will tell you approximately somewhere in the 0.5 range. So if the difference in electronegativity is greater than 0.5, you can go ahead and consider it to be mostly a polar covalent bond. If the difference in electronegativity is less than 0.5, we would consider that to be a nonpolar covalent bond. Now, I should point out that we're using the Pauling scale for electronegativity here. And there are several different scales for electronegativity."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So if the difference in electronegativity is greater than 0.5, you can go ahead and consider it to be mostly a polar covalent bond. If the difference in electronegativity is less than 0.5, we would consider that to be a nonpolar covalent bond. Now, I should point out that we're using the Pauling scale for electronegativity here. And there are several different scales for electronegativity. So these numbers are not absolute. These are more relative differences. And it's the relative difference in electronegativity that we care the most about."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And there are several different scales for electronegativity. So these numbers are not absolute. These are more relative differences. And it's the relative difference in electronegativity that we care the most about. Let's do another example. Let's compare oxygen to hydrogen. So let's think about what happens to the electrons between oxygen and hydrogen, so the electrons in red here."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And it's the relative difference in electronegativity that we care the most about. Let's do another example. Let's compare oxygen to hydrogen. So let's think about what happens to the electrons between oxygen and hydrogen, so the electrons in red here. All right, so we've already seen the electronegativity values for both of these atoms. Oxygen had a value of 3.5. And hydrogen had a value of 2.1."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So let's think about what happens to the electrons between oxygen and hydrogen, so the electrons in red here. All right, so we've already seen the electronegativity values for both of these atoms. Oxygen had a value of 3.5. And hydrogen had a value of 2.1. So that's an electronegativity difference of 1.4. So this is a polar covalent bond. Since oxygen is more electronegative than hydrogen, the electrons in red are going to move closer to the oxygen."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And hydrogen had a value of 2.1. So that's an electronegativity difference of 1.4. So this is a polar covalent bond. Since oxygen is more electronegative than hydrogen, the electrons in red are going to move closer to the oxygen. So the oxygen is going to get a partial negative charge. And the hydrogen is going to get a partial positive charge, like that. All right, let's do carbon and lithium now."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Since oxygen is more electronegative than hydrogen, the electrons in red are going to move closer to the oxygen. So the oxygen is going to get a partial negative charge. And the hydrogen is going to get a partial positive charge, like that. All right, let's do carbon and lithium now. So if I go ahead and draw a bond between carbon and lithium. And once again, we are concerned with the two electrons between carbon and lithium. The electronegativity value for carbon we've seen is 2.5."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "All right, let's do carbon and lithium now. So if I go ahead and draw a bond between carbon and lithium. And once again, we are concerned with the two electrons between carbon and lithium. The electronegativity value for carbon we've seen is 2.5. We need to go back up to our periodic table to find the electronegativity value for lithium. So I go up here, and I find lithium in group one of my periodic table has an electronegativity value of 1. So I go back down here, and I go ahead and put in a 1."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "The electronegativity value for carbon we've seen is 2.5. We need to go back up to our periodic table to find the electronegativity value for lithium. So I go up here, and I find lithium in group one of my periodic table has an electronegativity value of 1. So I go back down here, and I go ahead and put in a 1. And so that's a difference in electronegativity of 1.5. So we could consider this to be a polar covalent bond. This time, carbon is more electronegative than lithium."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So I go back down here, and I go ahead and put in a 1. And so that's a difference in electronegativity of 1.5. So we could consider this to be a polar covalent bond. This time, carbon is more electronegative than lithium. So the electrons in red are going to move closer to the carbon atom. And so the carbon is going to have a little bit more electron density than usual. So it's going to be partially negative."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "This time, carbon is more electronegative than lithium. So the electrons in red are going to move closer to the carbon atom. And so the carbon is going to have a little bit more electron density than usual. So it's going to be partially negative. And the lithium is losing electron density. So we're going to say that lithium is partially positive. Now here, I'm treating this bond as a polar covalent bond."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So it's going to be partially negative. And the lithium is losing electron density. So we're going to say that lithium is partially positive. Now here, I'm treating this bond as a polar covalent bond. But you'll see in a few minutes that we could also consider this to be an ionic bond. And that just depends on what electronegativity values you're dealing with, what type of chemical reaction that you're working with. So we could consider this to be an ionic bond."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Now here, I'm treating this bond as a polar covalent bond. But you'll see in a few minutes that we could also consider this to be an ionic bond. And that just depends on what electronegativity values you're dealing with, what type of chemical reaction that you're working with. So we could consider this to be an ionic bond. Let's go ahead and do an example of a compound that we know for sure is ionic. Sodium chloride, of course, would be the famous example. So to start with, I'm going to pretend like there's a covalent bond between the sodium and the chlorine."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So we could consider this to be an ionic bond. Let's go ahead and do an example of a compound that we know for sure is ionic. Sodium chloride, of course, would be the famous example. So to start with, I'm going to pretend like there's a covalent bond between the sodium and the chlorine. So I'm going to say there's a covalent bond to start with. And we'll put in our electrons. And we know that this bond consists of two electrons like that."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So to start with, I'm going to pretend like there's a covalent bond between the sodium and the chlorine. So I'm going to say there's a covalent bond to start with. And we'll put in our electrons. And we know that this bond consists of two electrons like that. Let's look at the differences in electronegativity between sodium and chlorine. So I'm going to go back up here. I'm going to find sodium, which has a value of 0.9, and chlorine, which has a value of 3."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And we know that this bond consists of two electrons like that. Let's look at the differences in electronegativity between sodium and chlorine. So I'm going to go back up here. I'm going to find sodium, which has a value of 0.9, and chlorine, which has a value of 3. So 0.9 for sodium and 3 for chlorine. So sodium's value is 0.9. Chlorine's is 3."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "I'm going to find sodium, which has a value of 0.9, and chlorine, which has a value of 3. So 0.9 for sodium and 3 for chlorine. So sodium's value is 0.9. Chlorine's is 3. That's a large difference in electronegativity. That's a difference of 2.1. And so chlorine is much more electronegative than sodium."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Chlorine's is 3. That's a large difference in electronegativity. That's a difference of 2.1. And so chlorine is much more electronegative than sodium. And it turns out it's so much more electronegative that it's no longer going to share electrons with sodium. It's going to steal those electrons. So when I redraw it here, I'm going to show chlorine being surrounded by eight electrons."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And so chlorine is much more electronegative than sodium. And it turns out it's so much more electronegative that it's no longer going to share electrons with sodium. It's going to steal those electrons. So when I redraw it here, I'm going to show chlorine being surrounded by eight electrons. So these two electrons in red, let me go ahead and show them, these two electrons in red here between the sodium and the chlorine, since chlorine is so much more electronegative, it's going to attract those two electrons in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine like that. And so the sodium is left over here."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So when I redraw it here, I'm going to show chlorine being surrounded by eight electrons. So these two electrons in red, let me go ahead and show them, these two electrons in red here between the sodium and the chlorine, since chlorine is so much more electronegative, it's going to attract those two electrons in red so strongly that it completely steals them. So those two electrons in red are going to be stolen by the chlorine like that. And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And so the sodium is left over here. And so chlorine has an extra electron, which gives it a negative 1 formal charge. So we're no longer talking about partial charges here. Chlorine gets a full negative 1 formal charge. Sodium lost an electron. So it ends up with a positive formal charge like that. And so we know that this is an ionic bond between these two ions."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "Chlorine gets a full negative 1 formal charge. Sodium lost an electron. So it ends up with a positive formal charge like that. And so we know that this is an ionic bond between these two ions. So this represents an ionic bond. And so the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks will see approximately somewhere around 1.7."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And so we know that this is an ionic bond between these two ions. So this represents an ionic bond. And so the difference in electronegativity is somewhere between 1.5 and 2.1, between a polar covalent bond and an ionic bond. So most textbooks will see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond, lower than 1.7 in the polar covalent range. But that doesn't always have to be the case. So we'll come back now to the example between carbon and lithium."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So most textbooks will see approximately somewhere around 1.7. So if you're higher than 1.7, it's generally considered to be mostly an ionic bond, lower than 1.7 in the polar covalent range. But that doesn't always have to be the case. So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So we'll come back now to the example between carbon and lithium. So if we go back up here to carbon and lithium, here we treat it like a polar covalent bond. But sometimes you might want to treat the bond in red as being an ionic bond. So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved onto the carbon atom. So it's no longer sharing it with the lithium."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw a picture of carbon and lithium where we're treating it as an ionic bond. So if carbon is more electronegative than lithium, carbon's going to steal the two electrons in red. So I'll go ahead and show the electrons in red have now moved onto the carbon atom. So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So it's no longer sharing it with the lithium. Carbon has stolen those electrons. And lithium is over here. So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here we're treating it like an ionic bond, full formal charges here. And this is useful for some organic chemistry reactions."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "So lithium lost one of its electrons, giving it a plus 1 formal charge. Carbon gained an electron, giving it a negative 1 formal charge. And so here we're treating it like an ionic bond, full formal charges here. And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute. It's a relative thing. You could draw the dot structure above."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "And this is useful for some organic chemistry reactions. And so what I'm trying to point out here is these divisions, 1.7, it's not absolute. It's a relative thing. You could draw the dot structure above. And this would be considered to be correct. You could draw it like this. Or you could treat it like an ionic bond down here, since it's relatively close to the cutoff."}, {"video_title": "Electronegativity and bonding Chemical bonds Chemistry Khan Academy.mp3", "Sentence": "You could draw the dot structure above. And this would be considered to be correct. You could draw it like this. Or you could treat it like an ionic bond down here, since it's relatively close to the cutoff. So this is an overview of electronegativity. And even though we've been dealing with numbers in this video, in future videos, we don't care so much about the numbers. We care about the relative differences in electronegativity."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Let's get some practice drawing Newman projections. Our goal is to look down the carbon two, carbon three bond for this compound, and in part A, our goal is to draw the most stable conformation. So let's number our carbons. This carbon, we could say, is number one. This carbon is number two. This carbon is number three. And this carbon is number four."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "This carbon, we could say, is number one. This carbon is number two. This carbon is number three. And this carbon is number four. If we're going to look down the C two, C three bond, that's this bond right here. So we're going to put our eye along this axis. If I draw this out right here, we're going to put our eye right here."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon is number four. If we're going to look down the C two, C three bond, that's this bond right here. So we're going to put our eye along this axis. If I draw this out right here, we're going to put our eye right here. And we're going to stare down the carbon two, carbon three bond. And we're going to draw what we see. Our goal is to find the most stable conformation."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "If I draw this out right here, we're going to put our eye right here. And we're going to stare down the carbon two, carbon three bond. And we're going to draw what we see. Our goal is to find the most stable conformation. And we know from earlier videos, that would be a staggered conformation. So we'll draw one staggered conformation, and then we'll draw the rest. And then we'll pick which one is the lowest energy, which one is the most stable."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Our goal is to find the most stable conformation. And we know from earlier videos, that would be a staggered conformation. So we'll draw one staggered conformation, and then we'll draw the rest. And then we'll pick which one is the lowest energy, which one is the most stable. So here's carbon one. And then we have carbon two, with the hydrogen coming out at us, and a methyl group going away from us in space. Attached to carbon three, we have a methyl group coming out at us in space, and a hydrogen going away from us in space."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll pick which one is the lowest energy, which one is the most stable. So here's carbon one. And then we have carbon two, with the hydrogen coming out at us, and a methyl group going away from us in space. Attached to carbon three, we have a methyl group coming out at us in space, and a hydrogen going away from us in space. And then we have carbon four. Our job is to stare down the carbon C two, C three bond, and draw our Newman projection. So if I rotate it, so we're staring down that C two, C three bond, we can see a Newman projection."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Attached to carbon three, we have a methyl group coming out at us in space, and a hydrogen going away from us in space. And then we have carbon four. Our job is to stare down the carbon C two, C three bond, and draw our Newman projection. So if I rotate it, so we're staring down that C two, C three bond, we can see a Newman projection. We can see a staggered conformation for our compound. To get another staggered conformation, I could rotate the front carbon and keep the back carbon stationary. And that gives us another staggered conformation."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So if I rotate it, so we're staring down that C two, C three bond, we can see a Newman projection. We can see a staggered conformation for our compound. To get another staggered conformation, I could rotate the front carbon and keep the back carbon stationary. And that gives us another staggered conformation. And I could do it again. I could rotate the front carbon to get another possible staggered conformation. In the video, we stared down our C two, C three bond."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And that gives us another staggered conformation. And I could do it again. I could rotate the front carbon to get another possible staggered conformation. In the video, we stared down our C two, C three bond. So this bond right here. We put our eye here. And we looked at the possible staggered conformations for our compound."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "In the video, we stared down our C two, C three bond. So this bond right here. We put our eye here. And we looked at the possible staggered conformations for our compound. It's important to be able to draw the Newman projections without using a model set. So let's go ahead and do that. So if we're staring at carbon two, so this is carbon two right here."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And we looked at the possible staggered conformations for our compound. It's important to be able to draw the Newman projections without using a model set. So let's go ahead and do that. So if we're staring at carbon two, so this is carbon two right here. And our Newman projection, that's represented by a point. So I draw a point here. And then we have a hydrogen going up and to the right."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So if we're staring at carbon two, so this is carbon two right here. And our Newman projection, that's represented by a point. So I draw a point here. And then we have a hydrogen going up and to the right. So if your eye is right here, your hydrogen goes up and to the right from this perspective. So we have a hydrogen going up and to the right. We have a methyl group going up and to the left."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a hydrogen going up and to the right. So if your eye is right here, your hydrogen goes up and to the right from this perspective. So we have a hydrogen going up and to the right. We have a methyl group going up and to the left. So a CH three up and to the left. And then finally, we have another methyl group going straight down. So a CH three going down."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "We have a methyl group going up and to the left. So a CH three up and to the left. And then finally, we have another methyl group going straight down. So a CH three going down. For carbon three, remember carbon three is this one right here. You can't see carbon three if your eye is right here. Carbon two is in the way."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So a CH three going down. For carbon three, remember carbon three is this one right here. You can't see carbon three if your eye is right here. Carbon two is in the way. But in our Newman projection, we represent that with a circle. So this circle represents carbon three. And what's attached to carbon three?"}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Carbon two is in the way. But in our Newman projection, we represent that with a circle. So this circle represents carbon three. And what's attached to carbon three? There's a methyl group going straight up. So again, if your eye is here, this methyl group is up. So we have a CH three going up."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And what's attached to carbon three? There's a methyl group going straight up. So again, if your eye is here, this methyl group is up. So we have a CH three going up. We have another CH three going down and to the right. So this is going down and to the right. And then we have a hydrogen going down and to the left."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So we have a CH three going up. We have another CH three going down and to the right. So this is going down and to the right. And then we have a hydrogen going down and to the left. So here is our Newman projection for a staggered conformation. Next, we rotated the front carbon and held the back carbon stationary to get another staggered conformation. So let's go ahead and do that."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a hydrogen going down and to the left. So here is our Newman projection for a staggered conformation. Next, we rotated the front carbon and held the back carbon stationary to get another staggered conformation. So let's go ahead and do that. Let's hold the back carbon stationary. We're going to take this methyl group. And we're going to rotate it all the way over to this position."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and do that. Let's hold the back carbon stationary. We're going to take this methyl group. And we're going to rotate it all the way over to this position. If we do that, then this hydrogen rotates all the way over to this position. And finally, this methyl group would rotate over to this position. The back carbon stationary, so let's go ahead and draw the back carbon in a circle."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to rotate it all the way over to this position. If we do that, then this hydrogen rotates all the way over to this position. And finally, this methyl group would rotate over to this position. The back carbon stationary, so let's go ahead and draw the back carbon in a circle. And the CH three is going to stay in the same spot. This CH three stays in the same spot. And this hydrogen stays in the same spot."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "The back carbon stationary, so let's go ahead and draw the back carbon in a circle. And the CH three is going to stay in the same spot. This CH three stays in the same spot. And this hydrogen stays in the same spot. So we rotated the front carbon. It doesn't really matter if you rotate the front or the back carbon. But let's go ahead and draw in our groups here on our front carbon."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And this hydrogen stays in the same spot. So we rotated the front carbon. It doesn't really matter if you rotate the front or the back carbon. But let's go ahead and draw in our groups here on our front carbon. So here's C two. And we moved the methyl group in magenta over to here. So let me go ahead and draw in that methyl group."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "But let's go ahead and draw in our groups here on our front carbon. So here's C two. And we moved the methyl group in magenta over to here. So let me go ahead and draw in that methyl group. That methyl group moved over to here. Next, the hydrogen in blue moved down here. So I can draw the hydrogen in blue here, my Newman projection."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw in that methyl group. That methyl group moved over to here. Next, the hydrogen in blue moved down here. So I can draw the hydrogen in blue here, my Newman projection. And then finally, the methyl group over here in red moved over to this position. So the methyl group in red moved over to here. And now we have another staggered conformation."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So I can draw the hydrogen in blue here, my Newman projection. And then finally, the methyl group over here in red moved over to this position. So the methyl group in red moved over to here. And now we have another staggered conformation. For our last staggered conformation, we're going to do the same thing. We're going to take the methyl group in magenta, rotate it over to here. And then that would cause the hydrogen right here to rotate over to this position and the methyl group in red to rotate over to this position."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And now we have another staggered conformation. For our last staggered conformation, we're going to do the same thing. We're going to take the methyl group in magenta, rotate it over to here. And then that would cause the hydrogen right here to rotate over to this position and the methyl group in red to rotate over to this position. So we keep the back carbon stationary. So we draw that in as a circle. Again, we put in our CH three."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then that would cause the hydrogen right here to rotate over to this position and the methyl group in red to rotate over to this position. So we keep the back carbon stationary. So we draw that in as a circle. Again, we put in our CH three. We put in our CH three. We put in our hydrogen. Those aren't moving."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Again, we put in our CH three. We put in our CH three. We put in our hydrogen. Those aren't moving. And the methyl group in magenta is now down. So this is the methyl group in magenta. The hydrogen moved over to this position."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Those aren't moving. And the methyl group in magenta is now down. So this is the methyl group in magenta. The hydrogen moved over to this position. And finally, the methyl group in red moved over to this position. So now we have our staggered conformations. We can double check ourselves by comparing these Newman projections to what we saw in the video."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen moved over to this position. And finally, the methyl group in red moved over to this position. So now we have our staggered conformations. We can double check ourselves by comparing these Newman projections to what we saw in the video. So here are the staggered conformations. These are stills from the video. And you can see they match the Newman projections that we drew."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "We can double check ourselves by comparing these Newman projections to what we saw in the video. So here are the staggered conformations. These are stills from the video. And you can see they match the Newman projections that we drew. Finally, we need to choose the most stable out of these three. So what is the most stable conformation? Well, we know that when we have this situation, let me go ahead and use blue for this."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And you can see they match the Newman projections that we drew. Finally, we need to choose the most stable out of these three. So what is the most stable conformation? Well, we know that when we have this situation, let me go ahead and use blue for this. So when we have a methyl group here that are 60 degrees apart, this is a Gauss interaction. So we have a methyl-methyl-Gauss interaction. And from the video on butane, conformations on butane, we saw that a Gauss interaction has 3.8 kilojoules per mole as an energy cost."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Well, we know that when we have this situation, let me go ahead and use blue for this. So when we have a methyl group here that are 60 degrees apart, this is a Gauss interaction. So we have a methyl-methyl-Gauss interaction. And from the video on butane, conformations on butane, we saw that a Gauss interaction has 3.8 kilojoules per mole as an energy cost. So we have 3.8 kilojoules per mole for this Gauss interaction. And we have another Gauss interaction here, so another 3.8 kilojoules per mole, and another Gauss interaction. So there are three Gauss interactions for this conformation for a total of 11.4."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And from the video on butane, conformations on butane, we saw that a Gauss interaction has 3.8 kilojoules per mole as an energy cost. So we have 3.8 kilojoules per mole for this Gauss interaction. And we have another Gauss interaction here, so another 3.8 kilojoules per mole, and another Gauss interaction. So there are three Gauss interactions for this conformation for a total of 11.4. So the total would be 11.4 kilojoules per mole as an energy cost, so 3.8 times 3. What about this conformation? Well, here's a Gauss interaction, here's a Gauss interaction, and here's a Gauss interaction."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So there are three Gauss interactions for this conformation for a total of 11.4. So the total would be 11.4 kilojoules per mole as an energy cost, so 3.8 times 3. What about this conformation? Well, here's a Gauss interaction, here's a Gauss interaction, and here's a Gauss interaction. So again, we have three Gauss interactions. So for this conformation, the total energy cost is also 11.4 kilojoules per mole. What about this conformation, our first one?"}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Well, here's a Gauss interaction, here's a Gauss interaction, and here's a Gauss interaction. So again, we have three Gauss interactions. So for this conformation, the total energy cost is also 11.4 kilojoules per mole. What about this conformation, our first one? Well, here's a Gauss interaction, so that's 3.8 kilojoules per mole, and here's a Gauss interaction. We have only two for this conformation, so only two Gauss interactions. That would be a total of 7.6 kilojoules per mole, which is the lowest energy."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "What about this conformation, our first one? Well, here's a Gauss interaction, so that's 3.8 kilojoules per mole, and here's a Gauss interaction. We have only two for this conformation, so only two Gauss interactions. That would be a total of 7.6 kilojoules per mole, which is the lowest energy. So this is the most stable conformation. In part B, our goal is to draw the least stable conformation. And we know that the least stable conformation is the one that's the highest in energy."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "That would be a total of 7.6 kilojoules per mole, which is the lowest energy. So this is the most stable conformation. In part B, our goal is to draw the least stable conformation. And we know that the least stable conformation is the one that's the highest in energy. So let's go to the video where I take the model set and I go from the staggered conformation to an eclipsed conformation, and then we look at all the possible eclipsed conformations. One of them is gonna be the least stable. Here we start with our staggered conformation, and if I rotate it a little bit, we get an eclipsed conformation."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And we know that the least stable conformation is the one that's the highest in energy. So let's go to the video where I take the model set and I go from the staggered conformation to an eclipsed conformation, and then we look at all the possible eclipsed conformations. One of them is gonna be the least stable. Here we start with our staggered conformation, and if I rotate it a little bit, we get an eclipsed conformation. I left it a little bit off so you could still see the back bonds. I rotate again and get another eclipsed conformation. Now for this one, if I turn it to the side, you can see these methyl groups are really close together in space."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Here we start with our staggered conformation, and if I rotate it a little bit, we get an eclipsed conformation. I left it a little bit off so you could still see the back bonds. I rotate again and get another eclipsed conformation. Now for this one, if I turn it to the side, you can see these methyl groups are really close together in space. And so that steric hindrance would destabilize this conformation. So if we go back to the eclipsed conformation, we rotate again, we get to the final eclipsed conformation. Here are the pictures of the eclipsed conformations from the video, and our goal is to pick the least stable."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Now for this one, if I turn it to the side, you can see these methyl groups are really close together in space. And so that steric hindrance would destabilize this conformation. So if we go back to the eclipsed conformation, we rotate again, we get to the final eclipsed conformation. Here are the pictures of the eclipsed conformations from the video, and our goal is to pick the least stable. But just for practice, let's try drawing all of them as Newman projections. So we'll start with the one on the left here. And we're staring at C2, so that's C2, which we represent by a point."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Here are the pictures of the eclipsed conformations from the video, and our goal is to pick the least stable. But just for practice, let's try drawing all of them as Newman projections. So we'll start with the one on the left here. And we're staring at C2, so that's C2, which we represent by a point. And then C2 has a methyl group going up, and I drew it a little bit to the right just so we can see what's going on behind it. And then we have a hydrogen going down and to the right, so there's a hydrogen bond to C2 going down and to the right. And then over here is another methyl group, so to the left is a CH3."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And we're staring at C2, so that's C2, which we represent by a point. And then C2 has a methyl group going up, and I drew it a little bit to the right just so we can see what's going on behind it. And then we have a hydrogen going down and to the right, so there's a hydrogen bond to C2 going down and to the right. And then over here is another methyl group, so to the left is a CH3. For the back carbon, even though we can't see it, we represent it with a circle. So the back carbon right there, which is C3. What's bonded to C3?"}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then over here is another methyl group, so to the left is a CH3. For the back carbon, even though we can't see it, we represent it with a circle. So the back carbon right there, which is C3. What's bonded to C3? Let me use a different color. And hopefully you can see there's a methyl group back here bonded to C3, so let's draw that in in red. And then there's another methyl group over here."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "What's bonded to C3? Let me use a different color. And hopefully you can see there's a methyl group back here bonded to C3, so let's draw that in in red. And then there's another methyl group over here. So this is a methyl group. I'll draw that one in, a CH3. And then we have a hydrogen down here."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then there's another methyl group over here. So this is a methyl group. I'll draw that one in, a CH3. And then we have a hydrogen down here. So here is the hydrogen. So that would be the Newman projection for this eclipsed conformation. Let's move on to the next picture, so this eclipsed conformation."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have a hydrogen down here. So here is the hydrogen. So that would be the Newman projection for this eclipsed conformation. Let's move on to the next picture, so this eclipsed conformation. So we're staring at C2, so that is our dot here. And then we can see there's a methyl group going up and to the right, so a little bit to the right. So there's a CH3 going that way."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Let's move on to the next picture, so this eclipsed conformation. So we're staring at C2, so that is our dot here. And then we can see there's a methyl group going up and to the right, so a little bit to the right. So there's a CH3 going that way. There's a CH3 going down, so this CH3 going down. And then we have a hydrogen going to the left, so hydrogen going to the left here. And then in the back, we have our carbon 3."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So there's a CH3 going that way. There's a CH3 going down, so this CH3 going down. And then we have a hydrogen going to the left, so hydrogen going to the left here. And then in the back, we have our carbon 3. And we held the back carbon stationary in the video, so there's no change. You can see there is a methyl group straight up here in the back, so CH3. And then there's a methyl group going to the right back here, so a CH3."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then in the back, we have our carbon 3. And we held the back carbon stationary in the video, so there's no change. You can see there is a methyl group straight up here in the back, so CH3. And then there's a methyl group going to the right back here, so a CH3. And then there's a hydrogen going to the left, so there's a hydrogen here. So here's the Newman projection for this conformation. And to go from the one on the left to the one on the right, we rotated the front carbon."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And then there's a methyl group going to the right back here, so a CH3. And then there's a hydrogen going to the left, so there's a hydrogen here. So here's the Newman projection for this conformation. And to go from the one on the left to the one on the right, we rotated the front carbon. So let me go ahead and show which carbon is which. So if you took this carbon and this methyl group, I should say, and rotated it over here, that methyl group in magenta becomes this methyl group. The hydrogen right here in blue gets rotated over to this position, so that's this hydrogen."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "And to go from the one on the left to the one on the right, we rotated the front carbon. So let me go ahead and show which carbon is which. So if you took this carbon and this methyl group, I should say, and rotated it over here, that methyl group in magenta becomes this methyl group. The hydrogen right here in blue gets rotated over to this position, so that's this hydrogen. And then finally, this methyl group right here in green would get rotated over to this position, so that's this methyl group. All right, for our last eclipsed conformation over here, we're staring at C2, so let's draw that C2 here. We have a hydrogen, a hydrogen going up a little bit to the right."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen right here in blue gets rotated over to this position, so that's this hydrogen. And then finally, this methyl group right here in green would get rotated over to this position, so that's this methyl group. All right, for our last eclipsed conformation over here, we're staring at C2, so let's draw that C2 here. We have a hydrogen, a hydrogen going up a little bit to the right. We have a methyl group going down, so we draw in that CH3. And then we have a methyl group going in this direction, so a CH3 here. We'll draw in the back carbon, so there is C3."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "We have a hydrogen, a hydrogen going up a little bit to the right. We have a methyl group going down, so we draw in that CH3. And then we have a methyl group going in this direction, so a CH3 here. We'll draw in the back carbon, so there is C3. Again, we didn't change anything. So again, there's a methyl group going straight up, so CH3, a methyl group going this way, and then, so that one right here, and then a hydrogen going this way. So here's our hydrogen."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "We'll draw in the back carbon, so there is C3. Again, we didn't change anything. So again, there's a methyl group going straight up, so CH3, a methyl group going this way, and then, so that one right here, and then a hydrogen going this way. So here's our hydrogen. If we show going from this conformation to this conformation, again, we'll start with the one in magenta. So this one in magenta was moved over to here, so that's this CH3. The hydrogen in blue was rotated over to this position, so that's this hydrogen."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "So here's our hydrogen. If we show going from this conformation to this conformation, again, we'll start with the one in magenta. So this one in magenta was moved over to here, so that's this CH3. The hydrogen in blue was rotated over to this position, so that's this hydrogen. And finally, the methyl group in green was rotated over to here. So now we have our eclipsed conformations, and let's analyze them and determine which one is the highest in energy. If we start with our first Newman projection, so this one right here, we have a methyl group eclipsing a hydrogen."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen in blue was rotated over to this position, so that's this hydrogen. And finally, the methyl group in green was rotated over to here. So now we have our eclipsed conformations, and let's analyze them and determine which one is the highest in energy. If we start with our first Newman projection, so this one right here, we have a methyl group eclipsing a hydrogen. And we know from earlier videos, that's an energy cost of 6 kilojoules per mole. Here we have another situation with a hydrogen and a methyl group eclipsing each other, so that's 6 kilojoules per mole. Finally, a methyl group eclipsing a methyl group, which we know is 11 kilojoules per mole."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "If we start with our first Newman projection, so this one right here, we have a methyl group eclipsing a hydrogen. And we know from earlier videos, that's an energy cost of 6 kilojoules per mole. Here we have another situation with a hydrogen and a methyl group eclipsing each other, so that's 6 kilojoules per mole. Finally, a methyl group eclipsing a methyl group, which we know is 11 kilojoules per mole. So that's a total of 23 kilojoules per mole, the energy cost for this eclipsed conformations. Let me go ahead and write that here. That's 23 kilojoules per mole."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "Finally, a methyl group eclipsing a methyl group, which we know is 11 kilojoules per mole. So that's a total of 23 kilojoules per mole, the energy cost for this eclipsed conformations. Let me go ahead and write that here. That's 23 kilojoules per mole. Let's go to the one on the right next, because you can see it's the same thing. We have a methyl group eclipsing a hydrogen, so that's 6. We have another situation with a hydrogen and a methyl group eclipsing each other, so that's 6."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "That's 23 kilojoules per mole. Let's go to the one on the right next, because you can see it's the same thing. We have a methyl group eclipsing a hydrogen, so that's 6. We have another situation with a hydrogen and a methyl group eclipsing each other, so that's 6. And we have a methyl group eclipsing a methyl group, so that's 11. So this one's also 23 kilojoules per mole. If we look at the one in the middle, it's a little bit different."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "We have another situation with a hydrogen and a methyl group eclipsing each other, so that's 6. And we have a methyl group eclipsing a methyl group, so that's 11. So this one's also 23 kilojoules per mole. If we look at the one in the middle, it's a little bit different. We have a pair of hydrogens eclipsing each other. We've already seen that's 4 kilojoules per mole, so that's 4 here. We have a methyl group eclipsing a methyl, which is 11."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "If we look at the one in the middle, it's a little bit different. We have a pair of hydrogens eclipsing each other. We've already seen that's 4 kilojoules per mole, so that's 4 here. We have a methyl group eclipsing a methyl, which is 11. And then we have another methyl group eclipsing a methyl, which is another 11. So what is 11 plus 11 plus 4? That's a total of 26 kilojoules per mole."}, {"video_title": "Newman projection practice 1 Organic chemistry Khan Academy.mp3", "Sentence": "We have a methyl group eclipsing a methyl, which is 11. And then we have another methyl group eclipsing a methyl, which is another 11. So what is 11 plus 11 plus 4? That's a total of 26 kilojoules per mole. So this is the highest in energy. This is the least stable conformation, the one where the methyl groups are the closest together in space, because these relatively bulky methyl groups, these destabilize this conformation. So you want to get the bulky groups as far away from each other as you possibly can."}, {"video_title": "The Moon.mp3", "Sentence": "Earth has a diameter of approximately 8,000 miles, while the moon has a diameter of approximately 2,200 miles, so a little bit more than 1 4th the diameter of Earth. Now the distance between the two is 239,000 miles, which you can imagine is incredibly far. Now one thing that is very interesting, and this is why the moon looks like it's the same size as the sun in the sky, even though it is 1 4 100th of the diameter, is the ratio between the distance to the moon and the diameter of the moon is roughly the same as the ratio of the distance of the sun to the diameter of the sun. So for the moon, that ratio, the distance to the moon, 239,000, we'll do everything in miles, over the radius of the moon, 2,200, it's approximately the same as the ratio between the distance from Earth to the sun, which would be 93 million miles over 865,000 miles in diameter. So this number is roughly 400 times larger than that number, and this number is roughly 400 times larger than that number, but what's neat is, and this depends on where Earth is in its orbit, but this is approximately equal to 108. This gives me goosebumps whenever I think about it, because it didn't have to be this way, that the moon and the sun, even though they have these vastly different diameters, that they look roughly the same size in the sky because the ratio of the distances is comparable to the ratio of their diameters. And this is why the number 108 is actually an auspicious number in some cultures, especially in Hinduism."}, {"video_title": "The Moon.mp3", "Sentence": "So for the moon, that ratio, the distance to the moon, 239,000, we'll do everything in miles, over the radius of the moon, 2,200, it's approximately the same as the ratio between the distance from Earth to the sun, which would be 93 million miles over 865,000 miles in diameter. So this number is roughly 400 times larger than that number, and this number is roughly 400 times larger than that number, but what's neat is, and this depends on where Earth is in its orbit, but this is approximately equal to 108. This gives me goosebumps whenever I think about it, because it didn't have to be this way, that the moon and the sun, even though they have these vastly different diameters, that they look roughly the same size in the sky because the ratio of the distances is comparable to the ratio of their diameters. And this is why the number 108 is actually an auspicious number in some cultures, especially in Hinduism. Now when you're discussing the moon, and especially the moon as viewed from Earth, one of the obvious questions is, why does the moon go through phases? Why do we see different portions of the moon lit up from day to day? This right over here is a diagram from NASA's website, and what you see here, clearly this is not at scale."}, {"video_title": "The Moon.mp3", "Sentence": "And this is why the number 108 is actually an auspicious number in some cultures, especially in Hinduism. Now when you're discussing the moon, and especially the moon as viewed from Earth, one of the obvious questions is, why does the moon go through phases? Why do we see different portions of the moon lit up from day to day? This right over here is a diagram from NASA's website, and what you see here, clearly this is not at scale. This picture over here, this inner picture, the size of the moon and the Earth is roughly at scale, but clearly the distances between them are not. You don't see that 239,000 miles between them. In this inner circle, what you see is that the moon and the Earth is always lit up from the right."}, {"video_title": "The Moon.mp3", "Sentence": "This right over here is a diagram from NASA's website, and what you see here, clearly this is not at scale. This picture over here, this inner picture, the size of the moon and the Earth is roughly at scale, but clearly the distances between them are not. You don't see that 239,000 miles between them. In this inner circle, what you see is that the moon and the Earth is always lit up from the right. So it's assuming that the sun is 93 million miles in that direction, and it is lighting up both the Earth and the moon from the right. Now as the moon rotates around the Earth in its 28-day cycle, and if you're wondering, gee, a 28-day cycle seems awfully close to a month, that is not a coincidence. The notion of a month comes from the cycles of the moon."}, {"video_title": "The Moon.mp3", "Sentence": "In this inner circle, what you see is that the moon and the Earth is always lit up from the right. So it's assuming that the sun is 93 million miles in that direction, and it is lighting up both the Earth and the moon from the right. Now as the moon rotates around the Earth in its 28-day cycle, and if you're wondering, gee, a 28-day cycle seems awfully close to a month, that is not a coincidence. The notion of a month comes from the cycles of the moon. In fact, even the word month and the word moon have the exact same root in Old English and in Proto-Germanic. They are essentially the same word. But what you see is the moon goes in this 28-day cycle."}, {"video_title": "The Moon.mp3", "Sentence": "The notion of a month comes from the cycles of the moon. In fact, even the word month and the word moon have the exact same root in Old English and in Proto-Germanic. They are essentially the same word. But what you see is the moon goes in this 28-day cycle. When the moon is roughly between the Earth and the sun, the lit up half of the moon is away from what we can see here on Earth, and so we see the non-lit up side, which would be a new moon. Now as the moon goes, this is viewing from above the Earth, this would be the North Pole right over here, as it goes counterclockwise, as viewed from above the Earth, we start to be able to see a little bit of that half of the moon that gets lit up. So when the moon is in this position, we see us from Earth, from this vantage point, we're able to see a little bit of the lit up side."}, {"video_title": "The Moon.mp3", "Sentence": "But what you see is the moon goes in this 28-day cycle. When the moon is roughly between the Earth and the sun, the lit up half of the moon is away from what we can see here on Earth, and so we see the non-lit up side, which would be a new moon. Now as the moon goes, this is viewing from above the Earth, this would be the North Pole right over here, as it goes counterclockwise, as viewed from above the Earth, we start to be able to see a little bit of that half of the moon that gets lit up. So when the moon is in this position, we see us from Earth, from this vantage point, we're able to see a little bit of the lit up side. When the moon is over here, we're able to see half the lit up side and half the non-lit up side, and that's called a first quarter moon. And that keeps on going. When we're halfway through our cycle here, at a full moon, the Earth is between the moon and the sun."}, {"video_title": "The Moon.mp3", "Sentence": "So when the moon is in this position, we see us from Earth, from this vantage point, we're able to see a little bit of the lit up side. When the moon is over here, we're able to see half the lit up side and half the non-lit up side, and that's called a first quarter moon. And that keeps on going. When we're halfway through our cycle here, at a full moon, the Earth is between the moon and the sun. And so from our vantage point, we are able to see the entire lit up side, and that's why it is a full moon. And then we keep going all the way until we get back to a new moon. Now one question that might be bothering you, it bothered me for many years, is as soon as I understood this, the cycle of the moon, how the moon has this 28-day cycle as Earth rotates around the sun, I always wondered, well in a new moon, it looks like the moon is between the Earth and the sun."}, {"video_title": "The Moon.mp3", "Sentence": "When we're halfway through our cycle here, at a full moon, the Earth is between the moon and the sun. And so from our vantage point, we are able to see the entire lit up side, and that's why it is a full moon. And then we keep going all the way until we get back to a new moon. Now one question that might be bothering you, it bothered me for many years, is as soon as I understood this, the cycle of the moon, how the moon has this 28-day cycle as Earth rotates around the sun, I always wondered, well in a new moon, it looks like the moon is between the Earth and the sun. Why doesn't it block out the sun every time we have a new moon? Why don't we have a solar eclipse every 28 days? Similarly, for the full moons, when we're in this position, it looks like the Earth is between the moon and the sun."}, {"video_title": "The Moon.mp3", "Sentence": "Now one question that might be bothering you, it bothered me for many years, is as soon as I understood this, the cycle of the moon, how the moon has this 28-day cycle as Earth rotates around the sun, I always wondered, well in a new moon, it looks like the moon is between the Earth and the sun. Why doesn't it block out the sun every time we have a new moon? Why don't we have a solar eclipse every 28 days? Similarly, for the full moons, when we're in this position, it looks like the Earth is between the moon and the sun. Why doesn't Earth's shadow block out the sun so that we have a lunar eclipse instead of a full moon, and we would have one of those every 28 days? Why don't we see that? Think about it yourself."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If we look at the molecule on the left and try to use our cis-trans terminology, we quickly realize that we can't use it. To use cis or trans, we would need to have two identical groups to compare. And here we have four different groups attached to our double bond. So we need to use a different system to find the configuration of our double bond. We're gonna use the E-Z system. So to use the E-Z system, you need to think about atomic number to assign priority to the groups attached to your double bond. So let's start with the carbon on the right side of our double bond."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we need to use a different system to find the configuration of our double bond. We're gonna use the E-Z system. So to use the E-Z system, you need to think about atomic number to assign priority to the groups attached to your double bond. So let's start with the carbon on the right side of our double bond. We look at the atoms directly bonded to that carbon. There's a hydrogen directly bonded to that carbon and there's an oxygen directly bonded to that carbon. Next, you need to compare those atoms in terms of atomic number."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with the carbon on the right side of our double bond. We look at the atoms directly bonded to that carbon. There's a hydrogen directly bonded to that carbon and there's an oxygen directly bonded to that carbon. Next, you need to compare those atoms in terms of atomic number. Hydrogen has an atomic number of one and oxygen has an atomic number of eight. The higher the atomic number, the higher the priority. So the group that contains oxygen is the higher priority group."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Next, you need to compare those atoms in terms of atomic number. Hydrogen has an atomic number of one and oxygen has an atomic number of eight. The higher the atomic number, the higher the priority. So the group that contains oxygen is the higher priority group. So this is the higher priority group and the hydrogen would be number two here. Now let's look at the left side of our double bond. So let's look at this carbon."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the group that contains oxygen is the higher priority group. So this is the higher priority group and the hydrogen would be number two here. Now let's look at the left side of our double bond. So let's look at this carbon. And we look at the atoms directly bonded to that carbon. There's a bromine directly bonded to that carbon and there's a carbon. We go over here and we see that carbon has atomic number of six and bromine has an atomic number of 35."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this carbon. And we look at the atoms directly bonded to that carbon. There's a bromine directly bonded to that carbon and there's a carbon. We go over here and we see that carbon has atomic number of six and bromine has an atomic number of 35. The higher the atomic number, the higher the priority. So the bromine gets higher priority. So the bromine would get a number one and we give a number two to the methyl group."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We go over here and we see that carbon has atomic number of six and bromine has an atomic number of 35. The higher the atomic number, the higher the priority. So the bromine gets higher priority. So the bromine would get a number one and we give a number two to the methyl group. Next, I like to draw a line to think about sides of our double bond. If our two higher priority groups are on opposite sides of our double bond, that is the E configuration. And the E comes from the German word for opposite."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the bromine would get a number one and we give a number two to the methyl group. Next, I like to draw a line to think about sides of our double bond. If our two higher priority groups are on opposite sides of our double bond, that is the E configuration. And the E comes from the German word for opposite. So if the two higher priority groups are on opposite sides of the double bond, it's the E configuration. Let's look at the example on the right. So it's very similar."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the E comes from the German word for opposite. So if the two higher priority groups are on opposite sides of the double bond, it's the E configuration. Let's look at the example on the right. So it's very similar. If we start on the right side of the double bond, this carbon is still bonded to an H and an OH. And the OH group gets a higher priority because the oxygen has the higher atomic number. So this gets a one and the hydrogen gets a two."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's very similar. If we start on the right side of the double bond, this carbon is still bonded to an H and an OH. And the OH group gets a higher priority because the oxygen has the higher atomic number. So this gets a one and the hydrogen gets a two. So that's the same as the previous example. When we go to the left side of the double bond and we look at this carbon, now I've switched the methyl group and the bromine. We know the bromine gets higher priority because the bromine has the higher atomic number."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this gets a one and the hydrogen gets a two. So that's the same as the previous example. When we go to the left side of the double bond and we look at this carbon, now I've switched the methyl group and the bromine. We know the bromine gets higher priority because the bromine has the higher atomic number. And so the methyl group gets a number two. So when I draw a line here, it's easier to see that the two higher priority groups are on the same sides, right? The OH and the bromine are on the same side of our double bond."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We know the bromine gets higher priority because the bromine has the higher atomic number. And so the methyl group gets a number two. So when I draw a line here, it's easier to see that the two higher priority groups are on the same sides, right? The OH and the bromine are on the same side of our double bond. This is the Z configuration. Z comes from the German word meaning together. So the two higher priority groups are together."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The OH and the bromine are on the same side of our double bond. This is the Z configuration. Z comes from the German word meaning together. So the two higher priority groups are together. They're on the same side. A good way to remember this is to think about the two higher priority groups being on Z same side. And that's one way to remember that the two higher priority groups being on the same side is Z."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the two higher priority groups are together. They're on the same side. A good way to remember this is to think about the two higher priority groups being on Z same side. And that's one way to remember that the two higher priority groups being on the same side is Z. So the EZ system is more inclusive than the cis-trans terminology. So EZ is often a better way to come up with the configuration of a double bond. Let's assign a configuration to this double bond."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And that's one way to remember that the two higher priority groups being on the same side is Z. So the EZ system is more inclusive than the cis-trans terminology. So EZ is often a better way to come up with the configuration of a double bond. Let's assign a configuration to this double bond. And let's start with the carbon on the left side. We look at the atoms directly bonded to that carbon. There's a bromine and there's a chlorine."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's assign a configuration to this double bond. And let's start with the carbon on the left side. We look at the atoms directly bonded to that carbon. There's a bromine and there's a chlorine. Since bromine has the higher atomic number, bromine gets the higher priority. So bromine gets a one and chlorine gets a two. If we go to the right side of our double bond, and we look at the atoms directly bonded to this carbon, we know that here's a carbon, and we know that here's a carbon."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "There's a bromine and there's a chlorine. Since bromine has the higher atomic number, bromine gets the higher priority. So bromine gets a one and chlorine gets a two. If we go to the right side of our double bond, and we look at the atoms directly bonded to this carbon, we know that here's a carbon, and we know that here's a carbon. So we have a tie because obviously carbon has the same atomic number. So to break a tie, we need to keep going. So what I'm gonna do is redraw this molecule, and I'm gonna do that over here on the right."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If we go to the right side of our double bond, and we look at the atoms directly bonded to this carbon, we know that here's a carbon, and we know that here's a carbon. So we have a tie because obviously carbon has the same atomic number. So to break a tie, we need to keep going. So what I'm gonna do is redraw this molecule, and I'm gonna do that over here on the right. So we have a carbon bonded to a bromine and a chlorine. And the carbon on the right is bonded to another carbon. And this carbon, let me change colors here."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what I'm gonna do is redraw this molecule, and I'm gonna do that over here on the right. So we have a carbon bonded to a bromine and a chlorine. And the carbon on the right is bonded to another carbon. And this carbon, let me change colors here. This carbon is this one, which is double bonded to an oxygen. So for the purpose of assigning priority, we're going to pretend like this carbon is bonded to two oxygens. Obviously, it has a double bond to only one oxygen, but this will help us to assign priority."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon, let me change colors here. This carbon is this one, which is double bonded to an oxygen. So for the purpose of assigning priority, we're going to pretend like this carbon is bonded to two oxygens. Obviously, it has a double bond to only one oxygen, but this will help us to assign priority. And then we also have this carbon bonded to a hydrogen. What about this carbon down here? Well, this carbon is bonded to two hydrogens and directly bonded to an oxygen, and that oxygen is bonded to a hydrogen."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Obviously, it has a double bond to only one oxygen, but this will help us to assign priority. And then we also have this carbon bonded to a hydrogen. What about this carbon down here? Well, this carbon is bonded to two hydrogens and directly bonded to an oxygen, and that oxygen is bonded to a hydrogen. So let's go back to thinking about priority. We started with the carbon on the right side, and we got to these two carbons, and then we had a tie. So next, we need to think about what atoms are directly bonded to those carbons."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, this carbon is bonded to two hydrogens and directly bonded to an oxygen, and that oxygen is bonded to a hydrogen. So let's go back to thinking about priority. We started with the carbon on the right side, and we got to these two carbons, and then we had a tie. So next, we need to think about what atoms are directly bonded to those carbons. So we'll start with this carbon up here. This carbon is directly bonded to an oxygen, an oxygen, and a hydrogen. So we write down oxygen, oxygen, hydrogen."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So next, we need to think about what atoms are directly bonded to those carbons. So we'll start with this carbon up here. This carbon is directly bonded to an oxygen, an oxygen, and a hydrogen. So we write down oxygen, oxygen, hydrogen. Down here, this carbon is directly bonded to an oxygen and two hydrogens. So oxygen, hydrogen, hydrogen. To assign priority, we look for the first point of difference."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we write down oxygen, oxygen, hydrogen. Down here, this carbon is directly bonded to an oxygen and two hydrogens. So oxygen, hydrogen, hydrogen. To assign priority, we look for the first point of difference. So we start with the two oxygens, and we have a tie. So we go to the next atom, and we have an oxygen versus a hydrogen. Obviously, oxygen has a higher atomic number than hydrogen."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "To assign priority, we look for the first point of difference. So we start with the two oxygens, and we have a tie. So we go to the next atom, and we have an oxygen versus a hydrogen. Obviously, oxygen has a higher atomic number than hydrogen. So this is our tiebreaker, right? Right here is our first point of difference. And so this group, this group gets higher priority."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Obviously, oxygen has a higher atomic number than hydrogen. So this is our tiebreaker, right? Right here is our first point of difference. And so this group, this group gets higher priority. So this group gets a number one for priority, and this group gets a number two. Next, we draw a line here for our double bond, and we look at our two higher priority groups. Our two higher priority groups are on the same side."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so this group, this group gets higher priority. So this group gets a number one for priority, and this group gets a number two. Next, we draw a line here for our double bond, and we look at our two higher priority groups. Our two higher priority groups are on the same side. So zee-zame-zide. This is the Z configuration for our double bond. So how do we incorporate the E-Z system into IUPAC nomenclature?"}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Our two higher priority groups are on the same side. So zee-zame-zide. This is the Z configuration for our double bond. So how do we incorporate the E-Z system into IUPAC nomenclature? Let's say our goal is to name this alkene. Well, we'd find the longest carbon chain that includes our double bond, and we'd give our lowest number possible to our substituents. So that means starting from the right side here."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So how do we incorporate the E-Z system into IUPAC nomenclature? Let's say our goal is to name this alkene. Well, we'd find the longest carbon chain that includes our double bond, and we'd give our lowest number possible to our substituents. So that means starting from the right side here. So this would be carbon one, carbon two, carbon three, four, five, six, and seven. So what do we call a seven-carbon alkene? That would be heptene."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that means starting from the right side here. So this would be carbon one, carbon two, carbon three, four, five, six, and seven. So what do we call a seven-carbon alkene? That would be heptene. So over here, I will write heptene. And our double bond starts at carbon three. So let me write three-heptene."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That would be heptene. So over here, I will write heptene. And our double bond starts at carbon three. So let me write three-heptene. Next, we think about the substituents coming off of our carbon chain. We have a methyl group coming off of carbon two, and an ethyl group coming off of carbon four. And we need to put those in alphabetical order."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me write three-heptene. Next, we think about the substituents coming off of our carbon chain. We have a methyl group coming off of carbon two, and an ethyl group coming off of carbon four. And we need to put those in alphabetical order. So 4-ethyl would come first. So let me go ahead and put that right here. So 4-ethyl, and then we have 2-methyl, 3-heptene."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we need to put those in alphabetical order. So 4-ethyl would come first. So let me go ahead and put that right here. So 4-ethyl, and then we have 2-methyl, 3-heptene. So that's the older way of naming it. You could have also have written here 4-ethyl, 2-methyl, hept-3-ene. So either one is accepted."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So 4-ethyl, and then we have 2-methyl, 3-heptene. So that's the older way of naming it. You could have also have written here 4-ethyl, 2-methyl, hept-3-ene. So either one is accepted. Now we have to think about our double bond. Is it an E configuration, or is it a Z configuration? So let's start with the carbon on the right side of our double bond here."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So either one is accepted. Now we have to think about our double bond. Is it an E configuration, or is it a Z configuration? So let's start with the carbon on the right side of our double bond here. We know there's a hydrogen coming up this way, and so we're comparing the atoms directly bonded to that carbon in red. So we have a hydrogen here, and then we have a carbon right here. Carbon has the higher atomic number."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with the carbon on the right side of our double bond here. We know there's a hydrogen coming up this way, and so we're comparing the atoms directly bonded to that carbon in red. So we have a hydrogen here, and then we have a carbon right here. Carbon has the higher atomic number. So this group gets higher priority. So we get a number one for this group, and hydrogen would get a number two. For the left side, so we're thinking about this carbon, we think about the atoms directly bonded to that carbon."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Carbon has the higher atomic number. So this group gets higher priority. So we get a number one for this group, and hydrogen would get a number two. For the left side, so we're thinking about this carbon, we think about the atoms directly bonded to that carbon. Well, there's a carbon here, and a carbon here. So there's a tie. So I'm actually gonna redraw part of the molecule here to help us figure out the higher priority group, to help us break that tie."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "For the left side, so we're thinking about this carbon, we think about the atoms directly bonded to that carbon. Well, there's a carbon here, and a carbon here. So there's a tie. So I'm actually gonna redraw part of the molecule here to help us figure out the higher priority group, to help us break that tie. So here's our double bond, here's our hydrogen, here's our carbon. So that's the right side of our double bond, again, only drawing in part of the molecule. On the left side of our double bond, we would have a carbon down here bonded to two hydrogens, and then this carbon has three hydrogens."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm actually gonna redraw part of the molecule here to help us figure out the higher priority group, to help us break that tie. So here's our double bond, here's our hydrogen, here's our carbon. So that's the right side of our double bond, again, only drawing in part of the molecule. On the left side of our double bond, we would have a carbon down here bonded to two hydrogens, and then this carbon has three hydrogens. So that's our ethyl group. And then up here, we would have a carbon bonded to two hydrogens, bonded to another carbon with two hydrogens, and finally bonded to a carbon with three hydrogens. All right, now we have to figure out which of those two groups on the left side is the higher priority group."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "On the left side of our double bond, we would have a carbon down here bonded to two hydrogens, and then this carbon has three hydrogens. So that's our ethyl group. And then up here, we would have a carbon bonded to two hydrogens, bonded to another carbon with two hydrogens, and finally bonded to a carbon with three hydrogens. All right, now we have to figure out which of those two groups on the left side is the higher priority group. So again, let's start with the carbon on the left side of our double bond. We look directly at the atom bonded to that, and it's a carbon, so we need to continue on. So this carbon is directly bonded to a carbon, and then a hydrogen, and then a hydrogen."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "All right, now we have to figure out which of those two groups on the left side is the higher priority group. So again, let's start with the carbon on the left side of our double bond. We look directly at the atom bonded to that, and it's a carbon, so we need to continue on. So this carbon is directly bonded to a carbon, and then a hydrogen, and then a hydrogen. So we have carbon, hydrogen, hydrogen. What about this carbon? That carbon's bonded to a carbon and two hydrogens."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon is directly bonded to a carbon, and then a hydrogen, and then a hydrogen. So we have carbon, hydrogen, hydrogen. What about this carbon? That carbon's bonded to a carbon and two hydrogens. So we have carbon, hydrogen, hydrogen. We look for first point of difference. We have carbon versus carbon, hydrogen versus hydrogen, and hydrogen versus hydrogen."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That carbon's bonded to a carbon and two hydrogens. So we have carbon, hydrogen, hydrogen. We look for first point of difference. We have carbon versus carbon, hydrogen versus hydrogen, and hydrogen versus hydrogen. So we don't have anything. We still don't know which is the higher priority group. So we need yet another tiebreaker."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have carbon versus carbon, hydrogen versus hydrogen, and hydrogen versus hydrogen. So we don't have anything. We still don't know which is the higher priority group. So we need yet another tiebreaker. So let's go to our next carbon. So that's this carbon right here versus this carbon down here. We'll go back up to this top carbon."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we need yet another tiebreaker. So let's go to our next carbon. So that's this carbon right here versus this carbon down here. We'll go back up to this top carbon. This carbon is bonded to a carbon, and then hydrogen, hydrogen. So we have carbon, hydrogen, hydrogen. Down here, let's look at this carbon."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We'll go back up to this top carbon. This carbon is bonded to a carbon, and then hydrogen, hydrogen. So we have carbon, hydrogen, hydrogen. Down here, let's look at this carbon. We have hydrogen, hydrogen, hydrogen. So hydrogen, hydrogen, hydrogen. Finally, we have our first point of difference because we're comparing a carbon to a hydrogen."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Down here, let's look at this carbon. We have hydrogen, hydrogen, hydrogen. So hydrogen, hydrogen, hydrogen. Finally, we have our first point of difference because we're comparing a carbon to a hydrogen. So carbon has the higher atomic number. So this group gets higher priority. So this group gets the higher priority."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Finally, we have our first point of difference because we're comparing a carbon to a hydrogen. So carbon has the higher atomic number. So this group gets higher priority. So this group gets the higher priority. Let's go back up to this drawing. We're talking about this group, higher priority, and our ethyl group gets a number two here. So finally, we go ahead and draw in our line so we can see which side are higher priority groups or which sides, I should say, are higher priority groups are on."}, {"video_title": "E\u2013Z system Alkenes and alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this group gets the higher priority. Let's go back up to this drawing. We're talking about this group, higher priority, and our ethyl group gets a number two here. So finally, we go ahead and draw in our line so we can see which side are higher priority groups or which sides, I should say, are higher priority groups are on. So our higher priority groups are on opposite sides of our double bond. So we know that is E. So the configuration of the double bond is E. So we put that in our name. So we put our E here, and we put it in parentheses."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So we have our alcohol over here on the left. And we react that with some concentrated nitric acid and concentrated sulfuric acid as our catalyst. And I believe this reaction is reversible, so we could think about that. We would form our nitrate ester over here on the right. We would also form water in the process. So the water molecule is going to come from this hydrogen on the alcohol and this OH on our nitric acid. Let's take a look at the mechanism to form nitrate esters."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "We would form our nitrate ester over here on the right. We would also form water in the process. So the water molecule is going to come from this hydrogen on the alcohol and this OH on our nitric acid. Let's take a look at the mechanism to form nitrate esters. And so I'll start by redrawing our nitric acid molecule here. I think this is the correct mechanism, although I tried to look up the mechanism in some textbooks, and I could not find this mechanism anywhere. So I hope that the one that I show you is the correct one."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "Let's take a look at the mechanism to form nitrate esters. And so I'll start by redrawing our nitric acid molecule here. I think this is the correct mechanism, although I tried to look up the mechanism in some textbooks, and I could not find this mechanism anywhere. So I hope that the one that I show you is the correct one. All right, well, if I think about nitric acid and sulfuric acid, sulfuric acid is actually the stronger acid. So sulfuric acid is going to donate protons, and the nitric acid is going to accept those protons. The nitric acid is actually going to function as a base."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So I hope that the one that I show you is the correct one. All right, well, if I think about nitric acid and sulfuric acid, sulfuric acid is actually the stronger acid. So sulfuric acid is going to donate protons, and the nitric acid is going to accept those protons. The nitric acid is actually going to function as a base. So a lone pair of electrons on the oxygen are going to pick up that proton. So now oxygen has three bonds, two to hydrogens and one still to this nitrogen here. It still has a lone pair of electrons on it, which give it a plus 1 formal charge."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "The nitric acid is actually going to function as a base. So a lone pair of electrons on the oxygen are going to pick up that proton. So now oxygen has three bonds, two to hydrogens and one still to this nitrogen here. It still has a lone pair of electrons on it, which give it a plus 1 formal charge. This nitrogen over here is still double bonded to one oxygen, and it's still bonded to this oxygen over here. A negative 1 formal charge and a plus 1 formal charge like that. So that was our first step of our mechanism and acid-base reaction."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "It still has a lone pair of electrons on it, which give it a plus 1 formal charge. This nitrogen over here is still double bonded to one oxygen, and it's still bonded to this oxygen over here. A negative 1 formal charge and a plus 1 formal charge like that. So that was our first step of our mechanism and acid-base reaction. In the next step, if we look closely, we can see that we kind of have water as a leaving group. If a lone pair of electrons on our oxygen move in here to form a new pi bond, that can kick these electrons in here off onto the oxygen, and water has now left. So let's go ahead and draw the results of that."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So that was our first step of our mechanism and acid-base reaction. In the next step, if we look closely, we can see that we kind of have water as a leaving group. If a lone pair of electrons on our oxygen move in here to form a new pi bond, that can kick these electrons in here off onto the oxygen, and water has now left. So let's go ahead and draw the results of that. So now we have H2O as a leaving group. So H2O has left. We now have a nitrogen, which was double bonded to an oxygen up here."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the results of that. So now we have H2O as a leaving group. So H2O has left. We now have a nitrogen, which was double bonded to an oxygen up here. It's now double bonded to another oxygen down here, and it still has a plus 1 formal charge now. So this is called the nitronium ion. And since the nitrogen is positively charged, it wants electrons."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "We now have a nitrogen, which was double bonded to an oxygen up here. It's now double bonded to another oxygen down here, and it still has a plus 1 formal charge now. So this is called the nitronium ion. And since the nitrogen is positively charged, it wants electrons. So it's going to function as an electrophile in the next step of the mechanism when an alcohol molecule comes along. Alcohols have these lone pairs of electrons here. We know that negatively charged electrons can function as nucleophiles."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "And since the nitrogen is positively charged, it wants electrons. So it's going to function as an electrophile in the next step of the mechanism when an alcohol molecule comes along. Alcohols have these lone pairs of electrons here. We know that negatively charged electrons can function as nucleophiles. So the negatively charged electron is attracted to the positively charged nitrogen, and nucleophilic attack will kick these electrons off onto your oxygen. Let's go ahead and draw the product of that nucleophilic attack. So now we have our R group bonded to an oxygen."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "We know that negatively charged electrons can function as nucleophiles. So the negatively charged electron is attracted to the positively charged nitrogen, and nucleophilic attack will kick these electrons off onto your oxygen. Let's go ahead and draw the product of that nucleophilic attack. So now we have our R group bonded to an oxygen. And our oxygen is now bonded to this nitrogen here. This oxygen is also still bonded to a hydrogen, giving it a plus 1 formal charge. And our nitrogen still has a double bond to the top oxygen here."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our R group bonded to an oxygen. And our oxygen is now bonded to this nitrogen here. This oxygen is also still bonded to a hydrogen, giving it a plus 1 formal charge. And our nitrogen still has a double bond to the top oxygen here. And it now has a single bond to this oxygen, giving this a negative 1 formal charge, giving this nitrogen a positive formal charge like that. So we've almost formed our nitrate ester. If we look at this, all we have to do now is an acid-base reaction to take this proton off of our oxygen."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "And our nitrogen still has a double bond to the top oxygen here. And it now has a single bond to this oxygen, giving this a negative 1 formal charge, giving this nitrogen a positive formal charge like that. So we've almost formed our nitrate ester. If we look at this, all we have to do now is an acid-base reaction to take this proton off of our oxygen. So water is a decent base. And a lone pair of electrons on our oxygen can take this proton, leaving these electrons behind on this oxygen here. And we would form our nitrate ester."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "If we look at this, all we have to do now is an acid-base reaction to take this proton off of our oxygen. So water is a decent base. And a lone pair of electrons on our oxygen can take this proton, leaving these electrons behind on this oxygen here. And we would form our nitrate ester. So now we have R with an oxygen, and then NO2 like that. So once again, I think this is the correct mechanism, but I'm not 100% sure. The formation of nitrate esters has a few very famous examples."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "And we would form our nitrate ester. So now we have R with an oxygen, and then NO2 like that. So once again, I think this is the correct mechanism, but I'm not 100% sure. The formation of nitrate esters has a few very famous examples. Let's look at the most famous reaction where a nitrate ester is formed. Let's look at what would happen if we started with this alcohol as our reactant. This is called glycerol or glycerin."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "The formation of nitrate esters has a few very famous examples. Let's look at the most famous reaction where a nitrate ester is formed. Let's look at what would happen if we started with this alcohol as our reactant. This is called glycerol or glycerin. It has three OH groups on it. And if you react glycerin with excess nitric acid and also sulfuric acid, all concentrated, well, we would form a nitrate ester at each one of our OH groups. So we're going to form a nitrate ester at each of our OH groups to form this as our product."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "This is called glycerol or glycerin. It has three OH groups on it. And if you react glycerin with excess nitric acid and also sulfuric acid, all concentrated, well, we would form a nitrate ester at each one of our OH groups. So we're going to form a nitrate ester at each of our OH groups to form this as our product. We put nitro groups onto glycerin, so this is called nitroglycerin. So of course, everyone knows about nitroglycerin. Very famous high explosive."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to form a nitrate ester at each of our OH groups to form this as our product. We put nitro groups onto glycerin, so this is called nitroglycerin. So of course, everyone knows about nitroglycerin. Very famous high explosive. It's a liquid. It's extremely shock sensitive, so it's extremely dangerous. Nitroglycerin is not something that anyone should attempt to make at home."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "Very famous high explosive. It's a liquid. It's extremely shock sensitive, so it's extremely dangerous. Nitroglycerin is not something that anyone should attempt to make at home. So it's not a good demonstration for chemistry instructors. A much better demonstration for chemistry instructors would be to form another nitrate ester starting from cotton or cellulose. So here we have the cellulose polymer."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "Nitroglycerin is not something that anyone should attempt to make at home. So it's not a good demonstration for chemistry instructors. A much better demonstration for chemistry instructors would be to form another nitrate ester starting from cotton or cellulose. So here we have the cellulose polymer. It's made up of a bunch of glucose molecules. So here's a glucose molecule. Here's a glucose molecule."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "So here we have the cellulose polymer. It's made up of a bunch of glucose molecules. So here's a glucose molecule. Here's a glucose molecule. And if you stick a whole bunch of glucose molecules together in some giant polymer, you form cellulose, otherwise known as cotton. So if you take a bunch of cotton balls and you mix them with concentrated nitric acid and concentrated sulfuric acid, you can put nitro groups at each one of those alcohol groups. You can form nitrate esters."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "Here's a glucose molecule. And if you stick a whole bunch of glucose molecules together in some giant polymer, you form cellulose, otherwise known as cotton. So if you take a bunch of cotton balls and you mix them with concentrated nitric acid and concentrated sulfuric acid, you can put nitro groups at each one of those alcohol groups. You can form nitrate esters. So each of these alcohol groups gets turned into a nitrate ester. So I can go ahead and put my nitrate esters in. You notice there are three nitrate esters for each glucose molecule."}, {"video_title": "Formation of nitrate esters Organic chemistry Khan Academy.mp3", "Sentence": "You can form nitrate esters. So each of these alcohol groups gets turned into a nitrate ester. So I can go ahead and put my nitrate esters in. You notice there are three nitrate esters for each glucose molecule. So we form nitrocellulose as our product, otherwise known as gun cotton. Gun cotton is much more stable than nitroglycerin is. You can store gun cotton overnight."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We've already talked about the life cycle of stars, roughly the same mass as our sun, give or take a little bit. What I want to do in this video is talk about more massive stars. Massive stars. And when I'm talking about massive stars, I'm talking about stars that have masses greater than 9 times the sun. So the general idea is exactly the same. You're going to start off with this huge cloud of mainly hydrogen, and now this cloud is going to have to be bigger than the clouds that condensed to form stars like our sun. But you're going to start with that, and eventually, gravity is going to pull it together, and the core of it is going to get hot and dense enough for hydrogen to ignite, for hydrogen to start fusing."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when I'm talking about massive stars, I'm talking about stars that have masses greater than 9 times the sun. So the general idea is exactly the same. You're going to start off with this huge cloud of mainly hydrogen, and now this cloud is going to have to be bigger than the clouds that condensed to form stars like our sun. But you're going to start with that, and eventually, gravity is going to pull it together, and the core of it is going to get hot and dense enough for hydrogen to ignite, for hydrogen to start fusing. So this is hydrogen, and it is now fusing. Let me write it. It is now fusing."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But you're going to start with that, and eventually, gravity is going to pull it together, and the core of it is going to get hot and dense enough for hydrogen to ignite, for hydrogen to start fusing. So this is hydrogen, and it is now fusing. Let me write it. It is now fusing. Hydrogen fusion. Let me write it like this. You now have hydrogen fusion in the middle, so it's ignited, and around it, you have just the other material of the cloud, so the rest of the hydrogen."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is now fusing. Hydrogen fusion. Let me write it like this. You now have hydrogen fusion in the middle, so it's ignited, and around it, you have just the other material of the cloud, so the rest of the hydrogen. And now, since it's so heated, it's really a plasma. It's really kind of a soup of electrons and nucleuses as opposed to well-formed atoms, especially close to the core. So now you have hydrogen fusion."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You now have hydrogen fusion in the middle, so it's ignited, and around it, you have just the other material of the cloud, so the rest of the hydrogen. And now, since it's so heated, it's really a plasma. It's really kind of a soup of electrons and nucleuses as opposed to well-formed atoms, especially close to the core. So now you have hydrogen fusion. We saw this happens at around 10 million Kelvin. And I want to make it very clear. Since we're talking about more massive stars, even at this stage, there's going to be more gravitational pressure, even at this stage, during the main sequence of the star, because it is more massive."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now you have hydrogen fusion. We saw this happens at around 10 million Kelvin. And I want to make it very clear. Since we're talking about more massive stars, even at this stage, there's going to be more gravitational pressure, even at this stage, during the main sequence of the star, because it is more massive. And so this is going to burn faster and hotter. So this is going to be faster and hotter than something the mass of our sun. Faster and hotter."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Since we're talking about more massive stars, even at this stage, there's going to be more gravitational pressure, even at this stage, during the main sequence of the star, because it is more massive. And so this is going to burn faster and hotter. So this is going to be faster and hotter than something the mass of our sun. Faster and hotter. And so even this stage is going to happen over a much shorter period of time than for a star the mass of our sun. Our sun's life is going to be 10 or 11 billion total years. Here, we're going to be talking about things in maybe the tens of millions of years, so a factor of 1,000 shorter lifespan."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Faster and hotter. And so even this stage is going to happen over a much shorter period of time than for a star the mass of our sun. Our sun's life is going to be 10 or 11 billion total years. Here, we're going to be talking about things in maybe the tens of millions of years, so a factor of 1,000 shorter lifespan. But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Here, we're going to be talking about things in maybe the tens of millions of years, so a factor of 1,000 shorter lifespan. But anyway, let's think about what happens. And so far, just the pattern of what happens, it's going to happen faster because we have more pressure, more gravity, more temperature. But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. A hydrogen fusion shell around it, and then you have the rest of the star around that. So let me label it."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it's going to happen in pretty much the same way as what we saw with a star the mass of the sun. Eventually, that hydrogen is going to fuse into a helium core that's going to have a hydrogen shell around it. A hydrogen fusion shell around it, and then you have the rest of the star around that. So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen and this shell fuses. And this is, in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me label it. This right here is our helium core. And more and more helium is going to be built up as this hydrogen and this shell fuses. And this is, in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, there's more and more gravitational pressure being put on this hydrogen shell out here where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is, in a star the size of our sun or the mass of our sun, this is when it starts to become a red giant. Because this core is getting denser and denser and denser as more and more helium is produced. And as it gets denser and denser and denser, there's more and more gravitational pressure being put on this hydrogen shell out here where we have fusion still happening. And so that's going to release more outward energy to push out the radius of the actual star. But then when you fast forward there, so the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out or with more and more energy."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so that's going to release more outward energy to push out the radius of the actual star. But then when you fast forward there, so the general process, and we're going to see this as the star gets more and more massive, is we're going to have heavier and heavier elements forming in the core. Those heavier and heavier elements, as the star gets denser and denser, will eventually ignite, kind of supporting the core. But because the core itself is getting denser and denser and denser, material is getting pushed further and further out or with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in a kind of a red giant with kind of a sun-like star. Let's just think about how this pattern is going to continue. So eventually, that helium is going to, once it gets dense enough, it's going to ignite."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But because the core itself is getting denser and denser and denser, material is getting pushed further and further out or with more and more energy. Although if the star is massive enough, it's not going to be able to be pushed out as far as you will have in a kind of a red giant with kind of a sun-like star. Let's just think about how this pattern is going to continue. So eventually, that helium is going to, once it gets dense enough, it's going to ignite. And it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So eventually, that helium is going to, once it gets dense enough, it's going to ignite. And it's going to fuse into carbon. And you're going to have a carbon core forming. So that is carbon core. That's a carbon core. Around that, you have a helium core. And near the center of the helium core, you have a shell of helium fusion."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that is carbon core. That's a carbon core. Around that, you have a helium core. And near the center of the helium core, you have a shell of helium fusion. That's helium, not hydrogen. Turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And near the center of the helium core, you have a shell of helium fusion. That's helium, not hydrogen. Turning into carbon, making that carbon core denser and hotter. And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then around that, you have hydrogen fusion. Have to be very careful. You have hydrogen fusion. And then around that, you have the rest of the star. And so this process is going to keep continuing eventually. That carbon is going to start fusing. And you're going to have heavier and heavier elements forming the core."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then around that, you have the rest of the star. And so this process is going to keep continuing eventually. That carbon is going to start fusing. And you're going to have heavier and heavier elements forming the core. And so this is a depiction off of Wikipedia of a fairly mature, massive star. And you keep forming these shells of heavier and heavier elements and cores of heavier and heavier elements until eventually you get to iron. And in particular, we're talking about iron 56."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you're going to have heavier and heavier elements forming the core. And so this is a depiction off of Wikipedia of a fairly mature, massive star. And you keep forming these shells of heavier and heavier elements and cores of heavier and heavier elements until eventually you get to iron. And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table, that 26 is its atomic number. It's how many protons it has."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And in particular, we're talking about iron 56. Iron with an atomic mass of 56. Here on this periodic table, that 26 is its atomic number. It's how many protons it has. 56 is kind of viewed as a count of the protons and neutrons, although it's not exact. But at this point, the reason why you stop here is that you cannot get energy by fusing iron. Fusing iron into heavier elements beyond iron actually requires energy."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's how many protons it has. 56 is kind of viewed as a count of the protons and neutrons, although it's not exact. But at this point, the reason why you stop here is that you cannot get energy by fusing iron. Fusing iron into heavier elements beyond iron actually requires energy. So it would actually be an endothermic process. So to fuse iron actually won't help support the core. So what I want to do in this, well, just to be very clear, this is how the heavy elements actually formed."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Fusing iron into heavier elements beyond iron actually requires energy. So it would actually be an endothermic process. So to fuse iron actually won't help support the core. So what I want to do in this, well, just to be very clear, this is how the heavy elements actually formed. We started with hydrogen. Hydrogen fusing into helium. Helium fusing into carbon."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what I want to do in this, well, just to be very clear, this is how the heavy elements actually formed. We started with hydrogen. Hydrogen fusing into helium. Helium fusing into carbon. And then all of these things in various combinations, and I won't go into all of the details, are fusing heavier and heavier elements. Neon, oxygen, and you see it right over here, silicon. And these aren't the only elements that are forming, but these are kind of the main core elements that are forming."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Helium fusing into carbon. And then all of these things in various combinations, and I won't go into all of the details, are fusing heavier and heavier elements. Neon, oxygen, and you see it right over here, silicon. And these aren't the only elements that are forming, but these are kind of the main core elements that are forming. But along the way, you have all this other stuff. Lithium, beryllium, boron, all of this other stuff is also forming. So this is how you form elements up to iron 56."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And these aren't the only elements that are forming, but these are kind of the main core elements that are forming. But along the way, you have all this other stuff. Lithium, beryllium, boron, all of this other stuff is also forming. So this is how you form elements up to iron 56. And also, actually, this is actually how you can form up to nickel 56, just to be exact. There will also be some nickel 56, which has the same mass as iron 56, just has two fewer neutrons and two more protons. So it's nickel 56."}, {"video_title": "Lifecycle of massive stars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is how you form elements up to iron 56. And also, actually, this is actually how you can form up to nickel 56, just to be exact. There will also be some nickel 56, which has the same mass as iron 56, just has two fewer neutrons and two more protons. So it's nickel 56. Will also form, can also kind of be a nickel iron core. But that's about how far a star can get, regardless of how massive it is, at least by going through traditional fusion, through the traditional ignition mechanism. What I want to do is leave you there, just so you can think about what might happen next, now that we can't fuse this star anymore."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here I have a molecule. Let's see if we can identify any chiral centers or any chiral atoms or asymmetric carbons. All words for the same thing. Although I guess you could have chiral centers that aren't necessarily carbon, but it tends to be carbon most of the time, especially in an organic chemistry class. So if we look here, the one that kind of jumps out, this carbon right here is bonded to three hydrogens and another carbon, so this is obviously not going to be a chiral center. It's bonded to three of the same group. These three guys are all bonded to two hydrogens each, so they're all bonded to two of the same group, so they can't be chiral centers."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Although I guess you could have chiral centers that aren't necessarily carbon, but it tends to be carbon most of the time, especially in an organic chemistry class. So if we look here, the one that kind of jumps out, this carbon right here is bonded to three hydrogens and another carbon, so this is obviously not going to be a chiral center. It's bonded to three of the same group. These three guys are all bonded to two hydrogens each, so they're all bonded to two of the same group, so they can't be chiral centers. This carbon right here is bonded to three hydrogens, once again, three of the same group, not going to be a chiral center. This one looks interesting. Looks like it could be a good candidate for a chiral center or a chiral carbon or an asymmetric carbon."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These three guys are all bonded to two hydrogens each, so they're all bonded to two of the same group, so they can't be chiral centers. This carbon right here is bonded to three hydrogens, once again, three of the same group, not going to be a chiral center. This one looks interesting. Looks like it could be a good candidate for a chiral center or a chiral carbon or an asymmetric carbon. Over here on the left, it's bonded to a methyl group, so this is a methyl group. And here on the right, it's bonded to a butyl group. Over here, it's bonded to an OH, and then over here, it's bonded to an H, so this is definitely a chiral carbon."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Looks like it could be a good candidate for a chiral center or a chiral carbon or an asymmetric carbon. Over here on the left, it's bonded to a methyl group, so this is a methyl group. And here on the right, it's bonded to a butyl group. Over here, it's bonded to an OH, and then over here, it's bonded to an H, so this is definitely a chiral carbon. We could put a little asterisk here. That's how they often denote that this is a chiral carbon. And if this doesn't make sense to you, because you might say, hey, Sal, look, this carbon is bonded to two other carbons."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Over here, it's bonded to an OH, and then over here, it's bonded to an H, so this is definitely a chiral carbon. We could put a little asterisk here. That's how they often denote that this is a chiral carbon. And if this doesn't make sense to you, because you might say, hey, Sal, look, this carbon is bonded to two other carbons. Isn't that the same thing? But the point here is that we're not looking at what atoms it's directly bonded to. We're looking at the groups that it's bonded to."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if this doesn't make sense to you, because you might say, hey, Sal, look, this carbon is bonded to two other carbons. Isn't that the same thing? But the point here is that we're not looking at what atoms it's directly bonded to. We're looking at the groups that it's bonded to. In this case, this hydrogen is a group and an atom. Over here, it's an entire group. It's an entire butyl group."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We're looking at the groups that it's bonded to. In this case, this hydrogen is a group and an atom. Over here, it's an entire group. It's an entire butyl group. We have four carbons here. We only have one over here. Another way to think about it, we could have drawn this molecule like this."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's an entire butyl group. We have four carbons here. We only have one over here. Another way to think about it, we could have drawn this molecule like this. We could have had a carbon in the center, and maybe this methyl group is popping out like this. So maybe this methyl group is popping out like this. You have your CH3, and then you would have this hydrogen coming out maybe in the plane."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Another way to think about it, we could have drawn this molecule like this. We could have had a carbon in the center, and maybe this methyl group is popping out like this. So maybe this methyl group is popping out like this. You have your CH3, and then you would have this hydrogen coming out maybe in the plane. And then behind it, you would have the butyl group. So kind of the back leg of the tripod, you'd have a butyl. And what is that?"}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You have your CH3, and then you would have this hydrogen coming out maybe in the plane. And then behind it, you would have the butyl group. So kind of the back leg of the tripod, you'd have a butyl. And what is that? That's C4H9. That's 6 plus C4H9. So it's C4H9 in the back."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And what is that? That's C4H9. That's 6 plus C4H9. So it's C4H9 in the back. And then above it, you have your OH. And when you look at it like this, it looks just like the other chiral carbons that we had identified in actually the last video. It looks very similar to something like this."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's C4H9 in the back. And then above it, you have your OH. And when you look at it like this, it looks just like the other chiral carbons that we had identified in actually the last video. It looks very similar to something like this. And when you take its mirror image, this and this is the same molecule here. I kind of made it a little bit more three-dimensional. But if you take the mirror image of either one, you're going to find that no matter how you try to rotate it or shift it, you won't be able to superimpose it on its mirror image for the same reasons as the other ones that I challenged you to if you can."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It looks very similar to something like this. And when you take its mirror image, this and this is the same molecule here. I kind of made it a little bit more three-dimensional. But if you take the mirror image of either one, you're going to find that no matter how you try to rotate it or shift it, you won't be able to superimpose it on its mirror image for the same reasons as the other ones that I challenged you to if you can. So this is a chiral carbon. This is a chiral center, we could say. Or we could even call it an asymmetric carbon."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But if you take the mirror image of either one, you're going to find that no matter how you try to rotate it or shift it, you won't be able to superimpose it on its mirror image for the same reasons as the other ones that I challenged you to if you can. So this is a chiral carbon. This is a chiral center, we could say. Or we could even call it an asymmetric carbon. It could be considered a stereocenter or a stereogenic center. All of those are valid things to call this carbon right there. And this is also a chiral molecule."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Or we could even call it an asymmetric carbon. It could be considered a stereocenter or a stereogenic center. All of those are valid things to call this carbon right there. And this is also a chiral molecule. This is also a chiral molecule. Now let's look at this blue example right here. And just as if we wanted to name it, just so we get a little bit of review, we could start it this fluorine right there."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And this is also a chiral molecule. This is also a chiral molecule. Now let's look at this blue example right here. And just as if we wanted to name it, just so we get a little bit of review, we could start it this fluorine right there. 1, 2, 3, 4, 5. This is what? This is 1, 3-difluorocyclopentane."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And just as if we wanted to name it, just so we get a little bit of review, we could start it this fluorine right there. 1, 2, 3, 4, 5. This is what? This is 1, 3-difluorocyclopentane. So that was a nice review of naming. But let's think about whether we have any chiral centers here and whether the molecule as a whole is chiral. So the immediate ones that we can kind of dismiss, and actually let me get rid of this numbering here just because I don't want you to think that there's somehow three hydrogens there."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is 1, 3-difluorocyclopentane. So that was a nice review of naming. But let's think about whether we have any chiral centers here and whether the molecule as a whole is chiral. So the immediate ones that we can kind of dismiss, and actually let me get rid of this numbering here just because I don't want you to think that there's somehow three hydrogens there. This is the number, that was the number 3 carbon, number 2 carbon, so on and so forth. But let me get rid of them now that we've named the molecule. Don't want to confuse how many hydrogens we have at any of these points."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the immediate ones that we can kind of dismiss, and actually let me get rid of this numbering here just because I don't want you to think that there's somehow three hydrogens there. This is the number, that was the number 3 carbon, number 2 carbon, so on and so forth. But let me get rid of them now that we've named the molecule. Don't want to confuse how many hydrogens we have at any of these points. So let's look at the carbons. Well, we can immediately dismiss that carbon, that carbon, and that carbon because each of those are bonded to two hydrogens. If we wanted to break it out, they would look like this."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Don't want to confuse how many hydrogens we have at any of these points. So let's look at the carbons. Well, we can immediately dismiss that carbon, that carbon, and that carbon because each of those are bonded to two hydrogens. If we wanted to break it out, they would look like this. So they're bonded to carbons, carbons, and then they're bonded to hydrogens. Now these might be different groups. These might be different types of alkane groups that it's bonded to, so that doesn't necessarily throw it out of the running."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we wanted to break it out, they would look like this. So they're bonded to carbons, carbons, and then they're bonded to hydrogens. Now these might be different groups. These might be different types of alkane groups that it's bonded to, so that doesn't necessarily throw it out of the running. But these two, the two hydrogens that it's bonded to, are definitely the same atom, the same group. We have an axis of symmetry through that atom. So it cannot be a stereogenic center."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These might be different types of alkane groups that it's bonded to, so that doesn't necessarily throw it out of the running. But these two, the two hydrogens that it's bonded to, are definitely the same atom, the same group. We have an axis of symmetry through that atom. So it cannot be a stereogenic center. It cannot be an asymmetric carbon. It cannot be a chiral center or a chiral atom. So we can knock those guys out of the running."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it cannot be a stereogenic center. It cannot be an asymmetric carbon. It cannot be a chiral center or a chiral atom. So we can knock those guys out of the running. But this guy and that guy seem pretty interesting. Because if we were to break it out a little bit, you could break it out like that and break it out like that instead of writing a CH, actually show the bond to the hydrogen. And this guy is bonded to one hydrogen, one fluorine."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can knock those guys out of the running. But this guy and that guy seem pretty interesting. Because if we were to break it out a little bit, you could break it out like that and break it out like that instead of writing a CH, actually show the bond to the hydrogen. And this guy is bonded to one hydrogen, one fluorine. And then if we were to work our way around the cycle, and these cyclic molecules are a little bit, it's sometimes a little tricky to identify whether you're bonded to the same group or different groups. But actually, let me not make it too messy while we try to figure this out. To figure out whether it's bonded to the same group, let's kind of take a walk around the cycle, around the cyclopentane ring."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And this guy is bonded to one hydrogen, one fluorine. And then if we were to work our way around the cycle, and these cyclic molecules are a little bit, it's sometimes a little tricky to identify whether you're bonded to the same group or different groups. But actually, let me not make it too messy while we try to figure this out. To figure out whether it's bonded to the same group, let's kind of take a walk around the cycle, around the cyclopentane ring. If we go this way, if we go in a counter, let me do a different color. If we go in a counterclockwise direction from the carbon in question, we're going to hit a CH2. Then we're going to hit a CH."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "To figure out whether it's bonded to the same group, let's kind of take a walk around the cycle, around the cyclopentane ring. If we go this way, if we go in a counter, let me do a different color. If we go in a counterclockwise direction from the carbon in question, we're going to hit a CH2. Then we're going to hit a CH. So we're going to hit a CH2, then we're going to hit a CH. If we go this way, we're going to hit a CH2, and then we're going to hit another CH2. So this guy is fundamentally, this bond is bonded to a different group than that bond up there is."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Then we're going to hit a CH. So we're going to hit a CH2, then we're going to hit a CH. If we go this way, we're going to hit a CH2, and then we're going to hit another CH2. So this guy is fundamentally, this bond is bonded to a different group than that bond up there is. And it's also bonded to a hydrogen, also bonded to a fluorine. So this is bonded to four different groups. So this is a chiral carbon."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this guy is fundamentally, this bond is bonded to a different group than that bond up there is. And it's also bonded to a hydrogen, also bonded to a fluorine. So this is bonded to four different groups. So this is a chiral carbon. So that is a chiral center. Now the exact same argument can be made for this carbon right here. You can make that exact same argument."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is a chiral carbon. So that is a chiral center. Now the exact same argument can be made for this carbon right here. You can make that exact same argument. That look, if you were to walk counterclockwise, you'd hit a CH2, then a CH2. If you were to go clockwise, you'd hit a CH2, then a CH, which happens to be connected to a fluorine. So you're actually going to see something different, depending whether you're going down that, into that group, or into that group."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can make that exact same argument. That look, if you were to walk counterclockwise, you'd hit a CH2, then a CH2. If you were to go clockwise, you'd hit a CH2, then a CH, which happens to be connected to a fluorine. So you're actually going to see something different, depending whether you're going down that, into that group, or into that group. And then it's also bonded to a hydrogen, and then it's bonded to a fluorine, four different groups. This is also a chiral center. Another way to think about it, and it's actually interesting to compare it to this molecule up here, which was not chiral and did not have a chiral center."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So you're actually going to see something different, depending whether you're going down that, into that group, or into that group. And then it's also bonded to a hydrogen, and then it's bonded to a fluorine, four different groups. This is also a chiral center. Another way to think about it, and it's actually interesting to compare it to this molecule up here, which was not chiral and did not have a chiral center. This molecule up here, and let me draw it a little different to make it a little bit more clear. Let me draw it a little bit different. So this one, I could draw it like this."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Another way to think about it, and it's actually interesting to compare it to this molecule up here, which was not chiral and did not have a chiral center. This molecule up here, and let me draw it a little different to make it a little bit more clear. Let me draw it a little bit different. So this one, I could draw it like this. I'm just making, if you have the chlorine like that, over here, we thought about this as a potential chiral center, and it's kind of playing the same role as in that example down here, as in this example of this carbon down here. But you see over here, this is not a chiral center, because there's actually an axis of symmetry for this molecule that goes through that carbon. So you can actually just draw an axis of symmetry that goes exactly through that carbon."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this one, I could draw it like this. I'm just making, if you have the chlorine like that, over here, we thought about this as a potential chiral center, and it's kind of playing the same role as in that example down here, as in this example of this carbon down here. But you see over here, this is not a chiral center, because there's actually an axis of symmetry for this molecule that goes through that carbon. So you can actually just draw an axis of symmetry that goes exactly through that carbon. The way I drew it, it's not completely neat, but you can see that that is the reflection of that, if I were to draw the bonds actually a little bit more symmetric. Over here, if we try to do the exact same thing, if we try to draw an axis of symmetry over here, we could make that bond to the fluorine go through our axis of symmetry. We'll see that this is still not the reflection of this, because we have a fluorine up here."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So you can actually just draw an axis of symmetry that goes exactly through that carbon. The way I drew it, it's not completely neat, but you can see that that is the reflection of that, if I were to draw the bonds actually a little bit more symmetric. Over here, if we try to do the exact same thing, if we try to draw an axis of symmetry over here, we could make that bond to the fluorine go through our axis of symmetry. We'll see that this is still not the reflection of this, because we have a fluorine up here. We don't have a fluorine over here. And we could do the same thing with this end. If you try to do an axis of symmetry, fluorine up there, no fluorine over here."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We'll see that this is still not the reflection of this, because we have a fluorine up here. We don't have a fluorine over here. And we could do the same thing with this end. If you try to do an axis of symmetry, fluorine up there, no fluorine over here. So each of these are definitely chiral centers, while this carbon up here was not a chiral center. Now, the next question is, well, this thing's got two chiral centers, two chiral carbons. It's probably a chiral molecule."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you try to do an axis of symmetry, fluorine up there, no fluorine over here. So each of these are definitely chiral centers, while this carbon up here was not a chiral center. Now, the next question is, well, this thing's got two chiral centers, two chiral carbons. It's probably a chiral molecule. Everything else we've seen so far, if you've had a chiral center, you had a chiral molecule. But let's take its mirror image. And to take its mirror image, let me clear out some real estate over here."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's probably a chiral molecule. Everything else we've seen so far, if you've had a chiral center, you had a chiral molecule. But let's take its mirror image. And to take its mirror image, let me clear out some real estate over here. So let me clear out this. So what's the mirror image going to look like? So we have, let me draw first the mirror."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And to take its mirror image, let me clear out some real estate over here. So let me clear out this. So what's the mirror image going to look like? So we have, let me draw first the mirror. So the mirror image, you're going to have a fluorine over there. Then you're going to bond to a carbon, which is also bonded to a hydrogen. And that's going to bond to a CH2."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have, let me draw first the mirror. So the mirror image, you're going to have a fluorine over there. Then you're going to bond to a carbon, which is also bonded to a hydrogen. And that's going to bond to a CH2. That's going to bond to a CH. That's the mirror image of that, which bonds to a fluorine. That's the mirror image of that."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And that's going to bond to a CH2. That's going to bond to a CH. That's the mirror image of that, which bonds to a fluorine. That's the mirror image of that. And then you go down. This is the mirror image of CH2 here. This is the mirror image of this."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's the mirror image of that. And then you go down. This is the mirror image of CH2 here. This is the mirror image of this. And you connect them. Now, these are mirror images of each other. But they are also the exact same molecule."}, {"video_title": "Chiral examples 2 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is the mirror image of this. And you connect them. Now, these are mirror images of each other. But they are also the exact same molecule. I could just literally move this guy over to the right, and it would be superimposed. They are exactly the same. So even though we have two chiral atoms, two chiral carbons, the molecule as a whole is not chiral."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "It's possible to have ethers in a ring system, and there are many different types of ring systems that you can have with ethers in them. The one that's studied most of the time would be the epoxides due to their reactivity. Here we have the simplest epoxide, and one name for this would be ethylene oxide, because this molecule is made from ethylene, and that's where you get your two carbons from, like that. So you could call this ethylene oxide, or you could give this an IUPAC name. And since there are two carbons in it, for the IUPAC name we start with ethane as our parent name. And we know that the epoxide forms between carbons 1 and 2, so we can go ahead and write 1, 2, epoxy, ethane for the IUPAC name. Let's do another one here."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So you could call this ethylene oxide, or you could give this an IUPAC name. And since there are two carbons in it, for the IUPAC name we start with ethane as our parent name. And we know that the epoxide forms between carbons 1 and 2, so we can go ahead and write 1, 2, epoxy, ethane for the IUPAC name. Let's do another one here. So let's put some more carbons on here. So we'll put a few extra carbons, and we'll name this using IUPAC nomenclature. So once again, find your longest carbon chain."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one here. So let's put some more carbons on here. So we'll put a few extra carbons, and we'll name this using IUPAC nomenclature. So once again, find your longest carbon chain. So we can go ahead and find how many carbons are our longest carbon chain. That would be 1, 2, 3, and 4, like that. So we can go ahead and write butane for our parent name."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So once again, find your longest carbon chain. So we can go ahead and find how many carbons are our longest carbon chain. That would be 1, 2, 3, and 4, like that. So we can go ahead and write butane for our parent name. So we can go ahead and put butane in here. I want to next number my carbon chain to give the lowest number possible to the substituents. So in this case, it makes more sense to number from the left."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and write butane for our parent name. So we can go ahead and put butane in here. I want to next number my carbon chain to give the lowest number possible to the substituents. So in this case, it makes more sense to number from the left. So I get 1, 2, 3, and 4 to give my substituents the lowest number possible. I can see now that my epoxide forms between carbons 2 and 3. So I'm going to write 2, 3, epoxy, butane, like that."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So in this case, it makes more sense to number from the left. So I get 1, 2, 3, and 4 to give my substituents the lowest number possible. I can see now that my epoxide forms between carbons 2 and 3. So I'm going to write 2, 3, epoxy, butane, like that. And I know that I also have a methyl group coming off of carbon 2. So to complete the name, all I have to do is write 2-methyl in the front here. So now I have 2-methyl, 2, 3, epoxy, butane for my IUPAC name."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to write 2, 3, epoxy, butane, like that. And I know that I also have a methyl group coming off of carbon 2. So to complete the name, all I have to do is write 2-methyl in the front here. So now I have 2-methyl, 2, 3, epoxy, butane for my IUPAC name. So how do we make epoxides? We've already seen one way to do it. And in an earlier video, we started with our alkene."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So now I have 2-methyl, 2, 3, epoxy, butane for my IUPAC name. So how do we make epoxides? We've already seen one way to do it. And in an earlier video, we started with our alkene. And to the alkene, we added a peroxy acid. And a peroxy acid looks a lot like a carboxylic acid, except it has an extra oxygen in there like that. And in the mechanism for the epoxidation of alkenes, we saw it was a concerted mechanism where one of those oxygens was added in here to form our epoxide like that."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And in an earlier video, we started with our alkene. And to the alkene, we added a peroxy acid. And a peroxy acid looks a lot like a carboxylic acid, except it has an extra oxygen in there like that. And in the mechanism for the epoxidation of alkenes, we saw it was a concerted mechanism where one of those oxygens was added in here to form our epoxide like that. So check out the earlier video to see the mechanism for epoxidation of alkenes. So in this video, we're going to cover another way to make epoxides. And that is using halohydrins."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And in the mechanism for the epoxidation of alkenes, we saw it was a concerted mechanism where one of those oxygens was added in here to form our epoxide like that. So check out the earlier video to see the mechanism for epoxidation of alkenes. So in this video, we're going to cover another way to make epoxides. And that is using halohydrins. So to make a halohydrin, you also start out with an alkene. And we also saw this mechanism in an earlier video. You add a halogen and water."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And that is using halohydrins. So to make a halohydrin, you also start out with an alkene. And we also saw this mechanism in an earlier video. You add a halogen and water. And we're going to add bromine and water. And we end up adding the OH and 1-bromine across our double bonds. So we ended up getting an anti-addition of an OH and a bromine."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "You add a halogen and water. And we're going to add bromine and water. And we end up adding the OH and 1-bromine across our double bonds. So we ended up getting an anti-addition of an OH and a bromine. So they're going to add on opposite sides from each other like that. So this molecule is called a halohydrin. And again, check out an earlier video for the mechanism to form a halohydrin."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So we ended up getting an anti-addition of an OH and a bromine. So they're going to add on opposite sides from each other like that. So this molecule is called a halohydrin. And again, check out an earlier video for the mechanism to form a halohydrin. Once you form a halohydrin, you can use that halohydrin to form an epoxide. So let's go ahead and take that halohydrin. And let's see the mechanism of how we can form an epoxide from that."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And again, check out an earlier video for the mechanism to form a halohydrin. Once you form a halohydrin, you can use that halohydrin to form an epoxide. So let's go ahead and take that halohydrin. And let's see the mechanism of how we can form an epoxide from that. So I'm going to redraw that halohydrin. So I'm going to go ahead and put in the OH here like that. Put my lone pairs of electrons."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And let's see the mechanism of how we can form an epoxide from that. So I'm going to redraw that halohydrin. So I'm going to go ahead and put in the OH here like that. Put my lone pairs of electrons. And then I have my bromine over here. And I'll go ahead and put in my lone pairs of electrons on bromine as well. And for right now, we can say anything could be attached to this."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "Put my lone pairs of electrons. And then I have my bromine over here. And I'll go ahead and put in my lone pairs of electrons on bromine as well. And for right now, we can say anything could be attached to this. And we'll go into stereochemistry in the next video. So what we need to do is add a base. So something like sodium hydroxide will work."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And for right now, we can say anything could be attached to this. And we'll go into stereochemistry in the next video. So what we need to do is add a base. So something like sodium hydroxide will work. So we're going to add in sodium hydroxide, Na plus, OH minus. So the hydroxide anion is going to function as a base. So a lone pair of electrons on the oxygen are going to take this proton on our alcohol."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So something like sodium hydroxide will work. So we're going to add in sodium hydroxide, Na plus, OH minus. So the hydroxide anion is going to function as a base. So a lone pair of electrons on the oxygen are going to take this proton on our alcohol. And these electrons in here are going to kick off onto our oxygen. So let's go ahead and draw the result of that acid-base reaction. So what do we make from that?"}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on the oxygen are going to take this proton on our alcohol. And these electrons in here are going to kick off onto our oxygen. So let's go ahead and draw the result of that acid-base reaction. So what do we make from that? Well, now we have our oxygen with three lone pairs of electrons around it like that, which give this oxygen a negative 1 formal charge. And we still have our bromine here like that. And then we still have these other groups attached to our carbon."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So what do we make from that? Well, now we have our oxygen with three lone pairs of electrons around it like that, which give this oxygen a negative 1 formal charge. And we still have our bromine here like that. And then we still have these other groups attached to our carbon. So in the next step, we need to think about, again, the polarization in the bonds between carbon and our halogen. Our halogen is more electronegative. So the halogen is going to take a little bit of this electron density in the bond between carbon and bromine and therefore give the bromine a partial negative charge."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And then we still have these other groups attached to our carbon. So in the next step, we need to think about, again, the polarization in the bonds between carbon and our halogen. Our halogen is more electronegative. So the halogen is going to take a little bit of this electron density in the bond between carbon and bromine and therefore give the bromine a partial negative charge. This carbon is going to lose a little bit of electron density. So this carbon is actually partially positive. And so the alkoxide that we formed when the alcohol was deprotonated has a negative charge."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So the halogen is going to take a little bit of this electron density in the bond between carbon and bromine and therefore give the bromine a partial negative charge. This carbon is going to lose a little bit of electron density. So this carbon is actually partially positive. And so the alkoxide that we formed when the alcohol was deprotonated has a negative charge. It's going to function as a nucleophile. The partially positive carbon, once electrons, it's going to function as an electrophile. And we're going to get a nucleophilic attack by our alkoxide anion on our partially positive carbon."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And so the alkoxide that we formed when the alcohol was deprotonated has a negative charge. It's going to function as a nucleophile. The partially positive carbon, once electrons, it's going to function as an electrophile. And we're going to get a nucleophilic attack by our alkoxide anion on our partially positive carbon. So this is actually an intramolecular Williamson ether synthesis. So if you think about it, if these electrons in here are going to attack this carbon, that would kick these electrons off onto your bromine like that. And it's an intramolecular Williamson ether synthesis, where your alkoxide is the nucleophile in an SN2 type mechanism."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to get a nucleophilic attack by our alkoxide anion on our partially positive carbon. So this is actually an intramolecular Williamson ether synthesis. So if you think about it, if these electrons in here are going to attack this carbon, that would kick these electrons off onto your bromine like that. And it's an intramolecular Williamson ether synthesis, where your alkoxide is the nucleophile in an SN2 type mechanism. So if we go ahead and draw the product, now this oxygen was bonded to the carbon on the right. Now it's also bonded to the carbon on the left and the bromine left. That was our leaving group."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And it's an intramolecular Williamson ether synthesis, where your alkoxide is the nucleophile in an SN2 type mechanism. So if we go ahead and draw the product, now this oxygen was bonded to the carbon on the right. Now it's also bonded to the carbon on the left and the bromine left. That was our leaving group. And so we can see that we're going to end up forming an epoxide with this mechanism. So let's go ahead and do a quick problem here, where we're forming an epoxide from an alkene. And we'll start with cyclohexene."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "That was our leaving group. And so we can see that we're going to end up forming an epoxide with this mechanism. So let's go ahead and do a quick problem here, where we're forming an epoxide from an alkene. And we'll start with cyclohexene. So here is our cyclohexene molecule. And we'll make an epoxide two ways. So in the first way, we'll add a peroxy acid."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with cyclohexene. So here is our cyclohexene molecule. And we'll make an epoxide two ways. So in the first way, we'll add a peroxy acid. And there are several that you can use. One of the most common ones would be peroxyacetic acid. So peroxyacetic acid looks very similar to acetic acid, except you have an extra oxygen in there like that."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So in the first way, we'll add a peroxy acid. And there are several that you can use. One of the most common ones would be peroxyacetic acid. So peroxyacetic acid looks very similar to acetic acid, except you have an extra oxygen in there like that. So it's epoxidation of an alkene. And when we draw our product, so let's go ahead and draw our product. We form an epoxide."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "So peroxyacetic acid looks very similar to acetic acid, except you have an extra oxygen in there like that. So it's epoxidation of an alkene. And when we draw our product, so let's go ahead and draw our product. We form an epoxide. And I'm going to go ahead and draw the product with a wedge here. So there's an oxygen coming out relative to that plane. And if we go ahead and name our product, so the parent name would be cyclohexane."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "We form an epoxide. And I'm going to go ahead and draw the product with a wedge here. So there's an oxygen coming out relative to that plane. And if we go ahead and name our product, so the parent name would be cyclohexane. And our epoxide would form between carbons one and two. So we could go ahead and name this as 1, 2-epoxycyclohexane, like that. Let's go ahead and to cyclohexene, let's do another reaction."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And if we go ahead and name our product, so the parent name would be cyclohexane. And our epoxide would form between carbons one and two. So we could go ahead and name this as 1, 2-epoxycyclohexane, like that. Let's go ahead and to cyclohexene, let's do another reaction. Let's start with cyclohexane. And this time in the first step, we'll add some bromine and some water. And that will form our halohydrin."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and to cyclohexene, let's do another reaction. Let's start with cyclohexane. And this time in the first step, we'll add some bromine and some water. And that will form our halohydrin. And in the second step, we'll add sodium hydroxide to act as our base. And we get an intramolecular Williamson ether synthesis. And so we're going to end up with the same product."}, {"video_title": "Nomenclature and preparation of epoxides Organic chemistry Khan Academy.mp3", "Sentence": "And that will form our halohydrin. And in the second step, we'll add sodium hydroxide to act as our base. And we get an intramolecular Williamson ether synthesis. And so we're going to end up with the same product. So we're going to end up with the same product here. We're going to end up with 1, 2-epoxycyclohexane. Now for this reaction, we don't have to worry about stereochemistry."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "In the last video, we learned that there are a class of stars called Cepheid variables. And these are these super giant stars, as much as 30,000 times as bright as the sun, as a mass 20 times the mass of the sun. And what's neat about them is, one, because they're so large and so bright, you can see them really, really far away. And what's even neater about them is that they're variable, that they pulsate. And because their pulsations are related to their actual luminosity, you know if you see a Cepheid variable star in some distant galaxy, you know what its luminosity actually is if you were kind of at the star, because you can see how its period of pulsation. And so if you know its actual luminosity, and then you know, obviously, its apparent luminosity, you know how much it's gotten dim. And the more dim it's gotten from its actual state, you know the farther away it is."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And what's even neater about them is that they're variable, that they pulsate. And because their pulsations are related to their actual luminosity, you know if you see a Cepheid variable star in some distant galaxy, you know what its luminosity actually is if you were kind of at the star, because you can see how its period of pulsation. And so if you know its actual luminosity, and then you know, obviously, its apparent luminosity, you know how much it's gotten dim. And the more dim it's gotten from its actual state, you know the farther away it is. So that's the real value of them. What I want to do in this video is to try to explain why they're variable, why they pulsate. And to do that, what we're going to think about is doubly and singly ionized helium."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And the more dim it's gotten from its actual state, you know the farther away it is. So that's the real value of them. What I want to do in this video is to try to explain why they're variable, why they pulsate. And to do that, what we're going to think about is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw neutral helium. Neutral helium's got two protons, two neutrons, and then two electrons. And obviously, this is not drawn to scale."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And to do that, what we're going to think about is doubly and singly ionized helium. And just to review, helium, so neutral helium, let me draw neutral helium. Neutral helium's got two protons, two neutrons, and then two electrons. And obviously, this is not drawn to scale. So this is neutral helium right over here. Now, if you singly ionize helium, you knock off one of these electrons. And these type of things happen in stars, when you have a lot of heat, easier to ionize things."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And obviously, this is not drawn to scale. So this is neutral helium right over here. Now, if you singly ionize helium, you knock off one of these electrons. And these type of things happen in stars, when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off, so now you only have one electron."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And these type of things happen in stars, when you have a lot of heat, easier to ionize things. So singly ionized helium will look like this. It'll have the same nucleus, two protons, two neutrons. One of the electrons gets knocked off, so now you only have one electron. And now you have a net positive charge. So here, let me do this in a different color. This helium now has a net charge."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "One of the electrons gets knocked off, so now you only have one electron. And now you have a net positive charge. So here, let me do this in a different color. This helium now has a net charge. We could write 1 plus here, but if you just write a plus, you implicitly mean a positive charge of 1. Now, you can also doubly ionize helium if the environment is hot enough. You can doubly ionize helium."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This helium now has a net charge. We could write 1 plus here, but if you just write a plus, you implicitly mean a positive charge of 1. Now, you can also doubly ionize helium if the environment is hot enough. You can doubly ionize helium. And doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus like this. This right here is doubly ionized helium."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "You can doubly ionize helium. And doubly ionizing helium is essentially knocking off both of the electrons. So then it's really just a helium nucleus like this. This right here is doubly ionized helium. Now, I just said, in order to do this, you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both. This electron really doesn't want to leave."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This right here is doubly ionized helium. Now, I just said, in order to do this, you have to have a hotter environment. There has to be a hotter environment in order to be able to knock off both. This electron really doesn't want to leave. To take an electron off of something that's already positive is difficult. You have to have a lot of, really, pressure and temperature. This is cooler."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This electron really doesn't want to leave. To take an electron off of something that's already positive is difficult. You have to have a lot of, really, pressure and temperature. This is cooler. And this is all relative. We're talking about the insides of stars. So this is a hotter part of the star versus a cooler part of the star, I guess, is the way you think about it."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This is cooler. And this is all relative. We're talking about the insides of stars. So this is a hotter part of the star versus a cooler part of the star, I guess, is the way you think about it. It's a very hot environment by our traditional everyday standards. Now, the other thing about the doubly ionized helium is that it is more opaque, which means it doesn't allow light to go through it. It actually absorbs light."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this is a hotter part of the star versus a cooler part of the star, I guess, is the way you think about it. It's a very hot environment by our traditional everyday standards. Now, the other thing about the doubly ionized helium is that it is more opaque, which means it doesn't allow light to go through it. It actually absorbs light. It is more opaque. It absorbs light. Or another way, it absorbs that light energy."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It actually absorbs light. It is more opaque. It absorbs light. Or another way, it absorbs that light energy. That energy will make it even hotter. So that's just something to think about. Now, the singly ionized helium is more transparent."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Or another way, it absorbs that light energy. That energy will make it even hotter. So that's just something to think about. Now, the singly ionized helium is more transparent. It allows the light to pass through it. So it doesn't get heated as much by photons that are kind of going near it or through it or whatever. It allows them to go through it."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Now, the singly ionized helium is more transparent. It allows the light to pass through it. So it doesn't get heated as much by photons that are kind of going near it or through it or whatever. It allows them to go through it. Here, the photons are going to actually heat up the ion. So let's think about how this might cause a Cepheid variable to pulsate. So assuming that Cepheid variables have large enough quantities, I should say, of these ions, we can imagine that when a Cepheid variable is dim."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It allows them to go through it. Here, the photons are going to actually heat up the ion. So let's think about how this might cause a Cepheid variable to pulsate. So assuming that Cepheid variables have large enough quantities, I should say, of these ions, we can imagine that when a Cepheid variable is dim. So let me draw a dim Cepheid variable. So I'll draw that like I'll draw this in a dim color. So this is a dim Cepheid variable right here."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So assuming that Cepheid variables have large enough quantities, I should say, of these ions, we can imagine that when a Cepheid variable is dim. So let me draw a dim Cepheid variable. So I'll draw that like I'll draw this in a dim color. So this is a dim Cepheid variable right here. In its dim state, just like this, you have a lot of the doubly ionized helium in the star, at least kind of the outer surface of the star. And so this does not allow a lot of light to pass through. So this is the dim part of the pulsation of the Cepheid variable."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this is a dim Cepheid variable right here. In its dim state, just like this, you have a lot of the doubly ionized helium in the star, at least kind of the outer surface of the star. And so this does not allow a lot of light to pass through. So this is the dim part of the pulsation of the Cepheid variable. Now, because this doubly ionized helium is opaque, it is absorbing the light. It is getting heated. It is getting heated."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So this is the dim part of the pulsation of the Cepheid variable. Now, because this doubly ionized helium is opaque, it is absorbing the light. It is getting heated. It is getting heated. And because it's getting heated, it'll cause the star to expand. So because it's getting heated, it'll become more energetic, and the star will actually expand. Now, as the star expands, because this doubly ionized helium is getting heated, what's going to happen?"}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It is getting heated. And because it's getting heated, it'll cause the star to expand. So because it's getting heated, it'll become more energetic, and the star will actually expand. Now, as the star expands, because this doubly ionized helium is getting heated, what's going to happen? The further away you are from the core of the star, the cooler it gets. So this expanded because it was getting heated, but then because it expanded, the outer layers of the star become cooler. And since they're cooler, helium won't be doubly ionized anymore."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Now, as the star expands, because this doubly ionized helium is getting heated, what's going to happen? The further away you are from the core of the star, the cooler it gets. So this expanded because it was getting heated, but then because it expanded, the outer layers of the star become cooler. And since they're cooler, helium won't be doubly ionized anymore. It'll get an electron from each helium atom. It can now get an electron from the plasma, I guess we can say, to become singly ionized helium. So now we have singly ionized helium."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And since they're cooler, helium won't be doubly ionized anymore. It'll get an electron from each helium atom. It can now get an electron from the plasma, I guess we can say, to become singly ionized helium. So now we have singly ionized helium. And now the star is going to be more transparent. It's going to allow more light to pass through it. So now this is the bright part of the pulsation."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So now we have singly ionized helium. And now the star is going to be more transparent. It's going to allow more light to pass through it. So now this is the bright part of the pulsation. It's going to allow more light through. So now it is bright. The star is bright."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So now this is the bright part of the pulsation. It's going to allow more light through. So now it is bright. The star is bright. But what's happening now? Because the light is no longer, or it's not being absorbed as well by the helium when it was a doubly ionized helium, now it's letting most of the light, or a lot more of the light, get through. It's not going to get heated as much."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "The star is bright. But what's happening now? Because the light is no longer, or it's not being absorbed as well by the helium when it was a doubly ionized helium, now it's letting most of the light, or a lot more of the light, get through. It's not going to get heated as much. And so it won't have the kinetic energy to kind of keep pushing out, to keep moving outward. And so it'll collapse back into the star. And so then this will cool down and collapse back in."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's not going to get heated as much. And so it won't have the kinetic energy to kind of keep pushing out, to keep moving outward. And so it'll collapse back into the star. And so then this will cool down and collapse back in. And when it collapses back in, what's going to happen? When it collapses back in, when these helium atoms get closer to the center of the star, to the core of the star, they're going to be heated again, because they're closer now to the core. And when they get heated, they're going to become doubly ionized."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And so then this will cool down and collapse back in. And when it collapses back in, what's going to happen? When it collapses back in, when these helium atoms get closer to the center of the star, to the core of the star, they're going to be heated again, because they're closer now to the core. And when they get heated, they're going to become doubly ionized. So then we have doubly ionized helium again. And then the cycle will go again. It is now opaque."}, {"video_title": "Why cepheids pulsate Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And when they get heated, they're going to become doubly ionized. So then we have doubly ionized helium again. And then the cycle will go again. It is now opaque. It will now absorb more energy. That'll cause it to have more kinetic energy to expand. Once it expands, it'll get cool again, and transparent, and bright."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's get some more practice with the RS system. So we'll start with this compound right here. We already know from earlier videos that this carbon is a chiral center. So let me go ahead and redraw everything because it's going to help us assign a configuration. So that's that carbon. We have a methyl group coming out at us. I'm going to draw in the carbon with the hydrogens here."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and redraw everything because it's going to help us assign a configuration. So that's that carbon. We have a methyl group coming out at us. I'm going to draw in the carbon with the hydrogens here. We have a hydrogen going away from us. And going to the right around the ring, we hit a CH2. So I'm drawing in a carbon with two hydrogens here."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to draw in the carbon with the hydrogens here. We have a hydrogen going away from us. And going to the right around the ring, we hit a CH2. So I'm drawing in a carbon with two hydrogens here. And then we hit CH. So this carbon bonded to a hydrogen. That's this carbon on the ring."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm drawing in a carbon with two hydrogens here. And then we hit CH. So this carbon bonded to a hydrogen. That's this carbon on the ring. Notice that this carbon that I just marked is double bonded to this carbon. And for the purposes of R and S system, we're going to pretend like this carbon is bonded to two different carbons, even though it's really one. So that's how to handle a double bond."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's this carbon on the ring. Notice that this carbon that I just marked is double bonded to this carbon. And for the purposes of R and S system, we're going to pretend like this carbon is bonded to two different carbons, even though it's really one. So that's how to handle a double bond. Going this way around the ring, we hit a CH2. So let me draw that in. So here's our CH2."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's how to handle a double bond. Going this way around the ring, we hit a CH2. So let me draw that in. So here's our CH2. And then we hit another CH2 right here. So a CH2. And then this carbon is bonded to this carbon."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's our CH2. And then we hit another CH2 right here. So a CH2. And then this carbon is bonded to this carbon. So I'm just going to draw a line in there like that. All right, let's think about priority. So this is our chiral center."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then this carbon is bonded to this carbon. So I'm just going to draw a line in there like that. All right, let's think about priority. So this is our chiral center. Let's look at the four groups attached to the chiral center. So this is step one. Prioritize the four groups using atomic number."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is our chiral center. Let's look at the four groups attached to the chiral center. So this is step one. Prioritize the four groups using atomic number. So what's directly attached to this carbon? There's a hydrogen, there's a carbon, there's a carbon, and there's a carbon. So a carbon beats hydrogen in terms of atomic number."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Prioritize the four groups using atomic number. So what's directly attached to this carbon? There's a hydrogen, there's a carbon, there's a carbon, and there's a carbon. So a carbon beats hydrogen in terms of atomic number. So hydrogen is the lowest priority group. So we assign that a group number four. All right, now we have a tie."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So a carbon beats hydrogen in terms of atomic number. So hydrogen is the lowest priority group. So we assign that a group number four. All right, now we have a tie. We have three carbons. We have three carbons. So we need to see what those carbons are directly bonded to."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, now we have a tie. We have three carbons. We have three carbons. So we need to see what those carbons are directly bonded to. Let's start with this top carbon here. This carbon is bonded to hydrogen, hydrogen, hydrogen. So let me write that down."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we need to see what those carbons are directly bonded to. Let's start with this top carbon here. This carbon is bonded to hydrogen, hydrogen, hydrogen. So let me write that down. So hydrogen, hydrogen, hydrogen. Let's move to this carbon on the right. This one's bonded to carbon, hydrogen, hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me write that down. So hydrogen, hydrogen, hydrogen. Let's move to this carbon on the right. This one's bonded to carbon, hydrogen, hydrogen. So carbon, hydrogen, hydrogen. We're thinking about the atoms directly bonded to it. And we're going in decreasing atomic number, which is why I put the carbon first."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This one's bonded to carbon, hydrogen, hydrogen. So carbon, hydrogen, hydrogen. We're thinking about the atoms directly bonded to it. And we're going in decreasing atomic number, which is why I put the carbon first. Now let's look at this carbon. So this carbon is bonded to a carbon, a hydrogen, and a hydrogen, so CHH. All right, let's compare now."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we're going in decreasing atomic number, which is why I put the carbon first. Now let's look at this carbon. So this carbon is bonded to a carbon, a hydrogen, and a hydrogen, so CHH. All right, let's compare now. Well, first point of difference. We look at the first atom here. This is a hydrogen, and this is a carbon."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's compare now. Well, first point of difference. We look at the first atom here. This is a hydrogen, and this is a carbon. So carbon beats hydrogen. And then over here, we have a carbon. So this one doesn't win."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a hydrogen, and this is a carbon. So carbon beats hydrogen. And then over here, we have a carbon. So this one doesn't win. This one must be third in terms of priority. So I put a three here for this methyl group. And now we continue on."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this one doesn't win. This one must be third in terms of priority. So I put a three here for this methyl group. And now we continue on. We have hydrogen versus hydrogen. So that's a tie. Another hydrogen, another hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And now we continue on. We have hydrogen versus hydrogen. So that's a tie. Another hydrogen, another hydrogen. So we have another tie. So we need to go to the next atom to break this tie. So we go to our next carbon."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Another hydrogen, another hydrogen. So we have another tie. So we need to go to the next atom to break this tie. So we go to our next carbon. So this one right here. What atoms is this carbon directly bonded to? Well, it's bonded to carbon, carbon, hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we go to our next carbon. So this one right here. What atoms is this carbon directly bonded to? Well, it's bonded to carbon, carbon, hydrogen. So carbon, carbon, hydrogen. This carbon is directly bonded to carbon, hydrogen, hydrogen. So carbon, hydrogen, hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, it's bonded to carbon, carbon, hydrogen. So carbon, carbon, hydrogen. This carbon is directly bonded to carbon, hydrogen, hydrogen. So carbon, hydrogen, hydrogen. We look for the first point of difference. This is a carbon versus a carbon. This is a carbon versus a hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So carbon, hydrogen, hydrogen. We look for the first point of difference. This is a carbon versus a carbon. This is a carbon versus a hydrogen. So the carbon wins. And that means that this way around the ring, this is the higher priority path around the ring. So this gets a number one."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a carbon versus a hydrogen. So the carbon wins. And that means that this way around the ring, this is the higher priority path around the ring. So this gets a number one. And then this path around the ring, going this way, gets a number two. So now we've assigned priority to our four groups. So now we're ready for step two."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this gets a number one. And then this path around the ring, going this way, gets a number two. So now we've assigned priority to our four groups. So now we're ready for step two. Orient the group so the lowest priority group is projecting away from us in space. So let's go back to our original dot structure here. We said that this way around the ring was the highest priority."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So now we're ready for step two. Orient the group so the lowest priority group is projecting away from us in space. So let's go back to our original dot structure here. We said that this way around the ring was the highest priority. So this got a number one. Next, going this way around the ring was second highest priority, so a number two. Our methyl group was a number three, and our hydrogen was a number four."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We said that this way around the ring was the highest priority. So this got a number one. Next, going this way around the ring was second highest priority, so a number two. Our methyl group was a number three, and our hydrogen was a number four. Our lowest priority group is going away from us. Our hydrogen is on a dash. So that's going away from us in space."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Our methyl group was a number three, and our hydrogen was a number four. Our lowest priority group is going away from us. Our hydrogen is on a dash. So that's going away from us in space. So now we have finally on to step three. And let me change colors again, because it's getting a little busy here. So step three, determine if the sequence one, two, three is clockwise or counterclockwise."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's going away from us in space. So now we have finally on to step three. And let me change colors again, because it's getting a little busy here. So step three, determine if the sequence one, two, three is clockwise or counterclockwise. So if I look one, two, three, and I go around a circle, here's one, here's two, and here's three. So going around one, two, three in a circle is this way. That is clockwise."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So step three, determine if the sequence one, two, three is clockwise or counterclockwise. So if I look one, two, three, and I go around a circle, here's one, here's two, and here's three. So going around one, two, three in a circle is this way. That is clockwise. You can see that's clockwise right here. And so clockwise is R. So the configuration of this chiral center is R. Now let's look at this compound. So we have only one chiral center to worry about."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That is clockwise. You can see that's clockwise right here. And so clockwise is R. So the configuration of this chiral center is R. Now let's look at this compound. So we have only one chiral center to worry about. It's this one right here. Let's think about the atoms that are directly bonded to our chiral center. Well, there's a bromine directly bonded to it, a chlorine, and over here would be a carbon, and then we have another carbon."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have only one chiral center to worry about. It's this one right here. Let's think about the atoms that are directly bonded to our chiral center. Well, there's a bromine directly bonded to it, a chlorine, and over here would be a carbon, and then we have another carbon. So we prioritize our groups in terms of atomic number. Bromine has the highest atomic number out of those atoms, so we give bromine a number one. Next would be chlorine with atomic number of 17, so that's number two."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, there's a bromine directly bonded to it, a chlorine, and over here would be a carbon, and then we have another carbon. So we prioritize our groups in terms of atomic number. Bromine has the highest atomic number out of those atoms, so we give bromine a number one. Next would be chlorine with atomic number of 17, so that's number two. Now we have a tie for our carbons. So we need to see what is directly bonded to those carbons. So for the carbon on the right here, this carbon is directly bonded to a carbon here, a carbon here, and then of course a hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next would be chlorine with atomic number of 17, so that's number two. Now we have a tie for our carbons. So we need to see what is directly bonded to those carbons. So for the carbon on the right here, this carbon is directly bonded to a carbon here, a carbon here, and then of course a hydrogen. So we write that in as carbon, carbon, hydrogen. And then for the carbon on the left, so this carbon, that carbon's bonded to a carbon and two hydrogens. So we write in carbon, hydrogen, hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So for the carbon on the right here, this carbon is directly bonded to a carbon here, a carbon here, and then of course a hydrogen. So we write that in as carbon, carbon, hydrogen. And then for the carbon on the left, so this carbon, that carbon's bonded to a carbon and two hydrogens. So we write in carbon, hydrogen, hydrogen. We look for the first point of difference. So we have carbon versus carbon, so that's a tie. So we keep going and we get carbon versus hydrogen."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we write in carbon, hydrogen, hydrogen. We look for the first point of difference. So we have carbon versus carbon, so that's a tie. So we keep going and we get carbon versus hydrogen. So the carbon wins and this group gets the higher priority. So this isopropyl group is a higher priority than this ethyl group. So that means the isopropyl group is gonna get a number three, so this is three, and the ethyl group is the lowest priority."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we keep going and we get carbon versus hydrogen. So the carbon wins and this group gets the higher priority. So this isopropyl group is a higher priority than this ethyl group. So that means the isopropyl group is gonna get a number three, so this is three, and the ethyl group is the lowest priority. It gets a number four. Now that we've assigned priority to our groups, we need to orient the molecules so the lowest priority group is pointing away from us, and the lowest priority group is group number four. So I'm gonna go to a video in a second, and in the video I'm gonna show you two different ways to think about putting this group going away from you in space."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that means the isopropyl group is gonna get a number three, so this is three, and the ethyl group is the lowest priority. It gets a number four. Now that we've assigned priority to our groups, we need to orient the molecules so the lowest priority group is pointing away from us, and the lowest priority group is group number four. So I'm gonna go to a video in a second, and in the video I'm gonna show you two different ways to think about putting this group going away from you in space. So one way would be to just think about an axis through this carbon here and then rotate, and then rotate around this axis until your lowest priority group is pointing away from you. Another way to think about it is like a Newman projection. If you stare down this bond, let me go ahead and change colors here, if you look down at this carbon-carbon bond here, so if you put your eye along this axis, so here's your eye, that would mean your lowest priority group, your ethyl group, would be going away from you."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna go to a video in a second, and in the video I'm gonna show you two different ways to think about putting this group going away from you in space. So one way would be to just think about an axis through this carbon here and then rotate, and then rotate around this axis until your lowest priority group is pointing away from you. Another way to think about it is like a Newman projection. If you stare down this bond, let me go ahead and change colors here, if you look down at this carbon-carbon bond here, so if you put your eye along this axis, so here's your eye, that would mean your lowest priority group, your ethyl group, would be going away from you. So that's another way to think about looking at the model, or the molecule I should say, in a way where the lowest priority group points away from you. So here's our compound. Let's say that red represents bromine."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you stare down this bond, let me go ahead and change colors here, if you look down at this carbon-carbon bond here, so if you put your eye along this axis, so here's your eye, that would mean your lowest priority group, your ethyl group, would be going away from you. So that's another way to think about looking at the model, or the molecule I should say, in a way where the lowest priority group points away from you. So here's our compound. Let's say that red represents bromine. So there's bromine, and yellow represents chlorine. So our goal is to rotate this to put our lowest priority group going away from us. So if we think about an axis through our chiral center, we'll rotate it so the ethyl group is pointing away."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's say that red represents bromine. So there's bromine, and yellow represents chlorine. So our goal is to rotate this to put our lowest priority group going away from us. So if we think about an axis through our chiral center, we'll rotate it so the ethyl group is pointing away. And now we can see that our red bromine is to the left, our yellow chlorine is to the right, and the isopropyl group is up in space. If we go back to where we were, this time let's think about a Newman projection. So we're staring down this carbon-carbon bond."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about an axis through our chiral center, we'll rotate it so the ethyl group is pointing away. And now we can see that our red bromine is to the left, our yellow chlorine is to the right, and the isopropyl group is up in space. If we go back to where we were, this time let's think about a Newman projection. So we're staring down this carbon-carbon bond. So let's rotate the molecule. And it's a little bit different perspective, but we're still able to see a red bromine to the left, a yellow chlorine to the right, and an isopropyl group up. So here's what we saw when we stared down our carbon-carbon bond."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we're staring down this carbon-carbon bond. So let's rotate the molecule. And it's a little bit different perspective, but we're still able to see a red bromine to the left, a yellow chlorine to the right, and an isopropyl group up. So here's what we saw when we stared down our carbon-carbon bond. So this carbon is our chiral center. So that's this one right here. And then I used red for bromine, and then I used yellow for chlorine."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here's what we saw when we stared down our carbon-carbon bond. So this carbon is our chiral center. So that's this one right here. And then I used red for bromine, and then I used yellow for chlorine. So we can see that our bromine is to the left here. And this was the highest priority group. The chlorine is to the right."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then I used red for bromine, and then I used yellow for chlorine. So we can see that our bromine is to the left here. And this was the highest priority group. The chlorine is to the right. That was the second highest priority. And then we have our isopropyl group here up. So that's one, two, three, with our lowest priority group pointing away from us."}, {"video_title": "R,S system practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The chlorine is to the right. That was the second highest priority. And then we have our isopropyl group here up. So that's one, two, three, with our lowest priority group pointing away from us. So all we have to do now is go around in a circle and see what we get. So we're going from one to two to three in a circle. So that means we're going this way around our circle."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, we looked at a Fischer projection of a compound that had only one chirality center. This molecule has two, and our goal is to determine the configuration of each chirality center. So at the intersection of these lines here, we know that this is a chiral center, and then we have another one down here. So those are the two chirality centers. Let's focus in on the top one here, and let's assign a configuration to that carbon. So let me go ahead and draw the carbon on the right. We know that in a Fischer projection, the horizontal lines represent a bond that's coming out of the page at us."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So those are the two chirality centers. Let's focus in on the top one here, and let's assign a configuration to that carbon. So let me go ahead and draw the carbon on the right. We know that in a Fischer projection, the horizontal lines represent a bond that's coming out of the page at us. So over here on the left, the bond to this hydrogen is coming out of the page, and so is the bond to this OH. So let's draw them on the right. We have a bond to a hydrogen coming out of the page on the left, and then we have a bond to OH coming out of the page on the right."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that in a Fischer projection, the horizontal lines represent a bond that's coming out of the page at us. So over here on the left, the bond to this hydrogen is coming out of the page, and so is the bond to this OH. So let's draw them on the right. We have a bond to a hydrogen coming out of the page on the left, and then we have a bond to OH coming out of the page on the right. So those are represented by a wedge. We know that the vertical lines in a Fischer projection represent a bond that's going away from us in space. So this line right here is a bond that's going away from us, so that would be a dash, so let's draw that in."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have a bond to a hydrogen coming out of the page on the left, and then we have a bond to OH coming out of the page on the right. So those are represented by a wedge. We know that the vertical lines in a Fischer projection represent a bond that's going away from us in space. So this line right here is a bond that's going away from us, so that would be a dash, so let's draw that in. That's going to a carboxylic acid, which we know is a carbon double bonded to an oxygen bonded to an OH. Let's look at the other vertical line, so this one down here. So that represents a bond going away from us in space, so we draw that one in."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this line right here is a bond that's going away from us, so that would be a dash, so let's draw that in. That's going to a carboxylic acid, which we know is a carbon double bonded to an oxygen bonded to an OH. Let's look at the other vertical line, so this one down here. So that represents a bond going away from us in space, so we draw that one in. So we're essentially staring down at our chiral center. And this carbon down here is directly bonded to an oxygen, a carbon, and a hydrogen. And those are the only atoms that we need to assign a configuration, so those are the only ones that I'm going to draw in."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that represents a bond going away from us in space, so we draw that one in. So we're essentially staring down at our chiral center. And this carbon down here is directly bonded to an oxygen, a carbon, and a hydrogen. And those are the only atoms that we need to assign a configuration, so those are the only ones that I'm going to draw in. Let's go back to our chiral center, so this carbon right here. We know to assign a configuration, we look at the atoms that are directly bonded to that carbon. There's a hydrogen, an oxygen, and then our two carbons."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And those are the only atoms that we need to assign a configuration, so those are the only ones that I'm going to draw in. Let's go back to our chiral center, so this carbon right here. We know to assign a configuration, we look at the atoms that are directly bonded to that carbon. There's a hydrogen, an oxygen, and then our two carbons. Out of those atoms, oxygen has the highest atomic number, so the OH group gets the highest priority, and we say that is a number one. The hydrogen has the lowest atomic number, so that's lowest priority, so we say that's a number four. Next we have carbon versus carbon, so this carbon versus this carbon."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's a hydrogen, an oxygen, and then our two carbons. Out of those atoms, oxygen has the highest atomic number, so the OH group gets the highest priority, and we say that is a number one. The hydrogen has the lowest atomic number, so that's lowest priority, so we say that's a number four. Next we have carbon versus carbon, so this carbon versus this carbon. And we know that carbon has the same atomic number. And so we've seen how to break our tie in earlier videos. We look at the atoms that are directly bonded to those carbons."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next we have carbon versus carbon, so this carbon versus this carbon. And we know that carbon has the same atomic number. And so we've seen how to break our tie in earlier videos. We look at the atoms that are directly bonded to those carbons. This bottom carbon here is directly bonded to an oxygen, a carbon, and a hydrogen, so we put those in order of decreasing atomic number, so that's oxygen, carbon, hydrogen. This top carbon here, the one in our carboxylic acid, has a double bond to an oxygen, and so we treat that like it's two bonds to two different oxygens, even though it's really only one double bond to one oxygen. And then we have another bond to an oxygen on the right here, so we could say, for the purposes of assigning a configuration, that would be oxygen, oxygen, oxygen."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We look at the atoms that are directly bonded to those carbons. This bottom carbon here is directly bonded to an oxygen, a carbon, and a hydrogen, so we put those in order of decreasing atomic number, so that's oxygen, carbon, hydrogen. This top carbon here, the one in our carboxylic acid, has a double bond to an oxygen, and so we treat that like it's two bonds to two different oxygens, even though it's really only one double bond to one oxygen. And then we have another bond to an oxygen on the right here, so we could say, for the purposes of assigning a configuration, that would be oxygen, oxygen, oxygen. Next we compare. So we have an oxygen versus an oxygen, so that's a tie. We go on to our next atom, which is oxygen versus carbon, and obviously oxygen wins."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we have another bond to an oxygen on the right here, so we could say, for the purposes of assigning a configuration, that would be oxygen, oxygen, oxygen. Next we compare. So we have an oxygen versus an oxygen, so that's a tie. We go on to our next atom, which is oxygen versus carbon, and obviously oxygen wins. It has the higher atomic number, so the carboxylic acid group gets the higher priority, so this must be group two. And this group down here must be group three. Once we've done this, our next goal is to put the hydrogen going away from us in space."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We go on to our next atom, which is oxygen versus carbon, and obviously oxygen wins. It has the higher atomic number, so the carboxylic acid group gets the higher priority, so this must be group two. And this group down here must be group three. Once we've done this, our next goal is to put the hydrogen going away from us in space. But I'm gonna use the trick that I talked about in the last video, because it means you don't have to move the molecule around in your mind. You can just use this little trick. And the trick was to ignore the hydrogen for the time being, and just look at one, two, and three."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Once we've done this, our next goal is to put the hydrogen going away from us in space. But I'm gonna use the trick that I talked about in the last video, because it means you don't have to move the molecule around in your mind. You can just use this little trick. And the trick was to ignore the hydrogen for the time being, and just look at one, two, and three. Here's one, here's two, and here's three. So if we go around in a circle from one to two to three, we're going around in this direction. That is counterclockwise, and counterclockwise, we know, is S. So this looks S, but the hydrogen is coming out at us in space, and so we know all we have to do is take the opposite."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the trick was to ignore the hydrogen for the time being, and just look at one, two, and three. Here's one, here's two, and here's three. So if we go around in a circle from one to two to three, we're going around in this direction. That is counterclockwise, and counterclockwise, we know, is S. So this looks S, but the hydrogen is coming out at us in space, and so we know all we have to do is take the opposite. So even though it looks S, since the hydrogen is coming out at us, we know it's actually R. So this is actually, actually R. So if you actually made this molecule, and rotated it, and put the hydrogen going away from you in space, then you would see that it's R directly. But this trick allows you not to worry about that. So I find this to be the easiest way to assign a configuration."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That is counterclockwise, and counterclockwise, we know, is S. So this looks S, but the hydrogen is coming out at us in space, and so we know all we have to do is take the opposite. So even though it looks S, since the hydrogen is coming out at us, we know it's actually R. So this is actually, actually R. So if you actually made this molecule, and rotated it, and put the hydrogen going away from you in space, then you would see that it's R directly. But this trick allows you not to worry about that. So I find this to be the easiest way to assign a configuration. For me, it's definitely the fastest. Let's move on to our other chirality center. So this carbon down here, so let me draw that one in."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I find this to be the easiest way to assign a configuration. For me, it's definitely the fastest. Let's move on to our other chirality center. So this carbon down here, so let me draw that one in. And let's look at our horizontal lines. So we have a bond to an OH on the left, and a bond to a hydrogen on the right, and those are coming out at us, since those are horizontal lines. So the OH is on the left, and we have a hydrogen on the right."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon down here, so let me draw that one in. And let's look at our horizontal lines. So we have a bond to an OH on the left, and a bond to a hydrogen on the right, and those are coming out at us, since those are horizontal lines. So the OH is on the left, and we have a hydrogen on the right. Let's look at our vertical lines. So this line right here means a bond going away from us in space. So we're staring down at our chiral center, and that's a bond to a carbon, and that carbon is directly bonded to an oxygen, a carbon, and a hydrogen."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the OH is on the left, and we have a hydrogen on the right. Let's look at our vertical lines. So this line right here means a bond going away from us in space. So we're staring down at our chiral center, and that's a bond to a carbon, and that carbon is directly bonded to an oxygen, a carbon, and a hydrogen. So let's draw those in. Again, we don't need all of the atoms, because this is all we need to assign a configuration, so that's all I'm drawing in. And then let's look at this other vertical line, so this one down here."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we're staring down at our chiral center, and that's a bond to a carbon, and that carbon is directly bonded to an oxygen, a carbon, and a hydrogen. So let's draw those in. Again, we don't need all of the atoms, because this is all we need to assign a configuration, so that's all I'm drawing in. And then let's look at this other vertical line, so this one down here. So that's a bond going away from us in space to this carbon, so let's draw that in. So we have a dash over here on the right going to a carbon, and this carbon is directly bonded to two hydrogens and an oxygen. So let's draw that in."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then let's look at this other vertical line, so this one down here. So that's a bond going away from us in space to this carbon, so let's draw that in. So we have a dash over here on the right going to a carbon, and this carbon is directly bonded to two hydrogens and an oxygen. So let's draw that in. So we have our two hydrogens and an oxygen. Going back to our chiral center, so this carbon right here, we look at our atoms that are directly bonded to it, and we have an oxygen, a hydrogen, and two carbons. So we know that the oxygen has the highest priority."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw that in. So we have our two hydrogens and an oxygen. Going back to our chiral center, so this carbon right here, we look at our atoms that are directly bonded to it, and we have an oxygen, a hydrogen, and two carbons. So we know that the oxygen has the highest priority. So once again, the OH group is the highest priority group, and the hydrogen is the lowest priority group, and we have a tie again between carbons, so carbon versus carbon. The top carbon is directly bonded to oxygen, carbon, hydrogen. So in order of decreasing atomic number, oxygen, carbon, hydrogen."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we know that the oxygen has the highest priority. So once again, the OH group is the highest priority group, and the hydrogen is the lowest priority group, and we have a tie again between carbons, so carbon versus carbon. The top carbon is directly bonded to oxygen, carbon, hydrogen. So in order of decreasing atomic number, oxygen, carbon, hydrogen. The bottom carbon would be oxygen, hydrogen, hydrogen. So oxygen, hydrogen, hydrogen. We look for the first point of difference, so we have oxygen versus oxygen, so that's a tie."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in order of decreasing atomic number, oxygen, carbon, hydrogen. The bottom carbon would be oxygen, hydrogen, hydrogen. So oxygen, hydrogen, hydrogen. We look for the first point of difference, so we have oxygen versus oxygen, so that's a tie. Next we'd have carbon versus hydrogen, and we know that carbon has the higher atomic number. So this top group here gets higher priority, so this should be a number two, and this group down here should be a number three. Let's use our trick again."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We look for the first point of difference, so we have oxygen versus oxygen, so that's a tie. Next we'd have carbon versus hydrogen, and we know that carbon has the higher atomic number. So this top group here gets higher priority, so this should be a number two, and this group down here should be a number three. Let's use our trick again. So we ignore the hydrogen for the time being, and we look at one, two, and three. So one to two to three is going around in this direction, which we know is clockwise. So we can say that this looks R, because clockwise is R, but because this is a Fischer projection, the hydrogen is coming out at us in space."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's use our trick again. So we ignore the hydrogen for the time being, and we look at one, two, and three. So one to two to three is going around in this direction, which we know is clockwise. So we can say that this looks R, because clockwise is R, but because this is a Fischer projection, the hydrogen is coming out at us in space. So we know our trick is just to take the opposite of how it looks. So if it looks R, we can say that it's actually S. So we write down here, actually, actually S. All right, so we've done it. We've determined the configuration of each chirality center."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can say that this looks R, because clockwise is R, but because this is a Fischer projection, the hydrogen is coming out at us in space. So we know our trick is just to take the opposite of how it looks. So if it looks R, we can say that it's actually S. So we write down here, actually, actually S. All right, so we've done it. We've determined the configuration of each chirality center. So this chiral center, let me use a different color, let me use a red over here. So this chiral center, we said that one was R, so it's R at this carbon, and then at this carbon down here, it is S. Let's also practice drawing the enantiomer of this compound. We saw how to do that in the last video."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We've determined the configuration of each chirality center. So this chiral center, let me use a different color, let me use a red over here. So this chiral center, we said that one was R, so it's R at this carbon, and then at this carbon down here, it is S. Let's also practice drawing the enantiomer of this compound. We saw how to do that in the last video. We put a mirror right here, and we reflected our groups in the mirror. So this OH here, I like to draw dashed lines. We just reflect that, and therefore we would have this."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We saw how to do that in the last video. We put a mirror right here, and we reflected our groups in the mirror. So this OH here, I like to draw dashed lines. We just reflect that, and therefore we would have this. And then on this side, we would have a hydrogen, so we're reflecting that hydrogen. And then let's go to this hydrogen down here. So we just reflect this in our mirror, and then we draw our horizontal line, and then we would have an OH."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We just reflect that, and therefore we would have this. And then on this side, we would have a hydrogen, so we're reflecting that hydrogen. And then let's go to this hydrogen down here. So we just reflect this in our mirror, and then we draw our horizontal line, and then we would have an OH. So we draw in an OH here. We can put in our vertical line like that, and then we have a carboxylic acid, but this carbon is not a chiral center, so we don't really need to worry about it. We can just write in COOH here, and then same with this carbon, not a chiral center, so we can just write CH2OH."}, {"video_title": "Fischer projection practice Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we just reflect this in our mirror, and then we draw our horizontal line, and then we would have an OH. So we draw in an OH here. We can put in our vertical line like that, and then we have a carboxylic acid, but this carbon is not a chiral center, so we don't really need to worry about it. We can just write in COOH here, and then same with this carbon, not a chiral center, so we can just write CH2OH. The important part is if you have an OH on the right, it'll be on the left in the mirror image. If you have an OH on the left in a Fischer projection, it'll be on the right in a mirror image. So now we have a pair of enantiomers."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Before we get to preparation of alcohols using sodium borohydride, let's take a look at a few of the other ways to make alcohols that we've already talked about in earlier videos. For example, you can make alcohols from alkenes, and you can add the OH on in a Markovnikov fashion, or you can add the OH on in an anti-Markovnikov fashion. So that's one type of way to make your alcohols. You could also do so from alkyl halides. The process could be an SN1 process, or it could be an SN2 process. And then what we haven't talked about yet is how to prepare alcohols from carbonyl compounds. And there are a couple different ways to do that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "You could also do so from alkyl halides. The process could be an SN1 process, or it could be an SN2 process. And then what we haven't talked about yet is how to prepare alcohols from carbonyl compounds. And there are a couple different ways to do that. You could use sodium borohydride, NaBH4, which is what we're going to talk about in this video. On the next video, we're going to talk about the use of lithium aluminum hydride, so LiAlH4. And you could also use an organometallic, something like a Grignard reagent."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And there are a couple different ways to do that. You could use sodium borohydride, NaBH4, which is what we're going to talk about in this video. On the next video, we're going to talk about the use of lithium aluminum hydride, so LiAlH4. And you could also use an organometallic, something like a Grignard reagent. So I'll put that down here as well, too. And we'll talk about that in a future video as well. So all we have time for in this video is sodium borohydride."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And you could also use an organometallic, something like a Grignard reagent. So I'll put that down here as well, too. And we'll talk about that in a future video as well. So all we have time for in this video is sodium borohydride. So let's check out the general reaction for the use of sodium borohydride to form an alcohol. And so you can see over here on the left, we're starting with either an aldehyde or a ketone, so aldehyde or ketone there. So there's either a hydrogen attached to your carbonyl or an R prime group like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So all we have time for in this video is sodium borohydride. So let's check out the general reaction for the use of sodium borohydride to form an alcohol. And so you can see over here on the left, we're starting with either an aldehyde or a ketone, so aldehyde or ketone there. So there's either a hydrogen attached to your carbonyl or an R prime group like that. We're going to add sodium borohydride. And then we're going to add a proton source. It could be just about anything."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So there's either a hydrogen attached to your carbonyl or an R prime group like that. We're going to add sodium borohydride. And then we're going to add a proton source. It could be just about anything. So H plus tends to work. And we're going to form either a primary or a secondary alcohol, depending on our starting materials. This is going to be a primary or a secondary alcohol."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "It could be just about anything. So H plus tends to work. And we're going to form either a primary or a secondary alcohol, depending on our starting materials. This is going to be a primary or a secondary alcohol. Let's look at the mechanism for the preparation of alcohols using sodium borohydride. So we'll start with our ketone here. So we're going to start with a ketone."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "This is going to be a primary or a secondary alcohol. Let's look at the mechanism for the preparation of alcohols using sodium borohydride. So we'll start with our ketone here. So we're going to start with a ketone. So we'll just do the reaction like that. So I have my carbonyl like that. Put in my lone pairs of electrons."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we're going to start with a ketone. So we'll just do the reaction like that. So I have my carbonyl like that. Put in my lone pairs of electrons. And we'll make this an R prime to make it our ketone. And sodium borohydride comes along. So let's go ahead and draw the structure for that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Put in my lone pairs of electrons. And we'll make this an R prime to make it our ketone. And sodium borohydride comes along. So let's go ahead and draw the structure for that. So Na plus, positive formal charge. And then we have boron bonded to four hydrogens like that. And there's a negative 1 formal charge on our boron."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So let's go ahead and draw the structure for that. So Na plus, positive formal charge. And then we have boron bonded to four hydrogens like that. And there's a negative 1 formal charge on our boron. So we'll go ahead and put that in there as well. So the mechanism I'm going to show you is a simplified mechanism. The actual mechanism is a little bit more complicated."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And there's a negative 1 formal charge on our boron. So we'll go ahead and put that in there as well. So the mechanism I'm going to show you is a simplified mechanism. The actual mechanism is a little bit more complicated. But this mechanism works. So let's just go with the simplified version. So the first thing you have to think about is this carbonyl here, this carbon double bonded to an oxygen."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "The actual mechanism is a little bit more complicated. But this mechanism works. So let's just go with the simplified version. So the first thing you have to think about is this carbonyl here, this carbon double bonded to an oxygen. There's an electronegativity difference between the carbon and the oxygen. So right here, this carbon and this oxygen. The oxygen's more electronegative, meaning that those bonds, the electrons in the double bond between the carbon and the oxygen are going to be pulled closer to the oxygen."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So the first thing you have to think about is this carbonyl here, this carbon double bonded to an oxygen. There's an electronegativity difference between the carbon and the oxygen. So right here, this carbon and this oxygen. The oxygen's more electronegative, meaning that those bonds, the electrons in the double bond between the carbon and the oxygen are going to be pulled closer to the oxygen. Therefore, oxygen has a partial negative charge, so increased electron density around it. Whereas the carbon here is losing some of the electrons around it. So it's partially positive like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "The oxygen's more electronegative, meaning that those bonds, the electrons in the double bond between the carbon and the oxygen are going to be pulled closer to the oxygen. Therefore, oxygen has a partial negative charge, so increased electron density around it. Whereas the carbon here is losing some of the electrons around it. So it's partially positive like that. So if carbon is partially positive, it's our electrophile. It wants electrons. So where can we get our electrons?"}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So it's partially positive like that. So if carbon is partially positive, it's our electrophile. It wants electrons. So where can we get our electrons? We can get our electrons from right here. So the two electrons on this hydrogen here are going to move out and attack this carbon. So that's going to be the nucleophilic portion of our molecule like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So where can we get our electrons? We can get our electrons from right here. So the two electrons on this hydrogen here are going to move out and attack this carbon. So that's going to be the nucleophilic portion of our molecule like that. And that would be too many bonds to carbon. So one of these, this pi bond here, is going to kick off onto the oxygen. So we're going to put two electrons on that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So that's going to be the nucleophilic portion of our molecule like that. And that would be too many bonds to carbon. So one of these, this pi bond here, is going to kick off onto the oxygen. So we're going to put two electrons on that. So let's go ahead and draw the result of that nucleophilic attack. So we have R, and then we have R carbon, and then R prime over here. And this top oxygen had two lone pairs of electrons."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we're going to put two electrons on that. So let's go ahead and draw the result of that nucleophilic attack. So we have R, and then we have R carbon, and then R prime over here. And this top oxygen had two lone pairs of electrons. It just picked up one more for a total of three. And that gives it a negative 1 formal charge like that. And we added on a hydrogen like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And this top oxygen had two lone pairs of electrons. It just picked up one more for a total of three. And that gives it a negative 1 formal charge like that. And we added on a hydrogen like that. So let's go ahead and color code some electrons here. So we can see where everything went. So I'm saying this hydrogen and the two electrons in this bond became this hydrogen and these two electrons right here."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And we added on a hydrogen like that. So let's go ahead and color code some electrons here. So we can see where everything went. So I'm saying this hydrogen and the two electrons in this bond became this hydrogen and these two electrons right here. So really what you're doing is sodium borohydride is a source of hydride anions. So a hydride anion would be hydrogen with two electrons around it and a negative 1 formal charge. So that's the way to think about these reagents."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So I'm saying this hydrogen and the two electrons in this bond became this hydrogen and these two electrons right here. So really what you're doing is sodium borohydride is a source of hydride anions. So a hydride anion would be hydrogen with two electrons around it and a negative 1 formal charge. So that's the way to think about these reagents. Hydride itself is not the best nucleophile because it isn't polarizable enough. It's such a small atom that it doesn't really work very well as a nucleophile by itself. So this, of course, is our simplified version."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So that's the way to think about these reagents. Hydride itself is not the best nucleophile because it isn't polarizable enough. It's such a small atom that it doesn't really work very well as a nucleophile by itself. So this, of course, is our simplified version. All right, in the next step of our mechanism, we get some acid-base chemistry. So we have protons floating around. So we added some H plus to our solution, and a lone pair of electrons picks up the proton."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So this, of course, is our simplified version. All right, in the next step of our mechanism, we get some acid-base chemistry. So we have protons floating around. So we added some H plus to our solution, and a lone pair of electrons picks up the proton. And we are done. So let's go ahead and draw the product. Would be our alcohol like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we added some H plus to our solution, and a lone pair of electrons picks up the proton. And we are done. So let's go ahead and draw the product. Would be our alcohol like that. And let's go ahead and put those lone pairs of electrons. And there's only one hydrogen added onto our carbon. So if we start with a ketone, we're going to end up with a secondary alcohol, as in this situation."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Would be our alcohol like that. And let's go ahead and put those lone pairs of electrons. And there's only one hydrogen added onto our carbon. So if we start with a ketone, we're going to end up with a secondary alcohol, as in this situation. All right, let's do an actual reaction here. And let's start with vanillin. So vanillin is a very nice smelling molecule."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So if we start with a ketone, we're going to end up with a secondary alcohol, as in this situation. All right, let's do an actual reaction here. And let's start with vanillin. So vanillin is a very nice smelling molecule. And if you look at the structure of vanillin, there's an aldehyde functional group right there on our ring. So there's some other functional groups. So we'll go ahead and put in the rest of the vanillin molecule like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So vanillin is a very nice smelling molecule. And if you look at the structure of vanillin, there's an aldehyde functional group right there on our ring. So there's some other functional groups. So we'll go ahead and put in the rest of the vanillin molecule like that. So we're going to add sodium borohydride in our first step. We're going to add sodium borohydride. And then once that's reacted, we're going to add a source of protons."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we'll go ahead and put in the rest of the vanillin molecule like that. So we're going to add sodium borohydride in our first step. We're going to add sodium borohydride. And then once that's reacted, we're going to add a source of protons. HCl works pretty well for this reaction. So we'll just say a source of H plus like that. All right, so when you're trying to figure out the product on a test, it's not even necessary to do that really simplified mechanism."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And then once that's reacted, we're going to add a source of protons. HCl works pretty well for this reaction. So we'll just say a source of H plus like that. All right, so when you're trying to figure out the product on a test, it's not even necessary to do that really simplified mechanism. You can pretend like the hydride ion is going to act as a nucleophile, even though we've already covered that it doesn't quite do that. So here we have our hydride nucleophile. So it's going to attack this carbon right here."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "All right, so when you're trying to figure out the product on a test, it's not even necessary to do that really simplified mechanism. You can pretend like the hydride ion is going to act as a nucleophile, even though we've already covered that it doesn't quite do that. So here we have our hydride nucleophile. So it's going to attack this carbon right here. This is our carbonyl carbon. So it's going to be partially positive. That's going to kick these electrons off onto our oxygen like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So it's going to attack this carbon right here. This is our carbonyl carbon. So it's going to be partially positive. That's going to kick these electrons off onto our oxygen like that. So let's go ahead and draw the intermediate for this reaction here. So we'll have our benzene ring like that. We'll put in our other functional groups over here, which don't participate in the reaction."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "That's going to kick these electrons off onto our oxygen like that. So let's go ahead and draw the intermediate for this reaction here. So we'll have our benzene ring like that. We'll put in our other functional groups over here, which don't participate in the reaction. And let's go ahead and draw what we have here. So we had one hydrogen already bonded to that carbon. We just added another hydrogen onto that carbon."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "We'll put in our other functional groups over here, which don't participate in the reaction. And let's go ahead and draw what we have here. So we had one hydrogen already bonded to that carbon. We just added another hydrogen onto that carbon. And then we have an oxygen still there with three lone pairs of electrons around it, giving it a negative 1 formal charge. So when we add our source of protons, a lone pair will pick up that proton there to form our alcohol. So we'll go ahead and draw our product right over here."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "We just added another hydrogen onto that carbon. And then we have an oxygen still there with three lone pairs of electrons around it, giving it a negative 1 formal charge. So when we add our source of protons, a lone pair will pick up that proton there to form our alcohol. So we'll go ahead and draw our product right over here. So once again, we have an OH coming off of our ring. We have this portion of the molecule, this ether portion right here. And then we have an alcohol like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we'll go ahead and draw our product right over here. So once again, we have an OH coming off of our ring. We have this portion of the molecule, this ether portion right here. And then we have an alcohol like that. So this molecule, since it has such a similar structure to vanillin, also has a nice vanilla smell. It's very pleasant. This is a very good undergraduate organic chemistry lab to do."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And then we have an alcohol like that. So this molecule, since it has such a similar structure to vanillin, also has a nice vanilla smell. It's very pleasant. This is a very good undergraduate organic chemistry lab to do. And let's classify the type of alcohol we got as either primary, secondary, or tertiary. So remember how to do that. You look at the carbon that's directly attached to the OH."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "This is a very good undergraduate organic chemistry lab to do. And let's classify the type of alcohol we got as either primary, secondary, or tertiary. So remember how to do that. You look at the carbon that's directly attached to the OH. That's this carbon. And you see how many carbons that carbon is attached to. That carbon is attached to one other carbon."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "You look at the carbon that's directly attached to the OH. That's this carbon. And you see how many carbons that carbon is attached to. That carbon is attached to one other carbon. So this is an example of a primary alcohol. So if you start with an aldehyde, so if you start with an aldehyde as we did over here, this is our aldehyde, you're going to end up with a primary alcohol as your product. Let's do another one."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "That carbon is attached to one other carbon. So this is an example of a primary alcohol. So if you start with an aldehyde, so if you start with an aldehyde as we did over here, this is our aldehyde, you're going to end up with a primary alcohol as your product. Let's do another one. This time, let's do a ketone. So let's go ahead and draw a ketone over here. So we'll do cyclohexanone."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Let's do another one. This time, let's do a ketone. So let's go ahead and draw a ketone over here. So we'll do cyclohexanone. So here's our ketone like that. And we're going to add sodium borohydride like this, NaBH4. And we'll add methanol."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we'll do cyclohexanone. So here's our ketone like that. And we're going to add sodium borohydride like this, NaBH4. And we'll add methanol. And we'll do this all in one step here. So once again, think about sodium borohydride being a source of hydride anions. So again, even though this isn't technically the mechanism, it's our simplified version."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And we'll add methanol. And we'll do this all in one step here. So once again, think about sodium borohydride being a source of hydride anions. So again, even though this isn't technically the mechanism, it's our simplified version. So if you're taking a test, this is the easiest way to do it. Attacks that carbon, kicks these electrons off onto your oxygen. So we have our intermediates right over here."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So again, even though this isn't technically the mechanism, it's our simplified version. So if you're taking a test, this is the easiest way to do it. Attacks that carbon, kicks these electrons off onto your oxygen. So we have our intermediates right over here. So let's go ahead and draw that. So we have the hydride anion attack to form a new bond to what used to be our carbonyl carbon. Now this oxygen has three lone pairs of electrons around it and a negative 1 formal charge."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we have our intermediates right over here. So let's go ahead and draw that. So we have the hydride anion attack to form a new bond to what used to be our carbonyl carbon. Now this oxygen has three lone pairs of electrons around it and a negative 1 formal charge. It's going to pick up a proton from methanol as its proton source this time. And therefore, our final product, if we're going to go ahead and draw it. I'll keep the hydrogen in there, and you'll see why in a second."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Now this oxygen has three lone pairs of electrons around it and a negative 1 formal charge. It's going to pick up a proton from methanol as its proton source this time. And therefore, our final product, if we're going to go ahead and draw it. I'll keep the hydrogen in there, and you'll see why in a second. And then I have my OH like that. So I form cyclohexanol as my product. This is a reduction reaction."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "I'll keep the hydrogen in there, and you'll see why in a second. And then I have my OH like that. So I form cyclohexanol as my product. This is a reduction reaction. So we are reducing the molecule to form our alcohol. So let's take the reactant and the product from this reaction and let's assign some oxidation states so we can see how this is an example of a reduction reaction. So I'm just going to redraw our starting ketone here."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "This is a reduction reaction. So we are reducing the molecule to form our alcohol. So let's take the reactant and the product from this reaction and let's assign some oxidation states so we can see how this is an example of a reduction reaction. So I'm just going to redraw our starting ketone here. So let's go ahead and put in some of the atoms this time. So I'm just going to redraw that starting ketone like this. And I'm going to go ahead and draw the product that we got as well."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So I'm just going to redraw our starting ketone here. So let's go ahead and put in some of the atoms this time. So I'm just going to redraw that starting ketone like this. And I'm going to go ahead and draw the product that we got as well. So put in these carbons here. And then we added on a hydrogen like that. And then we formed our alcohol over here like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And I'm going to go ahead and draw the product that we got as well. So put in these carbons here. And then we added on a hydrogen like that. And then we formed our alcohol over here like that. So let's go ahead and think about oxidation states. So when we're assigning oxidation states, we need to go ahead and draw in these electrons here. So let me go ahead and put in these electrons on these guys right in here like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And then we formed our alcohol over here like that. So let's go ahead and think about oxidation states. So when we're assigning oxidation states, we need to go ahead and draw in these electrons here. So let me go ahead and put in these electrons on these guys right in here like that. So let's think about how we did this in some of the earlier videos. It's actually thinking about electronegativity. And if I'm thinking about the four electrons between carbon and oxygen, oxygen is more electronegative."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So let me go ahead and put in these electrons on these guys right in here like that. So let's think about how we did this in some of the earlier videos. It's actually thinking about electronegativity. And if I'm thinking about the four electrons between carbon and oxygen, oxygen is more electronegative. So oxygen is going to get all of those electrons. So we go ahead and go like that. Carbon versus carbon, fighting over these two electrons, it's an equal electronegativity, obviously, since it's the same element."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And if I'm thinking about the four electrons between carbon and oxygen, oxygen is more electronegative. So oxygen is going to get all of those electrons. So we go ahead and go like that. Carbon versus carbon, fighting over these two electrons, it's an equal electronegativity, obviously, since it's the same element. So each carbon is going to get one of those electrons. And the same thing for over here. So when we assigned our oxidation states, we said that it's the number of valence electrons that carbon normally has, which of course is four."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Carbon versus carbon, fighting over these two electrons, it's an equal electronegativity, obviously, since it's the same element. So each carbon is going to get one of those electrons. And the same thing for over here. So when we assigned our oxidation states, we said that it's the number of valence electrons that carbon normally has, which of course is four. And from that, we subtract the number of electrons around it when we account for electronegativity, which is two in this case. So the oxidation state of that carbon is plus 2. So we have a plus 2 oxidation state on the left."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So when we assigned our oxidation states, we said that it's the number of valence electrons that carbon normally has, which of course is four. And from that, we subtract the number of electrons around it when we account for electronegativity, which is two in this case. So the oxidation state of that carbon is plus 2. So we have a plus 2 oxidation state on the left. Let's think about the right. Let's go ahead and put in our electrons. Each bond consists of two electrons."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So we have a plus 2 oxidation state on the left. Let's think about the right. Let's go ahead and put in our electrons. Each bond consists of two electrons. So we go ahead and put the fact that each bond consists of two electrons in here, like that. And then once again, we think about differences in electronegativity. So again, we know carbon versus carbon is a tie, so we can go like that."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "Each bond consists of two electrons. So we go ahead and put the fact that each bond consists of two electrons in here, like that. And then once again, we think about differences in electronegativity. So again, we know carbon versus carbon is a tie, so we can go like that. And carbon versus oxygen, oxygen's more electronegative, so it's going to get those electrons there. But carbon versus hydrogen, carbon is a little bit more electronegative. So now carbon's going to take those two electrons."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So again, we know carbon versus carbon is a tie, so we can go like that. And carbon versus oxygen, oxygen's more electronegative, so it's going to get those electrons there. But carbon versus hydrogen, carbon is a little bit more electronegative. So now carbon's going to take those two electrons. So carbon normally has four. In this case, carbon has four around it after we account for electronegativity. So the oxidation state of that carbon atom is zero."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So now carbon's going to take those two electrons. So carbon normally has four. In this case, carbon has four around it after we account for electronegativity. So the oxidation state of that carbon atom is zero. So let's think about what happened to the oxidation state. We started off with a plus 2 oxidation state, and that number was reduced to an oxidation state of zero. So this is a reduction reaction."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So the oxidation state of that carbon atom is zero. So let's think about what happened to the oxidation state. We started off with a plus 2 oxidation state, and that number was reduced to an oxidation state of zero. So this is a reduction reaction. This is reduction. So several different definitions you could use. You could think about reduction being a decrease in the oxidation state, or a reduction in the oxidation state."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "So this is a reduction reaction. This is reduction. So several different definitions you could use. You could think about reduction being a decrease in the oxidation state, or a reduction in the oxidation state. You could also think about that carbon gaining electrons. So it picked up two more electrons. And yet another way to think about it is carbon lost a bond to oxygen."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "You could think about reduction being a decrease in the oxidation state, or a reduction in the oxidation state. You could also think about that carbon gaining electrons. So it picked up two more electrons. And yet another way to think about it is carbon lost a bond to oxygen. Over here on the left, carbon had two bonds to oxygen. Over here on the right, carbon only has one bond to oxygen. And it formed a bond with hydrogen."}, {"video_title": "Preparation of alcohols using NaBH4 Alcohols, ethers, epoxides, sulfides Khan Academy.mp3", "Sentence": "And yet another way to think about it is carbon lost a bond to oxygen. Over here on the left, carbon had two bonds to oxygen. Over here on the right, carbon only has one bond to oxygen. And it formed a bond with hydrogen. So increased number of bonds to hydrogen, decreased number of bonds to oxygen is another way to think about this being a reduction reaction. But assigning your oxidation states is probably the best way to do it. In the next video, we'll take a look at how to prepare alcohols using lithium aluminum hydride, which is very similar to using sodium borohydride."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So if we look at the structure of our substrate, and we say it's secondary, we next need to look at the reagent. So we have NaCl, which we know is Na plus and Cl minus, and the chloride anion functions only as a nucleophile. So we would expect a substitution reaction, a nucleophilic substitution. So E1 and E2 are out. Between SN1 and SN2 with a secondary substrate, we're not sure until we look at the solvent. And DMSO is a polar aprotic solvent, which we saw in an earlier video, favors an SN2 mechanism. So SN1 is out, and we're gonna think about our chloride anion functioning as a nucleophile."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So E1 and E2 are out. Between SN1 and SN2 with a secondary substrate, we're not sure until we look at the solvent. And DMSO is a polar aprotic solvent, which we saw in an earlier video, favors an SN2 mechanism. So SN1 is out, and we're gonna think about our chloride anion functioning as a nucleophile. So let me draw it in over here. So this is with a negative one formal charge. In an SN2 mechanism, our nucleophile attacks the same time we get loss of a leaving group."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So SN1 is out, and we're gonna think about our chloride anion functioning as a nucleophile. So let me draw it in over here. So this is with a negative one formal charge. In an SN2 mechanism, our nucleophile attacks the same time we get loss of a leaving group. And our nucleophile is going to attack this carbon in red. So we're gonna form a bond between the chlorine and this carbon in red. And when the nucleophile attacks, we also get loss of our leaving group."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "In an SN2 mechanism, our nucleophile attacks the same time we get loss of a leaving group. And our nucleophile is going to attack this carbon in red. So we're gonna form a bond between the chlorine and this carbon in red. And when the nucleophile attacks, we also get loss of our leaving group. So these electrons come off onto the oxygen, and we know that tosylate is a good leaving group. So when we draw our product, let's draw this in here, and the carbon in red is this one, we know an SN2 mechanism means inversion of configuration. The nucleophile has to attack from the side opposite of the leaving group."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "And when the nucleophile attacks, we also get loss of our leaving group. So these electrons come off onto the oxygen, and we know that tosylate is a good leaving group. So when we draw our product, let's draw this in here, and the carbon in red is this one, we know an SN2 mechanism means inversion of configuration. The nucleophile has to attack from the side opposite of the leaving group. So we had a wedge here for our leaving group, so that means we're gonna have a dash for our chlorines. We're gonna put the chlorine right here, and that's the product of our SN2 reaction. For our next problem, we have a secondary alkyl halide."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "The nucleophile has to attack from the side opposite of the leaving group. So we had a wedge here for our leaving group, so that means we're gonna have a dash for our chlorines. We're gonna put the chlorine right here, and that's the product of our SN2 reaction. For our next problem, we have a secondary alkyl halide. So just looking at our reactions, we can't really rule any out here, so all four are possible, until we look at our reagent. And we saw in an earlier video that DBN is a strong base. It does not act like a nucleophile."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "For our next problem, we have a secondary alkyl halide. So just looking at our reactions, we can't really rule any out here, so all four are possible, until we look at our reagent. And we saw in an earlier video that DBN is a strong base. It does not act like a nucleophile. So SN1 and SN2 are out, and a strong base means an E2 reaction, so E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon, and the carbons directly bonded to the alpha carbon are the beta carbons."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "It does not act like a nucleophile. So SN1 and SN2 are out, and a strong base means an E2 reaction, so E1 is out. Now that we know we're doing an E2 mechanism, let's analyze the structure of our alkyl halide. The carbon that's directly bonded to our halogen is our alpha carbon, and the carbons directly bonded to the alpha carbon are the beta carbons. So I'll just do the beta carbon on the right, since they are the same, essentially. And we know that our base is gonna take a proton from that beta carbon. So let me just draw in a hydrogen here, and DBN is a neutral base, so I'll just draw a generic base here."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "The carbon that's directly bonded to our halogen is our alpha carbon, and the carbons directly bonded to the alpha carbon are the beta carbons. So I'll just do the beta carbon on the right, since they are the same, essentially. And we know that our base is gonna take a proton from that beta carbon. So let me just draw in a hydrogen here, and DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond, and these electrons come off to form our bromide anion. So our final product is an alkene, and our electrons in magenta in here moved in to form our double bond. For our next problem, we have another secondary alkyl halide, so right now all four of these are possible, until we look at our reagent, which is sodium hydroxide, Na plus OH minus."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So let me just draw in a hydrogen here, and DBN is a neutral base, so I'll just draw a generic base here. Our base is going to take this proton at the same time these electrons move in to form a double bond, and these electrons come off to form our bromide anion. So our final product is an alkene, and our electrons in magenta in here moved in to form our double bond. For our next problem, we have another secondary alkyl halide, so right now all four of these are possible, until we look at our reagent, which is sodium hydroxide, Na plus OH minus. And we know that the hydroxide ion can function as a strong nucleophile or a strong base. So a strong nucleophile makes us think an SN2 reaction and not an SN1, and a strong base makes us think about an E2 reaction and not an E1 reaction. Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "For our next problem, we have another secondary alkyl halide, so right now all four of these are possible, until we look at our reagent, which is sodium hydroxide, Na plus OH minus. And we know that the hydroxide ion can function as a strong nucleophile or a strong base. So a strong nucleophile makes us think an SN2 reaction and not an SN1, and a strong base makes us think about an E2 reaction and not an E1 reaction. Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here. So when we analyze our alkyl halide, the carbon bonded to the halogen is our alpha carbon, and the carbons directly bonded to that would be our beta carbons. So we have two beta carbons here. And let me number this ring."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "Since we have heat, heat favors an elimination reaction over a substitution, so E2 should be the major reaction here. So when we analyze our alkyl halide, the carbon bonded to the halogen is our alpha carbon, and the carbons directly bonded to that would be our beta carbons. So we have two beta carbons here. And let me number this ring. I'm gonna say the alpha carbon is carbon one, I'm gonna go around clockwise. So that's one, two, three, carbon four, carbon five, and then carbon six. And next we're going to translate this to our chair conformation over here."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "And let me number this ring. I'm gonna say the alpha carbon is carbon one, I'm gonna go around clockwise. So that's one, two, three, carbon four, carbon five, and then carbon six. And next we're going to translate this to our chair conformation over here. So carbon one would be this carbon, and then carbon two would be this one. So it'd be carbon three, four, five, and six. The bromine is coming out at us in space at carbon one, which means it's going up."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "And next we're going to translate this to our chair conformation over here. So carbon one would be this carbon, and then carbon two would be this one. So it'd be carbon three, four, five, and six. The bromine is coming out at us in space at carbon one, which means it's going up. So if I look at carbon one, we would have the bromine going up, which would be up axial. At carbon two, I have a methyl group going away from me in space, so that's going down. So at carbon two, we must have a methyl group going down, which makes it down axial."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "The bromine is coming out at us in space at carbon one, which means it's going up. So if I look at carbon one, we would have the bromine going up, which would be up axial. At carbon two, I have a methyl group going away from me in space, so that's going down. So at carbon two, we must have a methyl group going down, which makes it down axial. So we care about carbon two, let me highlight these again. So we care about carbon two, which is a beta carbon. We also care about carbon six, which is another beta carbon."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So at carbon two, we must have a methyl group going down, which makes it down axial. So we care about carbon two, let me highlight these again. So we care about carbon two, which is a beta carbon. We also care about carbon six, which is another beta carbon. So let's put in the hydrogens on those beta carbons. At carbon two, we would have a hydrogen that's up equatorial, and at carbon six, we would have a hydrogen that's down axial, and one that is up equatorial. So when we think about our E2 mechanism, we know our strong base is going to take a proton."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "We also care about carbon six, which is another beta carbon. So let's put in the hydrogens on those beta carbons. At carbon two, we would have a hydrogen that's up equatorial, and at carbon six, we would have a hydrogen that's down axial, and one that is up equatorial. So when we think about our E2 mechanism, we know our strong base is going to take a proton. And that proton must be antiperiplanar to our halogen. So our halogen, let me highlight our halogen here, which is bromine, that is in the axial position. So we need to take a proton that is antiperiplanar to that bromine."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So when we think about our E2 mechanism, we know our strong base is going to take a proton. And that proton must be antiperiplanar to our halogen. So our halogen, let me highlight our halogen here, which is bromine, that is in the axial position. So we need to take a proton that is antiperiplanar to that bromine. So at carbon two, let's look at carbon two first. At carbon two, I do not have a hydrogen that's antiperiplanar to my halogen. But I do have one at carbon six."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So we need to take a proton that is antiperiplanar to that bromine. So at carbon two, let's look at carbon two first. At carbon two, I do not have a hydrogen that's antiperiplanar to my halogen. But I do have one at carbon six. It's the one that is down axial. So our base is gonna take that proton. So let's draw in the hydroxide ion, which is a strong base."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "But I do have one at carbon six. It's the one that is down axial. So our base is gonna take that proton. So let's draw in the hydroxide ion, which is a strong base. And the hydroxide ion is going to take this proton, and then these electrons are gonna move into form a double bond. At the same time, we get these electrons coming off onto the bromine to form the bromide ion. So let's draw the product for this reaction."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So let's draw in the hydroxide ion, which is a strong base. And the hydroxide ion is going to take this proton, and then these electrons are gonna move into form a double bond. At the same time, we get these electrons coming off onto the bromine to form the bromide ion. So let's draw the product for this reaction. We would have our ring. And a double bond forms between carbon one and carbon six. So that means a double bond forms in here."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So let's draw the product for this reaction. We would have our ring. And a double bond forms between carbon one and carbon six. So that means a double bond forms in here. And then at carbon two, we still have a methyl group going away from us in space. Let me draw that in like that. So the electrons in red, it's hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So that means a double bond forms in here. And then at carbon two, we still have a methyl group going away from us in space. Let me draw that in like that. So the electrons in red, it's hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six. Let me label those again here. So carbon one and carbon six. Again, not IUPAC nomenclature, just so we can think about our product compared to our starting material."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So the electrons in red, it's hard to see, but if you think about these electrons in red back here, are gonna move in to form our double bond between what I've labeled as carbon one and carbon six. Let me label those again here. So carbon one and carbon six. Again, not IUPAC nomenclature, just so we can think about our product compared to our starting material. So this would be the major product of our reaction, which is an E2 reaction. It would also be possible to get some product from an SN2 mechanism. But since heat is here, an elimination reaction is favored over a substitution."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "Again, not IUPAC nomenclature, just so we can think about our product compared to our starting material. So this would be the major product of our reaction, which is an E2 reaction. It would also be possible to get some product from an SN2 mechanism. But since heat is here, an elimination reaction is favored over a substitution. Next we have a secondary alcohol with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So it's not gonna be SN1 or SN2."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "But since heat is here, an elimination reaction is favored over a substitution. Next we have a secondary alcohol with phosphoric acid and heat. And we saw a lot of these types of problems in the videos on elimination reactions. So it's not gonna be SN1 or SN2. And we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons, and we're going to protonate this oxygen for our first step."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So it's not gonna be SN1 or SN2. And we don't have a strong base, so don't think E2, think E1. And our first step would be to protonate our alcohol to form a better leaving group. So phosphoric acid is a source of protons, and we're going to protonate this oxygen for our first step. So let's draw in our ring. And we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons, and a plus one formal charge on the oxygen. So this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So phosphoric acid is a source of protons, and we're going to protonate this oxygen for our first step. So let's draw in our ring. And we protonate our oxygen, so now our oxygen has two bonds to hydrogen, one lone pair of electrons, and a plus one formal charge on the oxygen. So this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen, and we remove a bond from this carbon in red, which would give us a secondary carbocation. So let's draw in our secondary carbocation, and the carbon in red is this one."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So this lone pair of electrons on the oxygen picked up a proton from phosphoric acid to form this bond. And now we have a better leaving group than the hydroxide ion. These electrons come off onto the oxygen, and we remove a bond from this carbon in red, which would give us a secondary carbocation. So let's draw in our secondary carbocation, and the carbon in red is this one. And that carbon would have a plus one formal charge. So let me draw in a plus one formal charge here. And now we have water, which can function as a weak base in our E1 reaction, and take a proton from a carbon next to our carbon with a positive charge."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So let's draw in our secondary carbocation, and the carbon in red is this one. And that carbon would have a plus one formal charge. So let me draw in a plus one formal charge here. And now we have water, which can function as a weak base in our E1 reaction, and take a proton from a carbon next to our carbon with a positive charge. So let's say this carbon right here, it has two hydrogens on it, I'll just draw one hydrogen in. And water functions as a base, takes this proton, and these electrons move in to form a double bond. So let's draw our final product here."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "And now we have water, which can function as a weak base in our E1 reaction, and take a proton from a carbon next to our carbon with a positive charge. So let's say this carbon right here, it has two hydrogens on it, I'll just draw one hydrogen in. And water functions as a base, takes this proton, and these electrons move in to form a double bond. So let's draw our final product here. We would have a ring, we would have a double bond between these two carbons. So our electrons in, let's use magenta, our electrons in magenta moved in to form our double bond. So our product is cyclohexene."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So let's draw our final product here. We would have a ring, we would have a double bond between these two carbons. So our electrons in, let's use magenta, our electrons in magenta moved in to form our double bond. So our product is cyclohexene. So a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid or phosphoric acid and you heat it up. For this reaction, we have this secondary alkyl halide reacting with an aqueous solution of formic acid. Formic acid is a weak nucleophile, and water is a polar protic solvent."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So our product is cyclohexene. So a secondary alcohol undergoes an E1 reaction if you use something like sulfuric acid or phosphoric acid and you heat it up. For this reaction, we have this secondary alkyl halide reacting with an aqueous solution of formic acid. Formic acid is a weak nucleophile, and water is a polar protic solvent. A weak nucleophile and a polar protic solvent should make us think about an SN1 type mechanism, because water, as a polar protic solvent, can stabilize the formation of a carbocation. So let's draw the carbocation that would result. These electrons would come off onto our bromine, and we're taking a bond away from this carbon in red."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "Formic acid is a weak nucleophile, and water is a polar protic solvent. A weak nucleophile and a polar protic solvent should make us think about an SN1 type mechanism, because water, as a polar protic solvent, can stabilize the formation of a carbocation. So let's draw the carbocation that would result. These electrons would come off onto our bromine, and we're taking a bond away from this carbon in red. So the carbon in red gets a plus one formal charge, and let's draw our carbocation. So we have our benzene ring here. I'll put in my pi electrons."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "These electrons would come off onto our bromine, and we're taking a bond away from this carbon in red. So the carbon in red gets a plus one formal charge, and let's draw our carbocation. So we have our benzene ring here. I'll put in my pi electrons. And the carbon in red is this one. So that carbon gets a plus one formal charge. This is a secondary carbocation, but it's also a benzylic carbocation."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "I'll put in my pi electrons. And the carbon in red is this one. So that carbon gets a plus one formal charge. This is a secondary carbocation, but it's also a benzylic carbocation. So the positive charge is actually delocalized because of the pi electrons on the ring. So this is more stable than most secondary carbocations. Next, if we're thinking an SN1 type mechanism, this would be our electrophile."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "This is a secondary carbocation, but it's also a benzylic carbocation. So the positive charge is actually delocalized because of the pi electrons on the ring. So this is more stable than most secondary carbocations. Next, if we're thinking an SN1 type mechanism, this would be our electrophile. Our carbocation is our electrophile, and our nucleophile would be formic acid. And we saw in an earlier video how the carbonyl oxygen is actually more nucleophilic than this oxygen. So a lone pair of electrons on the carbonyl oxygen would attack our carbon in red."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "Next, if we're thinking an SN1 type mechanism, this would be our electrophile. Our carbocation is our electrophile, and our nucleophile would be formic acid. And we saw in an earlier video how the carbonyl oxygen is actually more nucleophilic than this oxygen. So a lone pair of electrons on the carbonyl oxygen would attack our carbon in red. And we would end up with, let's go ahead and draw in the result of our nucleophilic attack. And I won't go through all the steps of the mechanism since I covered this in great detail in an earlier video. So this is from our SN1, SN2 final summary video."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So a lone pair of electrons on the carbonyl oxygen would attack our carbon in red. And we would end up with, let's go ahead and draw in the result of our nucleophilic attack. And I won't go through all the steps of the mechanism since I covered this in great detail in an earlier video. So this is from our SN1, SN2 final summary video. So let me draw in what we would form in here. So this would be carbon double bonded to an oxygen, and this would be a hydrogen. So this is our product, and this carbon is a chiral center."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So this is from our SN1, SN2 final summary video. So let me draw in what we would form in here. So this would be carbon double bonded to an oxygen, and this would be a hydrogen. So this is our product, and this carbon is a chiral center. And because this is an SN1 type mechanism, and we have planar geometry in our carbocation, our nucleophile can attack from either side, and we're gonna end up with a mix of enantiomers. So again, for more details on this mechanism, I skipped a few steps here, please watch the SN1, SN2 final summary video. Next, let's think about what else could possibly happen."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So this is our product, and this carbon is a chiral center. And because this is an SN1 type mechanism, and we have planar geometry in our carbocation, our nucleophile can attack from either side, and we're gonna end up with a mix of enantiomers. So again, for more details on this mechanism, I skipped a few steps here, please watch the SN1, SN2 final summary video. Next, let's think about what else could possibly happen. So SN2 is out, we formed a carbocation. E1 is possible because we have a carbocation here, and we also have a weak base present. So our weak base could be something like water, and I'll just draw a generic base in here."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "Next, let's think about what else could possibly happen. So SN2 is out, we formed a carbocation. E1 is possible because we have a carbocation here, and we also have a weak base present. So our weak base could be something like water, and I'll just draw a generic base in here. And let's draw in a proton on this carbon. So our base could take this proton here, and these electrons would move in to form a double bond. So another possible product, we would have our benzene ring."}, {"video_title": "Elimination vs substitution secondary substrate.mp3", "Sentence": "So our weak base could be something like water, and I'll just draw a generic base in here. And let's draw in a proton on this carbon. So our base could take this proton here, and these electrons would move in to form a double bond. So another possible product, we would have our benzene ring. So I'll draw that in. And then we would have a double bond. So another possibility is an E1 mechanism."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so we'll start with the acid catalyzed. And so here we have an aldehyde or a ketone, and let's do hydration first. So we know that in a normal hydration reaction, you just have to add water, but in an acid catalyzed version, you would have to add a proton source, so H plus. And so you'd form hydronium or H3O plus. And so in an acid catalyzed reaction, the first thing that's going to happen is protonation of your carbonyl oxygen. So a lone pair of electrons on your oxygen here are going to pick up a proton from hydronium, right, leaving these electrons behind here. So let's go ahead and show what happens."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so you'd form hydronium or H3O plus. And so in an acid catalyzed reaction, the first thing that's going to happen is protonation of your carbonyl oxygen. So a lone pair of electrons on your oxygen here are going to pick up a proton from hydronium, right, leaving these electrons behind here. So let's go ahead and show what happens. All right, so we're going to protonate the carbonyl oxygen here. So we're going to have a hydrogen attached and give this a plus one formal charge on our oxygen now. Our carbon is still bonded to an R group and a hydrogen over here."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show what happens. All right, so we're going to protonate the carbonyl oxygen here. So we're going to have a hydrogen attached and give this a plus one formal charge on our oxygen now. Our carbon is still bonded to an R group and a hydrogen over here. And so we could draw a resonance structure for this, right? We could show these pi electrons here moving off onto the oxygen, so let's go ahead and do that. So now this top oxygen here would have two lone pairs of electrons around it, and we took a bond away from this carbon."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "Our carbon is still bonded to an R group and a hydrogen over here. And so we could draw a resonance structure for this, right? We could show these pi electrons here moving off onto the oxygen, so let's go ahead and do that. So now this top oxygen here would have two lone pairs of electrons around it, and we took a bond away from this carbon. So if we took a bond away from this carbon, we get a plus one formal charge. So let's go ahead and put resonance brackets in here, and then let's follow those electrons. So these pi electrons in here, right, move out onto that top oxygen, taking a bond away from your carbonyl carbon right here is going to give it a full positive charge in this resonance structure, so plus one formal charge."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So now this top oxygen here would have two lone pairs of electrons around it, and we took a bond away from this carbon. So if we took a bond away from this carbon, we get a plus one formal charge. So let's go ahead and put resonance brackets in here, and then let's follow those electrons. So these pi electrons in here, right, move out onto that top oxygen, taking a bond away from your carbonyl carbon right here is going to give it a full positive charge in this resonance structure, so plus one formal charge. And so this makes your carbonyl carbon more electrophilic, which means a nucleophile can attack it better, right? A nucleophile is going to be more attracted to it, and that makes your carbonyl more reactive. So acid catalysts work by making your carbonyl carbon more electrophilic, which makes it more reactive in a nucleophilic addition reaction."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons in here, right, move out onto that top oxygen, taking a bond away from your carbonyl carbon right here is going to give it a full positive charge in this resonance structure, so plus one formal charge. And so this makes your carbonyl carbon more electrophilic, which means a nucleophile can attack it better, right? A nucleophile is going to be more attracted to it, and that makes your carbonyl more reactive. So acid catalysts work by making your carbonyl carbon more electrophilic, which makes it more reactive in a nucleophilic addition reaction. And so another water molecule is going to come along here. Let's go ahead and make that yellow again. So a water molecule is going to come along and function as our nucleophile next in our mechanism."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So acid catalysts work by making your carbonyl carbon more electrophilic, which makes it more reactive in a nucleophilic addition reaction. And so another water molecule is going to come along here. Let's go ahead and make that yellow again. So a water molecule is going to come along and function as our nucleophile next in our mechanism. So a lone pair of electrons on the oxygen are going to attack our carbonyl carbon, and so let's go ahead and show that. So the oxygen adds on to our carbon here, and this oxygen still had two hydrogens attached to it, and it had used up one lone pair of electrons. It still has one more, which gives it a plus one formal charge."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So a water molecule is going to come along and function as our nucleophile next in our mechanism. So a lone pair of electrons on the oxygen are going to attack our carbonyl carbon, and so let's go ahead and show that. So the oxygen adds on to our carbon here, and this oxygen still had two hydrogens attached to it, and it had used up one lone pair of electrons. It still has one more, which gives it a plus one formal charge. This carbon is still bonded to this oxygen. This oxygen also has two lone pairs of electrons on it like that, and so we can show our intermediate. All right, so we have a hydrogen over here like that."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "It still has one more, which gives it a plus one formal charge. This carbon is still bonded to this oxygen. This oxygen also has two lone pairs of electrons on it like that, and so we can show our intermediate. All right, so we have a hydrogen over here like that. So let's show those electrons. So a lone pair of electrons on the oxygen here are going to attack our carbonyl carbon and form a bond. So those are these electrons in here."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "All right, so we have a hydrogen over here like that. So let's show those electrons. So a lone pair of electrons on the oxygen here are going to attack our carbonyl carbon and form a bond. So those are these electrons in here. And then once again, those pi electrons in here and our carbonyl had kicked off onto our oxygen like that. So that gives us our intermediates, and we're almost to the formation of a hydrate. The last step would be to deprotonate."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So those are these electrons in here. And then once again, those pi electrons in here and our carbonyl had kicked off onto our oxygen like that. So that gives us our intermediates, and we're almost to the formation of a hydrate. The last step would be to deprotonate. So a molecule of water comes along. This time it's going to function as a base and take up this proton away, leaving these two electrons behind on this oxygen. So let's go ahead and show our product, right?"}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "The last step would be to deprotonate. So a molecule of water comes along. This time it's going to function as a base and take up this proton away, leaving these two electrons behind on this oxygen. So let's go ahead and show our product, right? Formation of a hydrate. So we have a carbon bonded to two OH groups now, and so we form our hydrate or our gem diol. Let's follow those electrons now."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show our product, right? Formation of a hydrate. So we have a carbon bonded to two OH groups now, and so we form our hydrate or our gem diol. Let's follow those electrons now. So let's say that these electrons right in here moved off onto this top oxygen. It doesn't really matter which lone pair you say they are, and that's formation of our hydrate. We could have followed through as a ketone, right?"}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "Let's follow those electrons now. So let's say that these electrons right in here moved off onto this top oxygen. It doesn't really matter which lone pair you say they are, and that's formation of our hydrate. We could have followed through as a ketone, right? So instead of an aldehyde, we could have started off here as a ketone, and so this would have been R prime, and then this would have been R prime, and then this would have been R prime, and then finally this would have been R prime. So that's still formation of a hydrate. If we wanted to form a hemiacetal, we wouldn't have started with water."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "We could have followed through as a ketone, right? So instead of an aldehyde, we could have started off here as a ketone, and so this would have been R prime, and then this would have been R prime, and then this would have been R prime, and then finally this would have been R prime. So that's still formation of a hydrate. If we wanted to form a hemiacetal, we wouldn't have started with water. We would have started with an alcohol. So let's go ahead and show that. So let's make this R double prime OH, and of course an acid catalyzed version."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "If we wanted to form a hemiacetal, we wouldn't have started with water. We would have started with an alcohol. So let's go ahead and show that. So let's make this R double prime OH, and of course an acid catalyzed version. So R double prime OH and H plus. Instead of forming a hydrate, this is going to form a hemiacetal. So instead of H three O plus, which is what we had in our acid catalyzed hydration, we're going to protonate the alcohol."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's make this R double prime OH, and of course an acid catalyzed version. So R double prime OH and H plus. Instead of forming a hydrate, this is going to form a hemiacetal. So instead of H three O plus, which is what we had in our acid catalyzed hydration, we're going to protonate the alcohol. So let me go ahead and change this. This would no longer be an H here. This would be R double prime like that."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So instead of H three O plus, which is what we had in our acid catalyzed hydration, we're going to protonate the alcohol. So let me go ahead and change this. This would no longer be an H here. This would be R double prime like that. And so when we protonate our carbonyl oxygen here and make our carbon more electrophilic, this wouldn't be water as our nucleophile. This would be alcohol. So our alcohol would be, an alcohol molecule would be a nucleophile and attack our carbonyl carbon."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "This would be R double prime like that. And so when we protonate our carbonyl oxygen here and make our carbon more electrophilic, this wouldn't be water as our nucleophile. This would be alcohol. So our alcohol would be, an alcohol molecule would be a nucleophile and attack our carbonyl carbon. And so we could follow through that R double prime. So this wouldn't be an H. This would be R double prime, and therefore in our final product, this wouldn't be an H. This would be R double prime, and hopefully you can see that this is now a hemiacetal. It's very actually difficult to isolate hemiacetals most of the time, and if you're doing a reaction in acid, usually the hemiacetal is going to move on to form a full acetal."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So our alcohol would be, an alcohol molecule would be a nucleophile and attack our carbonyl carbon. And so we could follow through that R double prime. So this wouldn't be an H. This would be R double prime, and therefore in our final product, this wouldn't be an H. This would be R double prime, and hopefully you can see that this is now a hemiacetal. It's very actually difficult to isolate hemiacetals most of the time, and if you're doing a reaction in acid, usually the hemiacetal is going to move on to form a full acetal. So let's go ahead and show the structure of a full acetal. And this isn't just one step, and I won't go through the steps in this video. That'll be in the next video, but I just wanted to show you the structure of a full acetal here."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "It's very actually difficult to isolate hemiacetals most of the time, and if you're doing a reaction in acid, usually the hemiacetal is going to move on to form a full acetal. So let's go ahead and show the structure of a full acetal. And this isn't just one step, and I won't go through the steps in this video. That'll be in the next video, but I just wanted to show you the structure of a full acetal here. So we have R double prime and lone pairs of electrons on our oxygen, and we have over here R double prime and lone pairs of electrons on our oxygen. So this would be an acetal. And let's once again compare structures."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "That'll be in the next video, but I just wanted to show you the structure of a full acetal here. So we have R double prime and lone pairs of electrons on our oxygen, and we have over here R double prime and lone pairs of electrons on our oxygen. So this would be an acetal. And let's once again compare structures. So with an acetal, we have this OR double prime group and another OR double prime group on our acetal. For our hemiacetal, we had only one of those OR double prime groups. And so once again, we'll see much more details about acetals in the next video."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And let's once again compare structures. So with an acetal, we have this OR double prime group and another OR double prime group on our acetal. For our hemiacetal, we had only one of those OR double prime groups. And so once again, we'll see much more details about acetals in the next video. We can also form hydrates and hemiacetals using base to catalyze the reaction. So let's look at the mechanism for that. And so let's first start with hydration again."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, we'll see much more details about acetals in the next video. We can also form hydrates and hemiacetals using base to catalyze the reaction. So let's look at the mechanism for that. And so let's first start with hydration again. And over here on the right is our aldehyde or our ketone. In the uncatalyzed version, once again we used water as our nucleophile. We know that our carbonyl carbon right here is partially positive, our oxygen is partially negative."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And so let's first start with hydration again. And over here on the right is our aldehyde or our ketone. In the uncatalyzed version, once again we used water as our nucleophile. We know that our carbonyl carbon right here is partially positive, our oxygen is partially negative. And so our water molecule functioned as our nucleophile and attacked our carbonyl carbon. But in a base catalyzed version, a base can take off this proton here on water and make it into a stronger nucleophile. So it would form the hydroxide anion, which we know is a better nucleophile than water."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "We know that our carbonyl carbon right here is partially positive, our oxygen is partially negative. And so our water molecule functioned as our nucleophile and attacked our carbonyl carbon. But in a base catalyzed version, a base can take off this proton here on water and make it into a stronger nucleophile. So it would form the hydroxide anion, which we know is a better nucleophile than water. So a full negative one formal charge on our oxygen is going to make our hydroxide anion a better nucleophile. And this is what's going to attack in a base catalyzed version. So nucleophilic attack, our nucleophile attacks our electrophile, pushes these electrons off onto our oxygen."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So it would form the hydroxide anion, which we know is a better nucleophile than water. So a full negative one formal charge on our oxygen is going to make our hydroxide anion a better nucleophile. And this is what's going to attack in a base catalyzed version. So nucleophilic attack, our nucleophile attacks our electrophile, pushes these electrons off onto our oxygen. So let's go ahead and show the results of that. So now we would have our OH is now going to be bonded to our carbon. So let's go ahead and put that in there."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So nucleophilic attack, our nucleophile attacks our electrophile, pushes these electrons off onto our oxygen. So let's go ahead and show the results of that. So now we would have our OH is now going to be bonded to our carbon. So let's go ahead and put that in there. So now we have only two lone pairs of electrons on our oxygen. So let's follow some electrons once again. So these electrons right here are going to form a bond between that oxygen and that carbon."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and put that in there. So now we have only two lone pairs of electrons on our oxygen. So let's follow some electrons once again. So these electrons right here are going to form a bond between that oxygen and that carbon. So let's say it's this right here. And also bonded to that carbon we would have an oxygen, which had two lone pairs of electrons. It got another one, so now it has a negative one formal charge."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here are going to form a bond between that oxygen and that carbon. So let's say it's this right here. And also bonded to that carbon we would have an oxygen, which had two lone pairs of electrons. It got another one, so now it has a negative one formal charge. So let's follow those electrons too. So make those red. So right in here these electrons go off onto that oxygen."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "It got another one, so now it has a negative one formal charge. So let's follow those electrons too. So make those red. So right in here these electrons go off onto that oxygen. Once again, it doesn't really matter which lone pair you say they are. Let's just say it's those. And also bonded to our carbon we would have our R group and our hydrogen."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So right in here these electrons go off onto that oxygen. Once again, it doesn't really matter which lone pair you say they are. Let's just say it's those. And also bonded to our carbon we would have our R group and our hydrogen. So here is our intermediate. And in the next step, to form a hydrate, we would have to protonate. So we would have a water molecule right here could now function as an acid."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And also bonded to our carbon we would have our R group and our hydrogen. So here is our intermediate. And in the next step, to form a hydrate, we would have to protonate. So we would have a water molecule right here could now function as an acid. So let's go ahead and show that. So this is going to function as a base. So let's say these electrons in red pick up a proton from water, leaving these electrons behind on the oxygen."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So we would have a water molecule right here could now function as an acid. So let's go ahead and show that. So this is going to function as a base. So let's say these electrons in red pick up a proton from water, leaving these electrons behind on the oxygen. And that will form our hydrate product. Because now we have this oxygen over here on the left. And now we have this oxygen over here on the right has now been protonated."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's say these electrons in red pick up a proton from water, leaving these electrons behind on the oxygen. And that will form our hydrate product. Because now we have this oxygen over here on the left. And now we have this oxygen over here on the right has now been protonated. So let's go ahead and show those electrons in red. Grabbed this proton, and so it formed this bond right here. And we have formed our hydrate."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And now we have this oxygen over here on the right has now been protonated. So let's go ahead and show those electrons in red. Grabbed this proton, and so it formed this bond right here. And we have formed our hydrate. So let me go ahead and put in this R and this H. Once again, we could have carried this through starting with a ketone. So let's do that really fast. So instead of a hydrogen here, we could have had our R prime."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And we have formed our hydrate. So let me go ahead and put in this R and this H. Once again, we could have carried this through starting with a ketone. So let's do that really fast. So instead of a hydrogen here, we could have had our R prime. So it would have been an R prime here. It would have been an R prime here. And that's, of course, a hydrate as well."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So instead of a hydrogen here, we could have had our R prime. So it would have been an R prime here. It would have been an R prime here. And that's, of course, a hydrate as well. If we want to form a hemiacetal, once again, we would not start with water. We would start with an alcohol. So let's make this R double prime."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And that's, of course, a hydrate as well. If we want to form a hemiacetal, once again, we would not start with water. We would start with an alcohol. So let's make this R double prime. And if we use a base, we would take off the proton on our alcohol to form an alkoxide. So this would be R double prime here. And an alkoxide anion is more nucleophilic than alcohol just by itself."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "So let's make this R double prime. And if we use a base, we would take off the proton on our alcohol to form an alkoxide. So this would be R double prime here. And an alkoxide anion is more nucleophilic than alcohol just by itself. And so the alkoxide attacks. And so instead of this being a hydrogen, this would now be R double prime. And we could have shown this step, this wouldn't be water."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And an alkoxide anion is more nucleophilic than alcohol just by itself. And so the alkoxide attacks. And so instead of this being a hydrogen, this would now be R double prime. And we could have shown this step, this wouldn't be water. This would be our alcohol. So I could make this R double prime here. And then finally, for our hemiacetal product, this wouldn't be hydrogen."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And we could have shown this step, this wouldn't be water. This would be our alcohol. So I could make this R double prime here. And then finally, for our hemiacetal product, this wouldn't be hydrogen. This would be R double prime. And so we can see we have now formed a hemiacetal as our product. Now in base, a hemiacetal will not react to form an acetal."}, {"video_title": "Acid and base catalyzed formation of hydrates and hemiacetals Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, for our hemiacetal product, this wouldn't be hydrogen. This would be R double prime. And so we can see we have now formed a hemiacetal as our product. Now in base, a hemiacetal will not react to form an acetal. So that's kind of an important point when you're using acetals as protecting groups. They are stable in base. So you can't form acetals in base."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "Solvation can have a stabilizing effect on a conjugate base. So if we look at water, here's the water molecule, if water donates this proton, then our electrons in magenta are left behind on the oxygen, which gives the oxygen a negative one formal charge, and we form the conjugate base to water, which is the hydroxide anion. The hydroxide anion can be stabilized by our solvent, so if water is our solvent, let me go ahead and draw in a few water molecules here, so here is one water molecule, let me draw another one over here. We know that water is a polar solvent, it's a polar molecule. The oxygen has a partial negative charge, and the hydrogens have partial positive charges. That's because the oxygen is more electronegative than hydrogen. And now we have a situation where opposite charges attract."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "We know that water is a polar solvent, it's a polar molecule. The oxygen has a partial negative charge, and the hydrogens have partial positive charges. That's because the oxygen is more electronegative than hydrogen. And now we have a situation where opposite charges attract. The partially positive hydrogen on water is attracted to the negatively charged oxygen on hydroxide, so there is an attractive force between these opposite charges. And that attractive force stabilizes, or helps to stabilize the negative charge on the hydroxide anion. So over here we have another water molecule, so partial negative, partial positive."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "And now we have a situation where opposite charges attract. The partially positive hydrogen on water is attracted to the negatively charged oxygen on hydroxide, so there is an attractive force between these opposite charges. And that attractive force stabilizes, or helps to stabilize the negative charge on the hydroxide anion. So over here we have another water molecule, so partial negative, partial positive. We have another stabilizing force, another force to help stabilize that negative charge on the hydroxide anion. So the more water molecules you have, the more solvent molecules you have solvating your anion, the more stable the conjugate base. So if we look at our four compounds over here, so water, this proton on water has a pKa value of approximately 15.7."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "So over here we have another water molecule, so partial negative, partial positive. We have another stabilizing force, another force to help stabilize that negative charge on the hydroxide anion. So the more water molecules you have, the more solvent molecules you have solvating your anion, the more stable the conjugate base. So if we look at our four compounds over here, so water, this proton on water has a pKa value of approximately 15.7. For ethanol, this proton has a pKa value of about 16. For isopropanol, this proton has a pKa value of 17. And for tert-butanol, this proton has a pKa value of about 18."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "So if we look at our four compounds over here, so water, this proton on water has a pKa value of approximately 15.7. For ethanol, this proton has a pKa value of about 16. For isopropanol, this proton has a pKa value of 17. And for tert-butanol, this proton has a pKa value of about 18. The lower the pKa value, the stronger the acid. So water is the strongest acid out of these four, because water has the lowest value for the pKa. If water is the strongest acid, that means the conjugate base must be the most stable."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "And for tert-butanol, this proton has a pKa value of about 18. The lower the pKa value, the stronger the acid. So water is the strongest acid out of these four, because water has the lowest value for the pKa. If water is the strongest acid, that means the conjugate base must be the most stable. So the hydroxide anion is the most stable out of our conjugate bases, and that's because we can fit more water molecules around our hydroxide anion. The hydroxide anion is the best at being stabilized, the best at being solvated by our solvent. And that's because of something called steric hindrance."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "If water is the strongest acid, that means the conjugate base must be the most stable. So the hydroxide anion is the most stable out of our conjugate bases, and that's because we can fit more water molecules around our hydroxide anion. The hydroxide anion is the best at being stabilized, the best at being solvated by our solvent. And that's because of something called steric hindrance. So attached to this negatively charged oxygen, we have a really small little hydrogen. When we look at the other conjugate bases, so this is the ethoxide anion, which is the conjugate base to ethanol. We have a CH2 here, and then we have a CH3."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "And that's because of something called steric hindrance. So attached to this negatively charged oxygen, we have a really small little hydrogen. When we look at the other conjugate bases, so this is the ethoxide anion, which is the conjugate base to ethanol. We have a CH2 here, and then we have a CH3. CH2 and CH3 take up more space in this hydrogen. And as we go down, we see even more stuff, even more steric hindrance. This carbon right here is attached to a hydrogen, it's also attached to a CH3 and a CH3."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "We have a CH2 here, and then we have a CH3. CH2 and CH3 take up more space in this hydrogen. And as we go down, we see even more stuff, even more steric hindrance. This carbon right here is attached to a hydrogen, it's also attached to a CH3 and a CH3. So we have even more stuff to get in the way. And finally, for our last conjugate base, this is called the tert-butoxide anion. We have a carbon attached to a CH3, a CH3, and a CH3, so even more stuff to get in the way."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here is attached to a hydrogen, it's also attached to a CH3 and a CH3. So we have even more stuff to get in the way. And finally, for our last conjugate base, this is called the tert-butoxide anion. We have a carbon attached to a CH3, a CH3, and a CH3, so even more stuff to get in the way. And that steric hindrance, those bulky groups, prevents the interaction of solvent molecules with our conjugate base. So if I draw in a water molecule here, so there will be some stabilization, right? We know oxygen is partially negative, we know that hydrogens are partially positive, so there is an attractive force between our opposite charges and that would help to stabilize the conjugate base."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "We have a carbon attached to a CH3, a CH3, and a CH3, so even more stuff to get in the way. And that steric hindrance, those bulky groups, prevents the interaction of solvent molecules with our conjugate base. So if I draw in a water molecule here, so there will be some stabilization, right? We know oxygen is partially negative, we know that hydrogens are partially positive, so there is an attractive force between our opposite charges and that would help to stabilize the conjugate base. The problem is, these bulky CH3 groups prevent a lot of water molecules from being able to stabilize this conjugate base. And that means that this tert-butoxide anion is the least stable conjugate base. As we move up in this direction, we're decreasing in steric hindrance."}, {"video_title": "Stabilization of a conjugate base solvation Organic chemistry Khan Academy.mp3", "Sentence": "We know oxygen is partially negative, we know that hydrogens are partially positive, so there is an attractive force between our opposite charges and that would help to stabilize the conjugate base. The problem is, these bulky CH3 groups prevent a lot of water molecules from being able to stabilize this conjugate base. And that means that this tert-butoxide anion is the least stable conjugate base. As we move up in this direction, we're decreasing in steric hindrance. We go from three bulky methyl groups to two bulky methyl groups to one right here, and then finally to this hydrogen. So we're decreasing in the steric hindrance, we're increasing in the ability of our solvent to solvate our anion and stabilize our anion. And therefore, the hydroxide anion is the most stable in our solvent."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with this blue one. So the first thing we notice, it is an alkene. It has a double bond right there. So we can, right from the get-go, write that it's going to end with an ene. And then we need to figure out what the longest carbon chain is. And it looks like it's this one here. One, two, three, four, five, six, seven carbons."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "So we can, right from the get-go, write that it's going to end with an ene. And then we need to figure out what the longest carbon chain is. And it looks like it's this one here. One, two, three, four, five, six, seven carbons. It's going to be hept. It's going to be hept as our prefix on the alkene. It's going to be, let me write it out a little bit further to the left, hept."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six, seven carbons. It's going to be hept. It's going to be hept as our prefix on the alkene. It's going to be, let me write it out a little bit further to the left, hept. And then we have to figure out where the double bond is. And we always want to start numbering our carbons from the direction that bumps into the double bond first. And so the double bond is closer to this end of the chain than that end."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be, let me write it out a little bit further to the left, hept. And then we have to figure out where the double bond is. And we always want to start numbering our carbons from the direction that bumps into the double bond first. And so the double bond is closer to this end of the chain than that end. So we're going to start numbering here. So one, two, three, four, five, six, seven. And so the double bond starts on the three carbon."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "And so the double bond is closer to this end of the chain than that end. So we're going to start numbering here. So one, two, three, four, five, six, seven. And so the double bond starts on the three carbon. So it is hept-3-ene. And we also have a functional group. We have a methyl group sitting on the number four carbon."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "And so the double bond starts on the three carbon. So it is hept-3-ene. And we also have a functional group. We have a methyl group sitting on the number four carbon. So it is 4-methyl-hept-3-ene. Which is correct, but we still haven't considered what's going on on either side of the double bond. Whether the higher priority functional group is whether they're closer to each other on both sides of the double bond, or whether they're farther away from each other."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "We have a methyl group sitting on the number four carbon. So it is 4-methyl-hept-3-ene. Which is correct, but we still haven't considered what's going on on either side of the double bond. Whether the higher priority functional group is whether they're closer to each other on both sides of the double bond, or whether they're farther away from each other. And to do that, we first have to identify the higher priority group on each side of the double bond. So on the right side of the double bond, let me circle it in magenta on each side. So on the right side of the double bond, we only have one functional group."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "Whether the higher priority functional group is whether they're closer to each other on both sides of the double bond, or whether they're farther away from each other. And to do that, we first have to identify the higher priority group on each side of the double bond. So on the right side of the double bond, let me circle it in magenta on each side. So on the right side of the double bond, we only have one functional group. Right over here you could view this as an ethyl group. That's pretty obvious. There is no other functional group bonded to the carbon."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "So on the right side of the double bond, we only have one functional group. Right over here you could view this as an ethyl group. That's pretty obvious. There is no other functional group bonded to the carbon. There's only a hydrogen over here. Now on the left side, it's a little bit less obvious. We have two functional groups from the point of view of this carbon."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "There is no other functional group bonded to the carbon. There's only a hydrogen over here. Now on the left side, it's a little bit less obvious. We have two functional groups from the point of view of this carbon. You have the methyl that we already pointed out, and you also have these three carbons right over here. You could view that as a propyl group. And what we need to do to identify the highest priority group is to use the Kahn-Ingold prelog naming or priority scheme that we learned several videos ago."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "We have two functional groups from the point of view of this carbon. You have the methyl that we already pointed out, and you also have these three carbons right over here. You could view that as a propyl group. And what we need to do to identify the highest priority group is to use the Kahn-Ingold prelog naming or priority scheme that we learned several videos ago. And there, you literally go from this carbon, you look at what it's bonded to, and you compare the atomic numbers. But in both cases, it's a carbon to a carbon. This is a carbon to a carbon."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "And what we need to do to identify the highest priority group is to use the Kahn-Ingold prelog naming or priority scheme that we learned several videos ago. And there, you literally go from this carbon, you look at what it's bonded to, and you compare the atomic numbers. But in both cases, it's a carbon to a carbon. This is a carbon to a carbon. So their atomic numbers are the same. So then you go one bond further away, and you say, which one is bonded to a higher atomic number atom? This carbon bonds to a carbon, which is a higher atomic number."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "This is a carbon to a carbon. So their atomic numbers are the same. So then you go one bond further away, and you say, which one is bonded to a higher atomic number atom? This carbon bonds to a carbon, which is a higher atomic number. This carbon only bonds to three hydrogens. This one does two hydrogens and one carbon. Because it's bonded to another carbon, it takes priority."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "This carbon bonds to a carbon, which is a higher atomic number. This carbon only bonds to three hydrogens. This one does two hydrogens and one carbon. Because it's bonded to another carbon, it takes priority. This propyl group is a higher priority functional group. So now when we're trying to decide whether it is entgegen or zosamin, we look at these two groups, and we see that they are sitting on the same side of the double bond. They're both above the carbons."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "Because it's bonded to another carbon, it takes priority. This propyl group is a higher priority functional group. So now when we're trying to decide whether it is entgegen or zosamin, we look at these two groups, and we see that they are sitting on the same side of the double bond. They're both above the carbons. They're closer to each other. So this molecule is zosamin, which is, on some levels, you can think of the same thing as cis. But cis and trans stops applying when you start having more than two functional groups."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "They're both above the carbons. They're closer to each other. So this molecule is zosamin, which is, on some levels, you can think of the same thing as cis. But cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z for methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "But cis and trans stops applying when you start having more than two functional groups. In this case, we have three. So we would call this Z for methylhept-3-ene. And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. Someone pointed out, rightly, that I misspelled zosamin in the last video. It's actually spelled like this."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "And that's because the higher priority functional groups are on the same side of the double bond. Now let's do this one over here. Someone pointed out, rightly, that I misspelled zosamin in the last video. It's actually spelled like this. I had spelled it with two S's and one M. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "It's actually spelled like this. I had spelled it with two S's and one M. I guess you can forgive me. I don't speak German. But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "But anyway, I thought I would point that out. Now let's try to label this thing right over here. So the first thing, this once again is an alkene. Let's identify the longest carbon chain here. We have 1, 2, 3, 4, 5, 6, 7, 8 carbons. Double bonds are closer to the left-hand side, so we'll start numbering here. 1, 2, 3, 4, 5, 6, 7, 8."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "Let's identify the longest carbon chain here. We have 1, 2, 3, 4, 5, 6, 7, 8 carbons. Double bonds are closer to the left-hand side, so we'll start numbering here. 1, 2, 3, 4, 5, 6, 7, 8. So just the main chain is oct-3-ene. And then we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7, 8. So just the main chain is oct-3-ene. And then we have one functional group sitting off of the main chain. We have this bromine sitting right over there on the third carbon. So we would call this 3-bromo-oct-3-ene. And now we have to figure out, is it n-cogin or zosamin? So if we look on the carbon on the right-hand side, it's pretty obvious that this is the only functional group."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "We have this bromine sitting right over there on the third carbon. So we would call this 3-bromo-oct-3-ene. And now we have to figure out, is it n-cogin or zosamin? So if we look on the carbon on the right-hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there, so let me circle it in the magenta. Just like that. And then on the left-hand side, we have two functional groups."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "So if we look on the carbon on the right-hand side, it's pretty obvious that this is the only functional group. We just have a hydrogen there, so let me circle it in the magenta. Just like that. And then on the left-hand side, we have two functional groups. We have this bromine sitting right there, and then we have this propyl group. And because this propyl group is bigger, you might be tempted to say it takes higher priority. But remember, in the Kahn-Ingold prelog system, you give higher priority to the atom that has a higher atomic number."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "And then on the left-hand side, we have two functional groups. We have this bromine sitting right there, and then we have this propyl group. And because this propyl group is bigger, you might be tempted to say it takes higher priority. But remember, in the Kahn-Ingold prelog system, you give higher priority to the atom that has a higher atomic number. Bromine has an atomic number of 35. Carbon has an atomic number of only 6. So bromine is actually higher priority."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "But remember, in the Kahn-Ingold prelog system, you give higher priority to the atom that has a higher atomic number. Bromine has an atomic number of 35. Carbon has an atomic number of only 6. So bromine is actually higher priority. So this is the higher priority functional group right over here. So now, if we're deciding whether it's n-cogin or zosamin, we see that our higher priority groups are apart. They're on opposite sides of the double bond."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "So bromine is actually higher priority. So this is the higher priority functional group right over here. So now, if we're deciding whether it's n-cogin or zosamin, we see that our higher priority groups are apart. They're on opposite sides of the double bond. This one is on top. This one is below. We are apart."}, {"video_title": "Entgegen-Zusammen naming scheme for alkenes examples Organic chemistry Khan Academy.mp3", "Sentence": "They're on opposite sides of the double bond. This one is on top. This one is below. We are apart. So this is entgogen, or we would write E. This is E3-bromo-oct-3-ene. And E is for, just as a bit of a refresher, it's for entgogen, a word that I enjoy saying. Entgogen."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to start with a terminal alkyne over here. You can see there's a hydrogen on one side of our alkyne. And the other side of our alkyne, let's say it's an alkyl group bonded to this carbon on the right. So we're going to add water, sulfuric acid, and mercury to sulfate to our alkyne. We're going to hydrate it. Now we've seen the hydration reaction before. We did a hydration reaction with an alkene, and we added an H plus and an OH minus across our double bond."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to add water, sulfuric acid, and mercury to sulfate to our alkyne. We're going to hydrate it. Now we've seen the hydration reaction before. We did a hydration reaction with an alkene, and we added an H plus and an OH minus across our double bond. And we're going to do the same thing here. We're going to add an H plus and an OH minus, this time across our triple bond. And the OH minus always adds to the most substituted carbon."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We did a hydration reaction with an alkene, and we added an H plus and an OH minus across our double bond. And we're going to do the same thing here. We're going to add an H plus and an OH minus, this time across our triple bond. And the OH minus always adds to the most substituted carbon. So we saw that in our earlier video. So the regiochemistry shows Markovnikov's rule. So this is, I'll just write Mark's rule here."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the OH minus always adds to the most substituted carbon. So we saw that in our earlier video. So the regiochemistry shows Markovnikov's rule. So this is, I'll just write Mark's rule here. So we spent a lot of time talking about Markovnikov's rule in earlier videos, which means that this OH over here on the right is going to add to the most substituted carbon, which happens to be the carbon on the right, since that's going to be bonded to an alkyl group over here on the right. The H plus, right, added over here to this carbon on the left. So now there are two hydrogens on the left."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is, I'll just write Mark's rule here. So we spent a lot of time talking about Markovnikov's rule in earlier videos, which means that this OH over here on the right is going to add to the most substituted carbon, which happens to be the carbon on the right, since that's going to be bonded to an alkyl group over here on the right. The H plus, right, added over here to this carbon on the left. So now there are two hydrogens on the left. So we added H plus and OH minus. We added water, essentially, across our triple bond. And we get this molecule as our intermediate."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now there are two hydrogens on the left. So we added H plus and OH minus. We added water, essentially, across our triple bond. And we get this molecule as our intermediate. This molecule has a specific name. It's called an enol. So it's called an enol."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we get this molecule as our intermediate. This molecule has a specific name. It's called an enol. So it's called an enol. The en comes from the fact that there's a double bond present in this molecule. And the OL comes from the fact that there's an alcohol, right, this OH here. So this enol is the intermediate of this reaction."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's called an enol. The en comes from the fact that there's a double bond present in this molecule. And the OL comes from the fact that there's an alcohol, right, this OH here. So this enol is the intermediate of this reaction. However, the enol is not the most stable form of this molecule. So it's actually going to rearrange. And we're going to get a ketone for our product."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this enol is the intermediate of this reaction. However, the enol is not the most stable form of this molecule. So it's actually going to rearrange. And we're going to get a ketone for our product. And I'll show you the mechanism for this rearrangement in a few minutes here. So we get a methyl ketone from a terminal alkyne. So this reaction is best used when you're looking to make a methyl ketone."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to get a ketone for our product. And I'll show you the mechanism for this rearrangement in a few minutes here. So we get a methyl ketone from a terminal alkyne. So this reaction is best used when you're looking to make a methyl ketone. And you're starting with a triple bond here. So let's look at a reaction. So let's start with a terminal alkyne."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this reaction is best used when you're looking to make a methyl ketone. And you're starting with a triple bond here. So let's look at a reaction. So let's start with a terminal alkyne. So I'm going to put a methyl group on this side of the carbon like that. And I'm going to react this terminal alkyne with water and sulfuric acid. And with mercury 2 sulfate like that."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with a terminal alkyne. So I'm going to put a methyl group on this side of the carbon like that. And I'm going to react this terminal alkyne with water and sulfuric acid. And with mercury 2 sulfate like that. So if we think about what's going to happen, we're adding H plus and OH minus across our triple bond. We're going to add the OH minus to the most substituted carbon. So that would be the carbon on the right."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And with mercury 2 sulfate like that. So if we think about what's going to happen, we're adding H plus and OH minus across our triple bond. We're going to add the OH minus to the most substituted carbon. So that would be the carbon on the right. So we have carbon double bonded to another carbon. We're going to add the OH to the carbon on the right. We're going to add the H plus to the carbon on the left."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that would be the carbon on the right. So we have carbon double bonded to another carbon. We're going to add the OH to the carbon on the right. We're going to add the H plus to the carbon on the left. So there's water added across your triple bond. And then we have still another hydrogen on this carbon on the left here. And then the methyl group is still over here on the right."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to add the H plus to the carbon on the left. So there's water added across your triple bond. And then we have still another hydrogen on this carbon on the left here. And then the methyl group is still over here on the right. So that's our enol intermediate. Let me just draw this molecule again. So this molecule is equivalent to this molecule right here."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then the methyl group is still over here on the right. So that's our enol intermediate. Let me just draw this molecule again. So this molecule is equivalent to this molecule right here. So I'm going to draw in this form from now on. So this is my enol form. And my enol is going to rearrange."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this molecule is equivalent to this molecule right here. So I'm going to draw in this form from now on. So this is my enol form. And my enol is going to rearrange. This is not the most stable form of this molecule. So let's see the mechanism of how our enol rearranges. So let's go ahead and redraw our enol down here."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And my enol is going to rearrange. This is not the most stable form of this molecule. So let's see the mechanism of how our enol rearranges. So let's go ahead and redraw our enol down here. So let's get a little room like that. So here's my enol. Put in my double bond."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and redraw our enol down here. So let's get a little room like that. So here's my enol. Put in my double bond. Put in my lone pairs of electrons like that. So this enol is present in an acidic mixture. H2O and H2SO4 are going to give me hydronium ions in solution, H3O plus."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Put in my double bond. Put in my lone pairs of electrons like that. So this enol is present in an acidic mixture. H2O and H2SO4 are going to give me hydronium ions in solution, H3O plus. So there's going to be H3O plus floating around here in solution. So we have our hydronium ion looking like that. So this, of course, is capable of donating a proton."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "H2O and H2SO4 are going to give me hydronium ions in solution, H3O plus. So there's going to be H3O plus floating around here in solution. So we have our hydronium ion looking like that. So this, of course, is capable of donating a proton. And the pi electrons are going to actually move out here and pick up a proton like that. And then these electrons are going to kick off onto your oxygen. So this is an acid-base reaction as the first step."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this, of course, is capable of donating a proton. And the pi electrons are going to actually move out here and pick up a proton like that. And then these electrons are going to kick off onto your oxygen. So this is an acid-base reaction as the first step. So therefore, we're going to draw our equilibrium arrows here like that. And let's see what happens. Well, we're going to have this as our skeleton."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is an acid-base reaction as the first step. So therefore, we're going to draw our equilibrium arrows here like that. And let's see what happens. Well, we're going to have this as our skeleton. The OH didn't really do anything yet. And we're going to add a proton onto the carbon on the left side of your double bond like that. So let's go ahead and show what electrons did the moving."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, we're going to have this as our skeleton. The OH didn't really do anything yet. And we're going to add a proton onto the carbon on the left side of your double bond like that. So let's go ahead and show what electrons did the moving. The electrons in your pi bond, whichever one of these bonds here is your pi bond, those electrons are going to move and form a bond with this proton like this. So we ended up taking away a bond from this carbon right here. So we had a double bond that took a bond away from that carbon."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show what electrons did the moving. The electrons in your pi bond, whichever one of these bonds here is your pi bond, those electrons are going to move and form a bond with this proton like this. So we ended up taking away a bond from this carbon right here. So we had a double bond that took a bond away from that carbon. That's going to make that carbon positively charged. So there's a plus 1 formal charge on that carbon. And we can draw a resonance structure for this intermediate."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we had a double bond that took a bond away from that carbon. That's going to make that carbon positively charged. So there's a plus 1 formal charge on that carbon. And we can draw a resonance structure for this intermediate. So let's go ahead and draw our resonance bracket and our resonance arrows here. And what can we do with our lone pairs of electrons to share that positive charge? So we can take one of these lone pairs of electrons and we can move them in here to form a double bond between our carbon and our oxygen."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we can draw a resonance structure for this intermediate. So let's go ahead and draw our resonance bracket and our resonance arrows here. And what can we do with our lone pairs of electrons to share that positive charge? So we can take one of these lone pairs of electrons and we can move them in here to form a double bond between our carbon and our oxygen. So let's go ahead and do that. A double bond forms between our carbon and our oxygen like that. Now there's only one lone pair of electrons on this top oxygen."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we can take one of these lone pairs of electrons and we can move them in here to form a double bond between our carbon and our oxygen. So let's go ahead and do that. A double bond forms between our carbon and our oxygen like that. Now there's only one lone pair of electrons on this top oxygen. And if we think about what happened to our formal charge, it actually moves out here to this oxygen. So this oxygen now is a plus 1 formal charge. This is a resonance structure."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now there's only one lone pair of electrons on this top oxygen. And if we think about what happened to our formal charge, it actually moves out here to this oxygen. So this oxygen now is a plus 1 formal charge. This is a resonance structure. Let me go ahead and keep this hydrogen in here like that. So that's our resonance structure for this molecule on the left. Now oxygen does not like having a plus 1 formal charge either."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is a resonance structure. Let me go ahead and keep this hydrogen in here like that. So that's our resonance structure for this molecule on the left. Now oxygen does not like having a plus 1 formal charge either. So is there any way for this oxygen to get rid of its plus 1 formal charge? And of course there is. There's water floating around."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now oxygen does not like having a plus 1 formal charge either. So is there any way for this oxygen to get rid of its plus 1 formal charge? And of course there is. There's water floating around. So water is going to act as a base. So a lone pair of electrons on water is going to take this proton, just that proton, leaving these two electrons behind on my oxygen like that. So we have another acid-base reaction."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "There's water floating around. So water is going to act as a base. So a lone pair of electrons on water is going to take this proton, just that proton, leaving these two electrons behind on my oxygen like that. So we have another acid-base reaction. So let's go ahead and show this reaction being at equilibrium. So I draw my equilibrium arrows like that. And so now I have my oxygen."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have another acid-base reaction. So let's go ahead and show this reaction being at equilibrium. So I draw my equilibrium arrows like that. And so now I have my oxygen. It had one lone pair of electrons around it. Now it has two lone pairs of electrons around it. So now we're done."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so now I have my oxygen. It had one lone pair of electrons around it. Now it has two lone pairs of electrons around it. So now we're done. We formed our ketone version. So this is acetone, our simplest ketone. So we have the keto form of the product."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now we're done. We formed our ketone version. So this is acetone, our simplest ketone. So we have the keto form of the product. This is the keto form of the product in the form of a ketone. And then we have our enol over here. So the keto and the enol form are said to be tautomers of each other."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have the keto form of the product. This is the keto form of the product in the form of a ketone. And then we have our enol over here. So the keto and the enol form are said to be tautomers of each other. And they're in equilibrium with each other. And they can go back and forth. And in this case, we're using acid to catalyze this transition."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the keto and the enol form are said to be tautomers of each other. And they're in equilibrium with each other. And they can go back and forth. And in this case, we're using acid to catalyze this transition. So this is called acid-catalyzed tautomerization. So tautomerization like that. And we're going back and forth between the keto and the enol form."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And in this case, we're using acid to catalyze this transition. So this is called acid-catalyzed tautomerization. So tautomerization like that. And we're going back and forth between the keto and the enol form. So you'll also hear this called keto-enol tautomerization a lot like that. So the keto and the enol form are in equilibrium with each other. Now the equilibrium is actually going to favor the keto form."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we're going back and forth between the keto and the enol form. So you'll also hear this called keto-enol tautomerization a lot like that. So the keto and the enol form are in equilibrium with each other. Now the equilibrium is actually going to favor the keto form. So this is actually the more stable form. This is more stable. So that carbon double bonded to an oxygen for the keto form is more stable than the carbon double bonded to another carbon in the enol form."}, {"video_title": "Hydration of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now the equilibrium is actually going to favor the keto form. So this is actually the more stable form. This is more stable. So that carbon double bonded to an oxygen for the keto form is more stable than the carbon double bonded to another carbon in the enol form. And essentially, you're changing one proton, one pi bond in this equilibrium. And this is the acid-catalyzed form of this mechanism. So in the next video, we'll see a very similar reaction where we do a base-catalyzed tautomerization."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And one of those compounds is naphthalene. So over here on the left, we have the dot structure for naphthalene. Naphthalene is a white solid that is traditionally the component of mothballs. And so it has a very distinctive smell to it. Now, naphthalene is aromatic. However, it's not as stable as benzene. But we can think about it as two fused benzene-like rings."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it has a very distinctive smell to it. Now, naphthalene is aromatic. However, it's not as stable as benzene. But we can think about it as two fused benzene-like rings. And so if I were to analyze a naphthalene molecule using our rules, our criteria for aromaticity, I could look at each carbon and naphthalene. And I could see that each carbon has a double bond to it. So each carbon is sp2 hybridized."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But we can think about it as two fused benzene-like rings. And so if I were to analyze a naphthalene molecule using our rules, our criteria for aromaticity, I could look at each carbon and naphthalene. And I could see that each carbon has a double bond to it. So each carbon is sp2 hybridized. And therefore, each carbon has a p orbital, so an unhybridized p orbital. And so there are a total of 10 carbons in naphthalene. And so if I go over here to the drawing on the right, each of those carbons has a p orbital."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So each carbon is sp2 hybridized. And therefore, each carbon has a p orbital, so an unhybridized p orbital. And so there are a total of 10 carbons in naphthalene. And so if I go over here to the drawing on the right, each of those carbons has a p orbital. So I could draw in the p orbitals on each one of my carbons in here like that. Now, these p orbitals are right next to each other, which means they can overlap. And so if you think about the criteria for a compound to be aromatic, this would meet that first criteria there."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if I go over here to the drawing on the right, each of those carbons has a p orbital. So I could draw in the p orbitals on each one of my carbons in here like that. Now, these p orbitals are right next to each other, which means they can overlap. And so if you think about the criteria for a compound to be aromatic, this would meet that first criteria there. Because you can see that you have overlap of these p orbitals. And in this case, we have delocalization of electrons across two rings. Now, when we think about the second criteria, which was Huckel's rule in terms of the number of pi electrons our compound has, let's go ahead and analyze naphthalene, even though technically we can't use Huckel's rule."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if you think about the criteria for a compound to be aromatic, this would meet that first criteria there. Because you can see that you have overlap of these p orbitals. And in this case, we have delocalization of electrons across two rings. Now, when we think about the second criteria, which was Huckel's rule in terms of the number of pi electrons our compound has, let's go ahead and analyze naphthalene, even though technically we can't use Huckel's rule. But if we look at it, we can see there are 2, 4, 6, 8, and 10 pi electrons. So naphthalene has 10 pi electrons. And if you think about Huckel's rule, 4n plus 2, if n is equal to 2, 4 times 2 plus 2 is equal to 10 pi electrons."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, when we think about the second criteria, which was Huckel's rule in terms of the number of pi electrons our compound has, let's go ahead and analyze naphthalene, even though technically we can't use Huckel's rule. But if we look at it, we can see there are 2, 4, 6, 8, and 10 pi electrons. So naphthalene has 10 pi electrons. And if you think about Huckel's rule, 4n plus 2, if n is equal to 2, 4 times 2 plus 2 is equal to 10 pi electrons. And so 10 pi electrons is a Huckel number. And so it looks like naphthalene fulfills the two criteria, even though, again, technically we can't apply Huckel's rule to polycyclic compounds. But those 10 pi electrons are fully delocalized throughout both rings."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if you think about Huckel's rule, 4n plus 2, if n is equal to 2, 4 times 2 plus 2 is equal to 10 pi electrons. And so 10 pi electrons is a Huckel number. And so it looks like naphthalene fulfills the two criteria, even though, again, technically we can't apply Huckel's rule to polycyclic compounds. But those 10 pi electrons are fully delocalized throughout both rings. And one way to show that would be using resonance structures. So if we were to draw a resonance structure for naphthalene, I could take these electrons and move them in here. And then these electrons would go over here."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But those 10 pi electrons are fully delocalized throughout both rings. And one way to show that would be using resonance structures. So if we were to draw a resonance structure for naphthalene, I could take these electrons and move them in here. And then these electrons would go over here. And then these would go over there. So that would give me another resonance structure. So if I go ahead and draw the results of the movement of those electrons, I would now have my pi electrons over here like this."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons would go over here. And then these would go over there. So that would give me another resonance structure. So if I go ahead and draw the results of the movement of those electrons, I would now have my pi electrons over here like this. So it's like a benzene-like ring on the left. And then on the right, we still had these pi electrons in here like that. Now, in this case, I've shown these pi electrons right here."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I go ahead and draw the results of the movement of those electrons, I would now have my pi electrons over here like this. So it's like a benzene-like ring on the left. And then on the right, we still had these pi electrons in here like that. Now, in this case, I've shown these pi electrons right here. I've shown them on the left side. But in reality, those pi electrons are above and below our single bond in terms of the probability of finding those electrons. And so I don't have to draw it the way I did it here."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, in this case, I've shown these pi electrons right here. I've shown them on the left side. But in reality, those pi electrons are above and below our single bond in terms of the probability of finding those electrons. And so I don't have to draw it the way I did it here. I could draw it like this. So let me go ahead and put this is going to be equivalent to this structure. So these aren't different resonance structures."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I don't have to draw it the way I did it here. I could draw it like this. So let me go ahead and put this is going to be equivalent to this structure. So these aren't different resonance structures. I'm just drawing a different way of representing that resonance structure over here. So I could show those electrons in blue over here on this side like that. So let me go ahead and highlight those electrons."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these aren't different resonance structures. I'm just drawing a different way of representing that resonance structure over here. So I could show those electrons in blue over here on this side like that. So let me go ahead and highlight those electrons. So the electrons in blue are right here. And these two drawings are equivalent after I put in my other electrons right there. So the dot structures are just an imperfect way of representing the molecule."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and highlight those electrons. So the electrons in blue are right here. And these two drawings are equivalent after I put in my other electrons right there. So the dot structures are just an imperfect way of representing the molecule. So I could show those pi electrons on the left. I could show them on the right. It's really the same thing."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the dot structures are just an imperfect way of representing the molecule. So I could show those pi electrons on the left. I could show them on the right. It's really the same thing. Once I draw this picture, I'm now able to draw another resonance structure. So I can draw another resonance structure from this one right here. I could move these electrons over here, move these electrons over here, and then finally move these electrons over here."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's really the same thing. Once I draw this picture, I'm now able to draw another resonance structure. So I can draw another resonance structure from this one right here. I could move these electrons over here, move these electrons over here, and then finally move these electrons over here. And so when I go ahead and draw the resulting dot structure, now I would have, let's see, these pi electrons are still here. And then going around my ring, it would look like this. So there are a total of three resonance structures that you can draw for naphthalene, again, showing the delocalization of those 10 pi electrons and showing you a little bit about why naphthalene does exhibit some aromatic stability."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I could move these electrons over here, move these electrons over here, and then finally move these electrons over here. And so when I go ahead and draw the resulting dot structure, now I would have, let's see, these pi electrons are still here. And then going around my ring, it would look like this. So there are a total of three resonance structures that you can draw for naphthalene, again, showing the delocalization of those 10 pi electrons and showing you a little bit about why naphthalene does exhibit some aromatic stability. It's not quite as aromatic as benzene. All of benzene's bonds have the exact same length. But naphthalene is shown to have some aromatic stability."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there are a total of three resonance structures that you can draw for naphthalene, again, showing the delocalization of those 10 pi electrons and showing you a little bit about why naphthalene does exhibit some aromatic stability. It's not quite as aromatic as benzene. All of benzene's bonds have the exact same length. But naphthalene is shown to have some aromatic stability. Naphthalene is the simplest example of what's called a polycyclic aromatic hydrocarbon. And there are several different examples of polycyclic aromatic hydrocarbons. Another example would be something like anthracene."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But naphthalene is shown to have some aromatic stability. Naphthalene is the simplest example of what's called a polycyclic aromatic hydrocarbon. And there are several different examples of polycyclic aromatic hydrocarbons. Another example would be something like anthracene. And so there are many, many examples of ring systems that contain fused benzene-like rings throughout the system. But instead of focusing on those, I wanted to do another example, which is an isomer of naphthalene. And it's called azelene."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Another example would be something like anthracene. And so there are many, many examples of ring systems that contain fused benzene-like rings throughout the system. But instead of focusing on those, I wanted to do another example, which is an isomer of naphthalene. And it's called azelene. And azelene is a beautiful blue hydrocarbon, which is extremely rare in organic chemistry to have a hydrocarbon that's blue. It also has some other interesting properties. It has an increased dipole moment associated with the molecule."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it's called azelene. And azelene is a beautiful blue hydrocarbon, which is extremely rare in organic chemistry to have a hydrocarbon that's blue. It also has some other interesting properties. It has an increased dipole moment associated with the molecule. And there's also increased electron density on the five-membered ring. So over here on the left, we have azelene. And here's the five-membered ring over here on the left."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It has an increased dipole moment associated with the molecule. And there's also increased electron density on the five-membered ring. So over here on the left, we have azelene. And here's the five-membered ring over here on the left. And it turns out there are more electrons on the five-membered ring than we would expect, giving it a larger dipole moment. And if we think about a possible resonance structure for azelene, we can figure out why. So if I took these pi electrons right here and moved them in here, that would push these electrons off onto this carbon."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And here's the five-membered ring over here on the left. And it turns out there are more electrons on the five-membered ring than we would expect, giving it a larger dipole moment. And if we think about a possible resonance structure for azelene, we can figure out why. So if I took these pi electrons right here and moved them in here, that would push these electrons off onto this carbon. So if I go ahead and draw the resulting resonance structure, I would have an ion that looks like this. So if I think about my formal charges, if I think about these electrons in blue right here, those are going to go off onto that top carbon. So that top carbon is going to get a lone pair of electrons, which gives that top carbon a negative 1 formal charge."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I took these pi electrons right here and moved them in here, that would push these electrons off onto this carbon. So if I go ahead and draw the resulting resonance structure, I would have an ion that looks like this. So if I think about my formal charges, if I think about these electrons in blue right here, those are going to go off onto that top carbon. So that top carbon is going to get a lone pair of electrons, which gives that top carbon a negative 1 formal charge. So there's a negative formal charge on that carbon. And then if I think about this carbon over here, this carbon lost a bond. And so that's going to end up with a positive charge."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that top carbon is going to get a lone pair of electrons, which gives that top carbon a negative 1 formal charge. So there's a negative formal charge on that carbon. And then if I think about this carbon over here, this carbon lost a bond. And so that's going to end up with a positive charge. So we have a carbocation right here like that. And if I analyze this resonance structure, it has two formal charges in it. But if I look over on the right, I can see on the right, there's a seven-membered ring on the right."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's going to end up with a positive charge. So we have a carbocation right here like that. And if I analyze this resonance structure, it has two formal charges in it. But if I look over on the right, I can see on the right, there's a seven-membered ring on the right. And if I look at it, I can see there are six pi electrons. And then right here, I have a carbocation. And so this is one of the examples we did in the last video."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But if I look over on the right, I can see on the right, there's a seven-membered ring on the right. And if I look at it, I can see there are six pi electrons. And then right here, I have a carbocation. And so this is one of the examples we did in the last video. So every carbon is sp2 hybridized. And so we have overlapping p orbitals. And we have a total of six pi electrons."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this is one of the examples we did in the last video. So every carbon is sp2 hybridized. And so we have overlapping p orbitals. And we have a total of six pi electrons. And so this seven-membered ring is aromatic. If I look over here on the left, I can see that I have a five-membered ring. And I have some pi electrons right here."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have a total of six pi electrons. And so this seven-membered ring is aromatic. If I look over here on the left, I can see that I have a five-membered ring. And I have some pi electrons right here. And the pi electrons right here, as we saw in the example in naphthalene, are actually being shared by both rings. So I could pretend like those electrons are right here on my ring. And then this ring also has electrons like that with a negative 1 formal charge."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I have some pi electrons right here. And the pi electrons right here, as we saw in the example in naphthalene, are actually being shared by both rings. So I could pretend like those electrons are right here on my ring. And then this ring also has electrons like that with a negative 1 formal charge. And again, in the last video, we saw that this ion is aromatic. And it's a total of six pi electrons. So I'll go ahead and highlight those."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then this ring also has electrons like that with a negative 1 formal charge. And again, in the last video, we saw that this ion is aromatic. And it's a total of six pi electrons. So I'll go ahead and highlight those. So these, these, and these are all pi electrons when you think about resonance structures. And so six pi electrons and all the carbons turn out to be sp2 hybridized. Again, look at the previous video for a much more detailed explanation as to why these two ions are aromatic."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and highlight those. So these, these, and these are all pi electrons when you think about resonance structures. And so six pi electrons and all the carbons turn out to be sp2 hybridized. Again, look at the previous video for a much more detailed explanation as to why these two ions are aromatic. And so since these ions are aromatic, they have some aromatic stability. And this resonance structure over here on the right is a much greater contributor to the overall picture of the molecule. And the negative charge is delocalized throughout this 5-membered ring over here."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Again, look at the previous video for a much more detailed explanation as to why these two ions are aromatic. And so since these ions are aromatic, they have some aromatic stability. And this resonance structure over here on the right is a much greater contributor to the overall picture of the molecule. And the negative charge is delocalized throughout this 5-membered ring over here. And the positive charge is delocalized or spread out throughout this 7-membered ring. And that is what gives azoline its larger dipole moment. So there's a larger dipole moment in azoline than expected because of the fact that this would give us two aromatic rings, which confers, of course, extra stability."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the negative charge is delocalized throughout this 5-membered ring over here. And the positive charge is delocalized or spread out throughout this 7-membered ring. And that is what gives azoline its larger dipole moment. So there's a larger dipole moment in azoline than expected because of the fact that this would give us two aromatic rings, which confers, of course, extra stability. And so once again, this ion down here was the cyclopentadienyl anion. And then this cation over here was the cycloheptatrienyl cation from the previous video. The redistribution of these electrons allows azoline to absorb in the orange region, which is difficult for most organic molecules."}, {"video_title": "Aromatic stability V Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So there's a larger dipole moment in azoline than expected because of the fact that this would give us two aromatic rings, which confers, of course, extra stability. And so once again, this ion down here was the cyclopentadienyl anion. And then this cation over here was the cycloheptatrienyl cation from the previous video. The redistribution of these electrons allows azoline to absorb in the orange region, which is difficult for most organic molecules. Because it's a longer wavelength, and that allows it to reflect in the blue region, which is, again, rare, especially for a hydrocarbon. And the fact that it's blue is where this part of the name comes in there, like azure as in blue. So these are just two examples of some ring systems that also exhibit some form of aromatic stability."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Probably, and I would say it ranks their top three, because it really enabled people like Hubble to start realizing that the universe is expanding. Or even being able to think about how to measure distances to objects in space well beyond the reach of our tools with parallax. We saw with parallax, you have to have extremely sensitive instruments just to even measure distances to stars relatively close to us. Very sensitive instruments to get to stars maybe further out into our galaxy. And we don't have the instruments even today to measure things beyond our galaxy. But because of Henrietta Swan Leavitt, we were able to approximate or get good senses of the distance to objects beyond our galaxy. So let's just think about what she did."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Very sensitive instruments to get to stars maybe further out into our galaxy. And we don't have the instruments even today to measure things beyond our galaxy. But because of Henrietta Swan Leavitt, we were able to approximate or get good senses of the distance to objects beyond our galaxy. So let's just think about what she did. So her job was literally to classify stars in the large Magellanic, I have trouble saying that, Magellanic cloud and the small Magellanic clouds. And this is what they look like from the southern hemisphere. This is the large right over here."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So let's just think about what she did. So her job was literally to classify stars in the large Magellanic, I have trouble saying that, Magellanic cloud and the small Magellanic clouds. And this is what they look like from the southern hemisphere. This is the large right over here. And this is the small right over here. And remember, this is before Hubble realized or showed the world that there are stars beyond our galaxy, that there are galaxies beyond our galaxy. So at this point in time, people didn't even fully appreciate that these were separate galaxies."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This is the large right over here. And this is the small right over here. And remember, this is before Hubble realized or showed the world that there are stars beyond our galaxy, that there are galaxies beyond our galaxy. So at this point in time, people didn't even fully appreciate that these were separate galaxies. We just said, hey, these are kind of these blobs or these clusters of stars that we see in the southern hemisphere. And just to get a sense of where they are relative to our galaxy, the Milky Way galaxy, this is obviously not an actual picture. We can't take a picture from this vantage point."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So at this point in time, people didn't even fully appreciate that these were separate galaxies. We just said, hey, these are kind of these blobs or these clusters of stars that we see in the southern hemisphere. And just to get a sense of where they are relative to our galaxy, the Milky Way galaxy, this is obviously not an actual picture. We can't take a picture from this vantage point. This would have to be very, very far away. But this is the Milky Way right here. And this is the small Magellanic cloud."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "We can't take a picture from this vantage point. This would have to be very, very far away. But this is the Milky Way right here. And this is the small Magellanic cloud. And this is the large Magellanic cloud. I'm getting better at saying it. So her job was literally just to classify the different stars that she saw."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And this is the small Magellanic cloud. And this is the large Magellanic cloud. I'm getting better at saying it. So her job was literally just to classify the different stars that she saw. But while she was classifying, she looked at these things called variables. It turns out what she was looking at were a class of stars called Cepheid or Cepheid variable stars. And what's interesting about them is two things."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So her job was literally just to classify the different stars that she saw. But while she was classifying, she looked at these things called variables. It turns out what she was looking at were a class of stars called Cepheid or Cepheid variable stars. And what's interesting about them is two things. They're super duper bright. They're up to 30,000 times as luminous as the sun. And they're 5 to 20 times more massive than the sun."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And what's interesting about them is two things. They're super duper bright. They're up to 30,000 times as luminous as the sun. And they're 5 to 20 times more massive than the sun. 5 to 20 times the sun's mass. But what makes them interesting is one, they're really bright. So you can see them from really far away."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And they're 5 to 20 times more massive than the sun. 5 to 20 times the sun's mass. But what makes them interesting is one, they're really bright. So you can see them from really far away. You can see these Cepheid variable stars in other galaxies. In fact, we can see it well beyond even the small Magellanic cloud or the large Magellanic cloud. But you can see these stars in other galaxies."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So you can see them from really far away. You can see these Cepheid variable stars in other galaxies. In fact, we can see it well beyond even the small Magellanic cloud or the large Magellanic cloud. But you can see these stars in other galaxies. And what's even more interesting about them is that their intensity is variable, that they become brighter and dimmer with a well-defined period. So if you're looking at a Cepheid variable star, and this is just kind of a simulation, a very cheap simulation, it might look like this. And then over the course of the next three, four days, it might reduce in intensity to something like this."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But you can see these stars in other galaxies. And what's even more interesting about them is that their intensity is variable, that they become brighter and dimmer with a well-defined period. So if you're looking at a Cepheid variable star, and this is just kind of a simulation, a very cheap simulation, it might look like this. And then over the course of the next three, four days, it might reduce in intensity to something like this. And then after three, four days again, it will look like this. And then it'll look like this again. So its actual intensity is going up and down with a well-defined period."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And then over the course of the next three, four days, it might reduce in intensity to something like this. And then after three, four days again, it will look like this. And then it'll look like this again. So its actual intensity is going up and down with a well-defined period. So if this takes three days and this is another three days, then the period, one entire cycle of its going from low intensity back to high intensity is going to be six days. So this is a six-day period. And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she plotted, she assumed that things, everything in each of these clouds are roughly the same distance away."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So its actual intensity is going up and down with a well-defined period. So if this takes three days and this is another three days, then the period, one entire cycle of its going from low intensity back to high intensity is going to be six days. So this is a six-day period. And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she plotted, she assumed that things, everything in each of these clouds are roughly the same distance away. Everything in the large Magellanic cloud is roughly the same distance away. And it's obviously not exact. This is an entire galaxy, so you have obviously things further away in that galaxy and things closer up."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And what Henrietta Leavitt saw, and this wasn't an obvious thing to do, she plotted, she assumed that things, everything in each of these clouds are roughly the same distance away. Everything in the large Magellanic cloud is roughly the same distance away. And it's obviously not exact. This is an entire galaxy, so you have obviously things further away in that galaxy and things closer up. You have stars here and here, and their distance isn't going to be exactly the same to us, that we're sitting maybe over here someplace. But it's going to be close. It wasn't a bad approximation."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "This is an entire galaxy, so you have obviously things further away in that galaxy and things closer up. You have stars here and here, and their distance isn't going to be exactly the same to us, that we're sitting maybe over here someplace. But it's going to be close. It wasn't a bad approximation. And by making that assumption, she saw something pretty neat. So let me plot this right over here. So she plotted on the horizontal axis, she plotted the relative luminosity."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It wasn't a bad approximation. And by making that assumption, she saw something pretty neat. So let me plot this right over here. So she plotted on the horizontal axis, she plotted the relative luminosity. So really, the only way that she could measure this is just how bright did they look to her? And she's assuming that they're same distance. So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So she plotted on the horizontal axis, she plotted the relative luminosity. So really, the only way that she could measure this is just how bright did they look to her? And she's assuming that they're same distance. So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer. So if you assume that they're all roughly the same distance, then how bright it is will tell you how bright it is at the actual star. So she plotted relative luminosity of a star on one axis, and on the other axis, she plotted the period of these variable stars. And what I'm going to do is I'm going to do this on a logarithmic scale."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So obviously, if you have a brighter star, but it's much, much further away, it's going to look dimmer. So if you assume that they're all roughly the same distance, then how bright it is will tell you how bright it is at the actual star. So she plotted relative luminosity of a star on one axis, and on the other axis, she plotted the period of these variable stars. And what I'm going to do is I'm going to do this on a logarithmic scale. So let's say that this is in days. So this is one day, this is 10 days, this is 100 days, right over here. It's a logarithmic scale because I'm going up in powers of 10."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And what I'm going to do is I'm going to do this on a logarithmic scale. So let's say that this is in days. So this is one day, this is 10 days, this is 100 days, right over here. It's a logarithmic scale because I'm going up in powers of 10. I could say that if we take the log of these, this would be 0, this would be 1, this would be 2. And so that's what I'm using as a scale. So I'm using the log of the period, or I'm just marking them as 1, 10, 100, but I'm giving each of these factors of 10 an equal spacing."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "It's a logarithmic scale because I'm going up in powers of 10. I could say that if we take the log of these, this would be 0, this would be 1, this would be 2. And so that's what I'm using as a scale. So I'm using the log of the period, or I'm just marking them as 1, 10, 100, but I'm giving each of these factors of 10 an equal spacing. When you plot it on this scale, the relative luminosity versus the period, she got a plot that looks something like this. This is obviously not exact. She got a plot that looks something like this."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So I'm using the log of the period, or I'm just marking them as 1, 10, 100, but I'm giving each of these factors of 10 an equal spacing. When you plot it on this scale, the relative luminosity versus the period, she got a plot that looks something like this. This is obviously not exact. She got a plot that looks something like this. It was a fairly linear relationship when you plot the relative luminosity against the log of the period. So this is obviously a logarithmic scale over here. And so you could fit a line."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "She got a plot that looks something like this. It was a fairly linear relationship when you plot the relative luminosity against the log of the period. So this is obviously a logarithmic scale over here. And so you could fit a line. And why I'd argue, and I think most people would argue, this is one of the most important discoveries in astronomy, is if you know, because think about what the problem here is. We can look at all of these stars in space. Let's say you look at a fraction of the sky and you look at something that looks like that."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And so you could fit a line. And why I'd argue, and I think most people would argue, this is one of the most important discoveries in astronomy, is if you know, because think about what the problem here is. We can look at all of these stars in space. Let's say you look at a fraction of the sky and you look at something that looks like that. So it's really bright. And then you see something dim that looks like that. So if you have a very superficial understanding, you say, oh, this star is brighter."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Let's say you look at a fraction of the sky and you look at something that looks like that. So it's really bright. And then you see something dim that looks like that. So if you have a very superficial understanding, you say, oh, this star is brighter. You would say that this is a fundamentally brighter star. But how do you know that? Maybe instead of being brighter, maybe it's just a dimmer, closer star."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "So if you have a very superficial understanding, you say, oh, this star is brighter. You would say that this is a fundamentally brighter star. But how do you know that? Maybe instead of being brighter, maybe it's just a dimmer, closer star. Maybe this is an entire galaxy, but it's so far away that you can't even tell. But all of a sudden, by the work that Henrietta Leavitt did, if you see one of these Cepheid variable stars in another galaxy, you know its relative brightness compared to other Cepheid variable stars. And so if you can place just one of these Cepheid variable stars, if you know exactly the distance to one of them, and then you know its absolute luminosity, you then know the absolute luminosity of any other Cepheid variable stars."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "Maybe instead of being brighter, maybe it's just a dimmer, closer star. Maybe this is an entire galaxy, but it's so far away that you can't even tell. But all of a sudden, by the work that Henrietta Leavitt did, if you see one of these Cepheid variable stars in another galaxy, you know its relative brightness compared to other Cepheid variable stars. And so if you can place just one of these Cepheid variable stars, if you know exactly the distance to one of them, and then you know its absolute luminosity, you then know the absolute luminosity of any other Cepheid variable stars. So let's say using parallax, which is our other tool, we find, let's say there's some star in our galaxy. And let's say using parallax, we're able to come up with a pretty good measure that it is, I don't know, let's say it's 100 light years away. And this star is a Cepheid variable star."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And so if you can place just one of these Cepheid variable stars, if you know exactly the distance to one of them, and then you know its absolute luminosity, you then know the absolute luminosity of any other Cepheid variable stars. So let's say using parallax, which is our other tool, we find, let's say there's some star in our galaxy. And let's say using parallax, we're able to come up with a pretty good measure that it is, I don't know, let's say it's 100 light years away. And this star is a Cepheid variable star. And let's say its period is one day. So we now know something interesting. We know variable stars with a period of one day at 100 light years away will look like this."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "And this star is a Cepheid variable star. And let's say its period is one day. So we now know something interesting. We know variable stars with a period of one day at 100 light years away will look like this. Will look like this drawing right over here. So if we later on see a Cepheid variable star with a period of one day, so it gets brighter and dim over the course of one day, and maybe it's redshifted as well, but maybe it looks a little bit dimmer, it looks like this. We now know that if it was 100 light years away, it would have this luminosity."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "We know variable stars with a period of one day at 100 light years away will look like this. Will look like this drawing right over here. So if we later on see a Cepheid variable star with a period of one day, so it gets brighter and dim over the course of one day, and maybe it's redshifted as well, but maybe it looks a little bit dimmer, it looks like this. We now know that if it was 100 light years away, it would have this luminosity. So based on how much dimmer it is, we can then figure out how much further away this Cepheid variable star is. If that confuses you a little bit, I'll do a little bit more details in the next few videos so we can get a closer sense of how the math would work. But this was a big discovery, just discovering this class of stars, this Cepheid variable class."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "We now know that if it was 100 light years away, it would have this luminosity. So based on how much dimmer it is, we can then figure out how much further away this Cepheid variable star is. If that confuses you a little bit, I'll do a little bit more details in the next few videos so we can get a closer sense of how the math would work. But this was a big discovery, just discovering this class of stars, this Cepheid variable class. She wasn't the one who discovered them. People knew before her that there were these stars that got brighter and dimmer. But what her big discovery was is seeing this linear relationship between the relative luminosity of these stars and their period."}, {"video_title": "Cepheid variables 1 Stars, black holes and galaxies Cosmology & Astronomy Khan Academy (2).mp3", "Sentence": "But this was a big discovery, just discovering this class of stars, this Cepheid variable class. She wasn't the one who discovered them. People knew before her that there were these stars that got brighter and dimmer. But what her big discovery was is seeing this linear relationship between the relative luminosity of these stars and their period. Because then, if we see Cepheid variable stars in completely different galaxies or galactic clusters, by looking at their period, we know what their real relative luminosity is. And then we could guess how far those things really are. Or we could estimate how far those things really are."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So one pair of electrons on the oxygen pick up this proton, leaving these electrons behind on the A. So oxygen, oxygen is now bonded to three hydrogens, right? So it picked up a proton. That's gonna give this oxygen a plus one formal charge, and we can follow those electrons. So these two electrons in red here are gonna pick up this proton, forming this bond. So we make hydronium, H3O+. And these electrons in green right here are going to come off onto the A to make A minus."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "That's gonna give this oxygen a plus one formal charge, and we can follow those electrons. So these two electrons in red here are gonna pick up this proton, forming this bond. So we make hydronium, H3O+. And these electrons in green right here are going to come off onto the A to make A minus. Let's go ahead and draw that in. So we're gonna make A minus. Let me draw these electrons in green and give this a negative charge like that."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And these electrons in green right here are going to come off onto the A to make A minus. Let's go ahead and draw that in. So we're gonna make A minus. Let me draw these electrons in green and give this a negative charge like that. So let's analyze what happened. HA donated a proton. So this is our Bronsted-Lowry acid."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "Let me draw these electrons in green and give this a negative charge like that. So let's analyze what happened. HA donated a proton. So this is our Bronsted-Lowry acid. Once HA donates a proton, right, we're left with the conjugate base, which is A minus. Water, H2O, accepted a proton. So this is our Bronsted-Lowry base."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So this is our Bronsted-Lowry acid. Once HA donates a proton, right, we're left with the conjugate base, which is A minus. Water, H2O, accepted a proton. So this is our Bronsted-Lowry base. And then once H2O accepts a proton, we turn it into hydronium, H3O+. So this is the conjugate acid. So H3O+, the conjugate acid, and then A minus would be a base."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So this is our Bronsted-Lowry base. And then once H2O accepts a proton, we turn it into hydronium, H3O+. So this is the conjugate acid. So H3O+, the conjugate acid, and then A minus would be a base. So if you think about the reverse reaction, H3O+, donating a proton to A minus, then you would get back H2O and HA. Once this reaction reaches equilibrium, we can write an equilibrium expression. And we're gonna consider the stuff on the left to be the reactants."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So H3O+, the conjugate acid, and then A minus would be a base. So if you think about the reverse reaction, H3O+, donating a proton to A minus, then you would get back H2O and HA. Once this reaction reaches equilibrium, we can write an equilibrium expression. And we're gonna consider the stuff on the left to be the reactants. So we're gonna think about the forward reaction and the stuff on the right to be the products. So let's write our equilibrium expression. And so we write our equilibrium constant."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And we're gonna consider the stuff on the left to be the reactants. So we're gonna think about the forward reaction and the stuff on the right to be the products. So let's write our equilibrium expression. And so we write our equilibrium constant. And now we're gonna write Ka, which we call the acid ionization constant. So this is the acid ionization constant. Or you might hear acid dissociation constant."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And so we write our equilibrium constant. And now we're gonna write Ka, which we call the acid ionization constant. So this is the acid ionization constant. Or you might hear acid dissociation constant. So acid dissociation. So either one is fine. All right, and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "Or you might hear acid dissociation constant. So acid dissociation. So either one is fine. All right, and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants. So over here for our products, we have H3O+. So let's write the concentration of hydronium, H3O+, times the concentration of A minus. So times the concentration of A minus."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "All right, and we know when we're writing an equilibrium expression, we're gonna put the concentration of products over the concentration of reactants. So over here for our products, we have H3O+. So let's write the concentration of hydronium, H3O+, times the concentration of A minus. So times the concentration of A minus. All over the concentration of our reactants. So we have HA over here. So we have HA, so we write that in."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So times the concentration of A minus. All over the concentration of our reactants. So we have HA over here. So we have HA, so we write that in. And then for water, we leave water out of our equilibrium expression. It's a pure liquid, its concentration doesn't change, and so we leave H2O out of our equilibrium expression. All right, so let's use this idea of writing an ionization constant, and let's apply this to a strong acid."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So we have HA, so we write that in. And then for water, we leave water out of our equilibrium expression. It's a pure liquid, its concentration doesn't change, and so we leave H2O out of our equilibrium expression. All right, so let's use this idea of writing an ionization constant, and let's apply this to a strong acid. HCl is gonna function as a Bronsted-Lowry acid, and donate a proton to water, which is going to be our Bronsted-Lowry base. And so we could think about a lone pair of electrons in the oxygen taking our proton, right, leaving those electrons behind. And so the oxygen is now bonded to three hydrogens because it picked up a proton, giving this a plus one charge."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "All right, so let's use this idea of writing an ionization constant, and let's apply this to a strong acid. HCl is gonna function as a Bronsted-Lowry acid, and donate a proton to water, which is going to be our Bronsted-Lowry base. And so we could think about a lone pair of electrons in the oxygen taking our proton, right, leaving those electrons behind. And so the oxygen is now bonded to three hydrogens because it picked up a proton, giving this a plus one charge. And so once again, let's follow those electrons in red. So this electron pair picks up this proton to form this bond, so we form H3O+, or hydronium. And these electrons in green move off onto the chlorine."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And so the oxygen is now bonded to three hydrogens because it picked up a proton, giving this a plus one charge. And so once again, let's follow those electrons in red. So this electron pair picks up this proton to form this bond, so we form H3O+, or hydronium. And these electrons in green move off onto the chlorine. So let's show that. So we form the chloride anion. So let me go ahead and draw in the electrons in green."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And these electrons in green move off onto the chlorine. So let's show that. So we form the chloride anion. So let me go ahead and draw in the electrons in green. And let me go ahead and write a negative one charge here like that. So another way to represent this acid-base reaction would just be to write out H2O plus HCl gives us H3O+, plus Cl-. So this is just a faster way of doing it."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So let me go ahead and draw in the electrons in green. And let me go ahead and write a negative one charge here like that. So another way to represent this acid-base reaction would just be to write out H2O plus HCl gives us H3O+, plus Cl-. So this is just a faster way of doing it. And HCl is a strong acid. Strong acids donate protons very easily. And so we can say this process occurs 100%."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So this is just a faster way of doing it. And HCl is a strong acid. Strong acids donate protons very easily. And so we can say this process occurs 100%. So we get 100% ionization. The equilibrium is so far to the right that I just drew this one arrow down over here. So we get approximately 100% ionization."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And so we can say this process occurs 100%. So we get 100% ionization. The equilibrium is so far to the right that I just drew this one arrow down over here. So we get approximately 100% ionization. So everything turns into our products here. And let's go ahead and write our equilibrium expression. So Ka is equal to concentration of H3O+, so concentration of our products, times concentration of Cl-."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So we get approximately 100% ionization. So everything turns into our products here. And let's go ahead and write our equilibrium expression. So Ka is equal to concentration of H3O+, so concentration of our products, times concentration of Cl-. All over, we have HCl and we leave out water. So if we think about approximately 100% ionization, we have all products here. So we have a very, very large number in the numerator and an extremely small number in the denominator."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So Ka is equal to concentration of H3O+, so concentration of our products, times concentration of Cl-. All over, we have HCl and we leave out water. So if we think about approximately 100% ionization, we have all products here. So we have a very, very large number in the numerator and an extremely small number in the denominator. So if we think about what that does for your Ka, that's gonna give you an extremely high value for your Ka. So Ka is much, much, much greater than one here. And so that's how we recognize a strong acid."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So we have a very, very large number in the numerator and an extremely small number in the denominator. So if we think about what that does for your Ka, that's gonna give you an extremely high value for your Ka. So Ka is much, much, much greater than one here. And so that's how we recognize a strong acid. So an acid ionization constant that's much, much greater than one. Now let's think about the conjugate base. So let's go back up here."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "And so that's how we recognize a strong acid. So an acid ionization constant that's much, much greater than one. Now let's think about the conjugate base. So let's go back up here. So we had HCl and Cl- as our conjugate acid-base pair. And the stronger the acid, the weaker the conjugate base. So HCl is a strong acid."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So let's go back up here. So we had HCl and Cl- as our conjugate acid-base pair. And the stronger the acid, the weaker the conjugate base. So HCl is a strong acid. So Cl- is a weak conjugate base. So let me write that here. So the stronger the acid, so stronger the acid, weaker the conjugate base."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So HCl is a strong acid. So Cl- is a weak conjugate base. So let me write that here. So the stronger the acid, so stronger the acid, weaker the conjugate base. And one way to think about that is, if I look at this reaction, we can think about competing base strength. So here we have Bronsted-Lowry base. Water's acting as a Bronsted-Lowry base and accepting a proton."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So the stronger the acid, so stronger the acid, weaker the conjugate base. And one way to think about that is, if I look at this reaction, we can think about competing base strength. So here we have Bronsted-Lowry base. Water's acting as a Bronsted-Lowry base and accepting a proton. And over here, if you think about the reverse reaction, the chloride anion would be trying to pick up a proton from hydronium for the reverse reaction here. But since HCl is so good at donating protons, that means that the chloride anion is not very good at accepting them. So the stronger the acid, the weaker the conjugate base."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "Water's acting as a Bronsted-Lowry base and accepting a proton. And over here, if you think about the reverse reaction, the chloride anion would be trying to pick up a proton from hydronium for the reverse reaction here. But since HCl is so good at donating protons, that means that the chloride anion is not very good at accepting them. So the stronger the acid, the weaker the conjugate base. Water's a much stronger base than the chloride anion. And finally, let's look at acetic acid. So acetic acid is going to be our Bronsted-Lowry acid, and this is going to be the acidic proton."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So the stronger the acid, the weaker the conjugate base. Water's a much stronger base than the chloride anion. And finally, let's look at acetic acid. So acetic acid is going to be our Bronsted-Lowry acid, and this is going to be the acidic proton. So water's gonna function as a Bronsted-Lowry base. And a lone pair of electrons in the oxygen is going to take this acidic proton, leaving these electrons behind on the oxygen. So let's go ahead and draw our products."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So acetic acid is going to be our Bronsted-Lowry acid, and this is going to be the acidic proton. So water's gonna function as a Bronsted-Lowry base. And a lone pair of electrons in the oxygen is going to take this acidic proton, leaving these electrons behind on the oxygen. So let's go ahead and draw our products. So we would form the acetate anions. Let me go ahead and draw in the acetate anions. So negative one charge on the oxygen."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So let's go ahead and draw our products. So we would form the acetate anions. Let me go ahead and draw in the acetate anions. So negative one charge on the oxygen. And let's show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form hydronium."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So negative one charge on the oxygen. And let's show those electrons. These electrons in green move off onto the oxygen right here, giving it a negative charge. We're also gonna form hydronium. So H3O+. So let me go ahead and draw in hydronium. So a plus one formal charge on the oxygen."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "We're also gonna form hydronium. So H3O+. So let me go ahead and draw in hydronium. So a plus one formal charge on the oxygen. And let's show those electrons in red. So this electron pair picks up the acidic proton, forming this bond, and we get H3O+. So another way to write this acid-base reaction would be just to write acetic acid, CH3COOH plus H2O, gives us the acetate anion, CH3COO- plus H3O+."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So a plus one formal charge on the oxygen. And let's show those electrons in red. So this electron pair picks up the acidic proton, forming this bond, and we get H3O+. So another way to write this acid-base reaction would be just to write acetic acid, CH3COOH plus H2O, gives us the acetate anion, CH3COO- plus H3O+. Now, acetic acid is a weak acid, and weak acids don't donate protons very well. And so acetic acid's gonna stay mostly protonated. So when you think about this reaction coming to an equilibrium, you're gonna have a relatively high concentration of your reactants here."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So another way to write this acid-base reaction would be just to write acetic acid, CH3COOH plus H2O, gives us the acetate anion, CH3COO- plus H3O+. Now, acetic acid is a weak acid, and weak acids don't donate protons very well. And so acetic acid's gonna stay mostly protonated. So when you think about this reaction coming to an equilibrium, you're gonna have a relatively high concentration of your reactants here. So when we write the equilibrium expression, Ka is equal to the concentration of your product, so CH3COO- times the concentration of H3O+, all over the concentration of acetic acid, because we leave water out. So all over the concentration of acetic acid. The equilibrium lies to the left, because acetic acid is not good at donating this proton."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "So when you think about this reaction coming to an equilibrium, you're gonna have a relatively high concentration of your reactants here. So when we write the equilibrium expression, Ka is equal to the concentration of your product, so CH3COO- times the concentration of H3O+, all over the concentration of acetic acid, because we leave water out. So all over the concentration of acetic acid. The equilibrium lies to the left, because acetic acid is not good at donating this proton. And so we're going to get a very large number for the denominator, for this concentration. So this is a very large number. And a very small number for the numerator."}, {"video_title": "Ka and acid strength Chemical processes MCAT Khan Academy.mp3", "Sentence": "The equilibrium lies to the left, because acetic acid is not good at donating this proton. And so we're going to get a very large number for the denominator, for this concentration. So this is a very large number. And a very small number for the numerator. So this is a very small number. So if you think about what that does to the Ka, a very small number divided by a very large number, this gives you a Ka value, an ionization constant much less than one. So this value's going to be much less than one."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes when you're trying to synthesize a molecule, you have to use a protecting group. In this video, we're going to talk about how to protect alcohols using a trialkylsilyl group. Let's say our goal is to make this target compound over here on the right. We need to start with this compound over here on the left. You might think that this organolithium compound right here would function as a nucleophile. We have a negative one formal charge in this carbon, so this lone pair of electrons is going to be our nucleophile and attack this carbon, which is a little bit partially positive right here. These electrons will kick off onto your bromine and you would end up adding this carbon and this carbon on to form your target molecule, so like that."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We need to start with this compound over here on the left. You might think that this organolithium compound right here would function as a nucleophile. We have a negative one formal charge in this carbon, so this lone pair of electrons is going to be our nucleophile and attack this carbon, which is a little bit partially positive right here. These electrons will kick off onto your bromine and you would end up adding this carbon and this carbon on to form your target molecule, so like that. Unfortunately, this is not the reaction that occurs because not only can organolithium compounds be strong nucleophiles, they can also be strong bases. What would actually happen is this lone pair of electrons here would function as a base, take this proton, leaving these electrons behind in the oxygen to form an alkoxide. Really, you would form this product over here."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "These electrons will kick off onto your bromine and you would end up adding this carbon and this carbon on to form your target molecule, so like that. Unfortunately, this is not the reaction that occurs because not only can organolithium compounds be strong nucleophiles, they can also be strong bases. What would actually happen is this lone pair of electrons here would function as a base, take this proton, leaving these electrons behind in the oxygen to form an alkoxide. Really, you would form this product over here. We would have an oxygen. The oxygen would have now three lone pairs of electrons around it, giving it a negative one formal charge and we would have lithium plus. We would form an alkoxide product instead."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Really, you would form this product over here. We would have an oxygen. The oxygen would have now three lone pairs of electrons around it, giving it a negative one formal charge and we would have lithium plus. We would form an alkoxide product instead. The point of a protecting group is we need to protect this hydroxyl group to prevent it from reacting. If we can somehow protect this group, we can allow our reaction to occur at this portion of the molecule and then we could remove our protecting group to form our target compound. That's the idea behind it."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We would form an alkoxide product instead. The point of a protecting group is we need to protect this hydroxyl group to prevent it from reacting. If we can somehow protect this group, we can allow our reaction to occur at this portion of the molecule and then we could remove our protecting group to form our target compound. That's the idea behind it. Let's go ahead and show how we can use a protecting group. Over here on the right, we have, this would be t-butyl dimethylsilyl chloride. There's a tert-butyl group attached to a silicon and then there's two methyl groups attached to the silicon and also a chlorine."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That's the idea behind it. Let's go ahead and show how we can use a protecting group. Over here on the right, we have, this would be t-butyl dimethylsilyl chloride. There's a tert-butyl group attached to a silicon and then there's two methyl groups attached to the silicon and also a chlorine. This would be t-butyl or tert-butyl dimethylsilyl chloride. TBDMSCL. If you think about this silicon, the silicon is bonded to a carbon here, a carbon here, a carbon here and a chlorine."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "There's a tert-butyl group attached to a silicon and then there's two methyl groups attached to the silicon and also a chlorine. This would be t-butyl or tert-butyl dimethylsilyl chloride. TBDMSCL. If you think about this silicon, the silicon is bonded to a carbon here, a carbon here, a carbon here and a chlorine. All of those, the carbon and the chlorine are more electronegative than the silicon. They're going to withdraw some electron density from the silicon, making the silicon partially positive. The silicon can function as an electrophile."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If you think about this silicon, the silicon is bonded to a carbon here, a carbon here, a carbon here and a chlorine. All of those, the carbon and the chlorine are more electronegative than the silicon. They're going to withdraw some electron density from the silicon, making the silicon partially positive. The silicon can function as an electrophile. It's electrophilic. We can get some electrons from the oxygen. The alcohol over here is going to function as a nucleophile and the lone pair of electrons is going to attack the silicon and then these electrons would kick off onto your chlorine."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "The silicon can function as an electrophile. It's electrophilic. We can get some electrons from the oxygen. The alcohol over here is going to function as a nucleophile and the lone pair of electrons is going to attack the silicon and then these electrons would kick off onto your chlorine. We would lose HCl in the process. The imidazole, one of the things the imidazole does is help to remove the HCl and the mechanism is a little more complicated than what I've shown, but this is just a simple way of thinking about it. You have nucleophile, electrophile and you're going to put your protecting group onto your alcohol."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "The alcohol over here is going to function as a nucleophile and the lone pair of electrons is going to attack the silicon and then these electrons would kick off onto your chlorine. We would lose HCl in the process. The imidazole, one of the things the imidazole does is help to remove the HCl and the mechanism is a little more complicated than what I've shown, but this is just a simple way of thinking about it. You have nucleophile, electrophile and you're going to put your protecting group onto your alcohol. Let's go ahead and draw the product of this reaction. We would now have our oxygen bonded to the silicon and our oxygen would have two lone pairs of electrons around it. The silicon is bonded to two methyl groups and also a tert-butyl group like that."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "You have nucleophile, electrophile and you're going to put your protecting group onto your alcohol. Let's go ahead and draw the product of this reaction. We would now have our oxygen bonded to the silicon and our oxygen would have two lone pairs of electrons around it. The silicon is bonded to two methyl groups and also a tert-butyl group like that. We put on our protecting group. Sometimes you might see, instead of drawing out all of that stuff around the silicon, you might just see an oxygen and then you might see T-B-D-M-S for our protecting group, which is our T-butyl dimethyl silo protecting group like that. This would be another way of representing that portion of the molecule."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "The silicon is bonded to two methyl groups and also a tert-butyl group like that. We put on our protecting group. Sometimes you might see, instead of drawing out all of that stuff around the silicon, you might just see an oxygen and then you might see T-B-D-M-S for our protecting group, which is our T-butyl dimethyl silo protecting group like that. This would be another way of representing that portion of the molecule. Now that we've added our protecting group, we can go ahead and react with our organolithium compounds. Let me go ahead and draw in our organolithium compound again. We had a carbanion here, which now can function as a nucleophile."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This would be another way of representing that portion of the molecule. Now that we've added our protecting group, we can go ahead and react with our organolithium compounds. Let me go ahead and draw in our organolithium compound again. We had a carbanion here, which now can function as a nucleophile. This lone pair of electrons could attack this carbon right here and these electrons would kick off onto the bromine. We can go ahead and draw what we would get from that. Now we would have, we would add on our triple bond right here."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We had a carbanion here, which now can function as a nucleophile. This lone pair of electrons could attack this carbon right here and these electrons would kick off onto the bromine. We can go ahead and draw what we would get from that. Now we would have, we would add on our triple bond right here. Once again, let's highlight some carbons. This carbon would have added onto here. This carbon is right here."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now we would have, we would add on our triple bond right here. Once again, let's highlight some carbons. This carbon would have added onto here. This carbon is right here. Then we have these electrons formed this bond right here like that. We still have our protecting groups. Let's go ahead and draw that too."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is right here. Then we have these electrons formed this bond right here like that. We still have our protecting groups. Let's go ahead and draw that too. We have our oxygen and bonded to our oxygen, we have our silicon with our methyl groups and also our tert-butyl group like that. Now that we've done the desired reaction, now we can take off our protecting group. We can remove it to form our target compound."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw that too. We have our oxygen and bonded to our oxygen, we have our silicon with our methyl groups and also our tert-butyl group like that. Now that we've done the desired reaction, now we can take off our protecting group. We can remove it to form our target compound. We need to have something that reacts selectively with the silicon here. We're going to use tetrabutylammonium fluoride. Tetrabutylammonium fluoride, which is really just a good source of fluoride anions."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We can remove it to form our target compound. We need to have something that reacts selectively with the silicon here. We're going to use tetrabutylammonium fluoride. Tetrabutylammonium fluoride, which is really just a good source of fluoride anions. I'm going to go ahead and draw in fluoride anion here, which is normally a extremely poor nucleophile. But it's actually selective for silicon. If the fluoride functions as a nucleophile, it's going to attack the silicon here."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Tetrabutylammonium fluoride, which is really just a good source of fluoride anions. I'm going to go ahead and draw in fluoride anion here, which is normally a extremely poor nucleophile. But it's actually selective for silicon. If the fluoride functions as a nucleophile, it's going to attack the silicon here. It can do this for a couple of reasons. Let's talk about those reasons here. First of all, the silicon is bonded to some carbons."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "If the fluoride functions as a nucleophile, it's going to attack the silicon here. It can do this for a couple of reasons. Let's talk about those reasons here. First of all, the silicon is bonded to some carbons. Silicon is bigger than carbon if you look at where it is in the periodic table. The silicon-carbon bonds are longer than we're used to seeing. That means that there's decreased steric hindrance."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "First of all, the silicon is bonded to some carbons. Silicon is bigger than carbon if you look at where it is in the periodic table. The silicon-carbon bonds are longer than we're used to seeing. That means that there's decreased steric hindrance. The silicon is a little bit more exposed and that allows the fluoride anion to attack it a little more. Another factor that allows this is silicon is in the third period on the periodic table. It has vacant d-orbitals."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That means that there's decreased steric hindrance. The silicon is a little bit more exposed and that allows the fluoride anion to attack it a little more. Another factor that allows this is silicon is in the third period on the periodic table. It has vacant d-orbitals. We can go ahead and show a bond forming between the fluorine and the silicon. Let me go ahead and draw what we would get after the fluoride attacks the silicon. We would have this portion of the molecule."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It has vacant d-orbitals. We can go ahead and show a bond forming between the fluorine and the silicon. Let me go ahead and draw what we would get after the fluoride attacks the silicon. We would have this portion of the molecule. We would have our oxygen. Our oxygen bonded to our silicon in this intermediate. Now we could show the fluorine bonded to the silicon like that."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We would have this portion of the molecule. We would have our oxygen. Our oxygen bonded to our silicon in this intermediate. Now we could show the fluorine bonded to the silicon like that. The silicon is still bonded to two methyl groups and also a tert-butyl group like that. This would give the silicon a negative one formal charge. It looks a little bit weird because we see silicon has five bonds to it."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now we could show the fluorine bonded to the silicon like that. The silicon is still bonded to two methyl groups and also a tert-butyl group like that. This would give the silicon a negative one formal charge. It looks a little bit weird because we see silicon has five bonds to it. But that's again, it's okay because of where silicon is on the periodic table. It has those d-orbitals. Forming five bonds for an intermediate is okay."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It looks a little bit weird because we see silicon has five bonds to it. But that's again, it's okay because of where silicon is on the periodic table. It has those d-orbitals. Forming five bonds for an intermediate is okay. It's okay for it to have an expanded octet. Another reason why fluoride can attack the silicon very well is because the bond that forms between fluorine and silicon happens to be very strong. It's a very strong single bond here."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Forming five bonds for an intermediate is okay. It's okay for it to have an expanded octet. Another reason why fluoride can attack the silicon very well is because the bond that forms between fluorine and silicon happens to be very strong. It's a very strong single bond here. We can finish up by kicking these electrons back onto the oxygen and protonating and forming our target compound. We would go ahead and form our target compound here. We would get back our alcohol like that."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "It's a very strong single bond here. We can finish up by kicking these electrons back onto the oxygen and protonating and forming our target compound. We would go ahead and form our target compound here. We would get back our alcohol like that. We also successfully added on this portion of the molecule on the right. Then we would also form, we now have the fluorine bonded to the silicon like that. We selectively removed our protecting group and we formed our target compound."}, {"video_title": "Protection of alcohols Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "We would get back our alcohol like that. We also successfully added on this portion of the molecule on the right. Then we would also form, we now have the fluorine bonded to the silicon like that. We selectively removed our protecting group and we formed our target compound. That's the idea of a protecting group. It allows you to protect one area of the molecule and react with another area of the molecule. It's also nice to have it easily removed to get back your target molecule."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's a sulfur double bonded to an oxygen, and then it has single bonds to two separate chlorines. And then if you want to care, sulfur has six valence electrons, so it has another lone pair right over here. And we'll focus on this because this is really the simplest reaction of starting with the carboxylic acid and forming an acyl halide, and it can be generalized very easily if you turn this methyl group into just a larger chain, and then this will be a larger chain right here. And instead of having a chlorine here, you could do it with other halides as well. So let's think about how this might occur. So let's start, let's have our acetic acid right over here. And then you have this oxygen bonded to a hydrogen, just like that."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And instead of having a chlorine here, you could do it with other halides as well. So let's think about how this might occur. So let's start, let's have our acetic acid right over here. And then you have this oxygen bonded to a hydrogen, just like that. Now this guy's got two lone pairs. And then the thionyl chloride, I'll redraw the thionyl chloride right over here, and I'll set it up in the right position, the thionyl chloride looks like this. And this has a lone pair of electrons right over here."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then you have this oxygen bonded to a hydrogen, just like that. Now this guy's got two lone pairs. And then the thionyl chloride, I'll redraw the thionyl chloride right over here, and I'll set it up in the right position, the thionyl chloride looks like this. And this has a lone pair of electrons right over here. And all of these molecules, chlorine and oxygen, or all of these atoms, chlorine and oxygen, we can look at a periodic table up here. You see sulfur over here, and then to the right of sulfur is chlorine, and above sulfur is oxygen. So both chlorine and oxygen are more electronegative than sulfur."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this has a lone pair of electrons right over here. And all of these molecules, chlorine and oxygen, or all of these atoms, chlorine and oxygen, we can look at a periodic table up here. You see sulfur over here, and then to the right of sulfur is chlorine, and above sulfur is oxygen. So both chlorine and oxygen are more electronegative than sulfur. So sulfur is going to have a partial positive charge there, even though sulfur is a reasonably electronegative atom in its own right. But these other guys are going to be sucking electrons away from it. So you could imagine that this oxygen, or this entire molecule, could act as a nucleophile, with this oxygen right over here giving an electron to the sulfur."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So both chlorine and oxygen are more electronegative than sulfur. So sulfur is going to have a partial positive charge there, even though sulfur is a reasonably electronegative atom in its own right. But these other guys are going to be sucking electrons away from it. So you could imagine that this oxygen, or this entire molecule, could act as a nucleophile, with this oxygen right over here giving an electron to the sulfur. Sulfur is an electronegative atom, but it has a partial positive charge in just this thionyl chloride. So it'll be attracted there, and then this guy's going to have to give up an electron, and he'll give it back to this oxygen up there. And so that will be, and this reaction can go in either direction."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you could imagine that this oxygen, or this entire molecule, could act as a nucleophile, with this oxygen right over here giving an electron to the sulfur. Sulfur is an electronegative atom, but it has a partial positive charge in just this thionyl chloride. So it'll be attracted there, and then this guy's going to have to give up an electron, and he'll give it back to this oxygen up there. And so that will be, and this reaction can go in either direction. So once again, I'll draw it in equilibrium. But right after that happens, we'll have this, I guess we could call it a complex, that will look like this. This oxygen, it had two lone pairs."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so that will be, and this reaction can go in either direction. So once again, I'll draw it in equilibrium. But right after that happens, we'll have this, I guess we could call it a complex, that will look like this. This oxygen, it had two lone pairs. It had one pair, two pairs, and now it has a third pair, because it got this electron. It always had that other electron in the covalent bond. And now it has a negative charge."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This oxygen, it had two lone pairs. It had one pair, two pairs, and now it has a third pair, because it got this electron. It always had that other electron in the covalent bond. And now it has a negative charge. And now the sulfur is bonded to those two chlorines. And then it is also bonded to this oxygen over here. So let me draw it over here."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And now it has a negative charge. And now the sulfur is bonded to those two chlorines. And then it is also bonded to this oxygen over here. So let me draw it over here. So we have this oxygen, which is bonded to this carbonyl carbon, just like that. And then it is bonded to a hydrogen right over there. It has one lone pair now."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw it over here. So we have this oxygen, which is bonded to this carbonyl carbon, just like that. And then it is bonded to a hydrogen right over there. It has one lone pair now. And then the other lone pair is now turned into a covalent bond with the sulfur. And so this guy gave away an electron, so he now has a positive charge. Now the next step, you could imagine that this oxygen says, hey, I liked having a double bond with the sulfur."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It has one lone pair now. And then the other lone pair is now turned into a covalent bond with the sulfur. And so this guy gave away an electron, so he now has a positive charge. Now the next step, you could imagine that this oxygen says, hey, I liked having a double bond with the sulfur. The sulfur is still bonded to a bunch of things that are more electronegative to it. It still has a slightly positive charge there. So you could imagine that this electron gets given back to the sulfur, but then the sulfur needs to bump an electron."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Now the next step, you could imagine that this oxygen says, hey, I liked having a double bond with the sulfur. The sulfur is still bonded to a bunch of things that are more electronegative to it. It still has a slightly positive charge there. So you could imagine that this electron gets given back to the sulfur, but then the sulfur needs to bump an electron. Chlorine is pretty electronegative. So one of these chlorines will be able to take away an electron. So this chlorine could take away an electron."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you could imagine that this electron gets given back to the sulfur, but then the sulfur needs to bump an electron. Chlorine is pretty electronegative. So one of these chlorines will be able to take away an electron. So this chlorine could take away an electron. So then after that happens, our situation, and once again, this is in equilibrium, our situation will look like this. So let me draw the original acetic acid, or the part that was part of the original acetic acid. And so you have this oxygen right over here bonded to a hydrogen, bonded to that same sulfur right over here, which now has a double bond with this oxygen up here."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this chlorine could take away an electron. So then after that happens, our situation, and once again, this is in equilibrium, our situation will look like this. So let me draw the original acetic acid, or the part that was part of the original acetic acid. And so you have this oxygen right over here bonded to a hydrogen, bonded to that same sulfur right over here, which now has a double bond with this oxygen up here. So I'll do it in that same color. It's still bonded to that chlorine up there. But now this chlorine is left."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so you have this oxygen right over here bonded to a hydrogen, bonded to that same sulfur right over here, which now has a double bond with this oxygen up here. So I'll do it in that same color. It's still bonded to that chlorine up there. But now this chlorine is left. It's taken that electron with it. So the chlorine had 1, 2, 3, 4, 5, 6, 7 valence electrons. It gained this electron, this orange electron."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But now this chlorine is left. It's taken that electron with it. So the chlorine had 1, 2, 3, 4, 5, 6, 7 valence electrons. It gained this electron, this orange electron. So now it has a negative charge. And by the way, this guy never lost his positive charge. Still has a positive charge right like that."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It gained this electron, this orange electron. So now it has a negative charge. And by the way, this guy never lost his positive charge. Still has a positive charge right like that. Now, the chlorine could act as a nucleophile on the carbonyl carbon. This guy is bonded to two oxygens, more electronegative, slightly or partially positive charge. This guy can give an electron to the carbonyl carbon."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Still has a positive charge right like that. Now, the chlorine could act as a nucleophile on the carbonyl carbon. This guy is bonded to two oxygens, more electronegative, slightly or partially positive charge. This guy can give an electron to the carbonyl carbon. And right as that happens, this is a nucleophilic attack, right as that happens, the carbonyl carbon can give up an electron to the carbonyl oxygen. And then we will be in equilibrium. So make sure you realize we're going to the right, then down."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This guy can give an electron to the carbonyl carbon. And right as that happens, this is a nucleophilic attack, right as that happens, the carbonyl carbon can give up an electron to the carbonyl oxygen. And then we will be in equilibrium. So make sure you realize we're going to the right, then down. Now we're going to the left with this thing. So this was the carbonyl carbon bonded to this oxygen right now, which now just took another electron. So now it has a negative charge."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So make sure you realize we're going to the right, then down. Now we're going to the left with this thing. So this was the carbonyl carbon bonded to this oxygen right now, which now just took another electron. So now it has a negative charge. It had one, two lone pairs. Now it will have one more lone pair because it took that electron. It always had this end of that bond."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So now it has a negative charge. It had one, two lone pairs. Now it will have one more lone pair because it took that electron. It always had this end of that bond. So now it has both of these electrons over here. It has a negative charge. It gained an electron."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It always had this end of that bond. So now it has both of these electrons over here. It has a negative charge. It gained an electron. And then it is bonded to this OH group right over here. So this OH group, just like that, and then that is bonded to the sulfur, which is bonded to the chlorine and now double-bonded to that other oxygen. Of course, we have this chlorine over here that did the nucleophilic attack."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It gained an electron. And then it is bonded to this OH group right over here. So this OH group, just like that, and then that is bonded to the sulfur, which is bonded to the chlorine and now double-bonded to that other oxygen. Of course, we have this chlorine over here that did the nucleophilic attack. So you have this chlorine over here. And this nucleophilic attack, it gave an electron to what was this carbonyl carbon. It gave an electron."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Of course, we have this chlorine over here that did the nucleophilic attack. So you have this chlorine over here. And this nucleophilic attack, it gave an electron to what was this carbonyl carbon. It gave an electron. So it is now neutral. And you can kind of imagine that this negative charge got transferred to this oxygen up here. And this oxygen right here still has a positive charge."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It gave an electron. So it is now neutral. And you can kind of imagine that this negative charge got transferred to this oxygen up here. And this oxygen right here still has a positive charge. Don't want to forget that. Now the next step that could happen, once again, all of these can go in either direction, is that this guy doesn't like having a negative charge. And this guy is still going to have a partially positive charge because he's bonded to a bunch of more electronegative atoms in itself."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this oxygen right here still has a positive charge. Don't want to forget that. Now the next step that could happen, once again, all of these can go in either direction, is that this guy doesn't like having a negative charge. And this guy is still going to have a partially positive charge because he's bonded to a bunch of more electronegative atoms in itself. So he wants to reform the double bond. And that kicks off all of this business over here. So then this, let me do this in a new color."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this guy is still going to have a partially positive charge because he's bonded to a bunch of more electronegative atoms in itself. So he wants to reform the double bond. And that kicks off all of this business over here. So then this, let me do this in a new color. I've already used the magenta. I'll use the pink. This electron right here gets taken back by this oxygen that had a positive charge anyway."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So then this, let me do this in a new color. I've already used the magenta. I'll use the pink. This electron right here gets taken back by this oxygen that had a positive charge anyway. So it would want to take it back. And then we are left with, and this is we're getting pretty close to the punch line. We are then in equilibrium."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This electron right here gets taken back by this oxygen that had a positive charge anyway. So it would want to take it back. And then we are left with, and this is we're getting pretty close to the punch line. We are then in equilibrium. This part over here will then look like this. We now have reformed our double bond. It is only going to be bonded to the chlorine just like this, and this whole part over here on the right has broken off."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We are then in equilibrium. This part over here will then look like this. We now have reformed our double bond. It is only going to be bonded to the chlorine just like this, and this whole part over here on the right has broken off. So you have this oxygen bonded. Let me do it in the same colors. You have this oxygen bonded to a hydrogen."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It is only going to be bonded to the chlorine just like this, and this whole part over here on the right has broken off. So you have this oxygen bonded. Let me do it in the same colors. You have this oxygen bonded to a hydrogen. It now gained an electron. It now gained an electron. So it is now neutral."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "You have this oxygen bonded to a hydrogen. It now gained an electron. It now gained an electron. So it is now neutral. And it is bonded to this sulfur. The sulfur is in green. It's bonded to the sulfur, which is bonded to a chlorine."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So it is now neutral. And it is bonded to this sulfur. The sulfur is in green. It's bonded to the sulfur, which is bonded to a chlorine. And then it has a double bond to this oxygen right over here. So we've already formed our acetyl chloride. And we've really kind of finished with the hard part of the reaction to see how you could get an acetyl chloride."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It's bonded to the sulfur, which is bonded to a chlorine. And then it has a double bond to this oxygen right over here. So we've already formed our acetyl chloride. And we've really kind of finished with the hard part of the reaction to see how you could get an acetyl chloride. And then to kind of complete this reaction, because if you actually perform this in a beaker, you'll end up with some hydrogen chloride and some sulfur oxide as a byproduct. You could imagine that this oxygen right over here takes back its electron from this proton, gives it to this sulfur, and then the sulfur, since it got an electron, will allow the electronegative chlorine to take back its electron. And then if we just focus on this, the acetyl chloride at this point isn't doing anything anymore."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And we've really kind of finished with the hard part of the reaction to see how you could get an acetyl chloride. And then to kind of complete this reaction, because if you actually perform this in a beaker, you'll end up with some hydrogen chloride and some sulfur oxide as a byproduct. You could imagine that this oxygen right over here takes back its electron from this proton, gives it to this sulfur, and then the sulfur, since it got an electron, will allow the electronegative chlorine to take back its electron. And then if we just focus on this, the acetyl chloride at this point isn't doing anything anymore. This would be in equilibrium with something that looks like this. You now have a hydrogen proton floating around. You now have this hydrogen proton floating around, just a naked proton really."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then if we just focus on this, the acetyl chloride at this point isn't doing anything anymore. This would be in equilibrium with something that looks like this. You now have a hydrogen proton floating around. You now have this hydrogen proton floating around, just a naked proton really. There's not even any neutrons there. You have this oxygen, which is now double bonded. It is now double bonded with this sulfur."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "You now have this hydrogen proton floating around, just a naked proton really. There's not even any neutrons there. You have this oxygen, which is now double bonded. It is now double bonded with this sulfur. So I'm going to try to keep all of the colors the same. This right here was yellow, and then that original bond that it had with it was magenta. And then you have the chlorine nabbed that electron."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "It is now double bonded with this sulfur. So I'm going to try to keep all of the colors the same. This right here was yellow, and then that original bond that it had with it was magenta. And then you have the chlorine nabbed that electron. The chlorine, I'll do this chlorine in blue now, the chlorine nabbed an electron. And it already had seven valence electrons. 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then you have the chlorine nabbed that electron. The chlorine, I'll do this chlorine in blue now, the chlorine nabbed an electron. And it already had seven valence electrons. 1, 2, 3, 4, 5, 6, 7. It now has a negative charge. And then the final step, so we already have our sulfur dioxide, now the final step is just the chlorine giving one of its electrons to the hydrogen to form hydrogen chloride. So then the final step is just this."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7. It now has a negative charge. And then the final step, so we already have our sulfur dioxide, now the final step is just the chlorine giving one of its electrons to the hydrogen to form hydrogen chloride. So then the final step is just this. And then we end up with, so when all is said and done, we end up with some acetyl chloride. We end up with some hydrogen chloride. And we end up with some sulfur dioxide."}, {"video_title": "Acid chloride formation Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So then the final step is just this. And then we end up with, so when all is said and done, we end up with some acetyl chloride. We end up with some hydrogen chloride. And we end up with some sulfur dioxide. And once again, you can generalize this starting with any carboxylic acid and forming the acyl halide version of it, or the acyl chloride version if you want to stick with a chloride right over here. Oh, actually this right here is acetyl chloride. Anyway, hopefully you found that entertaining."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the reason why beta carotene has a color is because it absorbs light in the visible region of the electromagnetic spectrum. The visible region starts at approximately 400 nanometers. So if I draw a line right here, to the left of that line would be the ultraviolet, the UV region of the electromagnetic spectrum, and on the right would be the visible region. So we see that beta carotene absorbs light with wavelengths of approximately 450 to 500 nanometers for a range, and it absorbs strongly in the visible region. To explain why beta carotene is orange, we need to look in a little bit more detail at the visible region of the electromagnetic spectrum. And so here we have the different colors. We have the colors in the visible region, so essentially the colors of the rainbow."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we see that beta carotene absorbs light with wavelengths of approximately 450 to 500 nanometers for a range, and it absorbs strongly in the visible region. To explain why beta carotene is orange, we need to look in a little bit more detail at the visible region of the electromagnetic spectrum. And so here we have the different colors. We have the colors in the visible region, so essentially the colors of the rainbow. Approximately 400 nanometers, we're talking about violet light. So we have violet light right here. If you go beyond violet light, you're in the ultraviolet region, or the UV region."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have the colors in the visible region, so essentially the colors of the rainbow. Approximately 400 nanometers, we're talking about violet light. So we have violet light right here. If you go beyond violet light, you're in the ultraviolet region, or the UV region. So the visible region goes to somewhere around 700 nanometers, or a little bit beyond that. So when you're in 700 nanometers, you're talking about red light. And if you go just past red light, you are in the infrared region of the electromagnetic spectrum."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you go beyond violet light, you're in the ultraviolet region, or the UV region. So the visible region goes to somewhere around 700 nanometers, or a little bit beyond that. So when you're in 700 nanometers, you're talking about red light. And if you go just past red light, you are in the infrared region of the electromagnetic spectrum. And here I have six colors. I have red, orange, yellow, green, blue, and violet. And when Isaac Newton did his famous experiment with a prism, and he wrote down seven colors, so he included indigo because he wanted to have seven colors in the visible region."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if you go just past red light, you are in the infrared region of the electromagnetic spectrum. And here I have six colors. I have red, orange, yellow, green, blue, and violet. And when Isaac Newton did his famous experiment with a prism, and he wrote down seven colors, so he included indigo because he wanted to have seven colors in the visible region. So you usually memorize ROYGBIV for the colors of the rainbow. But the reason I've left out indigo here is because this allows us to better see a color wheel. So Isaac Newton was the first to represent a color wheel."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And when Isaac Newton did his famous experiment with a prism, and he wrote down seven colors, so he included indigo because he wanted to have seven colors in the visible region. So you usually memorize ROYGBIV for the colors of the rainbow. But the reason I've left out indigo here is because this allows us to better see a color wheel. So Isaac Newton was the first to represent a color wheel. And you get a color wheel by taking the violet and moving it over here, and taking the red and moving it over here. And so you put the violet right next to the red, and so you get a color wheel. And it's useful to look at a color wheel because it allows you to see the relationship between complementary colors."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So Isaac Newton was the first to represent a color wheel. And you get a color wheel by taking the violet and moving it over here, and taking the red and moving it over here. And so you put the violet right next to the red, and so you get a color wheel. And it's useful to look at a color wheel because it allows you to see the relationship between complementary colors. For example, if I wanted to know the complementary color for red, all I have to do is look across on my color wheel, and I can see that the complementary color is green. The complementary color for violet, if I look directly across, that would be yellow. And then finally, the complementary color for blue would be orange."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it's useful to look at a color wheel because it allows you to see the relationship between complementary colors. For example, if I wanted to know the complementary color for red, all I have to do is look across on my color wheel, and I can see that the complementary color is green. The complementary color for violet, if I look directly across, that would be yellow. And then finally, the complementary color for blue would be orange. And this is useful because it allows you to think about why things appear to be a certain color. For example, if I look at this orange sheet of paper here, and we try to understand why this sheet of paper is orange, we know that white light consists of all of these different wavelengths. We know white light consists of all of the different colors of the rainbow here."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, the complementary color for blue would be orange. And this is useful because it allows you to think about why things appear to be a certain color. For example, if I look at this orange sheet of paper here, and we try to understand why this sheet of paper is orange, we know that white light consists of all of these different wavelengths. We know white light consists of all of the different colors of the rainbow here. We can simplify that even further, and we can think about white light being two complementary colors. So we could say, oh, okay, so this part consists of blue wavelengths of light, and then on the right here, this part of the color wheel consists of orange wavelengths of light. And we could think about white light consisting of blue wavelengths and orange wavelengths."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We know white light consists of all of the different colors of the rainbow here. We can simplify that even further, and we can think about white light being two complementary colors. So we could say, oh, okay, so this part consists of blue wavelengths of light, and then on the right here, this part of the color wheel consists of orange wavelengths of light. And we could think about white light consisting of blue wavelengths and orange wavelengths. So if we have white light coming in, here are the blue wavelengths of light, and then we have the orange wavelengths of light. So this is an oversimplified way to think about white light striking our orange object here. So if the object absorbs the blue wavelengths of light, so we are absorbing the blue wavelengths of light, therefore we are reflecting the orange wavelengths."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we could think about white light consisting of blue wavelengths and orange wavelengths. So if we have white light coming in, here are the blue wavelengths of light, and then we have the orange wavelengths of light. So this is an oversimplified way to think about white light striking our orange object here. So if the object absorbs the blue wavelengths of light, so we are absorbing the blue wavelengths of light, therefore we are reflecting the orange wavelengths. And so if we reflect the orange wavelengths of light, and our eye happens to be right here, so we see the object as being orange. So our brains perceive the object as being orange because we are seeing the reflected orange light. And so that's how to think about why something appears to be a certain color."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if the object absorbs the blue wavelengths of light, so we are absorbing the blue wavelengths of light, therefore we are reflecting the orange wavelengths. And so if we reflect the orange wavelengths of light, and our eye happens to be right here, so we see the object as being orange. So our brains perceive the object as being orange because we are seeing the reflected orange light. And so that's how to think about why something appears to be a certain color. If I go back up here to beta-carotene again, so I look and see where beta-carotene is absorbing. Beta-carotene is absorbing somewhere in the range of 450 to 500 nanometers, and those are blue wavelengths of light. If I look at it down here, so 450 to 500 nanometers, we're absorbing the blue wavelengths of light."}, {"video_title": "Absorption in the visible region Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that's how to think about why something appears to be a certain color. If I go back up here to beta-carotene again, so I look and see where beta-carotene is absorbing. Beta-carotene is absorbing somewhere in the range of 450 to 500 nanometers, and those are blue wavelengths of light. If I look at it down here, so 450 to 500 nanometers, we're absorbing the blue wavelengths of light. And therefore we are reflecting the orange wavelengths. And so we perceive beta-carotene to be orange. And so that's a little bit of the theory behind why we perceive something as having a certain color."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And the first thing to do is just to see if there are any chiral centers in this molecule. If there aren't, then we don't even have to use the RS system. We could just use our standard nomenclature rules and we'd be done. So if we look here, this carbon is attached to three hydrogens, so it's definitely not attached to four different groups. Same thing about this carbon right here. This carbon right here is attached to a fluorine, but then it's attached to two, you could view them as two methyl groups. So it's the same group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So if we look here, this carbon is attached to three hydrogens, so it's definitely not attached to four different groups. Same thing about this carbon right here. This carbon right here is attached to a fluorine, but then it's attached to two, you could view them as two methyl groups. So it's the same group. So this is also not a chiral carbon or an asymmetric carbon. This carbon right here is attached to a hydrogen and three other carbons, but each of these three carbons look like different groups. This carbon is attached to two methyls and a fluorine."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So it's the same group. So this is also not a chiral carbon or an asymmetric carbon. This carbon right here is attached to a hydrogen and three other carbons, but each of these three carbons look like different groups. This carbon is attached to two methyls and a fluorine. This carbon is attached to two hydrogens and a bromine. This carbon is just a methyl group. So this right here does look like a chiral center."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is attached to two methyls and a fluorine. This carbon is attached to two hydrogens and a bromine. This carbon is just a methyl group. So this right here does look like a chiral center. The other ones don't. This is just a methyl group, has three hydrogens, so definitely not attached to four different groups. And this is attached to two hydrogens and those are obviously the same group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this right here does look like a chiral center. The other ones don't. This is just a methyl group, has three hydrogens, so definitely not attached to four different groups. And this is attached to two hydrogens and those are obviously the same group. So this is also not a chiral center. So we have one chiral center, so the RS naming system will apply. But a good starting point will just to be naming it using our standard nomenclature rules."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And this is attached to two hydrogens and those are obviously the same group. So this is also not a chiral center. So we have one chiral center, so the RS naming system will apply. But a good starting point will just to be naming it using our standard nomenclature rules. And to do that, we look for the longest carbon chain here. And let's see, we have, if we start over here, and I don't know what direction I'm going to name it from yet, but I just want to identify the longest chain. If we went from here, we have one, two, three, we could either go to four or to four there."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "But a good starting point will just to be naming it using our standard nomenclature rules. And to do that, we look for the longest carbon chain here. And let's see, we have, if we start over here, and I don't know what direction I'm going to name it from yet, but I just want to identify the longest chain. If we went from here, we have one, two, three, we could either go to four or to four there. So we definitely have four carbons, four carbon longest chain, so that tells us that we will be using the prefix bute or it will be a butane because they're all single bonds here. So it is a butane. But to decide whether we branch off, it doesn't matter whether we use this CH3 or this CH3, they're the same group, but decide whether we use this as part of the longest chain or we use that, we think about the rule that the core chain to use should have as many simple groups attached to it as possible as opposed to as few complex groups."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "If we went from here, we have one, two, three, we could either go to four or to four there. So we definitely have four carbons, four carbon longest chain, so that tells us that we will be using the prefix bute or it will be a butane because they're all single bonds here. So it is a butane. But to decide whether we branch off, it doesn't matter whether we use this CH3 or this CH3, they're the same group, but decide whether we use this as part of the longest chain or we use that, we think about the rule that the core chain to use should have as many simple groups attached to it as possible as opposed to as few complex groups. So if we used this carbon as part of our longest chain, then this will be a group that's attached to it, which would be kind of a, what is this, a bromo methyl group, which is not as simple as maybe it could be, but if we use this carbon in our longest chain, we'll have two groups. We'll have a bromo attached and we'll also have a methyl group. And that's what we want."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "But to decide whether we branch off, it doesn't matter whether we use this CH3 or this CH3, they're the same group, but decide whether we use this as part of the longest chain or we use that, we think about the rule that the core chain to use should have as many simple groups attached to it as possible as opposed to as few complex groups. So if we used this carbon as part of our longest chain, then this will be a group that's attached to it, which would be kind of a, what is this, a bromo methyl group, which is not as simple as maybe it could be, but if we use this carbon in our longest chain, we'll have two groups. We'll have a bromo attached and we'll also have a methyl group. And that's what we want. We want more simple groups attached to the longest chain. So what we're going to do is we're going to use this carbon, this carbon, this carbon, and that carbon as our longest chain. And we want to start from the end that's closest to something being attached to it, and that bromine's right there."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And that's what we want. We want more simple groups attached to the longest chain. So what we're going to do is we're going to use this carbon, this carbon, this carbon, and that carbon as our longest chain. And we want to start from the end that's closest to something being attached to it, and that bromine's right there. So this is going to be our number one carbon, our number two carbon, our number three carbon, and our number four carbon. And then we can label the different groups and then figure out what order they should be listed in. So this is a 1-bromo, and then this will be a 2-methyl right here, and then let's see, just a hydrogen, then 3, we have a fluoro."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And we want to start from the end that's closest to something being attached to it, and that bromine's right there. So this is going to be our number one carbon, our number two carbon, our number three carbon, and our number four carbon. And then we can label the different groups and then figure out what order they should be listed in. So this is a 1-bromo, and then this will be a 2-methyl right here, and then let's see, just a hydrogen, then 3, we have a fluoro. So on carbon 3, we have a fluoro, and then on carbon 3, we also have a methyl group right here. So we also have a 3-methyl. So when we name it, we put it in alphabetical order."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this is a 1-bromo, and then this will be a 2-methyl right here, and then let's see, just a hydrogen, then 3, we have a fluoro. So on carbon 3, we have a fluoro, and then on carbon 3, we also have a methyl group right here. So we also have a 3-methyl. So when we name it, we put it in alphabetical order. Bromo comes first. So this thing right here is 1-bromo, and then alphabetically, fluoro comes next. 1-bromo, 3-fluoro, we have 2-methyls."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So when we name it, we put it in alphabetical order. Bromo comes first. So this thing right here is 1-bromo, and then alphabetically, fluoro comes next. 1-bromo, 3-fluoro, we have 2-methyls. So it's going to be 2, 3-dimethyl. And remember, the d doesn't count in alphabetical order. Dimethyl butane, because we have the longest chain is 4 carbons, dimethyl butane."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "1-bromo, 3-fluoro, we have 2-methyls. So it's going to be 2, 3-dimethyl. And remember, the d doesn't count in alphabetical order. Dimethyl butane, because we have the longest chain is 4 carbons, dimethyl butane. So that's just the standard nomenclature rules. We still haven't used the RS system. Now we can do that."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "Dimethyl butane, because we have the longest chain is 4 carbons, dimethyl butane. So that's just the standard nomenclature rules. We still haven't used the RS system. Now we can do that. Now to think about it, we already said that this is our chiral center. So we just have to essentially rank the groups attached to it in order of atomic number, and then use the Con-Ingold-Prelog rules. We'll do all of that in this example."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "Now we can do that. Now to think about it, we already said that this is our chiral center. So we just have to essentially rank the groups attached to it in order of atomic number, and then use the Con-Ingold-Prelog rules. We'll do all of that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has 3 carbons and a hydrogen. Carbon is definitely higher in atomic number on the periodic table."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "We'll do all of that in this example. So let's look at the different groups attached to it. So when you look at it, this guy has 3 carbons and a hydrogen. Carbon is definitely higher in atomic number on the periodic table. It has atomic number of 6, hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number 4."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "Carbon is definitely higher in atomic number on the periodic table. It has atomic number of 6, hydrogen is 1. You probably know that already. So hydrogen is definitely going to be number 4. So let me put number 4 there next to the hydrogen. So hydrogen, and let me find a nice color to do it. I'll do it in white."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So hydrogen is definitely going to be number 4. So let me put number 4 there next to the hydrogen. So hydrogen, and let me find a nice color to do it. I'll do it in white. So hydrogen is definitely the number 4 group. But we have to differentiate between this carbon group, that carbon group, and that carbon group. And the way you do it, if there's a tie on the 3 carbons, you then look at what is attached to those carbons."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "I'll do it in white. So hydrogen is definitely the number 4 group. But we have to differentiate between this carbon group, that carbon group, and that carbon group. And the way you do it, if there's a tie on the 3 carbons, you then look at what is attached to those carbons. And you compare the highest thing attached to each of those carbons to the highest things attached to the other carbons, and then you do the same ranking. And if that's a tie, then you keep going on and on and on. So in this carbon right here, we have a bromine."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And the way you do it, if there's a tie on the 3 carbons, you then look at what is attached to those carbons. And you compare the highest thing attached to each of those carbons to the highest things attached to the other carbons, and then you do the same ranking. And if that's a tie, then you keep going on and on and on. So in this carbon right here, we have a bromine. Bromine has an atomic number of 35, which is higher than fluorine, or which is higher than carbon. So this guy has a bromine attached to it. This guy only has hydrogens attached to it."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So in this carbon right here, we have a bromine. Bromine has an atomic number of 35, which is higher than fluorine, or which is higher than carbon. So this guy has a bromine attached to it. This guy only has hydrogens attached to it. This guy has a fluorine attached to it. That's the highest thing. So this is going to be the third lowest, or I should say the second to lowest, because it only has hydrogens attached to it."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "This guy only has hydrogens attached to it. This guy has a fluorine attached to it. That's the highest thing. So this is going to be the third lowest, or I should say the second to lowest, because it only has hydrogens attached to it. So that is number 3. The one that has a bromine attached to it is going to be number 1, and the one that has a fluorine attached to it is number 2. So this is the number 2 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be the third lowest, or I should say the second to lowest, because it only has hydrogens attached to it. So that is number 3. The one that has a bromine attached to it is going to be number 1, and the one that has a fluorine attached to it is number 2. So this is the number 2 group. And just as a reminder, we were tied with the carbon, so we had to look at the next highest constituent. And even if this had 3 fluorines attached to it, the bromine would still trump it. You compare the highest to the highest."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this is the number 2 group. And just as a reminder, we were tied with the carbon, so we had to look at the next highest constituent. And even if this had 3 fluorines attached to it, the bromine would still trump it. You compare the highest to the highest. So now that we've done that, let me redraw this molecule so it's a little bit easier to visualize. So I'll draw our chiral carbon in the middle. So let's draw our chiral carbon."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "You compare the highest to the highest. So now that we've done that, let me redraw this molecule so it's a little bit easier to visualize. So I'll draw our chiral carbon in the middle. So let's draw our chiral carbon. And I'm just doing this for visualization purposes. And right here we have our number 1 group. I'll literally just call that our number 1 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw our chiral carbon. And I'm just doing this for visualization purposes. And right here we have our number 1 group. I'll literally just call that our number 1 group. So right there, that is our number 1 group. It's in the plane of the screen. So I'll just call that our number 1 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "I'll literally just call that our number 1 group. So right there, that is our number 1 group. It's in the plane of the screen. So I'll just call that our number 1 group. That's our number 1 group. Over here, also in the plane of the screen, I have our number 2 group. So let me do it like that."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So I'll just call that our number 1 group. That's our number 1 group. Over here, also in the plane of the screen, I have our number 2 group. So let me do it like that. So then you have your number 2 group. And then you have your number 3 group behind the molecule right now, the way it's drawn. So then you have your number 3 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let me do it like that. So then you have your number 2 group. And then you have your number 3 group behind the molecule right now, the way it's drawn. So then you have your number 3 group. It's behind the molecule, so I'll do it like this. This is our number 3 group. And then we have our number 4 group, which is the hydrogen pointing out right now."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So then you have your number 3 group. It's behind the molecule, so I'll do it like this. This is our number 3 group. And then we have our number 4 group, which is the hydrogen pointing out right now. And I'll just do that in yellow. We have our number 4 group pointing out in front right now. So that is number 4."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our number 4 group, which is the hydrogen pointing out right now. And I'll just do that in yellow. We have our number 4 group pointing out in front right now. So that is number 4. Just like that. And actually, maybe this one, actually let me draw it a little bit clearer. So it looks a little bit more like the tripod structure that it's supposed to be."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So that is number 4. Just like that. And actually, maybe this one, actually let me draw it a little bit clearer. So it looks a little bit more like the tripod structure that it's supposed to be. So let me redraw the number 3 group. The number 3 group should look like, it should look a little bit. So this is our number 3 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So it looks a little bit more like the tripod structure that it's supposed to be. So let me redraw the number 3 group. The number 3 group should look like, it should look a little bit. So this is our number 3 group. Let me draw it a little bit more like this. So number 3 group is behind us. Number 3."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this is our number 3 group. Let me draw it a little bit more like this. So number 3 group is behind us. Number 3. And then finally, you have your number 4 group in yellow. Your number 4 group, which is just a hydrogen. And that's coming straight out."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "Number 3. And then finally, you have your number 4 group in yellow. Your number 4 group, which is just a hydrogen. And that's coming straight out. So that is coming straight out of, well, not straight out, but at an angle out of the page. So that's our number 4 group. I'll just label it number 4."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "And that's coming straight out. So that is coming straight out of, well, not straight out, but at an angle out of the page. So that's our number 4 group. I'll just label it number 4. It really is just a hydrogen, so I really didn't have to simplify it much there. Now, by the RS system, or by the Kahn-Ingold prelog system, we want our number 4 group to be behind, to be the one furthest back. So we really want it kind of where the number 3 position is."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "I'll just label it number 4. It really is just a hydrogen, so I really didn't have to simplify it much there. Now, by the RS system, or by the Kahn-Ingold prelog system, we want our number 4 group to be behind, to be the one furthest back. So we really want it kind of where the number 3 position is. And so the easiest way I can think of doing that is you can imagine this is kind of a tripod that's leaning kind of upside down. Or another way to view it is you can kind of view it as an umbrella, where this is the handle of the umbrella, and that's kind of the top of the umbrella that would block the rain, I guess. But the easiest way to get the number 4 group that's actually a hydrogen in the number 3 position would be to rotate it, you could imagine, rotate it around the axis defined by the number 1 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we really want it kind of where the number 3 position is. And so the easiest way I can think of doing that is you can imagine this is kind of a tripod that's leaning kind of upside down. Or another way to view it is you can kind of view it as an umbrella, where this is the handle of the umbrella, and that's kind of the top of the umbrella that would block the rain, I guess. But the easiest way to get the number 4 group that's actually a hydrogen in the number 3 position would be to rotate it, you could imagine, rotate it around the axis defined by the number 1 group. So the number 1 group is just going to stay where it is. The number 4 is going to rotate to the number 3 group. Number 3 is going to rotate around to the number 2 group."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "But the easiest way to get the number 4 group that's actually a hydrogen in the number 3 position would be to rotate it, you could imagine, rotate it around the axis defined by the number 1 group. So the number 1 group is just going to stay where it is. The number 4 is going to rotate to the number 3 group. Number 3 is going to rotate around to the number 2 group. And then the number 2 group is going to rotate to where the number 4 group is right now. So if we were to redraw that, let's redraw our chiral carbon. So let me scroll over a little bit."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "Number 3 is going to rotate around to the number 2 group. And then the number 2 group is going to rotate to where the number 4 group is right now. So if we were to redraw that, let's redraw our chiral carbon. So let me scroll over a little bit. So we have our chiral carbon. I put a little asterisk there to say that that's our chiral carbon. The number 4 group is now behind."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let me scroll over a little bit. So we have our chiral carbon. I put a little asterisk there to say that that's our chiral carbon. The number 4 group is now behind. I'll do it with the circles. It makes it look a little bit more like atoms. So the number 4 group is now behind where the number 3 group used to be."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "The number 4 group is now behind. I'll do it with the circles. It makes it look a little bit more like atoms. So the number 4 group is now behind where the number 3 group used to be. So number 4 is now there. Number 1 hasn't changed. That's kind of the axis that we rotated around."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So the number 4 group is now behind where the number 3 group used to be. So number 4 is now there. Number 1 hasn't changed. That's kind of the axis that we rotated around. So the number 1 group has not changed. Number 1 is still there. Number 2 is now where number 4 used to be."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "That's kind of the axis that we rotated around. So the number 1 group has not changed. Number 1 is still there. Number 2 is now where number 4 used to be. So number 2 is now jutting out of the page. And then we have number 3 is now where number 2 was. So number 3 is there."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "Number 2 is now where number 4 used to be. So number 2 is now jutting out of the page. And then we have number 3 is now where number 2 was. So number 3 is there. And now that we've put our fourth group behind the molecule, we literally just figure out whether we have to go clockwise or counterclockwise to go from 1, 2, to 3. And that's pretty straightforward. To go from 1 to 2 to 3, we have to go counterclockwise."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "So number 3 is there. And now that we've put our fourth group behind the molecule, we literally just figure out whether we have to go clockwise or counterclockwise to go from 1, 2, to 3. And that's pretty straightforward. To go from 1 to 2 to 3, we have to go counterclockwise. Or another way to think of it, we're going to the left counterclockwise, at least on the top of the clock. We're going to the left. And so since we're going to the left, this is S or sinister."}, {"video_title": "R,S (Cahn-Ingold-Prelog) naming system example 2 Organic chemistry Khan Academy.mp3", "Sentence": "To go from 1 to 2 to 3, we have to go counterclockwise. Or another way to think of it, we're going to the left counterclockwise, at least on the top of the clock. We're going to the left. And so since we're going to the left, this is S or sinister. This is S, which stands for sinister, which is Latin for left. So we're done. We've named it using the RS system."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "In this video, we're gonna look at the SN1 mechanism. And we'll start with our alkyl halide. In the first step of our SN1 mechanism, we get loss of a leaving group. So the electrons in this bond come off onto the bromine to form the bromide ion. When that happens, we take a bond away from this carbon in red. So the carbon in red gets a plus one formal charge. So let me draw that in here."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So the electrons in this bond come off onto the bromine to form the bromide ion. When that happens, we take a bond away from this carbon in red. So the carbon in red gets a plus one formal charge. So let me draw that in here. So the carbon in red is this one. It now has a plus one formal charge. And we have a carbocation."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So let me draw that in here. So the carbon in red is this one. It now has a plus one formal charge. And we have a carbocation. The carbon in red went from being SP3 hybridized in our alkyl halide to being SP2 hybridized in our carbocation, which means the geometry directly around the carbon in red is planar. We also have our bromine, so let me draw that in here. It has four lone pairs of electrons around it now, which gives it a negative one formal charge."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And we have a carbocation. The carbon in red went from being SP3 hybridized in our alkyl halide to being SP2 hybridized in our carbocation, which means the geometry directly around the carbon in red is planar. We also have our bromine, so let me draw that in here. It has four lone pairs of electrons around it now, which gives it a negative one formal charge. So that's the bromide ion. So the electrons in this bond, these electrons in here, come off onto the bromine to form our bromide ion, which is a good leaving group. So in the first step of our SN1 mechanism, we get loss of a leaving group."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "It has four lone pairs of electrons around it now, which gives it a negative one formal charge. So that's the bromide ion. So the electrons in this bond, these electrons in here, come off onto the bromine to form our bromide ion, which is a good leaving group. So in the first step of our SN1 mechanism, we get loss of a leaving group. And when that happens, we form our carbocation. And our carbocation has a plus one formal charge on this carbon. This is gonna function as our electrophile."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So in the first step of our SN1 mechanism, we get loss of a leaving group. And when that happens, we form our carbocation. And our carbocation has a plus one formal charge on this carbon. This is gonna function as our electrophile. And our nucleophile will be the hydrosulfide ion with a negative charge. So opposite charges attract, and in the second step of our mechanism, our nucleophile attacks our electrophile. So in our second step, we get nucleophilic attack."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "This is gonna function as our electrophile. And our nucleophile will be the hydrosulfide ion with a negative charge. So opposite charges attract, and in the second step of our mechanism, our nucleophile attacks our electrophile. So in our second step, we get nucleophilic attack. So nucleophilic attack, and a lone pair of electrons on our sulfur form a bond with our carbon in red. So in our final product, this is our carbon in red, and I'll highlight a lone pair of electrons on this sulfur. And that lone pair forms this bond to give us our product."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So in our second step, we get nucleophilic attack. So nucleophilic attack, and a lone pair of electrons on our sulfur form a bond with our carbon in red. So in our final product, this is our carbon in red, and I'll highlight a lone pair of electrons on this sulfur. And that lone pair forms this bond to give us our product. Let's go to the video so we can see this mechanism using a model set. Here's our alkyl halide, and I'm saying that the green is our bromine. So in the first step of our mechanism, we get loss of a leaving group."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And that lone pair forms this bond to give us our product. Let's go to the video so we can see this mechanism using a model set. Here's our alkyl halide, and I'm saying that the green is our bromine. So in the first step of our mechanism, we get loss of a leaving group. So these electrons come off onto our bromine to form the bromide ion, and we form our carbocation. But using this model set, it doesn't look like we have a planar carbocation. Those carbons are not in the same plane."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So in the first step of our mechanism, we get loss of a leaving group. So these electrons come off onto our bromine to form the bromide ion, and we form our carbocation. But using this model set, it doesn't look like we have a planar carbocation. Those carbons are not in the same plane. So let me grab another model set here. So you can see that actually those carbons are in the same plane, and we have an sp2 hybridized carbon in the center. If we look at the carbocation from a top view, we can see the drawing."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "Those carbons are not in the same plane. So let me grab another model set here. So you can see that actually those carbons are in the same plane, and we have an sp2 hybridized carbon in the center. If we look at the carbocation from a top view, we can see the drawing. That's how we draw it in our mechanism. So next, our nucleophile comes along, which is our hydrosulfide ion. And our nucleophile can attack from either above or below, since we have a planar carbocation that is flat."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "If we look at the carbocation from a top view, we can see the drawing. That's how we draw it in our mechanism. So next, our nucleophile comes along, which is our hydrosulfide ion. And our nucleophile can attack from either above or below, since we have a planar carbocation that is flat. Either way, we get the same product. So let's show the final product here. We're back to an sp3 hybridized carbon, so we have tetrahedral geometry in our final product."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And our nucleophile can attack from either above or below, since we have a planar carbocation that is flat. Either way, we get the same product. So let's show the final product here. We're back to an sp3 hybridized carbon, so we have tetrahedral geometry in our final product. The first step of our mechanism, loss of a leaving group, turns out to be the slow step. And the second step, nucleophilic attack, turns out to be the fast step. And this mechanism is consistent with the experimentally determined rate law for this reaction."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "We're back to an sp3 hybridized carbon, so we have tetrahedral geometry in our final product. The first step of our mechanism, loss of a leaving group, turns out to be the slow step. And the second step, nucleophilic attack, turns out to be the fast step. And this mechanism is consistent with the experimentally determined rate law for this reaction. The rate of the reaction is equal to the rate constant, k, times the concentration of our alkyl halide. So experiments have determined this to be our rate law, and this is the concentration of our alkyl halide to the first power. So the rate of the reaction depends on the concentration of the alkyl halide, but not on the concentration of the nucleophile."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And this mechanism is consistent with the experimentally determined rate law for this reaction. The rate of the reaction is equal to the rate constant, k, times the concentration of our alkyl halide. So experiments have determined this to be our rate law, and this is the concentration of our alkyl halide to the first power. So the rate of the reaction depends on the concentration of the alkyl halide, but not on the concentration of the nucleophile. And that's because our first step is our slow step, and this is our rate determining step, if you remember this stuff from general chemistry. And that means if you increase the concentration of our alkyl halide, if you increase the concentration of our alkyl halide by a factor of two, you increase the rate by a factor of two, since it's first order with respect to our alkyl halide. But if you try to increase the concentration of your nucleophile, so let's say you increase the concentration of your nucleophile by a factor of two, there's no effect on the rate."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So the rate of the reaction depends on the concentration of the alkyl halide, but not on the concentration of the nucleophile. And that's because our first step is our slow step, and this is our rate determining step, if you remember this stuff from general chemistry. And that means if you increase the concentration of our alkyl halide, if you increase the concentration of our alkyl halide by a factor of two, you increase the rate by a factor of two, since it's first order with respect to our alkyl halide. But if you try to increase the concentration of your nucleophile, so let's say you increase the concentration of your nucleophile by a factor of two, there's no effect on the rate. So this is zero order with respect to our nucleophile. Our nucleophile can't attack until our carbocation is formed, and that's dependent only on the concentration of our alkyl halide. And so that's why this reaction is first order with respect to our alkyl halide."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "But if you try to increase the concentration of your nucleophile, so let's say you increase the concentration of your nucleophile by a factor of two, there's no effect on the rate. So this is zero order with respect to our nucleophile. Our nucleophile can't attack until our carbocation is formed, and that's dependent only on the concentration of our alkyl halide. And so that's why this reaction is first order with respect to our alkyl halide. So we call this an SN1 reaction. So the S stands for substitution, the N stands for nucleophilic, and the one refers to the fact that this is a unimolecular, this is a unimolecular reaction, which means that the rate of the reaction depends on the concentration of only one thing, which is our substrate, our alkyl halide. So it's first order with respect to our alkyl halide."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And so that's why this reaction is first order with respect to our alkyl halide. So we call this an SN1 reaction. So the S stands for substitution, the N stands for nucleophilic, and the one refers to the fact that this is a unimolecular, this is a unimolecular reaction, which means that the rate of the reaction depends on the concentration of only one thing, which is our substrate, our alkyl halide. So it's first order with respect to our alkyl halide. And the nucleophilic substitution means that our nucleophile has substituted for our leaving group in our product. So that's an SN1 mechanism. The structure of the substrate also affects the rate of the reaction."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So it's first order with respect to our alkyl halide. And the nucleophilic substitution means that our nucleophile has substituted for our leaving group in our product. So that's an SN1 mechanism. The structure of the substrate also affects the rate of the reaction. If we start with a tertiary alkyl halide, like we did in the example above, we're gonna get a tertiary carbocation. So if these electrons come off onto the bromine, we're left with a tertiary carbocation, a plus one formal charge on this carbon. And we know that tertiary carbocations are the most stable."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "The structure of the substrate also affects the rate of the reaction. If we start with a tertiary alkyl halide, like we did in the example above, we're gonna get a tertiary carbocation. So if these electrons come off onto the bromine, we're left with a tertiary carbocation, a plus one formal charge on this carbon. And we know that tertiary carbocations are the most stable. We saw this in an earlier video. The more alkyl groups you have, the more electron density you can donate to help stabilize this positive charge. If we started with a secondary alkyl halide, when these electrons came off onto the bromine, we'd be left with a secondary carbocation."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And we know that tertiary carbocations are the most stable. We saw this in an earlier video. The more alkyl groups you have, the more electron density you can donate to help stabilize this positive charge. If we started with a secondary alkyl halide, when these electrons came off onto the bromine, we'd be left with a secondary carbocation. So let me draw this in here. And a secondary carbocation is only stabilized by two alkyl groups. So in this case, it'd be these two methyl groups here."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "If we started with a secondary alkyl halide, when these electrons came off onto the bromine, we'd be left with a secondary carbocation. So let me draw this in here. And a secondary carbocation is only stabilized by two alkyl groups. So in this case, it'd be these two methyl groups here. And since a secondary carbocation is not as stable as a tertiary, a tertiary carbocation would form a lot faster. So tertiary alkyl halides are the most reactive in an SN1 mechanism. A primary alkyl halide or a methyl halide, these wouldn't have a very stable carbocation."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So in this case, it'd be these two methyl groups here. And since a secondary carbocation is not as stable as a tertiary, a tertiary carbocation would form a lot faster. So tertiary alkyl halides are the most reactive in an SN1 mechanism. A primary alkyl halide or a methyl halide, these wouldn't have a very stable carbocation. So the carbocation is too unstable to exist. So generally, a primary alkyl halide does not react via an SN1 mechanism. And same with a methyl halide."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "A primary alkyl halide or a methyl halide, these wouldn't have a very stable carbocation. So the carbocation is too unstable to exist. So generally, a primary alkyl halide does not react via an SN1 mechanism. And same with a methyl halide. In the previous example, we had a nucleophile with a negative one formal charge on it. So what do you do if your nucleophile is neutral? In this case, the water molecule has no charge on the oxygen, but it can still act as a nucleophile in our SN1 mechanism."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And same with a methyl halide. In the previous example, we had a nucleophile with a negative one formal charge on it. So what do you do if your nucleophile is neutral? In this case, the water molecule has no charge on the oxygen, but it can still act as a nucleophile in our SN1 mechanism. So our first step is loss of a leaving group. So these electrons come off onto chlorine to form the chloride anion. And we're taking a bond away from this carbon in red."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "In this case, the water molecule has no charge on the oxygen, but it can still act as a nucleophile in our SN1 mechanism. So our first step is loss of a leaving group. So these electrons come off onto chlorine to form the chloride anion. And we're taking a bond away from this carbon in red. So that carbon in red is gonna get a plus one formal charge. We draw our carbocation to try to show the planar geometry around that central carbon here, which has a plus one formal charge. And in the next step, we know our nucleophile attacks our electrophile."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And we're taking a bond away from this carbon in red. So that carbon in red is gonna get a plus one formal charge. We draw our carbocation to try to show the planar geometry around that central carbon here, which has a plus one formal charge. And in the next step, we know our nucleophile attacks our electrophile. So our nucleophile has this oxygen with a partial negative charge. The oxygen's more electronegative than these hydrogens, so it withdraws some electron density. And a lone pair of electrons here on this oxygen can form a bond with this carbon."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And in the next step, we know our nucleophile attacks our electrophile. So our nucleophile has this oxygen with a partial negative charge. The oxygen's more electronegative than these hydrogens, so it withdraws some electron density. And a lone pair of electrons here on this oxygen can form a bond with this carbon. So our nucleophile attacks our electrophile. And let's draw what we would form. So let me draw in these groups here."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And a lone pair of electrons here on this oxygen can form a bond with this carbon. So our nucleophile attacks our electrophile. And let's draw what we would form. So let me draw in these groups here. And then we have our oxygen, which is now bonded to our carbon. And our oxygen is still bonded to two hydrogens. So let me show the electrons in magenta here, forming a bond between this oxygen and our carbon."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So let me draw in these groups here. And then we have our oxygen, which is now bonded to our carbon. And our oxygen is still bonded to two hydrogens. So let me show the electrons in magenta here, forming a bond between this oxygen and our carbon. So the electrons in magenta would be these electrons. We still have a lone pair of electrons on the oxygen that's left over. So that's this lone pair."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So let me show the electrons in magenta here, forming a bond between this oxygen and our carbon. So the electrons in magenta would be these electrons. We still have a lone pair of electrons on the oxygen that's left over. So that's this lone pair. So let me draw those in. And that would give this oxygen a plus one formal charge. So if we compare this with our final product, notice we only need to do a proton transfer, an acid-base reaction."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So that's this lone pair. So let me draw those in. And that would give this oxygen a plus one formal charge. So if we compare this with our final product, notice we only need to do a proton transfer, an acid-base reaction. Another molecule of water could come along and act as a base. Let me draw that in here. And it could take one of these protons."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So if we compare this with our final product, notice we only need to do a proton transfer, an acid-base reaction. Another molecule of water could come along and act as a base. Let me draw that in here. And it could take one of these protons. So it could take this and leave these electrons behind on the oxygen. So let's highlight those electrons in red. So these electrons in red here end up on that oxygen."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "And it could take one of these protons. So it could take this and leave these electrons behind on the oxygen. So let's highlight those electrons in red. So these electrons in red here end up on that oxygen. And our final product is neutral. So we have no charge and we form tert-butanol. This is called a solvolysis reaction."}, {"video_title": "Sn1 mechanism kinetics and substrate.mp3", "Sentence": "So these electrons in red here end up on that oxygen. And our final product is neutral. So we have no charge and we form tert-butanol. This is called a solvolysis reaction. So let me write that in here. So this is a solvolysis reaction, which just means that our solvent is the nucleophile. So our solvent is water, which also functions as the nucleophile in this mechanism."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So a Bronsted-Lowry acid is a proton donor. And a Bronsted-Lowry base is a proton acceptor. So let's really quickly review what this definition means by proton. So if I look at this diagram right here, I'm going to draw the hydrogen atom, or the most common isotope. So hydrogen has one proton in the nucleus, and one electron somewhere around our nucleus, so a negative charge like that. And so we would say this is hydrogen, and then we'd put its one valence electron right there to represent the hydrogen atom, or the most common isotope. If we were to somehow take away this electron, we would only be left with the proton here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at this diagram right here, I'm going to draw the hydrogen atom, or the most common isotope. So hydrogen has one proton in the nucleus, and one electron somewhere around our nucleus, so a negative charge like that. And so we would say this is hydrogen, and then we'd put its one valence electron right there to represent the hydrogen atom, or the most common isotope. If we were to somehow take away this electron, we would only be left with the proton here. We'd only be left with the proton and nucleus. And so when we're talking about a proton, we're talking about the nucleus of a hydrogen atom, which is equal to H+. So no longer are we talking about the electron."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we were to somehow take away this electron, we would only be left with the proton here. We'd only be left with the proton and nucleus. And so when we're talking about a proton, we're talking about the nucleus of a hydrogen atom, which is equal to H+. So no longer are we talking about the electron. So let's see how this applies to an acid-base reaction. And so we start over here with water, and then we have HCl over here on the right. Now in this bond between the H and the Cl, one of those electrons came from the hydrogen, and one of them came from the chlorine."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So no longer are we talking about the electron. So let's see how this applies to an acid-base reaction. And so we start over here with water, and then we have HCl over here on the right. Now in this bond between the H and the Cl, one of those electrons came from the hydrogen, and one of them came from the chlorine. So let me just go ahead and draw those in. So the one from the hydrogen I'm going to put in blue here. And that's this electron from hydrogen right here in blue."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now in this bond between the H and the Cl, one of those electrons came from the hydrogen, and one of them came from the chlorine. So let me just go ahead and draw those in. So the one from the hydrogen I'm going to put in blue here. And that's this electron from hydrogen right here in blue. And then for chlorine, I'm going to make that electron green. So right in here like that. And so for this acid-base reaction, a lone pair of electrons in the oxygen is going to take this proton, so just the nucleus of the hydrogen atom, leaving the electron in blue behind."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And that's this electron from hydrogen right here in blue. And then for chlorine, I'm going to make that electron green. So right in here like that. And so for this acid-base reaction, a lone pair of electrons in the oxygen is going to take this proton, so just the nucleus of the hydrogen atom, leaving the electron in blue behind. And that electron in blue stays behind and ends up on the chlorine. And so let's go ahead and draw what we would form from that. We would have oxygen here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so for this acid-base reaction, a lone pair of electrons in the oxygen is going to take this proton, so just the nucleus of the hydrogen atom, leaving the electron in blue behind. And that electron in blue stays behind and ends up on the chlorine. And so let's go ahead and draw what we would form from that. We would have oxygen here. The oxygen had two bonds to hydrogen, and the oxygen just picked up another bond to hydrogen. And so let me go ahead and mark those electrons. So these electrons in here in magenta formed a new bond with that proton, so that's this bond right here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We would have oxygen here. The oxygen had two bonds to hydrogen, and the oxygen just picked up another bond to hydrogen. And so let me go ahead and mark those electrons. So these electrons in here in magenta formed a new bond with that proton, so that's this bond right here. And then we had some electrons on oxygen. Let me go ahead and make those in red. So these electrons in red on the oxygen didn't do anything, so they're still there, so they're right here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here in magenta formed a new bond with that proton, so that's this bond right here. And then we had some electrons on oxygen. Let me go ahead and make those in red. So these electrons in red on the oxygen didn't do anything, so they're still there, so they're right here. And that's going to give that oxygen a plus one formal charge. And so this is the hydronium ion, H3O+. Our other product, we would also make, we would have our chlorine, which had three lone pairs of electrons around it already."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in red on the oxygen didn't do anything, so they're still there, so they're right here. And that's going to give that oxygen a plus one formal charge. And so this is the hydronium ion, H3O+. Our other product, we would also make, we would have our chlorine, which had three lone pairs of electrons around it already. And then it picked up both of those electrons. Let me go ahead and mark them. The one in green that it had originally brought to the dot structure, and also the one in blue, the one it took from hydrogen like that."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Our other product, we would also make, we would have our chlorine, which had three lone pairs of electrons around it already. And then it picked up both of those electrons. Let me go ahead and mark them. The one in green that it had originally brought to the dot structure, and also the one in blue, the one it took from hydrogen like that. So chlorine now has a negative charge, so it's really the chloride anion, so this would be Cl minus like that. And so let's identify our Bronsted-Lowry acid and our Bronsted-Lowry base for this reaction. So let's go back over here and see what happened."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The one in green that it had originally brought to the dot structure, and also the one in blue, the one it took from hydrogen like that. So chlorine now has a negative charge, so it's really the chloride anion, so this would be Cl minus like that. And so let's identify our Bronsted-Lowry acid and our Bronsted-Lowry base for this reaction. So let's go back over here and see what happened. So the H2O, the water, acted as a proton acceptor. It accepted a proton from HCl, so water would be our Bronsted-Lowry base. And HCl donated a proton to water, so HCl would therefore be our Bronsted-Lowry acid."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go back over here and see what happened. So the H2O, the water, acted as a proton acceptor. It accepted a proton from HCl, so water would be our Bronsted-Lowry base. And HCl donated a proton to water, so HCl would therefore be our Bronsted-Lowry acid. So let's go ahead and identify conjugate acid-base pairs here. So if HCl is our Bronsted-Lowry acid, and I can think about its conjugate base over here, would be the chloride anion. So this would be the conjugate base over here, so conjugate base."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And HCl donated a proton to water, so HCl would therefore be our Bronsted-Lowry acid. So let's go ahead and identify conjugate acid-base pairs here. So if HCl is our Bronsted-Lowry acid, and I can think about its conjugate base over here, would be the chloride anion. So this would be the conjugate base over here, so conjugate base. So H2O was our Bronsted-Lowry base, and then over here we can find its conjugate acid. That's H3O+, so this would be the conjugate acid over here. So when you're looking for conjugate acid-base pairs, you're looking for one proton difference."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this would be the conjugate base over here, so conjugate base. So H2O was our Bronsted-Lowry base, and then over here we can find its conjugate acid. That's H3O+, so this would be the conjugate acid over here. So when you're looking for conjugate acid-base pairs, you're looking for one proton difference. So H2O and H3O+, are a conjugate acid-base pair, and HCl and Cl- are a conjugate acid-base pair. And if we look at what we have on the right here, we're now saying H3O+, is an acid, and Cl- is a base. And so one thing you'd think about is H3O+, donating a proton to Cl-, and so we'll draw a little tiny arrow going back to the left, because the equilibrium for this reaction lies far to the right, so you're gonna get a lot more of your products on the right here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So when you're looking for conjugate acid-base pairs, you're looking for one proton difference. So H2O and H3O+, are a conjugate acid-base pair, and HCl and Cl- are a conjugate acid-base pair. And if we look at what we have on the right here, we're now saying H3O+, is an acid, and Cl- is a base. And so one thing you'd think about is H3O+, donating a proton to Cl-, and so we'll draw a little tiny arrow going back to the left, because the equilibrium for this reaction lies far to the right, so you're gonna get a lot more of your products on the right here. But just thinking about these definitions, H3O+, would be donating a proton, and Cl- would be accepting a proton, the chloride anion would be accepting a proton. But again, we know HCl is a strong acid, so we know the equilibrium lies far to the right. So that's the idea about Bronsted-Lowry."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so one thing you'd think about is H3O+, donating a proton to Cl-, and so we'll draw a little tiny arrow going back to the left, because the equilibrium for this reaction lies far to the right, so you're gonna get a lot more of your products on the right here. But just thinking about these definitions, H3O+, would be donating a proton, and Cl- would be accepting a proton, the chloride anion would be accepting a proton. But again, we know HCl is a strong acid, so we know the equilibrium lies far to the right. So that's the idea about Bronsted-Lowry. Let's look at another definition, which is actually a little bit more broad. So this is Lewis acid and Lewis base. So a Lewis acid is an electron pair acceptor."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's the idea about Bronsted-Lowry. Let's look at another definition, which is actually a little bit more broad. So this is Lewis acid and Lewis base. So a Lewis acid is an electron pair acceptor. And so an easy way to remember this is acid acceptor. And a Lewis base is an electron pair donor. And so one way to remember that this Lewis base is an electron pair donor is to, if you think about this B, right, being lowercase, and then just flipping it around, you would get a D here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So a Lewis acid is an electron pair acceptor. And so an easy way to remember this is acid acceptor. And a Lewis base is an electron pair donor. And so one way to remember that this Lewis base is an electron pair donor is to, if you think about this B, right, being lowercase, and then just flipping it around, you would get a D here. So you would get a D. So a base is a donor. So let's look at this reaction here. And we have this cyclic ether over here on the left, and then we have borane over here on the right."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so one way to remember that this Lewis base is an electron pair donor is to, if you think about this B, right, being lowercase, and then just flipping it around, you would get a D here. So you would get a D. So a base is a donor. So let's look at this reaction here. And we have this cyclic ether over here on the left, and then we have borane over here on the right. Now, notice there's no octet of electrons around boron. Boron's only surrounded by six electrons here. And that makes it very reactive."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we have this cyclic ether over here on the left, and then we have borane over here on the right. Now, notice there's no octet of electrons around boron. Boron's only surrounded by six electrons here. And that makes it very reactive. Boron is sp2 hybridized, which means it has an empty p orbital. And so let me go ahead and represent that empty p orbital like this. So it's able to accept a pair of electrons."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And that makes it very reactive. Boron is sp2 hybridized, which means it has an empty p orbital. And so let me go ahead and represent that empty p orbital like this. So it's able to accept a pair of electrons. And the ether over here is going to donate a pair of electrons. And so let's go ahead and show what happens. The oxygen here is going to donate a pair of electrons into the empty orbital."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's able to accept a pair of electrons. And the ether over here is going to donate a pair of electrons. And so let's go ahead and show what happens. The oxygen here is going to donate a pair of electrons into the empty orbital. And there's going to be a bond that forms between the oxygen and the boron. So the ether over here is donating a pair of electrons. So that must be our Lewis base."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen here is going to donate a pair of electrons into the empty orbital. And there's going to be a bond that forms between the oxygen and the boron. So the ether over here is donating a pair of electrons. So that must be our Lewis base. And borane over here is accepting a pair of electrons. So that's our Lewis acid. Let's go ahead and draw the product for our Lewis acid base reaction here."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that must be our Lewis base. And borane over here is accepting a pair of electrons. So that's our Lewis acid. Let's go ahead and draw the product for our Lewis acid base reaction here. So we have our oxygen is now bonded to the boron. The boron is still bonded to three hydrogens. So we draw those in there like that."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw the product for our Lewis acid base reaction here. So we have our oxygen is now bonded to the boron. The boron is still bonded to three hydrogens. So we draw those in there like that. And let's follow some of our electrons here before we finish drawing everything in. So these electrons in magenta formed this bond between the oxygen and the boron. And then we also had some other electrons on that oxygen."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we draw those in there like that. And let's follow some of our electrons here before we finish drawing everything in. So these electrons in magenta formed this bond between the oxygen and the boron. And then we also had some other electrons on that oxygen. Let me go ahead and identify those. So these electrons right here in red are still on that oxygen. So they are right here on that oxygen."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we also had some other electrons on that oxygen. Let me go ahead and identify those. So these electrons right here in red are still on that oxygen. So they are right here on that oxygen. That oxygen therefore has a plus one formal charge. So plus one formal charge on oxygen. And boron gets a negative one formal charge now like that."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So they are right here on that oxygen. That oxygen therefore has a plus one formal charge. So plus one formal charge on oxygen. And boron gets a negative one formal charge now like that. And so that's one Lewis acid base reaction here. Now the Lewis acid base definition is once again more inclusive than Bronsted-Lowry. If we actually go up here to the previous reaction, we can actually classify these using the definition for Lewis acid and Lewis base."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And boron gets a negative one formal charge now like that. And so that's one Lewis acid base reaction here. Now the Lewis acid base definition is once again more inclusive than Bronsted-Lowry. If we actually go up here to the previous reaction, we can actually classify these using the definition for Lewis acid and Lewis base. And so let's look again at what's happening here. So water is donating a pair of electrons. Well according to Lewis base, electron pair donor."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we actually go up here to the previous reaction, we can actually classify these using the definition for Lewis acid and Lewis base. And so let's look again at what's happening here. So water is donating a pair of electrons. Well according to Lewis base, electron pair donor. So we could say that water, we could say this is a Lewis base. And HCl is accepting a pair of electrons. So electron pair acceptor is Lewis acid."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well according to Lewis base, electron pair donor. So we could say that water, we could say this is a Lewis base. And HCl is accepting a pair of electrons. So electron pair acceptor is Lewis acid. So we could call this a Lewis acid. So notice it doesn't matter what definition you use. If you use Bronsted-Lowry, this is your acid."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So electron pair acceptor is Lewis acid. So we could call this a Lewis acid. So notice it doesn't matter what definition you use. If you use Bronsted-Lowry, this is your acid. If you use Lewis, this is your acid. Or if you use, over here for base, this is your base according to Bronsted-Lowry. This is also a base according to Lewis."}, {"video_title": "Acid-base definitions Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you use Bronsted-Lowry, this is your acid. If you use Lewis, this is your acid. Or if you use, over here for base, this is your base according to Bronsted-Lowry. This is also a base according to Lewis. And Lewis acid and base also have particular importance in organic chemistry because you can talk about the term Lewis acid as being synonymous with electrophile. So you could say this is an electrophile. And then you could say a Lewis base is an electron pair donor."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to ask some people why we have seasons, they might say that maybe it's due to how far we are from the sun at different points in the year, at different points in Earth's orbit. And what I want to do in this video is show you why that isn't the case. So the line of reasoning would go something like this. This is the sun at the center of our solar system and roughly at the center of Earth's orbit. And let me draw Earth's orbit over here. And so the line of reasoning is that there are certain points in Earth's orbit where we're closer to the sun. And let me draw a better job than that."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the sun at the center of our solar system and roughly at the center of Earth's orbit. And let me draw Earth's orbit over here. And so the line of reasoning is that there are certain points in Earth's orbit where we're closer to the sun. And let me draw a better job than that. So let's say this is the point where we're closer to the sun. We get a little further, then we get a lot further, and then we get a little bit closer, and then we get a little bit closer, and then this is the closest point. So maybe Earth's orbit looks something like this."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me draw a better job than that. So let's say this is the point where we're closer to the sun. We get a little further, then we get a lot further, and then we get a little bit closer, and then we get a little bit closer, and then this is the closest point. So maybe Earth's orbit looks something like this. So the argument would go, look, there are points in Earth's orbit where we're closer to the sun and points where we are further from the sun. And actually that part of the argument is true. Earth's orbit is not a perfect circle, and there are points in Earth's orbit where we are closer or further away from the sun."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So maybe Earth's orbit looks something like this. So the argument would go, look, there are points in Earth's orbit where we're closer to the sun and points where we are further from the sun. And actually that part of the argument is true. Earth's orbit is not a perfect circle, and there are points in Earth's orbit where we are closer or further away from the sun. And actually when we are closest to the sun, so if Earth is right over here, there's a word for that. It's called perihelion. It just means the closest point in orbit."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Earth's orbit is not a perfect circle, and there are points in Earth's orbit where we are closer or further away from the sun. And actually when we are closest to the sun, so if Earth is right over here, there's a word for that. It's called perihelion. It just means the closest point in orbit. Perihelion. Closest point in orbit to the sun. And there is a furthest point from the sun, and this is called aphelion or aphelion."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It just means the closest point in orbit. Perihelion. Closest point in orbit to the sun. And there is a furthest point from the sun, and this is called aphelion or aphelion. I've sometimes seen it called aphelion, pronounced aphelion. Aphelion. So it is true that Earth's orbit is not a perfect circle around the sun, although it's pretty close, but it's not a perfect circle."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there is a furthest point from the sun, and this is called aphelion or aphelion. I've sometimes seen it called aphelion, pronounced aphelion. Aphelion. So it is true that Earth's orbit is not a perfect circle around the sun, although it's pretty close, but it's not a perfect circle. It has a slightly elliptical shape, and because of that, there are times in the year where we are closest to the sun, and there are times in the year where we are furthest to the sun. And the difference is about 3%, so it's not a huge difference in distance. I've really exaggerated the difference in this diagram right over here."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it is true that Earth's orbit is not a perfect circle around the sun, although it's pretty close, but it's not a perfect circle. It has a slightly elliptical shape, and because of that, there are times in the year where we are closest to the sun, and there are times in the year where we are furthest to the sun. And the difference is about 3%, so it's not a huge difference in distance. I've really exaggerated the difference in this diagram right over here. But based on this reasoning, people would say, and this is the flawed part, that when we're close to the sun, this must be the summer. And when we are furthest away from the sun, this must be the winter. And the most obvious point of evidence why this is not the case is that when it is summer at one point in the planet, it is not summer throughout the planet at that time."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I've really exaggerated the difference in this diagram right over here. But based on this reasoning, people would say, and this is the flawed part, that when we're close to the sun, this must be the summer. And when we are furthest away from the sun, this must be the winter. And the most obvious point of evidence why this is not the case is that when it is summer at one point in the planet, it is not summer throughout the planet at that time. In particular, when it is summer in the northern hemisphere, it is winter in the southern hemisphere. And when it is summer in the southern hemisphere, it is winter in the northern hemisphere. So the entire planet does not experience the seasons at the same time."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the most obvious point of evidence why this is not the case is that when it is summer at one point in the planet, it is not summer throughout the planet at that time. In particular, when it is summer in the northern hemisphere, it is winter in the southern hemisphere. And when it is summer in the southern hemisphere, it is winter in the northern hemisphere. So the entire planet does not experience the seasons at the same time. So that's probably, I guess you could say, the biggest point of data that we observe on our planet, why this by itself cannot explain the change in seasons. And in particular, it really goes against what we experience in the northern hemisphere because our perihelion right now is occurring in January. It is occurring during the winter, the northern hemisphere winter."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the entire planet does not experience the seasons at the same time. So that's probably, I guess you could say, the biggest point of data that we observe on our planet, why this by itself cannot explain the change in seasons. And in particular, it really goes against what we experience in the northern hemisphere because our perihelion right now is occurring in January. It is occurring during the winter, the northern hemisphere winter. Perihelion right now is during the northern hemisphere winter. And when we are furthest away from the sun, this is actually the northern hemisphere summer. Northern hemisphere summer."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is occurring during the winter, the northern hemisphere winter. Perihelion right now is during the northern hemisphere winter. And when we are furthest away from the sun, this is actually the northern hemisphere summer. Northern hemisphere summer. So although it might seem like a fairly intuitive idea, hey, if we are close to the sun, the whole planet is getting warmer, maybe that is summer. When we are further away, the whole planet is getting a little less energy, that is winter. The evidence we see on earth goes directly against that."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Northern hemisphere summer. So although it might seem like a fairly intuitive idea, hey, if we are close to the sun, the whole planet is getting warmer, maybe that is summer. When we are further away, the whole planet is getting a little less energy, that is winter. The evidence we see on earth goes directly against that. In particular, we don't have the same seasons in both the northern and southern hemispheres at the same time. And in particular, in the northern hemisphere, when we are closest to the sun, it is actually in January. It is actually in the middle of winter."}, {"video_title": "Seasons aren't dictated by closeness to sun Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The evidence we see on earth goes directly against that. In particular, we don't have the same seasons in both the northern and southern hemispheres at the same time. And in particular, in the northern hemisphere, when we are closest to the sun, it is actually in January. It is actually in the middle of winter. So I will leave you there in this video. I have left you just saying, okay, so the closeness to the sun does not dictate what season we are in. It is just saying, what is the reason?"}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And one thing I do want to emphasize, in the last video, these numbers that I came up with, I picked these numbers to make the math fairly simple and to give you the general idea. I don't want you to think that the most that a square kilometer can support today is exactly 1,000 people. It depends hugely on what the land is like, how much of the land you're actually using for agriculture, what crops you're planting, et cetera, et cetera, how the people are living, how many calories they need. The whole point of the last video was just to give you a framework that, wow, there is this upper bound based on how much productivity you actually get from the land. Now, what we want to think about in this video, that was kind of the last video was this axis right here, getting more and more out of the land. What I want to think about in this video is how did humans, through different technologies, how did we get by doing less and less of the labor for getting those calories out of the land? Obviously, you just don't have land and things spontaneously grow."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The whole point of the last video was just to give you a framework that, wow, there is this upper bound based on how much productivity you actually get from the land. Now, what we want to think about in this video, that was kind of the last video was this axis right here, getting more and more out of the land. What I want to think about in this video is how did humans, through different technologies, how did we get by doing less and less of the labor for getting those calories out of the land? Obviously, you just don't have land and things spontaneously grow. Well, I guess that would happen in the wild. But if you're doing agriculture, you need to put some energy into the land. You've got to work the land."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Obviously, you just don't have land and things spontaneously grow. Well, I guess that would happen in the wild. But if you're doing agriculture, you need to put some energy into the land. You've got to work the land. And so what we have over here in this chart, and this chart is derived from information from this book right over here. This is Energy and Society. And what we want to think about here is the different ways that humans have gone about to till soil."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You've got to work the land. And so what we have over here in this chart, and this chart is derived from information from this book right over here. This is Energy and Society. And what we want to think about here is the different ways that humans have gone about to till soil. So we're not even going to think about the total process or the total energy required to grow a certain crop. What we're just going to focus is one step of the agricultural process, and that is tilling the soil. And in case you're like me and you have never worked on a farm, which that's one thing I would like to change one day, is actually go through that process."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what we want to think about here is the different ways that humans have gone about to till soil. So we're not even going to think about the total process or the total energy required to grow a certain crop. What we're just going to focus is one step of the agricultural process, and that is tilling the soil. And in case you're like me and you have never worked on a farm, which that's one thing I would like to change one day, is actually go through that process. But tilling the soil is kind of churning it up. So you get the nutrients from the bottom layers to the surface. You bury all of the remains from the last harvest."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And in case you're like me and you have never worked on a farm, which that's one thing I would like to change one day, is actually go through that process. But tilling the soil is kind of churning it up. So you get the nutrients from the bottom layers to the surface. You bury all of the remains from the last harvest. You bury all of the weeds so that they die. And you also kind of get air in the soil. What it does is essentially it prepares the soil for the next agricultural cycle."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You bury all of the remains from the last harvest. You bury all of the weeds so that they die. And you also kind of get air in the soil. What it does is essentially it prepares the soil for the next agricultural cycle. So it's a process that humans have been doing since antiquity. And what I want to do in this video is think about the different ways to do it and how much energy is required to do it. And we're going to think about the energy in two ways."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What it does is essentially it prepares the soil for the next agricultural cycle. So it's a process that humans have been doing since antiquity. And what I want to do in this video is think about the different ways to do it and how much energy is required to do it. And we're going to think about the energy in two ways. How much of that energy comes from humans and how much of that energy comes from things other than humans. So just as an example, when we talk about human power, we're talking about someone literally hand plowing this field. So this woman right over here is literally she has this little cart that's digging up the soil behind her."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we're going to think about the energy in two ways. How much of that energy comes from humans and how much of that energy comes from things other than humans. So just as an example, when we talk about human power, we're talking about someone literally hand plowing this field. So this woman right over here is literally she has this little cart that's digging up the soil behind her. When we talk about oxen power, we're talking about the oxen doing most of the work. They're the ones dragging this plow, which is digging up all of the soil. And this gentleman has to kind of be there to supervise."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this woman right over here is literally she has this little cart that's digging up the soil behind her. When we talk about oxen power, we're talking about the oxen doing most of the work. They're the ones dragging this plow, which is digging up all of the soil. And this gentleman has to kind of be there to supervise. But this still is fairly intense labor that this gentleman is doing right over here. And then when we talk about tractors, we're talking about a scenario like this, where the tractor is doing most of the work of actually digging up, dragging this plow behind it, and digging up all of the soil. And from this book right over here, that's where we got these numbers."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this gentleman has to kind of be there to supervise. But this still is fairly intense labor that this gentleman is doing right over here. And then when we talk about tractors, we're talking about a scenario like this, where the tractor is doing most of the work of actually digging up, dragging this plow behind it, and digging up all of the soil. And from this book right over here, that's where we got these numbers. I'll tell you which numbers I got from them and which numbers I kind of reasoned through because I wasn't fully comfortable with the numbers that they had. But these are their numbers, that if you're of human power to till one hectare of soil, it'll take 400 hours. Oxen power, and this should be a pair, not a parry."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And from this book right over here, that's where we got these numbers. I'll tell you which numbers I got from them and which numbers I kind of reasoned through because I wasn't fully comfortable with the numbers that they had. But these are their numbers, that if you're of human power to till one hectare of soil, it'll take 400 hours. Oxen power, and this should be a pair, not a parry. That should be a pair of oxen. 65 hours. A 6-horsepower tractor, 25 hours."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Oxen power, and this should be a pair, not a parry. That should be a pair of oxen. 65 hours. A 6-horsepower tractor, 25 hours. A 50-horsepower tractor, 4 hours. And in case y'all are wondering, what is a hectare of soil? It is literally a plot of land, a hectare of soil."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "A 6-horsepower tractor, 25 hours. A 50-horsepower tractor, 4 hours. And in case y'all are wondering, what is a hectare of soil? It is literally a plot of land, a hectare of soil. Let me write it. A hectare of land, I should say, is a plot of land that is 100 meters by 100 meters. And it's roughly equal to 2 and 1\u20442 acres."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is literally a plot of land, a hectare of soil. Let me write it. A hectare of land, I should say, is a plot of land that is 100 meters by 100 meters. And it's roughly equal to 2 and 1\u20442 acres. Not exactly 2 and 1\u20442 acres. It's like 2.4, I think 2.47 something, but roughly 2 and 1\u20442 acres. So we're just thinking about how many hours to essentially dig up all of the soil for a plot of land 100 by 100 meters."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's roughly equal to 2 and 1\u20442 acres. Not exactly 2 and 1\u20442 acres. It's like 2.4, I think 2.47 something, but roughly 2 and 1\u20442 acres. So we're just thinking about how many hours to essentially dig up all of the soil for a plot of land 100 by 100 meters. So what we have over here, so clearly human takes a lot longer. Oxen, they can do a little bit faster. 6-horsepower tractor, even faster."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're just thinking about how many hours to essentially dig up all of the soil for a plot of land 100 by 100 meters. So what we have over here, so clearly human takes a lot longer. Oxen, they can do a little bit faster. 6-horsepower tractor, even faster. 50-horsepower tractor, very powerful tractor, even faster than that. Now this column right over here is the amount of energy required to actually produce and maintain the machinery used. So this is a very unintuitive thing."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "6-horsepower tractor, even faster. 50-horsepower tractor, very powerful tractor, even faster than that. Now this column right over here is the amount of energy required to actually produce and maintain the machinery used. So this is a very unintuitive thing. Whenever you think about, like for example, whenever you think about the amount of energy to plow this land over here, you tend to think, OK, well this individual is going to have to expend a lot of her energy. You don't think about the amount of energy required to actually maintain the tool, to one, build the tool that she's using, in this case a hand plow, and then to maintain that as she does it. And so this estimate, and I got these two from these fellows right over here."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is a very unintuitive thing. Whenever you think about, like for example, whenever you think about the amount of energy to plow this land over here, you tend to think, OK, well this individual is going to have to expend a lot of her energy. You don't think about the amount of energy required to actually maintain the tool, to one, build the tool that she's using, in this case a hand plow, and then to maintain that as she does it. And so this estimate, and I got these two from these fellows right over here. Actually, one of them might be a gal. But it's about 6,000 kilocalories. And this is kilocalories for lowercase c. And one thing I want to emphasize here, 1 kcal is the same thing as 1 calorie with a capital C, which is the same thing as 1,000 calories with a lowercase c. And we talked about this in the last video."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this estimate, and I got these two from these fellows right over here. Actually, one of them might be a gal. But it's about 6,000 kilocalories. And this is kilocalories for lowercase c. And one thing I want to emphasize here, 1 kcal is the same thing as 1 calorie with a capital C, which is the same thing as 1,000 calories with a lowercase c. And we talked about this in the last video. But these, when people talk about food calories, they're really talking about a calorie with a capital C. Or you could say they're talking about kilocalories. So your candy bar, 200 calories, they're talking about this right over here. In chemistry class, when you talk about the amount of energy to raise a gram of water 1 degree Celsius, you're talking about these calories here."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is kilocalories for lowercase c. And one thing I want to emphasize here, 1 kcal is the same thing as 1 calorie with a capital C, which is the same thing as 1,000 calories with a lowercase c. And we talked about this in the last video. But these, when people talk about food calories, they're really talking about a calorie with a capital C. Or you could say they're talking about kilocalories. So your candy bar, 200 calories, they're talking about this right over here. In chemistry class, when you talk about the amount of energy to raise a gram of water 1 degree Celsius, you're talking about these calories here. So in all of these numbers in this chart right over here, you can either view them as this unit, kcals, kilocalories, or calories with a capital C. They're essentially the same units that we used in the last video. And these are the same numbers that you were used to from a dietary calorie point of view. So for example, 6,000 calories, that's about how much a typical male would expend in three days."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In chemistry class, when you talk about the amount of energy to raise a gram of water 1 degree Celsius, you're talking about these calories here. So in all of these numbers in this chart right over here, you can either view them as this unit, kcals, kilocalories, or calories with a capital C. They're essentially the same units that we used in the last video. And these are the same numbers that you were used to from a dietary calorie point of view. So for example, 6,000 calories, that's about how much a typical male would expend in three days. So this is to maintain it over the course of these 400 hours in the case of the hand plow. And the total amount of calories that were needed to make the plow divided by the total number of hours. So whatever fraction of the plow's life is being used here, you use that fraction right over here to put this 6,000 calories."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So for example, 6,000 calories, that's about how much a typical male would expend in three days. So this is to maintain it over the course of these 400 hours in the case of the hand plow. And the total amount of calories that were needed to make the plow divided by the total number of hours. So whatever fraction of the plow's life is being used here, you use that fraction right over here to put this 6,000 calories. But needless to say, for at least the plow, for either the human or the oxen scenario, this isn't a significant amount of the total calories. So obviously, if you're doing it either with human power or oxen power, you're not using any gasoline. You're not using any petroleum."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So whatever fraction of the plow's life is being used here, you use that fraction right over here to put this 6,000 calories. But needless to say, for at least the plow, for either the human or the oxen scenario, this isn't a significant amount of the total calories. So obviously, if you're doing it either with human power or oxen power, you're not using any gasoline. You're not using any petroleum. In all of these scenarios, we're going to assume that someone has 10 working hours in the day. And that right over here, this is kind of a measure of how hard that person's work is. And I kind of estimated these numbers here."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You're not using any petroleum. In all of these scenarios, we're going to assume that someone has 10 working hours in the day. And that right over here, this is kind of a measure of how hard that person's work is. And I kind of estimated these numbers here. They're slightly different than what the original numbers were in this book right over here. But we're saying, look, if you are actually walking along using this hand plow, that is actually very, very vigorous activity. So it's going to require about 400 calories per hour to do this type of activity."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I kind of estimated these numbers here. They're slightly different than what the original numbers were in this book right over here. But we're saying, look, if you are actually walking along using this hand plow, that is actually very, very vigorous activity. So it's going to require about 400 calories per hour to do this type of activity. You do it over 10 hours, it's going to require 4,000 calories just to do that over 10 hours. And then we're assuming that the rest of the day, you're going to walk around, and maybe you're going to cook dinner, eat breakfast, you're going to sleep, some of it. We're assuming that the other 14 hours of the day are going to be at about 100 calories per hour."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's going to require about 400 calories per hour to do this type of activity. You do it over 10 hours, it's going to require 4,000 calories just to do that over 10 hours. And then we're assuming that the rest of the day, you're going to walk around, and maybe you're going to cook dinner, eat breakfast, you're going to sleep, some of it. We're assuming that the other 14 hours of the day are going to be at about 100 calories per hour. And so this is the total. If someone were to, using this technique, work for a total of 10 hours, this is how many calories they would consume in the day. And you could see this is the most labor intensive."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're assuming that the other 14 hours of the day are going to be at about 100 calories per hour. And so this is the total. If someone were to, using this technique, work for a total of 10 hours, this is how many calories they would consume in the day. And you could see this is the most labor intensive. So it looks like that they would consume the most calories per day. These two are the least labor intensive. You're sitting on a tractor, although that still requires more calories than sleeping or watching TV."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you could see this is the most labor intensive. So it looks like that they would consume the most calories per day. These two are the least labor intensive. You're sitting on a tractor, although that still requires more calories than sleeping or watching TV. And so that's the number of calories they would consume. Now this right over here, and this is the interesting number, or one of the really interesting numbers. Based on all of these assumptions, this is the total human input in calories to do this task, to till this one hectare of soil."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You're sitting on a tractor, although that still requires more calories than sleeping or watching TV. And so that's the number of calories they would consume. Now this right over here, and this is the interesting number, or one of the really interesting numbers. Based on all of these assumptions, this is the total human input in calories to do this task, to till this one hectare of soil. So over here, you're using 5,400 calories a day. If you're working 10 hours per day and it requires 400 total hours, you're going to be working 40 days, 400 divided by 10. 40 days times 5,400 calories per day, it's going to take a human, just the human part, not even thinking about the 6,000 calories necessary to maintain and make that plow."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Based on all of these assumptions, this is the total human input in calories to do this task, to till this one hectare of soil. So over here, you're using 5,400 calories a day. If you're working 10 hours per day and it requires 400 total hours, you're going to be working 40 days, 400 divided by 10. 40 days times 5,400 calories per day, it's going to take a human, just the human part, not even thinking about the 6,000 calories necessary to maintain and make that plow. The human is going to spend 216,000 calories to till, to plow that one hectare of land. And if you add the other 6,000 in for the actual plow, and you could debate what this number should be, but it's not a significant number compared to this, you get about 222,000 total calories. When you go to the oxen situation, you're acquiring fewer hours, and each hour it requires a little less calories."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "40 days times 5,400 calories per day, it's going to take a human, just the human part, not even thinking about the 6,000 calories necessary to maintain and make that plow. The human is going to spend 216,000 calories to till, to plow that one hectare of land. And if you add the other 6,000 in for the actual plow, and you could debate what this number should be, but it's not a significant number compared to this, you get about 222,000 total calories. When you go to the oxen situation, you're acquiring fewer hours, and each hour it requires a little less calories. This is still labor intensive, but not as labor intensive as what this woman right over here is doing. So on a daily basis, you're using a little bit fewer calories, but since you're only doing 6 and 1 1 days of this, 65 hours divided by 10, you've significantly reduced the number of calories, the total number of calories, that the human needs to put into this task. Now, there still is other energy being done."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When you go to the oxen situation, you're acquiring fewer hours, and each hour it requires a little less calories. This is still labor intensive, but not as labor intensive as what this woman right over here is doing. So on a daily basis, you're using a little bit fewer calories, but since you're only doing 6 and 1 1 days of this, 65 hours divided by 10, you've significantly reduced the number of calories, the total number of calories, that the human needs to put into this task. Now, there still is other energy being done. Now all of a sudden, the oxen have gotten involved. And if you assume that each oxen consumes about 20,000 calories a day, and you have two of them, so 40,000 calories per day just to feed the oxen, and you're going to do that for 6 and 1 1 days, 65 divided by 10, the oxen are going to consume 260,000 calories to do this task. So the total energy input here now has gone up."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, there still is other energy being done. Now all of a sudden, the oxen have gotten involved. And if you assume that each oxen consumes about 20,000 calories a day, and you have two of them, so 40,000 calories per day just to feed the oxen, and you're going to do that for 6 and 1 1 days, 65 divided by 10, the oxen are going to consume 260,000 calories to do this task. So the total energy input here now has gone up. So this is an interesting phenomenon that is going on right over here. What the human is putting in as we get better and better technology goes down substantially, 216,000 to 33,000. And we'll see with the tractor it goes down even more."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the total energy input here now has gone up. So this is an interesting phenomenon that is going on right over here. What the human is putting in as we get better and better technology goes down substantially, 216,000 to 33,000. And we'll see with the tractor it goes down even more. But the total energy, if you include the amount of energy that the oxen have to put in, or if you include the amount of energy due to the gasoline that has to be used for the tractor, the total amount of energy is going up to plow that field. But the human energy goes down dramatically. Now, the last thing I want to highlight here, and these are where my numbers depart a little bit, or fairly significantly, from this original study right over here, this original estimate, is the machinery input on the tractors."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we'll see with the tractor it goes down even more. But the total energy, if you include the amount of energy that the oxen have to put in, or if you include the amount of energy due to the gasoline that has to be used for the tractor, the total amount of energy is going up to plow that field. But the human energy goes down dramatically. Now, the last thing I want to highlight here, and these are where my numbers depart a little bit, or fairly significantly, from this original study right over here, this original estimate, is the machinery input on the tractors. So if you look this up, and you can Google search it, they have much larger numbers here. But I did a little research, and it looks like for most petroleum-based, combustion-based engines or vehicles, roughly 20% of the total energy that's used in fuel, 20% of that energy is used for the actual production and maintenance of that vehicle over its life. So what we did over here is we said, OK, for a 6-horsepower tractor, I used their numbers, where you're going to have to use 25 hours to do it, it's going to use this much petroleum, assuming that it uses 23.5 liters of gas or petroleum over 25 hours."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, the last thing I want to highlight here, and these are where my numbers depart a little bit, or fairly significantly, from this original study right over here, this original estimate, is the machinery input on the tractors. So if you look this up, and you can Google search it, they have much larger numbers here. But I did a little research, and it looks like for most petroleum-based, combustion-based engines or vehicles, roughly 20% of the total energy that's used in fuel, 20% of that energy is used for the actual production and maintenance of that vehicle over its life. So what we did over here is we said, OK, for a 6-horsepower tractor, I used their numbers, where you're going to have to use 25 hours to do it, it's going to use this much petroleum, assuming that it uses 23.5 liters of gas or petroleum over 25 hours. And then I just took 20% of that number for saying, well, how much energy had to be used to maintain that vehicle over that amount of time? And if you think about what fraction of this vehicle's life that 25 hours represents, that fraction times the total amount of energy required to produce that. So remember, these things are made out of metal."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what we did over here is we said, OK, for a 6-horsepower tractor, I used their numbers, where you're going to have to use 25 hours to do it, it's going to use this much petroleum, assuming that it uses 23.5 liters of gas or petroleum over 25 hours. And then I just took 20% of that number for saying, well, how much energy had to be used to maintain that vehicle over that amount of time? And if you think about what fraction of this vehicle's life that 25 hours represents, that fraction times the total amount of energy required to produce that. So remember, these things are made out of metal. They had to be made in furnaces. So just producing a vehicle requires a lot of energy. And so this right over here is 20%."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So remember, these things are made out of metal. They had to be made in furnaces. So just producing a vehicle requires a lot of energy. And so this right over here is 20%. And I just use that rule of thumb for most petroleum-based or combustion-based vehicles. The 20% of the total energy expenditure over the course of that vehicle's life is roughly equal to the amount of energy used to produce that vehicle. But either way, you go all the way over here, the human has to spend less calories sitting on the vehicle, so they spend less calories per day."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this right over here is 20%. And I just use that rule of thumb for most petroleum-based or combustion-based vehicles. The 20% of the total energy expenditure over the course of that vehicle's life is roughly equal to the amount of energy used to produce that vehicle. But either way, you go all the way over here, the human has to spend less calories sitting on the vehicle, so they spend less calories per day. And then the total human input right over here for the 6-horsepower tractor is going to take them 2 and 1 half days, 25 hours at 10 hours a day, is going to be 8,500 calories. But of course, you have the petroleum used and then some estimate of the amount of energy used to produce that tractor. And you're just taking the fraction over that 25 hours."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But either way, you go all the way over here, the human has to spend less calories sitting on the vehicle, so they spend less calories per day. And then the total human input right over here for the 6-horsepower tractor is going to take them 2 and 1 half days, 25 hours at 10 hours a day, is going to be 8,500 calories. But of course, you have the petroleum used and then some estimate of the amount of energy used to produce that tractor. And you're just taking the fraction over that 25 hours. You're not taking the entire life of that 6-horsepower tractor. To produce a 6-horsepower tractor, this number would be much, much larger. If you talked about the total number of energy, we're just taking the small fraction of its life that we're using it right over here."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you're just taking the fraction over that 25 hours. You're not taking the entire life of that 6-horsepower tractor. To produce a 6-horsepower tractor, this number would be much, much larger. If you talked about the total number of energy, we're just taking the small fraction of its life that we're using it right over here. Same thing for the 50-horsepower tractor. But either way you look at it, the human, and this is the really interesting thing, humans, by going from human power all the way to a 50-horsepower tractor, you're getting almost a factor of 200 improvement in terms of how little energy has to be put in by the human to till that land. But you actually get a total increase if you factor in things like the petroleum and then definitely the amount of energy to actually produce that machine."}, {"video_title": "Energy inputs for tilling a hectare of land Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you talked about the total number of energy, we're just taking the small fraction of its life that we're using it right over here. Same thing for the 50-horsepower tractor. But either way you look at it, the human, and this is the really interesting thing, humans, by going from human power all the way to a 50-horsepower tractor, you're getting almost a factor of 200 improvement in terms of how little energy has to be put in by the human to till that land. But you actually get a total increase if you factor in things like the petroleum and then definitely the amount of energy to actually produce that machine. So anyway, hopefully you found that interesting. I find this kind of, it's something that you don't think a lot about. How much energy input has to be put in?"}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Different molecules can absorb different wavelengths of light. And if a molecule happens to absorb light in the ultraviolet or the visible region of the electromagnetic spectrum, we can find the wavelength or wavelengths of light that are absorbed by that compound by using a UV-Vis spectrophotometer. And essentially what that does is it shines light with a range of wavelengths. So the wavelengths range from approximately 200 nanometers all the way up to 800 nanometers. And so we shine that range of wavelengths of light through a sample of the compound and you get an absorption spectrum. And so here is an absorption spectrum for this molecule, for 1,3-butadiene. And if we look over here, we can see that this molecule absorbs most strongly right about here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the wavelengths range from approximately 200 nanometers all the way up to 800 nanometers. And so we shine that range of wavelengths of light through a sample of the compound and you get an absorption spectrum. And so here is an absorption spectrum for this molecule, for 1,3-butadiene. And if we look over here, we can see that this molecule absorbs most strongly right about here. And if we drop down, we can see what wavelength of light is absorbed most strongly by the compound. And so we see that's just under 220 nanometers. And it turns out to be 217 nanometers."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if we look over here, we can see that this molecule absorbs most strongly right about here. And if we drop down, we can see what wavelength of light is absorbed most strongly by the compound. And so we see that's just under 220 nanometers. And it turns out to be 217 nanometers. So we call this lambda max. So the wavelength of light absorbed by this molecule is about 217 nanometers. It absorbs in the UV region."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it turns out to be 217 nanometers. So we call this lambda max. So the wavelength of light absorbed by this molecule is about 217 nanometers. It absorbs in the UV region. Therefore, butadiene does not have any color. It's colorless. Let's look at the dot structure a little bit more carefully here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It absorbs in the UV region. Therefore, butadiene does not have any color. It's colorless. Let's look at the dot structure a little bit more carefully here. We have four carbons. And all four of these carbons, each one is sp2 hybridized, which means each one of those carbons has a p orbital. So we're talking about four p orbitals here, or four atomic orbitals."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the dot structure a little bit more carefully here. We have four carbons. And all four of these carbons, each one is sp2 hybridized, which means each one of those carbons has a p orbital. So we're talking about four p orbitals here, or four atomic orbitals. And when you're dealing with molecular orbital theory, four atomic orbitals recombine to form four molecular orbitals, two bonding molecular orbitals and two anti-bonding molecular orbitals. So let's go over here and let's look at the four molecular orbitals. And we're going to focus in on the left side first."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're talking about four p orbitals here, or four atomic orbitals. And when you're dealing with molecular orbital theory, four atomic orbitals recombine to form four molecular orbitals, two bonding molecular orbitals and two anti-bonding molecular orbitals. So let's go over here and let's look at the four molecular orbitals. And we're going to focus in on the left side first. The bonding molecular orbitals are lower in energy than the anti-bonding ones. So this orbital and this orbital, these are our bonding molecular orbitals here. And this one and this one are the anti-bonding molecular orbitals."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to focus in on the left side first. The bonding molecular orbitals are lower in energy than the anti-bonding ones. So this orbital and this orbital, these are our bonding molecular orbitals here. And this one and this one are the anti-bonding molecular orbitals. And you could see energy. So energy is increasing. And so the anti-bonding molecular orbitals are higher in energy."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this one and this one are the anti-bonding molecular orbitals. And you could see energy. So energy is increasing. And so the anti-bonding molecular orbitals are higher in energy. Let's look at the dot structure again for butadiene. And let's see how many pi electrons we have. So here are two pi electrons."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the anti-bonding molecular orbitals are higher in energy. Let's look at the dot structure again for butadiene. And let's see how many pi electrons we have. So here are two pi electrons. And here are two pi electrons. So a total of four pi electrons. And when you're thinking about molecular orbitals, you can think about electron configurations."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here are two pi electrons. And here are two pi electrons. So a total of four pi electrons. And when you're thinking about molecular orbitals, you can think about electron configurations. So we have four electrons. And where do we put those electrons? We're going to put them in the lowest energy orbitals first."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And when you're thinking about molecular orbitals, you can think about electron configurations. So we have four electrons. And where do we put those electrons? We're going to put them in the lowest energy orbitals first. And we're also going to pair our spins. So four electrons. We're going to put two into this bonding molecular orbital."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're going to put them in the lowest energy orbitals first. And we're also going to pair our spins. So four electrons. We're going to put two into this bonding molecular orbital. And we paired our spins. And then two into this bonding molecular orbitals. So the four pi electrons go into the bonding molecular orbitals when you're talking about the ground state."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're going to put two into this bonding molecular orbital. And we paired our spins. And then two into this bonding molecular orbitals. So the four pi electrons go into the bonding molecular orbitals when you're talking about the ground state. So here's the ground state of butadiene. So next, we shine light on butadiene. And the molecule is going to absorb energy from the light."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the four pi electrons go into the bonding molecular orbitals when you're talking about the ground state. So here's the ground state of butadiene. So next, we shine light on butadiene. And the molecule is going to absorb energy from the light. And let's look at that here. So there's a difference in energy. There's a difference in energy between the orbitals."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the molecule is going to absorb energy from the light. And let's look at that here. So there's a difference in energy. There's a difference in energy between the orbitals. And in particular, we're concerned about these two orbitals right here. So there's a difference in energy between these two orbitals. This orbital down here, this is occupied by electrons."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "There's a difference in energy between the orbitals. And in particular, we're concerned about these two orbitals right here. So there's a difference in energy between these two orbitals. This orbital down here, this is occupied by electrons. And it's higher in energy than this orbital. So this is the highest occupied molecular orbital. So highest occupied molecular orbital, or HOMO."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This orbital down here, this is occupied by electrons. And it's higher in energy than this orbital. So this is the highest occupied molecular orbital. So highest occupied molecular orbital, or HOMO. This orbital right here is unoccupied. The antibonding molecular orbital right now is unoccupied. And it's lower in energy than this antibonding molecular orbital."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So highest occupied molecular orbital, or HOMO. This orbital right here is unoccupied. The antibonding molecular orbital right now is unoccupied. And it's lower in energy than this antibonding molecular orbital. So this is the lowest unoccupied molecular orbital. So when you're talking about a molecule absorbing energy, we're concerned about the HOMO, the highest occupied molecular orbital, and the LUMO, the lowest unoccupied molecular orbital. The energy difference between those two orbitals is what we're thinking about."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it's lower in energy than this antibonding molecular orbital. So this is the lowest unoccupied molecular orbital. So when you're talking about a molecule absorbing energy, we're concerned about the HOMO, the highest occupied molecular orbital, and the LUMO, the lowest unoccupied molecular orbital. The energy difference between those two orbitals is what we're thinking about. So the molecule absorbs energy. And a pi electron absorbs energy from the light and is promoted to a higher energy level. So let me go ahead and write over here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The energy difference between those two orbitals is what we're thinking about. So the molecule absorbs energy. And a pi electron absorbs energy from the light and is promoted to a higher energy level. So let me go ahead and write over here. Now we're talking about the excited state. So we shine light on the molecule. So this is the excited state of butadiene."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write over here. Now we're talking about the excited state. So we shine light on the molecule. So this is the excited state of butadiene. And these two pi electrons stay there. One of these pi electrons stays here. And one of the pi electrons absorbs the energy from the light and is promoted to a higher energy level."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is the excited state of butadiene. And these two pi electrons stay there. One of these pi electrons stays here. And one of the pi electrons absorbs the energy from the light and is promoted to a higher energy level. So I'm saying this one right here was promoted to a higher energy level. It goes from the HOMO to the LUMO. And it had to absorb a specific amount of energy in order to do that."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And one of the pi electrons absorbs the energy from the light and is promoted to a higher energy level. So I'm saying this one right here was promoted to a higher energy level. It goes from the HOMO to the LUMO. And it had to absorb a specific amount of energy in order to do that. So it had to absorb the right amount of energy in order to make that transition. And we know that energy came from the light. And we also know the energy of a photon of light is equal to h, where h is Planck's constant, times the frequency of light, which is nu."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And it had to absorb a specific amount of energy in order to do that. So it had to absorb the right amount of energy in order to make that transition. And we know that energy came from the light. And we also know the energy of a photon of light is equal to h, where h is Planck's constant, times the frequency of light, which is nu. And over here for the absorption spectrum, we have everything in wavelengths. So we need to write the energy in terms of a wavelength. And we know that the frequency of light and the wavelength of light are related by the speed of light is equal to the wavelength times the frequency."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we also know the energy of a photon of light is equal to h, where h is Planck's constant, times the frequency of light, which is nu. And over here for the absorption spectrum, we have everything in wavelengths. So we need to write the energy in terms of a wavelength. And we know that the frequency of light and the wavelength of light are related by the speed of light is equal to the wavelength times the frequency. And so the frequency is equal to the speed of light over the wavelength. And we can take that. So frequency is equal to c over lambda and plug it into here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we know that the frequency of light and the wavelength of light are related by the speed of light is equal to the wavelength times the frequency. And so the frequency is equal to the speed of light over the wavelength. And we can take that. So frequency is equal to c over lambda and plug it into here. So now we have the energy. The energy is equal to h times c over lambda. And this is really important."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So frequency is equal to c over lambda and plug it into here. So now we have the energy. The energy is equal to h times c over lambda. And this is really important. So energy and wavelength are inversely proportional to each other. And you can think about one wavelength giving you a specific amount of energy. And so this energy difference, this energy difference between the HOMO and the LUMO, this corresponds to a wavelength."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this is really important. So energy and wavelength are inversely proportional to each other. And you can think about one wavelength giving you a specific amount of energy. And so this energy difference, this energy difference between the HOMO and the LUMO, this corresponds to a wavelength. And if we go over here to the absorption spectrum for butadiene, we're talking about a wavelength of 217 nanometers. And so at first it might be a little bit confusing because it looks like we have a very broad range of wavelengths that are absorbed here. And don't worry about that too much."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this energy difference, this energy difference between the HOMO and the LUMO, this corresponds to a wavelength. And if we go over here to the absorption spectrum for butadiene, we're talking about a wavelength of 217 nanometers. And so at first it might be a little bit confusing because it looks like we have a very broad range of wavelengths that are absorbed here. And don't worry about that too much. This just results from the different vibrations and rotations of the molecule, which can change the energy differences slightly. And so we don't see one exact wavelength. We end up seeing this broad band of wavelengths being absorbed here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And don't worry about that too much. This just results from the different vibrations and rotations of the molecule, which can change the energy differences slightly. And so we don't see one exact wavelength. We end up seeing this broad band of wavelengths being absorbed here. So what you do is you just look for the one that's absorbed most strongly and think about that as being the wavelength that corresponds to the energy difference between these two orbitals here. So that's how to think about it. All right, let's look at another molecule here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We end up seeing this broad band of wavelengths being absorbed here. So what you do is you just look for the one that's absorbed most strongly and think about that as being the wavelength that corresponds to the energy difference between these two orbitals here. So that's how to think about it. All right, let's look at another molecule here. So instead of butadiene, let's look at this molecule. So we have ethanol. So here's our dot structure."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at another molecule here. So instead of butadiene, let's look at this molecule. So we have ethanol. So here's our dot structure. And if we look at this molecule, we know we have two pi electrons here for ethanol. So two pi electrons. We know that those electrons are going to go into the bonding molecular orbital."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's our dot structure. And if we look at this molecule, we know we have two pi electrons here for ethanol. So two pi electrons. We know that those electrons are going to go into the bonding molecular orbital. So let me draw a line right here on this diagram. So this is our bonding molecular orbital down here. And so we're talking about two pi electrons."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We know that those electrons are going to go into the bonding molecular orbital. So let me draw a line right here on this diagram. So this is our bonding molecular orbital down here. And so we're talking about two pi electrons. So let's put in our two pi electrons into here. And then let me just go ahead and change colors up here. So this up here is our antibonding molecular orbital, which we call pi star."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so we're talking about two pi electrons. So let's put in our two pi electrons into here. And then let me just go ahead and change colors up here. So this up here is our antibonding molecular orbital, which we call pi star. So there's an energy difference between the bonding molecular orbital and the antibonding molecular orbital. So this is delta E. And we talked about the fact that this corresponds to a certain wavelength of light. And so ethanol can have, when it promotes one of these pi electrons up, it can have a pi."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this up here is our antibonding molecular orbital, which we call pi star. So there's an energy difference between the bonding molecular orbital and the antibonding molecular orbital. So this is delta E. And we talked about the fact that this corresponds to a certain wavelength of light. And so ethanol can have, when it promotes one of these pi electrons up, it can have a pi. We call this a pi to pi star transition. So the molecule is going to absorb energy. And the energy, let me use a different color here, the energy corresponds to a wavelength of light."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so ethanol can have, when it promotes one of these pi electrons up, it can have a pi. We call this a pi to pi star transition. So the molecule is going to absorb energy. And the energy, let me use a different color here, the energy corresponds to a wavelength of light. So this energy difference between our two orbitals. And it turns out that this pi to pi star transition is approximately 180 nanometers, which is below the range of what you're usually measuring when you're using a UV-Vis spectrophotometer. But we have another possibility here, too."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the energy, let me use a different color here, the energy corresponds to a wavelength of light. So this energy difference between our two orbitals. And it turns out that this pi to pi star transition is approximately 180 nanometers, which is below the range of what you're usually measuring when you're using a UV-Vis spectrophotometer. But we have another possibility here, too. So let me go ahead and highlight a lone pair of electrons here on the oxygen. So we have a lone pair. So we have non-bonding electrons."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But we have another possibility here, too. So let me go ahead and highlight a lone pair of electrons here on the oxygen. So we have a lone pair. So we have non-bonding electrons. And non-bonding electrons occupy a non-bonding orbital, which is actually a little bit higher in energy than our bonding molecular orbital. So another possibility, we call this n right here. So this is a non-bonding orbital."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have non-bonding electrons. And non-bonding electrons occupy a non-bonding orbital, which is actually a little bit higher in energy than our bonding molecular orbital. So another possibility, we call this n right here. So this is a non-bonding orbital. So non-bonding orbital here. And we can put some electrons into that orbital. So we put those two electrons into the non-bonding orbital."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is a non-bonding orbital. So non-bonding orbital here. And we can put some electrons into that orbital. So we put those two electrons into the non-bonding orbital. And we can have a different type of transition. So we're still talking about a pi star, an antibonding molecular orbital right here. We can have a n to pi star transition."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we put those two electrons into the non-bonding orbital. And we can have a different type of transition. So we're still talking about a pi star, an antibonding molecular orbital right here. We can have a n to pi star transition. So we can have an n to pi star transition as well, since we have a carbonyl compound. So we're not just talking about pi electrons here. The carbonyl, we can think about a non-bonding electron here."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We can have a n to pi star transition. So we can have an n to pi star transition as well, since we have a carbonyl compound. So we're not just talking about pi electrons here. The carbonyl, we can think about a non-bonding electron here. And let's think about this energy difference. So this energy difference is smaller than before. So this energy difference is smaller than this energy difference."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The carbonyl, we can think about a non-bonding electron here. And let's think about this energy difference. So this energy difference is smaller than before. So this energy difference is smaller than this energy difference. And so what would happen to the wavelength of light that's absorbed? So if we have a smaller energy difference, energy and wavelength are inversely proportional. So this must be a longer wavelength."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this energy difference is smaller than this energy difference. And so what would happen to the wavelength of light that's absorbed? So if we have a smaller energy difference, energy and wavelength are inversely proportional. So this must be a longer wavelength. So this absorbs light at a different wavelength, so a higher wavelength. And it turns out to be, let me go ahead and change colors here, so this energy transition corresponds to a wavelength of light that's approximately 290 nanometers. And so this n to pi star transition, a lower difference, a smaller difference in energy, I should say, corresponding to a higher wavelength."}, {"video_title": "UV Vis spectroscopy Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this must be a longer wavelength. So this absorbs light at a different wavelength, so a higher wavelength. And it turns out to be, let me go ahead and change colors here, so this energy transition corresponds to a wavelength of light that's approximately 290 nanometers. And so this n to pi star transition, a lower difference, a smaller difference in energy, I should say, corresponding to a higher wavelength. This is an important concept. So as you decrease the energy difference between your orbitals, you're going to increase the wavelength of light that's absorbed. And we'll talk much more about that in the next few videos, because that's where the idea of color comes in."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And since we're not concerned with things like mechanisms and reactions in this video, we don't need to worry about details like if this oxygen is really this oxygen. It depends on how you're making the ester. All we're concerned about is how to name the ester. And so let's think about how to approach it. The first thing you would do is look at this R prime group, so the one on this oxygen. And you're going to name this R prime group as an alkyl group. And then you're going to look over here and think about the carboxylic acid over here on the left to name this portion of your ester."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And so let's think about how to approach it. The first thing you would do is look at this R prime group, so the one on this oxygen. And you're going to name this R prime group as an alkyl group. And then you're going to look over here and think about the carboxylic acid over here on the left to name this portion of your ester. Let's go ahead and do an example. So if we're going to name this ester down here, once again, we look at the R prime group and name that as an alkyl group. So that's a two-carbon alkyl group."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And then you're going to look over here and think about the carboxylic acid over here on the left to name this portion of your ester. Let's go ahead and do an example. So if we're going to name this ester down here, once again, we look at the R prime group and name that as an alkyl group. So that's a two-carbon alkyl group. So here's one carbon, and here's the second carbon. And so a two-carbon alkyl group would be ethyl. So let's go ahead and write that down."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So that's a two-carbon alkyl group. So here's one carbon, and here's the second carbon. And so a two-carbon alkyl group would be ethyl. So let's go ahead and write that down. So we would have ethyl to start with. Next, we think about this portion of the ester. We can think about that as coming from this carboxylic acid."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that down. So we would have ethyl to start with. Next, we think about this portion of the ester. We can think about that as coming from this carboxylic acid. And we know this carboxylic acid is called acetic acid. So let's go ahead and write that. So this is acetic acid here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "We can think about that as coming from this carboxylic acid. And we know this carboxylic acid is called acetic acid. So let's go ahead and write that. So this is acetic acid here. And to finish naming off our ester, we're going to drop the ending for our acetic acid. So we're going to drop. Let me go ahead and change colors here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this is acetic acid here. And to finish naming off our ester, we're going to drop the ending for our acetic acid. So we're going to drop. Let me go ahead and change colors here. So we're going to drop the IC part and the acid. And then we're going to add 8 instead. So we would have this portion plus 8."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and change colors here. So we're going to drop the IC part and the acid. And then we're going to add 8 instead. So we would have this portion plus 8. So let's go ahead and write that. And let's go ahead and use red for that. So we would have this portion."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So we would have this portion plus 8. So let's go ahead and write that. And let's go ahead and use red for that. So we would have this portion. We drop the IC and the acid, and we add 8. So the name of this ester would be ethyl acetate. So instead of ethyl acetate, if we had named this using IUPAC nomenclature, if we had called this ethanoic acid, which most people would not do that, but if we called it ethanoic acid, we could do the same thing."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So we would have this portion. We drop the IC and the acid, and we add 8. So the name of this ester would be ethyl acetate. So instead of ethyl acetate, if we had named this using IUPAC nomenclature, if we had called this ethanoic acid, which most people would not do that, but if we called it ethanoic acid, we could do the same thing. We could go ahead and drop the IC and then the acid part and add 8. So once again, we would still have the ethyl portion. So this would still be ethyl."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So instead of ethyl acetate, if we had named this using IUPAC nomenclature, if we had called this ethanoic acid, which most people would not do that, but if we called it ethanoic acid, we could do the same thing. We could go ahead and drop the IC and then the acid part and add 8. So once again, we would still have the ethyl portion. So this would still be ethyl. And then when we're naming the rest of it, we would take this portion and add 8. So it would be ethanoate. So another name for this would be ethyl ethanoate."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this would still be ethyl. And then when we're naming the rest of it, we would take this portion and add 8. So it would be ethanoate. So another name for this would be ethyl ethanoate. So two different ways to name this ester, although you are most likely going to see this form. Ethyl acetate is what everyone says. So let's go ahead and name another ester."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So another name for this would be ethyl ethanoate. So two different ways to name this ester, although you are most likely going to see this form. Ethyl acetate is what everyone says. So let's go ahead and name another ester. So once again, our goal is to name the ester over here on the right. And we start with the R prime group. The R prime group is the one that's attached to our oxygen."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and name another ester. So once again, our goal is to name the ester over here on the right. And we start with the R prime group. The R prime group is the one that's attached to our oxygen. So here's our R prime group attached to this oxygen. And notice the symmetry. We have another one over here as well."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "The R prime group is the one that's attached to our oxygen. So here's our R prime group attached to this oxygen. And notice the symmetry. We have another one over here as well. So we have two ethyl groups this time. So we're going to go ahead and write diethyl to start naming our ester. And to finish naming our ester, let's think about the carboxylic acid portion."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "We have another one over here as well. So we have two ethyl groups this time. So we're going to go ahead and write diethyl to start naming our ester. And to finish naming our ester, let's think about the carboxylic acid portion. So if we're thinking about this portion right here, so this was derived. You could think about this as being derived from this dicarboxylic acid over here. The IUPAC name would be propane dioic acid."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And to finish naming our ester, let's think about the carboxylic acid portion. So if we're thinking about this portion right here, so this was derived. You could think about this as being derived from this dicarboxylic acid over here. The IUPAC name would be propane dioic acid. The common name would be malonic acid. So this would be malonic acid. So let's go ahead and write that."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "The IUPAC name would be propane dioic acid. The common name would be malonic acid. So this would be malonic acid. So let's go ahead and write that. So malonic acid. And to finish the name of our ester, remember we're going to drop the IC and the acid. So we drop all of this and add 8."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write that. So malonic acid. And to finish the name of our ester, remember we're going to drop the IC and the acid. So we drop all of this and add 8. So this would be malonate. So I'll go ahead and write that in here. So the name of this molecule, this ester, would be diethyl malonate, like that."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So we drop all of this and add 8. So this would be malonate. So I'll go ahead and write that in here. So the name of this molecule, this ester, would be diethyl malonate, like that. Let's do another one. Let's look at another ester. So a very famous ester over here on the right."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So the name of this molecule, this ester, would be diethyl malonate, like that. Let's do another one. Let's look at another ester. So a very famous ester over here on the right. So you might recognize this as being wintergreen. So let's go ahead and name it. Let's find the ester portion of our molecule."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So a very famous ester over here on the right. So you might recognize this as being wintergreen. So let's go ahead and name it. Let's find the ester portion of our molecule. So the ester portion of our molecule is over here. And we start by naming the R prime group. So here's our R prime group on this oxygen."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "Let's find the ester portion of our molecule. So the ester portion of our molecule is over here. And we start by naming the R prime group. So here's our R prime group on this oxygen. That's one carbon, so that's a methyl group. So let's go ahead and write that. So we have methyl."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So here's our R prime group on this oxygen. That's one carbon, so that's a methyl group. So let's go ahead and write that. So we have methyl. And then we think about the carboxylic acid. The corresponding carboxylic acid would be this portion. And we see it over here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So we have methyl. And then we think about the carboxylic acid. The corresponding carboxylic acid would be this portion. And we see it over here. And we know this is salicylic acid. So let's go ahead and write that. So this would be salicylic acid, which we've talked about in earlier videos."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And we see it over here. And we know this is salicylic acid. So let's go ahead and write that. So this would be salicylic acid, which we've talked about in earlier videos. So once again, if you want to finish the name of your ester, you would have to drop your IC and your acid ending and add 8, so salicylate. So let's go ahead and write that. So this would be methyl salicylate as the name for this ester, which is, of course, oil of wintergreen."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this would be salicylic acid, which we've talked about in earlier videos. So once again, if you want to finish the name of your ester, you would have to drop your IC and your acid ending and add 8, so salicylate. So let's go ahead and write that. So this would be methyl salicylate as the name for this ester, which is, of course, oil of wintergreen. And of course, one thing that esters have is a lot of esters have really nice smells to them. It's one of the properties of esters. And so wintergreen is one of the best esters."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this would be methyl salicylate as the name for this ester, which is, of course, oil of wintergreen. And of course, one thing that esters have is a lot of esters have really nice smells to them. It's one of the properties of esters. And so wintergreen is one of the best esters. And there are many, many, many more that you can make using something like a Fischer esterification reaction. Let's look at one more ester to name. So let's look at how to name this one over here on the right."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And so wintergreen is one of the best esters. And there are many, many, many more that you can make using something like a Fischer esterification reaction. Let's look at one more ester to name. So let's look at how to name this one over here on the right. So once again, we first start with our alkyl group coming off of this oxygen. And once again, we have a methyl group. So let's go ahead and write methyl here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at how to name this one over here on the right. So once again, we first start with our alkyl group coming off of this oxygen. And once again, we have a methyl group. So let's go ahead and write methyl here. And when we're thinking about our carboxylic acid portion, so this portion right here, so I've drawn the carboxylic acid over here on the left. And if you remember how to name that, we have a cyclohexane ring. So this part would be cyclohexane."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write methyl here. And when we're thinking about our carboxylic acid portion, so this portion right here, so I've drawn the carboxylic acid over here on the left. And if you remember how to name that, we have a cyclohexane ring. So this part would be cyclohexane. And then we have a carboxylic acid coming off of that. So cyclohexane carboxylic acid, kind of long here. And once again, we're going to drop our ending."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this part would be cyclohexane. And then we have a carboxylic acid coming off of that. So cyclohexane carboxylic acid, kind of long here. And once again, we're going to drop our ending. We're going to drop the IC and the acid and add 8. So cyclohexane carboxylate would be the name. So let's hopefully have enough room over here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we're going to drop our ending. We're going to drop the IC and the acid and add 8. So cyclohexane carboxylate would be the name. So let's hopefully have enough room over here. So methyl cyclohexane carboxylate. See if I can squeeze this in. So methyl cyclohexane carboxylate would be the name for our ester."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's hopefully have enough room over here. So methyl cyclohexane carboxylate. See if I can squeeze this in. So methyl cyclohexane carboxylate would be the name for our ester. Finally, let's look at physical properties of esters. And here I have some different molecules. So let's look at our ester here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So methyl cyclohexane carboxylate would be the name for our ester. Finally, let's look at physical properties of esters. And here I have some different molecules. So let's look at our ester here. So if we were to name this ester, we would have an alkyl group. That's a methyl group. So we have methyl."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at our ester here. So if we were to name this ester, we would have an alkyl group. That's a methyl group. So we have methyl. And then we think about the carboxylic acid portion. So what kind of a carboxylic acid would this be, a two-carbon carboxylic acid? So that would be acetic acid."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So we have methyl. And then we think about the carboxylic acid portion. So what kind of a carboxylic acid would this be, a two-carbon carboxylic acid? So that would be acetic acid. So we drop the ending and add 8. So that would be acetate. And so we could call this ester methyl acetate."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So that would be acetic acid. So we drop the ending and add 8. So that would be acetate. And so we could call this ester methyl acetate. So let's compare methyl acetate to some similar sized molecules. And over here on the left, we have 2-methylbutane. So if we just go ahead and number it really fast, we see this is 2-methylbutane."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And so we could call this ester methyl acetate. So let's compare methyl acetate to some similar sized molecules. And over here on the left, we have 2-methylbutane. So if we just go ahead and number it really fast, we see this is 2-methylbutane. And then over here on the right, we have a 4-carbon alcohol. So 1, 2, 3, 4. So this would be 2-butanol."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So if we just go ahead and number it really fast, we see this is 2-methylbutane. And then over here on the right, we have a 4-carbon alcohol. So 1, 2, 3, 4. So this would be 2-butanol. So let's compare methyl acetate to these other molecules in terms of boiling points. So let's start over here with 2-methylbutane. When we think about boiling points, we have to think about our intermolecular forces."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this would be 2-butanol. So let's compare methyl acetate to these other molecules in terms of boiling points. So let's start over here with 2-methylbutane. When we think about boiling points, we have to think about our intermolecular forces. The only intermolecular forces present between these two non-polar, well, the same molecule, 2-methylbutane is a non-polar molecule. So the only forces present are London dispersion forces, which we know are the weakest. And so it's the easiest to pull these two molecules apart."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "When we think about boiling points, we have to think about our intermolecular forces. The only intermolecular forces present between these two non-polar, well, the same molecule, 2-methylbutane is a non-polar molecule. So the only forces present are London dispersion forces, which we know are the weakest. And so it's the easiest to pull these two molecules apart. So the boiling point of 2-methylbutane is approximately 28 degrees Celsius. When we think about methyl acetate, we have some polarity here. So this oxygen is partial negative."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And so it's the easiest to pull these two molecules apart. So the boiling point of 2-methylbutane is approximately 28 degrees Celsius. When we think about methyl acetate, we have some polarity here. So this oxygen is partial negative. This carbon right here is a partial positive. So something like methyl acetate, a small ester, is moderately polar. So we have a little bit of polarity here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen is partial negative. This carbon right here is a partial positive. So something like methyl acetate, a small ester, is moderately polar. So we have a little bit of polarity here. Same thing for this molecule of methyl acetate. And so the attractive intermolecular force between these two molecules would be dipole-dipole. Dipole-dipole, we know is a stronger intermolecular force than London dispersion."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So we have a little bit of polarity here. Same thing for this molecule of methyl acetate. And so the attractive intermolecular force between these two molecules would be dipole-dipole. Dipole-dipole, we know is a stronger intermolecular force than London dispersion. And so the boiling point of methyl acetate should be higher than the boiling point of 2-methylbutane. So the boiling point turns out to be approximately 57 degrees Celsius. So it takes more energy, more heat, to pull apart molecules of methyl acetate because dipole-dipole is a stronger attractive force."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "Dipole-dipole, we know is a stronger intermolecular force than London dispersion. And so the boiling point of methyl acetate should be higher than the boiling point of 2-methylbutane. So the boiling point turns out to be approximately 57 degrees Celsius. So it takes more energy, more heat, to pull apart molecules of methyl acetate because dipole-dipole is a stronger attractive force. Finally, 2-butanol, the boiling point for 2-butanol is about 99 degrees Celsius because of the hydrogen bonding that's present. So we have some hydrogen bonding right here. It's the strongest intermolecular force, so it takes increased energy to pull apart molecules of 2-butanol."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So it takes more energy, more heat, to pull apart molecules of methyl acetate because dipole-dipole is a stronger attractive force. Finally, 2-butanol, the boiling point for 2-butanol is about 99 degrees Celsius because of the hydrogen bonding that's present. So we have some hydrogen bonding right here. It's the strongest intermolecular force, so it takes increased energy to pull apart molecules of 2-butanol. So methyl acetate, in terms of boiling point, is somewhat in between that for an alkane of a similar size and that of an alcohol of a similar size because of its polarity. In terms of solubility in water, so let's go ahead and draw out water here. So if we have water, we know that water is, of course, a polar molecule, right?"}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "It's the strongest intermolecular force, so it takes increased energy to pull apart molecules of 2-butanol. So methyl acetate, in terms of boiling point, is somewhat in between that for an alkane of a similar size and that of an alcohol of a similar size because of its polarity. In terms of solubility in water, so let's go ahead and draw out water here. So if we have water, we know that water is, of course, a polar molecule, right? Partial negative on the oxygen, partial positive on the hydrogen. So we have a polar water molecule. And so there could be some hydrogen bonding right here."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So if we have water, we know that water is, of course, a polar molecule, right? Partial negative on the oxygen, partial positive on the hydrogen. So we have a polar water molecule. And so there could be some hydrogen bonding right here. So between this oxygen on methyl acetate and water, there could be some hydrogen bonding. We have a polar molecule and a polar molecule. So methyl acetate is soluble in water."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "And so there could be some hydrogen bonding right here. So between this oxygen on methyl acetate and water, there could be some hydrogen bonding. We have a polar molecule and a polar molecule. So methyl acetate is soluble in water. But of course, as you increase the number of carbons that you have for your ester, so let's think about this ester. Here we have two carbons and then one carbon. As we increase the number of carbons, we increase the nonpolar character of our ester."}, {"video_title": "Nomenclature and properties of esters Organic chemistry Khan Academy.mp3", "Sentence": "So methyl acetate is soluble in water. But of course, as you increase the number of carbons that you have for your ester, so let's think about this ester. Here we have two carbons and then one carbon. As we increase the number of carbons, we increase the nonpolar character of our ester. And we would therefore decrease the solubility of our ester in water. So although methyl acetate turns out to be soluble in water, if you go up to something like ethyl acetate, you decrease the solubility. And ethyl acetate can actually be used to do something like extract a nonpolar molecule like caffeine from water."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we saw that the nitro group is a meta-director because of this plus 1 formal charge on the nitrogen right next to your ring. We also said that the nitro group is a deactivator, which meant that the nitrobenzene molecule reacted more slowly than benzene itself in a nitration reaction. And so let's see if we can figure out why the nitro group is a deactivator in terms of both the inductive and a resonance effect. And we'll start with induction first, which we know is due to differences in electronegativity. And so we could think about this nitrogen here compared to this carbon on the ring, the nitrogen being much more electronegative. And so it can pull electrons closer to itself. And so we can show the movement of electrons towards nitrogen."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with induction first, which we know is due to differences in electronegativity. And so we could think about this nitrogen here compared to this carbon on the ring, the nitrogen being much more electronegative. And so it can pull electrons closer to itself. And so we can show the movement of electrons towards nitrogen. A positively charged nitrogen is extremely electron withdrawing. And it can even withdraw electron density from your ring itself. That makes the ring more positive, which would destabilize the sigma complex that results in the mechanism for electrophilic aromatic substitution."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we can show the movement of electrons towards nitrogen. A positively charged nitrogen is extremely electron withdrawing. And it can even withdraw electron density from your ring itself. That makes the ring more positive, which would destabilize the sigma complex that results in the mechanism for electrophilic aromatic substitution. Since the sigma complex is less likely to form, that would deactivate the ring towards electrophilic aromatic substitution. Also, you could think about the electron withdrawing effect of this nitro group, making the ring less nucleophilic. And so both of those factors mean that the nitro group is a deactivator towards electrophilic aromatic substitution."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That makes the ring more positive, which would destabilize the sigma complex that results in the mechanism for electrophilic aromatic substitution. Since the sigma complex is less likely to form, that would deactivate the ring towards electrophilic aromatic substitution. Also, you could think about the electron withdrawing effect of this nitro group, making the ring less nucleophilic. And so both of those factors mean that the nitro group is a deactivator towards electrophilic aromatic substitution. So induction says that the nitro group is a deactivator. Let's look at resonance next. So we could draw a resonance structure, where we take these pi electrons and move them in here, which would push these electrons off onto your oxygens."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so both of those factors mean that the nitro group is a deactivator towards electrophilic aromatic substitution. So induction says that the nitro group is a deactivator. Let's look at resonance next. So we could draw a resonance structure, where we take these pi electrons and move them in here, which would push these electrons off onto your oxygens. Let's go ahead and draw that. We have our ring. We have pi electrons in our ring."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we could draw a resonance structure, where we take these pi electrons and move them in here, which would push these electrons off onto your oxygens. Let's go ahead and draw that. We have our ring. We have pi electrons in our ring. Now the nitrogen is double bonded to our ring. And we have an oxygen on the right, which has a negative 1 formal charge. The nitrogen is bonded to an oxygen on the left, which now has three lone pairs of electrons, and therefore has a negative 1 formal charge."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have pi electrons in our ring. Now the nitrogen is double bonded to our ring. And we have an oxygen on the right, which has a negative 1 formal charge. The nitrogen is bonded to an oxygen on the left, which now has three lone pairs of electrons, and therefore has a negative 1 formal charge. Let me go ahead and highlight those electrons here. So if these electrons in blue move off onto the oxygen, that gives that oxygen a negative 1 formal charge. This nitrogen has a plus 1 formal charge."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The nitrogen is bonded to an oxygen on the left, which now has three lone pairs of electrons, and therefore has a negative 1 formal charge. Let me go ahead and highlight those electrons here. So if these electrons in blue move off onto the oxygen, that gives that oxygen a negative 1 formal charge. This nitrogen has a plus 1 formal charge. And let's go ahead and follow these pi electrons too. So in red, these pi electrons move out here to form this pi bond, taking a bond away from that carbon, therefore giving that carbon a plus 1 formal charge. And you could keep going and draw more resonance structures, and they would have a positive charge in the ring."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This nitrogen has a plus 1 formal charge. And let's go ahead and follow these pi electrons too. So in red, these pi electrons move out here to form this pi bond, taking a bond away from that carbon, therefore giving that carbon a plus 1 formal charge. And you could keep going and draw more resonance structures, and they would have a positive charge in the ring. But I'm going to stop there, because the point I'm trying to make is that the nitro group is electron withdrawing. And it makes the ring more positive, which would, of course, deactivate the ring towards electrophilic aromatic substitution. So both the inductive effect and the resonance effect say that the nitro group is a deactivator."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you could keep going and draw more resonance structures, and they would have a positive charge in the ring. But I'm going to stop there, because the point I'm trying to make is that the nitro group is electron withdrawing. And it makes the ring more positive, which would, of course, deactivate the ring towards electrophilic aromatic substitution. So both the inductive effect and the resonance effect say that the nitro group is a deactivator. And so we can say it's a strong deactivator, because of both of those effects. Let's look at another molecule that's considered to be a strong deactivator, so this molecule down here. And we can see that there's a carbon bonded to our benzene ring, and that carbon is bonded to three fluorines."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So both the inductive effect and the resonance effect say that the nitro group is a deactivator. And so we can say it's a strong deactivator, because of both of those effects. Let's look at another molecule that's considered to be a strong deactivator, so this molecule down here. And we can see that there's a carbon bonded to our benzene ring, and that carbon is bonded to three fluorines. So you could call this trifluoromethyl group. You could call it CF3 group, whatever you want to call it. In terms of electronegativity difference, obviously fluorine is much more electronegative than carbon."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we can see that there's a carbon bonded to our benzene ring, and that carbon is bonded to three fluorines. So you could call this trifluoromethyl group. You could call it CF3 group, whatever you want to call it. In terms of electronegativity difference, obviously fluorine is much more electronegative than carbon. So you could think about these fluorine atoms withdrawing a lot of electron density away from that carbon, which would give that carbon a partial positive charge. And so that's an explanation for why the CF3 group is a meta-director. So once again, look at the last video for more details about why a positive charge next to your ring favors a meta-attack."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In terms of electronegativity difference, obviously fluorine is much more electronegative than carbon. So you could think about these fluorine atoms withdrawing a lot of electron density away from that carbon, which would give that carbon a partial positive charge. And so that's an explanation for why the CF3 group is a meta-director. So once again, look at the last video for more details about why a positive charge next to your ring favors a meta-attack. In terms of the inductive effect on the ring, these are very electronegative atoms. And so they can withdraw some electron density away from your ring. So you could show the electrons of the ring moving in that direction towards your electronegative fluorines."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, look at the last video for more details about why a positive charge next to your ring favors a meta-attack. In terms of the inductive effect on the ring, these are very electronegative atoms. And so they can withdraw some electron density away from your ring. So you could show the electrons of the ring moving in that direction towards your electronegative fluorines. And so that withdraws electron density from the ring, making the ring more positive, which would, of course, deactivate the ring towards electrophilic aromatic substitution. And so this trifluoromethyl group is both a meta-director and deactivating. And this inductive effect is so powerful, this CF3 group turns out to be a strong deactivator."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could show the electrons of the ring moving in that direction towards your electronegative fluorines. And so that withdraws electron density from the ring, making the ring more positive, which would, of course, deactivate the ring towards electrophilic aromatic substitution. And so this trifluoromethyl group is both a meta-director and deactivating. And this inductive effect is so powerful, this CF3 group turns out to be a strong deactivator. We can't have any kind of resonance effect here, because these lone pairs of electrons on one of the fluorines, I'll just choose this one right here, those are too far away from the ring to participate in resonance. So only an inductive effect. But it turns out to be a very powerful one."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this inductive effect is so powerful, this CF3 group turns out to be a strong deactivator. We can't have any kind of resonance effect here, because these lone pairs of electrons on one of the fluorines, I'll just choose this one right here, those are too far away from the ring to participate in resonance. So only an inductive effect. But it turns out to be a very powerful one. Let's look at one more example of a meta-director. And this one is a moderate deactivator. So first, let's try to figure out why this is a meta-director."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But it turns out to be a very powerful one. Let's look at one more example of a meta-director. And this one is a moderate deactivator. So first, let's try to figure out why this is a meta-director. And so if we look at the atom that is directly bonded to the ring, that is double bonded to an oxygen. And we know that oxygen is much more electronegative than this carbon. So we could think about this oxygen being partially negative, pulling some electron density towards it, withdrawing some electron density away from this carbon, giving this carbon a partial positive charge."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So first, let's try to figure out why this is a meta-director. And so if we look at the atom that is directly bonded to the ring, that is double bonded to an oxygen. And we know that oxygen is much more electronegative than this carbon. So we could think about this oxygen being partially negative, pulling some electron density towards it, withdrawing some electron density away from this carbon, giving this carbon a partial positive charge. And so once again, you have a partial positive next to a ring, which makes this group a meta-director. In terms of being deactivating, we need to think about resonance structures. So let's think about moving these electrons in this pi bond off onto your oxygen."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we could think about this oxygen being partially negative, pulling some electron density towards it, withdrawing some electron density away from this carbon, giving this carbon a partial positive charge. And so once again, you have a partial positive next to a ring, which makes this group a meta-director. In terms of being deactivating, we need to think about resonance structures. So let's think about moving these electrons in this pi bond off onto your oxygen. So let's go ahead and draw that one. So we have our benzene ring. And we have now our carbon with only a single bond to this oxygen."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about moving these electrons in this pi bond off onto your oxygen. So let's go ahead and draw that one. So we have our benzene ring. And we have now our carbon with only a single bond to this oxygen. This oxygen now has a negative 1 formal charge on it like that. And then this carbon is still bonded to this R group here. We took a bond away from that carbon."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have now our carbon with only a single bond to this oxygen. This oxygen now has a negative 1 formal charge on it like that. And then this carbon is still bonded to this R group here. We took a bond away from that carbon. So now this carbon has a full plus 1 formal charge like that. So we can now show a resonance structure for this. We have these pi electrons here."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We took a bond away from that carbon. So now this carbon has a full plus 1 formal charge like that. So we can now show a resonance structure for this. We have these pi electrons here. We can move out into here to form a pi bond between those two carbons. Let's go ahead and show this resonance structure. So we have our ring, these pi electrons."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have these pi electrons here. We can move out into here to form a pi bond between those two carbons. Let's go ahead and show this resonance structure. So we have our ring, these pi electrons. Now there's double bond here. And this carbon is still bonded to this oxygen. So this oxygen has a negative 1 formal charge."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our ring, these pi electrons. Now there's double bond here. And this carbon is still bonded to this oxygen. So this oxygen has a negative 1 formal charge. And there's an R group bonded to that carbon. Let's show those pi electrons moving here. So these pi electrons moved into here, taking a bond away from this carbon."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen has a negative 1 formal charge. And there's an R group bonded to that carbon. Let's show those pi electrons moving here. So these pi electrons moved into here, taking a bond away from this carbon. So we have a plus 1 formal charge right here. And of course, once again, you could keep going and show more resonance structures. But I'm only trying to point out that the presence of this group increases the positive charge on the ring."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons moved into here, taking a bond away from this carbon. So we have a plus 1 formal charge right here. And of course, once again, you could keep going and show more resonance structures. But I'm only trying to point out that the presence of this group increases the positive charge on the ring. And so the ring is more positive, which means it is deactivating towards electrophilic aromatic substitution. And so this group is both a metadirector and a deactivator. Now let's think about a general way to recognize these moderate deactivators here."}, {"video_title": "Meta directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But I'm only trying to point out that the presence of this group increases the positive charge on the ring. And so the ring is more positive, which means it is deactivating towards electrophilic aromatic substitution. And so this group is both a metadirector and a deactivator. Now let's think about a general way to recognize these moderate deactivators here. So if I go back to this first drawing on the left, we have an atom bonded to a benzene ring. And that atom has a pi bond to an electronegative atom. In this case, it's oxygen."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What I want to do in this video is try to get a better understanding of the structure of the Earth. And we're actually going to think about it in two different ways. So let me just draw half of the Earth over here. That's my best shot at drawing half of a circle. And what we're going to do is think about it in two ways. And on the left-hand side, we're going to think about it as the compositional layers, or the chemical layers. So over here we're going to think about the chemical structure, or the composition."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's my best shot at drawing half of a circle. And what we're going to do is think about it in two ways. And on the left-hand side, we're going to think about it as the compositional layers, or the chemical layers. So over here we're going to think about the chemical structure, or the composition. And on the right-hand side, we're going to think about the mechanical properties of the layer. And when I say the mechanical properties, I'm really just saying, is that layer kind of a solid, rigid layer? Is it kind of a liquid layer?"}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So over here we're going to think about the chemical structure, or the composition. And on the right-hand side, we're going to think about the mechanical properties of the layer. And when I say the mechanical properties, I'm really just saying, is that layer kind of a solid, rigid layer? Is it kind of a liquid layer? Or is it something in between? A kind of a putty-like, non-rigid, solid layer? So let's think about it on the chemical or the compositional side first."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Is it kind of a liquid layer? Or is it something in between? A kind of a putty-like, non-rigid, solid layer? So let's think about it on the chemical or the compositional side first. Because to some degree, this is simpler. So the outermost layer is the crust. That's the layer that we're sitting on right here, right now, I'm assuming."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's think about it on the chemical or the compositional side first. Because to some degree, this is simpler. So the outermost layer is the crust. That's the layer that we're sitting on right here, right now, I'm assuming. Assuming you're on the planet. So this right here is the crust. It is the outermost."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's the layer that we're sitting on right here, right now, I'm assuming. Assuming you're on the planet. So this right here is the crust. It is the outermost. It's obviously solid. We'll think about that when we talk about the mechanical side of things. And it's also the thinnest layer."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is the outermost. It's obviously solid. We'll think about that when we talk about the mechanical side of things. And it's also the thinnest layer. And crust is not uniform. There is both oceanic crust and continental crust. Let me draw the crust on this side as well."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's also the thinnest layer. And crust is not uniform. There is both oceanic crust and continental crust. Let me draw the crust on this side as well. So let me draw some crust over here. I've got some crust right over there. And there is both oceanic crust and continental crust."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me draw the crust on this side as well. So let me draw some crust over here. I've got some crust right over there. And there is both oceanic crust and continental crust. So oceanic crust is thinner crust. So let's say that this part right here, let me draw some thicker crust. Let me draw some thicker crust right over here."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there is both oceanic crust and continental crust. So oceanic crust is thinner crust. So let's say that this part right here, let me draw some thicker crust. Let me draw some thicker crust right over here. We'll call the thicker stuff the continental crust. Which is thicker and less dense than the oceanic crust. So what I'm doing in this light green color, this is continental."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me draw some thicker crust right over here. We'll call the thicker stuff the continental crust. Which is thicker and less dense than the oceanic crust. So what I'm doing in this light green color, this is continental. So this right here is continental. And then in this kind of more fluorescent green, this is oceanic. Oceanic crust."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what I'm doing in this light green color, this is continental. So this right here is continental. And then in this kind of more fluorescent green, this is oceanic. Oceanic crust. And the oceanic crust is pretty thin. It's on the order of about 5 or 10 kilometers. So let's just call this approximately 5 to 10 kilometers thick."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Oceanic crust. And the oceanic crust is pretty thin. It's on the order of about 5 or 10 kilometers. So let's just call this approximately 5 to 10 kilometers thick. And when I talk about oceanic crust, I'm not talking about the oceans. I'm not talking about the liquid part, the water. I'm talking about the rock that kind of holds the water."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's just call this approximately 5 to 10 kilometers thick. And when I talk about oceanic crust, I'm not talking about the oceans. I'm not talking about the liquid part, the water. I'm talking about the rock that kind of holds the water. The rock underneath the oceans. And so this is 5 to 10 kilometers thick. If you were to go to the bottom of the ocean and you were to kind of sit on the rock and then drill, you'd have to drill about 5 to 10 kilometers to get through that layer."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'm talking about the rock that kind of holds the water. The rock underneath the oceans. And so this is 5 to 10 kilometers thick. If you were to go to the bottom of the ocean and you were to kind of sit on the rock and then drill, you'd have to drill about 5 to 10 kilometers to get through that layer. This compositional layer. So this is 5 to 10 kilometers. And the continental crust is about 10 to 70 kilometers thick."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to go to the bottom of the ocean and you were to kind of sit on the rock and then drill, you'd have to drill about 5 to 10 kilometers to get through that layer. This compositional layer. So this is 5 to 10 kilometers. And the continental crust is about 10 to 70 kilometers thick. And obviously they are both rigid. They are both solid, solid rock. Now below, when you think about composition or what the layers are made up of, the next layer below that, and this is actually the biggest layer of the earth by volume, is the mantle."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the continental crust is about 10 to 70 kilometers thick. And obviously they are both rigid. They are both solid, solid rock. Now below, when you think about composition or what the layers are made up of, the next layer below that, and this is actually the biggest layer of the earth by volume, is the mantle. Let me draw it like that. I always have trouble drawing the right-hand side of this circle. So let me draw."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now below, when you think about composition or what the layers are made up of, the next layer below that, and this is actually the biggest layer of the earth by volume, is the mantle. Let me draw it like that. I always have trouble drawing the right-hand side of this circle. So let me draw. So this is the mantle right over here. Let me write down. This is all the mantle."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me draw. So this is the mantle right over here. Let me write down. This is all the mantle. And once again, we differentiate it from the crust because it's composed of different types of rock. Now you go even deeper, and let me give you the depths here. So the mantle starts right below the crust, right below the oceanic and the continental crust, oceanic and continental, and it goes about 2,900 kilometers deeper."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is all the mantle. And once again, we differentiate it from the crust because it's composed of different types of rock. Now you go even deeper, and let me give you the depths here. So the mantle starts right below the crust, right below the oceanic and the continental crust, oceanic and continental, and it goes about 2,900 kilometers deeper. So it's much, much, much thicker than the crust. The crust is on the order of 5 to maybe 70 kilometers thick. This is much, much thicker."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the mantle starts right below the crust, right below the oceanic and the continental crust, oceanic and continental, and it goes about 2,900 kilometers deeper. So it's much, much, much thicker than the crust. The crust is on the order of 5 to maybe 70 kilometers thick. This is much, much thicker. So even though I've drawn the crust fairly thin, I didn't draw it thin enough relative to how thick I've drawn the mantle. This isn't drawn to scale. Now you go even deeper than that, you get kind of the densest part of the earth, and that is the core."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is much, much thicker. So even though I've drawn the crust fairly thin, I didn't draw it thin enough relative to how thick I've drawn the mantle. This isn't drawn to scale. Now you go even deeper than that, you get kind of the densest part of the earth, and that is the core. And there's going to be a couple of themes here, especially when we think about the mechanical properties of the earth, is the deeper you get, you're going to get denser elements and you're going to have more heat and more pressure. And the reason why you're going to have denser elements is when earth was first forming and it was kind of in its molten state, the denser elements just kind of sunk to the bottom and the lighter elements would kind of rise to the top. They would have this buoyancy because they're less dense than everything around it."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now you go even deeper than that, you get kind of the densest part of the earth, and that is the core. And there's going to be a couple of themes here, especially when we think about the mechanical properties of the earth, is the deeper you get, you're going to get denser elements and you're going to have more heat and more pressure. And the reason why you're going to have denser elements is when earth was first forming and it was kind of in its molten state, the denser elements just kind of sunk to the bottom and the lighter elements would kind of rise to the top. They would have this buoyancy because they're less dense than everything around it. And really, even the gases would kind of bubble up, would essentially bubble up and form our atmosphere. So that's why in general, the densest things are in the center and the least dense things are on the outside. They're in our atmosphere."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They would have this buoyancy because they're less dense than everything around it. And really, even the gases would kind of bubble up, would essentially bubble up and form our atmosphere. So that's why in general, the densest things are in the center and the least dense things are on the outside. They're in our atmosphere. And the core, once again, its composition is fundamentally different than the mantle and the crust. We believe that it's mainly metals, and in particular, iron and nickel. So that's the structure, the layers of the earth from a composition point of view, from a chemical point of view."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're in our atmosphere. And the core, once again, its composition is fundamentally different than the mantle and the crust. We believe that it's mainly metals, and in particular, iron and nickel. So that's the structure, the layers of the earth from a composition point of view, from a chemical point of view. Now let's kind of think about the same layers, but we're going to think more in terms of what's liquid, what's rigid and solid, and what's in between. So the outermost rigid layer of the earth is made up of the crust, both the continental and the oceanic crust, and the coolest top layer of the mantle. So let me draw that in pink."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's the structure, the layers of the earth from a composition point of view, from a chemical point of view. Now let's kind of think about the same layers, but we're going to think more in terms of what's liquid, what's rigid and solid, and what's in between. So the outermost rigid layer of the earth is made up of the crust, both the continental and the oceanic crust, and the coolest top layer of the mantle. So let me draw that in pink. So this layer right over here. So what I'm drawing in pink is the cool, rigid, solid part of the mantle. So this is the cool, so it is solid rock, the part of the mantle that's solid rock."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me draw that in pink. So this layer right over here. So what I'm drawing in pink is the cool, rigid, solid part of the mantle. So this is the cool, so it is solid rock, the part of the mantle that's solid rock. Its composition is different than, say, the continental crust, but they are both rigid. So if you combine this topmost layer of the mantle with the crust, then you're talking about the lithosphere. So this is the lithosphere."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the cool, so it is solid rock, the part of the mantle that's solid rock. Its composition is different than, say, the continental crust, but they are both rigid. So if you combine this topmost layer of the mantle with the crust, then you're talking about the lithosphere. So this is the lithosphere. And this essentially gets you about the lithosphere, depending on where you are on the surface of the earth, is 10 to 200 kilometers thick. And most of the time it's closer to the high end of this range. The 10 is kind of where you have hot spots in the mantle, and it's essentially been able to kind of dissolve parts of the lithosphere and essentially create new, well we'll talk more about that when we talk about the actual plate tectonics of it all."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is the lithosphere. And this essentially gets you about the lithosphere, depending on where you are on the surface of the earth, is 10 to 200 kilometers thick. And most of the time it's closer to the high end of this range. The 10 is kind of where you have hot spots in the mantle, and it's essentially been able to kind of dissolve parts of the lithosphere and essentially create new, well we'll talk more about that when we talk about the actual plate tectonics of it all. And when we talk about plate tectonics, the plates are actually lithospheric plates. It's actually the lithosphere that's moving on top of the lower layer, the lower layers of the mantle. So the lithosphere, it is rigid, it is solid, it's made up of the crust and the upper layer of the mantle, the uppermost layer of the mantle."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The 10 is kind of where you have hot spots in the mantle, and it's essentially been able to kind of dissolve parts of the lithosphere and essentially create new, well we'll talk more about that when we talk about the actual plate tectonics of it all. And when we talk about plate tectonics, the plates are actually lithospheric plates. It's actually the lithosphere that's moving on top of the lower layer, the lower layers of the mantle. So the lithosphere, it is rigid, it is solid, it's made up of the crust and the upper layer of the mantle, the uppermost layer of the mantle. Now you go a little bit deeper, the temperatures and the pressures increase, but now the temperatures have increased enough, you have the same composition as the uppermost, the rigid part of the mantle, but the temperatures have now gone up enough that it now turns into not quite a liquid, we won't call it a liquid, it actually still transmits the type of waves that liquids would not transmit. It's more of like a putty type texture, something, it has fluid properties, it can flow, it's way more viscous than what we would associate with most fluids, so it's not rigid and solid, it can have convection going on in it, but it's not a liquid, it still will transmit certain types of waves that liquids won't. And this is called the asthenosphere, kind of this jelly putty layer."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the lithosphere, it is rigid, it is solid, it's made up of the crust and the upper layer of the mantle, the uppermost layer of the mantle. Now you go a little bit deeper, the temperatures and the pressures increase, but now the temperatures have increased enough, you have the same composition as the uppermost, the rigid part of the mantle, but the temperatures have now gone up enough that it now turns into not quite a liquid, we won't call it a liquid, it actually still transmits the type of waves that liquids would not transmit. It's more of like a putty type texture, something, it has fluid properties, it can flow, it's way more viscous than what we would associate with most fluids, so it's not rigid and solid, it can have convection going on in it, but it's not a liquid, it still will transmit certain types of waves that liquids won't. And this is called the asthenosphere, kind of this jelly putty layer. And it's like that because it's so hot that the rock is somewhat melted. So this is, this layer right here in magenta is the asthenosphere. I've seen some spellings where there's an E after the A, I think that's maybe the European spelling."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is called the asthenosphere, kind of this jelly putty layer. And it's like that because it's so hot that the rock is somewhat melted. So this is, this layer right here in magenta is the asthenosphere. I've seen some spellings where there's an E after the A, I think that's maybe the European spelling. And the asthenosphere obviously starts right below the lithosphere, it's what the lithospheric plates, when we talk about plate tectonics, are riding on top of. It's kind of the gummy material that allows it to actually move, that allows the rigid layer to actually move on top. And it goes, so it starts below the lithosphere and it ends at around 660 kilometers deep."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I've seen some spellings where there's an E after the A, I think that's maybe the European spelling. And the asthenosphere obviously starts right below the lithosphere, it's what the lithospheric plates, when we talk about plate tectonics, are riding on top of. It's kind of the gummy material that allows it to actually move, that allows the rigid layer to actually move on top. And it goes, so it starts below the lithosphere and it ends at around 660 kilometers deep. So this right here is 660 kilometers deep. And then you go even deeper than that, and now the pressures are so big that even though the temperatures are even higher, even though the temperatures are even higher, the pressures are so big that the same material can't have fluid motion anymore. It's essentially been jammed together."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it goes, so it starts below the lithosphere and it ends at around 660 kilometers deep. So this right here is 660 kilometers deep. And then you go even deeper than that, and now the pressures are so big that even though the temperatures are even higher, even though the temperatures are even higher, the pressures are so big that the same material can't have fluid motion anymore. It's essentially been jammed together. So you can imagine, if you have things that are somewhat fluid, that means that the molecules can kind of slide past each other, maybe very slowly. But if you increase the pressure enough, they'll be jammed into each other. And that's essentially what happens in the next layer of the mantle."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's essentially been jammed together. So you can imagine, if you have things that are somewhat fluid, that means that the molecules can kind of slide past each other, maybe very slowly. But if you increase the pressure enough, they'll be jammed into each other. And that's essentially what happens in the next layer of the mantle. All of these layers of the mantle are made up of the same thing, it's just a difference of temperature and pressure. And so that next layer of the mantle is called the mesosphere. This is called the mesosphere, but there's also a layer of our atmosphere, the layer right above the stratosphere, that's called the mesosphere."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's essentially what happens in the next layer of the mantle. All of these layers of the mantle are made up of the same thing, it's just a difference of temperature and pressure. And so that next layer of the mantle is called the mesosphere. This is called the mesosphere, but there's also a layer of our atmosphere, the layer right above the stratosphere, that's called the mesosphere. And so don't get confused here, these are two different mesospheres. And this layer, the pressure is so big that now we are rigid again. We are kind of definitely solid."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is called the mesosphere, but there's also a layer of our atmosphere, the layer right above the stratosphere, that's called the mesosphere. And so don't get confused here, these are two different mesospheres. And this layer, the pressure is so big that now we are rigid again. We are kind of definitely solid. None of this debate about a little bit of fluid motion because the pressures are so big. Now you go a little bit deeper. We are now in the core, the metallic core, and the temperatures are so high that even though the pressures are high, because we have a compositional change, we're at pressures where this type of mesospheric rock is rigid, but metals at these temperatures actually can be fluid, can actually be liquid."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We are kind of definitely solid. None of this debate about a little bit of fluid motion because the pressures are so big. Now you go a little bit deeper. We are now in the core, the metallic core, and the temperatures are so high that even though the pressures are high, because we have a compositional change, we're at pressures where this type of mesospheric rock is rigid, but metals at these temperatures actually can be fluid, can actually be liquid. And so we actually have a liquid outer core. The entire core, as far as we know, is made up of the same stuff, just the outer part of the core, the temperatures are high enough to melt the metal, but the pressures aren't so high enough to make them solid. The pressures are definitely high enough to make more rocky material solid, but not the metals."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We are now in the core, the metallic core, and the temperatures are so high that even though the pressures are high, because we have a compositional change, we're at pressures where this type of mesospheric rock is rigid, but metals at these temperatures actually can be fluid, can actually be liquid. And so we actually have a liquid outer core. The entire core, as far as we know, is made up of the same stuff, just the outer part of the core, the temperatures are high enough to melt the metal, but the pressures aren't so high enough to make them solid. The pressures are definitely high enough to make more rocky material solid, but not the metals. And then you go even deeper, now the pressure, even though the temperature keeps going up, the pressure is so strong that even the metals are solid. So this is the solid inner core. So when you think about the mechanical properties, the innermost, and just so you know the total distances we're talking about, the outer core starts at, I actually didn't tell you where the mesosphere ends, the mantle ends at about 2,900 kilometers deep, so that's clearly where the mesosphere ends as well, because the mesosphere is kind of the lower mantle."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The pressures are definitely high enough to make more rocky material solid, but not the metals. And then you go even deeper, now the pressure, even though the temperature keeps going up, the pressure is so strong that even the metals are solid. So this is the solid inner core. So when you think about the mechanical properties, the innermost, and just so you know the total distances we're talking about, the outer core starts at, I actually didn't tell you where the mesosphere ends, the mantle ends at about 2,900 kilometers deep, so that's clearly where the mesosphere ends as well, because the mesosphere is kind of the lower mantle. So this is 2,900 kilometers deep. Then you go even deeper, you're in the liquid outer core, and that extends from about 2,900 kilometers deep to about 5,100 kilometers deep. So I really should, I frankly should make the liquid core in my drawing even wider."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So when you think about the mechanical properties, the innermost, and just so you know the total distances we're talking about, the outer core starts at, I actually didn't tell you where the mesosphere ends, the mantle ends at about 2,900 kilometers deep, so that's clearly where the mesosphere ends as well, because the mesosphere is kind of the lower mantle. So this is 2,900 kilometers deep. Then you go even deeper, you're in the liquid outer core, and that extends from about 2,900 kilometers deep to about 5,100 kilometers deep. So I really should, I frankly should make the liquid core in my drawing even wider. So this extends to about, this depth right over here is about 5,100 kilometers deep. And then obviously then you have the center of the earth, and the entire radius of the earth is about 6,400 kilometers. So hopefully that clarifies things when you hear people talking about the lithosphere or the mantle, that they're really talking about mechanical versus composition."}, {"video_title": "Compositional and mechanical layers of the earth Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I really should, I frankly should make the liquid core in my drawing even wider. So this extends to about, this depth right over here is about 5,100 kilometers deep. And then obviously then you have the center of the earth, and the entire radius of the earth is about 6,400 kilometers. So hopefully that clarifies things when you hear people talking about the lithosphere or the mantle, that they're really talking about mechanical versus composition. When we talk about mechanical, solid inner core, liquid outer core, essentially solid mesosphere, it's rigid. Then you have something kind of a spongy, somewhat fluid, not solid, not liquid, a stenosphere that the lithospheric plates can ride on top of. And then you have your actual rigid, solid lithosphere made up of the uppermost part of the mantle and the crust."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so if we have five carbons here, the maximum number of hydrogens we could have is two N plus two, where N is equal to five. So two times five plus two is equal to 12. So 12 is the maximum number of hydrogens. Here we have only 10 hydrogens, so we're missing two hydrogens. We're missing one pair of hydrogens, and so therefore the hydrogen deficiency index is equal to one. Immediately, that makes me think about a double bond might be present in this molecule, or one ring. Let's go down to the integration value."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Here we have only 10 hydrogens, so we're missing two hydrogens. We're missing one pair of hydrogens, and so therefore the hydrogen deficiency index is equal to one. Immediately, that makes me think about a double bond might be present in this molecule, or one ring. Let's go down to the integration value. So for this signal, there's an integration value of 27. For this signal, the integration value is 40.2. For this signal, it's 28.4."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's go down to the integration value. So for this signal, there's an integration value of 27. For this signal, the integration value is 40.2. For this signal, it's 28.4. And for this signal, it's 42.2. Remember what you do. You divide all four integration values by the lowest one."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For this signal, it's 28.4. And for this signal, it's 42.2. Remember what you do. You divide all four integration values by the lowest one. So the lowest one is, of course, 27. So 27 divided by 27 is, of course, one. 40.2 divided by 27 is pretty close to 1.5."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "You divide all four integration values by the lowest one. So the lowest one is, of course, 27. So 27 divided by 27 is, of course, one. 40.2 divided by 27 is pretty close to 1.5. 28.4 divided by 27, that's pretty close to one. And 42.2 divided by 27 is, once again, pretty close to 1.5. Remember, these are just the relative numbers of protons giving you these signals, but you can't have 1.5 protons giving you a signal."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "40.2 divided by 27 is pretty close to 1.5. 28.4 divided by 27, that's pretty close to one. And 42.2 divided by 27 is, once again, pretty close to 1.5. Remember, these are just the relative numbers of protons giving you these signals, but you can't have 1.5 protons giving you a signal. We need a whole number, and we need to account for 10 total protons here. So if we multiply one by two, then we get two protons. So this signal represents two protons."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Remember, these are just the relative numbers of protons giving you these signals, but you can't have 1.5 protons giving you a signal. We need a whole number, and we need to account for 10 total protons here. So if we multiply one by two, then we get two protons. So this signal represents two protons. If we multiply 1.5 by two, we get three. So this signal represents three protons. If we multiply one by two, we get two."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this signal represents two protons. If we multiply 1.5 by two, we get three. So this signal represents three protons. If we multiply one by two, we get two. And if we multiply 1.5 by two, we get three. So if you add all those up, two plus three plus two plus three, that's, of course, 10 protons. So now we've accounted for all 10 protons using our integration values."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we multiply one by two, we get two. And if we multiply 1.5 by two, we get three. So if you add all those up, two plus three plus two plus three, that's, of course, 10 protons. So now we've accounted for all 10 protons using our integration values. Next, let's look at each signal one by one. So we'll start with this signal. So we have a CH2, since we have two protons."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So now we've accounted for all 10 protons using our integration values. Next, let's look at each signal one by one. So we'll start with this signal. So we have a CH2, since we have two protons. So this signal represents a CH2. How many neighboring protons do those CH2 protons have? So we can figure that out by the number of peaks on the signal."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have a CH2, since we have two protons. So this signal represents a CH2. How many neighboring protons do those CH2 protons have? So we can figure that out by the number of peaks on the signal. So this signal here has one, two, three peaks. And if you think about the n plus one rule, if you have n neighbors, you get n plus one peaks. So if we have three peaks, all we have to do is subtract one to find out how many neighboring protons."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we can figure that out by the number of peaks on the signal. So this signal here has one, two, three peaks. And if you think about the n plus one rule, if you have n neighbors, you get n plus one peaks. So if we have three peaks, all we have to do is subtract one to find out how many neighboring protons. So three minus one is equal to two. So this CH2, these CH2 protons have two neighboring protons. Let's think about the chemical shift for this signal."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we have three peaks, all we have to do is subtract one to find out how many neighboring protons. So three minus one is equal to two. So this CH2, these CH2 protons have two neighboring protons. Let's think about the chemical shift for this signal. So the chemical shift for this signal is between two parts per million and 2.5. And that's in the region for a proton next to a carbonyl. And that would make a lot of sense because we calculated an HDI of one, indicating there might be a double bond present."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about the chemical shift for this signal. So the chemical shift for this signal is between two parts per million and 2.5. And that's in the region for a proton next to a carbonyl. And that would make a lot of sense because we calculated an HDI of one, indicating there might be a double bond present. And we need to account for an oxygen in our molecular formula. So let's go ahead and draw in a carbonyl. And these CH2 protons are next to the carbonyl."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And that would make a lot of sense because we calculated an HDI of one, indicating there might be a double bond present. And we need to account for an oxygen in our molecular formula. So let's go ahead and draw in a carbonyl. And these CH2 protons are next to the carbonyl. We know that because of the shift, right? The carbonyl, the oxygen deshields these two protons a little bit and gives us a higher value for the chemical shift compared to something in an alkane-type region. All right, let's color these protons in here magenta."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And these CH2 protons are next to the carbonyl. We know that because of the shift, right? The carbonyl, the oxygen deshields these two protons a little bit and gives us a higher value for the chemical shift compared to something in an alkane-type region. All right, let's color these protons in here magenta. So we're saying this signal is due to those two protons. Let's move on to the next signal. So we have three protons."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's color these protons in here magenta. So we're saying this signal is due to those two protons. Let's move on to the next signal. So we have three protons. So that would be a methyl group, so CH3. How many neighboring protons for those methyl protons? Well, there's only one peak here."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have three protons. So that would be a methyl group, so CH3. How many neighboring protons for those methyl protons? Well, there's only one peak here. So one minus one is zero. So zero neighboring protons. What about the chemical shift?"}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, there's only one peak here. So one minus one is zero. So zero neighboring protons. What about the chemical shift? So the chemical shift for this signal is once again past two parts per million. So this signal is deshielded. So these protons must be deshielded a little bit."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What about the chemical shift? So the chemical shift for this signal is once again past two parts per million. So this signal is deshielded. So these protons must be deshielded a little bit. And those must be next to our carbonyl. So we'll draw in our methyl protons, being deshielded because they're next to the carbonyl. And let's make these blue."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So these protons must be deshielded a little bit. And those must be next to our carbonyl. So we'll draw in our methyl protons, being deshielded because they're next to the carbonyl. And let's make these blue. So these protons in blue here, these three protons are giving us this signal. And we should expect zero neighboring protons. So those protons are on this carbon, and the next-door carbon has no protons on it."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's make these blue. So these protons in blue here, these three protons are giving us this signal. And we should expect zero neighboring protons. So those protons are on this carbon, and the next-door carbon has no protons on it. So zero neighboring protons makes sense. Move on to the next signal. So this signal represents two protons, so that's a CH2."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So those protons are on this carbon, and the next-door carbon has no protons on it. So zero neighboring protons makes sense. Move on to the next signal. So this signal represents two protons, so that's a CH2. How many neighboring protons? Well, let's count up how many peaks we have. One, two, three, four, five, six."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this signal represents two protons, so that's a CH2. How many neighboring protons? Well, let's count up how many peaks we have. One, two, three, four, five, six. So six peaks. So we subtract one from that. So six minus one is five."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six. So six peaks. So we subtract one from that. So six minus one is five. So we would expect five neighboring protons. Let's move on to the next signal. Three protons, so that's a methyl group."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So six minus one is five. So we would expect five neighboring protons. Let's move on to the next signal. Three protons, so that's a methyl group. So CH3. How many neighbors? Well, we have one, two, three peaks."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Three protons, so that's a methyl group. So CH3. How many neighbors? Well, we have one, two, three peaks. So three minus one is two. So we would expect two neighboring protons. Now these two signals, this signal here and this signal here, we're talking about under two parts per million now."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have one, two, three peaks. So three minus one is two. So we would expect two neighboring protons. Now these two signals, this signal here and this signal here, we're talking about under two parts per million now. So these must be the furthest away from the carbonyl. So they're not being deshielded as much as the two signals in this direction. So these two neighboring protons for the methyl group must be these two protons right here."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Now these two signals, this signal here and this signal here, we're talking about under two parts per million now. So these must be the furthest away from the carbonyl. So they're not being deshielded as much as the two signals in this direction. So these two neighboring protons for the methyl group must be these two protons right here. So let's go ahead and draw that in. So we have our methyl protons right here. And let me make these protons red."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So these two neighboring protons for the methyl group must be these two protons right here. So let's go ahead and draw that in. So we have our methyl protons right here. And let me make these protons red. So these protons right here in red are giving us this signal. And from this signal, we know that these methyl protons are next to two neighbors. And so we must have a CH2 next to that methyl."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let me make these protons red. So these protons right here in red are giving us this signal. And from this signal, we know that these methyl protons are next to two neighbors. And so we must have a CH2 next to that methyl. And so let's go ahead and draw in our CH2. And this CH2 must be this signal right here. So let's think about how many neighbors."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so we must have a CH2 next to that methyl. And so let's go ahead and draw in our CH2. And this CH2 must be this signal right here. So let's think about how many neighbors. We expected five neighbors for these CH2 protons. Let's count them up. One, two, three, four, five."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about how many neighbors. We expected five neighbors for these CH2 protons. Let's count them up. One, two, three, four, five. So five neighboring protons matches what we see on the NMR spectrum. Now those protons, those magenta and red protons, are actually in different environments. And so the simplified version of the N plus one rule isn't quite true, but it works for this example."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five. So five neighboring protons matches what we see on the NMR spectrum. Now those protons, those magenta and red protons, are actually in different environments. And so the simplified version of the N plus one rule isn't quite true, but it works for this example. It works for this example, so we're going to go with it because all we care about is getting the structure of our molecule here. All right, so let's think about the red protons again. So the red protons are supposed to have two neighbors."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so the simplified version of the N plus one rule isn't quite true, but it works for this example. It works for this example, so we're going to go with it because all we care about is getting the structure of our molecule here. All right, so let's think about the red protons again. So the red protons are supposed to have two neighbors. So how many neighbors do the red protons have? Here's one and here's two. So this makes sense."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the red protons are supposed to have two neighbors. So how many neighbors do the red protons have? Here's one and here's two. So this makes sense. So that one makes sense. This one makes sense right here. We just talked about that."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this makes sense. So that one makes sense. This one makes sense right here. We just talked about that. Let's think about the magenta protons. So the magenta protons were supposed to have two neighbors. So let's look at the carbon next door to this one."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We just talked about that. Let's think about the magenta protons. So the magenta protons were supposed to have two neighbors. So let's look at the carbon next door to this one. So here's the carbon next door. Here we have a neighbor and here we have a neighbor. So this makes sense."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at the carbon next door to this one. So here's the carbon next door. Here we have a neighbor and here we have a neighbor. So this makes sense. And then we already talked about the blue protons, right, having zero neighbors. So everything seems to make sense here. And to sum everything up, make sure to count all of your atoms and you will get, of course, five carbons, ten hydrogens, and one oxygen when you do that."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this makes sense. And then we already talked about the blue protons, right, having zero neighbors. So everything seems to make sense here. And to sum everything up, make sure to count all of your atoms and you will get, of course, five carbons, ten hydrogens, and one oxygen when you do that. So this is the dot structure that we were trying to find. Let's take a look at another one here. So we have a molecular formula of C8H10."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And to sum everything up, make sure to count all of your atoms and you will get, of course, five carbons, ten hydrogens, and one oxygen when you do that. So this is the dot structure that we were trying to find. Let's take a look at another one here. So we have a molecular formula of C8H10. And let's go ahead and calculate the hydrogen deficiency index. So if we have eight carbons, then we can have a maximum of two times eight plus two hydrogens. So two times eight is 16 plus two is 18."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have a molecular formula of C8H10. And let's go ahead and calculate the hydrogen deficiency index. So if we have eight carbons, then we can have a maximum of two times eight plus two hydrogens. So two times eight is 16 plus two is 18. So for eight carbons, 18 hydrogens is the max. Here we have ten hydrogens. So we're missing eight hydrogens, or we're missing four pairs of hydrogens."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So two times eight is 16 plus two is 18. So for eight carbons, 18 hydrogens is the max. Here we have ten hydrogens. So we're missing eight hydrogens, or we're missing four pairs of hydrogens. So the hydrogen deficiency index is equal to four. Remember, any time your HDI is equal to four, you should think about a benzene ring. So I'm going to go ahead and draw a benzene ring in here."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're missing eight hydrogens, or we're missing four pairs of hydrogens. So the hydrogen deficiency index is equal to four. Remember, any time your HDI is equal to four, you should think about a benzene ring. So I'm going to go ahead and draw a benzene ring in here. And let's look at the integration. So sometimes you'll see the integration given like this. So this represents five protons for this very complex-looking signal right here."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and draw a benzene ring in here. And let's look at the integration. So sometimes you'll see the integration given like this. So this represents five protons for this very complex-looking signal right here. This signal over here represents two protons, and this signal represents three protons. So let's go back to these five protons with this complex signal. This is in the aromatic range, so approximately 6.528."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this represents five protons for this very complex-looking signal right here. This signal over here represents two protons, and this signal represents three protons. So let's go back to these five protons with this complex signal. This is in the aromatic range, so approximately 6.528. Those must be five aromatic protons. So we can go ahead and draw in five protons off of our benzene ring. So even though those protons are in slightly different environments, because this integration value is five here, and we know a benzene ring is present, we're done."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is in the aromatic range, so approximately 6.528. Those must be five aromatic protons. So we can go ahead and draw in five protons off of our benzene ring. So even though those protons are in slightly different environments, because this integration value is five here, and we know a benzene ring is present, we're done. We don't have to worry about the slightly different environments. We know that these five protons are giving us this complicated signal over here. Next, let's look at this signal."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So even though those protons are in slightly different environments, because this integration value is five here, and we know a benzene ring is present, we're done. We don't have to worry about the slightly different environments. We know that these five protons are giving us this complicated signal over here. Next, let's look at this signal. So two protons, so that must be a CH2. How many neighboring protons for the CH2? Well, there's one, two, three, four peaks."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at this signal. So two protons, so that must be a CH2. How many neighboring protons for the CH2? Well, there's one, two, three, four peaks. So if there's four peaks, four minus one is three. So these CH2 protons have three neighbors. All right, let's look at this next signal."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, there's one, two, three, four peaks. So if there's four peaks, four minus one is three. So these CH2 protons have three neighbors. All right, let's look at this next signal. So three protons, so this must be a CH3. How many neighboring protons do we have for the CH3 protons? Well, we have one, two, three peaks."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at this next signal. So three protons, so this must be a CH3. How many neighboring protons do we have for the CH3 protons? Well, we have one, two, three peaks. So three peaks, three minus one is two. So two neighbors. And so two neighboring protons, that must be these two neighboring protons."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have one, two, three peaks. So three peaks, three minus one is two. So two neighbors. And so two neighboring protons, that must be these two neighboring protons. And so we can go ahead and draw an ethyl group. So this is an ethyl group pattern over here. So there's a CH2, and there's your CH3."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so two neighboring protons, that must be these two neighboring protons. And so we can go ahead and draw an ethyl group. So this is an ethyl group pattern over here. So there's a CH2, and there's your CH3. So let's draw in all of our protons. So we have these protons, and we have these protons. Let's color coordinate here, just to make sure everything makes sense."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So there's a CH2, and there's your CH3. So let's draw in all of our protons. So we have these protons, and we have these protons. Let's color coordinate here, just to make sure everything makes sense. So these red protons right here, they must be giving us this signal. So these methyl protons, right? We'd expect two neighbors, because we have three peaks on our signal."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's color coordinate here, just to make sure everything makes sense. So these red protons right here, they must be giving us this signal. So these methyl protons, right? We'd expect two neighbors, because we have three peaks on our signal. And here are the two neighbors. One, two, so two neighboring protons. Let's look at these two protons next."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We'd expect two neighbors, because we have three peaks on our signal. And here are the two neighbors. One, two, so two neighboring protons. Let's look at these two protons next. So these two protons are giving us this signal. We would expect three neighbors, because we have four peaks. One, two, three, four."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at these two protons next. So these two protons are giving us this signal. We would expect three neighbors, because we have four peaks. One, two, three, four. So three neighbors. So here's a next-door carbon. So one, two, three, three protons."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four. So three neighbors. So here's a next-door carbon. So one, two, three, three protons. And then this is also a next-door carbon, but there are no protons on this carbon. And so we see only these three neighbors. And so everything seems to make sense."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So one, two, three, three protons. And then this is also a next-door carbon, but there are no protons on this carbon. And so we see only these three neighbors. And so everything seems to make sense. And once again, count up all of your atoms, and make sure you've accounted for everything. For example, you have eight carbons you have to worry about. So on your ring, we know there are six."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so everything seems to make sense. And once again, count up all of your atoms, and make sure you've accounted for everything. For example, you have eight carbons you have to worry about. So on your ring, we know there are six. One, two, three, four, five, six. And then we get seven, and then we get eight. So the eight carbons is correct."}, {"video_title": "Proton NMR practice 1 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So on your ring, we know there are six. One, two, three, four, five, six. And then we get seven, and then we get eight. So the eight carbons is correct. And then if you count all those up, we have our 10 hydrogens accounted for, too. So this is the NMR spectrum for ethyl benzene. And this one was a little bit easier than the previous example."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And our best sense of what actually happened is that there was a supernova in our vicinity of the galaxy. And this right here is a picture of a supernova remnant, actually the remnant for Kepler's supernova. The supernova in this picture actually happened 400 years ago in 1604. So right at the center, a star essentially exploded and for a few weeks was the brightest object in the night sky. And it was observed by Kepler and other people in 1604. And this is what it looks like now. So this is what we see is kind of the shock wave that's been traveling out for the past 400 years."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So right at the center, a star essentially exploded and for a few weeks was the brightest object in the night sky. And it was observed by Kepler and other people in 1604. And this is what it looks like now. So this is what we see is kind of the shock wave that's been traveling out for the past 400 years. And so now it must be many light years across. It wasn't obviously, matter wasn't traveling at the speed of light, but it must have been traveling pretty, pretty fast, at least relativistic speeds where a reasonable fraction of the speed of light. So this has traveled a good bit out now."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is what we see is kind of the shock wave that's been traveling out for the past 400 years. And so now it must be many light years across. It wasn't obviously, matter wasn't traveling at the speed of light, but it must have been traveling pretty, pretty fast, at least relativistic speeds where a reasonable fraction of the speed of light. So this has traveled a good bit out now. But what you can imagine is when you have the shock wave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shock wave came by, it just wasn't dense enough. It wasn't dense enough for gravity to take over and for it to accrete essentially into a solar system. But when the shock wave passes by, it compresses all of this gas and all of this material and all of these molecules."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this has traveled a good bit out now. But what you can imagine is when you have the shock wave traveling out from a supernova, let's say you had a cloud of molecules, a cloud of gas, that before the shock wave came by, it just wasn't dense enough. It wasn't dense enough for gravity to take over and for it to accrete essentially into a solar system. But when the shock wave passes by, it compresses all of this gas and all of this material and all of these molecules. So it now does have that critical density to form, to accrete into a star and a solar system. And so we think that's what's happening. The reason why we feel pretty strongly that it must have been caused by a supernova is that the only way that the really heavy elements can form or the only way that we know that they can form is in kind of the heat of a supernova."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But when the shock wave passes by, it compresses all of this gas and all of this material and all of these molecules. So it now does have that critical density to form, to accrete into a star and a solar system. And so we think that's what's happening. The reason why we feel pretty strongly that it must have been caused by a supernova is that the only way that the really heavy elements can form or the only way that we know that they can form is in kind of the heat of a supernova. And our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed roughly at the time of the formation of Earth, at about 4 and 1 half billion years ago. And we'll talk in a little bit more depth in future videos on exactly how people figure that out. But since the uranium seems about the same age as our solar system, it must have been formed at around the same time."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The reason why we feel pretty strongly that it must have been caused by a supernova is that the only way that the really heavy elements can form or the only way that we know that they can form is in kind of the heat of a supernova. And our uranium, the uranium that seems to be in our solar system on Earth, seems to have formed roughly at the time of the formation of Earth, at about 4 and 1 half billion years ago. And we'll talk in a little bit more depth in future videos on exactly how people figure that out. But since the uranium seems about the same age as our solar system, it must have been formed at around the same time. And so it must have been formed by a supernova. And it must be coming from a supernova. So a supernova shock wave must have passed through our part of the universe."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But since the uranium seems about the same age as our solar system, it must have been formed at around the same time. And so it must have been formed by a supernova. And it must be coming from a supernova. So a supernova shock wave must have passed through our part of the universe. And that's a good reason for gas to get compressed and begin to accrete. So you fast forward a few million years ago that gas would have accreted into something like this. It would have reached the critical temperature, critical density, and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So a supernova shock wave must have passed through our part of the universe. And that's a good reason for gas to get compressed and begin to accrete. So you fast forward a few million years ago that gas would have accreted into something like this. It would have reached the critical temperature, critical density, and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium. This right here is our early sun. Around the sun, you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. They were actually supported by a little bit of pressure, too."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would have reached the critical temperature, critical density, and pressure at the center for ignition to occur, for fusion to start to happen, for hydrogen to start fusing into helium. This right here is our early sun. Around the sun, you have all of the gases and particles and molecules that had enough angular velocity to not fall into the sun, to go into orbit around the sun. They were actually supported by a little bit of pressure, too. Because you can kind of view this as kind of a big cloud of gas. So they're always bumping into each other. But for the most part, it was their angular velocity."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They were actually supported by a little bit of pressure, too. Because you can kind of view this as kind of a big cloud of gas. So they're always bumping into each other. But for the most part, it was their angular velocity. And over the next tens of millions of years, they'll slowly bump into each other and clump into each other. Even small particles have gravity. And they're going to slowly become rocks and asteroids and eventually what we'd call planetesimals, which are really kind of view them as seeds of planets or early planets."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But for the most part, it was their angular velocity. And over the next tens of millions of years, they'll slowly bump into each other and clump into each other. Even small particles have gravity. And they're going to slowly become rocks and asteroids and eventually what we'd call planetesimals, which are really kind of view them as seeds of planets or early planets. And then those would have a reasonable amount of gravity. And other things would be attracted to them and slowly clump up to them. But this wasn't like a simple process."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they're going to slowly become rocks and asteroids and eventually what we'd call planetesimals, which are really kind of view them as seeds of planets or early planets. And then those would have a reasonable amount of gravity. And other things would be attracted to them and slowly clump up to them. But this wasn't like a simple process. You could imagine you might have one planetesimal form. And maybe there's another planetesimal form. And instead of having a nice, gentle, those two guys accreting into each other, they might have huge relative velocities and ram into each other and then just shatter."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this wasn't like a simple process. You could imagine you might have one planetesimal form. And maybe there's another planetesimal form. And instead of having a nice, gentle, those two guys accreting into each other, they might have huge relative velocities and ram into each other and then just shatter. So this wasn't just a nice, gentle process of constant accretion. It would actually have been a very violent process. It actually happened early in Earth's history."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And instead of having a nice, gentle, those two guys accreting into each other, they might have huge relative velocities and ram into each other and then just shatter. So this wasn't just a nice, gentle process of constant accretion. It would actually have been a very violent process. It actually happened early in Earth's history. And we actually think this is why the moon formed. So at some point, you fast forward a little bit from this. Earth would have formed, or I should say the mass that eventually becomes our modern Earth would have been forming."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It actually happened early in Earth's history. And we actually think this is why the moon formed. So at some point, you fast forward a little bit from this. Earth would have formed, or I should say the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So let's say that that is our modern Earth. And what we think happened is that another protoplanet, or it was actually a planet because it was roughly the size of Mars, ran into what is eventually going to become our Earth."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Earth would have formed, or I should say the mass that eventually becomes our modern Earth would have been forming. Let me draw it over here. So let's say that that is our modern Earth. And what we think happened is that another protoplanet, or it was actually a planet because it was roughly the size of Mars, ran into what is eventually going to become our Earth. And this is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars. And it ran into what eventually would become Earth."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what we think happened is that another protoplanet, or it was actually a planet because it was roughly the size of Mars, ran into what is eventually going to become our Earth. And this is actually a picture of it. This is an artist's depiction of that collision, where this planet right here is the size of Mars. And it ran into what eventually would become Earth. And this we call Theia. This is Theia. And what we believe happened, and if you go onto the internet, you'll see some simulations that talk about this, is that we think it was a glancing blow, that it wasn't a direct hit that would have just kind of shattered each of them and turned them into one big molten ball."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it ran into what eventually would become Earth. And this we call Theia. This is Theia. And what we believe happened, and if you go onto the internet, you'll see some simulations that talk about this, is that we think it was a glancing blow, that it wasn't a direct hit that would have just kind of shattered each of them and turned them into one big molten ball. We think it was a glancing blow, something like this. So if this was essentially Earth, obviously Earth got changed dramatically once Theia ran into it. But Theia is right over here."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what we believe happened, and if you go onto the internet, you'll see some simulations that talk about this, is that we think it was a glancing blow, that it wasn't a direct hit that would have just kind of shattered each of them and turned them into one big molten ball. We think it was a glancing blow, something like this. So if this was essentially Earth, obviously Earth got changed dramatically once Theia ran into it. But Theia is right over here. And we think it was a glancing blow, where it came and it hit Earth at kind of an angle. And then obviously the combined energies from that interaction would have made both of them molten. And frankly, they probably already were molten, because you had a bunch of smaller collisions and accretion events and little things hitting the surface of probably both of them during this entire period."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But Theia is right over here. And we think it was a glancing blow, where it came and it hit Earth at kind of an angle. And then obviously the combined energies from that interaction would have made both of them molten. And frankly, they probably already were molten, because you had a bunch of smaller collisions and accretion events and little things hitting the surface of probably both of them during this entire period. But this would have had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. So it would have just come in, had a glancing blow on Earth, and then splashed a bunch of molten material. Some of it would have been captured by Earth."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And frankly, they probably already were molten, because you had a bunch of smaller collisions and accretion events and little things hitting the surface of probably both of them during this entire period. But this would have had a glancing blow on Earth and essentially splashed a bunch of molten material out into orbit. So it would have just come in, had a glancing blow on Earth, and then splashed a bunch of molten material. Some of it would have been captured by Earth. So this is the before, and then the after. You could imagine Earth is kind of this molten, super hot ball. And some of it just gets splashed into orbit from the collision."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some of it would have been captured by Earth. So this is the before, and then the after. You could imagine Earth is kind of this molten, super hot ball. And some of it just gets splashed into orbit from the collision. And let me see if I can draw Theia here. So Theia has collided. And it's also molten now, because huge energies."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And some of it just gets splashed into orbit from the collision. And let me see if I can draw Theia here. So Theia has collided. And it's also molten now, because huge energies. And it splashes some of it into orbit. And if we fast forward a little bit, this stuff that got splashed into orbit, it's going in that direction. That becomes our moon."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's also molten now, because huge energies. And it splashes some of it into orbit. And if we fast forward a little bit, this stuff that got splashed into orbit, it's going in that direction. That becomes our moon. And then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the moon actually formed. And even after this happened, the Earth still had a lot more, I guess, violence to experience."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That becomes our moon. And then the rest of this material eventually kind of condenses back into a spherical shape and is what we now call our Earth. So that's how we actually think right now that the moon actually formed. And even after this happened, the Earth still had a lot more, I guess, violence to experience. So just to get a sense of where we are in the history of Earth, we're going to refer to this time clock a lot over the next few videos. This time clock starts right here at the formation of our solar system 4.6 billion years ago, probably coinciding with some type of supernova. And as we go clockwise on this diagram, we're moving forward in time."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even after this happened, the Earth still had a lot more, I guess, violence to experience. So just to get a sense of where we are in the history of Earth, we're going to refer to this time clock a lot over the next few videos. This time clock starts right here at the formation of our solar system 4.6 billion years ago, probably coinciding with some type of supernova. And as we go clockwise on this diagram, we're moving forward in time. And we're going to go all the way forward to the present period. And just so you understand some of the terminology, GA means billions of years ago, G for giga. MA means millions of years ago, M for mega."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as we go clockwise on this diagram, we're moving forward in time. And we're going to go all the way forward to the present period. And just so you understand some of the terminology, GA means billions of years ago, G for giga. MA means millions of years ago, M for mega. So where we are right now, the moon has formed. We're in what we call the Hadean period. Or actually, I shouldn't say period."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "MA means millions of years ago, M for mega. So where we are right now, the moon has formed. We're in what we call the Hadean period. Or actually, I shouldn't say period. It's the Hadean eon of Earth. Period is actually another time period. So let me make this very clear."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or actually, I shouldn't say period. It's the Hadean eon of Earth. Period is actually another time period. So let me make this very clear. It's the Hadean. We are in the Hadean eon. And an eon is kind of the largest period of time that we talk about, especially relative to Earth."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me make this very clear. It's the Hadean. We are in the Hadean eon. And an eon is kind of the largest period of time that we talk about, especially relative to Earth. And it's roughly 500 million to a billion years is an eon. And what makes the Hadean eon distinctive, well, from a geological point of view, what makes it distinctive is it's really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic scale rocks from the Hadean period."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And an eon is kind of the largest period of time that we talk about, especially relative to Earth. And it's roughly 500 million to a billion years is an eon. And what makes the Hadean eon distinctive, well, from a geological point of view, what makes it distinctive is it's really we don't have any rocks from the Hadean period. We don't have any kind of macroscopic scale rocks from the Hadean period. And that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava. And it was molten because it was the product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. So if you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma or whatever, it would look like this."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We don't have any kind of macroscopic scale rocks from the Hadean period. And that's because at that time, we believe, the Earth was just this molten ball of kind of magma and lava. And it was molten because it was the product of all of these accretion events and all of these collisions and all this kinetic energy turning into heat. So if you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma or whatever, it would look like this. And you wouldn't be able to breathe anyway. This is what the surface of the Earth would look like. It would look like a big magma pool."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you were to look at the surface of the Earth, if you were to be on the surface of the Earth during the Hadean eon, which you probably wouldn't want to be because you might get hit by a falling meteorite or probably burned by some magma or whatever, it would look like this. And you wouldn't be able to breathe anyway. This is what the surface of the Earth would look like. It would look like a big magma pool. And that's why we don't have any rocks from there. Because the rocks were just constantly being recycled, being dissolved, and churned inside of this giant molten ball. And frankly, the Earth still is a giant molten ball."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It would look like a big magma pool. And that's why we don't have any rocks from there. Because the rocks were just constantly being recycled, being dissolved, and churned inside of this giant molten ball. And frankly, the Earth still is a giant molten ball. It's just we live on the super thin, cooled crust of that molten ball. If you go right below that crust, and we'll talk a little bit more about that in future videos, you will get magma. And if you go dig deeper, you'll have liquid iron."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And frankly, the Earth still is a giant molten ball. It's just we live on the super thin, cooled crust of that molten ball. If you go right below that crust, and we'll talk a little bit more about that in future videos, you will get magma. And if you go dig deeper, you'll have liquid iron. So I mean, it still is a molten ball. And this whole period is just a violent, not only was Earth itself volcanic, molten ball, it began to harden as you get to the late Hadean eon. But we also had stuff falling from the sky and constantly colliding with Earth and really just continuing to add to the heat of this molten ball."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if you go dig deeper, you'll have liquid iron. So I mean, it still is a molten ball. And this whole period is just a violent, not only was Earth itself volcanic, molten ball, it began to harden as you get to the late Hadean eon. But we also had stuff falling from the sky and constantly colliding with Earth and really just continuing to add to the heat of this molten ball. Anyway, I'll leave you there. And as you can imagine, at this point, as far as we can tell, there was no life on Earth. Some people believe that maybe some life could have formed in the late Hadean eon."}, {"video_title": "Earth formation Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we also had stuff falling from the sky and constantly colliding with Earth and really just continuing to add to the heat of this molten ball. Anyway, I'll leave you there. And as you can imagine, at this point, as far as we can tell, there was no life on Earth. Some people believe that maybe some life could have formed in the late Hadean eon. But for the most part, this is completely inhospitable for any life forming. So I'll leave you there. And where we take up the next video, we'll talk a little bit about the Archean eon."}, {"video_title": "What causes precession and other orbital changes Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But they're really a byproduct of the Earth and Sun's interactions with Earth and with the fact that Earth is not a perfect sphere. So if I draw Earth, so this is my little drawing of Earth, and let me put the poles over here, North Pole and South Pole, it actually turns out that Earth is fatter than it is taller. So if you were to measure Earth's diameter along the equator, it is 43 kilometers, or about 27, 43 kilometers, which is approximately 27 miles longer than if you were to measure its diameter from pole to pole. So longer than the pole to pole diameter. And the fact that Earth has this equatorial bulge, that it's not a perfect sphere, and once again I'm not going to go into the math here, is the interactions between that, I guess you could call it that one asymmetry of the Earth, it's that interaction between that and the pull of gravity between the Earth and the Sun and the Moon that causes these long-term cycles, this axial precession. And other, I guess, less noticeable changes in Earth's orbit. And as we'll see in the next video, these aren't the only types of changes in orbits we have."}, {"video_title": "What causes precession and other orbital changes Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So longer than the pole to pole diameter. And the fact that Earth has this equatorial bulge, that it's not a perfect sphere, and once again I'm not going to go into the math here, is the interactions between that, I guess you could call it that one asymmetry of the Earth, it's that interaction between that and the pull of gravity between the Earth and the Sun and the Moon that causes these long-term cycles, this axial precession. And other, I guess, less noticeable changes in Earth's orbit. And as we'll see in the next video, these aren't the only types of changes in orbits we have. We also have changes in the actual ellipse that Earth's orbit has actually rotates over time. But that's due more to interactions with Earth's orbit and the orbit of other planets in our solar system. And once again, it's one of those things that happen over thousands and thousands of years."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if you take methane and react it with oxygen, you'll get carbon dioxide and water as the products. And for all hydrocarbons, you're gonna get CO2 and H2O as the products of a combustion reaction. Delta H zero for this reaction is negative 890 kilojoules per one mole of methane, which means one mole of methane, if you combust one mole of methane, you're gonna get 890 kilojoules of heat. And so this is an exothermic reaction, heat is given off. So the heat of combustion is the heat that's released on the complete combustion of one mole of a substance. In this case, we're talking about methane. Let's talk about hexane next."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And so this is an exothermic reaction, heat is given off. So the heat of combustion is the heat that's released on the complete combustion of one mole of a substance. In this case, we're talking about methane. Let's talk about hexane next. Let me draw out the structure for hexane. So we have six carbons and notice we have two CH3 groups. So here's a CH3 and here's a CH3."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Let's talk about hexane next. Let me draw out the structure for hexane. So we have six carbons and notice we have two CH3 groups. So here's a CH3 and here's a CH3. So CH3 and CH3, and then we have four CH2 groups. So here's one, two, three, and four. So that's why we have a four right here."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So here's a CH3 and here's a CH3. So CH3 and CH3, and then we have four CH2 groups. So here's one, two, three, and four. So that's why we have a four right here. So there are four CH2 groups in hexane. And here we have the heat of combustion. And now we're gonna talk about the heat of combustion as the negative change in the enthalpy."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So that's why we have a four right here. So there are four CH2 groups in hexane. And here we have the heat of combustion. And now we're gonna talk about the heat of combustion as the negative change in the enthalpy. So negative delta H zero in terms of kilojoules per mole. That gives us a positive value here. So for hexane, it's 4,163 kilojoules for every one mole of hexane that we combust."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And now we're gonna talk about the heat of combustion as the negative change in the enthalpy. So negative delta H zero in terms of kilojoules per mole. That gives us a positive value here. So for hexane, it's 4,163 kilojoules for every one mole of hexane that we combust. Notice what happens as we move on to heptane here. We're increasing in one CH2 group. We're going from four CH2 groups to five CH2 groups."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So for hexane, it's 4,163 kilojoules for every one mole of hexane that we combust. Notice what happens as we move on to heptane here. We're increasing in one CH2 group. We're going from four CH2 groups to five CH2 groups. So now we have five CH2 groups. We get an increase in the heat of combustion. So more heat is released."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "We're going from four CH2 groups to five CH2 groups. So now we have five CH2 groups. We get an increase in the heat of combustion. So more heat is released. And that makes sense. If you increase the number of carbons, you're gonna increase in the heat of combustion. So if we increase by one CH2 group, how much do we increase in terms of the heat of combustion?"}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So more heat is released. And that makes sense. If you increase the number of carbons, you're gonna increase in the heat of combustion. So if we increase by one CH2 group, how much do we increase in terms of the heat of combustion? Well, if you take 4,163 and subtract that from 4,817, that's a difference of 654 kilojoules. The pattern continues when we move on to octane. So octane, now we have six CH2 groups."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if we increase by one CH2 group, how much do we increase in terms of the heat of combustion? Well, if you take 4,163 and subtract that from 4,817, that's a difference of 654 kilojoules. The pattern continues when we move on to octane. So octane, now we have six CH2 groups. So we've added one more. We've increased our heat of combustion to 5,471 kilojoules per one mole. And that is also an increase of 654 kilojoules."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So octane, now we have six CH2 groups. So we've added one more. We've increased our heat of combustion to 5,471 kilojoules per one mole. And that is also an increase of 654 kilojoules. So for nonane, with one more CH2, we could predict, we could estimate the heat of combustion. And we could just add 654 to 5,471. So we can go ahead and set that up over here."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And that is also an increase of 654 kilojoules. So for nonane, with one more CH2, we could predict, we could estimate the heat of combustion. And we could just add 654 to 5,471. So we can go ahead and set that up over here. So we have 5,471. We're adding 654 to that. So that would give us a 5 here."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and set that up over here. So we have 5,471. We're adding 654 to that. So that would give us a 5 here. And then 7 and 5 is 12. So we carry the 1. That gives us 11."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So that would give us a 5 here. And then 7 and 5 is 12. So we carry the 1. That gives us 11. So we get 6,125. So if I write that in here, 6,125 kilojoules per mole is a pretty good estimation for the heat of combustion of nonane. Heats of combustion are really useful when you're trying to compare the stability of isomers."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "That gives us 11. So we get 6,125. So if I write that in here, 6,125 kilojoules per mole is a pretty good estimation for the heat of combustion of nonane. Heats of combustion are really useful when you're trying to compare the stability of isomers. For example, here we have three isomers, octane, 2-methylheptane, and 2,2-dimethylhexane. So all of these molecules have the molecular formula C8H18. So if we write out the combustion reaction, so we have plus O2, we know the products are CO2 and H2O."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Heats of combustion are really useful when you're trying to compare the stability of isomers. For example, here we have three isomers, octane, 2-methylheptane, and 2,2-dimethylhexane. So all of these molecules have the molecular formula C8H18. So if we write out the combustion reaction, so we have plus O2, we know the products are CO2 and H2O. So we have CO2 plus H2O. Let's go ahead and balance this. We have eight carbons on the left side, so we need an eight over here on the right."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if we write out the combustion reaction, so we have plus O2, we know the products are CO2 and H2O. So we have CO2 plus H2O. Let's go ahead and balance this. We have eight carbons on the left side, so we need an eight over here on the right. We'll put that in front of the CO2. And now let's look at hydrogens. There's 18 on the left, and so we need 18 on the right."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "We have eight carbons on the left side, so we need an eight over here on the right. We'll put that in front of the CO2. And now let's look at hydrogens. There's 18 on the left, and so we need 18 on the right. And we can do that by putting a nine here, because nine and two give us 18 hydrogens. Now let's see how many oxygens are on the right side. Well, we have eight and two give us 16 oxygens here, and then nine gives us nine for the water."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "There's 18 on the left, and so we need 18 on the right. And we can do that by putting a nine here, because nine and two give us 18 hydrogens. Now let's see how many oxygens are on the right side. Well, we have eight and two give us 16 oxygens here, and then nine gives us nine for the water. So that's a total of 25 oxygens on the right side. So we need 25 oxygens on the left side. Now if we pretend like this two isn't here for a second, we could write a 25 right here, but there is a two, so we need to divide by two."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have eight and two give us 16 oxygens here, and then nine gives us nine for the water. So that's a total of 25 oxygens on the right side. So we need 25 oxygens on the left side. Now if we pretend like this two isn't here for a second, we could write a 25 right here, but there is a two, so we need to divide by two. And if we're talking about one mole of C8H18, we're gonna leave this as a fraction. So 25 over two is the amount of oxygen that's necessary to completely react with one mole of C8H18. So now we have our balance equation for our combustion reaction, and if we compare these three isomers, octane, 2-methylheptane, and 2,2-dimethylhexane, in terms of their heats of combustion, we can figure out which one is the most stable isomer."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Now if we pretend like this two isn't here for a second, we could write a 25 right here, but there is a two, so we need to divide by two. And if we're talking about one mole of C8H18, we're gonna leave this as a fraction. So 25 over two is the amount of oxygen that's necessary to completely react with one mole of C8H18. So now we have our balance equation for our combustion reaction, and if we compare these three isomers, octane, 2-methylheptane, and 2,2-dimethylhexane, in terms of their heats of combustion, we can figure out which one is the most stable isomer. Here we have an energy diagram for our three isomers. So on the left here, we'll put increasing energy this way. All three of our isomers would produce the same products if you combust one mole of them."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our balance equation for our combustion reaction, and if we compare these three isomers, octane, 2-methylheptane, and 2,2-dimethylhexane, in terms of their heats of combustion, we can figure out which one is the most stable isomer. Here we have an energy diagram for our three isomers. So on the left here, we'll put increasing energy this way. All three of our isomers would produce the same products if you combust one mole of them. Each one would give us eight CO2 and nine H2O. So there's an energy level associated with our products, and we'll say this is the energy level associated with our products. It's the same for all three of our isomers."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "All three of our isomers would produce the same products if you combust one mole of them. Each one would give us eight CO2 and nine H2O. So there's an energy level associated with our products, and we'll say this is the energy level associated with our products. It's the same for all three of our isomers. Next, we look up the heat of combustion data for the isomers, and this is determined experimentally. For example, octane has a heat of combustion of about 5,471 kilojoules per mole, so that's how much heat is released, 5,471 kilojoules per mole. And if we know the energy level of the products, and we know how much energy was released, we can find the energy level of our reactants."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "It's the same for all three of our isomers. Next, we look up the heat of combustion data for the isomers, and this is determined experimentally. For example, octane has a heat of combustion of about 5,471 kilojoules per mole, so that's how much heat is released, 5,471 kilojoules per mole. And if we know the energy level of the products, and we know how much energy was released, we can find the energy level of our reactants. So there is the energy level for octane. Next, let's think about 2-methylheptane. So again, the energy level of the products is here, and if we look up the heat of combustion for 2-methylheptane, we'll find a value somewhere around 5,466."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And if we know the energy level of the products, and we know how much energy was released, we can find the energy level of our reactants. So there is the energy level for octane. Next, let's think about 2-methylheptane. So again, the energy level of the products is here, and if we look up the heat of combustion for 2-methylheptane, we'll find a value somewhere around 5,466. So that's how much energy is released when you combust one mole of 2-methylheptane, 5,466 kilojoules per mole. Since the energy level of the products is the same as the one before, but we have a smaller heat of combustion this time, that must mean the energy level for 2-methylheptane is lower than the energy level for octane. And finally, we move on to 2,2-dimethylhexane."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So again, the energy level of the products is here, and if we look up the heat of combustion for 2-methylheptane, we'll find a value somewhere around 5,466. So that's how much energy is released when you combust one mole of 2-methylheptane, 5,466 kilojoules per mole. Since the energy level of the products is the same as the one before, but we have a smaller heat of combustion this time, that must mean the energy level for 2-methylheptane is lower than the energy level for octane. And finally, we move on to 2,2-dimethylhexane. The energy level of the products is the same as before. The heat of combustion is 5,458. So somewhere around 5,458 kilojoules per mole is released, right?"}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And finally, we move on to 2,2-dimethylhexane. The energy level of the products is the same as before. The heat of combustion is 5,458. So somewhere around 5,458 kilojoules per mole is released, right? That's how much energy or heat is released when you combust one mole of 2,2-dimethylhexane. And since we have a smaller value for the heat of combustion, that must mean the energy is lower for 2,2-dimethylhexane. Now we can finally compare the stabilities of our isomers."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So somewhere around 5,458 kilojoules per mole is released, right? That's how much energy or heat is released when you combust one mole of 2,2-dimethylhexane. And since we have a smaller value for the heat of combustion, that must mean the energy is lower for 2,2-dimethylhexane. Now we can finally compare the stabilities of our isomers. The higher the energy, the less stable the compound. So octane has the highest energy level. It has the highest heat of combustion."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Now we can finally compare the stabilities of our isomers. The higher the energy, the less stable the compound. So octane has the highest energy level. It has the highest heat of combustion. So this is the least stable out of these three. This is the least stable. The lower the energy for our starting compound, the more stable it is."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "It has the highest heat of combustion. So this is the least stable out of these three. This is the least stable. The lower the energy for our starting compound, the more stable it is. So that must mean that 2,2-dimethylhexane is the most stable isomer out of these three. So now we can think about trends. As we go from octane to 2-methylheptane to 2,2-dimethylhexane, we're increasing in branching."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "The lower the energy for our starting compound, the more stable it is. So that must mean that 2,2-dimethylhexane is the most stable isomer out of these three. So now we can think about trends. As we go from octane to 2-methylheptane to 2,2-dimethylhexane, we're increasing in branching. So we're increasing in the amount of branching that we have. What happened to the heats of combustion? If we define heat of combustion as the negative of the change in the enthalpy, the heats of combustion decreased."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "As we go from octane to 2-methylheptane to 2,2-dimethylhexane, we're increasing in branching. So we're increasing in the amount of branching that we have. What happened to the heats of combustion? If we define heat of combustion as the negative of the change in the enthalpy, the heats of combustion decreased. We went from 5,471 to 5,466 to 5,458. So there was a decrease in the heat of combustion. And what about stability?"}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "If we define heat of combustion as the negative of the change in the enthalpy, the heats of combustion decreased. We went from 5,471 to 5,466 to 5,458. So there was a decrease in the heat of combustion. And what about stability? We got more stable as we branched more. So increased branching in general means increased stability. So it's important to be able to analyze heat of combustion data."}, {"video_title": "Heats of combustion of alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And what about stability? We got more stable as we branched more. So increased branching in general means increased stability. So it's important to be able to analyze heat of combustion data. Just remember, the lower the energy, the more stable the compound. So the compound that had the highest heat of combustion was the least stable. And the compound with the lowest heat of combustion was the most stable."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "What I want to do in this video is go through a bunch of examples and see if we can identify if there are any chiral atoms and to also see if we're dealing with a chiral molecule. So let's look at our examples here. So here I have, what is this? This is chlorocyclopentane. So the first question is, do we have any chiral atoms? And when we look at our definition that we thought of chiral atoms, and it all comes from this notion of handedness and not being able to be superimposable on your mirror image. But we said that they're usually carbons bonded to four different groups."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is chlorocyclopentane. So the first question is, do we have any chiral atoms? And when we look at our definition that we thought of chiral atoms, and it all comes from this notion of handedness and not being able to be superimposable on your mirror image. But we said that they're usually carbons bonded to four different groups. So let's see, do we have any carbons here bonded to four different groups? Well, all the CH2s, they're bonded to another CH2 and then two H's. I could draw it like this, H and H. So they're bonded to two of the same group."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But we said that they're usually carbons bonded to four different groups. So let's see, do we have any carbons here bonded to four different groups? Well, all the CH2s, they're bonded to another CH2 and then two H's. I could draw it like this, H and H. So they're bonded to two of the same group. So none of these CH2s are good candidates for being a chiral center or a chiral carbon. They're both bonded to, or all of them are bonded to two hydrogens and then two other very similar looking CH2 groups, although you have to look at the entire group that it's bonded to. But they're all definitely bonded to two hydrogens."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I could draw it like this, H and H. So they're bonded to two of the same group. So none of these CH2s are good candidates for being a chiral center or a chiral carbon. They're both bonded to, or all of them are bonded to two hydrogens and then two other very similar looking CH2 groups, although you have to look at the entire group that it's bonded to. But they're all definitely bonded to two hydrogens. So it's not four different groups. If we look at this CH right here, we could separate it out like this. So it's bonded to a hydrogen."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But they're all definitely bonded to two hydrogens. So it's not four different groups. If we look at this CH right here, we could separate it out like this. So it's bonded to a hydrogen. This carbon is bonded to a chlorine. And then it's bonded to, well, it's not clear when you look at it right from the get-go whether this group is different than this group. But if you were to split it halfway like this, or maybe another better way to think about it, is if you were to go around this molecule in that direction, the counter clockwise direction, you would encounter a CH2 group, and then you would encounter a CH2 group, and then you would encounter a third."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's bonded to a hydrogen. This carbon is bonded to a chlorine. And then it's bonded to, well, it's not clear when you look at it right from the get-go whether this group is different than this group. But if you were to split it halfway like this, or maybe another better way to think about it, is if you were to go around this molecule in that direction, the counter clockwise direction, you would encounter a CH2 group, and then you would encounter a CH2 group, and then you would encounter a third. And then you would encounter a fourth CH2 group, then you would come back to where you were before. Se you would encounter 4 CH 2s, then you would come back to where you were before. If you go in this direction, what happens?"}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But if you were to split it halfway like this, or maybe another better way to think about it, is if you were to go around this molecule in that direction, the counter clockwise direction, you would encounter a CH2 group, and then you would encounter a CH2 group, and then you would encounter a third. And then you would encounter a fourth CH2 group, then you would come back to where you were before. Se you would encounter 4 CH 2s, then you would come back to where you were before. If you go in this direction, what happens? You'd encounter 1, 2, 3, 4 CH2s, and you come back to where you were before. So all of this bottom group depending on how far you want to extend it, and this top group are really the same group. So this is not a chiral center or not a chiral carbon."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you go in this direction, what happens? You'd encounter 1, 2, 3, 4 CH2s, and you come back to where you were before. So all of this bottom group depending on how far you want to extend it, and this top group are really the same group. So this is not a chiral center or not a chiral carbon. It's not bonded to four different groups. And this is also not a chiral molecule because it does not have a chiral center. And to see that it's not a chiral molecule, let me see if I can backtrack this back to the way I wrote it right before."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is not a chiral center or not a chiral carbon. It's not bonded to four different groups. And this is also not a chiral molecule because it does not have a chiral center. And to see that it's not a chiral molecule, let me see if I can backtrack this back to the way I wrote it right before. So to see that it's not a chiral molecule, there's a couple of ways you could think about it. The easiest way, or the way my brain likes to think about it, is just to think about its mirror image. Its mirror image will look like this."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And to see that it's not a chiral molecule, let me see if I can backtrack this back to the way I wrote it right before. So to see that it's not a chiral molecule, there's a couple of ways you could think about it. The easiest way, or the way my brain likes to think about it, is just to think about its mirror image. Its mirror image will look like this. So if that's the mirror, you would have a chlorine, and then you have a CH, CH2, CH2, and you have a CH2, CH2, and then you complete your cyclopentane. Now in this situation, is there any way to rotate this to get this over there? Well, if you just took this molecule right here and you just rotated it 180 degrees, what would it look like?"}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Its mirror image will look like this. So if that's the mirror, you would have a chlorine, and then you have a CH, CH2, CH2, and you have a CH2, CH2, and then you complete your cyclopentane. Now in this situation, is there any way to rotate this to get this over there? Well, if you just took this molecule right here and you just rotated it 180 degrees, what would it look like? If you just complete, well, maybe a little over, yeah, well, not quite 180 degrees, but if you were to rotate it so that the chlorine goes about that far, you would get this exact molecule. You would get something. It would look a little bit different."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, if you just took this molecule right here and you just rotated it 180 degrees, what would it look like? If you just complete, well, maybe a little over, yeah, well, not quite 180 degrees, but if you were to rotate it so that the chlorine goes about that far, you would get this exact molecule. You would get something. It would look a little bit different. It would look like this. Let me see if I can do it justice. It would look like this."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It would look a little bit different. It would look like this. Let me see if I can do it justice. It would look like this. You would have a CH2, so let me do it up here where we have a little bit more space. If I were to rotate this about that far, I would get a CH, you get the chlorine, and then you have your CH2, and then you have another CH2, and then you would have your CH2 up there. If you were to rotate this all the way around, or actually this is almost 180 degrees, it would look like this."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It would look like this. You would have a CH2, so let me do it up here where we have a little bit more space. If I were to rotate this about that far, I would get a CH, you get the chlorine, and then you have your CH2, and then you have another CH2, and then you would have your CH2 up there. If you were to rotate this all the way around, or actually this is almost 180 degrees, it would look like this. And the only difference between this and this is just how we drew this bond here. I could have easily, instead of drawing that bond like that, I could draw it facing up like that. And these are the exact same molecule."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you were to rotate this all the way around, or actually this is almost 180 degrees, it would look like this. And the only difference between this and this is just how we drew this bond here. I could have easily, instead of drawing that bond like that, I could draw it facing up like that. And these are the exact same molecule. So this molecule is also not chiral. Now let's go to this one over here. So what is this?"}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And these are the exact same molecule. So this molecule is also not chiral. Now let's go to this one over here. So what is this? This is bromochlorofluoromethane, just to practice our naming a little bit. But it's very clear that we are bonded to four different groups. All of the different groups, or the atoms in this case, that are bonded to this carbon are different."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So what is this? This is bromochlorofluoromethane, just to practice our naming a little bit. But it's very clear that we are bonded to four different groups. All of the different groups, or the atoms in this case, that are bonded to this carbon are different. So this carbon is a chiral center, and it should also be pretty clear that it is also a chiral molecule. If you were to take its mirror image, and this is very similar to the example we did in the first video on chirality, but its mirror image will look like this. You have the bromine on the right now, the hydrogen is still in back, and then you have the fluorine above it."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "All of the different groups, or the atoms in this case, that are bonded to this carbon are different. So this carbon is a chiral center, and it should also be pretty clear that it is also a chiral molecule. If you were to take its mirror image, and this is very similar to the example we did in the first video on chirality, but its mirror image will look like this. You have the bromine on the right now, the hydrogen is still in back, and then you have the fluorine above it. No matter how you try to rotate this thing, if you try to get the bromine all the way over there, all the way to that position, then the hydrogen would be in this position, and the chlorine would be in that position. And no matter how you try to flip this around, or rotate it, or shift it, you will never be able to superimpose this molecule on that molecule right there. So that is a chiral center, and this is a chiral molecule."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You have the bromine on the right now, the hydrogen is still in back, and then you have the fluorine above it. No matter how you try to rotate this thing, if you try to get the bromine all the way over there, all the way to that position, then the hydrogen would be in this position, and the chlorine would be in that position. And no matter how you try to flip this around, or rotate it, or shift it, you will never be able to superimpose this molecule on that molecule right there. So that is a chiral center, and this is a chiral molecule. And there's a word for these two versions. We're going to go into the naming of them later on. It's a little bit more involved."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that is a chiral center, and this is a chiral molecule. And there's a word for these two versions. We're going to go into the naming of them later on. It's a little bit more involved. We'll have a whole separate video on it. But these two versions of bromochlorofluoromethane, they sometimes have different chemical properties, and these are called enantiomers. And enantiomers are just the mirror images."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's a little bit more involved. We'll have a whole separate video on it. But these two versions of bromochlorofluoromethane, they sometimes have different chemical properties, and these are called enantiomers. And enantiomers are just the mirror images. Each enantiomer is a mirror image of each other, but they are stereoisomers. These are all just terminology. Stereoisomers."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And enantiomers are just the mirror images. Each enantiomer is a mirror image of each other, but they are stereoisomers. These are all just terminology. Stereoisomers. You're familiar with the word isomer. An isomer just means that you have the same atoms in your molecule, but then you have different types of isomers. You have constitutional isomers that say, OK, different things are connected to different things."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Stereoisomers. You're familiar with the word isomer. An isomer just means that you have the same atoms in your molecule, but then you have different types of isomers. You have constitutional isomers that say, OK, different things are connected to different things. Stereoisomers are the same things are all connected to the same things. You have a carbon connected to only a fluorine, a chlorine connected to the carbon, a hydrogen connected to the carbon, a bromine connected to the carbon. So all of the same things are connected to the same things, but they're three-dimensional configuration."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You have constitutional isomers that say, OK, different things are connected to different things. Stereoisomers are the same things are all connected to the same things. You have a carbon connected to only a fluorine, a chlorine connected to the carbon, a hydrogen connected to the carbon, a bromine connected to the carbon. So all of the same things are connected to the same things, but they're three-dimensional configuration. That's where we're dealing with the stereo part. Stereochemistry is the study of three-dimensional chemistry, essentially, understanding the actual three-dimensional structure of things. So stereoisomers mean that we have the same constituents, the same atoms."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So all of the same things are connected to the same things, but they're three-dimensional configuration. That's where we're dealing with the stereo part. Stereochemistry is the study of three-dimensional chemistry, essentially, understanding the actual three-dimensional structure of things. So stereoisomers mean that we have the same constituents, the same atoms. They have the same connections to each other. Bromine is still connected to carbon, which is still connected to hydrogen. That's all true over here."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So stereoisomers mean that we have the same constituents, the same atoms. They have the same connections to each other. Bromine is still connected to carbon, which is still connected to hydrogen. That's all true over here. But their three-dimensional orientation is still different. And in this case, where they are mirror images of each other, we call them enantiomers. And I should probably make one clarification."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's all true over here. But their three-dimensional orientation is still different. And in this case, where they are mirror images of each other, we call them enantiomers. And I should probably make one clarification. In the last few videos, I've been a little bit, you know, sometimes I'll say configuration, and sometimes I'll use the word confirmation. So sometimes I'll use the word configuration, and sometimes I use the word confirmation. And I actually should be a little bit more, or I should have been, a little bit more exact about these."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I should probably make one clarification. In the last few videos, I've been a little bit, you know, sometimes I'll say configuration, and sometimes I'll use the word confirmation. So sometimes I'll use the word configuration, and sometimes I use the word confirmation. And I actually should be a little bit more, or I should have been, a little bit more exact about these. When you're talking about a configuration, you're actually talking about a different structure. To go from one configuration to another configuration, you would actually have to break bonds and kind of reassemble them. So these are different configurations."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And I actually should be a little bit more, or I should have been, a little bit more exact about these. When you're talking about a configuration, you're actually talking about a different structure. To go from one configuration to another configuration, you would actually have to break bonds and kind of reassemble them. So these are different configurations. Because in order for them to be able to be the same thing, you would have to swap maybe the bromine and the hydrogen where they are relative to the carbon. So these are different configurations. Confirmations are really just different shapes or different orientations of the same molecule."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these are different configurations. Because in order for them to be able to be the same thing, you would have to swap maybe the bromine and the hydrogen where they are relative to the carbon. So these are different configurations. Confirmations are really just different shapes or different orientations of the same molecule. So when we talked about cyclohexane being in a boat, so this is cyclohexane being in a boat confirmation, or this is the cyclohexane being in a chair confirmation. It's the exact same molecule with the exact same connections. We didn't detach any bonds or reattach any bonds."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Confirmations are really just different shapes or different orientations of the same molecule. So when we talked about cyclohexane being in a boat, so this is cyclohexane being in a boat confirmation, or this is the cyclohexane being in a chair confirmation. It's the exact same molecule with the exact same connections. We didn't detach any bonds or reattach any bonds. They just flipped around a little bit. So these are two different confirmations. These are two different configurations."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We didn't detach any bonds or reattach any bonds. They just flipped around a little bit. So these are two different confirmations. These are two different configurations. To go from one configuration to another, you have to rearrange bonds. Now let's look at this molecule over here. Can we identify any stereocenters or chiral carbons or chiral atoms?"}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "These are two different configurations. To go from one configuration to another, you have to rearrange bonds. Now let's look at this molecule over here. Can we identify any stereocenters or chiral carbons or chiral atoms? And you have this carbon right here. It's bonded to a chlorine, a hydrogen, a bromine. And then another carbon."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Can we identify any stereocenters or chiral carbons or chiral atoms? And you have this carbon right here. It's bonded to a chlorine, a hydrogen, a bromine. And then another carbon. So this is bonded to four different things. So this is a chiral carbon. And sometimes they put little asterisks there."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then another carbon. So this is bonded to four different things. So this is a chiral carbon. And sometimes they put little asterisks there. But if we look at this carbon right here, well, it's bonded to a fluorine, another carbon. But it's bonded to two hydrogens. So it's not chiral."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And sometimes they put little asterisks there. But if we look at this carbon right here, well, it's bonded to a fluorine, another carbon. But it's bonded to two hydrogens. So it's not chiral. It has two of the same things that it's bonded to. And you can even see a little axis of symmetry through it. If you look at that, you can kind of flip it over, and it's going to be the same thing."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's not chiral. It has two of the same things that it's bonded to. And you can even see a little axis of symmetry through it. If you look at that, you can kind of flip it over, and it's going to be the same thing. But this one right here, that is a chiral center or chiral carbon or chiral atom or asymmetric carbon. You'll see it used in different ways. And because this molecule has got that chiral center, you'll see that if you were to take its mirror image, it would be an enantiomer."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you look at that, you can kind of flip it over, and it's going to be the same thing. But this one right here, that is a chiral center or chiral carbon or chiral atom or asymmetric carbon. You'll see it used in different ways. And because this molecule has got that chiral center, you'll see that if you were to take its mirror image, it would be an enantiomer. This is not superimposable on its mirror image. We could even try to draw it. And just to show you, you don't always have to do the mirror image on the right side."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And because this molecule has got that chiral center, you'll see that if you were to take its mirror image, it would be an enantiomer. This is not superimposable on its mirror image. We could even try to draw it. And just to show you, you don't always have to do the mirror image on the right side. We could draw the mirror image on the left side. So if we were to draw its mirror image, it would look like this. You would have a fluorine carbon carbon chlorine."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And just to show you, you don't always have to do the mirror image on the right side. We could draw the mirror image on the left side. So if we were to draw its mirror image, it would look like this. You would have a fluorine carbon carbon chlorine. You have your two hydrogens. And then you would have a hydrogen here. And then you would have your bromine here."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You would have a fluorine carbon carbon chlorine. You have your two hydrogens. And then you would have a hydrogen here. And then you would have your bromine here. No matter what you try to do, if you try to flip this around or whatever, you will never be able to superimpose this on top of this. So these are enantiomers. These are both stereoisomers, really, relative to each other."}, {"video_title": "Chiral examples 1 Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then you would have your bromine here. No matter what you try to do, if you try to flip this around or whatever, you will never be able to superimpose this on top of this. So these are enantiomers. These are both stereoisomers, really, relative to each other. And either of these, regardless of which one you pick, are chiral molecules. So these are chiral molecules. I'm over the time that I normally want to go into video."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So what I want to do in this video is just familiarize you with some of these molecules and how to name them so you know what you're looking at. Or I guess so you can name what you're looking at. So you've seen this multiple times. Benzene just looks like this. It's a six-carbon ring with three double bonds like that. And it's not its only configuration. Sometimes it's shown to be in resonance with this form right here."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Benzene just looks like this. It's a six-carbon ring with three double bonds like that. And it's not its only configuration. Sometimes it's shown to be in resonance with this form right here. So the double bonds all flip around the circle like that. Or sometimes it's simply drawn like this. And I talked about this in the last video."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes it's shown to be in resonance with this form right here. So the double bonds all flip around the circle like that. Or sometimes it's simply drawn like this. And I talked about this in the last video. Just to show that really neither of these configurations is exactly right, that these pi electrons are just moving around the entire ring. So sometimes you'll just have the hexagon with a circle on the inside to show that the pi electrons are just floating around the entire ring. Now, if I were to add something to the benzene ring, it's pretty straightforward to name it."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I talked about this in the last video. Just to show that really neither of these configurations is exactly right, that these pi electrons are just moving around the entire ring. So sometimes you'll just have the hexagon with a circle on the inside to show that the pi electrons are just floating around the entire ring. Now, if I were to add something to the benzene ring, it's pretty straightforward to name it. So say I have this molecule right here. So let's say I have that. That's benzene."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, if I were to add something to the benzene ring, it's pretty straightforward to name it. So say I have this molecule right here. So let's say I have that. That's benzene. But let's say this carbon over here, it has one, two, three bonds. If I didn't draw anything else, you just assume that there's also a hydrogen here. But maybe there's no hydrogen there."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That's benzene. But let's say this carbon over here, it has one, two, three bonds. If I didn't draw anything else, you just assume that there's also a hydrogen here. But maybe there's no hydrogen there. Maybe, let me do this in a different color, maybe you have a chlorine there. Well, this is just chlorobenzene. If that was a bromine, it would be bromobenzene."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But maybe there's no hydrogen there. Maybe, let me do this in a different color, maybe you have a chlorine there. Well, this is just chlorobenzene. If that was a bromine, it would be bromobenzene. So pretty straightforward. If, let me change it a little bit. Let's say you had a chloro there, and let's say you had a bromo over here."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If that was a bromine, it would be bromobenzene. So pretty straightforward. If, let me change it a little bit. Let's say you had a chloro there, and let's say you had a bromo over here. Let's say you have a bromo over there. Draw the bromo. So you could just start numbering."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's say you had a chloro there, and let's say you had a bromo over here. Let's say you have a bromo over there. Draw the bromo. So you could just start numbering. You could actually start numbering in either place. Let's start with the bromine. It's alphabetically in front."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you could just start numbering. You could actually start numbering in either place. Let's start with the bromine. It's alphabetically in front. So this would be 1-bromo-2-chlorobenzene. Pretty straightforward. Now it gets a little bit more involved."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's alphabetically in front. So this would be 1-bromo-2-chlorobenzene. Pretty straightforward. Now it gets a little bit more involved. The IUPAC, which we know is kind of the group that named most things, everything else we've seen so far, they came up with a separate naming mechanism from the common names, but benzene, it's so ingrained in the organic chemistry community that all of the benzene or benzene derivative molecules, they just kind of said, well, you know what, we're just going to use the names that everyone uses. So here's a couple of common benzene derivatives. So if you have this molecule right here, I'll actually draw the pi electrons as a circle."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now it gets a little bit more involved. The IUPAC, which we know is kind of the group that named most things, everything else we've seen so far, they came up with a separate naming mechanism from the common names, but benzene, it's so ingrained in the organic chemistry community that all of the benzene or benzene derivative molecules, they just kind of said, well, you know what, we're just going to use the names that everyone uses. So here's a couple of common benzene derivatives. So if you have this molecule right here, I'll actually draw the pi electrons as a circle. Actually, I'll draw it with the double bonds. So let's say you have this molecule right here, and over here you have an OH right there. This is called a phenol."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if you have this molecule right here, I'll actually draw the pi electrons as a circle. Actually, I'll draw it with the double bonds. So let's say you have this molecule right here, and over here you have an OH right there. This is called a phenol. So it's not called benzene anymore. And if you had a molecule that looked like this, so just like a phenol, so essentially it is a phenol, but let's say you have a bromine right there. What you do is you start numbering at the group that's making this a phenol."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This is called a phenol. So it's not called benzene anymore. And if you had a molecule that looked like this, so just like a phenol, so essentially it is a phenol, but let's say you have a bromine right there. What you do is you start numbering at the group that's making this a phenol. So you start numbering there, 1, and then you get to the 2. So this is 2-bromophenol. And unfortunately, this is one of those things that you just kind of have to memorize, that a phenol is really just a benzene ring with an OH group."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "What you do is you start numbering at the group that's making this a phenol. So you start numbering there, 1, and then you get to the 2. So this is 2-bromophenol. And unfortunately, this is one of those things that you just kind of have to memorize, that a phenol is really just a benzene ring with an OH group. Another one that is probably a good idea to memorize, I mean, at least for me in organic chemistry, the most confusing thing is when someone says the name of a molecule that sounds like they expect you to understand, but you don't understand it. So it's, I guess, a good idea to understand as many names as possible. So if you just have a benzene ring and you just have a methyl group attached to that benzene ring, this is called toluene."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And unfortunately, this is one of those things that you just kind of have to memorize, that a phenol is really just a benzene ring with an OH group. Another one that is probably a good idea to memorize, I mean, at least for me in organic chemistry, the most confusing thing is when someone says the name of a molecule that sounds like they expect you to understand, but you don't understand it. So it's, I guess, a good idea to understand as many names as possible. So if you just have a benzene ring and you just have a methyl group attached to that benzene ring, this is called toluene. And once again, if you had a fluorine right over there, this would be 1, 2, 3, fluoro-toluene. I always have trouble saying that. Because 1, 2, 3."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if you just have a benzene ring and you just have a methyl group attached to that benzene ring, this is called toluene. And once again, if you had a fluorine right over there, this would be 1, 2, 3, fluoro-toluene. I always have trouble saying that. Because 1, 2, 3. You start at the group that's kind of making this the toluene. And so let me make it a little bit clearer. So this would be 3-fluorotoluene."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Because 1, 2, 3. You start at the group that's kind of making this the toluene. And so let me make it a little bit clearer. So this would be 3-fluorotoluene. This would all be in one word. I just wrote it up here to save some space. Now, a couple of other ones that you'll see."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this would be 3-fluorotoluene. This would all be in one word. I just wrote it up here to save some space. Now, a couple of other ones that you'll see. And once again, as you know, if you watched many, many Khan Academy videos, I hate memorizing things. But these are just names. And these aren't systematically derived."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, a couple of other ones that you'll see. And once again, as you know, if you watched many, many Khan Academy videos, I hate memorizing things. But these are just names. And these aren't systematically derived. The chemistry community just uses them because that's the way they've heard them. So it is one of those things you kind of have to memorize. So if you have a benzene ring where the functional group that's kind of defining the benzene ring is essentially becoming an amine."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And these aren't systematically derived. The chemistry community just uses them because that's the way they've heard them. So it is one of those things you kind of have to memorize. So if you have a benzene ring where the functional group that's kind of defining the benzene ring is essentially becoming an amine. We haven't actually covered videos on amines yet. And I'll do that in a future video. But it's replaced one of the hydrogens in ammonia."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if you have a benzene ring where the functional group that's kind of defining the benzene ring is essentially becoming an amine. We haven't actually covered videos on amines yet. And I'll do that in a future video. But it's replaced one of the hydrogens in ammonia. Ammonia has three hydrogens and one lone pair. Here we've replaced one of the hydrogens. Actually, there's two hydrogens here."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But it's replaced one of the hydrogens in ammonia. Ammonia has three hydrogens and one lone pair. Here we've replaced one of the hydrogens. Actually, there's two hydrogens here. We've replaced one of the hydrogens with a benzene ring. This type of thing, you call it aniline. They're in the same idea."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Actually, there's two hydrogens here. We've replaced one of the hydrogens with a benzene ring. This type of thing, you call it aniline. They're in the same idea. If I put a fluoro here, this would be 2-fluoroaniline. So you use whatever the base molecule is. And then you just name it really the way we've named a lot of things before."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "They're in the same idea. If I put a fluoro here, this would be 2-fluoroaniline. So you use whatever the base molecule is. And then you just name it really the way we've named a lot of things before. And I'll just introduce you to two more just so that you have them under your belt so that you will have seen them before if anyone introduces them to you. And they're really introducing groups that we'll study in more detail in the future. But if you have a benzene ring, once again, and then you have a carboxyl group."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then you just name it really the way we've named a lot of things before. And I'll just introduce you to two more just so that you have them under your belt so that you will have seen them before if anyone introduces them to you. And they're really introducing groups that we'll study in more detail in the future. But if you have a benzene ring, once again, and then you have a carboxyl group. And we'll talk more about carboxyl groups in the future. But it creates carboxylic acid. This thing right here, this hydrogen, can actually be released quite easily."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But if you have a benzene ring, once again, and then you have a carboxyl group. And we'll talk more about carboxyl groups in the future. But it creates carboxylic acid. This thing right here, this hydrogen, can actually be released quite easily. And we'll talk about that in the future. This right here is called benzoic acid. And this name, actually, there is a little bit more logic to this, you have the benz part for the benzene."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This thing right here, this hydrogen, can actually be released quite easily. And we'll talk about that in the future. This right here is called benzoic acid. And this name, actually, there is a little bit more logic to this, you have the benz part for the benzene. So let me make this in different colors. The benzene part, that gives us the benz right there. And then in general, whenever you have a carboxyl group like this, it becomes a carboxylic acid."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this name, actually, there is a little bit more logic to this, you have the benz part for the benzene. So let me make this in different colors. The benzene part, that gives us the benz right there. And then in general, whenever you have a carboxyl group like this, it becomes a carboxylic acid. And you tend to add the oic acid to the end. So this actually has some logical naming to it. Now the last one I'll introduce you to is very similar to this."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then in general, whenever you have a carboxyl group like this, it becomes a carboxylic acid. And you tend to add the oic acid to the end. So this actually has some logical naming to it. Now the last one I'll introduce you to is very similar to this. Instead of having an OH right here, we just have a hydrogen. And it would be an aldehyde. So let me draw that."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now the last one I'll introduce you to is very similar to this. Instead of having an OH right here, we just have a hydrogen. And it would be an aldehyde. So let me draw that. And you could almost imagine what that's going to be called. It's benzaldehyde. So let me draw it."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw that. And you could almost imagine what that's going to be called. It's benzaldehyde. So let me draw it. So you have your benzene ring. And then instead of this carboxyl group, you have an aldehyde group. You have your carbon."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw it. So you have your benzene ring. And then instead of this carboxyl group, you have an aldehyde group. You have your carbon. There's implicitly a carbon here. Let me make that just in case it's the first time you're seeing it. You have a carbon there."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You have your carbon. There's implicitly a carbon here. Let me make that just in case it's the first time you're seeing it. You have a carbon there. Instead of an OH group, you have just a hydrogen. And so once again, you have a benzene. So it actually makes sense to put the benz over there."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "You have a carbon there. Instead of an OH group, you have just a hydrogen. And so once again, you have a benzene. So it actually makes sense to put the benz over there. And then you have the aldehyde group. And then you have an aldehyde group right over here. So this is called a benzaldehyde."}, {"video_title": "Naming benzene derivatives introduction Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it actually makes sense to put the benz over there. And then you have the aldehyde group. And then you have an aldehyde group right over here. So this is called a benzaldehyde. Or this molecule is called benzaldehyde. And we'll study amines and aldehydes and carboxylic acids in much more detail in future videos and actually see reactions that involve them. But I just wanted to expose you to this."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "There was nothing that would imply that this is more stable than that, or vice versa. But what we want to do in this video is address what happens if, instead of just a pure cyclohexane, what if we added a methyl group to it? And so instead of just a cyclohexane, let's think about a methyl cyclohexane. And so we know what to do when we're naming things. So a methyl cyclohexane will look like that. It'll look like we'll have a hexagon here. This is the way we've been drawing it historically."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And so we know what to do when we're naming things. So a methyl cyclohexane will look like that. It'll look like we'll have a hexagon here. This is the way we've been drawing it historically. We've been drawing it just like that. And then it'll have a methyl group. And you could literally just draw a methyl group like this, or maybe like this."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This is the way we've been drawing it historically. We've been drawing it just like that. And then it'll have a methyl group. And you could literally just draw a methyl group like this, or maybe like this. And by implication, you have a carbon here, essentially a CH3 group here, and of course you have another hydrogen bonded here. Now if you were to draw this methyl group in kind of three dimensions like this, it would look just like our different chair positions before. But instead of this H being just an H, we could make this entire methyl group."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And you could literally just draw a methyl group like this, or maybe like this. And by implication, you have a carbon here, essentially a CH3 group here, and of course you have another hydrogen bonded here. Now if you were to draw this methyl group in kind of three dimensions like this, it would look just like our different chair positions before. But instead of this H being just an H, we could make this entire methyl group. So we could turn this CH3 right here. So this, let me draw it out in a different color. We could make this carbon that implicitly has three hydrogens on it."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "But instead of this H being just an H, we could make this entire methyl group. So we could turn this CH3 right here. So this, let me draw it out in a different color. We could make this carbon that implicitly has three hydrogens on it. So this is a CH3 group. That is a methyl group that's right there. We could turn one of these, we could substitute one of these hydrogens with it, and then this would be one of the shapes of this methyl cyclohexane."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We could make this carbon that implicitly has three hydrogens on it. So this is a CH3 group. That is a methyl group that's right there. We could turn one of these, we could substitute one of these hydrogens with it, and then this would be one of the shapes of this methyl cyclohexane. Let me write it down just to practice our naming. So this is methyl cyclohexane. Now, this is one of their chair positions."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We could turn one of these, we could substitute one of these hydrogens with it, and then this would be one of the shapes of this methyl cyclohexane. Let me write it down just to practice our naming. So this is methyl cyclohexane. Now, this is one of their chair positions. And then if this were to flip down and the other side were to flip up, the other chair position would take this methyl group from being in an axial position, and it would put it in an equatorial position. So this is the same group right here. Let me put a circle around it."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now, this is one of their chair positions. And then if this were to flip down and the other side were to flip up, the other chair position would take this methyl group from being in an axial position, and it would put it in an equatorial position. So this is the same group right here. Let me put a circle around it. It's a whole group. That's the CH3. Now, the question we want to answer in this video, is one of these two configurations going to be more stable?"}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let me put a circle around it. It's a whole group. That's the CH3. Now, the question we want to answer in this video, is one of these two configurations going to be more stable? Now, you might just be able to eyeball it, looking at this diagram, and say, hey, when we're in the position where the methyl group is in the axial position, it's going to be closer to all of these other carbons over here, closer to the electron clouds. Maybe it has higher potential energy. Maybe it'll want to spring away."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now, the question we want to answer in this video, is one of these two configurations going to be more stable? Now, you might just be able to eyeball it, looking at this diagram, and say, hey, when we're in the position where the methyl group is in the axial position, it's going to be closer to all of these other carbons over here, closer to the electron clouds. Maybe it has higher potential energy. Maybe it'll want to spring away. And if that was what you're guessing or what you're kind of eyeballing, you'd be right. Because in this position, this methyl is much further away from all of the stuff out here. So there'll be less, I guess you can consider it, electron cloud crowding, if you will."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Maybe it'll want to spring away. And if that was what you're guessing or what you're kind of eyeballing, you'd be right. Because in this position, this methyl is much further away from all of the stuff out here. So there'll be less, I guess you can consider it, electron cloud crowding, if you will. Now, to see that a little bit more clear, I want to do what we call a double Newman diagram. And really, that's the whole motivation of this video, to kind of expose you that a Newman diagram isn't just useful for simple things like a butane or an ethane. You can actually do it with cyclical rings."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So there'll be less, I guess you can consider it, electron cloud crowding, if you will. Now, to see that a little bit more clear, I want to do what we call a double Newman diagram. And really, that's the whole motivation of this video, to kind of expose you that a Newman diagram isn't just useful for simple things like a butane or an ethane. You can actually do it with cyclical rings. And to do the Newman diagram, let me number these carbons. So we could number them like this. So this is carbon 1, 2, 3, 4, 5, 6."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "You can actually do it with cyclical rings. And to do the Newman diagram, let me number these carbons. So we could number them like this. So this is carbon 1, 2, 3, 4, 5, 6. And you don't have to call this one methylcyclohexane, because whenever you have only one group attached to the ring, you implicitly start the numbering at the carbon that the group is attached. So that is the one carbon over here. This is the one carbon, 2, 3, 4, 5, 6."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this is carbon 1, 2, 3, 4, 5, 6. And you don't have to call this one methylcyclohexane, because whenever you have only one group attached to the ring, you implicitly start the numbering at the carbon that the group is attached. So that is the one carbon over here. This is the one carbon, 2, 3, 4, 5, 6. And over here, once again, is the one carbon, 2, 3, 4, 5, 6. Now, what I want to do is draw two Newman projections. And both of them will involve, well, actually I'll draw four, but you'll see what I'm talking about in a second."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This is the one carbon, 2, 3, 4, 5, 6. And over here, once again, is the one carbon, 2, 3, 4, 5, 6. Now, what I want to do is draw two Newman projections. And both of them will involve, well, actually I'll draw four, but you'll see what I'm talking about in a second. So the first Newman projection, I'm going to start at that carbon right over there. So let's think about what that would look like. So we're at the front, we're staring straight onto that carbon, so this carbon's in the front, this carbon over here, this carbon, carbon number 2, is in the back."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And both of them will involve, well, actually I'll draw four, but you'll see what I'm talking about in a second. So the first Newman projection, I'm going to start at that carbon right over there. So let's think about what that would look like. So we're at the front, we're staring straight onto that carbon, so this carbon's in the front, this carbon over here, this carbon, carbon number 2, is in the back. So let me label this. This is carbon number 1 that we're looking head on. Now, right in the axial position, you have that methyl group."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we're at the front, we're staring straight onto that carbon, so this carbon's in the front, this carbon over here, this carbon, carbon number 2, is in the back. So let me label this. This is carbon number 1 that we're looking head on. Now, right in the axial position, you have that methyl group. So let me draw that. So in the axial position, you have that methyl group right over there. And then in the equatorial position right here, you have this hydrogen."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now, right in the axial position, you have that methyl group. So let me draw that. So in the axial position, you have that methyl group right over there. And then in the equatorial position right here, you have this hydrogen. So let me draw that. So you have this hydrogen. And then over here, this bond right over here, this gets us to another CH2 group."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then in the equatorial position right here, you have this hydrogen. So let me draw that. So you have this hydrogen. And then over here, this bond right over here, this gets us to another CH2 group. So let me draw that like this. So that gets us to another CH2. That's a CH2 right there, right?"}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then over here, this bond right over here, this gets us to another CH2 group. So let me draw that like this. So that gets us to another CH2. That's a CH2 right there, right? Or maybe I should circle it like this. This whole thing is a CH2. So it's bonded to the carbon, but the carbon has two hydrogens on it."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "That's a CH2 right there, right? Or maybe I should circle it like this. This whole thing is a CH2. So it's bonded to the carbon, but the carbon has two hydrogens on it. And this is actually carbon number 6. So this is the front. Now, if we're staring straight and carbon number 2 is right behind it, let me draw carbon number 2."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So it's bonded to the carbon, but the carbon has two hydrogens on it. And this is actually carbon number 6. So this is the front. Now, if we're staring straight and carbon number 2 is right behind it, let me draw carbon number 2. I'll do it in this blue color. So carbon number 2, that bigger circle. Now, carbon number 2, what's going on over there?"}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now, if we're staring straight and carbon number 2 is right behind it, let me draw carbon number 2. I'll do it in this blue color. So carbon number 2, that bigger circle. Now, carbon number 2, what's going on over there? It has a hydrogen in the axial position going straight down. This hydrogen right over there. And then it has another hydrogen going equatorial."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Now, carbon number 2, what's going on over there? It has a hydrogen in the axial position going straight down. This hydrogen right over there. And then it has another hydrogen going equatorial. So this green hydrogen right here is going equatorial. It looks like that. And then it connects to carbon number 3, which is another CH2."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then it has another hydrogen going equatorial. So this green hydrogen right here is going equatorial. It looks like that. And then it connects to carbon number 3, which is another CH2. It has two hydrogens branching off of it. So this right here. So in front we have carbon number 1."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then it connects to carbon number 3, which is another CH2. It has two hydrogens branching off of it. So this right here. So in front we have carbon number 1. In the back we have carbon number 2. Let me color code it a little better. Carbon number 2."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So in front we have carbon number 1. In the back we have carbon number 2. Let me color code it a little better. Carbon number 2. And then carbon number 2 branches off to, so if you think of this branch right here, if you think of that branch right here, that's carbon number 2 branching off to carbon number 3, or CH3 right there. Let me do that in a new color. So this is a, actually a CH2 group."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Carbon number 2. And then carbon number 2 branches off to, so if you think of this branch right here, if you think of that branch right here, that's carbon number 2 branching off to carbon number 3, or CH3 right there. Let me do that in a new color. So this is a, actually a CH2 group. That's a CH2. This is a carbon. It has two hydrogens."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this is a, actually a CH2 group. That's a CH2. This is a carbon. It has two hydrogens. So this is a CH2. This is carbon number 3 right there. And then that goes, and actually I'll pause there."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It has two hydrogens. So this is a CH2. This is carbon number 3 right there. And then that goes, and actually I'll pause there. What I'll do now is I'll draw another axial, another Newman projection, but for this Newman projection, we'll be looking straight on our carbon number 5 right here. And we're going to see, we're going to form a ring. Because the first Newman projection we just did essentially covers this bond."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then that goes, and actually I'll pause there. What I'll do now is I'll draw another axial, another Newman projection, but for this Newman projection, we'll be looking straight on our carbon number 5 right here. And we're going to see, we're going to form a ring. Because the first Newman projection we just did essentially covers this bond. This bond is sitting straight into the screen the way I did it right now. Carbon number 2 is directly behind carbon number 1. So I guess the opposite side of the ring is 5 to 4."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Because the first Newman projection we just did essentially covers this bond. This bond is sitting straight into the screen the way I did it right now. Carbon number 2 is directly behind carbon number 1. So I guess the opposite side of the ring is 5 to 4. If you look at here, 1 to 2 is there, and then 5 to 4 is just like that. So you can imagine when we're doing the Newman projections, we're looking straight on here on the left Newman projection. We're going to look straight in that direction on the right Newman projection."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So I guess the opposite side of the ring is 5 to 4. If you look at here, 1 to 2 is there, and then 5 to 4 is just like that. So you can imagine when we're doing the Newman projections, we're looking straight on here on the left Newman projection. We're going to look straight in that direction on the right Newman projection. So if we have carbon number 5 in front, what are its bonds going to look like? Well, it's going to bond to carbon number 6 right over here, so we can make this right here, this bond right here, is this bond right here, to that same CH2 that our first Newman projection bonded to. And then he's going to have 2 hydrogens."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We're going to look straight in that direction on the right Newman projection. So if we have carbon number 5 in front, what are its bonds going to look like? Well, it's going to bond to carbon number 6 right over here, so we can make this right here, this bond right here, is this bond right here, to that same CH2 that our first Newman projection bonded to. And then he's going to have 2 hydrogens. I haven't drawn them here. Let me draw them just so you can see them. Going to have 2 hydrogens, one in an axial position and one in an equatorial position."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then he's going to have 2 hydrogens. I haven't drawn them here. Let me draw them just so you can see them. Going to have 2 hydrogens, one in an axial position and one in an equatorial position. It's hard to see now, but he's going to have 1 hydrogen in axial and 1 in equatorial position. And then in the back, we're going to draw carbon number 4. And carbon number 4 I will do in this green color."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Going to have 2 hydrogens, one in an axial position and one in an equatorial position. It's hard to see now, but he's going to have 1 hydrogen in axial and 1 in equatorial position. And then in the back, we're going to draw carbon number 4. And carbon number 4 I will do in this green color. So carbon number 4, or actually, I'll do carbon number 4 in that green color. And this is really just an exercise in visualization. That's why I wanted to do it with you."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And carbon number 4 I will do in this green color. So carbon number 4, or actually, I'll do carbon number 4 in that green color. And this is really just an exercise in visualization. That's why I wanted to do it with you. So carbon number 4 has an axial hydrogen, so it has a hydrogen pointing straight down. That hydrogen is that hydrogen. It has an equatorial hydrogen going out like that."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "That's why I wanted to do it with you. So carbon number 4 has an axial hydrogen, so it has a hydrogen pointing straight down. That hydrogen is that hydrogen. It has an equatorial hydrogen going out like that. And then it bonds to carbon number 3. So this bond right here is this bond, just like that. So what do we see immediately when we draw this chair position, when our methyl group is in the axial position, what do we see?"}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It has an equatorial hydrogen going out like that. And then it bonds to carbon number 3. So this bond right here is this bond, just like that. So what do we see immediately when we draw this chair position, when our methyl group is in the axial position, what do we see? Methyl in axial position. We see that it's gauche, or gauche, I don't know the best way to pronounce it. It's only 60 degrees away from carbon number 3."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So what do we see immediately when we draw this chair position, when our methyl group is in the axial position, what do we see? Methyl in axial position. We see that it's gauche, or gauche, I don't know the best way to pronounce it. It's only 60 degrees away from carbon number 3. This is only 60 degrees, or our dihedral angle, when you use the Newman projection, this is only a 60 degree angle. It is gauche to carbon number 3. So maybe they're crowding each other a little bit."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It's only 60 degrees away from carbon number 3. This is only 60 degrees, or our dihedral angle, when you use the Newman projection, this is only a 60 degree angle. It is gauche to carbon number 3. So maybe they're crowding each other a little bit. Let's compare it to the situation where our methyl group is equatorial, where the carbon that it's attached to is on the down part of the chair. Let's see what that Newman projection looks like. So same thing."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So maybe they're crowding each other a little bit. Let's compare it to the situation where our methyl group is equatorial, where the carbon that it's attached to is on the down part of the chair. Let's see what that Newman projection looks like. So same thing. Let me scroll over to the right a little bit. So this is this configuration, and it's in equilibrium with this configuration right here. We'll do the exact same exercise."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So same thing. Let me scroll over to the right a little bit. So this is this configuration, and it's in equilibrium with this configuration right here. We'll do the exact same exercise. Carbon number 1. But now carbon number 1, in carbon number 1 right here, the hydrogen is now axial, and it's pointing straight down. Let me draw that."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "We'll do the exact same exercise. Carbon number 1. But now carbon number 1, in carbon number 1 right here, the hydrogen is now axial, and it's pointing straight down. Let me draw that. So we have a hydrogen pointing straight down now. So hydrogen's pointing down, and now the CH3 is in an equatorial position, which you can see more clearly on this than over here. So you have a CH3."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw that. So we have a hydrogen pointing straight down now. So hydrogen's pointing down, and now the CH3 is in an equatorial position, which you can see more clearly on this than over here. So you have a CH3. The methyl group is right there. And now this bond to carbon number 6 will look like this. So you have a CH2."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So you have a CH3. The methyl group is right there. And now this bond to carbon number 6 will look like this. So you have a CH2. This is number 6 right over there. And now if we were to go to back, if we were to go to carbon number 2 in back, which we had done in the blue color before, so I'll do it in the blue color again. Carbon number 2 in black."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So you have a CH2. This is number 6 right over there. And now if we were to go to back, if we were to go to carbon number 2 in back, which we had done in the blue color before, so I'll do it in the blue color again. Carbon number 2 in black. It has a hydrogen in the axial position going straight up. That's that hydrogen right there. It has a hydrogen going straight up."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Carbon number 2 in black. It has a hydrogen in the axial position going straight up. That's that hydrogen right there. It has a hydrogen going straight up. It has another hydrogen over here. And then it bonds to carbon number 3 in the back, so CH2 over here. So this guy's the same thing as this guy, but now we've flipped configurations."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It has a hydrogen going straight up. It has another hydrogen over here. And then it bonds to carbon number 3 in the back, so CH2 over here. So this guy's the same thing as this guy, but now we've flipped configurations. So he bonds to that guy in the back. And now we do the Newman diagram looking straight on to carbon number 5. So we're going to look straight in this direction for this Newman projection."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this guy's the same thing as this guy, but now we've flipped configurations. So he bonds to that guy in the back. And now we do the Newman diagram looking straight on to carbon number 5. So we're going to look straight in this direction for this Newman projection. Now we're going to look straight in this direction for our other Newman projection. So carbon number 5, if we draw it in the front, I haven't drawn it here, but it has a hydrogen in the axial position and then it has a hydrogen in the equatorial position. So carbon number 5, if we look at this, and this is really, if you're getting a little stressed out about this because it's a little hard to understand, you might want to re-watch."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to look straight in this direction for this Newman projection. Now we're going to look straight in this direction for our other Newman projection. So carbon number 5, if we draw it in the front, I haven't drawn it here, but it has a hydrogen in the axial position and then it has a hydrogen in the equatorial position. So carbon number 5, if we look at this, and this is really, if you're getting a little stressed out about this because it's a little hard to understand, you might want to re-watch. This is really just an example of visualization. So I really hope this isn't confusing you. If you find this really daunting, this isn't going to really trip you up in the rest of organic chemistry."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So carbon number 5, if we look at this, and this is really, if you're getting a little stressed out about this because it's a little hard to understand, you might want to re-watch. This is really just an example of visualization. So I really hope this isn't confusing you. If you find this really daunting, this isn't going to really trip you up in the rest of organic chemistry. But if you can get it, it's even better. It'll be that much better at visualizing some of these molecules. So if we look straight on carbon number 5, we have a hydrogen in the axial position coming straight down."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "If you find this really daunting, this isn't going to really trip you up in the rest of organic chemistry. But if you can get it, it's even better. It'll be that much better at visualizing some of these molecules. So if we look straight on carbon number 5, we have a hydrogen in the axial position coming straight down. It is bonded, carbon number 5 is bonded to carbon number 6, which is right over there. So let me make it very clear. So this bond right here."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So if we look straight on carbon number 5, we have a hydrogen in the axial position coming straight down. It is bonded, carbon number 5 is bonded to carbon number 6, which is right over there. So let me make it very clear. So this bond right here. I want to do it in a different color. This bond right here is the same thing as this one. This is a really good way to, if you can do this, then your brain is pretty good at translating between Newman projections and these kind of seat diagrams that we have up here."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So this bond right here. I want to do it in a different color. This bond right here is the same thing as this one. This is a really good way to, if you can do this, then your brain is pretty good at translating between Newman projections and these kind of seat diagrams that we have up here. So this is this bond. This axial hydrogen is this axial hydrogen. And then we have another hydrogen."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This is a really good way to, if you can do this, then your brain is pretty good at translating between Newman projections and these kind of seat diagrams that we have up here. So this is this bond. This axial hydrogen is this axial hydrogen. And then we have another hydrogen. We have this hydrogen right here, which would be that like that. And then behind it, you have carbon number 4. Carbon number 4 is like that."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And then we have another hydrogen. We have this hydrogen right here, which would be that like that. And then behind it, you have carbon number 4. Carbon number 4 is like that. So carbon number 4. Draw a circle. It has a hydrogen in its axial position, another hydrogen like that."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Carbon number 4 is like that. So carbon number 4. Draw a circle. It has a hydrogen in its axial position, another hydrogen like that. And then it bonds to carbon number 3. So this was number 3 right here. So just like that."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It has a hydrogen in its axial position, another hydrogen like that. And then it bonds to carbon number 3. So this was number 3 right here. So just like that. So what do we see about this methyl group here? In this situation, this methyl group is anti- there's two ways to think about it. Its dihedral angle versus carbon number 3 is now 180 degrees."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So just like that. So what do we see about this methyl group here? In this situation, this methyl group is anti- there's two ways to think about it. Its dihedral angle versus carbon number 3 is now 180 degrees. Over here, it was gauch. It was 60 degrees. It was dihedral angle to carbon number 3 was 60 degrees."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "Its dihedral angle versus carbon number 3 is now 180 degrees. Over here, it was gauch. It was 60 degrees. It was dihedral angle to carbon number 3 was 60 degrees. Now it's 180 degrees. So it's much further. And now its dihedral angle to the carbon number 6 is also 120 degrees."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "It was dihedral angle to carbon number 3 was 60 degrees. Now it's 180 degrees. So it's much further. And now its dihedral angle to the carbon number 6 is also 120 degrees. So in this situation, where our methyl group is equatorial, it's not axial here, it's equatorial here when we jumped back down, because notice, the bond is parallel to parts of the ring. In this situation, we are farther away from the other methyl groups. There's less crowding."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "And now its dihedral angle to the carbon number 6 is also 120 degrees. So in this situation, where our methyl group is equatorial, it's not axial here, it's equatorial here when we jumped back down, because notice, the bond is parallel to parts of the ring. In this situation, we are farther away from the other methyl groups. There's less crowding. And so this is a more stable situation. So you could say it is an anti-configuration relative to carbon number 3. While over here, it was gauch to carbon number 3."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "There's less crowding. And so this is a more stable situation. So you could say it is an anti-configuration relative to carbon number 3. While over here, it was gauch to carbon number 3. I don't know if I'm saying it's gauche, gauch. I don't know the best way to pronounce it. So in this case, there's less crowding."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "While over here, it was gauch to carbon number 3. I don't know if I'm saying it's gauche, gauch. I don't know the best way to pronounce it. So in this case, there's less crowding. This is more stable, lower potential energy. More stable. So I hope you found that interesting."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So in this case, there's less crowding. This is more stable, lower potential energy. More stable. So I hope you found that interesting. This was really a way, just an exercise in being able to go from this visualization to kind of this double Newman diagram. And if it makes it any easier, the way you could think about it is, we're viewing in this Newman diagram right here, this carbon number 6 is this carbon number 6. This CH2 is this one back there."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "So I hope you found that interesting. This was really a way, just an exercise in being able to go from this visualization to kind of this double Newman diagram. And if it makes it any easier, the way you could think about it is, we're viewing in this Newman diagram right here, this carbon number 6 is this carbon number 6. This CH2 is this one back there. So when we're looking at it from this, we're kind of looking at this hexane ring from this direction. We see this in front. This is this."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This CH2 is this one back there. So when we're looking at it from this, we're kind of looking at this hexane ring from this direction. We see this in front. This is this. And we see that in back. That is that. Over here, same thing."}, {"video_title": "Double Newman diagram for methylcyclohexane Organic chemistry Khan Academy.mp3", "Sentence": "This is this. And we see that in back. That is that. Over here, same thing. We're looking at it directly from this direction. We see this guy on top is over here. And this guy on the bottom is over here."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So the first thing we want to do is identify the longest chain of carbons. So let's see, it could be one, two, three, four, five, six, seven, eight, or let's see, maybe it's one, two, three, four, five, six, seven, eight, nine, 10, yes, indeed, that's the longest chain. And if you go one, two, three, four, five, six, seven, or one, two, three, four, five, six, seven, so those aren't the longest, so the longest one is this one. One, two, three, four, five, six, seven, eight, nine, 10, 10 carbons. And this is going to be an alkane because it's all single bonds. And an alkane that's a chain of 10 carbons, we would use the prefix dec for 10. So this is a decane."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six, seven, eight, nine, 10, 10 carbons. And this is going to be an alkane because it's all single bonds. And an alkane that's a chain of 10 carbons, we would use the prefix dec for 10. So this is a decane. Let me write that right over here. This is a decane, decane. Now let's think about the groups that are attached to this decane."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So this is a decane. Let me write that right over here. This is a decane, decane. Now let's think about the groups that are attached to this decane. So we have this group right over here. This has two carbons in it, one carbon, two carbons. And so, because it has two carbons, we would use the prefix eth."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "Now let's think about the groups that are attached to this decane. So we have this group right over here. This has two carbons in it, one carbon, two carbons. And so, because it has two carbons, we would use the prefix eth. Remember, meth is one carbon, eth is two carbons. And since it's a group, we're not talking about the backbone, this is an ethyl group. This is an ethyl group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And so, because it has two carbons, we would use the prefix eth. Remember, meth is one carbon, eth is two carbons. And since it's a group, we're not talking about the backbone, this is an ethyl group. This is an ethyl group. And we have another ethyl group right over here, two carbons attached right over here. This is also an ethyl group. And now this group right over here is interesting."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "This is an ethyl group. And we have another ethyl group right over here, two carbons attached right over here. This is also an ethyl group. And now this group right over here is interesting. We can count the carbons in it. So it has one, two, three carbons. So you could think about, well, this has three carbons."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And now this group right over here is interesting. We can count the carbons in it. So it has one, two, three carbons. So you could think about, well, this has three carbons. Our prefix for three carbons is prop. So you could say, hey, maybe this is a propyl group. This right over here, this right over here, you could say, maybe this is a propyl group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about, well, this has three carbons. Our prefix for three carbons is prop. So you could say, hey, maybe this is a propyl group. This right over here, this right over here, you could say, maybe this is a propyl group. And you wouldn't be completely off base by saying that. But we have to be a little bit more careful when we name it. Because a propyl group, you would assume that you're attaching to one end of the propyl group, but we're not attaching to one end of the propyl group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "This right over here, this right over here, you could say, maybe this is a propyl group. And you wouldn't be completely off base by saying that. But we have to be a little bit more careful when we name it. Because a propyl group, you would assume that you're attaching to one end of the propyl group, but we're not attaching to one end of the propyl group. We're attaching, essentially, to the second carbon, to the middle carbon. And this is a secondary carbon. The reason why it's called a secondary carbon is because it's attached to two other carbons."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "Because a propyl group, you would assume that you're attaching to one end of the propyl group, but we're not attaching to one end of the propyl group. We're attaching, essentially, to the second carbon, to the middle carbon. And this is a secondary carbon. The reason why it's called a secondary carbon is because it's attached to two other carbons. If it was attached to three other carbons, it would be a tertiary carbon. If it was attached to only one carbon, it would be a primary carbon. So you could call this, since we're attached to the secondary carbon right over here, this is sometimes called a secpropyl group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "The reason why it's called a secondary carbon is because it's attached to two other carbons. If it was attached to three other carbons, it would be a tertiary carbon. If it was attached to only one carbon, it would be a primary carbon. So you could call this, since we're attached to the secondary carbon right over here, this is sometimes called a secpropyl group. And it's also sometimes called isopropyl, an isopropyl group. And you'll actually see isopropyl a little bit more frequently. And these would both be referred to as common names for this group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So you could call this, since we're attached to the secondary carbon right over here, this is sometimes called a secpropyl group. And it's also sometimes called isopropyl, an isopropyl group. And you'll actually see isopropyl a little bit more frequently. And these would both be referred to as common names for this group. Now, if you wanted to name this systematically, then you would do it very similar to the way that you would name the entire molecule. You would look for the longest chain here. And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And these would both be referred to as common names for this group. Now, if you wanted to name this systematically, then you would do it very similar to the way that you would name the entire molecule. You would look for the longest chain here. And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons. So that the backbone right over here is ethyl. It's an ethyl backbone here. And then you could view this carbon as a group attached to that ethyl backbone."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And the longest chain in this molecule, starting with where you are attached, is a chain of two carbons. So that the backbone right over here is ethyl. It's an ethyl backbone here. And then you could view this carbon as a group attached to that ethyl backbone. And we would start counting right where we are attached to the main chain. So this is the one carbon. This is the two carbon."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And then you could view this carbon as a group attached to that ethyl backbone. And we would start counting right where we are attached to the main chain. So this is the one carbon. This is the two carbon. So this right over here, this is just one carbon group. This right over here is an methyl group or a methyl group. This is a methyl group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "This is the two carbon. So this right over here, this is just one carbon group. This right over here is an methyl group or a methyl group. This is a methyl group. So you have a methyl group attached to the one carbon of an ethyl group. So the systematic name for this, and this is a little bit less typical for a group as small as a propyl group, but you could call this 1-methyl-ethyl is the systematic name for this. Now, the systematic name, you might say, hey, why go through the pain of doing this for something so simple that we could just call isopropyl?"}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "This is a methyl group. So you have a methyl group attached to the one carbon of an ethyl group. So the systematic name for this, and this is a little bit less typical for a group as small as a propyl group, but you could call this 1-methyl-ethyl is the systematic name for this. Now, the systematic name, you might say, hey, why go through the pain of doing this for something so simple that we could just call isopropyl? This is useful if this was a much larger or a much more complex group that was attached to this main chain. But more typically, and this is why it's called the common name, you'll see this thing right over here just called isopropyl. And sometimes you would see it called secpropyl or even s-propyl."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "Now, the systematic name, you might say, hey, why go through the pain of doing this for something so simple that we could just call isopropyl? This is useful if this was a much larger or a much more complex group that was attached to this main chain. But more typically, and this is why it's called the common name, you'll see this thing right over here just called isopropyl. And sometimes you would see it called secpropyl or even s-propyl. Now that we've named all of the groups, let's think about what carbons they are attached to and where we can start counting from. And the way that this is done is that you would start counting from the end of your carbon chain, your main, this kind of this decane backbone. And you would count from the end that bumps into the most groups faster."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And sometimes you would see it called secpropyl or even s-propyl. Now that we've named all of the groups, let's think about what carbons they are attached to and where we can start counting from. And the way that this is done is that you would start counting from the end of your carbon chain, your main, this kind of this decane backbone. And you would count from the end that bumps into the most groups faster. So for example, if you count from this end, this would be the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon. On the 5-carbon, we bump into two groups. If we started over here, this would be the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And you would count from the end that bumps into the most groups faster. So for example, if you count from this end, this would be the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon. On the 5-carbon, we bump into two groups. If we started over here, this would be the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon. On the 5-carbon, we do bump into a group, but only one group. And we'd have to wait until the 6-carbon to bump into two groups. So we get to the two groups faster, we would start counting on this end."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "If we started over here, this would be the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon. On the 5-carbon, we do bump into a group, but only one group. And we'd have to wait until the 6-carbon to bump into two groups. So we get to the two groups faster, we would start counting on this end. So this is the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon, 6-carbon, 7-carbon, 8, 9, and 10-carbon. And so when we think about which groups are we going to refer to first, do we refer to the ethyl groups first? Or do we refer to this isopropyl group first?"}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So we get to the two groups faster, we would start counting on this end. So this is the 1-carbon, 2-carbon, 3-carbon, 4-carbon, 5-carbon, 6-carbon, 7-carbon, 8, 9, and 10-carbon. And so when we think about which groups are we going to refer to first, do we refer to the ethyl groups first? Or do we refer to this isopropyl group first? We just think about what letter they start with in alphabetical order. So for example, these ethyl groups, they clearly start with an E. They clearly start with an E. E comes before in the alphabet, then the I in isopropyl, before the P in secpropyl. You would normally ignore the sec or the tert when you're thinking about alphabetical order."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "Or do we refer to this isopropyl group first? We just think about what letter they start with in alphabetical order. So for example, these ethyl groups, they clearly start with an E. They clearly start with an E. E comes before in the alphabet, then the I in isopropyl, before the P in secpropyl. You would normally ignore the sec or the tert when you're thinking about alphabetical order. And E also comes before the M in methyl ethyl. So we will talk about the ethyl groups first. So remember, our backbone, 10 carbons, all single bonds."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "You would normally ignore the sec or the tert when you're thinking about alphabetical order. And E also comes before the M in methyl ethyl. So we will talk about the ethyl groups first. So remember, our backbone, 10 carbons, all single bonds. It's an alkane. Since there's 10 of them, it is decane. And so let's talk about the ethyls first."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So remember, our backbone, 10 carbons, all single bonds. It's an alkane. Since there's 10 of them, it is decane. And so let's talk about the ethyls first. And since we have two ethyls, we can say that this is diethyl. So let me write that down. So we can write diethyl."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "And so let's talk about the ethyls first. And since we have two ethyls, we can say that this is diethyl. So let me write that down. So we can write diethyl. And of course, we need to specify where those two ethyls are. One is at the 5-carbon. One is at the 6-carbon."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So we can write diethyl. And of course, we need to specify where those two ethyls are. One is at the 5-carbon. One is at the 6-carbon. So we could call this 5, 6-diethyl. That refers to the 5 refers to this ethyl group. 6 refers to this ethyl group."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "One is at the 6-carbon. So we could call this 5, 6-diethyl. That refers to the 5 refers to this ethyl group. 6 refers to this ethyl group. And now we could talk about the isopropyl group. The isopropyl group is also on the 5-carbon. So we could say 5, 6-diethyl, 5-isopropyl decane."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "6 refers to this ethyl group. And now we could talk about the isopropyl group. The isopropyl group is also on the 5-carbon. So we could say 5, 6-diethyl, 5-isopropyl decane. Or if we wanted to use secpropyl instead of isopropyl, we could write secpropyl here. Or if we wanted to do the systematic naming, we could call this 5, 6-diethyl, 5. Instead of writing isopropyl here, we could write all of this, 1-methyl ethyl here."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "So we could say 5, 6-diethyl, 5-isopropyl decane. Or if we wanted to use secpropyl instead of isopropyl, we could write secpropyl here. Or if we wanted to do the systematic naming, we could call this 5, 6-diethyl, 5. Instead of writing isopropyl here, we could write all of this, 1-methyl ethyl here. So actually, let me just copy and paste that. Let me just copy this 1-methyl ethyl right over here. Copy and paste."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "Instead of writing isopropyl here, we could write all of this, 1-methyl ethyl here. So actually, let me just copy and paste that. Let me just copy this 1-methyl ethyl right over here. Copy and paste. Let me stick that there. 1-methyl ethyl. And of course, the main backbone is decane."}, {"video_title": "Alkane with isopropyl group Organic chemistry Khan Academy.mp3", "Sentence": "Copy and paste. Let me stick that there. 1-methyl ethyl. And of course, the main backbone is decane. So all of these are reasonable ways to name it. This would be the common way where we use isopropyl. Here we're doing it more systematically by calling that group 1-methyl ethyl."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "For example, this base right here is the ethoxide anion, which you could get from sodium ethoxide. So Na plus, OEt minus. And this could act as a base and take a proton from an aldehyde or a ketone. In this case, we have an aldehyde. So this is acetaldehyde here. And we need to find the alpha carbon. So the alpha carbon is the one next to the carbonyl carbon."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "In this case, we have an aldehyde. So this is acetaldehyde here. And we need to find the alpha carbon. So the alpha carbon is the one next to the carbonyl carbon. So this is the alpha carbon right here on acetaldehyde. There are three hydrogens attached to that alpha carbon. So we have three alpha protons."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the alpha carbon is the one next to the carbonyl carbon. So this is the alpha carbon right here on acetaldehyde. There are three hydrogens attached to that alpha carbon. So we have three alpha protons. So our base could take any one of those three alpha protons. I'm just gonna draw one in here to simplify things. And so we could show our base taking this proton and leaving these electrons behind on our carbon."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have three alpha protons. So our base could take any one of those three alpha protons. I'm just gonna draw one in here to simplify things. And so we could show our base taking this proton and leaving these electrons behind on our carbon. So we can draw the enolate anion that would form. So we have our carbonyl here. And then we could show these electrons on this carbon now."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so we could show our base taking this proton and leaving these electrons behind on our carbon. So we can draw the enolate anion that would form. So we have our carbonyl here. And then we could show these electrons on this carbon now. So let me go ahead and follow those electrons. So in magenta, these electrons move off onto this carbon to form a carb anion. Remember, there's also two other hydrogens attached to that carbon there."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we could show these electrons on this carbon now. So let me go ahead and follow those electrons. So in magenta, these electrons move off onto this carbon to form a carb anion. Remember, there's also two other hydrogens attached to that carbon there. And we could draw a resonance structure for this. So we could show these electrons in magenta moving in here to form a double bond, push these electrons off onto the oxygen. And if we do that, we'll have a resonance structure showing the negative charge, this time on the oxygen."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Remember, there's also two other hydrogens attached to that carbon there. And we could draw a resonance structure for this. So we could show these electrons in magenta moving in here to form a double bond, push these electrons off onto the oxygen. And if we do that, we'll have a resonance structure showing the negative charge, this time on the oxygen. So the oxygen gets a negative one formal charge now. And then we have a double bond over here on the right. And so this would be our other resonance structure."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And if we do that, we'll have a resonance structure showing the negative charge, this time on the oxygen. So the oxygen gets a negative one formal charge now. And then we have a double bond over here on the right. And so this would be our other resonance structure. So the electrons in magenta moved in here to form our double bond. And then let's make these electrons in here blue, move off onto the oxygen to make an oxyanion. So this is our enolate anion here."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so this would be our other resonance structure. So the electrons in magenta moved in here to form our double bond. And then let's make these electrons in here blue, move off onto the oxygen to make an oxyanion. So this is our enolate anion here. So we have the carb anion form. And then the oxyanion form of our enolate anion. So your enolate anion here, which is, once again, extremely important for reactions we'll talk about later."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is our enolate anion here. So we have the carb anion form. And then the oxyanion form of our enolate anion. So your enolate anion here, which is, once again, extremely important for reactions we'll talk about later. So our base has formed our enolate anion. If you think about what happens to the base, if you protonate ethoxide, you're going to form ethanol. So we'll go ahead and show ethanol also formed here in this reaction."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So your enolate anion here, which is, once again, extremely important for reactions we'll talk about later. So our base has formed our enolate anion. If you think about what happens to the base, if you protonate ethoxide, you're going to form ethanol. So we'll go ahead and show ethanol also formed here in this reaction. All right, so we have an equilibrium here between our aldehydes and our enolate anion. And to figure out which direction is favored, we need to know some pKa values. So the pKa value for this aldehyde is approximately 17."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we'll go ahead and show ethanol also formed here in this reaction. All right, so we have an equilibrium here between our aldehydes and our enolate anion. And to figure out which direction is favored, we need to know some pKa values. So the pKa value for this aldehyde is approximately 17. And the pKa value for ethanol is approximately 16. And so one way to figure out which direction is favored is to use these equations down here. We could first find the pKaEQ by taking the pKa of the acid on the left."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the pKa value for this aldehyde is approximately 17. And the pKa value for ethanol is approximately 16. And so one way to figure out which direction is favored is to use these equations down here. We could first find the pKaEQ by taking the pKa of the acid on the left. So the acid on the left is our aldehyde. So the pKa is 17. And from that number, we subtract the pKa of the acid on the right."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could first find the pKaEQ by taking the pKa of the acid on the left. So the acid on the left is our aldehyde. So the pKa is 17. And from that number, we subtract the pKa of the acid on the right. The pKa of the acid on the right is 16, which is ethanol. So 17 minus 16 gives us one. And then to find the Keq, we can take 10 to the negative pKaEQ."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And from that number, we subtract the pKa of the acid on the right. The pKa of the acid on the right is 16, which is ethanol. So 17 minus 16 gives us one. And then to find the Keq, we can take 10 to the negative pKaEQ. So 10 to the negative one is equal to.1, which is obviously less than one. And so we know that the equilibrium favors the reactants, right? The equilibrium is back in this direction."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then to find the Keq, we can take 10 to the negative pKaEQ. So 10 to the negative one is equal to.1, which is obviously less than one. And so we know that the equilibrium favors the reactants, right? The equilibrium is back in this direction. So the equilibrium favors formation of the aldehyde. So at equilibrium, we're gonna have some aldehyde present. We're also gonna have some enolate anion present here."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The equilibrium is back in this direction. So the equilibrium favors formation of the aldehyde. So at equilibrium, we're gonna have some aldehyde present. We're also gonna have some enolate anion present here. So if you choose sodium ethoxide as your base, you're gonna have both the aldehyde and your enolate anion present. Another way to figure out which direction the equilibrium goes is to think about which one is the weaker acid. So we have these two acids here."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We're also gonna have some enolate anion present here. So if you choose sodium ethoxide as your base, you're gonna have both the aldehyde and your enolate anion present. Another way to figure out which direction the equilibrium goes is to think about which one is the weaker acid. So we have these two acids here. Let me go ahead and change colors. We have a pKa of 17 and a pKa of 16. The lower the pKa, the more acidic something is."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have these two acids here. Let me go ahead and change colors. We have a pKa of 17 and a pKa of 16. The lower the pKa, the more acidic something is. So ethanol is more acidic than our aldehyde. And the equilibrium favors formation of the weaker acid. And so since the aldehyde is the weaker acid, the equilibrium favors formation of this weaker acid."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The lower the pKa, the more acidic something is. So ethanol is more acidic than our aldehyde. And the equilibrium favors formation of the weaker acid. And so since the aldehyde is the weaker acid, the equilibrium favors formation of this weaker acid. So that's another way to think about which direction for the equilibrium. Alright, what if you wanted to completely make your enolate anion? So one thing you could do is add a base like hydride."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so since the aldehyde is the weaker acid, the equilibrium favors formation of this weaker acid. So that's another way to think about which direction for the equilibrium. Alright, what if you wanted to completely make your enolate anion? So one thing you could do is add a base like hydride. So here we have, once again, acid aldehyde here. So we have acid aldehyde. This time our base is the hydride anion."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So one thing you could do is add a base like hydride. So here we have, once again, acid aldehyde here. So we have acid aldehyde. This time our base is the hydride anion. So we could get that from something like sodium hydride, Na plus H minus, or potassium hydride, K plus H minus. And so for, once again, we identify our alpha carbon, which is this one, with three alpha protons. So we can just go ahead and draw one of them in there."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This time our base is the hydride anion. So we could get that from something like sodium hydride, Na plus H minus, or potassium hydride, K plus H minus. And so for, once again, we identify our alpha carbon, which is this one, with three alpha protons. So we can just go ahead and draw one of them in there. And we could show our hydride anion functioning as a base, taking this proton, leaving these electrons behind on that carbon. So let's go ahead and show the enolate anion that results. Alright, so we have our carbonyl here."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we can just go ahead and draw one of them in there. And we could show our hydride anion functioning as a base, taking this proton, leaving these electrons behind on that carbon. So let's go ahead and show the enolate anion that results. Alright, so we have our carbonyl here. And then we have our electrons on this carbon, giving it a negative one formal charge. So electrons in magenta move out onto this carbon, forming the carb anion. And just to save time, I won't draw in the oxyanion, but that's the one that's actually a greater contributor to the resonance hybrid here."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Alright, so we have our carbonyl here. And then we have our electrons on this carbon, giving it a negative one formal charge. So electrons in magenta move out onto this carbon, forming the carb anion. And just to save time, I won't draw in the oxyanion, but that's the one that's actually a greater contributor to the resonance hybrid here. So we have our carb anion. And then we would also form hydrogen gas, right? So if this hydride anion picks up a proton, we would form H2."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And just to save time, I won't draw in the oxyanion, but that's the one that's actually a greater contributor to the resonance hybrid here. So we have our carb anion. And then we would also form hydrogen gas, right? So if this hydride anion picks up a proton, we would form H2. And so let's show those electrons. So the electrons in red here on our hydride anion pick up this proton, and forming this bond, and so we get hydrogen gas, which would bubble out of solution. And since this is going to bubble out of solution, we're going to drive the reaction to completion."}, {"video_title": "Enolate formation from aldehydes Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if this hydride anion picks up a proton, we would form H2. And so let's show those electrons. So the electrons in red here on our hydride anion pick up this proton, and forming this bond, and so we get hydrogen gas, which would bubble out of solution. And since this is going to bubble out of solution, we're going to drive the reaction to completion. So we're going to push the reaction to completion. And so we're gonna get our enolate anion. So we're pretty much gonna get complete formation of our enolate anion here."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "What I want to do in this video is show that you can really visualize longer chains, or even, we'll see in future videos, even, cyclical ring-based carbon molecules with Newman projections as well. And I guess the next most complex molecule to study would be butane. We could do propane, but butane will be interesting. This was ethane right here. Butane will have four carbons. And if I were to draw it as a ball and stick model, it would look something like this. So this would be one carbon right there."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "This was ethane right here. Butane will have four carbons. And if I were to draw it as a ball and stick model, it would look something like this. So this would be one carbon right there. Then you would have another carbon right over there, and another carbon right over there. And then you would have your fourth carbon. And then your hydrogens, you would have a hydrogen coming out like this, like that, and then up like that."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this would be one carbon right there. Then you would have another carbon right over there, and another carbon right over there. And then you would have your fourth carbon. And then your hydrogens, you would have a hydrogen coming out like this, like that, and then up like that. This guy would have two hydrogens that would stick out like that. This guy would have two hydrogens that stick out like that, and then finally, this guy will also have three hydrogens, the CH3, just like that. Now, if we try to draw a Newman projection here, it's like, well, what do you consider the front or the back carbon and all of that?"}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then your hydrogens, you would have a hydrogen coming out like this, like that, and then up like that. This guy would have two hydrogens that would stick out like that. This guy would have two hydrogens that stick out like that, and then finally, this guy will also have three hydrogens, the CH3, just like that. Now, if we try to draw a Newman projection here, it's like, well, what do you consider the front or the back carbon and all of that? And you actually can pick. And what's interesting in a butane molecule is if you pick this guy, so this is the one, two, three, four carbons. If you pick the two carbon as our front and our three carbon as our back, and then we view this carbon, the CH3, as kind of one of the add-ons onto that carbon, you can then do a Newman projection."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "Now, if we try to draw a Newman projection here, it's like, well, what do you consider the front or the back carbon and all of that? And you actually can pick. And what's interesting in a butane molecule is if you pick this guy, so this is the one, two, three, four carbons. If you pick the two carbon as our front and our three carbon as our back, and then we view this carbon, the CH3, as kind of one of the add-ons onto that carbon, you can then do a Newman projection. So let's try to do this. So this will be the front one. So we'll put this carbon in the front, and we'll put this carbon over here in the back."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "If you pick the two carbon as our front and our three carbon as our back, and then we view this carbon, the CH3, as kind of one of the add-ons onto that carbon, you can then do a Newman projection. So let's try to do this. So this will be the front one. So we'll put this carbon in the front, and we'll put this carbon over here in the back. And before I even draw the Newman projection, let me redraw this. But I'm just going to draw this, instead of with the hydrogens, the bonds explicitly defined, I'm just going to call this a CH3. So let me redraw this."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we'll put this carbon in the front, and we'll put this carbon over here in the back. And before I even draw the Newman projection, let me redraw this. But I'm just going to draw this, instead of with the hydrogens, the bonds explicitly defined, I'm just going to call this a CH3. So let me redraw this. So I'll do this in orange. So you have this carbon. I'll do this kind of as a modified ball and stick."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let me redraw this. So I'll do this in orange. So you have this carbon. I'll do this kind of as a modified ball and stick. So that carbon, it has that hydrogen and that hydrogen. And instead of drawing this out, I'm going to just draw this whole thing right here, and I'll do it in magenta. I'm going to draw this whole thing as just a CH3."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "I'll do this kind of as a modified ball and stick. So that carbon, it has that hydrogen and that hydrogen. And instead of drawing this out, I'm going to just draw this whole thing right here, and I'll do it in magenta. I'm going to draw this whole thing as just a CH3. So I'm going to draw this whole thing as just a CH3. So I'll draw it really big, because it's not just one atom, it's four atoms. So this is our CH3."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to draw this whole thing as just a CH3. So I'm going to draw this whole thing as just a CH3. So I'll draw it really big, because it's not just one atom, it's four atoms. So this is our CH3. And well, to do ball and stick, everything really should be a ball. So I'll draw a ball there, a ball there. So that's our carbon number two."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this is our CH3. And well, to do ball and stick, everything really should be a ball. So I'll draw a ball there, a ball there. So that's our carbon number two. And then it has this bond. It has this bond over here to this carbon number three, which when we do our Newman projection, we'll put in the back. So our carbon number three is like that."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So that's our carbon number two. And then it has this bond. It has this bond over here to this carbon number three, which when we do our Newman projection, we'll put in the back. So our carbon number three is like that. And then the carbon number three, it has two hydrogens, and then it has this. You can kind of view it as this methyl group attached to it if you want. It has this CH3 attached to it right there."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So our carbon number three is like that. And then the carbon number three, it has two hydrogens, and then it has this. You can kind of view it as this methyl group attached to it if you want. It has this CH3 attached to it right there. So I'll do the CH3. I'll do it in this blue color. And so we could draw it like this."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "It has this CH3 attached to it right there. So I'll do the CH3. I'll do it in this blue color. And so we could draw it like this. So the CH3 is coming off. So I'll draw it really big, because it's not just one atom. So it's a CH3."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And so we could draw it like this. So the CH3 is coming off. So I'll draw it really big, because it's not just one atom. So it's a CH3. And then you have your two hydrogens down here. So let me be very clear here. So let me, this hydrogen, that hydrogen, that's that hydrogen, and that hydrogen."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So it's a CH3. And then you have your two hydrogens down here. So let me be very clear here. So let me, this hydrogen, that hydrogen, that's that hydrogen, and that hydrogen. This thing here is that thing there. That big ball right there is this whole ball. And then let me find a, I'll do green."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let me, this hydrogen, that hydrogen, that's that hydrogen, and that hydrogen. This thing here is that thing there. That big ball right there is this whole ball. And then let me find a, I'll do green. This hydrogen and this hydrogen is this hydrogen and this hydrogen. And when you look at it this way, now you say, oh, now I can see how I would draw a Newman projection. I put this in the front, that in the back."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then let me find a, I'll do green. This hydrogen and this hydrogen is this hydrogen and this hydrogen. And when you look at it this way, now you say, oh, now I can see how I would draw a Newman projection. I put this in the front, that in the back. I treat this whole part of the molecule as just a group, if you will. It is a group. So let's do the Newman projection here."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "I put this in the front, that in the back. I treat this whole part of the molecule as just a group, if you will. It is a group. So let's do the Newman projection here. And then we can think about where it's most stable. So the way I've drawn it right here, so you have your C, I'll do this as the front. This carbon 2 is going to be the front molecule."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So let's do the Newman projection here. And then we can think about where it's most stable. So the way I've drawn it right here, so you have your C, I'll do this as the front. This carbon 2 is going to be the front molecule. So you have this CH3 group going down. And then you have these two hydrogens. That's the front."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "This carbon 2 is going to be the front molecule. So you have this CH3 group going down. And then you have these two hydrogens. That's the front. And in the back, you have this blue one. You can imagine in the front, if we want to, maybe I'll do a little small orange thing to show this is the orange carbon. And then the blue carbon is going to be in the back."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "That's the front. And in the back, you have this blue one. You can imagine in the front, if we want to, maybe I'll do a little small orange thing to show this is the orange carbon. And then the blue carbon is going to be in the back. I'll draw it like this. So that's my blue carbon. And the way we've done it here, we have a CH3 pointing straight up, and then we have our two hydrogens."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then the blue carbon is going to be in the back. I'll draw it like this. So that's my blue carbon. And the way we've done it here, we have a CH3 pointing straight up, and then we have our two hydrogens. Now, just like we talked about in the first video on Newman projections, all of these groups, these hydrogens have electron clouds around them. This whole CH3 group has a larger electron cloud around it. It's a carbon atom plus hydrogens."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And the way we've done it here, we have a CH3 pointing straight up, and then we have our two hydrogens. Now, just like we talked about in the first video on Newman projections, all of these groups, these hydrogens have electron clouds around them. This whole CH3 group has a larger electron cloud around it. It's a carbon atom plus hydrogens. They all want to get away from each other. And the CH3 is even a bigger molecule. So to some degree, it's going to play a bigger role in whether something has a higher or lower potential energy or whether it's wound or not."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "It's a carbon atom plus hydrogens. They all want to get away from each other. And the CH3 is even a bigger molecule. So to some degree, it's going to play a bigger role in whether something has a higher or lower potential energy or whether it's wound or not. So I guess the most obvious, or maybe it's not obvious, but the CH3 group, since they have the biggest electron crowd, they're kind of crowding the molecule, this CH3 group and this CH3 group, where they're going to want to get as far away from each other as possible. So the way we did this, it looks kind of like our staggered conformation. But when we're dealing with actual methyl groups that are separated as far as they can from each other, we call this the anti-conformation."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So to some degree, it's going to play a bigger role in whether something has a higher or lower potential energy or whether it's wound or not. So I guess the most obvious, or maybe it's not obvious, but the CH3 group, since they have the biggest electron crowd, they're kind of crowding the molecule, this CH3 group and this CH3 group, where they're going to want to get as far away from each other as possible. So the way we did this, it looks kind of like our staggered conformation. But when we're dealing with actual methyl groups that are separated as far as they can from each other, we call this the anti-conformation. And if we think about dihedral angles between the two methyl groups, the dihedral angle here is 180 degrees. And this is the lowest potential energy or the most stable. And that confuses you when I talk about lowest potential energy."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "But when we're dealing with actual methyl groups that are separated as far as they can from each other, we call this the anti-conformation. And if we think about dihedral angles between the two methyl groups, the dihedral angle here is 180 degrees. And this is the lowest potential energy or the most stable. And that confuses you when I talk about lowest potential energy. Just think about it. A rock on the ground has a lower potential energy than a rock that is 50 feet in the air. A rock on the ground is also more stable."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And that confuses you when I talk about lowest potential energy. Just think about it. A rock on the ground has a lower potential energy than a rock that is 50 feet in the air. A rock on the ground is also more stable. It's less likely to do something. Something 50 feet in the air, maybe if you nudge it a little bit, it'll fall off the cliff or wherever it is. Or maybe it's already falling."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "A rock on the ground is also more stable. It's less likely to do something. Something 50 feet in the air, maybe if you nudge it a little bit, it'll fall off the cliff or wherever it is. Or maybe it's already falling. Who knows? It's going to move when you have higher potential energy. Or it takes very little for it to release energy."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "Or maybe it's already falling. Who knows? It's going to move when you have higher potential energy. Or it takes very little for it to release energy. But when you have lower potential energy, you're more stable. So this is the most stable conformation. Now, what are the other situations you could do here?"}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "Or it takes very little for it to release energy. But when you have lower potential energy, you're more stable. So this is the most stable conformation. Now, what are the other situations you could do here? Well, you could keep rotating these. Let's say we rotated the back carbon around clockwise. What are the other conformations we could get?"}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "Now, what are the other situations you could do here? Well, you could keep rotating these. Let's say we rotated the back carbon around clockwise. What are the other conformations we could get? And so let me just draw the front portion right here. So you have your CH3. And then you have your two hydrogens."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "What are the other conformations we could get? And so let me just draw the front portion right here. So you have your CH3. And then you have your two hydrogens. Hydrogen and hydrogen. And let me copy and paste this. There's two other real."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then you have your two hydrogens. Hydrogen and hydrogen. And let me copy and paste this. There's two other real. I mean, there's everything in between. But these are the ones that are interesting. And then let me copy it."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "There's two other real. I mean, there's everything in between. But these are the ones that are interesting. And then let me copy it. Copy and then paste. So I'll actually draw three of these. So then you have that."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then let me copy it. Copy and then paste. So I'll actually draw three of these. So then you have that. And then let me paste it one more time. Then you have that. So obviously, this would be the front carbon in every situation."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So then you have that. And then let me paste it one more time. Then you have that. So obviously, this would be the front carbon in every situation. If I want, I could make it a little orange dot to show that that's the front carbon. And then let me draw the back carbon. I should have copied and pasted this as well."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So obviously, this would be the front carbon in every situation. If I want, I could make it a little orange dot to show that that's the front carbon. And then let me draw the back carbon. I should have copied and pasted this as well. So you have your back carbon in every situation. Now, if we were to rotate this character by 60 degrees, actually, if we were to rotate the back by 60 degrees, what would it look like? Well, then we would have this hydrogen would move up there."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "I should have copied and pasted this as well. So you have your back carbon in every situation. Now, if we were to rotate this character by 60 degrees, actually, if we were to rotate the back by 60 degrees, what would it look like? Well, then we would have this hydrogen would move up there. So then you would have this hydrogen. Actually, if we were to move it by 120 degrees, I should say, this would be 60 and then another 120 degrees. So this hydrogen would go up there."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "Well, then we would have this hydrogen would move up there. So then you would have this hydrogen. Actually, if we were to move it by 120 degrees, I should say, this would be 60 and then another 120 degrees. So this hydrogen would go up there. This methyl group would now be over here. And then this hydrogen would go over here. So we've just rotated the whole thing by 120 degrees."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen would go up there. This methyl group would now be over here. And then this hydrogen would go over here. So we've just rotated the whole thing by 120 degrees. Now, this conformation, this was called the anti-conformation. It's the most stable because the methyl groups are as far away from each other as possible. This right here is called the Gauss conformation."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So we've just rotated the whole thing by 120 degrees. Now, this conformation, this was called the anti-conformation. It's the most stable because the methyl groups are as far away from each other as possible. This right here is called the Gauss conformation. And you can view this as the second most stable. At least the methyls are staggered. They're not directly behind each other."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "This right here is called the Gauss conformation. And you can view this as the second most stable. At least the methyls are staggered. They're not directly behind each other. So here the methyls are as far apart from each other as possible. If you look at the ball and stick model, I actually drew it in that conformation right here. They're as far apart from each other."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "They're not directly behind each other. So here the methyls are as far apart from each other as possible. If you look at the ball and stick model, I actually drew it in that conformation right here. They're as far apart from each other. So if you were to flip this molecule, this methyl would get closer to this methyl and their electron clouds would start to crowd each other. So in this situation, this is anti-most stable. If you rotate a little bit, they'll get a little bit closer."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "They're as far apart from each other. So if you were to flip this molecule, this methyl would get closer to this methyl and their electron clouds would start to crowd each other. So in this situation, this is anti-most stable. If you rotate a little bit, they'll get a little bit closer. But they'll still be staggered. You get the Gauss conformation. Now, if we rotate this, if we rotate the back guy now 60 degrees clockwise, so if we rotate it just 60 degrees clockwise, what's going to happen?"}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "If you rotate a little bit, they'll get a little bit closer. But they'll still be staggered. You get the Gauss conformation. Now, if we rotate this, if we rotate the back guy now 60 degrees clockwise, so if we rotate it just 60 degrees clockwise, what's going to happen? Well, then you're going to have an eclipsed conformation where the carbons are directly, but where the methyl groups are directly behind each other. And that's going to be your least stable situation. So you would have this guy."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "Now, if we rotate this, if we rotate the back guy now 60 degrees clockwise, so if we rotate it just 60 degrees clockwise, what's going to happen? Well, then you're going to have an eclipsed conformation where the carbons are directly, but where the methyl groups are directly behind each other. And that's going to be your least stable situation. So you would have this guy. And I'll draw it slightly. So you'd have this guy, CH3 there. And then you would have your hydrogens that are right behind each other."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "So you would have this guy. And I'll draw it slightly. So you'd have this guy, CH3 there. And then you would have your hydrogens that are right behind each other. So a hydrogen and a hydrogen. So in this situation, we're eclipsed. This is the least stable."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "And then you would have your hydrogens that are right behind each other. So a hydrogen and a hydrogen. So in this situation, we're eclipsed. This is the least stable. And also the most potential energy. And then if we were to go another 60 degrees from this, then we'd go to another Gauss conformation. If you rotate this another 60 degrees, then you'd have a CH3 here, and then you would have this hydrogen would be up here, and then this hydrogen here."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "This is the least stable. And also the most potential energy. And then if we were to go another 60 degrees from this, then we'd go to another Gauss conformation. If you rotate this another 60 degrees, then you'd have a CH3 here, and then you would have this hydrogen would be up here, and then this hydrogen here. So this is staggered. The methyl groups are, at least they're not directly behind each other, but they're not as far as they could be if we were to rotate another 120 degrees and get to the anti-conformation. So this one right here is also a Gauss conformation."}, {"video_title": "Newman projections 2 Organic chemistry Khan Academy.mp3", "Sentence": "If you rotate this another 60 degrees, then you'd have a CH3 here, and then you would have this hydrogen would be up here, and then this hydrogen here. So this is staggered. The methyl groups are, at least they're not directly behind each other, but they're not as far as they could be if we were to rotate another 120 degrees and get to the anti-conformation. So this one right here is also a Gauss conformation. So hopefully you understand now that if you just have to pick two carbons, and then if there's kind of big things attached to each of those carbons, you can just represent them as groups. And when you do that, you can really use a Newman projection for any part of a molecule. And when you do that, you can start to think about how it can rotate and what parts or what versions of it will be more or less stable."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "We know that the first step of our SN1 mechanism should be loss of a leaving group. So if these electrons come off onto the bromine, we would form the bromide anion. And we're taking a bond away from the carbon in red, so the carbon in red should get a plus one formal charge. So let's draw the resulting carbocation here, so let me sketch that in. The carbon in red is this carbon, so that carbon should have a plus one formal charge. In the next step of our mechanism, our nucleophile will attack. So the nucleophile attacks the electrophile, and a bond will form between the sulfur and the carbon in red."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So let's draw the resulting carbocation here, so let me sketch that in. The carbon in red is this carbon, so that carbon should have a plus one formal charge. In the next step of our mechanism, our nucleophile will attack. So the nucleophile attacks the electrophile, and a bond will form between the sulfur and the carbon in red. But remember, the geometry directly around that carbon in red, the carbons that are bonded to it, so this carbon in magenta, this carbon in magenta, and this carbon in magenta, are in the same plane as the carbon in red. And so the nucleophile could attack from either side of that plane. At this point, I think it's really helpful to look at this reaction using the model set."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So the nucleophile attacks the electrophile, and a bond will form between the sulfur and the carbon in red. But remember, the geometry directly around that carbon in red, the carbons that are bonded to it, so this carbon in magenta, this carbon in magenta, and this carbon in magenta, are in the same plane as the carbon in red. And so the nucleophile could attack from either side of that plane. At this point, I think it's really helpful to look at this reaction using the model set. So here's a screenshot from the video I'm gonna show you in a second. And in that video, I make bromine green, so here you can see this green bromine. This methyl group coming out at us in space is gonna be red in the video."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "At this point, I think it's really helpful to look at this reaction using the model set. So here's a screenshot from the video I'm gonna show you in a second. And in that video, I make bromine green, so here you can see this green bromine. This methyl group coming out at us in space is gonna be red in the video. On the right side, this ethyl group here will be yellow. And finally, on the left side, this propyl group will be gray. So here's our alkyl halides with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "This methyl group coming out at us in space is gonna be red in the video. On the right side, this ethyl group here will be yellow. And finally, on the left side, this propyl group will be gray. So here's our alkyl halides with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left. So I'll just turn this a little bit so we get a different viewpoint. And we know that the first step is loss of our leaving group. So I'm gonna show these electrons coming off onto our bromine and leaving to form a carbocation."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So here's our alkyl halides with our bromine going away from us, our methyl group coming out at us, our ethyl group on the right, and the propyl group on the left. So I'll just turn this a little bit so we get a different viewpoint. And we know that the first step is loss of our leaving group. So I'm gonna show these electrons coming off onto our bromine and leaving to form a carbocation. But that's not what the carbocation should look like. We need planar geometry around that central carbon. So here's another model which is more accurate."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So I'm gonna show these electrons coming off onto our bromine and leaving to form a carbocation. But that's not what the carbocation should look like. We need planar geometry around that central carbon. So here's another model which is more accurate. Now, the nucleophile could attack from the left or from the right. And first, let's look at what happens when the nucleophile attacks from the left. So we form a bond between the sulfur and the carbon."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So here's another model which is more accurate. Now, the nucleophile could attack from the left or from the right. And first, let's look at what happens when the nucleophile attacks from the left. So we form a bond between the sulfur and the carbon. And let's go ahead and look at a model set of one of our products. So here's the product that results when the nucleophile attacks from the left side of the carbocation. Here's our carbocation again."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So we form a bond between the sulfur and the carbon. And let's go ahead and look at a model set of one of our products. So here's the product that results when the nucleophile attacks from the left side of the carbocation. Here's our carbocation again. And this time, let's say the nucleophile approaches from the right side. So we're gonna form a bond between this sulfur and this carbon. Let's make a model of the product that forms when the nucleophile attacks from the right."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "Here's our carbocation again. And this time, let's say the nucleophile approaches from the right side. So we're gonna form a bond between this sulfur and this carbon. Let's make a model of the product that forms when the nucleophile attacks from the right. So here is that product. And let me hold up the carbocation so we can compare the two. Now let's compare this product with the product when the nucleophile attacked from the left side."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "Let's make a model of the product that forms when the nucleophile attacks from the right. So here is that product. And let me hold up the carbocation so we can compare the two. Now let's compare this product with the product when the nucleophile attacked from the left side. So in my left hand, I'm holding the product when the nucleophile attacked from the left. And on the right, I'm holding when the nucleophile attacks from the right. So what's the relationship between these two?"}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "Now let's compare this product with the product when the nucleophile attacked from the left side. So in my left hand, I'm holding the product when the nucleophile attacked from the left. And on the right, I'm holding when the nucleophile attacks from the right. So what's the relationship between these two? Well, they're mirror images of each other. But if I try to superimpose one on top of the other, you can see I can't do it. So these are non-superimposable mirror images of each other."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So what's the relationship between these two? Well, they're mirror images of each other. But if I try to superimpose one on top of the other, you can see I can't do it. So these are non-superimposable mirror images of each other. These are enantiomers. Now let's look at our products from a different viewpoint. So I'm gonna take the product on the left and I'm going to turn it so that the SH is coming out at me in space."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So these are non-superimposable mirror images of each other. These are enantiomers. Now let's look at our products from a different viewpoint. So I'm gonna take the product on the left and I'm going to turn it so that the SH is coming out at me in space. So here we can see the SH coming out at us in space, the methyl group going away from us, the ethyl group on the right, and the propyl group on the left. So now let's look at our other product. And this time, if we're gonna keep the same carbon chain, the methyl group's coming out at us in space, the SH is going away from us, the ethyl group is on the right, and the propyl group is still on the left."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So I'm gonna take the product on the left and I'm going to turn it so that the SH is coming out at me in space. So here we can see the SH coming out at us in space, the methyl group going away from us, the ethyl group on the right, and the propyl group on the left. So now let's look at our other product. And this time, if we're gonna keep the same carbon chain, the methyl group's coming out at us in space, the SH is going away from us, the ethyl group is on the right, and the propyl group is still on the left. Here are the two products that we got from the video. But since you won't always have a model set, let's go back to the drawings over here and pretend like we don't have a model set. We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "And this time, if we're gonna keep the same carbon chain, the methyl group's coming out at us in space, the SH is going away from us, the ethyl group is on the right, and the propyl group is still on the left. Here are the two products that we got from the video. But since you won't always have a model set, let's go back to the drawings over here and pretend like we don't have a model set. We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon. So if I draw in my carbon chain here, I know a bond formed between the sulfur and the carbon. Let me highlight the electrons. So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "We know that our nucleophile attacks our electrophile and a bond forms between the sulfur and that carbon. So if I draw in my carbon chain here, I know a bond formed between the sulfur and the carbon. Let me highlight the electrons. So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbon is a chiral center. There are four different groups attached to that carbon. So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So let's say a lone pair of electrons in magenta on the sulfur form this bond, and here's our product. But if you look at our product, notice that this carbon is a chiral center. There are four different groups attached to that carbon. So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product. So if I draw in my carbon chain here, I could represent one enantiomer by putting the SH on a wedge. So let me just draw that in here. So here's our SH on a wedge, which means the methyl group must be going away from us in space."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So if you think about the stereochemistry of this mechanism, with the nucleophile approaching the electrophile from either side of that plane, you should get a mixture of enantiomers as your product. So if I draw in my carbon chain here, I could represent one enantiomer by putting the SH on a wedge. So let me just draw that in here. So here's our SH on a wedge, which means the methyl group must be going away from us in space. And if I'm going to draw the other enantiomer, I would have to show the SH going away from us in space, which means the methyl group is coming out at us. And notice that these two products match the model sets that we drew here. Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "So here's our SH on a wedge, which means the methyl group must be going away from us in space. And if I'm going to draw the other enantiomer, I would have to show the SH going away from us in space, which means the methyl group is coming out at us. And notice that these two products match the model sets that we drew here. Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products. So I'm gonna say here approximately 50% is this enantiomer, and approximately 50% of our products is this enantiomer. Finally, let's go through the hybridization states of this carbon in red one more time. So for our starting alkyl halide, the carbon in red is tetrahedral."}, {"video_title": "Stereochemistry of Sn1 reactions.mp3", "Sentence": "Since there's an equal likelihood that the nucleophile could attack from one side or the other, we would expect to see an equal mixture of our products. So I'm gonna say here approximately 50% is this enantiomer, and approximately 50% of our products is this enantiomer. Finally, let's go through the hybridization states of this carbon in red one more time. So for our starting alkyl halide, the carbon in red is tetrahedral. It's sp3 hybridized, so it has tetrahedral geometry. When we formed our carbocation, the carbon in red is now sp2 hybridized, so it has planar geometry. But for our products, we're back to an sp3 hybridized carbon with tetrahedral geometry, so we have to think about the stereochemistry."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to start with the first criteria. So does it contain a ring of continuously overlapping p orbitals? Well, if we analyze these carbons here, this carbon is double bonded. So it's sp2 hybridized and therefore has a free p orbital. Same with this carbon and same with these other two here. So those four carbons are sp2 hybridized. If I look at this carbon, however, it's sp3 hybridized."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's sp2 hybridized and therefore has a free p orbital. Same with this carbon and same with these other two here. So those four carbons are sp2 hybridized. If I look at this carbon, however, it's sp3 hybridized. That's a little bit easier to see if I go ahead and draw in some hydrogens on there. So since this carbon right here has four single bonds to it, it is sp3 hybridized, which means that that carbon does not have a p orbital. And so cyclopentadiene is not aromatic."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If I look at this carbon, however, it's sp3 hybridized. That's a little bit easier to see if I go ahead and draw in some hydrogens on there. So since this carbon right here has four single bonds to it, it is sp3 hybridized, which means that that carbon does not have a p orbital. And so cyclopentadiene is not aromatic. It violates the first criteria. It does not contain a ring of continuously overlapping p orbitals. Since it violates the first criteria, we can go ahead and say that cyclopentadiene is non-aromatic."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so cyclopentadiene is not aromatic. It violates the first criteria. It does not contain a ring of continuously overlapping p orbitals. Since it violates the first criteria, we can go ahead and say that cyclopentadiene is non-aromatic. So you don't even need to worry about the second criteria. It doesn't fulfill the first criteria for something to be aromatic. However, cyclopentadiene has an interesting property."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Since it violates the first criteria, we can go ahead and say that cyclopentadiene is non-aromatic. So you don't even need to worry about the second criteria. It doesn't fulfill the first criteria for something to be aromatic. However, cyclopentadiene has an interesting property. It's extremely acidic for a hydrocarbon. It actually has a pKa of approximately 16 for one of these two protons here that I drew in yellow. And so there must be some sort of stability associated with the conjugate base in order for cyclopentadiene to be so acidic."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "However, cyclopentadiene has an interesting property. It's extremely acidic for a hydrocarbon. It actually has a pKa of approximately 16 for one of these two protons here that I drew in yellow. And so there must be some sort of stability associated with the conjugate base in order for cyclopentadiene to be so acidic. And so if we think about a base coming along, so I'm giving a lone pair of electrons, a negative charge, just some generic base, it's going to take this proton right here, leaving these two electrons behind on that top carbon. So if we go ahead and draw the structure of the conjugate base, we have our pi electrons here. And now we have a lone pair of electrons on our top carbon, which makes this negatively charged."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so there must be some sort of stability associated with the conjugate base in order for cyclopentadiene to be so acidic. And so if we think about a base coming along, so I'm giving a lone pair of electrons, a negative charge, just some generic base, it's going to take this proton right here, leaving these two electrons behind on that top carbon. So if we go ahead and draw the structure of the conjugate base, we have our pi electrons here. And now we have a lone pair of electrons on our top carbon, which makes this negatively charged. So here's the conjugate base. And there must be some sort of stabilization associated with this conjugate base because of the fact that cyclopentadiene is so acidic. Remember, the more stable the conjugate base, the more acidic the compound."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now we have a lone pair of electrons on our top carbon, which makes this negatively charged. So here's the conjugate base. And there must be some sort of stabilization associated with this conjugate base because of the fact that cyclopentadiene is so acidic. Remember, the more stable the conjugate base, the more acidic the compound. And so in this case, the extra stability is associated with the fact that this ion is aromatic. And it might not look like that because if we look at this top carbon, this top carbon still looks like it's sp3 hybridized because it's a carb anion. However, that lone pair of electrons can participate in resonance."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Remember, the more stable the conjugate base, the more acidic the compound. And so in this case, the extra stability is associated with the fact that this ion is aromatic. And it might not look like that because if we look at this top carbon, this top carbon still looks like it's sp3 hybridized because it's a carb anion. However, that lone pair of electrons can participate in resonance. So if we take this lone pair of electrons and we move them in here, and that would kick these electrons in here off onto this carbon. If we draw one of the possible resonance structures for this, so I'm going to go ahead and draw and show that these electrons have now moved into here. These electrons are now off on this carbon right here."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "However, that lone pair of electrons can participate in resonance. So if we take this lone pair of electrons and we move them in here, and that would kick these electrons in here off onto this carbon. If we draw one of the possible resonance structures for this, so I'm going to go ahead and draw and show that these electrons have now moved into here. These electrons are now off on this carbon right here. So it's a negative 1 formal charge. And then we have these pi electrons over here like that. Now, if we analyze this top carbon here, now we can see that it's actually sp2 hybridized."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "These electrons are now off on this carbon right here. So it's a negative 1 formal charge. And then we have these pi electrons over here like that. Now, if we analyze this top carbon here, now we can see that it's actually sp2 hybridized. So in this resonance structure, it looks like it's sp2 hybridized. And that lone pair of electrons, this lone pair of electrons in here in magenta, is now occupying a p orbital because it's participating in resonance. And so there are actually many more resonance structures you can draw."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, if we analyze this top carbon here, now we can see that it's actually sp2 hybridized. So in this resonance structure, it looks like it's sp2 hybridized. And that lone pair of electrons, this lone pair of electrons in here in magenta, is now occupying a p orbital because it's participating in resonance. And so there are actually many more resonance structures you can draw. And we're not going to do that here in this video. Here I'm just trying to point out that the electrons in magenta, the lone pair of electrons in magenta participate in resonance. And therefore, that lone pair of electrons is actually delocalized and occupying a p orbital in the ring."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so there are actually many more resonance structures you can draw. And we're not going to do that here in this video. Here I'm just trying to point out that the electrons in magenta, the lone pair of electrons in magenta participate in resonance. And therefore, that lone pair of electrons is actually delocalized and occupying a p orbital in the ring. And so now we have a ring of continuously overlapping p orbitals. So over here on the left, we'd already said that these carbons are sp2 hybridized. And if you think about these resonance structures, you could have all those carbons in that ring are now sp2 hybridized."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, that lone pair of electrons is actually delocalized and occupying a p orbital in the ring. And so now we have a ring of continuously overlapping p orbitals. So over here on the left, we'd already said that these carbons are sp2 hybridized. And if you think about these resonance structures, you could have all those carbons in that ring are now sp2 hybridized. And so you fulfill the first criteria. You have a ring of continuously overlapping p orbitals. Let's analyze this anion a little bit more using our frost circles."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if you think about these resonance structures, you could have all those carbons in that ring are now sp2 hybridized. And so you fulfill the first criteria. You have a ring of continuously overlapping p orbitals. Let's analyze this anion a little bit more using our frost circles. So if I go ahead and sketch in the fact that this is our anion with a negative 1 formal charge. When I'm looking for pi electrons in this molecule, so I will use this color here. So here are two pi electrons and then four pi electrons."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's analyze this anion a little bit more using our frost circles. So if I go ahead and sketch in the fact that this is our anion with a negative 1 formal charge. When I'm looking for pi electrons in this molecule, so I will use this color here. So here are two pi electrons and then four pi electrons. And then we've seen that this lone pair participates in resonance. So this lone pair of electrons occupies a p orbital. And those are actually pi electrons now."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So here are two pi electrons and then four pi electrons. And then we've seen that this lone pair participates in resonance. So this lone pair of electrons occupies a p orbital. And those are actually pi electrons now. So we have a total of six pi electrons. We have six pi electrons for this anion. And we've now seen that all of these carbons in here are sp2 hybridized when you draw all of your resonance structures."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And those are actually pi electrons now. So we have a total of six pi electrons. We have six pi electrons for this anion. And we've now seen that all of these carbons in here are sp2 hybridized when you draw all of your resonance structures. And so each one of those carbons has a p orbital. And so I can sketch in a p orbital on this diagram. So we have five carbons in the ring."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we've now seen that all of these carbons in here are sp2 hybridized when you draw all of your resonance structures. And so each one of those carbons has a p orbital. And so I can sketch in a p orbital on this diagram. So we have five carbons in the ring. Each carbon has a p orbital. And so there are five p orbitals in this ring. And you can see that those p orbitals in the ring can overlap to delocalize those electrons."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have five carbons in the ring. Each carbon has a p orbital. And so there are five p orbitals in this ring. And you can see that those p orbitals in the ring can overlap to delocalize those electrons. So the overlap of those p orbitals satisfies our first criteria. It contains a ring of continuously overlapping p orbitals. I have a total of five p orbitals."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you can see that those p orbitals in the ring can overlap to delocalize those electrons. So the overlap of those p orbitals satisfies our first criteria. It contains a ring of continuously overlapping p orbitals. I have a total of five p orbitals. And I know that those p orbitals are atomic orbitals. So five atomic orbitals, according to MO theory, are going to give me five molecular orbitals. And I can represent those molecular orbitals in terms of their energy on my frost circles."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I have a total of five p orbitals. And I know that those p orbitals are atomic orbitals. So five atomic orbitals, according to MO theory, are going to give me five molecular orbitals. And I can represent those molecular orbitals in terms of their energy on my frost circles. I'm going to go ahead and draw this line in here. When you're doing frost circles, remember to always start at the bottom here. A five-membered ring means I'm going to try to inscribe a five-sided polygon in my circle."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I can represent those molecular orbitals in terms of their energy on my frost circles. I'm going to go ahead and draw this line in here. When you're doing frost circles, remember to always start at the bottom here. A five-membered ring means I'm going to try to inscribe a five-sided polygon in my circle. So I'm going to try to draw a pentagon in here. So let's see if we can do it. So we'll put those lines in here like that."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "A five-membered ring means I'm going to try to inscribe a five-sided polygon in my circle. So I'm going to try to draw a pentagon in here. So let's see if we can do it. So we'll put those lines in here like that. And then you can see that here is my pentagon inscribed in my circle. Once again, the important thing is where that polygon intersects with my circle. That represents the energy level of my molecular orbitals."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we'll put those lines in here like that. And then you can see that here is my pentagon inscribed in my circle. Once again, the important thing is where that polygon intersects with my circle. That represents the energy level of my molecular orbitals. And so you can see I'm going to have three bonding molecular orbitals, the ones below the center line, and two anti-bonding molecular orbitals like that. I need to fill my molecular orbitals with my 6 pi electrons. And so I go ahead and put in my 6 pi electrons like that."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That represents the energy level of my molecular orbitals. And so you can see I'm going to have three bonding molecular orbitals, the ones below the center line, and two anti-bonding molecular orbitals like that. I need to fill my molecular orbitals with my 6 pi electrons. And so I go ahead and put in my 6 pi electrons like that. And you can see that I have filled my bonding molecular orbitals, analogous to having a full outer shell. And I have Huckel's rule. I have 4n plus 2 pi electrons."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I go ahead and put in my 6 pi electrons like that. And you can see that I have filled my bonding molecular orbitals, analogous to having a full outer shell. And I have Huckel's rule. I have 4n plus 2 pi electrons. And once again, we can see that using this energy diagram. Here would be the two electrons. And then 4 times 1, so n is equal to 1 here, gives me a total of 6 pi electrons."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I have 4n plus 2 pi electrons. And once again, we can see that using this energy diagram. Here would be the two electrons. And then 4 times 1, so n is equal to 1 here, gives me a total of 6 pi electrons. And so this ion satisfies both criteria. It contains a ring of continuously overlapping p orbitals. It also has 4n plus 2 pi electrons in the ring."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then 4 times 1, so n is equal to 1 here, gives me a total of 6 pi electrons. And so this ion satisfies both criteria. It contains a ring of continuously overlapping p orbitals. It also has 4n plus 2 pi electrons in the ring. And so this conjugate base up here is stable because it's an aromatic anion, which is the reason for such a low pKa value for cyclopentadiene. So even though cyclopentadiene itself is non-aromatic, this ion over here turns out to be aromatic, which explains this stability. Let's analyze one more ion here."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It also has 4n plus 2 pi electrons in the ring. And so this conjugate base up here is stable because it's an aromatic anion, which is the reason for such a low pKa value for cyclopentadiene. So even though cyclopentadiene itself is non-aromatic, this ion over here turns out to be aromatic, which explains this stability. Let's analyze one more ion here. So here we have the cycloheptatrienyl cation. And if I'm looking for my pi electrons, I have 2, 4, and 6 pi electrons in this ion, so a total of 6 pi electrons. When I'm looking for sp2 hybridized carbons, everything that has a double bond, these are all sp2 hybridized."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's analyze one more ion here. So here we have the cycloheptatrienyl cation. And if I'm looking for my pi electrons, I have 2, 4, and 6 pi electrons in this ion, so a total of 6 pi electrons. When I'm looking for sp2 hybridized carbons, everything that has a double bond, these are all sp2 hybridized. And then my carbocation right here, this carbon, is also sp2 hybridized. So all seven carbons are sp2 hybridized, which means that each one of those carbons has a p orbital. And so a little bit difficult to sketch in here, but you can see that each one of my seven carbons has a p orbital, which makes it relatively easy for those orbitals to overlap in my ring to delocalize those pi electrons."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "When I'm looking for sp2 hybridized carbons, everything that has a double bond, these are all sp2 hybridized. And then my carbocation right here, this carbon, is also sp2 hybridized. So all seven carbons are sp2 hybridized, which means that each one of those carbons has a p orbital. And so a little bit difficult to sketch in here, but you can see that each one of my seven carbons has a p orbital, which makes it relatively easy for those orbitals to overlap in my ring to delocalize those pi electrons. And so I satisfy my first criteria, a ring of continuously overlapping p orbitals here. So a total of seven p orbitals, so seven atomic orbitals. And once again, those seven atomic orbitals give me seven molecular orbitals, which I can represent using my Frost circle."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so a little bit difficult to sketch in here, but you can see that each one of my seven carbons has a p orbital, which makes it relatively easy for those orbitals to overlap in my ring to delocalize those pi electrons. And so I satisfy my first criteria, a ring of continuously overlapping p orbitals here. So a total of seven p orbitals, so seven atomic orbitals. And once again, those seven atomic orbitals give me seven molecular orbitals, which I can represent using my Frost circle. So I go over here to my Frost circle. I draw my dividing line between my bonding and my antibonding molecular orbitals. And now, since I have a seven-membered ring, I have to attempt to sketch in a seven-sided polygon into my Frost circle."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, those seven atomic orbitals give me seven molecular orbitals, which I can represent using my Frost circle. So I go over here to my Frost circle. I draw my dividing line between my bonding and my antibonding molecular orbitals. And now, since I have a seven-membered ring, I have to attempt to sketch in a seven-sided polygon into my Frost circle. And so this is probably one of the trickiest ones to draw here. And so let's see if we can do something that approximates a seven-sided figure inside of my circle here. So something like this gives me a seven-sided figure."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And now, since I have a seven-membered ring, I have to attempt to sketch in a seven-sided polygon into my Frost circle. And so this is probably one of the trickiest ones to draw here. And so let's see if we can do something that approximates a seven-sided figure inside of my circle here. So something like this gives me a seven-sided figure. And once again, we care about the points of intersection, because that represents the energy levels of my molecular orbitals. And so now it's easy to see that we have three bonding molecular orbitals and four antibonding molecular orbitals like that. And we're going to fill six pi electrons in our diagram."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So something like this gives me a seven-sided figure. And once again, we care about the points of intersection, because that represents the energy levels of my molecular orbitals. And so now it's easy to see that we have three bonding molecular orbitals and four antibonding molecular orbitals like that. And we're going to fill six pi electrons in our diagram. So once again, electrons fill the lowest orbitals first. And so we can take care of our six pi electrons like that. And you can see that we've filled all of our bonding molecular orbitals."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to fill six pi electrons in our diagram. So once again, electrons fill the lowest orbitals first. And so we can take care of our six pi electrons like that. And you can see that we've filled all of our bonding molecular orbitals. That's the extra stability that goes along with an ion being aromatic here. So n is equal to 1. So 4 times 1 gives me 4 plus the 2 gives me a total of six pi electrons."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you can see that we've filled all of our bonding molecular orbitals. That's the extra stability that goes along with an ion being aromatic here. So n is equal to 1. So 4 times 1 gives me 4 plus the 2 gives me a total of six pi electrons. And so both criteria have been fulfilled. We have a ring of continuously overlapping p orbitals, which is over here. So over here on the left is our first criteria."}, {"video_title": "Aromatic stability IV Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So 4 times 1 gives me 4 plus the 2 gives me a total of six pi electrons. And so both criteria have been fulfilled. We have a ring of continuously overlapping p orbitals, which is over here. So over here on the left is our first criteria. So this is our ring. And then we also have 4n plus 2 pi electrons. And so the cycloheptatrienal cation is aromatic."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, we named this molecule using the common names for this group right over here, and I thought it would be fun to also use, to do the same thing, but use the systematic name. So in the last video, we called this isobutyl, but if we wanted to do it systematically, we would look at this group, we would start at where it is attached to the main backbone, and we would think about the longest chain of carbons from there. So if we start there, we can get one, two, three carbons, so if we're dealing with three carbons, then this is going to be a propyl group, and we would number it one, two, three, and we see on the two carbon of the propyl group, the two carbon of the propyl group, we have a one carbon group right over here. So this is a methyl group branching off of the two carbon of the propyl group. So we could call this, we could call this thing, the systematic name, instead of calling it isobutyl, we could call it two methyl, so that's the methyl group right over there. So let me write this down, two methyl propyl, two methyl propyl, two methyl propyl. And so this is the systematic name, and of course there's two of them."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "So this is a methyl group branching off of the two carbon of the propyl group. So we could call this, we could call this thing, the systematic name, instead of calling it isobutyl, we could call it two methyl, so that's the methyl group right over there. So let me write this down, two methyl propyl, two methyl propyl, two methyl propyl. And so this is the systematic name, and of course there's two of them. This is a two methyl propyl right over here, instead of an isobutyl, we'll call it two methyl propyl, and this is another two methyl propyl. So instead of writing the five, seven diisobutyl here, we can instead substitute that with the systematic name. So let's do that, let me copy and paste everything else that comes before it."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "And so this is the systematic name, and of course there's two of them. This is a two methyl propyl right over here, instead of an isobutyl, we'll call it two methyl propyl, and this is another two methyl propyl. So instead of writing the five, seven diisobutyl here, we can instead substitute that with the systematic name. So let's do that, let me copy and paste everything else that comes before it. So let me copy, copy and paste it. Whoops, I'm in the wrong layer of my program. Let me go one layer down."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "So let's do that, let me copy and paste everything else that comes before it. So let me copy, copy and paste it. Whoops, I'm in the wrong layer of my program. Let me go one layer down. So let me copy and paste it again. Copy and paste, there we go. So I got that part, but now I'm gonna write this part differently."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "Let me go one layer down. So let me copy and paste it again. Copy and paste, there we go. So I got that part, but now I'm gonna write this part differently. So we still have stuff, it's still on the five carbon and the seven carbon of our main chain. So five comma seven, and we have two of them, but when we're using systematic naming, we won't say di this thing, instead we say bis. So five comma seven bis, that says that, hey, we got two of what I'm about to say."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "So I got that part, but now I'm gonna write this part differently. So we still have stuff, it's still on the five carbon and the seven carbon of our main chain. So five comma seven, and we have two of them, but when we're using systematic naming, we won't say di this thing, instead we say bis. So five comma seven bis, that says that, hey, we got two of what I'm about to say. Bis this thing. So let me copy, actually that's in a different layer. Let me copy and let me paste that."}, {"video_title": "Naming two isobutyl groups systematically Organic chemistry Khan Academy.mp3", "Sentence": "So five comma seven bis, that says that, hey, we got two of what I'm about to say. Bis this thing. So let me copy, actually that's in a different layer. Let me copy and let me paste that. Bis this stuff right over here, and then of course we have cyclooctane. And then of course we have cyclooctane. Cyclooctane, and we are done, we have named it systematically as well."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "When I was at MIT, I always used to go to this Chinese food truck, and I always used to show up at the food truck like two seconds before Alan Guth. Like he was always one or two people behind me in line. But anyway, he was the founder of the cosmic inflation theory, which is basically this idea that in the very early moments, or the very early period after the Big Bang, we went through kind of this major inflation in the expansion of space. But anyway, if based on the theory of cosmic inflation, then the observable universe is on the order of, or maybe another way to say it, the entire universe is on the order of 10 to the 23 times the size of the observable universe. So that would mean, this is just a tiny, tiny fraction. I mean, this is an unimaginably large number. In fact, it is unimaginable."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "But anyway, if based on the theory of cosmic inflation, then the observable universe is on the order of, or maybe another way to say it, the entire universe is on the order of 10 to the 23 times the size of the observable universe. So that would mean, this is just a tiny, tiny fraction. I mean, this is an unimaginably large number. In fact, it is unimaginable. So already everything we've talked about, this itself is a huge, this is an incomprehensible amount of space, but this is an incomprehensible multiple of this incomprehensible amount of space. And that's just based on that theory. But it is possible, we cannot rule out even the idea that the actual universe is smaller than the observable universe."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "In fact, it is unimaginable. So already everything we've talked about, this itself is a huge, this is an incomprehensible amount of space, but this is an incomprehensible multiple of this incomprehensible amount of space. And that's just based on that theory. But it is possible, we cannot rule out even the idea that the actual universe is smaller than the observable universe. And that one is in some ways even more mind blowing than the idea that the universe is this big. The fact that what we're observing is actually larger than the actual universe. And so you might say, well, Sal, that's impossible."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "But it is possible, we cannot rule out even the idea that the actual universe is smaller than the observable universe. And that one is in some ways even more mind blowing than the idea that the universe is this big. The fact that what we're observing is actually larger than the actual universe. And so you might say, well, Sal, that's impossible. But just think about it a little bit. This is the observable universe. And the way we've depicted it is based on how long the light has taken to reach us."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And so you might say, well, Sal, that's impossible. But just think about it a little bit. This is the observable universe. And the way we've depicted it is based on how long the light has taken to reach us. We've already covered before that this point in space is now 46 billion light years away, not 13.7 the way it looks right over here. It just took 13.7 billion years to reach us. If there's any photon that would take longer than 13.7 billion years to reach us, it hasn't reached us yet."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And the way we've depicted it is based on how long the light has taken to reach us. We've already covered before that this point in space is now 46 billion light years away, not 13.7 the way it looks right over here. It just took 13.7 billion years to reach us. If there's any photon that would take longer than 13.7 billion years to reach us, it hasn't reached us yet. Because it could have only started 13.7 billion years ago, so they're on their way. And they started at some point outside of our observable universe, so our observable universe will grow over time. But with that said, let's imagine that the actual universe is a subset of this observable universe."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "If there's any photon that would take longer than 13.7 billion years to reach us, it hasn't reached us yet. Because it could have only started 13.7 billion years ago, so they're on their way. And they started at some point outside of our observable universe, so our observable universe will grow over time. But with that said, let's imagine that the actual universe is a subset of this observable universe. Let's say it's roughly half the diameter. So let's say it looks like this. Maybe I'll make it a little bit of an oval."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "But with that said, let's imagine that the actual universe is a subset of this observable universe. Let's say it's roughly half the diameter. So let's say it looks like this. Maybe I'll make it a little bit of an oval. Maybe the actual universe, and this is just to be a little bit provocative, and it's not impossible. Let's say that this is the actual universe. And the way I drew it, it makes it look like Earth is the center, that we're the center of it."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "Maybe I'll make it a little bit of an oval. Maybe the actual universe, and this is just to be a little bit provocative, and it's not impossible. Let's say that this is the actual universe. And the way I drew it, it makes it look like Earth is the center, that we're the center of it. But remember, this is very likely to be the surface, or it is curved. It has a slight curvature, but it could very well be the surface of a four-dimensional object. And maybe the simplest one to visualize is a four-dimensional sphere."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And the way I drew it, it makes it look like Earth is the center, that we're the center of it. But remember, this is very likely to be the surface, or it is curved. It has a slight curvature, but it could very well be the surface of a four-dimensional object. And maybe the simplest one to visualize is a four-dimensional sphere. So if you really wanted to visualize this right, this whole volume, and remember, this whole picture, it keeps looking two-dimensional, but it has depth. It is a volume of space, an incredibly vast volume of space. And so what I've done here is, this is an ellipsoid right here."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And maybe the simplest one to visualize is a four-dimensional sphere. So if you really wanted to visualize this right, this whole volume, and remember, this whole picture, it keeps looking two-dimensional, but it has depth. It is a volume of space, an incredibly vast volume of space. And so what I've done here is, this is an ellipsoid right here. It's an elliptical volume of space that I've bubbled out right over here. But if this was really the entire universe, and if the entire universe really were the surface of a four-dimensional sphere, then the reality is that this entire space could be represented like this. It could be represented as the surface."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And so what I've done here is, this is an ellipsoid right here. It's an elliptical volume of space that I've bubbled out right over here. But if this was really the entire universe, and if the entire universe really were the surface of a four-dimensional sphere, then the reality is that this entire space could be represented like this. It could be represented as the surface. If this was a four-dimensional sphere, obviously the way I can only draw three-dimensional spheres, but let me show you that it has some, that it's not just a circle, that it actually has some depth to it. And I can even shade it right over here. And so you can imagine that this point over here is actually the same thing as that point over there, that they have wrapped around, that they're connected right at the back over here."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "It could be represented as the surface. If this was a four-dimensional sphere, obviously the way I can only draw three-dimensional spheres, but let me show you that it has some, that it's not just a circle, that it actually has some depth to it. And I can even shade it right over here. And so you can imagine that this point over here is actually the same thing as that point over there, that they have wrapped around, that they're connected right at the back over here. Let me show the draw, go behind. That they're connected right over there. And that this point and this point are actually the same point, that they've wrapped around."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And so you can imagine that this point over here is actually the same thing as that point over there, that they have wrapped around, that they're connected right at the back over here. Let me show the draw, go behind. That they're connected right over there. And that this point and this point are actually the same point, that they've wrapped around. Maybe they've wrapped around, actually the way I've drawn it right here, they would actually all wrap around right back there at that point, if I'm visualizing properly. But if you go in any one direction, you would come back on the other side of the surface. If you go, let's say that Earth is right here."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And that this point and this point are actually the same point, that they've wrapped around. Maybe they've wrapped around, actually the way I've drawn it right here, they would actually all wrap around right back there at that point, if I'm visualizing properly. But if you go in any one direction, you would come back on the other side of the surface. If you go, let's say that Earth is right here. The way we've depicted it, Earth is the center. But we see that when you look at it like this, there is no center to the surface of a sphere, even a four-dimensional sphere. So in this sense, if you go in any one direction, you'll come back out the other side."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "If you go, let's say that Earth is right here. The way we've depicted it, Earth is the center. But we see that when you look at it like this, there is no center to the surface of a sphere, even a four-dimensional sphere. So in this sense, if you go in any one direction, you'll come back out the other side. So if you start from Earth and you go in that direction, once you get there, you're really here again. And then you would come back to Earth. And so if this were the case, if the actual volume of the true universe was smaller than what it looks like, the observable universe, then what's all this stuff on the outside?"}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "So in this sense, if you go in any one direction, you'll come back out the other side. So if you start from Earth and you go in that direction, once you get there, you're really here again. And then you would come back to Earth. And so if this were the case, if the actual volume of the true universe was smaller than what it looks like, the observable universe, then what's all this stuff on the outside? And to think about it, think about what would happen. If 13.7 billion years ago, when we were in that primitive state, where that background radiation, those photons, that electromagnetic waves are being released. Let's say they get released."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And so if this were the case, if the actual volume of the true universe was smaller than what it looks like, the observable universe, then what's all this stuff on the outside? And to think about it, think about what would happen. If 13.7 billion years ago, when we were in that primitive state, where that background radiation, those photons, that electromagnetic waves are being released. Let's say they get released. And those photons, on their first pass, and I think you know where this is going, on their first pass, they would get to us in about, this looks like a distance of about, I don't know, this looks like about 6 billion years. Then they would pass us up, and then they would get back to this point again in another 6 billion years, and then they would come back here. And so that very first pass photon are going to be right over here."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "Let's say they get released. And those photons, on their first pass, and I think you know where this is going, on their first pass, they would get to us in about, this looks like a distance of about, I don't know, this looks like about 6 billion years. Then they would pass us up, and then they would get back to this point again in another 6 billion years, and then they would come back here. And so that very first pass photon are going to be right over here. And from our point of view, we're not going to see them for a couple of billion years. And so when we do see them, we're going to perceive them, we're going to say, wow, it took 15, 16 billion years for that photon to get to me. That must be from something out here."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And so that very first pass photon are going to be right over here. And from our point of view, we're not going to see them for a couple of billion years. And so when we do see them, we're going to perceive them, we're going to say, wow, it took 15, 16 billion years for that photon to get to me. That must be from something out here. But the reality is, it's a photon from something within a smaller physical universe, within a smaller actual universe, that's just taken several passes by us. And we're just seeing a pass after 14 billion years. We just think it's from something further out."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "That must be from something out here. But the reality is, it's a photon from something within a smaller physical universe, within a smaller actual universe, that's just taken several passes by us. And we're just seeing a pass after 14 billion years. We just think it's from something further out. Now the other thing is, if this was the case, if we could just go in one direction of the universe and then come out of the other side, and if all of that was within the observable universe, wouldn't we be able to tell? Wouldn't we be able to look in two directions and see the same thing from a different perspective? And the answer there is to think about what happens."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "We just think it's from something further out. Now the other thing is, if this was the case, if we could just go in one direction of the universe and then come out of the other side, and if all of that was within the observable universe, wouldn't we be able to tell? Wouldn't we be able to look in two directions and see the same thing from a different perspective? And the answer there is to think about what happens. Or actually, wouldn't we even be able to see ourselves? Because if we emit some light, and it would take maybe, I don't know how far that is, let's say that's 6 or 7 billion light years to get right over here, which would be right over there. And then it would take another 6 or 7 billion light years to get over there."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And the answer there is to think about what happens. Or actually, wouldn't we even be able to see ourselves? Because if we emit some light, and it would take maybe, I don't know how far that is, let's say that's 6 or 7 billion light years to get right over here, which would be right over there. And then it would take another 6 or 7 billion light years to get over there. So maybe that background radiation we're seeing is actually background radiation emitted from that exact point in space that we are right now, or from a very similar point in space to where we are right now. Or part of the background radiation is from a similar point in space that we are right now. So how come we can't just see ourselves?"}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "And then it would take another 6 or 7 billion light years to get over there. So maybe that background radiation we're seeing is actually background radiation emitted from that exact point in space that we are right now, or from a very similar point in space to where we are right now. Or part of the background radiation is from a similar point in space that we are right now. So how come we can't just see ourselves? Well, I kind of just answered the question. That second path, if you're observing the same point in space, if you're observing light from the same point in space on a previous path, that light was emitted a long, long, long time ago. Maybe 13 billion years ago."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "So how come we can't just see ourselves? Well, I kind of just answered the question. That second path, if you're observing the same point in space, if you're observing light from the same point in space on a previous path, that light was emitted a long, long, long time ago. Maybe 13 billion years ago. And so it would be unrecognizable. The region, this region of space, the region of space that we are in right now, if we saw the same region of space 13 billion years ago, we just wouldn't recognize it. Now there are people attempting to see if there's some patterns, see if you can model how the universe would change."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "Maybe 13 billion years ago. And so it would be unrecognizable. The region, this region of space, the region of space that we are in right now, if we saw the same region of space 13 billion years ago, we just wouldn't recognize it. Now there are people attempting to see if there's some patterns, see if you can model how the universe would change. And if you see patterns, and maybe the actual universe is a subset of the observable. We just haven't seen it yet. But it's completely a possibility."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "Now there are people attempting to see if there's some patterns, see if you can model how the universe would change. And if you see patterns, and maybe the actual universe is a subset of the observable. We just haven't seen it yet. But it's completely a possibility. Hopefully I didn't confuse you. I actually find this kind of an interesting idea. That these things that we think are, this light that has taken 13, let's say the light that's taken us 8 billion years to reach us, we think it's from something based on this scale, 8 billion light years out."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "But it's completely a possibility. Hopefully I didn't confuse you. I actually find this kind of an interesting idea. That these things that we think are, this light that has taken 13, let's say the light that's taken us 8 billion years to reach us, we think it's from something based on this scale, 8 billion light years out. It's actually further because the universe is expanding. So it would have actually traversed more space than that, but we think it's from something like that. But it could have been something further in if the actual universe is smaller and it's just on its second path."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "That these things that we think are, this light that has taken 13, let's say the light that's taken us 8 billion years to reach us, we think it's from something based on this scale, 8 billion light years out. It's actually further because the universe is expanding. So it would have actually traversed more space than that, but we think it's from something like that. But it could have been something further in if the actual universe is smaller and it's just on its second path. It's actually coming back again. And that's why it took 8 billion years to reach us. And we don't even recognize it because it looks very different than that region of space right now."}, {"video_title": "A Universe Smaller than the Observable.mp3", "Sentence": "But it could have been something further in if the actual universe is smaller and it's just on its second path. It's actually coming back again. And that's why it took 8 billion years to reach us. And we don't even recognize it because it looks very different than that region of space right now. Or that region of space after 4 billion years. Looks completely different than it did when it first released. Anyway, hopefully I didn't confuse you too much."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Remember from general chemistry that mechanisms show the steps by which a reaction occurs. And so for this reaction, let's look at this alkyl halide on the left here. We know that chlorine is more electronegative than this carbon, so the chlorine is going to withdraw some electron density away from that carbon, which makes it partially positive. From the last video, we know that since this carbon is partially positive, this is the electrophilic center of this compound. If we look at hydroxide ion, which we could've gotten from something like sodium hydroxide, we know that this negatively charged oxygen would be the nucleophilic portion. So hydroxide is gonna act like a nucleophile, and this carbon on our alkyl halide is going to act like an electrophile. We know that opposite charges attract, so the negatively charged oxygen is going to be attracted to the partially positively charged carbon on the alkyl halide."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "From the last video, we know that since this carbon is partially positive, this is the electrophilic center of this compound. If we look at hydroxide ion, which we could've gotten from something like sodium hydroxide, we know that this negatively charged oxygen would be the nucleophilic portion. So hydroxide is gonna act like a nucleophile, and this carbon on our alkyl halide is going to act like an electrophile. We know that opposite charges attract, so the negatively charged oxygen is going to be attracted to the partially positively charged carbon on the alkyl halide. And we say that the nucleophile attacks the electrophile. So I could draw a curved arrow showing the movement of two electrons over here to this carbon. At the same time, these two electrons come off onto the chlorine."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We know that opposite charges attract, so the negatively charged oxygen is going to be attracted to the partially positively charged carbon on the alkyl halide. And we say that the nucleophile attacks the electrophile. So I could draw a curved arrow showing the movement of two electrons over here to this carbon. At the same time, these two electrons come off onto the chlorine. So the chlorine had three lone pairs of electrons around it. Let me go ahead and draw those in. We're gonna add in an extra lone pair of electrons, and let me highlight those electrons in magenta."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "At the same time, these two electrons come off onto the chlorine. So the chlorine had three lone pairs of electrons around it. Let me go ahead and draw those in. We're gonna add in an extra lone pair of electrons, and let me highlight those electrons in magenta. So these two electrons come off onto the chlorine, so I'll make these this pair, and that gives the chlorine a negative one charge. So this is the chloride anion, and we call this a leaving group. So we're gonna form a bond between the oxygen and this partially positive carbon."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We're gonna add in an extra lone pair of electrons, and let me highlight those electrons in magenta. So these two electrons come off onto the chlorine, so I'll make these this pair, and that gives the chlorine a negative one charge. So this is the chloride anion, and we call this a leaving group. So we're gonna form a bond between the oxygen and this partially positive carbon. So let me say that this lone pair of electrons on the oxygen is gonna form a bond between the oxygen and that carbon. So on our product, on this alcohol, those two electrons must be these two electrons right here in this bond that formed. So that's a simple organic chemistry mechanism."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So we're gonna form a bond between the oxygen and this partially positive carbon. So let me say that this lone pair of electrons on the oxygen is gonna form a bond between the oxygen and that carbon. So on our product, on this alcohol, those two electrons must be these two electrons right here in this bond that formed. So that's a simple organic chemistry mechanism. We had only one step, the nucleophile attacked, and the leaving group left, all in the same step. And the goal is not to understand every single thing about this organic chemistry mechanism in great detail right now. Our goal right now is just to appreciate how nucleophiles and electrophiles are used in organic chemistry mechanisms, and start to get a feeling for how these curved arrows show the movement or flow of electrons."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So that's a simple organic chemistry mechanism. We had only one step, the nucleophile attacked, and the leaving group left, all in the same step. And the goal is not to understand every single thing about this organic chemistry mechanism in great detail right now. Our goal right now is just to appreciate how nucleophiles and electrophiles are used in organic chemistry mechanisms, and start to get a feeling for how these curved arrows show the movement or flow of electrons. Let's look at another organic chemistry mechanism, and we're gonna start by identifying our nucleophile and our electrophile. So let's look at this compound first. We know that oxygen is more electronegative than carbon, so this oxygen is going to withdraw some electron density from this carbon, and this chlorine is gonna do the same thing, because chlorine is more electronegative than carbon too."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Our goal right now is just to appreciate how nucleophiles and electrophiles are used in organic chemistry mechanisms, and start to get a feeling for how these curved arrows show the movement or flow of electrons. Let's look at another organic chemistry mechanism, and we're gonna start by identifying our nucleophile and our electrophile. So let's look at this compound first. We know that oxygen is more electronegative than carbon, so this oxygen is going to withdraw some electron density from this carbon, and this chlorine is gonna do the same thing, because chlorine is more electronegative than carbon too. So this carbon is electron deficient, right? It is partially positive, and that can act as an electrophile. On the right, we have the acetate anion, which could come from sodium acetate, and this oxygen has a negative one formal charge, so the oxygen is the nucleophilic center of our acetate anion, and our acetate anion can act as a nucleophile."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We know that oxygen is more electronegative than carbon, so this oxygen is going to withdraw some electron density from this carbon, and this chlorine is gonna do the same thing, because chlorine is more electronegative than carbon too. So this carbon is electron deficient, right? It is partially positive, and that can act as an electrophile. On the right, we have the acetate anion, which could come from sodium acetate, and this oxygen has a negative one formal charge, so the oxygen is the nucleophilic center of our acetate anion, and our acetate anion can act as a nucleophile. Opposite charges attract, so this negatively charged oxygen is attracted to this positively charged carbon, and we can say that the nucleophile attacks the electrophile and I draw a curved arrow to show the movement of these two electrons. Now I can't show a bond directly from this oxygen to this carbon until I take these pi electrons and move them off onto the top oxygen here, because remember, carbon can never exceed an octet of electrons around it. So let me draw the movement of all of those electrons here."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "On the right, we have the acetate anion, which could come from sodium acetate, and this oxygen has a negative one formal charge, so the oxygen is the nucleophilic center of our acetate anion, and our acetate anion can act as a nucleophile. Opposite charges attract, so this negatively charged oxygen is attracted to this positively charged carbon, and we can say that the nucleophile attacks the electrophile and I draw a curved arrow to show the movement of these two electrons. Now I can't show a bond directly from this oxygen to this carbon until I take these pi electrons and move them off onto the top oxygen here, because remember, carbon can never exceed an octet of electrons around it. So let me draw the movement of all of those electrons here. So let's draw everything in. So we now would have an oxygen at the top here with three lone pairs of electrons, which would give this oxygen a negative one formal charge. So if I'm showing movement of electrons, I'm saying that two electrons from here moved off onto the oxygen, which gives it a negative one formal charge."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So let me draw the movement of all of those electrons here. So let's draw everything in. So we now would have an oxygen at the top here with three lone pairs of electrons, which would give this oxygen a negative one formal charge. So if I'm showing movement of electrons, I'm saying that two electrons from here moved off onto the oxygen, which gives it a negative one formal charge. I still have this chlorine attached to this carbon, so let me draw in that chlorine down here. So this had three lone pairs of electrons around it, and now we've formed a new bond. We've formed a bond between this oxygen and this carbon, and let me highlight these two electrons in red."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So if I'm showing movement of electrons, I'm saying that two electrons from here moved off onto the oxygen, which gives it a negative one formal charge. I still have this chlorine attached to this carbon, so let me draw in that chlorine down here. So this had three lone pairs of electrons around it, and now we've formed a new bond. We've formed a bond between this oxygen and this carbon, and let me highlight these two electrons in red. So those two electrons in red form a bond between that oxygen and this carbon, and this oxygen still has two lone pairs of electrons around it like that. And then let's draw in the rest of this over here. So we have a carbonyl, so let me draw that in, put in my lone pairs of electrons, and then we have a methyl group coming off like that."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We've formed a bond between this oxygen and this carbon, and let me highlight these two electrons in red. So those two electrons in red form a bond between that oxygen and this carbon, and this oxygen still has two lone pairs of electrons around it like that. And then let's draw in the rest of this over here. So we have a carbonyl, so let me draw that in, put in my lone pairs of electrons, and then we have a methyl group coming off like that. So the first step of this mechanism is a nucleophilic attack, so let me write first step here. The nucleophile attacks the electrophile. And it turns out this is a two-step mechanism, and in the second step of this mechanism, we're gonna get loss of a leaving group."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So we have a carbonyl, so let me draw that in, put in my lone pairs of electrons, and then we have a methyl group coming off like that. So the first step of this mechanism is a nucleophilic attack, so let me write first step here. The nucleophile attacks the electrophile. And it turns out this is a two-step mechanism, and in the second step of this mechanism, we're gonna get loss of a leaving group. So let's say a lone pair of electrons on this oxygen moves back in to reform a carbonyl, but we cannot exceed an octet of electrons to this carbon, so that must mean that these two electrons come off onto chlorine, come off onto our leaving group, which would be the chloride anion. Let me draw in those electrons here, which gives the chlorine a negative charge. So the electrons in, let's make them blue here."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "And it turns out this is a two-step mechanism, and in the second step of this mechanism, we're gonna get loss of a leaving group. So let's say a lone pair of electrons on this oxygen moves back in to reform a carbonyl, but we cannot exceed an octet of electrons to this carbon, so that must mean that these two electrons come off onto chlorine, come off onto our leaving group, which would be the chloride anion. Let me draw in those electrons here, which gives the chlorine a negative charge. So the electrons in, let's make them blue here. So these electrons in blue come off onto the chlorine to form the chloride anion, which is our leaving group. So we get loss of a leaving group, and we reform our carbonyl, which gives us our final product. Again, don't worry too much about the details of the mechanism, our goal is to identify our nucleophile electrophile, and appreciate our electron flow, and start to think about different steps of a mechanism."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So the electrons in, let's make them blue here. So these electrons in blue come off onto the chlorine to form the chloride anion, which is our leaving group. So we get loss of a leaving group, and we reform our carbonyl, which gives us our final product. Again, don't worry too much about the details of the mechanism, our goal is to identify our nucleophile electrophile, and appreciate our electron flow, and start to think about different steps of a mechanism. Nucleophilic attack is a very common one, so is loss of a leaving group. Let's look at one more organic chemistry mechanism, and we're gonna start by identifying our nucleophile and our electrophile. Let's look at this compound first."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Again, don't worry too much about the details of the mechanism, our goal is to identify our nucleophile electrophile, and appreciate our electron flow, and start to think about different steps of a mechanism. Nucleophilic attack is a very common one, so is loss of a leaving group. Let's look at one more organic chemistry mechanism, and we're gonna start by identifying our nucleophile and our electrophile. Let's look at this compound first. We know that oxygen is more electronegative than this carbon, so the oxygen is going to withdraw some electron density away from that carbon, which makes this carbon partially positive. So that's the electrophilic center of this compound. In the first step, we're adding propyl lithium, and we know that carbon is more electronegative than lithium, so carbon's gonna pull these two electrons closer to it, giving this carbon a partially negative charge."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Let's look at this compound first. We know that oxygen is more electronegative than this carbon, so the oxygen is going to withdraw some electron density away from that carbon, which makes this carbon partially positive. So that's the electrophilic center of this compound. In the first step, we're adding propyl lithium, and we know that carbon is more electronegative than lithium, so carbon's gonna pull these two electrons closer to it, giving this carbon a partially negative charge. Or, since the electronegativity difference is so great, we could take those two electrons in that bond and put them on that carbon. So we have three carbons here, let me draw them in, and we have two electrons on that carbon. Let me highlight those electrons in magenta."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "In the first step, we're adding propyl lithium, and we know that carbon is more electronegative than lithium, so carbon's gonna pull these two electrons closer to it, giving this carbon a partially negative charge. Or, since the electronegativity difference is so great, we could take those two electrons in that bond and put them on that carbon. So we have three carbons here, let me draw them in, and we have two electrons on that carbon. Let me highlight those electrons in magenta. So these two electrons in magenta go onto that carbon here, which gives this carbon a negative one formal charge, and we have a carbanion. Since we have a negatively charged carbon, this carbanion is an excellent nucleophile, and this is going to attack the electrophilic portion that we already identified. So opposite charges attract, the negative charge is attracted to the positive charge, and in the first step of our mechanism, our nucleophile attacks our electrophile."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Let me highlight those electrons in magenta. So these two electrons in magenta go onto that carbon here, which gives this carbon a negative one formal charge, and we have a carbanion. Since we have a negatively charged carbon, this carbanion is an excellent nucleophile, and this is going to attack the electrophilic portion that we already identified. So opposite charges attract, the negative charge is attracted to the positive charge, and in the first step of our mechanism, our nucleophile attacks our electrophile. Now we can't just show a bond between those two carbons, because that would give 10 electrons around this carbon, that would exceed an octet of electrons. So we have to show some electrons going somewhere else. We could take these two electrons and move them off onto our oxygen."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So opposite charges attract, the negative charge is attracted to the positive charge, and in the first step of our mechanism, our nucleophile attacks our electrophile. Now we can't just show a bond between those two carbons, because that would give 10 electrons around this carbon, that would exceed an octet of electrons. So we have to show some electrons going somewhere else. We could take these two electrons and move them off onto our oxygen. So in our first step, our nucleophile attacks our electrophile and we form a carbon-carbon bond. So we also have an oxygen here with three lone pairs of electrons around it, which give it a negative one formal charge. So we could follow some of those electrons."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We could take these two electrons and move them off onto our oxygen. So in our first step, our nucleophile attacks our electrophile and we form a carbon-carbon bond. So we also have an oxygen here with three lone pairs of electrons around it, which give it a negative one formal charge. So we could follow some of those electrons. Let me make them blue here. So these electrons come off onto our oxygen, giving the oxygen a negative one formal charge. We're going to form a bond between our two carbons, and this is where people get messed up a little bit, because counting carbons can sometimes be difficult."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So we could follow some of those electrons. Let me make them blue here. So these electrons come off onto our oxygen, giving the oxygen a negative one formal charge. We're going to form a bond between our two carbons, and this is where people get messed up a little bit, because counting carbons can sometimes be difficult. We need to account for three carbons on our nucleophile, and we also are going to form a bond. So if you look at our product over here, it gives us a clue as to how to draw that. We have three carbons, one, two, and three, and the electrons in magenta are forming this bond in here."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We're going to form a bond between our two carbons, and this is where people get messed up a little bit, because counting carbons can sometimes be difficult. We need to account for three carbons on our nucleophile, and we also are going to form a bond. So if you look at our product over here, it gives us a clue as to how to draw that. We have three carbons, one, two, and three, and the electrons in magenta are forming this bond in here. So let's go ahead and draw that on our intermediate. So we have our three carbons accounted for. Let me highlight those."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "We have three carbons, one, two, and three, and the electrons in magenta are forming this bond in here. So let's go ahead and draw that on our intermediate. So we have our three carbons accounted for. Let me highlight those. So one, two, and three, and then the electrons in magenta, these electrons, form a bond between this carbon and this carbon. So the electrons in magenta form a bond between those two carbons. This is our intermediate."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Let me highlight those. So one, two, and three, and then the electrons in magenta, these electrons, form a bond between this carbon and this carbon. So the electrons in magenta form a bond between those two carbons. This is our intermediate. So our first step is nucleophilic attack, and the second step, we have a source of protons here. So I'm using the hydronium ion, so let me draw that in really fast. So we have H3O+, so a positive formal charge on the oxygen."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "This is our intermediate. So our first step is nucleophilic attack, and the second step, we have a source of protons here. So I'm using the hydronium ion, so let me draw that in really fast. So we have H3O+, so a positive formal charge on the oxygen. And the second step of this mechanism is acid-base chemistry. It's a proton transfer. This negatively charged oxygen on our intermediate acts as a base and takes a proton."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So we have H3O+, so a positive formal charge on the oxygen. And the second step of this mechanism is acid-base chemistry. It's a proton transfer. This negatively charged oxygen on our intermediate acts as a base and takes a proton. So let's say it's this lone pair of electrons, takes a proton from H3O+, leaving these electrons behind on the oxygen. So let me highlight those electrons in red. So these electrons in red are gonna take this proton."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "This negatively charged oxygen on our intermediate acts as a base and takes a proton. So let's say it's this lone pair of electrons, takes a proton from H3O+, leaving these electrons behind on the oxygen. So let me highlight those electrons in red. So these electrons in red are gonna take this proton. So let's say that the electrons in red form this bond, and here was the proton that we took. So we form an alcohol as our final product here. Again, our goal is just to appreciate nucleophiles and electrophiles, and this is a reaction that comes much later in the course."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So these electrons in red are gonna take this proton. So let's say that the electrons in red form this bond, and here was the proton that we took. So we form an alcohol as our final product here. Again, our goal is just to appreciate nucleophiles and electrophiles, and this is a reaction that comes much later in the course. The last thing that I wanted to talk about in this video is something I'm calling the Schwarz Principles. And Dr. Schwarz was my organic chemistry professor in college, and he was by far the best teacher that I've ever had. One day he said to me, organic chemistry is only five things."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "Again, our goal is just to appreciate nucleophiles and electrophiles, and this is a reaction that comes much later in the course. The last thing that I wanted to talk about in this video is something I'm calling the Schwarz Principles. And Dr. Schwarz was my organic chemistry professor in college, and he was by far the best teacher that I've ever had. One day he said to me, organic chemistry is only five things. You need to know valence electrons, you need to understand electronegativity, you need to know your acid-base chemistry, you need to know about oxidation reduction reactions, or redox, and finally, you need to understand nucleophiles and electrophiles. Let's go back to that previous mechanism and see how it actually has all five of the Schwarz Principles. First, let's talk about valence electrons."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "One day he said to me, organic chemistry is only five things. You need to know valence electrons, you need to understand electronegativity, you need to know your acid-base chemistry, you need to know about oxidation reduction reactions, or redox, and finally, you need to understand nucleophiles and electrophiles. Let's go back to that previous mechanism and see how it actually has all five of the Schwarz Principles. First, let's talk about valence electrons. So when we're showing these curved arrows in our mechanisms, like this curved arrow or that curved arrow, we're showing the movement of valence electrons. We used electronegativity a lot. That's how we figured out our nucleophile and our electrophile."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "First, let's talk about valence electrons. So when we're showing these curved arrows in our mechanisms, like this curved arrow or that curved arrow, we're showing the movement of valence electrons. We used electronegativity a lot. That's how we figured out our nucleophile and our electrophile. The third Schwarz Principle was acid-base chemistry. Well, that was the second step of this mechanism, right? So the second step of the mechanism was acid-base chemistry, and you see acid-base reactions a lot in organic chemistry mechanisms."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "That's how we figured out our nucleophile and our electrophile. The third Schwarz Principle was acid-base chemistry. Well, that was the second step of this mechanism, right? So the second step of the mechanism was acid-base chemistry, and you see acid-base reactions a lot in organic chemistry mechanisms. The fourth principle was redox. This is actually a redox reaction. If you assign some oxidation states, you'll see that our starting compound is reduced, right?"}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "So the second step of the mechanism was acid-base chemistry, and you see acid-base reactions a lot in organic chemistry mechanisms. The fourth principle was redox. This is actually a redox reaction. If you assign some oxidation states, you'll see that our starting compound is reduced, right? This ketone is reduced to an alcohol. And finally, nucleophile and electrophile. Obviously, that was the first step of our mechanism and something that we've been focusing on."}, {"video_title": "Intro to organic mechanisms.mp3", "Sentence": "If you assign some oxidation states, you'll see that our starting compound is reduced, right? This ketone is reduced to an alcohol. And finally, nucleophile and electrophile. Obviously, that was the first step of our mechanism and something that we've been focusing on. So you don't necessarily have to have all five things in one mechanism. This one just happens to have all five things, and so I wanted to talk about the Schwarz Principles. And if you understand those five concepts really well, the mechanisms will be a lot easier for you."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The ketone we have here is acetone. To find our alpha carbon, we just look at the carbon next to our carbonyl carbon. This could be an alpha carbon, and this could be an alpha carbon. Each one of those alpha carbons has three alpha protons, and so there's a total of six. I'm just going to draw one in here, and this is the one that we're going to show being deprotonated here. The base that's going to deprotonate acetone, we're going to use LDA, which is lithium diisopropyl amide. I could go ahead and draw in the lithium here, so Li plus, and then we see the two isopropyl groups like that, a negative one charge on our nitrogen."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Each one of those alpha carbons has three alpha protons, and so there's a total of six. I'm just going to draw one in here, and this is the one that we're going to show being deprotonated here. The base that's going to deprotonate acetone, we're going to use LDA, which is lithium diisopropyl amide. I could go ahead and draw in the lithium here, so Li plus, and then we see the two isopropyl groups like that, a negative one charge on our nitrogen. This is a very strong base. It's also very bulky and sterically hindered. You can think about a lone pair of electrons in the nitrogen taking that proton, leaving these electrons behind on this carbon."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I could go ahead and draw in the lithium here, so Li plus, and then we see the two isopropyl groups like that, a negative one charge on our nitrogen. This is a very strong base. It's also very bulky and sterically hindered. You can think about a lone pair of electrons in the nitrogen taking that proton, leaving these electrons behind on this carbon. We can go ahead and draw the conjugate base here. We would have electrons on this carbon now. That's a carbanion."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can think about a lone pair of electrons in the nitrogen taking that proton, leaving these electrons behind on this carbon. We can go ahead and draw the conjugate base here. We would have electrons on this carbon now. That's a carbanion. Let me go ahead and show those electrons. These electrons in here, magenta, are going to come off onto this carbon. This carbon is a carbanion because remember, there's also two other hydrogens attached to it."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's a carbanion. Let me go ahead and show those electrons. These electrons in here, magenta, are going to come off onto this carbon. This carbon is a carbanion because remember, there's also two other hydrogens attached to it. That's what gives it a negative one formal charge here. We could draw a resonance structure. We could show these electrons in magenta moving in here, and these electrons coming off onto our oxygen."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This carbon is a carbanion because remember, there's also two other hydrogens attached to it. That's what gives it a negative one formal charge here. We could draw a resonance structure. We could show these electrons in magenta moving in here, and these electrons coming off onto our oxygen. For our resonance structure, we would show the negative charge is now on our oxygen. This would be a negative one formal charge like that now. The electrons in magenta moved into here to form our double bond."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could show these electrons in magenta moving in here, and these electrons coming off onto our oxygen. For our resonance structure, we would show the negative charge is now on our oxygen. This would be a negative one formal charge like that now. The electrons in magenta moved into here to form our double bond. Then we could show the electrons in here in blue moving out onto the oxygen. This is our enolate anion. This represents our enolate anion."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in magenta moved into here to form our double bond. Then we could show the electrons in here in blue moving out onto the oxygen. This is our enolate anion. This represents our enolate anion. For our two resonance structures, we have one with a negative charge on the carbon, so that's our carbanion, and one with our negative charge on our oxygen. That's our oxyanion. Remember, the oxyanion contributes more to the overall hybrid because oxygen is more electronegative than carbon."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This represents our enolate anion. For our two resonance structures, we have one with a negative charge on the carbon, so that's our carbanion, and one with our negative charge on our oxygen. That's our oxyanion. Remember, the oxyanion contributes more to the overall hybrid because oxygen is more electronegative than carbon. That's our enolate anion that is formed. If we form our enolate anion, we're also going to get another product. If I think about adding a proton onto our base, we're going to form an amine."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Remember, the oxyanion contributes more to the overall hybrid because oxygen is more electronegative than carbon. That's our enolate anion that is formed. If we form our enolate anion, we're also going to get another product. If I think about adding a proton onto our base, we're going to form an amine. Let me go ahead and draw the amine that we would make. We would have, it would now be this amine. Let's show those electrons."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If I think about adding a proton onto our base, we're going to form an amine. Let me go ahead and draw the amine that we would make. We would have, it would now be this amine. Let's show those electrons. The electrons over here in red, these electrons right here pick up this proton forming this bond, and we'd form an amine. This reaction is at equilibrium. To figure out which direction is favored, one way to do it would be to calculate the Keq."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's show those electrons. The electrons over here in red, these electrons right here pick up this proton forming this bond, and we'd form an amine. This reaction is at equilibrium. To figure out which direction is favored, one way to do it would be to calculate the Keq. You can see over here on the left these ways for calculating the Keq which we talked about in the last video. We first need to calculate the pKeq which would be the pKa of the acid on the left. The acid on the left would be acetone with a pKa of approximately 19, or sometimes you'll see 20."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "To figure out which direction is favored, one way to do it would be to calculate the Keq. You can see over here on the left these ways for calculating the Keq which we talked about in the last video. We first need to calculate the pKeq which would be the pKa of the acid on the left. The acid on the left would be acetone with a pKa of approximately 19, or sometimes you'll see 20. From that number, we're going to subtract the pKa of the acid on the right which would be our amine. The pKa of this is approximately 36. 19 minus 36 gives us negative 17."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The acid on the left would be acetone with a pKa of approximately 19, or sometimes you'll see 20. From that number, we're going to subtract the pKa of the acid on the right which would be our amine. The pKa of this is approximately 36. 19 minus 36 gives us negative 17. To find the Keq, all we do is take 10 to the negative of that number. 10 to the negative of negative 17 is the same thing as 10 to the 17th which is obviously a huge number, much, much greater than one. We know that the equilibrium lies to the right."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "19 minus 36 gives us negative 17. To find the Keq, all we do is take 10 to the negative of that number. 10 to the negative of negative 17 is the same thing as 10 to the 17th which is obviously a huge number, much, much greater than one. We know that the equilibrium lies to the right. The equilibrium favors formation of the enolate anions. For all practical purposes, this is a huge number. You're pretty much going to get complete formation of your enolate anions."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that the equilibrium lies to the right. The equilibrium favors formation of the enolate anions. For all practical purposes, this is a huge number. You're pretty much going to get complete formation of your enolate anions. If you add LDA to acetone, you're going to get, again, pretty much all enolate anion and none of your acetone will remain here. Another way of figuring that out is to know that the equilibrium favors formation of the weaker acid. The weaker acid is the one with the higher value for the pKa."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "You're pretty much going to get complete formation of your enolate anions. If you add LDA to acetone, you're going to get, again, pretty much all enolate anion and none of your acetone will remain here. Another way of figuring that out is to know that the equilibrium favors formation of the weaker acid. The weaker acid is the one with the higher value for the pKa. Remember, the lower the pKa, the more acidic something is. Acetone is more acidic than this amine. Since the amine has the higher value for the pKa, the equilibrium favors formation of this weaker acid also to the right like that."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The weaker acid is the one with the higher value for the pKa. Remember, the lower the pKa, the more acidic something is. Acetone is more acidic than this amine. Since the amine has the higher value for the pKa, the equilibrium favors formation of this weaker acid also to the right like that. Let's talk about the pKa of acetone or ketones in general compared to aldehydes. This pKa is higher than that for our aldehyde like we saw in the last video. We can think about why by looking at the enolate anion and thinking about the fact that now we have a methyl group here."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Since the amine has the higher value for the pKa, the equilibrium favors formation of this weaker acid also to the right like that. Let's talk about the pKa of acetone or ketones in general compared to aldehydes. This pKa is higher than that for our aldehyde like we saw in the last video. We can think about why by looking at the enolate anion and thinking about the fact that now we have a methyl group here. Now, acrylic groups are electron donating. If it's donating a little bit of electron density, it already has a negative charge on it. Donating a little bit of electron density would destabilize this negative charge a little bit."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can think about why by looking at the enolate anion and thinking about the fact that now we have a methyl group here. Now, acrylic groups are electron donating. If it's donating a little bit of electron density, it already has a negative charge on it. Donating a little bit of electron density would destabilize this negative charge a little bit. If the conjugate base is destabilized, that means that acetone is not quite as likely to donate a proton. That's the situation when you have a ketone here. When we had an aldehyde, we didn't have this electron donating factor."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Donating a little bit of electron density would destabilize this negative charge a little bit. If the conjugate base is destabilized, that means that acetone is not quite as likely to donate a proton. That's the situation when you have a ketone here. When we had an aldehyde, we didn't have this electron donating factor. That's just one way to think about why a ketone is not as acidic as an aldehyde in general. Here's an example where we have a very acidic ketone. This is a special kind of ketone called a beta diketone."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "When we had an aldehyde, we didn't have this electron donating factor. That's just one way to think about why a ketone is not as acidic as an aldehyde in general. Here's an example where we have a very acidic ketone. This is a special kind of ketone called a beta diketone. This is a beta diketone here. If we look for alpha carbons, we look for the carbon next to our carbonyl. This is an alpha carbon."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a special kind of ketone called a beta diketone. This is a beta diketone here. If we look for alpha carbons, we look for the carbon next to our carbonyl. This is an alpha carbon. This is an alpha carbon. This is an alpha carbon. The question is, which one of those alpha carbons is the one that has the most acidic protons?"}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is an alpha carbon. This is an alpha carbon. This is an alpha carbon. The question is, which one of those alpha carbons is the one that has the most acidic protons? It turns out to be the alpha carbon in the center, the one between our two carbonyls. There are two alpha protons on that carbon. The pKa for one of those acidic protons is about nine."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The question is, which one of those alpha carbons is the one that has the most acidic protons? It turns out to be the alpha carbon in the center, the one between our two carbonyls. There are two alpha protons on that carbon. The pKa for one of those acidic protons is about nine. That's very acidic, much more acidic than acetone or acetaldehyde like we talked about in the last video. Since a beta diketone has such acidic protons, we don't need a super strong base like LDA. We could use something like sodium ethoxide, so a negative one formal charge here on this oxygen."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The pKa for one of those acidic protons is about nine. That's very acidic, much more acidic than acetone or acetaldehyde like we talked about in the last video. Since a beta diketone has such acidic protons, we don't need a super strong base like LDA. We could use something like sodium ethoxide, so a negative one formal charge here on this oxygen. Sodium ethoxide could be used to deprotonate this beta diketone. We think about a lone pair of electrons taking this proton, leaving these electrons behind on this carbon. Let's go ahead and draw the conjugate base."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We could use something like sodium ethoxide, so a negative one formal charge here on this oxygen. Sodium ethoxide could be used to deprotonate this beta diketone. We think about a lone pair of electrons taking this proton, leaving these electrons behind on this carbon. Let's go ahead and draw the conjugate base. We would have our carbonyls here. We took a proton away. Now we have our electrons on this carbon, so a negative one formal charge."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw the conjugate base. We would have our carbonyls here. We took a proton away. Now we have our electrons on this carbon, so a negative one formal charge. These electrons right in here, magenta, moved on to this carbon, which gives this carbon a negative one formal charge because there's still a hydrogen bonded to that carbon. I'm just not drawing it in so we can see a little bit better. For resonance structures, we could show these electrons in magenta moving into here, pushing these electrons off onto the oxygen."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now we have our electrons on this carbon, so a negative one formal charge. These electrons right in here, magenta, moved on to this carbon, which gives this carbon a negative one formal charge because there's still a hydrogen bonded to that carbon. I'm just not drawing it in so we can see a little bit better. For resonance structures, we could show these electrons in magenta moving into here, pushing these electrons off onto the oxygen. We could draw a resonance structure for that. For our resonance structure, we would form a double bond here. Then this oxygen would get a negative one formal charge, like that."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "For resonance structures, we could show these electrons in magenta moving into here, pushing these electrons off onto the oxygen. We could draw a resonance structure for that. For our resonance structure, we would form a double bond here. Then this oxygen would get a negative one formal charge, like that. Then our carbonyl on the right is still here, like that. That's one of our possible resonance structures. Electrons in magenta moved in here to form our double bond."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Then this oxygen would get a negative one formal charge, like that. Then our carbonyl on the right is still here, like that. That's one of our possible resonance structures. Electrons in magenta moved in here to form our double bond. Then we could show these electrons in here moving off onto our oxygen, like that. We could have shown our electrons moving over on this side as well. We can push those electrons off."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Electrons in magenta moved in here to form our double bond. Then we could show these electrons in here moving off onto our oxygen, like that. We could have shown our electrons moving over on this side as well. We can push those electrons off. Let's go ahead and draw another resonance structure here. Let's get some room and let's show the formation of another resonance structure. This time our carbonyl on the left is still there."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We can push those electrons off. Let's go ahead and draw another resonance structure here. Let's get some room and let's show the formation of another resonance structure. This time our carbonyl on the left is still there. We move some electrons into here. Then we would form a negative one charge on this oxygen, so a negative one charge right here, like that. Let's show those electrons."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This time our carbonyl on the left is still there. We move some electrons into here. Then we would form a negative one charge on this oxygen, so a negative one charge right here, like that. Let's show those electrons. Electrons in red move in to form our double bond. Then let's show these electrons in here coming off onto this oxygen, like that. We would total up three resonance structures for this enolate anion."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's show those electrons. Electrons in red move in to form our double bond. Then let's show these electrons in here coming off onto this oxygen, like that. We would total up three resonance structures for this enolate anion. You can see this negative one formal charge is delocalized. It's delocalized over this carbon. It's delocalized on this oxygen."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We would total up three resonance structures for this enolate anion. You can see this negative one formal charge is delocalized. It's delocalized over this carbon. It's delocalized on this oxygen. It's delocalized on this oxygen. The more you delocalize or spread out a negative charge, the more you stabilize the anion. This is a very stable anion because of resonance and also because of conjugation."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's delocalized on this oxygen. It's delocalized on this oxygen. The more you delocalize or spread out a negative charge, the more you stabilize the anion. This is a very stable anion because of resonance and also because of conjugation. If you think about the conjugation present here, so here's a double bond, and then here's a single bond, and then here's a double bond. You have some conjugation stabilizing it as well. We have a very stable conjugate base for all those reasons that we just talked about."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a very stable anion because of resonance and also because of conjugation. If you think about the conjugation present here, so here's a double bond, and then here's a single bond, and then here's a double bond. You have some conjugation stabilizing it as well. We have a very stable conjugate base for all those reasons that we just talked about. Since we have a stable conjugate base, this beta diketone is likely to donate one of these protons. That's why its pKa value is so low. If ethoxide takes one of these acidic protons, we're going to form ethanol as our other product."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have a very stable conjugate base for all those reasons that we just talked about. Since we have a stable conjugate base, this beta diketone is likely to donate one of these protons. That's why its pKa value is so low. If ethoxide takes one of these acidic protons, we're going to form ethanol as our other product. We can go ahead and draw ethanol in here like that. Then we could calculate the Keq for this reaction. Ethanol's pKa we have already seen is approximately 16."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "If ethoxide takes one of these acidic protons, we're going to form ethanol as our other product. We can go ahead and draw ethanol in here like that. Then we could calculate the Keq for this reaction. Ethanol's pKa we have already seen is approximately 16. If I want to calculate the Keq, first we find the pKaq. The pKa of the acid on the left, the acid on the left is this beta diketone, so that'd be nine. From that number, we subtract the pKa of the acid on the right, which is ethanol."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Ethanol's pKa we have already seen is approximately 16. If I want to calculate the Keq, first we find the pKaq. The pKa of the acid on the left, the acid on the left is this beta diketone, so that'd be nine. From that number, we subtract the pKa of the acid on the right, which is ethanol. Nine minus 16 gives us negative seven. That's the pKeq. To find the Keq, we take 10 to the negative of that number."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "From that number, we subtract the pKa of the acid on the right, which is ethanol. Nine minus 16 gives us negative seven. That's the pKeq. To find the Keq, we take 10 to the negative of that number. 10 to the negative of negative seven is of course the same thing as 10 to the seventh. Obviously, that's a number much, much greater than one. We know the equilibrium lies to the right, favoring formation of the enolate anion here."}, {"video_title": "Enolate formation from ketones Alpha Carbon Chemistry Organic chemistry Khan Academy.mp3", "Sentence": "To find the Keq, we take 10 to the negative of that number. 10 to the negative of negative seven is of course the same thing as 10 to the seventh. Obviously, that's a number much, much greater than one. We know the equilibrium lies to the right, favoring formation of the enolate anion here. Favoring formation of our enolate anion. For all practical purposes, we're going to get pretty much nearly complete formation of our enolate anion. We're going to get our enolate anion."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's start by calculating the hydrogen deficiency index. If we have five carbons, we can have a maximum of two times five plus two hydrogens. That's equal to 12. 12 hydrogens is the maximum number for five carbons. Here we have only 10 hydrogens. We're missing two hydrogens. We're missing one pair of hydrogens."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "12 hydrogens is the maximum number for five carbons. Here we have only 10 hydrogens. We're missing two hydrogens. We're missing one pair of hydrogens. Therefore, we can say the HDI is equal to one. When the HDI is equal to one, I immediately think a double bond is present in this molecule or a ring is present in this molecule. Next, you go to the integration values."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're missing one pair of hydrogens. Therefore, we can say the HDI is equal to one. When the HDI is equal to one, I immediately think a double bond is present in this molecule or a ring is present in this molecule. Next, you go to the integration values. For this signal, an integration value of 33.2. For this one, 48.4. For this one, 33.3 and then 48.7."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Next, you go to the integration values. For this signal, an integration value of 33.2. For this one, 48.4. For this one, 33.3 and then 48.7. You divide all four integration values by the lowest one. The lowest one would be 33.2. 33.2 divided by 33.2 is obviously equal to one."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "For this one, 33.3 and then 48.7. You divide all four integration values by the lowest one. The lowest one would be 33.2. 33.2 divided by 33.2 is obviously equal to one. 48.4 divided by 33.2 is approximately 1.5. 33.3 divided by 33.2 is obviously very close to one. Finally, 48.7 divided by 33.2 gives us approximately 1.5."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "33.2 divided by 33.2 is obviously equal to one. 48.4 divided by 33.2 is approximately 1.5. 33.3 divided by 33.2 is obviously very close to one. Finally, 48.7 divided by 33.2 gives us approximately 1.5. These numbers tell us the relative ratio of protons that are giving us these four signals. We need to think about 10 protons. We have 10 hydrogens here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Finally, 48.7 divided by 33.2 gives us approximately 1.5. These numbers tell us the relative ratio of protons that are giving us these four signals. We need to think about 10 protons. We have 10 hydrogens here. This is only relative. To get the absolute number of hydrogens, we need to multiply these numbers in this ratio by two. If we multiply one by two, we obviously get two."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have 10 hydrogens here. This is only relative. To get the absolute number of hydrogens, we need to multiply these numbers in this ratio by two. If we multiply one by two, we obviously get two. This signal represents two protons. Multiply 1.5 by two and we get three. This signal represents three protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we multiply one by two, we obviously get two. This signal represents two protons. Multiply 1.5 by two and we get three. This signal represents three protons. Multiply one by two, this signal is two protons. Multiply 1.5 by two, this signal is three protons. Let's go through and look at each signal one by one."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This signal represents three protons. Multiply one by two, this signal is two protons. Multiply 1.5 by two, this signal is three protons. Let's go through and look at each signal one by one. We start with the signal that has two protons on the left over here. Two protons, we're talking about a CH2 group. I go ahead and draw in a CH2 group over here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's go through and look at each signal one by one. We start with the signal that has two protons on the left over here. Two protons, we're talking about a CH2 group. I go ahead and draw in a CH2 group over here. How many neighboring protons for this CH2 group? We can find that by thinking about the n plus one rule. If we have n neighboring protons, we would expect to see n plus one peaks."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I go ahead and draw in a CH2 group over here. How many neighboring protons for this CH2 group? We can find that by thinking about the n plus one rule. If we have n neighboring protons, we would expect to see n plus one peaks. How many peaks do we have? One, two, three. Three peaks."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we have n neighboring protons, we would expect to see n plus one peaks. How many peaks do we have? One, two, three. Three peaks. To find the number of neighbors, just subtract one. Three minus one is two. These two CH2 protons have two neighboring protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Three peaks. To find the number of neighbors, just subtract one. Three minus one is two. These two CH2 protons have two neighboring protons. Next, let's think about the chemical shift. This signal has a chemical shift of approximately four parts per million. That's the highest value for the chemical shift out of all four of these signals."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These two CH2 protons have two neighboring protons. Next, let's think about the chemical shift. This signal has a chemical shift of approximately four parts per million. That's the highest value for the chemical shift out of all four of these signals. That's the region for a proton that's connected to a carbon that's bonded to an electronegative atom. If we look at our molecular formula, we have two oxygens here. I'm going to take one of those oxygens and put that oxygen on that carbon."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's the highest value for the chemical shift out of all four of these signals. That's the region for a proton that's connected to a carbon that's bonded to an electronegative atom. If we look at our molecular formula, we have two oxygens here. I'm going to take one of those oxygens and put that oxygen on that carbon. Once again, when you're thinking about the signal for a proton on a carbon that's bonded to an electronegative atom, you get a chemical shift somewhere, in this case, close to four parts per million because the oxygen is deshielding those protons. The oxygen is withdrawing electron density. That's one piece of the puzzle."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to take one of those oxygens and put that oxygen on that carbon. Once again, when you're thinking about the signal for a proton on a carbon that's bonded to an electronegative atom, you get a chemical shift somewhere, in this case, close to four parts per million because the oxygen is deshielding those protons. The oxygen is withdrawing electron density. That's one piece of the puzzle. Let's move on to our next signal. This will be the next piece of the puzzle. We have three protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's one piece of the puzzle. Let's move on to our next signal. This will be the next piece of the puzzle. We have three protons. That's a methyl group. Let me go ahead and draw in a methyl group here, CH3. How many neighboring protons for those methyl protons?"}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have three protons. That's a methyl group. Let me go ahead and draw in a methyl group here, CH3. How many neighboring protons for those methyl protons? We see one peak here. One minus one is zero, so zero neighboring protons. Let's think about the chemical shift."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "How many neighboring protons for those methyl protons? We see one peak here. One minus one is zero, so zero neighboring protons. Let's think about the chemical shift. The chemical shift is just past two parts per million. That's in the region for a proton that's next to a carbonyl. Let's go ahead and draw in a carbonyl here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about the chemical shift. The chemical shift is just past two parts per million. That's in the region for a proton that's next to a carbonyl. Let's go ahead and draw in a carbonyl here. We're assuming that the carbonyl carbon is bonded to another carbon over here. That makes sense in terms of numbers of neighbors. Let's use blue for this."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw in a carbonyl here. We're assuming that the carbonyl carbon is bonded to another carbon over here. That makes sense in terms of numbers of neighbors. Let's use blue for this. The signal for these three protons, zero neighbors. If we go to this carbon, if there's a carbon over here, there are no protons on this carbonyl carbon. That explains this piece of the puzzle."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's use blue for this. The signal for these three protons, zero neighbors. If we go to this carbon, if there's a carbon over here, there are no protons on this carbonyl carbon. That explains this piece of the puzzle. Next signal, a CH2. We draw in a CH2 here. How many neighbors?"}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That explains this piece of the puzzle. Next signal, a CH2. We draw in a CH2 here. How many neighbors? A little bit hard to see. One, two, three, four, five, six peaks. Six minus one is five."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "How many neighbors? A little bit hard to see. One, two, three, four, five, six peaks. Six minus one is five. We would expect five neighbors using the oversimplified n plus one rule. We'll come back to that CH2. Finally, we have a signal with three protons, a CH3."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Six minus one is five. We would expect five neighbors using the oversimplified n plus one rule. We'll come back to that CH2. Finally, we have a signal with three protons, a CH3. How many neighbors for these methyl protons? We have one, two, three peaks. Three minus one is two, so two neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Finally, we have a signal with three protons, a CH3. How many neighbors for these methyl protons? We have one, two, three peaks. Three minus one is two, so two neighbors. Let's put all of the pieces of the puzzle together. Let's draw the final dot structure. Let's start with this piece of the puzzle right here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Three minus one is two, so two neighbors. Let's put all of the pieces of the puzzle together. Let's draw the final dot structure. Let's start with this piece of the puzzle right here. We have a carbonyl. I'm going to draw in my carbonyl here. Then we have a methyl group bonded to that carbonyl."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with this piece of the puzzle right here. We have a carbonyl. I'm going to draw in my carbonyl here. Then we have a methyl group bonded to that carbonyl. I'm going to draw in our methyl group like that. We used blue for these methyl protons. The signal for these methyl protons appears right here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Then we have a methyl group bonded to that carbonyl. I'm going to draw in our methyl group like that. We used blue for these methyl protons. The signal for these methyl protons appears right here. Next, let's think about a possible functional group. We have a carbonyl. Then over here, we have an oxygen."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The signal for these methyl protons appears right here. Next, let's think about a possible functional group. We have a carbonyl. Then over here, we have an oxygen. If we put the oxygen next to the carbonyl, that gives us an ester. Let's go ahead and do that. Let's put the oxygen next to the carbonyl."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Then over here, we have an oxygen. If we put the oxygen next to the carbonyl, that gives us an ester. Let's go ahead and do that. Let's put the oxygen next to the carbonyl. Then bonded to that oxygen, we had a CH2. I draw in my CH2 there. Let's identify those protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's put the oxygen next to the carbonyl. Then bonded to that oxygen, we had a CH2. I draw in my CH2 there. Let's identify those protons. The protons in magenta are giving us this signal over here. Next, let's think about... Let's use red. Let's think about these protons, these methyl protons right here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's identify those protons. The protons in magenta are giving us this signal over here. Next, let's think about... Let's use red. Let's think about these protons, these methyl protons right here. Those methyl protons have two neighbors. It makes sense that there would be these two neighbors right here. These chemical shifts are under two."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about these protons, these methyl protons right here. Those methyl protons have two neighbors. It makes sense that there would be these two neighbors right here. These chemical shifts are under two. Those are relatively far away from the oxygens. Let's draw in the methyl protons next. We have our methyl protons right here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These chemical shifts are under two. Those are relatively far away from the oxygens. Let's draw in the methyl protons next. We have our methyl protons right here. Let's make them red. Let's highlight those. These methyl protons right here are giving us this signal."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have our methyl protons right here. Let's make them red. Let's highlight those. These methyl protons right here are giving us this signal. Then finally, we have a CH2 left right here. We draw in our CH2. Let's make these green."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These methyl protons right here are giving us this signal. Then finally, we have a CH2 left right here. We draw in our CH2. Let's make these green. These two protons right here in green are giving us this signal. Let's go through and see if everything makes sense. We'll start with the green protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's make these green. These two protons right here in green are giving us this signal. Let's go through and see if everything makes sense. We'll start with the green protons. We predicted five neighbors. We look at the carbon that the green protons are on. We go to the next door carbon."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We'll start with the green protons. We predicted five neighbors. We look at the carbon that the green protons are on. We go to the next door carbon. One, two, three neighboring protons. We go to the other next door carbon. One, two, so a total of five neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We go to the next door carbon. One, two, three neighboring protons. We go to the other next door carbon. One, two, so a total of five neighbors. That's what we predicted using the N plus one rule. In reality, the magenta and red protons are in different environments. This oversimplified N plus one rule isn't exactly correct, but it corresponds to what we see on the NMR spectrum."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "One, two, so a total of five neighbors. That's what we predicted using the N plus one rule. In reality, the magenta and red protons are in different environments. This oversimplified N plus one rule isn't exactly correct, but it corresponds to what we see on the NMR spectrum. We can just go with it here. It helped us figure out the structure of the molecule. This makes sense, five neighbors here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This oversimplified N plus one rule isn't exactly correct, but it corresponds to what we see on the NMR spectrum. We can just go with it here. It helped us figure out the structure of the molecule. This makes sense, five neighbors here. Let's move on to the red protons. We predicted two neighbors for the red protons. We go to the carbon that's next door."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This makes sense, five neighbors here. Let's move on to the red protons. We predicted two neighbors for the red protons. We go to the carbon that's next door. This carbon has the red protons. We go to the next door carbon. We see two neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We go to the carbon that's next door. This carbon has the red protons. We go to the next door carbon. We see two neighbors. This makes sense. Let's look at the magenta protons. The magenta protons, we predicted two neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We see two neighbors. This makes sense. Let's look at the magenta protons. The magenta protons, we predicted two neighbors. The magenta protons are on this carbon. We go to the carbon next door. I see one, two neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The magenta protons, we predicted two neighbors. The magenta protons are on this carbon. We go to the carbon next door. I see one, two neighbors. This makes sense. Then finally, the blue protons over here. We predicted zero neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "I see one, two neighbors. This makes sense. Then finally, the blue protons over here. We predicted zero neighbors. We go to the carbon that's next door to this carbon. There are no protons on this carbon. Zero neighbors, and that's why we got our singlet over here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We predicted zero neighbors. We go to the carbon that's next door to this carbon. There are no protons on this carbon. Zero neighbors, and that's why we got our singlet over here. This makes sense. The chemical shifts make sense. The splitting makes sense."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Zero neighbors, and that's why we got our singlet over here. This makes sense. The chemical shifts make sense. The splitting makes sense. Everything seems to make sense for this ester. One thing I've noticed when students get a problem with an ester, sometimes they reverse the ester. Let me show you what I mean."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The splitting makes sense. Everything seems to make sense for this ester. One thing I've noticed when students get a problem with an ester, sometimes they reverse the ester. Let me show you what I mean. I've seen a lot of students do this on exams. What they'll do is they identify the fact they have an ester, but in this case, they would put the methyl group on the oxygen and then draw in the rest of the molecule over here. We have three carbons, so one, two, three."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me show you what I mean. I've seen a lot of students do this on exams. What they'll do is they identify the fact they have an ester, but in this case, they would put the methyl group on the oxygen and then draw in the rest of the molecule over here. We have three carbons, so one, two, three. They'll put this down for their answer on the test. Why is this wrong? This is wrong because the signal for these methyl protons, that's right next to this oxygen, this oxygen is deshielding."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have three carbons, so one, two, three. They'll put this down for their answer on the test. Why is this wrong? This is wrong because the signal for these methyl protons, that's right next to this oxygen, this oxygen is deshielding. We would get a singlet for these methyl protons, but the signal would be closer to four parts per million and not right here, not right here at two parts per million. That's a clue as to how to assemble the pieces for an ester. Think about what's next to this oxygen here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is wrong because the signal for these methyl protons, that's right next to this oxygen, this oxygen is deshielding. We would get a singlet for these methyl protons, but the signal would be closer to four parts per million and not right here, not right here at two parts per million. That's a clue as to how to assemble the pieces for an ester. Think about what's next to this oxygen here. Think about the chemical shifts and you won't make that mistake. For this NMR spectrum, our molecule has a molecular formula of C4H10O. Let's calculate the hydrogen deficiency index."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Think about what's next to this oxygen here. Think about the chemical shifts and you won't make that mistake. For this NMR spectrum, our molecule has a molecular formula of C4H10O. Let's calculate the hydrogen deficiency index. If we have four carbons, the maximum number of hydrogens we can have is two times four plus two, which is equal to 10. That's how many hydrogens we have in our molecular formula. This time we're not missing any hydrogens."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's calculate the hydrogen deficiency index. If we have four carbons, the maximum number of hydrogens we can have is two times four plus two, which is equal to 10. That's how many hydrogens we have in our molecular formula. This time we're not missing any hydrogens. The HDI is equal to zero. We wouldn't expect any double bonds or rings. No double bonds or rings for the dot structure."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This time we're not missing any hydrogens. The HDI is equal to zero. We wouldn't expect any double bonds or rings. No double bonds or rings for the dot structure. For integration, this signal is one proton, this signal is three protons, and this signal over here is six protons. Let's focus in on this signal. We have one proton."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "No double bonds or rings for the dot structure. For integration, this signal is one proton, this signal is three protons, and this signal over here is six protons. Let's focus in on this signal. We have one proton. I'm going to go ahead and draw that proton on a carbon like that. Let's think about how many neighbors that proton has. Look at the peaks."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have one proton. I'm going to go ahead and draw that proton on a carbon like that. Let's think about how many neighbors that proton has. Look at the peaks. I see one, two, three, four, five, six, seven peaks. Seven minus one is six. That proton is going to have six neighboring protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Look at the peaks. I see one, two, three, four, five, six, seven peaks. Seven minus one is six. That proton is going to have six neighboring protons. Let me go ahead and give that proton a color. This is the signal for the magenta proton. The chemical shift is somewhere around four parts per million and that's the chemical shift that we would expect if that proton was on a carbon bonded to an electronegative atom."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That proton is going to have six neighboring protons. Let me go ahead and give that proton a color. This is the signal for the magenta proton. The chemical shift is somewhere around four parts per million and that's the chemical shift that we would expect if that proton was on a carbon bonded to an electronegative atom. We have an oxygen here. I'm going to go ahead and put an oxygen right here bonded to that carbon because the oxygen is withdrawing electron density away from this proton. It's deshielding this proton, giving it a higher chemical shift."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The chemical shift is somewhere around four parts per million and that's the chemical shift that we would expect if that proton was on a carbon bonded to an electronegative atom. We have an oxygen here. I'm going to go ahead and put an oxygen right here bonded to that carbon because the oxygen is withdrawing electron density away from this proton. It's deshielding this proton, giving it a higher chemical shift. Let's move on to our next signal. We have three protons. This would be a methyl group."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's deshielding this proton, giving it a higher chemical shift. Let's move on to our next signal. We have three protons. This would be a methyl group. We have three methyl protons. How many neighboring protons for those methyl protons? We have only one peak here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This would be a methyl group. We have three methyl protons. How many neighboring protons for those methyl protons? We have only one peak here. One minus one is zero. Zero neighboring protons. The chemical shift is about 3.5 parts per million."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have only one peak here. One minus one is zero. Zero neighboring protons. The chemical shift is about 3.5 parts per million. Those methyl protons must be really close to that oxygen. We get a higher value for the chemical shift. We can put the methyl protons over here on the left side of the oxygen."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The chemical shift is about 3.5 parts per million. Those methyl protons must be really close to that oxygen. We get a higher value for the chemical shift. We can put the methyl protons over here on the left side of the oxygen. That makes sense in terms of the chemical shift. It also makes sense in terms of the number of neighbors. These three protons right here give us this signal and we have zero neighbors."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We can put the methyl protons over here on the left side of the oxygen. That makes sense in terms of the chemical shift. It also makes sense in terms of the number of neighbors. These three protons right here give us this signal and we have zero neighbors. There are no neighbors for these three protons. Let's move on to the last signal. We have six protons."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "These three protons right here give us this signal and we have zero neighbors. There are no neighbors for these three protons. Let's move on to the last signal. We have six protons. That's like two methyl groups. We have two methyl groups here and the signal is split into a doublet. We see these two peaks here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have six protons. That's like two methyl groups. We have two methyl groups here and the signal is split into a doublet. We see these two peaks here. Two minus one is one. We expect one neighbor for these two methyl groups. Obviously, it would have to be this proton in magenta because we have only these two places left."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We see these two peaks here. Two minus one is one. We expect one neighbor for these two methyl groups. Obviously, it would have to be this proton in magenta because we have only these two places left. We put these methyl groups right here and we can see we get one neighbor. Those two methyl groups are further away from the oxygen relatively because of the signal having a lower value for the chemical shift. Let's use red here."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Obviously, it would have to be this proton in magenta because we have only these two places left. We put these methyl groups right here and we can see we get one neighbor. Those two methyl groups are further away from the oxygen relatively because of the signal having a lower value for the chemical shift. Let's use red here. Red for these six protons. It's giving us this signal. We have one neighboring proton."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's use red here. Red for these six protons. It's giving us this signal. We have one neighboring proton. It's this proton in magenta right here. Let's think about the magenta proton. We said that the magenta proton would have six neighbors and that's what we see."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We have one neighboring proton. It's this proton in magenta right here. Let's think about the magenta proton. We said that the magenta proton would have six neighbors and that's what we see. Three from this methyl and then three from this methyl. Six neighbors for the magenta proton. Then finally, we said there'd be zero neighbors for the protons in blue."}, {"video_title": "Proton NMR practice 2 Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We said that the magenta proton would have six neighbors and that's what we see. Three from this methyl and then three from this methyl. Six neighbors for the magenta proton. Then finally, we said there'd be zero neighbors for the protons in blue. That's what we see for our dot structure. This molecule is an ether. This NMR spectrum represents this ether."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "And so from the Earth's point of view, or from the point of view of someone standing on Earth, you're seeing the side of the moon that is not lit up. But an obvious question is, is why doesn't that block out the sun every time we are in a new moon position? After all, the sun would be out here, 93 million miles away, and so wouldn't the moon block out the sun in that scenario? And we also talked about when the moon is on the other side of the Earth. When we would typically see a full moon, the Earth is between the moon and the sun. Why doesn't the Earth block the light from the sun that's being reflected on the moon? So the big question is, why don't we see a solar eclipse every new moon, and why don't we see a lunar eclipse every full moon?"}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "And we also talked about when the moon is on the other side of the Earth. When we would typically see a full moon, the Earth is between the moon and the sun. Why doesn't the Earth block the light from the sun that's being reflected on the moon? So the big question is, why don't we see a solar eclipse every new moon, and why don't we see a lunar eclipse every full moon? If we wanna ask the same question, looking at the scale of the Earth and the moon, we can see it right over here. And so this would be the new moon position, where the moon is between the Earth and the sun. The key explanation for why we do not have a solar eclipse every time we are in the new moon position is that the rotation of the moon around the Earth is not in the exact same plane as the rotation of the Earth around the sun."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "So the big question is, why don't we see a solar eclipse every new moon, and why don't we see a lunar eclipse every full moon? If we wanna ask the same question, looking at the scale of the Earth and the moon, we can see it right over here. And so this would be the new moon position, where the moon is between the Earth and the sun. The key explanation for why we do not have a solar eclipse every time we are in the new moon position is that the rotation of the moon around the Earth is not in the exact same plane as the rotation of the Earth around the sun. It actually turns out that the plane of the moon's rotation is at a five-degree angle with the plane of the Earth and the sun. If this is the sun here, and this is not drawn to scale by any means, this is the Earth, and so the Earth is in its orbit, something like that. The moon's orbit does not sit exactly in this plane."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "The key explanation for why we do not have a solar eclipse every time we are in the new moon position is that the rotation of the moon around the Earth is not in the exact same plane as the rotation of the Earth around the sun. It actually turns out that the plane of the moon's rotation is at a five-degree angle with the plane of the Earth and the sun. If this is the sun here, and this is not drawn to scale by any means, this is the Earth, and so the Earth is in its orbit, something like that. The moon's orbit does not sit exactly in this plane. It does not sit exactly in this plane. The moon's orbit is at a five-degree angle, is at a five-degree angle to this plane. So depending what time of year you are at and where the moon is in its cycle, the moon will often sit above or below the plane of the Earth's orbit."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "The moon's orbit does not sit exactly in this plane. It does not sit exactly in this plane. The moon's orbit is at a five-degree angle, is at a five-degree angle to this plane. So depending what time of year you are at and where the moon is in its cycle, the moon will often sit above or below the plane of the Earth's orbit. So for example, from the vantage point of the Earth, the moon will vary up to five degrees above the plane of Earth's rotation around the sun, often known as the ecliptic, and five degrees below that. And so you can see here, the shadow of the moon will only fall on the Earth when the moon is crossing through the plane of the rotation of Earth around the sun. Many times, the moon might be here or here or here, and in those times, the shadow will not hit the Earth."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "So depending what time of year you are at and where the moon is in its cycle, the moon will often sit above or below the plane of the Earth's orbit. So for example, from the vantage point of the Earth, the moon will vary up to five degrees above the plane of Earth's rotation around the sun, often known as the ecliptic, and five degrees below that. And so you can see here, the shadow of the moon will only fall on the Earth when the moon is crossing through the plane of the rotation of Earth around the sun. Many times, the moon might be here or here or here, and in those times, the shadow will not hit the Earth. Now what you do see here, depicted in this picture, would be a solar eclipse. Now there's two things that I've drawn here, and this is important for understanding a solar eclipse. What you see in those yellow lines, that's the umbra of the moon."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "Many times, the moon might be here or here or here, and in those times, the shadow will not hit the Earth. Now what you do see here, depicted in this picture, would be a solar eclipse. Now there's two things that I've drawn here, and this is important for understanding a solar eclipse. What you see in those yellow lines, that's the umbra of the moon. So if you are at that point right over here in this picture, then the moon will completely block out the sun. And it turns out, depending on the solar eclipse, where that umbra is hitting the Earth, it might only be a few hundred miles where the moon completely blocks out the sun. What you see in this dark blue color, that is the penumbra."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "What you see in those yellow lines, that's the umbra of the moon. So if you are at that point right over here in this picture, then the moon will completely block out the sun. And it turns out, depending on the solar eclipse, where that umbra is hitting the Earth, it might only be a few hundred miles where the moon completely blocks out the sun. What you see in this dark blue color, that is the penumbra. That is the outer shadow. And if you're in one of those areas, you will see the moon partially block out the sun. Now this is the exact same reason, the fact that the plane of its rotation is at a five degree angle with the plane of rotation of the Earth around the sun."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "What you see in this dark blue color, that is the penumbra. That is the outer shadow. And if you're in one of those areas, you will see the moon partially block out the sun. Now this is the exact same reason, the fact that the plane of its rotation is at a five degree angle with the plane of rotation of the Earth around the sun. That's also the reason why we don't see a lunar eclipse every 28 days. Here we see different scenarios where the moon is on the other side of Earth from the sun, when the moon is what we would call a full moon position. And as we can see, it can vary five degrees above the plane of the Earth and the sun, or five degrees below that."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "Now this is the exact same reason, the fact that the plane of its rotation is at a five degree angle with the plane of rotation of the Earth around the sun. That's also the reason why we don't see a lunar eclipse every 28 days. Here we see different scenarios where the moon is on the other side of Earth from the sun, when the moon is what we would call a full moon position. And as we can see, it can vary five degrees above the plane of the Earth and the sun, or five degrees below that. And you're only going to get a scenario of a lunar eclipse when the moon happens to fall in the shadow of the Earth. And here, once again, between the yellow lines, you see the umbra, and between the blue lines, you see the penumbra, which would be kind of a partial shadow. So next time you look up at the moon, hopefully you are armed with a lot more information to think about how it's oriented with respect to the Earth."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "And as we can see, it can vary five degrees above the plane of the Earth and the sun, or five degrees below that. And you're only going to get a scenario of a lunar eclipse when the moon happens to fall in the shadow of the Earth. And here, once again, between the yellow lines, you see the umbra, and between the blue lines, you see the penumbra, which would be kind of a partial shadow. So next time you look up at the moon, hopefully you are armed with a lot more information to think about how it's oriented with respect to the Earth. In fact, now when you look at the moon at night, you can usually look at the moon and tell which direction the sun is in. And based on that, you can even think about what time of day it is, and you could figure out east, west, north, and south. And we've talked a lot about the sun."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "So next time you look up at the moon, hopefully you are armed with a lot more information to think about how it's oriented with respect to the Earth. In fact, now when you look at the moon at night, you can usually look at the moon and tell which direction the sun is in. And based on that, you can even think about what time of day it is, and you could figure out east, west, north, and south. And we've talked a lot about the sun. In fact, this entire video and the one before it is all about light from the sun. And just to finish off with a little bit of awe, let's just appreciate how large and how far the sun is. As we mentioned before, this is a scaled representation of the Earth and the moon, the moon being roughly 240,000 miles away from Earth."}, {"video_title": "Solar and lunar eclipses.mp3", "Sentence": "And we've talked a lot about the sun. In fact, this entire video and the one before it is all about light from the sun. And just to finish off with a little bit of awe, let's just appreciate how large and how far the sun is. As we mentioned before, this is a scaled representation of the Earth and the moon, the moon being roughly 240,000 miles away from Earth. If we were to try to model the sun here, it is 400 times as far. Depending on your screen size, it would be a football field to the right of your screen, and you would have something roughly the size of a beach ball to represent the sun. If you wanted to just compare the sizes of the sun to the Earth and the moon, here you go."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And we'll start with our substrate, so on the left. Let's say we're dealing with an alkyl halide. So the carbon that's bonded to our halogen would be the alpha carbon, and the carbon next to that carbon would be the beta carbon. So we need a beta hydrogen for this reaction. The first step of an E1 elimination mechanism is loss of our leaving group. So loss of leaving group. Let me just write that in here really quickly."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So we need a beta hydrogen for this reaction. The first step of an E1 elimination mechanism is loss of our leaving group. So loss of leaving group. Let me just write that in here really quickly. And in this case, the electrons would come off onto our leaving group in the first step of the mechanism. So we're taking a bond away from this carbon, the one that I've circled in red here. So that carbon is going from being sp3 hybridized to being sp2 hybridized."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "Let me just write that in here really quickly. And in this case, the electrons would come off onto our leaving group in the first step of the mechanism. So we're taking a bond away from this carbon, the one that I've circled in red here. So that carbon is going from being sp3 hybridized to being sp2 hybridized. So now we have a carbocation. And we know that carbocations, sp2 hybridized carbons, have planar geometry around them, so I've attempted to show the planar geometry around this carbocation. So that's the first step, loss of the leaving group to form a carbocation."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So that carbon is going from being sp3 hybridized to being sp2 hybridized. So now we have a carbocation. And we know that carbocations, sp2 hybridized carbons, have planar geometry around them, so I've attempted to show the planar geometry around this carbocation. So that's the first step, loss of the leaving group to form a carbocation. In the second step, our base comes along and takes this proton, which leaves these electrons behind, and those electrons move in to form our alkene. So this is the second step of the mechanism, which is the base takes or abstracts a proton. So base takes a proton to form our alkene."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So that's the first step, loss of the leaving group to form a carbocation. In the second step, our base comes along and takes this proton, which leaves these electrons behind, and those electrons move in to form our alkene. So this is the second step of the mechanism, which is the base takes or abstracts a proton. So base takes a proton to form our alkene. And let me go ahead and highlight those electrons. So these electrons here in magenta moved in to form our double bond, and we form our product, we form our alkene. So the first step of the mechanism, the loss of the leaving group, this turns out to be the rate-determining step."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So base takes a proton to form our alkene. And let me go ahead and highlight those electrons. So these electrons here in magenta moved in to form our double bond, and we form our product, we form our alkene. So the first step of the mechanism, the loss of the leaving group, this turns out to be the rate-determining step. So this is the slow step of the mechanism. So if you're writing a rate law, the rate of this reaction would be equal to the rate constant, K, times the concentration of your substrate. So that's what studies have shown, that these mechanisms depend on the concentration of only your substrate, this over here on the left."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the first step of the mechanism, the loss of the leaving group, this turns out to be the rate-determining step. So this is the slow step of the mechanism. So if you're writing a rate law, the rate of this reaction would be equal to the rate constant, K, times the concentration of your substrate. So that's what studies have shown, that these mechanisms depend on the concentration of only your substrate, this over here on the left. So it's first order with respect to the substrate. And that's because of this rate-determining step. The loss of the leaving group is the rate-determining step, and so the concentration of your substrate, your starting material, that's what matters."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So that's what studies have shown, that these mechanisms depend on the concentration of only your substrate, this over here on the left. So it's first order with respect to the substrate. And that's because of this rate-determining step. The loss of the leaving group is the rate-determining step, and so the concentration of your substrate, your starting material, that's what matters. Your base can't do anything until you lose your leaving group. And so, since the base does not participate in the rate-determining step, it participates in the second step, the concentration of the base has no effect on the rate of the reaction. So it's the concentration of the substrate only."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "The loss of the leaving group is the rate-determining step, and so the concentration of your substrate, your starting material, that's what matters. Your base can't do anything until you lose your leaving group. And so, since the base does not participate in the rate-determining step, it participates in the second step, the concentration of the base has no effect on the rate of the reaction. So it's the concentration of the substrate only. And since it's only dependent on the concentration of the substrate, that's where the one comes from, an E1. So let me go ahead and write this out here. So an E1 mechanism, the one comes from the fact that this is a unimolecular, a unimolecular rate law here."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So it's the concentration of the substrate only. And since it's only dependent on the concentration of the substrate, that's where the one comes from, an E1. So let me go ahead and write this out here. So an E1 mechanism, the one comes from the fact that this is a unimolecular, a unimolecular rate law here. And the E comes from the fact that this is an elimination reaction. So when you see E1, that's what you're thinking about. It's an elimination reaction, and it's unimolecular."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So an E1 mechanism, the one comes from the fact that this is a unimolecular, a unimolecular rate law here. And the E comes from the fact that this is an elimination reaction. So when you see E1, that's what you're thinking about. It's an elimination reaction, and it's unimolecular. The overall rate of the reaction only depends on the concentration of your substrate. So if you increase, let's say you had, let's say this was your substrate right here, and you increase the concentration of your substrate, let me just write this down. So if you increase the concentration of your substrate by a factor of two, you would also increase the rate of reaction by a factor of two."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "It's an elimination reaction, and it's unimolecular. The overall rate of the reaction only depends on the concentration of your substrate. So if you increase, let's say you had, let's say this was your substrate right here, and you increase the concentration of your substrate, let me just write this down. So if you increase the concentration of your substrate by a factor of two, you would also increase the rate of reaction by a factor of two. So it's first order with respect to the substrate. So this is some general chemistry here. If you increase the concentration of your base by a factor of two, you would have no effect on the overall rate of the reaction."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So if you increase the concentration of your substrate by a factor of two, you would also increase the rate of reaction by a factor of two. So it's first order with respect to the substrate. So this is some general chemistry here. If you increase the concentration of your base by a factor of two, you would have no effect on the overall rate of the reaction. So let's talk about one more point here in the mechanism, and that is the formation of this carbocation. Since we have a carbocation in this mechanism, you need to think about the possibility of rearrangements in the mechanism, and you need to think what would form, what substrate would form a stable carbocation. So something like a tertiary substrate forming a tertiary carbocation would be favorable for an E1 mechanism."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "If you increase the concentration of your base by a factor of two, you would have no effect on the overall rate of the reaction. So let's talk about one more point here in the mechanism, and that is the formation of this carbocation. Since we have a carbocation in this mechanism, you need to think about the possibility of rearrangements in the mechanism, and you need to think what would form, what substrate would form a stable carbocation. So something like a tertiary substrate forming a tertiary carbocation would be favorable for an E1 mechanism. Here we have a tertiary alkyl halide. And let's say this tertiary alkyl halide undergoes an E1 elimination reaction. So the carbon that's bonded to the iodine must be our alpha carbon, and then we would have three beta carbons."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So something like a tertiary substrate forming a tertiary carbocation would be favorable for an E1 mechanism. Here we have a tertiary alkyl halide. And let's say this tertiary alkyl halide undergoes an E1 elimination reaction. So the carbon that's bonded to the iodine must be our alpha carbon, and then we would have three beta carbons. So that's a beta carbon, that's a beta carbon, and that's a beta carbon. So the first step in an E1 mechanism is loss of our leaving group. So if I draw in the lone pairs of electrons in here on iodine, I know that these electrons in this bond would come off onto iodine to form the iodide anions."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the carbon that's bonded to the iodine must be our alpha carbon, and then we would have three beta carbons. So that's a beta carbon, that's a beta carbon, and that's a beta carbon. So the first step in an E1 mechanism is loss of our leaving group. So if I draw in the lone pairs of electrons in here on iodine, I know that these electrons in this bond would come off onto iodine to form the iodide anions. So let me draw that in here. So we would make the iodide anion. And let me highlight our electrons."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So if I draw in the lone pairs of electrons in here on iodine, I know that these electrons in this bond would come off onto iodine to form the iodide anions. So let me draw that in here. So we would make the iodide anion. And let me highlight our electrons. So the electrons in this bond come off onto the iodine to form the iodide anion. And this is an excellent leaving group. Iodide is an excellent leaving group, and you know that by looking at pKa values."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And let me highlight our electrons. So the electrons in this bond come off onto the iodine to form the iodide anion. And this is an excellent leaving group. Iodide is an excellent leaving group, and you know that by looking at pKa values. The iodide anion is the conjugate base of a very strong acid, HI, with an approximate pKa value of negative 11. So HI is very good at donating a proton, which must mean that the conjugate base is very stable. So the iodide anion is an excellent leaving group."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "Iodide is an excellent leaving group, and you know that by looking at pKa values. The iodide anion is the conjugate base of a very strong acid, HI, with an approximate pKa value of negative 11. So HI is very good at donating a proton, which must mean that the conjugate base is very stable. So the iodide anion is an excellent leaving group. So if we lose the iodide anion, that means we're gonna have a carbocation. So we lost a bond to this carbon in red. So we're gonna form a carbocation."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the iodide anion is an excellent leaving group. So if we lose the iodide anion, that means we're gonna have a carbocation. So we lost a bond to this carbon in red. So we're gonna form a carbocation. Let me go ahead and draw that in. So this is a planar carbocation, and so the carbon, let me go ahead and highlight it here, the carbon in red has a plus one formal charge. It lost a bond."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So we're gonna form a carbocation. Let me go ahead and draw that in. So this is a planar carbocation, and so the carbon, let me go ahead and highlight it here, the carbon in red has a plus one formal charge. It lost a bond. So that's the first step of an E1 elimination mechanism. The second step of an E1 elimination mechanism is the base comes along, and it takes a proton from a beta carbon. So our base in this case would be ethanol."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "It lost a bond. So that's the first step of an E1 elimination mechanism. The second step of an E1 elimination mechanism is the base comes along, and it takes a proton from a beta carbon. So our base in this case would be ethanol. So let me go ahead and draw in lone pairs of electrons on the oxygen. So notice we're also heating this reaction. So the ethanol's gonna function as a base."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So our base in this case would be ethanol. So let me go ahead and draw in lone pairs of electrons on the oxygen. So notice we're also heating this reaction. So the ethanol's gonna function as a base. So ethanol's not a strong base, but it can take a proton. So let me go ahead and draw in a proton right here. And a lone pair of electrons on the oxygen is going to take this proton, and the electrons would move into here to form our alkene."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the ethanol's gonna function as a base. So ethanol's not a strong base, but it can take a proton. So let me go ahead and draw in a proton right here. And a lone pair of electrons on the oxygen is going to take this proton, and the electrons would move into here to form our alkene. So let me go ahead and draw our product. So let me put that in here, and let me highlight some electrons. So the electrons in blue moved in here to form our double bond."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And a lone pair of electrons on the oxygen is going to take this proton, and the electrons would move into here to form our alkene. So let me go ahead and draw our product. So let me put that in here, and let me highlight some electrons. So the electrons in blue moved in here to form our double bond. So a couple of points about this reaction. One point is when you're looking at SN1 mechanisms, the first step is loss of a leaving group to form your carbocation. So when you get to this carbocation, you might think, well, why is ethanol acting as a base here?"}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the electrons in blue moved in here to form our double bond. So a couple of points about this reaction. One point is when you're looking at SN1 mechanisms, the first step is loss of a leaving group to form your carbocation. So when you get to this carbocation, you might think, well, why is ethanol acting as a base here? Why couldn't it act as a nucleophile? And the answer is the ethanol certainly can act as a nucleophile, and it would attack the positively charged carbon, and you would definitely get a substitution product for this reaction as well. So if ethanol acts as a nucleophile, you're gonna get a substitution reaction, an SN1 mechanism."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So when you get to this carbocation, you might think, well, why is ethanol acting as a base here? Why couldn't it act as a nucleophile? And the answer is the ethanol certainly can act as a nucleophile, and it would attack the positively charged carbon, and you would definitely get a substitution product for this reaction as well. So if ethanol acts as a nucleophile, you're gonna get a substitution reaction, an SN1 mechanism. If the ethanol acts as a base, you're gonna get an E1 elimination mechanism. So here, we're just gonna focus on the elimination product. We won't worry about the substitution product, but we will talk about this stuff in a later video, because that would definitely happen."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So if ethanol acts as a nucleophile, you're gonna get a substitution reaction, an SN1 mechanism. If the ethanol acts as a base, you're gonna get an E1 elimination mechanism. So here, we're just gonna focus on the elimination product. We won't worry about the substitution product, but we will talk about this stuff in a later video, because that would definitely happen. All right, something else I want to talk about is we had three beta carbons over here. If I look at these three beta carbons, and I just picked one of them, right? I just said that this carbon right here, let me highlight it, I just took a proton from this carbon."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "We won't worry about the substitution product, but we will talk about this stuff in a later video, because that would definitely happen. All right, something else I want to talk about is we had three beta carbons over here. If I look at these three beta carbons, and I just picked one of them, right? I just said that this carbon right here, let me highlight it, I just took a proton from this carbon. But it doesn't matter which of those carbons that we take a proton off because of symmetry. Let me go ahead and draw this in over here. So this is my carbocation, let's say."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "I just said that this carbon right here, let me highlight it, I just took a proton from this carbon. But it doesn't matter which of those carbons that we take a proton off because of symmetry. Let me go ahead and draw this in over here. So this is my carbocation, let's say. Let's say we took a proton from this carbon. So our weak base comes along, and takes a proton from here, and these electrons would move into here. That would give us the same product, right?"}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So this is my carbocation, let's say. Let's say we took a proton from this carbon. So our weak base comes along, and takes a proton from here, and these electrons would move into here. That would give us the same product, right? So this would be, let me go ahead and highlight those electrons. So these electrons in dark blue would move in to form our double bond, but this is the same as that product. Alcohols can also react via an E1 mechanism."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "That would give us the same product, right? So this would be, let me go ahead and highlight those electrons. So these electrons in dark blue would move in to form our double bond, but this is the same as that product. Alcohols can also react via an E1 mechanism. The carbon that's bonded to the OH would be the alpha carbon, and the carbon next to that would be the beta carbon. So reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene. And sometimes phosphoric acid is used instead of sulfuric acid."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "Alcohols can also react via an E1 mechanism. The carbon that's bonded to the OH would be the alpha carbon, and the carbon next to that would be the beta carbon. So reacting an alcohol with sulfuric acid and heating up your reaction mixture will give you an alkene. And sometimes phosphoric acid is used instead of sulfuric acid. So we saw the first step of an E1 mechanism was loss of a leaving group. But if that happens here, if these electrons come off onto the oxygen, that would form hydroxide as your leaving group. And the hydroxide anion is a poor leaving group."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And sometimes phosphoric acid is used instead of sulfuric acid. So we saw the first step of an E1 mechanism was loss of a leaving group. But if that happens here, if these electrons come off onto the oxygen, that would form hydroxide as your leaving group. And the hydroxide anion is a poor leaving group. And we know that by looking at pKa values. Down here is the hydroxide anion. It is the conjugate base to water."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And the hydroxide anion is a poor leaving group. And we know that by looking at pKa values. Down here is the hydroxide anion. It is the conjugate base to water. But water is not a great acid. And we know that from the pKa value here. So water is not great at donating a proton, which means that the hydroxide anion is not that stable."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "It is the conjugate base to water. But water is not a great acid. And we know that from the pKa value here. So water is not great at donating a proton, which means that the hydroxide anion is not that stable. And since the hydroxide anion is not that stable, it's not a great leaving group. So let's go ahead and take off this arrow here because the first step is not loss of a leaving group. The first step is a proton transfer."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So water is not great at donating a proton, which means that the hydroxide anion is not that stable. And since the hydroxide anion is not that stable, it's not a great leaving group. So let's go ahead and take off this arrow here because the first step is not loss of a leaving group. The first step is a proton transfer. We have a strong acid here, sulfuric acid. And the alcohol will act as a base and take a proton from sulfuric acid. And that would form water as your leaving group."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "The first step is a proton transfer. We have a strong acid here, sulfuric acid. And the alcohol will act as a base and take a proton from sulfuric acid. And that would form water as your leaving group. And water is a much better leaving group than the hydroxide anion. And again, we know that by pKa values. Water is the conjugate base to the hydronium ion, H3O+, which is much better at donating a proton."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And that would form water as your leaving group. And water is a much better leaving group than the hydroxide anion. And again, we know that by pKa values. Water is the conjugate base to the hydronium ion, H3O+, which is much better at donating a proton. The pKa value is much, much lower. And that means that water is stable. So the first step when you are doing an E1 mechanism with an alcohol is to protonate the OH group."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "Water is the conjugate base to the hydronium ion, H3O+, which is much better at donating a proton. The pKa value is much, much lower. And that means that water is stable. So the first step when you are doing an E1 mechanism with an alcohol is to protonate the OH group. So here's our alcohol. And the carbon bonded to the OH is our alpha carbon. And then these carbons next to the alpha carbon would all be beta carbons."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the first step when you are doing an E1 mechanism with an alcohol is to protonate the OH group. So here's our alcohol. And the carbon bonded to the OH is our alpha carbon. And then these carbons next to the alpha carbon would all be beta carbons. We just saw the first step is a proton transfer. A lone pair of electrons on the oxygen take a proton from sulfuric acid. So we transfer a proton."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And then these carbons next to the alpha carbon would all be beta carbons. We just saw the first step is a proton transfer. A lone pair of electrons on the oxygen take a proton from sulfuric acid. So we transfer a proton. And let's go ahead and draw in what we would have now. So there would be a plus 1 formal charge on the oxygen. So let's highlight our electrons."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So we transfer a proton. And let's go ahead and draw in what we would have now. So there would be a plus 1 formal charge on the oxygen. So let's highlight our electrons. In magenta, these electrons took this proton to form this bond. And now we have water as a leaving group. Let me just fix this hydrogen here really fast."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So let's highlight our electrons. In magenta, these electrons took this proton to form this bond. And now we have water as a leaving group. Let me just fix this hydrogen here really fast. And these electrons can come off onto our oxygen. So that gives us water as our leaving group. And let me go ahead and draw in the water molecule here."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "Let me just fix this hydrogen here really fast. And these electrons can come off onto our oxygen. So that gives us water as our leaving group. And let me go ahead and draw in the water molecule here. And let me highlight electrons. The electrons in light blue in this bond came off onto the oxygen, which formed water. And we know water is a good leaving group."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And let me go ahead and draw in the water molecule here. And let me highlight electrons. The electrons in light blue in this bond came off onto the oxygen, which formed water. And we know water is a good leaving group. We took a bond away from this carbon in red. So that carbon would now be a carbocation. So let me draw in our carbocation here."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And we know water is a good leaving group. We took a bond away from this carbon in red. So that carbon would now be a carbocation. So let me draw in our carbocation here. So the carbon in red is now positively charged. So let me draw in a plus 1 formal charge on that carbon. The next step of our mechanism, we know a weak base comes along and takes a proton, one of the protons on one of the beta carbons over here."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So let me draw in our carbocation here. So the carbon in red is now positively charged. So let me draw in a plus 1 formal charge on that carbon. The next step of our mechanism, we know a weak base comes along and takes a proton, one of the protons on one of the beta carbons over here. So let's just say it's this one. And I'm just going to draw in a generic base, so a generic base right here, which is going to take this proton. And then these electrons are going to move in to form our product."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "The next step of our mechanism, we know a weak base comes along and takes a proton, one of the protons on one of the beta carbons over here. So let's just say it's this one. And I'm just going to draw in a generic base, so a generic base right here, which is going to take this proton. And then these electrons are going to move in to form our product. So let me draw our product in here. And let me highlight those electrons. So the electrons in, let's use green this time, the electrons in green moved in here to form our double bond, to form our alkene."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "And then these electrons are going to move in to form our product. So let me draw our product in here. And let me highlight those electrons. So the electrons in, let's use green this time, the electrons in green moved in here to form our double bond, to form our alkene. So I just put a generic base. Let me go ahead and talk about the base for a second here. I just put in a generic base."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "So the electrons in, let's use green this time, the electrons in green moved in here to form our double bond, to form our alkene. So I just put a generic base. Let me go ahead and talk about the base for a second here. I just put in a generic base. Sometimes you might see water acting as the base. Sometimes you might see HSO4 minus, the conjugate base to sulfuric acid acting as the base. Different textbooks give you different things."}, {"video_title": "E1 mechanism kinetics and substrate.mp3", "Sentence": "I just put in a generic base. Sometimes you might see water acting as the base. Sometimes you might see HSO4 minus, the conjugate base to sulfuric acid acting as the base. Different textbooks give you different things. I don't think it really matters. But one of those acting as a weak base, it's probably water, takes this proton to form your alkene. Sometimes this reaction is called a dehydration reaction since we lost water in the process."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And this mechanism occurs via an SN2 type mechanism, which means that it's only going to work with primary or secondary alcohols. And it's possible to get, and you will get inversion of configuration if you have a chirality center present in your final product. So let's take a look at the mechanism. And we'll start with our alcohol. And so the oxygen is going to have to leave somehow. But by itself, OH is not the best leaving group. And so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with our alcohol. And so the oxygen is going to have to leave somehow. But by itself, OH is not the best leaving group. And so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group. And so if we draw the dot structure for thionyl chloride, we would have the sulfur double bonded to an oxygen here. And then the sulfur is also bonded to chlorine. So I'll go ahead and put in those lone pairs of electrons on the chlorines like this."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group. And so if we draw the dot structure for thionyl chloride, we would have the sulfur double bonded to an oxygen here. And then the sulfur is also bonded to chlorine. So I'll go ahead and put in those lone pairs of electrons on the chlorines like this. And if you count up your valence electrons, turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and put in those lone pairs of electrons on the chlorines like this. And if you count up your valence electrons, turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also bonded to a hydrogen."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also bonded to a hydrogen. One lone pair of electrons form that new bond. So one pair of electrons is left behind, which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons, picked up one more lone pair, which gives it a negative 1 formal charge."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen is also bonded to a hydrogen. One lone pair of electrons form that new bond. So one pair of electrons is left behind, which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons, picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorines. So we can go ahead and draw those in. And we can go ahead and draw in that lone pair of electrons on that sulfur like that."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "Connected to the sulfur, this top oxygen here had two lone pairs of electrons, picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorines. So we can go ahead and draw those in. And we can go ahead and draw in that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. So these electrons are going to kick in here. And these electrons would kick off onto that chlorine."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And we can go ahead and draw in that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. So these electrons are going to kick in here. And these electrons would kick off onto that chlorine. So when we draw the next intermediate here, we would now have our oxygen still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again, with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And these electrons would kick off onto that chlorine. So when we draw the next intermediate here, we would now have our oxygen still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again, with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now. So one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "The sulfur is also bonded to one chlorine now. So one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. And it's a negatively charged chloride anion. And then still, there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So one of the chlorines left here. And it's a negatively charged chloride anion. And then still, there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine, it's a base. And so it looks like a benzene ring, except we have a nitrogen here instead. And there would be a lone pair of electrons on this nitrogen."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine, it's a base. And so it looks like a benzene ring, except we have a nitrogen here instead. And there would be a lone pair of electrons on this nitrogen. And so that lone pair of electrons is going to function as a Bronsted-Lowry base and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that, let's go ahead and get some more room here."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And there would be a lone pair of electrons on this nitrogen. And so that lone pair of electrons is going to function as a Bronsted-Lowry base and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that, let's go ahead and get some more room here. So what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So when we go ahead and draw that, let's go ahead and get some more room here. So what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it. And we have our sulfur and our chlorine and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "Our oxygen now has two lone pairs of electrons around it. And we have our sulfur and our chlorine and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, oxygen being more electrically charged. So negative will be partially negative. And this carbon here will be partially positive."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, oxygen being more electrically charged. So negative will be partially negative. And this carbon here will be partially positive. So carbon will be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon here will be partially positive. So carbon will be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism. So that's going to be the nucleophile. And it's going to attack our partially positive carbon, an SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "Our nucleophile will be this chloride anion up here that we formed in the mechanism. So that's going to be the nucleophile. And it's going to attack our partially positive carbon, an SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here. And then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added onto our carbon on the left."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here. And then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added onto our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxide, so SO2, and also the chloride anion, so Cl minus like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So when we draw the product, we can go ahead and show the chlorine has now added onto our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxide, so SO2, and also the chloride anion, so Cl minus like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here. And if we react ethanol with thionyl chloride, SOCl2, and we add some pyridine as our base, we're going to replace the OH with our chlorine like that."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here. And if we react ethanol with thionyl chloride, SOCl2, and we add some pyridine as our base, we're going to replace the OH with our chlorine like that. And so once again, if we look at our alcohol, this is a primary alcohol. And so primary alcohols will work the best because there's decreased steric hindrance. And we don't have to worry about stereochemistry since we don't have any chirality centers in our product."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And if we react ethanol with thionyl chloride, SOCl2, and we add some pyridine as our base, we're going to replace the OH with our chlorine like that. And so once again, if we look at our alcohol, this is a primary alcohol. And so primary alcohols will work the best because there's decreased steric hindrance. And we don't have to worry about stereochemistry since we don't have any chirality centers in our product. Let's look at a way to form an alkyl bromide. So we just formed an alkyl chloride. Let's look at the general reaction for forming an alkyl bromide here."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And we don't have to worry about stereochemistry since we don't have any chirality centers in our product. Let's look at a way to form an alkyl bromide. So we just formed an alkyl chloride. Let's look at the general reaction for forming an alkyl bromide here. So I go ahead and have my alcohol. And I react that with phosphorus tribromide, so PBr3. And I'm going to get the OH group is going to leave."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the general reaction for forming an alkyl bromide here. So I go ahead and have my alcohol. And I react that with phosphorus tribromide, so PBr3. And I'm going to get the OH group is going to leave. And I'm going to put a bromine in its place. And once again, this mechanism is an SN2 type mechanism, so primary or secondary alcohols only. And possible inversion of configuration for your products depending on whether chirality centers are present or not."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to get the OH group is going to leave. And I'm going to put a bromine in its place. And once again, this mechanism is an SN2 type mechanism, so primary or secondary alcohols only. And possible inversion of configuration for your products depending on whether chirality centers are present or not. So another SN2 mechanism. And again, we need to use phosphorus tribromide because the OH group is not the best leaving group. So we look at the mechanism."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And possible inversion of configuration for your products depending on whether chirality centers are present or not. So another SN2 mechanism. And again, we need to use phosphorus tribromide because the OH group is not the best leaving group. So we look at the mechanism. When we look at this mechanism here, let's go ahead and show that lone pair of electrons better like that. We have phosphorus tribromide. So I'm going to go ahead and draw the dot structure."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So we look at the mechanism. When we look at this mechanism here, let's go ahead and show that lone pair of electrons better like that. We have phosphorus tribromide. So I'm going to go ahead and draw the dot structure. So we would have these bromines here with lone pairs of electrons. And there's three of them. So I'll go ahead and put in those bromines."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to go ahead and draw the dot structure. So we would have these bromines here with lone pairs of electrons. And there's three of them. So I'll go ahead and put in those bromines. And we still have two more valence electrons to account for. And those would go on our phosphorus like that. So the first step, it's analogous to our previous mechanism."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and put in those bromines. And we still have two more valence electrons to account for. And those would go on our phosphorus like that. So the first step, it's analogous to our previous mechanism. Lone pair of electrons and oxygen are going to form a bond with phosphorus. And that would kick these electrons off onto one of the bromines. So I just chose that one."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So the first step, it's analogous to our previous mechanism. Lone pair of electrons and oxygen are going to form a bond with phosphorus. And that would kick these electrons off onto one of the bromines. So I just chose that one. It doesn't really matter which one you choose. And so when we show the result of that, we would now have our oxygen bonded to a phosphorus. The oxygen is still bonded to a hydrogen."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So I just chose that one. It doesn't really matter which one you choose. And so when we show the result of that, we would now have our oxygen bonded to a phosphorus. The oxygen is still bonded to a hydrogen. There's still a lone pair of electrons left behind in this oxygen, which give this oxygen a plus informal charge. And the phosphorus is now bonded to only two bromines. So we can show the phosphorus bonded to only two bromines here."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen is still bonded to a hydrogen. There's still a lone pair of electrons left behind in this oxygen, which give this oxygen a plus informal charge. And the phosphorus is now bonded to only two bromines. So we can show the phosphorus bonded to only two bromines here. So I can go ahead and put those in. And I can put in the lone pair of electrons on phosphorus as well. And so we lost one of our bromines."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So we can show the phosphorus bonded to only two bromines here. So I can go ahead and put those in. And I can put in the lone pair of electrons on phosphorus as well. And so we lost one of our bromines. And that formed a bromide anion. So we go ahead and draw in our bromide anion here. And once again, we've made a better leaving group."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And so we lost one of our bromines. And that formed a bromide anion. So we go ahead and draw in our bromide anion here. And once again, we've made a better leaving group. So all this stuff here on the right is a better leaving group than the OH. And so we can think about our SN2 type mechanism, where once again, the carbon, this carbon right here, is going to be electrophilic. So it wants negatively charged electrons, which it could get from the bromide anion, so nucleophilic attack."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we've made a better leaving group. So all this stuff here on the right is a better leaving group than the OH. And so we can think about our SN2 type mechanism, where once again, the carbon, this carbon right here, is going to be electrophilic. So it wants negatively charged electrons, which it could get from the bromide anion, so nucleophilic attack. And then that would kick these electrons in here off onto your oxygen. And we can go ahead and show our products. So when we draw the products, all we need to do now is show we have substituted."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So it wants negatively charged electrons, which it could get from the bromide anion, so nucleophilic attack. And then that would kick these electrons in here off onto your oxygen. And we can go ahead and show our products. So when we draw the products, all we need to do now is show we have substituted. The bromine is now attached to the carbon like that. And our other product would be a hydrogen attached to an oxygen. And the oxygen we attach to phosphorus."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "So when we draw the products, all we need to do now is show we have substituted. The bromine is now attached to the carbon like that. And our other product would be a hydrogen attached to an oxygen. And the oxygen we attach to phosphorus. And the phosphorus is attached to two bromines like that. So we've formed our alkyl bromide. So if we just show a quick example, once again, we'll start with ethanol."}, {"video_title": "Preparation of alkyl halides from alcohols Organic chemistry Khan Academy.mp3", "Sentence": "And the oxygen we attach to phosphorus. And the phosphorus is attached to two bromines like that. So we've formed our alkyl bromide. So if we just show a quick example, once again, we'll start with ethanol. If we wanted to convert ethanol into ethyl bromides, all we would have to do is add phosphorus tribromide like that. And of course, we're going to replace the OH with a bromine. And so that's one way to prepare alkyl halides from alcohols."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And just as a review, acetic acid looks like this. And the common name, as I just said, is acetic acid. And if you want to use the systematic name, you look for the longest chain, which is right over there. There's two carbons, so we use the eth prefix. So it's ethane. And since this is a carboxylic acid, it is ethanoic acid. Now, the derivatives of acetic acid, and we can later generalize this to all carboxylic acids."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "There's two carbons, so we use the eth prefix. So it's ethane. And since this is a carboxylic acid, it is ethanoic acid. Now, the derivatives of acetic acid, and we can later generalize this to all carboxylic acids. We really just have to change what's going on on this carbon chain right here. It won't have to necessarily just be two carbons. It could just keep going, have benzene rings, whatever, and that would change the name."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "Now, the derivatives of acetic acid, and we can later generalize this to all carboxylic acids. We really just have to change what's going on on this carbon chain right here. It won't have to necessarily just be two carbons. It could just keep going, have benzene rings, whatever, and that would change the name. But really, I just want to give you the gist and the gist of the naming. So if we were to replace this hydroxyl group with an amine, and in future videos, we'll see how that is done. So let me just draw the acyl group."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "It could just keep going, have benzene rings, whatever, and that would change the name. But really, I just want to give you the gist and the gist of the naming. So if we were to replace this hydroxyl group with an amine, and in future videos, we'll see how that is done. So let me just draw the acyl group. So the acyl group is just that right over there. And we're just going to keep changing what's bonded to the acyl group right over here. So if this is bonded to an amine, so let me draw, well, this would be the simplest amine right over here, which would be NH2."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So let me just draw the acyl group. So the acyl group is just that right over there. And we're just going to keep changing what's bonded to the acyl group right over here. So if this is bonded to an amine, so let me draw, well, this would be the simplest amine right over here, which would be NH2. This thing right here is called an amide or an amide. And if we were to give this its common name, this would be acetamide. This particular example would be acetamide."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if this is bonded to an amine, so let me draw, well, this would be the simplest amine right over here, which would be NH2. This thing right here is called an amide or an amide. And if we were to give this its common name, this would be acetamide. This particular example would be acetamide. And if we wanted the systematic name for it, it would be ethanamide. You have two carbons right there. So it is ethanamide."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "This particular example would be acetamide. And if we wanted the systematic name for it, it would be ethanamide. You have two carbons right there. So it is ethanamide. Now, the natural question is, all amines won't just be primary. You might have other things other than hydrogens attached to it, other radical groups, other carbon chains. So how do you name those?"}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So it is ethanamide. Now, the natural question is, all amines won't just be primary. You might have other things other than hydrogens attached to it, other radical groups, other carbon chains. So how do you name those? And so if you had a molecule that looked like this, and actually, let me just change things up a little bit so that we diverge a little bit from the ethane root. So let's say you had three carbons bonded or part of the acyl group right there. And then we are bonded to a nitrogen, which is bonded to a methyl group and then another hydrogen."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So how do you name those? And so if you had a molecule that looked like this, and actually, let me just change things up a little bit so that we diverge a little bit from the ethane root. So let's say you had three carbons bonded or part of the acyl group right there. And then we are bonded to a nitrogen, which is bonded to a methyl group and then another hydrogen. In this case, you start naming with this methyl group right here. And to show that that methyl group is attached to the nitrogen, you call this N-methyl. And then you look at the chain that forms the acyl group, the carbon chain."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then we are bonded to a nitrogen, which is bonded to a methyl group and then another hydrogen. In this case, you start naming with this methyl group right here. And to show that that methyl group is attached to the nitrogen, you call this N-methyl. And then you look at the chain that forms the acyl group, the carbon chain. We have one, two, three carbons. So it is propanamide. If you had another methyl here, you would say N, N-dimethyl."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then you look at the chain that forms the acyl group, the carbon chain. We have one, two, three carbons. So it is propanamide. If you had another methyl here, you would say N, N-dimethyl. If you had a methyl here and a propyl group here, you would call it N-methyl, N-propyl propanamide. So hopefully that gives you a sense of amides. Now, and this is something we've seen before, so it's a little bit of a review."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "If you had another methyl here, you would say N, N-dimethyl. If you had a methyl here and a propyl group here, you would call it N-methyl, N-propyl propanamide. So hopefully that gives you a sense of amides. Now, and this is something we've seen before, so it's a little bit of a review. If you have something that looks like this, I'll have it attached to a methyl group right over here. We've seen this before. This is an ester."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "Now, and this is something we've seen before, so it's a little bit of a review. If you have something that looks like this, I'll have it attached to a methyl group right over here. We've seen this before. This is an ester. And let me actually make the part that makes it an ester in blue to differentiate it. We keep substituting what is attached to the acyl group. Let me label it."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "This is an ester. And let me actually make the part that makes it an ester in blue to differentiate it. We keep substituting what is attached to the acyl group. Let me label it. This right here is called an acyl group. That right there is an acyl group. So right over here for the ester, if we were to give it its common name, and we've seen this ester before, it is acetate, and if we wanted to give it its systematic name, you look at the longest chain, one, two carbons, so it is ethane, and you don't call it ethanoic acid anymore."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "Let me label it. This right here is called an acyl group. That right there is an acyl group. So right over here for the ester, if we were to give it its common name, and we've seen this ester before, it is acetate, and if we wanted to give it its systematic name, you look at the longest chain, one, two carbons, so it is ethane, and you don't call it ethanoic acid anymore. You call it ethanoate, just like that. Now, the next one, and we haven't seen this one before, and it looks complex, but when you really break it down into its constituents, it's not so bad. So let's say we have a molecule that looks like this."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So right over here for the ester, if we were to give it its common name, and we've seen this ester before, it is acetate, and if we wanted to give it its systematic name, you look at the longest chain, one, two carbons, so it is ethane, and you don't call it ethanoic acid anymore. You call it ethanoate, just like that. Now, the next one, and we haven't seen this one before, and it looks complex, but when you really break it down into its constituents, it's not so bad. So let's say we have a molecule that looks like this. So we have one acyl group, and then you have bonded to an oxygen, which is bonded to another acyl group. So it's almost like you have two carboxylic acids that have been joined together. And you really do have two acyl groups joined by an oxygen here."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So let's say we have a molecule that looks like this. So we have one acyl group, and then you have bonded to an oxygen, which is bonded to another acyl group. So it's almost like you have two carboxylic acids that have been joined together. And you really do have two acyl groups joined by an oxygen here. This is called an anhydride. Anhydride. And they look very complex, but you just have to realize they're two carboxylic acids attached to each other, and usually the same one."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And you really do have two acyl groups joined by an oxygen here. This is called an anhydride. Anhydride. And they look very complex, but you just have to realize they're two carboxylic acids attached to each other, and usually the same one. Most anhydrides you're going to see in organic chemistry are formed from the same carboxylic acid. So however many carbons you have on this end, you're normally going to have on this end. So the way to name these is you name it just the same way that you would have named the carboxylic acid, but instead of writing the word acid, you write the word anhydride."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And they look very complex, but you just have to realize they're two carboxylic acids attached to each other, and usually the same one. Most anhydrides you're going to see in organic chemistry are formed from the same carboxylic acid. So however many carbons you have on this end, you're normally going to have on this end. So the way to name these is you name it just the same way that you would have named the carboxylic acid, but instead of writing the word acid, you write the word anhydride. So this right here would be acetic anhydride. It's derived from acetic acid. This right here is acetic anhydride."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So the way to name these is you name it just the same way that you would have named the carboxylic acid, but instead of writing the word acid, you write the word anhydride. So this right here would be acetic anhydride. It's derived from acetic acid. This right here is acetic anhydride. Or the systematic name is you have one, two carbons, so it's ethanoic anhydride. Ethanoic anhydride. And just to make things clear, if this molecule, instead of that, if we had something that looked like this, where the carbon chains on either end had three carbons, one, two, three, we would call this propanoic anhydride."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "This right here is acetic anhydride. Or the systematic name is you have one, two carbons, so it's ethanoic anhydride. Ethanoic anhydride. And just to make things clear, if this molecule, instead of that, if we had something that looked like this, where the carbon chains on either end had three carbons, one, two, three, we would call this propanoic anhydride. In the unusual circumstance, and it is unusual, where you would see different carbon chains here, you would list each of them. So if this had two here and three here, it would be ethanoic propanoic anhydride. But that is very, very unusual."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And just to make things clear, if this molecule, instead of that, if we had something that looked like this, where the carbon chains on either end had three carbons, one, two, three, we would call this propanoic anhydride. In the unusual circumstance, and it is unusual, where you would see different carbon chains here, you would list each of them. So if this had two here and three here, it would be ethanoic propanoic anhydride. But that is very, very unusual. Normally, these carbon chains on either end of the, or both acyl groups, will contain the same number of carbons. Now the last carboxylic acid derivative that you should know about, and we've already seen it, are the acyl halides, and in particular, the acyl chlorides. So let me draw it right over here."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "But that is very, very unusual. Normally, these carbon chains on either end of the, or both acyl groups, will contain the same number of carbons. Now the last carboxylic acid derivative that you should know about, and we've already seen it, are the acyl halides, and in particular, the acyl chlorides. So let me draw it right over here. So you have your acyl group right there, and then it is bonded to a chlorine. And this right here is an acyl chloride. Maybe the most intuitive name."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw it right over here. So you have your acyl group right there, and then it is bonded to a chlorine. And this right here is an acyl chloride. Maybe the most intuitive name. This right here is an acyl group, and then you have a chlorine. So it's an acyl chloride. And we've seen this exact acyl chloride."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "Maybe the most intuitive name. This right here is an acyl group, and then you have a chlorine. So it's an acyl chloride. And we've seen this exact acyl chloride. It's derived from acetic acid. So this is acetyl chloride. And this right here is acetyl chloride."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And we've seen this exact acyl chloride. It's derived from acetic acid. So this is acetyl chloride. And this right here is acetyl chloride. But if you wanted to give it its systematic name, and we haven't seen its systematic name before, we have one, two carbons. So it is ethanoil. This tells us that we are dealing with an acyl group."}, {"video_title": "Amides, anhydrides, esters, and acyl chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And this right here is acetyl chloride. But if you wanted to give it its systematic name, and we haven't seen its systematic name before, we have one, two carbons. So it is ethanoil. This tells us that we are dealing with an acyl group. Ethanoil chloride is how we would name this. And if this had three carbons, it would be propanoil chloride. So hopefully that gives you at least a good introduction to the differences in structures of all these groups and an introduction to naming them."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Right now, the prevailing theory of how the universe came about is commonly called the Big Bang Theory. Big Bang Theory. And really, it's just this idea that the universe started as kind of this infinitely small point, this infinitely small singularity, and then it just had a big bang, or it just expanded from that state to the universe that we know right now. And when I first imagined this, it's also a byproduct of how it's named Big Bang. You kind of imagine this type of explosion. You kind of imagine this type of explosion, that everything was infinitely packed in together, and then it exploded, and then it exploded outward. And then as all of the matter exploded outward, it started to condense, and then you have these little galaxies and super clusters of galaxies, and they started to condense, and within them, planets condensed and stars condensed, and we have the type of universe that we have right now."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And when I first imagined this, it's also a byproduct of how it's named Big Bang. You kind of imagine this type of explosion. You kind of imagine this type of explosion, that everything was infinitely packed in together, and then it exploded, and then it exploded outward. And then as all of the matter exploded outward, it started to condense, and then you have these little galaxies and super clusters of galaxies, and they started to condense, and within them, planets condensed and stars condensed, and we have the type of universe that we have right now. But this model for visualizing the Big Bang has a couple of problems. One is, when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then as all of the matter exploded outward, it started to condense, and then you have these little galaxies and super clusters of galaxies, and they started to condense, and within them, planets condensed and stars condensed, and we have the type of universe that we have right now. But this model for visualizing the Big Bang has a couple of problems. One is, when we talk about the Big Bang, we're not talking about the matter, just the mass or just the matter in the universe being in one point. We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding, but you're like, well, look, isn't it expanding into something else?"}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking about actual space expanding. So we're not just talking about something inside of space, like the physical mass, the physical matter expanding. We're talking about space itself. And so when you have this type of model, you have all of this stuff expanding, but you're like, well, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding?"}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so when you have this type of model, you have all of this stuff expanding, but you're like, well, look, isn't it expanding into something else? Maybe if the furthest out parts of this matter is right over here, what's this stuff over here? And so you say, well, wouldn't that be space? So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, is that one, the universe does not have an edge, and two, there is no outside space."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So how can you say space itself is expanding? And another idea that a Big Bang also implies is if this is the furthest stuff out there, would this be the edge of the universe? Does the universe have an edge? And the answer to either of those questions, and that's what we're going to try to tackle in this, is that one, the universe does not have an edge, and two, there is no outside space. We are not expanding into another space, and I'm going to explain that. Hopefully we'll see why that is the case right now. So the best way to view it, and we're going to view it by analogy, if I were to tell you that I have a two-dimensional space that has a finite area, so it's not infinite, and it also has no edge, this once again, when you first look at it, it seems difficult."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the answer to either of those questions, and that's what we're going to try to tackle in this, is that one, the universe does not have an edge, and two, there is no outside space. We are not expanding into another space, and I'm going to explain that. Hopefully we'll see why that is the case right now. So the best way to view it, and we're going to view it by analogy, if I were to tell you that I have a two-dimensional space that has a finite area, so it's not infinite, and it also has no edge, this once again, when you first look at it, it seems difficult. How do I construct something that has a finite area but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved?"}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the best way to view it, and we're going to view it by analogy, if I were to tell you that I have a two-dimensional space that has a finite area, so it's not infinite, and it also has no edge, this once again, when you first look at it, it seems difficult. How do I construct something that has a finite area but still has no edge? Every time I try to draw an area, it looks like I have to have some edges. And then you might remember, what if that two-dimensional space is curved? What happens? And I think the easiest example of that is the surface of a sphere. The surface of a sphere, let me draw a sphere over here."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then you might remember, what if that two-dimensional space is curved? What happens? And I think the easiest example of that is the surface of a sphere. The surface of a sphere, let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitude and lines on this sphere. On this sphere, all of a sudden, and I'll shade it in a little bit, make it look nice."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The surface of a sphere, let me draw a sphere over here. So this right here is a sphere. Let me draw some longitude and latitude and lines on this sphere. On this sphere, all of a sudden, and I'll shade it in a little bit, make it look nice. This type of a sphere, you have a finite area. You could imagine the surface of a balloon or the surface of a bubble or the surface of the Earth. You have a finite area, but you have no edge."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "On this sphere, all of a sudden, and I'll shade it in a little bit, make it look nice. This type of a sphere, you have a finite area. You could imagine the surface of a balloon or the surface of a bubble or the surface of the Earth. You have a finite area, but you have no edge. If you keep going forever in one direction, you're just going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and, and I don't want to say finite area anymore because we're not talking about a three-dimensional space. Let me draw it over here."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have a finite area, but you have no edge. If you keep going forever in one direction, you're just going to go all the way around and come back to the other side. Now, to imagine a three-dimensional space that has these same properties, a finite area and, and I don't want to say finite area anymore because we're not talking about a three-dimensional space. Let me draw it over here. So let's think about a three-dimensional space. So three-dimensional space. Instead of area, since we're in three dimensions now, I want to talk about a finite volume, finite volume, and no edge."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me draw it over here. So let's think about a three-dimensional space. So three-dimensional space. Instead of area, since we're in three dimensions now, I want to talk about a finite volume, finite volume, and no edge. How do I do that? When you think about it superficially, well, look, if I have a finite volume, it's going to be in, maybe it'll be contained in some type of a cube, and then we clearly have edges in those situations, or you could even think about a finite volume as being the inside of a sphere, and that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has finite volume and no edge?"}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Instead of area, since we're in three dimensions now, I want to talk about a finite volume, finite volume, and no edge. How do I do that? When you think about it superficially, well, look, if I have a finite volume, it's going to be in, maybe it'll be contained in some type of a cube, and then we clearly have edges in those situations, or you could even think about a finite volume as being the inside of a sphere, and that clearly has an edge, this entire surface over there. So how do you construct a three-dimensional space that has finite volume and no edge? And that, I'm going to tell you right now, it's very hard for us to visualize it, but in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere, but, so let me make it clear. This is a two-dimensional surface, right, on the surface of the sphere, you can only move in two directions, two perpendicular directions."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So how do you construct a three-dimensional space that has finite volume and no edge? And that, I'm going to tell you right now, it's very hard for us to visualize it, but in order to visualize it, I'm essentially going to draw the same thing as I drew right here. What you have to imagine, and you almost have to imagine it by analogy unless you have some type of a profound brain wired for more than three spatial dimensions, is a sphere, but, so let me make it clear. This is a two-dimensional surface, right, on the surface of the sphere, you can only move in two directions, two perpendicular directions. You can move like that, or you can move like that. You can move left and right, or you can move up and down. So it's a two-dimensional surface of a three-dimensional sphere."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is a two-dimensional surface, right, on the surface of the sphere, you can only move in two directions, two perpendicular directions. You can move like that, or you can move like that. You can move left and right, or you can move up and down. So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface, and you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's a two-dimensional surface of a three-dimensional sphere. So if we take it by analogy, let's imagine, and it's hard to imagine, a three-dimensional surface, and you can do it mathematically. The math here is actually not that difficult. It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way, so if we kind of view those three dimensions as just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that, and I'm not saying that this is actually the shape of the universe."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's a three-dimensional surface of a four-dimensional sphere. And I'm going to draw it the same way, so if we kind of view those three dimensions as just these two dimensions of the surface, the same thing. It's the same thing. And if you imagine that, and I'm not saying that this is actually the shape of the universe. We don't know the actual shape, but we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if you imagine that, and I'm not saying that this is actually the shape of the universe. We don't know the actual shape, but we do know that it does have a slight curvature. We don't know the actual shape, but a sphere is the simplest. There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear. We actually don't even know whether it has just a finite volume."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's other ones we could do. A toroid would also fit the bill of having a finite volume with no edge. And another thing, I want to make it clear. We actually don't even know whether it has just a finite volume. That's still an open question, but what I want to do is show you that it can have a finite volume and also have no edge. And most people believe, and I want to say believe here because we can just go based on evidence and all of that, that we are talking about something with a finite volume, especially when you talk about the Big Bang Theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We actually don't even know whether it has just a finite volume. That's still an open question, but what I want to do is show you that it can have a finite volume and also have no edge. And most people believe, and I want to say believe here because we can just go based on evidence and all of that, that we are talking about something with a finite volume, especially when you talk about the Big Bang Theory. That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite. Now, if you have this, let's imagine this sphere. Once again, if you're on this surface of this four-dimensional sphere, I obviously cannot draw a four-dimensional sphere, but if you're on the surface of this four-dimensional sphere, if you go in any direction, you will come back out and come back to where you started. If you go that way, you'll come back around here."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That kind of, on some dimension, implies a finite volume, although it could be a super large, unfathomably large volume, it is finite. Now, if you have this, let's imagine this sphere. Once again, if you're on this surface of this four-dimensional sphere, I obviously cannot draw a four-dimensional sphere, but if you're on the surface of this four-dimensional sphere, if you go in any direction, you will come back out and come back to where you started. If you go that way, you'll come back around here. Now, the universe is super huge, so even light, maybe light itself would take an unbelievable amount of time to traverse it. And if this sphere itself is expanding, it might be expanding so fast that light might not ever be able to come back around it, but in theory, if something were fast enough, if something were to keep going around, it could eventually go back to this point. Now, when we talk about a three-dimensional surface, it's a three-dimensional surface of a four-dimensional sphere, that means that any of the three dimensions, over here on the surface, I can only draw two, but that means if this is true, if the universe is a three-dimensional surface of a four-dimensional sphere, that means that if you go up and you just keep going up, you'll eventually come back from the bottom."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you go that way, you'll come back around here. Now, the universe is super huge, so even light, maybe light itself would take an unbelievable amount of time to traverse it. And if this sphere itself is expanding, it might be expanding so fast that light might not ever be able to come back around it, but in theory, if something were fast enough, if something were to keep going around, it could eventually go back to this point. Now, when we talk about a three-dimensional surface, it's a three-dimensional surface of a four-dimensional sphere, that means that any of the three dimensions, over here on the surface, I can only draw two, but that means if this is true, if the universe is a three-dimensional surface of a four-dimensional sphere, that means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were, and if you were to go into the page, so if you were to go into the page, let me draw it that way, if you were to go into the page, you would eventually come back from above the page and come back to the point that you are."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, when we talk about a three-dimensional surface, it's a three-dimensional surface of a four-dimensional sphere, that means that any of the three dimensions, over here on the surface, I can only draw two, but that means if this is true, if the universe is a three-dimensional surface of a four-dimensional sphere, that means that if you go up and you just keep going up, you'll eventually come back from the bottom. So if you keep going all the way up, you'll eventually come back to the point that you were. It might be an unbelievably large distance, but you'll eventually get back where you were. If you go to the right, you'll eventually come back all the way around to the point where you were, and if you were to go into the page, so if you were to go into the page, let me draw it that way, if you were to go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be, that you would eventually get back to where you are. So let's go back to the question of an expanding universe. An expanding universe that's not expanding into any other space, that is all of the space, but it's still expanding."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you go to the right, you'll eventually come back all the way around to the point where you were, and if you were to go into the page, so if you were to go into the page, let me draw it that way, if you were to go into the page, you would eventually come back from above the page and come back to the point that you are. So that's what this implication would be, that you would eventually get back to where you are. So let's go back to the question of an expanding universe. An expanding universe that's not expanding into any other space, that is all of the space, but it's still expanding. Well, this is the model. So you could imagine, shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe it was a little small four-dimensional sphere."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "An expanding universe that's not expanding into any other space, that is all of the space, but it's still expanding. Well, this is the model. So you could imagine, shortly after the Big Bang, our four-dimensional sphere looked like this. Maybe it was a little small four-dimensional sphere. At some, you know, maybe, you know, right at the Big Bang, it was like this little unbelievably small sphere, then a little bit later, it's this larger sphere. Let me just shade it in to show you that it has three, that it has, that it's kind of popping out of the page, that it's a sphere. And then at a later time, the sphere might look like this."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe it was a little small four-dimensional sphere. At some, you know, maybe, you know, right at the Big Bang, it was like this little unbelievably small sphere, then a little bit later, it's this larger sphere. Let me just shade it in to show you that it has three, that it has, that it's kind of popping out of the page, that it's a sphere. And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of the sphere? Isn't that some type of a space that it's expanding into?"}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then at a later time, the sphere might look like this. The sphere might look like this. Now, your temptation might be to say, wait, Sal, isn't this stuff outside of the sphere? Isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say, if you're talking in three dimensions, no, it's not. The entire universe is this surface."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Isn't that some type of a space that it's expanding into? Isn't that somehow part of the universe? And I would say, if you're talking in three dimensions, no, it's not. The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space-time, so you can, you know, one way to view the fourth dimension is it is time itself, things are just getting further and further apart."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The entire universe is this surface. It is this surface of this four-dimensional sphere. If you start talking about more dimensions, then yes, you could talk about maybe things outside of our three-dimensional universe. So as this expands in space-time, so you can, you know, one way to view the fourth dimension is it is time itself, things are just getting further and further apart. And I'll talk about more evidence in future videos for why we, why the Big Bang is the best theory we have out there right now. But as you can imagine, if we have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just, let me draw three points. Let's say those are three points."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So as this expands in space-time, so you can, you know, one way to view the fourth dimension is it is time itself, things are just getting further and further apart. And I'll talk about more evidence in future videos for why we, why the Big Bang is the best theory we have out there right now. But as you can imagine, if we have two points on this sphere that are that far apart, as this sphere expands, this four-dimensional sphere, as this bubble blows up or this balloon blows up, those two points are just, let me draw three points. Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that, or one of the first reasons why it made sense to believe in the Big Bang is that everything is expanding not from some central point, but everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away."}, {"video_title": "Big bang introduction Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say those are three points. Those three points are just going to get further and further apart. And that's actually one of the main points that, or one of the first reasons why it made sense to believe in the Big Bang is that everything is expanding not from some central point, but everything is expanding from everything. That if you go in any direction from any point in the universe, everything else is expanding away. And the further away you go, it looks like the faster it's expanding away from you. So I'll leave you there. Something for you to kind of think about a little bit."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the left, I have my thiol. And to that thiol, I'm going to add sodium hydroxide. And the sodium hydroxide is going to deprotonate the thiol, which is then going to react with this alkyl halide in the second step of the reaction to produce my sulfide as my product. So here's my sulfide right here. This reaction is the analog of the Williamson ether synthesis, which we've seen in an earlier video. So in that video, we started off with an alcohol. And we reacted our alcohol with a strong base in the first step and an alkyl halide in the second step."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So here's my sulfide right here. This reaction is the analog of the Williamson ether synthesis, which we've seen in an earlier video. So in that video, we started off with an alcohol. And we reacted our alcohol with a strong base in the first step and an alkyl halide in the second step. And we formed an ether as our product. So we can go ahead and draw our ether in here like that. So the thiol is the sulfur analog to an alcohol."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we reacted our alcohol with a strong base in the first step and an alkyl halide in the second step. And we formed an ether as our product. So we can go ahead and draw our ether in here like that. So the thiol is the sulfur analog to an alcohol. And a sulfide is the sulfur analog to an ether. Let's look at the mechanism to make sulfides. So if I start with my thiol right here, so I have carbon bonded to sulfur, bonded to a hydrogen, and then two lone pairs of electrons on that sulfur."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the thiol is the sulfur analog to an alcohol. And a sulfide is the sulfur analog to an ether. Let's look at the mechanism to make sulfides. So if I start with my thiol right here, so I have carbon bonded to sulfur, bonded to a hydrogen, and then two lone pairs of electrons on that sulfur. And if I think about the difference in electronegativity between carbon and sulfur, there's actually not much of a difference in terms of numbers. So this is not a very polar bond, actually. That's different from what we see with an alcohol."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if I start with my thiol right here, so I have carbon bonded to sulfur, bonded to a hydrogen, and then two lone pairs of electrons on that sulfur. And if I think about the difference in electronegativity between carbon and sulfur, there's actually not much of a difference in terms of numbers. So this is not a very polar bond, actually. That's different from what we see with an alcohol. So up here, if we look at the alcohol, I know that oxygen's much more electronegative than carbon. So this oxygen here would get a partial negative. And this carbon on the left, it's bonded to, would get a partial positive."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That's different from what we see with an alcohol. So up here, if we look at the alcohol, I know that oxygen's much more electronegative than carbon. So this oxygen here would get a partial negative. And this carbon on the left, it's bonded to, would get a partial positive. So there's much more of an electronegativity difference in alcohols. And thiols, there's not really that much of a difference. But thiols can still function as nucleophiles because these lone pairs of electrons are located further away from the nucleus than the lone pair of electrons in oxygen because sulfur is a larger atom."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And this carbon on the left, it's bonded to, would get a partial positive. So there's much more of an electronegativity difference in alcohols. And thiols, there's not really that much of a difference. But thiols can still function as nucleophiles because these lone pairs of electrons are located further away from the nucleus than the lone pair of electrons in oxygen because sulfur is a larger atom. So those electrons are more polarizable. And so thiols are actually excellent nucleophiles. So if we go back here to our mechanism, we're now going to add sodium hydroxide, which is a base."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "But thiols can still function as nucleophiles because these lone pairs of electrons are located further away from the nucleus than the lone pair of electrons in oxygen because sulfur is a larger atom. So those electrons are more polarizable. And so thiols are actually excellent nucleophiles. So if we go back here to our mechanism, we're now going to add sodium hydroxide, which is a base. So we can go ahead and put OH minus over here. So the hydroxide anion is going to function as a base. And a lone pair of electrons are going to take this proton and leave these electrons behind on the sulfur."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So if we go back here to our mechanism, we're now going to add sodium hydroxide, which is a base. So we can go ahead and put OH minus over here. So the hydroxide anion is going to function as a base. And a lone pair of electrons are going to take this proton and leave these electrons behind on the sulfur. So let's go ahead and draw the conjugate base to the thiol. So we now have carbon bonded to sulfur, and this sulfur now has three lone pairs of electrons, giving it a negative 1 formal charge. So this is called a thiolate anion."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And a lone pair of electrons are going to take this proton and leave these electrons behind on the sulfur. So let's go ahead and draw the conjugate base to the thiol. So we now have carbon bonded to sulfur, and this sulfur now has three lone pairs of electrons, giving it a negative 1 formal charge. So this is called a thiolate anion. So let me just go ahead and write that. A thiolate anion in here. And thiolate anions are very stable."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this is called a thiolate anion. So let me just go ahead and write that. A thiolate anion in here. And thiolate anions are very stable. That negative charge in the sulfur, since sulfur is a large atom, you can spread out that negative charge over a very large area. So the thiolate anion is relatively stable, and that makes thiols more acidic than alcohols. Alcohols don't have the same type of stabilization since alcohols are smaller."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And thiolate anions are very stable. That negative charge in the sulfur, since sulfur is a large atom, you can spread out that negative charge over a very large area. So the thiolate anion is relatively stable, and that makes thiols more acidic than alcohols. Alcohols don't have the same type of stabilization since alcohols are smaller. So thiols are actually very acidic, and that's why we can use sodium hydroxide here to deprotonate our thiol to form the thiolate anion. In the second step, we add our alkyl halide. And so here's my alkyl halide."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Alcohols don't have the same type of stabilization since alcohols are smaller. So thiols are actually very acidic, and that's why we can use sodium hydroxide here to deprotonate our thiol to form the thiolate anion. In the second step, we add our alkyl halide. And so here's my alkyl halide. And the alkyl halide does have a polarized bond. The difference in electronegativity between a halogen and a carbon atom is fairly large. So this halogen here is going to get a partial negative charge, and this carbon is going to get a partial positive charge."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so here's my alkyl halide. And the alkyl halide does have a polarized bond. The difference in electronegativity between a halogen and a carbon atom is fairly large. So this halogen here is going to get a partial negative charge, and this carbon is going to get a partial positive charge. So thiols are good nucleophiles. Thiolate anions are even better nucleophiles. And so the thiolate anion is going to function as the nucleophile."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So this halogen here is going to get a partial negative charge, and this carbon is going to get a partial positive charge. So thiols are good nucleophiles. Thiolate anions are even better nucleophiles. And so the thiolate anion is going to function as the nucleophile. The partially positive carbon is going to function as the electrophile. And we're going to get an SN2 type mechanism, where our strong nucleophile attacks our electrophile and kicks these electrons off here onto the halogen. And we can go ahead and form our product."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And so the thiolate anion is going to function as the nucleophile. The partially positive carbon is going to function as the electrophile. And we're going to get an SN2 type mechanism, where our strong nucleophile attacks our electrophile and kicks these electrons off here onto the halogen. And we can go ahead and form our product. So this is an SN2 type mechanism. And we end up with the sulfur now bonded to two R groups. They could obviously be the same R groups, or they could be different R groups."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "And we can go ahead and form our product. So this is an SN2 type mechanism. And we end up with the sulfur now bonded to two R groups. They could obviously be the same R groups, or they could be different R groups. And so we formed our sulfide like that. Let's do an example of the preparation of sulfide. So we're going to start with this one right here."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "They could obviously be the same R groups, or they could be different R groups. And so we formed our sulfide like that. Let's do an example of the preparation of sulfide. So we're going to start with this one right here. So we start with this molecule. And to that thiol, we're going to add sodium hydroxide in the first step. So let's go ahead and write sodium hydroxide here."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to start with this one right here. So we start with this molecule. And to that thiol, we're going to add sodium hydroxide in the first step. So let's go ahead and write sodium hydroxide here. Na plus, and then OH minus, like that. And then in the second step, we're going to add an alkyl halide. So let's add this as our alkyl halide, so ethyl bromide."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and write sodium hydroxide here. Na plus, and then OH minus, like that. And then in the second step, we're going to add an alkyl halide. So let's add this as our alkyl halide, so ethyl bromide. So when I think about the mechanism, I know the first step is an acid-base reaction. The electrons on the hydroxide anions, so one of these electron pairs here, are going to take this proton. That's the acidic proton on my thiol, leaving these electrons behind on the sulfur."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So let's add this as our alkyl halide, so ethyl bromide. So when I think about the mechanism, I know the first step is an acid-base reaction. The electrons on the hydroxide anions, so one of these electron pairs here, are going to take this proton. That's the acidic proton on my thiol, leaving these electrons behind on the sulfur. So I can go ahead and draw the resulting thiolate anion. So I can go ahead and draw that. Let me see what we have here."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "That's the acidic proton on my thiol, leaving these electrons behind on the sulfur. So I can go ahead and draw the resulting thiolate anion. So I can go ahead and draw that. Let me see what we have here. We have our ring, and then we have our sulfur. And our sulfur now has three lone pairs of electrons around it like that. Thiolate anions are excellent nucleophiles."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Let me see what we have here. We have our ring, and then we have our sulfur. And our sulfur now has three lone pairs of electrons around it like that. Thiolate anions are excellent nucleophiles. And when I look at my alkyl halide, once again, I know the electronegativity difference between bromine and this carbon here. I'm going to give bromine a partial negative charge, and this carbon is going to be partially positive. So the thiolate anion is going to act as a nucleophile in an SN2 type mechanism."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Thiolate anions are excellent nucleophiles. And when I look at my alkyl halide, once again, I know the electronegativity difference between bromine and this carbon here. I'm going to give bromine a partial negative charge, and this carbon is going to be partially positive. So the thiolate anion is going to act as a nucleophile in an SN2 type mechanism. And a lone pair of electrons here are going to attack my electrophile. And we can go ahead and form our product. So now we have our ring here, which is connected to our sulfur."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So the thiolate anion is going to act as a nucleophile in an SN2 type mechanism. And a lone pair of electrons here are going to attack my electrophile. And we can go ahead and form our product. So now we have our ring here, which is connected to our sulfur. And our sulfur now just picked up two more carbons, because these electrons in here are going to kick off onto the bromine. And we end up putting an ethyl group onto that sulfur. So there are two carbons now on that sulfur like that."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our ring here, which is connected to our sulfur. And our sulfur now just picked up two more carbons, because these electrons in here are going to kick off onto the bromine. And we end up putting an ethyl group onto that sulfur. So there are two carbons now on that sulfur like that. And the sulfur has two lone pairs of electrons. And so we formed our sulfide. Now, if I were to name this sulfide, it's a lot like naming ethers."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So there are two carbons now on that sulfur like that. And the sulfur has two lone pairs of electrons. And so we formed our sulfide. Now, if I were to name this sulfide, it's a lot like naming ethers. So I could use the common way of naming this and treat those as alkyl groups. And if I look on the right, this would be an ethyl group. So I could go ahead and start naming it."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "Now, if I were to name this sulfide, it's a lot like naming ethers. So I could use the common way of naming this and treat those as alkyl groups. And if I look on the right, this would be an ethyl group. So I could go ahead and start naming it. I could say it's ethyl. And if I look at the alkyl group on the left side, this is a phenyl group right here. So it's ethyl."}, {"video_title": "Preparation of sulfides Alcohols, ethers, epoxides, sulfides Organic chemistry Khan Academy.mp3", "Sentence": "So I could go ahead and start naming it. I could say it's ethyl. And if I look at the alkyl group on the left side, this is a phenyl group right here. So it's ethyl. And then phenyl, since I'm following the alphabet rule here, E before P. And then finally, I know it's a sulfide. So I can just go ahead and finish the nomenclature by saying sulfide here. So ethyl phenyl sulfide is the sulfide produced in this analog of the Williamson ether synthesis."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And if we rotate, if we rotate the front carbon and keep the back carbon stationary, I'm gonna rotate 60 degrees, and we're gonna get an eclipsed conformation. So I left it a little bit off so you could still see the bonds in the back. So from the eclipsed conformation of butane, I rotate again, and we get a staggered conformation. I rotate another 60 degrees here, and we get an eclipsed conformation. And I'm gonna turn to its side so we can see how close these methyl groups are in space. If I rotate around, you can see with the model set, these hydrogens actually hit. So those hydrogens are close enough where they hit in the model set, and that's called steric hindrance or steric strain."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "I rotate another 60 degrees here, and we get an eclipsed conformation. And I'm gonna turn to its side so we can see how close these methyl groups are in space. If I rotate around, you can see with the model set, these hydrogens actually hit. So those hydrogens are close enough where they hit in the model set, and that's called steric hindrance or steric strain. So back to the eclipsed conformation of butane. If we rotate again, then we get a staggered conformation. We're gonna rotate again to get another eclipsed, again slightly, a little bit off so you can still see the bonds in the back here."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So those hydrogens are close enough where they hit in the model set, and that's called steric hindrance or steric strain. So back to the eclipsed conformation of butane. If we rotate again, then we get a staggered conformation. We're gonna rotate again to get another eclipsed, again slightly, a little bit off so you can still see the bonds in the back here. And we rotate one more time to get back to our staggered conformation. Here's an energy diagram showing the different conformations we saw in the video. And these pictures are just stills from the actual video."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "We're gonna rotate again to get another eclipsed, again slightly, a little bit off so you can still see the bonds in the back here. And we rotate one more time to get back to our staggered conformation. Here's an energy diagram showing the different conformations we saw in the video. And these pictures are just stills from the actual video. We started with the staggered conformation of butane right here, which has a certain potential energy. And we went from this staggered conformation to this eclipsed conformation right here by rotating 60 degrees. It takes energy to go from the staggered conformation to this eclipsed conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And these pictures are just stills from the actual video. We started with the staggered conformation of butane right here, which has a certain potential energy. And we went from this staggered conformation to this eclipsed conformation right here by rotating 60 degrees. It takes energy to go from the staggered conformation to this eclipsed conformation. The eclipsed conformation is higher in energy. The eclipsed conformation is less stable. Remember, the higher the potential energy, the less stable the conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "It takes energy to go from the staggered conformation to this eclipsed conformation. The eclipsed conformation is higher in energy. The eclipsed conformation is less stable. Remember, the higher the potential energy, the less stable the conformation. The lower the potential energy, the more stable. So the staggered is more stable than the eclipsed. The energy difference between these two conformations, let me go ahead and draw a dashed line here."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "Remember, the higher the potential energy, the less stable the conformation. The lower the potential energy, the more stable. So the staggered is more stable than the eclipsed. The energy difference between these two conformations, let me go ahead and draw a dashed line here. So this energy difference between these two conformations turns out to be approximately 16 kilojoules per mole. So it takes energy to go from the staggered conformation to this eclipsed conformation. From the eclipsed conformation, we rotated 60 degrees and we got this staggered conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "The energy difference between these two conformations, let me go ahead and draw a dashed line here. So this energy difference between these two conformations turns out to be approximately 16 kilojoules per mole. So it takes energy to go from the staggered conformation to this eclipsed conformation. From the eclipsed conformation, we rotated 60 degrees and we got this staggered conformation. Notice this staggered conformation is a little higher in energy than our first staggered conformation. So if I draw a line right here, we can see there's an energy difference between our two staggered conformations. So the energy difference turns out to be approximately 3.8 kilojoules per mole."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "From the eclipsed conformation, we rotated 60 degrees and we got this staggered conformation. Notice this staggered conformation is a little higher in energy than our first staggered conformation. So if I draw a line right here, we can see there's an energy difference between our two staggered conformations. So the energy difference turns out to be approximately 3.8 kilojoules per mole. Going from this staggered conformation up here to this eclipsed conformation takes energy. So if I draw a line here, so indicating the bottom, this energy difference is approximately 19 kilojoules per mole, so approximately 19 kilojoules per mole. And notice that this eclipsed conformation, this eclipsed conformation is higher in energy."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So the energy difference turns out to be approximately 3.8 kilojoules per mole. Going from this staggered conformation up here to this eclipsed conformation takes energy. So if I draw a line here, so indicating the bottom, this energy difference is approximately 19 kilojoules per mole, so approximately 19 kilojoules per mole. And notice that this eclipsed conformation, this eclipsed conformation is higher in energy. Let me change colors here. This eclipsed conformation is higher in energy than this eclipsed conformation. So if we draw a line right here, we can see there's an energy difference of approximately 3 kilojoules per mole."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And notice that this eclipsed conformation, this eclipsed conformation is higher in energy. Let me change colors here. This eclipsed conformation is higher in energy than this eclipsed conformation. So if we draw a line right here, we can see there's an energy difference of approximately 3 kilojoules per mole. So this is the least stable conformation. This eclipsed conformation has the highest potential energy. From this eclipsed conformation, we could go to this staggered, so that's a decrease in potential energy."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So if we draw a line right here, we can see there's an energy difference of approximately 3 kilojoules per mole. So this is the least stable conformation. This eclipsed conformation has the highest potential energy. From this eclipsed conformation, we could go to this staggered, so that's a decrease in potential energy. Notice this staggered has the same energy as this staggered conformation, so they are degenerate. Going from this staggered conformation up here to this eclipsed would take energy. And notice this eclipsed conformation is the same in energy as this one over here."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "From this eclipsed conformation, we could go to this staggered, so that's a decrease in potential energy. Notice this staggered has the same energy as this staggered conformation, so they are degenerate. Going from this staggered conformation up here to this eclipsed would take energy. And notice this eclipsed conformation is the same in energy as this one over here. So if I draw a line, you can see it's the same energy. So these two are degenerate. These two eclipsed conformations are degenerate."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And notice this eclipsed conformation is the same in energy as this one over here. So if I draw a line, you can see it's the same energy. So these two are degenerate. These two eclipsed conformations are degenerate. And finally, going from this eclipsed conformation back down to our staggered conformation, this is lower in energy. Now let's look at the conformations in more detail. And we'll start with this staggered conformation of butane."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "These two eclipsed conformations are degenerate. And finally, going from this eclipsed conformation back down to our staggered conformation, this is lower in energy. Now let's look at the conformations in more detail. And we'll start with this staggered conformation of butane. And let's go ahead and number the carbons. If this carbon is number one, as we called it in the video, that carbon is attached to this carbon, which is carbon number two. We stare down the carbon 2, 3 bond to get our Newman projection."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with this staggered conformation of butane. And let's go ahead and number the carbons. If this carbon is number one, as we called it in the video, that carbon is attached to this carbon, which is carbon number two. We stare down the carbon 2, 3 bond to get our Newman projection. And in the video, you can't see carbon number three, because carbon number two is in front of it. But when you're drawing a Newman projection, you represent the carbon in back, in this case, carbon number three, with a circle. So the circle here represents carbon number three."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "We stare down the carbon 2, 3 bond to get our Newman projection. And in the video, you can't see carbon number three, because carbon number two is in front of it. But when you're drawing a Newman projection, you represent the carbon in back, in this case, carbon number three, with a circle. So the circle here represents carbon number three. And finally, this would be carbon number four. Let's think about the dihedral angle between our two methyl groups, so between this methyl group and this methyl group. Well, that would be 180 degrees."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So the circle here represents carbon number three. And finally, this would be carbon number four. Let's think about the dihedral angle between our two methyl groups, so between this methyl group and this methyl group. Well, that would be 180 degrees. So hopefully you can see there's 180 degrees between our two groups. So the dihedral angle is 180 degrees. This conformation is called the anti-conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "Well, that would be 180 degrees. So hopefully you can see there's 180 degrees between our two groups. So the dihedral angle is 180 degrees. This conformation is called the anti-conformation. And the anti-conformation is lowest in potential energy. Therefore, the anti-conformation is the most stable conformation for butane. And that's because we take these bulky methyl groups, and we put them as far away from each other as we possibly can."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "This conformation is called the anti-conformation. And the anti-conformation is lowest in potential energy. Therefore, the anti-conformation is the most stable conformation for butane. And that's because we take these bulky methyl groups, and we put them as far away from each other as we possibly can. And all of our bonds are staggered. So if we think about these bonds here, everything is staggered. So that makes this the most stable conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And that's because we take these bulky methyl groups, and we put them as far away from each other as we possibly can. And all of our bonds are staggered. So if we think about these bonds here, everything is staggered. So that makes this the most stable conformation. If we rotate 60 degrees from the anti-conformation, in the video I kept the back carbon stationary, and I rotated the front carbon, we would get this conformation. And this is an eclipsed conformation. So think about this bond eclipsing this bond, and this hydrogen eclipsing this hydrogen."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So that makes this the most stable conformation. If we rotate 60 degrees from the anti-conformation, in the video I kept the back carbon stationary, and I rotated the front carbon, we would get this conformation. And this is an eclipsed conformation. So think about this bond eclipsing this bond, and this hydrogen eclipsing this hydrogen. I didn't draw them as being completely eclipsed, just so we could actually see what's going on here. And remember, in some of the earlier videos, we talked about the energy costs associated with a pair of eclipsed hydrogens as 4 kilojoules per mole. So this energy cost is 4 kilojoules per mole."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So think about this bond eclipsing this bond, and this hydrogen eclipsing this hydrogen. I didn't draw them as being completely eclipsed, just so we could actually see what's going on here. And remember, in some of the earlier videos, we talked about the energy costs associated with a pair of eclipsed hydrogens as 4 kilojoules per mole. So this energy cost is 4 kilojoules per mole. We also talked about the energy costs from a methyl group eclipsing a hydrogen, so in the video on propane. And this was approximately 6 kilojoules per mole. So 6 kilojoules per mole for a methyl group eclipsing a hydrogen."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So this energy cost is 4 kilojoules per mole. We also talked about the energy costs from a methyl group eclipsing a hydrogen, so in the video on propane. And this was approximately 6 kilojoules per mole. So 6 kilojoules per mole for a methyl group eclipsing a hydrogen. That's the same situation we have down here, a hydrogen and a methyl group eclipsing each other. So this should be an energy cost of 6 kilojoules per mole, too. If we add all of those up, so this would be 4 plus 6 is 10, plus another 6 is 16."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So 6 kilojoules per mole for a methyl group eclipsing a hydrogen. That's the same situation we have down here, a hydrogen and a methyl group eclipsing each other. So this should be an energy cost of 6 kilojoules per mole, too. If we add all of those up, so this would be 4 plus 6 is 10, plus another 6 is 16. We can see that's the energy difference between these two conformations, so 16 kilojoules per mole higher. So this eclipsed conformation is higher in potential energy. Let's look at the other eclipsed conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "If we add all of those up, so this would be 4 plus 6 is 10, plus another 6 is 16. We can see that's the energy difference between these two conformations, so 16 kilojoules per mole higher. So this eclipsed conformation is higher in potential energy. Let's look at the other eclipsed conformation. So that's the one over here. This is the highest in potential energy. So this must be the least stable conformation for butane."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the other eclipsed conformation. So that's the one over here. This is the highest in potential energy. So this must be the least stable conformation for butane. If we look, here we have a pair of hydrogens eclipsing each other, so that should be 4 kilojoules per mole. We have another pair of hydrogens eclipsing each other, so that's another 4 kilojoules per mole. And then we have two methyl groups eclipsing each other."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So this must be the least stable conformation for butane. If we look, here we have a pair of hydrogens eclipsing each other, so that should be 4 kilojoules per mole. We have another pair of hydrogens eclipsing each other, so that's another 4 kilojoules per mole. And then we have two methyl groups eclipsing each other. So think about this bond and this bond eclipsing each other and these methyl groups being right on top of each other. So what's the energy cost associated with two methyl groups, a methyl group eclipsing a methyl group? We can figure that out because we know the total energy cost is 19 kilojoules per mole."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And then we have two methyl groups eclipsing each other. So think about this bond and this bond eclipsing each other and these methyl groups being right on top of each other. So what's the energy cost associated with two methyl groups, a methyl group eclipsing a methyl group? We can figure that out because we know the total energy cost is 19 kilojoules per mole. So this is what we don't know, so I'll call that x. So what is x? Well, if we add everything up, it should get 19."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "We can figure that out because we know the total energy cost is 19 kilojoules per mole. So this is what we don't know, so I'll call that x. So what is x? Well, if we add everything up, it should get 19. So 19 is the total, and we have 4 and 4 and x. So 4 plus 4 plus x is equal to 19. So obviously x is equal to 11."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "Well, if we add everything up, it should get 19. So 19 is the total, and we have 4 and 4 and x. So 4 plus 4 plus x is equal to 19. So obviously x is equal to 11. So 11 kilojoules per mole is the energy cost associated with a methyl group eclipsing another methyl group. And so there's torsional strain there, but there's also steric strain or steric hindrance, which we saw in the video. These two methyl groups, the hydrogens can actually get close enough to touch in the video when you're using a model set, and that is destabilizing."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So obviously x is equal to 11. So 11 kilojoules per mole is the energy cost associated with a methyl group eclipsing another methyl group. And so there's torsional strain there, but there's also steric strain or steric hindrance, which we saw in the video. These two methyl groups, the hydrogens can actually get close enough to touch in the video when you're using a model set, and that is destabilizing. So if you have increased steric hindrance, that destabilizes your conformation. And that's why this conformation, this eclipse conformation, is the highest in potential energy. It's the least stable."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "These two methyl groups, the hydrogens can actually get close enough to touch in the video when you're using a model set, and that is destabilizing. So if you have increased steric hindrance, that destabilizes your conformation. And that's why this conformation, this eclipse conformation, is the highest in potential energy. It's the least stable. So if I draw a line over here, just remember that this eclipse conformation is even higher in potential energy than this eclipse conformation because of these two methyl groups being so close together. And finally, let's look at our staggered conformation. So our other staggered conformation is right here."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "It's the least stable. So if I draw a line over here, just remember that this eclipse conformation is even higher in potential energy than this eclipse conformation because of these two methyl groups being so close together. And finally, let's look at our staggered conformation. So our other staggered conformation is right here. Notice this staggered conformation is higher in energy than our anti. So if we look at our methyl groups, we think about the dihedral angle. So if I think about the angle between these methyl groups here, that is 60 degrees."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So our other staggered conformation is right here. Notice this staggered conformation is higher in energy than our anti. So if we look at our methyl groups, we think about the dihedral angle. So if I think about the angle between these methyl groups here, that is 60 degrees. And we say that this is the gauche conformation. So let me go ahead and write this as the gauche conformation. And for the gauche conformation, this is a little bit higher in energy than for the anti-conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "So if I think about the angle between these methyl groups here, that is 60 degrees. And we say that this is the gauche conformation. So let me go ahead and write this as the gauche conformation. And for the gauche conformation, this is a little bit higher in energy than for the anti-conformation. And that's because these two methyl groups are closer together in space. We don't really have to worry about torsional strain here, but we do have to worry about steric hindrance. So these hydrogens on these methyl groups can get pretty close to each other in the gauche conformation."}, {"video_title": "Conformational analysis of butane Organic chemistry Khan Academy.mp3", "Sentence": "And for the gauche conformation, this is a little bit higher in energy than for the anti-conformation. And that's because these two methyl groups are closer together in space. We don't really have to worry about torsional strain here, but we do have to worry about steric hindrance. So these hydrogens on these methyl groups can get pretty close to each other in the gauche conformation. And that has a destabilizing effect, therefore having a higher potential energy for this conformation. So the gauche conformation, while it's staggered, this is more stable than our eclipse conformations. The gauche conformation is not as stable as the anti-conformation because these methyl groups are relatively close together in space."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we get a stretching vibration, so if this bond stretches, if it stretches this way, we're obviously gonna change the distance, and if we're changing the distance, we're changing the dipole moment. And that's important, because only stretching vibrations that produce a change in the dipole moment are observed as signals on your IR spectrum. And so let's look at an example of this. Let's look at cyclohexanone, so here's cyclohexanone. We know that the carbonyl has a dipole moment. The oxygen is more electronegative than the carbon, so the oxygen gets a partial negative, this carbon here gets a partial positive, and so we have a pretty large dipole moment associated with our carbonyl, and so we would expect to see a pretty strong signal for that carbonyl bond stretch. If we look at our spectrum here, and we go ahead and divide our diagnostic from our fingerprint region, here is a very, very strong signal, so just past 1700 wave numbers."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at cyclohexanone, so here's cyclohexanone. We know that the carbonyl has a dipole moment. The oxygen is more electronegative than the carbon, so the oxygen gets a partial negative, this carbon here gets a partial positive, and so we have a pretty large dipole moment associated with our carbonyl, and so we would expect to see a pretty strong signal for that carbonyl bond stretch. If we look at our spectrum here, and we go ahead and divide our diagnostic from our fingerprint region, here is a very, very strong signal, so just past 1700 wave numbers. This is approximately 17, 17, 15, and this is in our double bond region that we talked about earlier. And so this must be the carbonyl bond stretch. So this represents, let me go ahead and use a different color, so this signal on our spectrum represents the carbonyl bond stretch, and this is a partial negative, partial positive."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we look at our spectrum here, and we go ahead and divide our diagnostic from our fingerprint region, here is a very, very strong signal, so just past 1700 wave numbers. This is approximately 17, 17, 15, and this is in our double bond region that we talked about earlier. And so this must be the carbonyl bond stretch. So this represents, let me go ahead and use a different color, so this signal on our spectrum represents the carbonyl bond stretch, and this is a partial negative, partial positive. So large dipole moment means a strong signal. And if a large dipole moment means a strong signal, that means that a smaller dipole moment would be a weaker signal. So let's look at an example of that next."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this represents, let me go ahead and use a different color, so this signal on our spectrum represents the carbonyl bond stretch, and this is a partial negative, partial positive. So large dipole moment means a strong signal. And if a large dipole moment means a strong signal, that means that a smaller dipole moment would be a weaker signal. So let's look at an example of that next. So we're gonna compare these two IR spectra. And so now we're looking at a different molecule, we're looking at the IR spectrum for one hexene here. Once again, let's go ahead and divide our regions."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at an example of that next. So we're gonna compare these two IR spectra. And so now we're looking at a different molecule, we're looking at the IR spectrum for one hexene here. Once again, let's go ahead and divide our regions. And if we look in the double bond region, we see this signal right here. And so if we drop down, the signal is about halfway between 1600 and 1700, so we'll say approximately 1650. So that's the signal in the double bond region, and of course that is, that's the carbon-carbon double bond stretch."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Once again, let's go ahead and divide our regions. And if we look in the double bond region, we see this signal right here. And so if we drop down, the signal is about halfway between 1600 and 1700, so we'll say approximately 1650. So that's the signal in the double bond region, and of course that is, that's the carbon-carbon double bond stretch. So that's this double bond here on one hexene. And notice that it's not as intense as the one that we talked about before, right? So this signal, let me go ahead and use a different color here, this signal is much stronger than this signal, right?"}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's the signal in the double bond region, and of course that is, that's the carbon-carbon double bond stretch. So that's this double bond here on one hexene. And notice that it's not as intense as the one that we talked about before, right? So this signal, let me go ahead and use a different color here, this signal is much stronger than this signal, right? So this is a weaker signal. So this must not have as strong of a dipole moment, and indeed that's the case. So if we think about this double bond right here, put some hydrogens on, so a little bit easier to think about, it's gonna have a very small dipole moment."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this signal, let me go ahead and use a different color here, this signal is much stronger than this signal, right? So this is a weaker signal. So this must not have as strong of a dipole moment, and indeed that's the case. So if we think about this double bond right here, put some hydrogens on, so a little bit easier to think about, it's gonna have a very small dipole moment. It has a dipole moment because this alkyl group right here, remember alkyl groups are electron donating, and so because it's not symmetric, you're gonna get a weak dipole moment, and so because you have a weak dipole moment, you're not gonna get an intense signal, you get this weaker signal here. So this, going back up here to this carbonyl again, this is really important when you're looking at IR spectrum, this super intense signal for a carbonyl is often helps you figure out what functional groups that you're dealing with. Alright, let's do one more."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about this double bond right here, put some hydrogens on, so a little bit easier to think about, it's gonna have a very small dipole moment. It has a dipole moment because this alkyl group right here, remember alkyl groups are electron donating, and so because it's not symmetric, you're gonna get a weak dipole moment, and so because you have a weak dipole moment, you're not gonna get an intense signal, you get this weaker signal here. So this, going back up here to this carbonyl again, this is really important when you're looking at IR spectrum, this super intense signal for a carbonyl is often helps you figure out what functional groups that you're dealing with. Alright, let's do one more. Let's compare this alkene to another alkene. So let's look at this one now. So over here, we have 2,3-dimethyl-2-butene."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Alright, let's do one more. Let's compare this alkene to another alkene. So let's look at this one now. So over here, we have 2,3-dimethyl-2-butene. So we go one, two, three, four, that's 2,3-dimethyl-2-butene. And if we once again draw our line at around 1500, and we look in the double bond region, so somewhere in here, we don't see a signal. And the reason we don't see a signal is because this is a symmetrical alkene."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So over here, we have 2,3-dimethyl-2-butene. So we go one, two, three, four, that's 2,3-dimethyl-2-butene. And if we once again draw our line at around 1500, and we look in the double bond region, so somewhere in here, we don't see a signal. And the reason we don't see a signal is because this is a symmetrical alkene. Alright, this is symmetric about the double bond here, so it's the same on both sides. And since it's a symmetrical alkene, there's no dipole moment, right? The electron donating effect of the alkyl groups would cancel, there's no dipole moment, and so therefore there's no change in dipole moment when the carbon-carbon double bond stretches."}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the reason we don't see a signal is because this is a symmetrical alkene. Alright, this is symmetric about the double bond here, so it's the same on both sides. And since it's a symmetrical alkene, there's no dipole moment, right? The electron donating effect of the alkyl groups would cancel, there's no dipole moment, and so therefore there's no change in dipole moment when the carbon-carbon double bond stretches. And so therefore we don't see a signal. And so this signal is absent on our IR spectrum. And so this is important to think about if you have something that's symmetrical, right?"}, {"video_title": "Signal characteristics - intensity Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The electron donating effect of the alkyl groups would cancel, there's no dipole moment, and so therefore there's no change in dipole moment when the carbon-carbon double bond stretches. And so therefore we don't see a signal. And so this signal is absent on our IR spectrum. And so this is important to think about if you have something that's symmetrical, right? You could be talking about an alkyne too. You're not gonna see a signal on your IR spectrum. So that's something to think about."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So if we just let this be and we didn't heat it up or put any UV light into this reaction, pretty much nothing will happen. Both of these molecules are reasonably happy being the way they are. But if we were to add heat into it, if we were to start making all the atoms and molecules vibrate more and bump into each other more, or we were to add energy in the form of UV light, what we could start doing is breaking some of these chlorine-chlorine bonds. Out of all of the bonds here, those are the weakest. That would be the most susceptible to breakage. So let's say we were to add some heat. What would happen?"}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Out of all of the bonds here, those are the weakest. That would be the most susceptible to breakage. So let's say we were to add some heat. What would happen? Let me draw the valence electrons of each of these chlorines. Chlorine has 1, 2, 3, 4, 5, 6, 7 valence electrons. And this chlorine over here has 1, 2, 3, 4, 5, 6, 7 valence electrons."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "What would happen? Let me draw the valence electrons of each of these chlorines. Chlorine has 1, 2, 3, 4, 5, 6, 7 valence electrons. And this chlorine over here has 1, 2, 3, 4, 5, 6, 7 valence electrons. Now, when you add heat to this reaction, enough for these guys to vibrate away from each other, for this bond to break, what's going to happen, and we haven't drawn an arrow like this just yet, but what's going to happen is that each of these chlorines, this bond is going to break. Each of these chlorines are just going to take their part of the bond. So this guy on the left, he's just going to take his electron, and notice I draw it with this half arrow."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And this chlorine over here has 1, 2, 3, 4, 5, 6, 7 valence electrons. Now, when you add heat to this reaction, enough for these guys to vibrate away from each other, for this bond to break, what's going to happen, and we haven't drawn an arrow like this just yet, but what's going to happen is that each of these chlorines, this bond is going to break. Each of these chlorines are just going to take their part of the bond. So this guy on the left, he's just going to take his electron, and notice I draw it with this half arrow. It looks like a fish hook. It's just half an arrowhead. This means that this electron is just going to go back to this chlorine, and this other magenta electron is going to go back to the right chlorine."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So this guy on the left, he's just going to take his electron, and notice I draw it with this half arrow. It looks like a fish hook. It's just half an arrowhead. This means that this electron is just going to go back to this chlorine, and this other magenta electron is going to go back to the right chlorine. So we could draw it like that. If it was up to me, I would have drawn it more like this to show that that electron just goes back to the chlorine, but the convention shows that you could show that half of the bond is going back to the entire atom. Now, after this happens, what will everything look like?"}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This means that this electron is just going to go back to this chlorine, and this other magenta electron is going to go back to the right chlorine. So we could draw it like that. If it was up to me, I would have drawn it more like this to show that that electron just goes back to the chlorine, but the convention shows that you could show that half of the bond is going back to the entire atom. Now, after this happens, what will everything look like? Well, we're still going to have our methane here. It hasn't really reacted. We still have our methane."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Now, after this happens, what will everything look like? Well, we're still going to have our methane here. It hasn't really reacted. We still have our methane. Let me draw it a little bit. So we still have our methane here, and all that's happened, because we've put energy into the system, we've been able to break this bond. The molecular chlorine has broken up into two chlorine atoms."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We still have our methane. Let me draw it a little bit. So we still have our methane here, and all that's happened, because we've put energy into the system, we've been able to break this bond. The molecular chlorine has broken up into two chlorine atoms. So we have the one on the left over here, and then we have the one on the right. Let me draw the left valence electron. It has one, two, three, four, five, six, seven."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The molecular chlorine has broken up into two chlorine atoms. So we have the one on the left over here, and then we have the one on the right. Let me draw the left valence electron. It has one, two, three, four, five, six, seven. I just flipped it over so that the lone electron is on the left-hand side right here. Then you have the guy on the right. He has one, two, three, four, five, six, seven valence electrons."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It has one, two, three, four, five, six, seven. I just flipped it over so that the lone electron is on the left-hand side right here. Then you have the guy on the right. He has one, two, three, four, five, six, seven valence electrons. Now that each of these guys have an unpaired electron, they're actually very, very, very reactive. We actually call any molecule that has an unpaired electron and is very reactive a free radical. So both of these guys now are free radicals."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "He has one, two, three, four, five, six, seven valence electrons. Now that each of these guys have an unpaired electron, they're actually very, very, very reactive. We actually call any molecule that has an unpaired electron and is very reactive a free radical. So both of these guys now are free radicals. Actually, the whole topic of this video is free radical reactions. Both of these guys are free radicals. You've probably heard the word free radical before in the context of nutrition, that you don't want free radicals running around."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So both of these guys now are free radicals. Actually, the whole topic of this video is free radical reactions. Both of these guys are free radicals. You've probably heard the word free radical before in the context of nutrition, that you don't want free radicals running around. It's the exact same idea. It's not necessarily chlorine that they're talking about, but they're talking about molecules that have unpaired electrons. They'll react with some of your cell's machinery, maybe even with your DNA, maybe cause mutations that might lead to things like cancer."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "You've probably heard the word free radical before in the context of nutrition, that you don't want free radicals running around. It's the exact same idea. It's not necessarily chlorine that they're talking about, but they're talking about molecules that have unpaired electrons. They'll react with some of your cell's machinery, maybe even with your DNA, maybe cause mutations that might lead to things like cancer. That's why people think you shouldn't have free radicals in your body. But as soon as we form these free radicals, in this step right here where we put energy in the system to break this bond, we call this the initiation step. Let me put this."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "They'll react with some of your cell's machinery, maybe even with your DNA, maybe cause mutations that might lead to things like cancer. That's why people think you shouldn't have free radicals in your body. But as soon as we form these free radicals, in this step right here where we put energy in the system to break this bond, we call this the initiation step. Let me put this. We used energy here. This was endothermic. We used energy."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let me put this. We used energy here. This was endothermic. We used energy. This right here is the initiation step. What we're going to see in general with free radical reactions is you need some energy to get it started, but once it gets started, it kind of starts this chain reaction. As one free radical reacts with something else, it creates another free radical, and that keeps propagating until really everything has reacted."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We used energy. This right here is the initiation step. What we're going to see in general with free radical reactions is you need some energy to get it started, but once it gets started, it kind of starts this chain reaction. As one free radical reacts with something else, it creates another free radical, and that keeps propagating until really everything has reacted. That's why these can be so dangerous or so bad for biological systems. I've told you that they react a lot, so how will they react now? This guy wants to form a pair with someone else, and maybe if he swipes by this methane in just the right way, with just enough energy, what will happen is he could take the hydrogen off of the carbon, and not just the proton, the entire hydrogen."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "As one free radical reacts with something else, it creates another free radical, and that keeps propagating until really everything has reacted. That's why these can be so dangerous or so bad for biological systems. I've told you that they react a lot, so how will they react now? This guy wants to form a pair with someone else, and maybe if he swipes by this methane in just the right way, with just enough energy, what will happen is he could take the hydrogen off of the carbon, and not just the proton, the entire hydrogen. He will form a bond with the hydrogen using the hydrogen's electrons. They'll get together and they'll form a bond. The hydrogen will contribute one electron."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This guy wants to form a pair with someone else, and maybe if he swipes by this methane in just the right way, with just enough energy, what will happen is he could take the hydrogen off of the carbon, and not just the proton, the entire hydrogen. He will form a bond with the hydrogen using the hydrogen's electrons. They'll get together and they'll form a bond. The hydrogen will contribute one electron. Notice I'm drawing the half arrow again. The hydrogen isn't giving away the electron to someone else. That would be a full arrow."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen will contribute one electron. Notice I'm drawing the half arrow again. The hydrogen isn't giving away the electron to someone else. That would be a full arrow. The hydrogen is just contributing its electron to half of a bond. Then the carbon would do the same. I'll do that in blue."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "That would be a full arrow. The hydrogen is just contributing its electron to half of a bond. Then the carbon would do the same. I'll do that in blue. The carbon, this valence electron right here, could be contributed to half of a bond. Then they will bond, and this bond over here will break. The carbon over here on the left will take back its electron."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I'll do that in blue. The carbon, this valence electron right here, could be contributed to half of a bond. Then they will bond, and this bond over here will break. The carbon over here on the left will take back its electron. What does it look like? What does everything look like after that's done? Our methane now, it's no longer methane."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The carbon over here on the left will take back its electron. What does it look like? What does everything look like after that's done? Our methane now, it's no longer methane. It is now, if you think about it, we have three hydrogens. It took its electron back. It is now a free radical."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Our methane now, it's no longer methane. It is now, if you think about it, we have three hydrogens. It took its electron back. It is now a free radical. It now has an unpaired reactive electron. The hydrogen and this chlorine have bonded. The hydrogen and the chlorine have bonded."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It is now a free radical. It now has an unpaired reactive electron. The hydrogen and this chlorine have bonded. The hydrogen and the chlorine have bonded. Let me draw the chlorine. It has this electron right over here. It has the other six valence electrons."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen and the chlorine have bonded. Let me draw the chlorine. It has this electron right over here. It has the other six valence electrons. One, two, three, four, five, six. We have the hydrogen with its pink electron that it's contributing to the bond. We have them bonded now."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It has the other six valence electrons. One, two, three, four, five, six. We have the hydrogen with its pink electron that it's contributing to the bond. We have them bonded now. This chlorine is no longer a free radical, although this one out here is still a free radical. We copy and paste it. It's just hanging around."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We have them bonded now. This chlorine is no longer a free radical, although this one out here is still a free radical. We copy and paste it. It's just hanging around. Copy and paste. Notice, we had one free radical react, but it formed another free radical. That's why we call this a propagation step."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It's just hanging around. Copy and paste. Notice, we had one free radical react, but it formed another free radical. That's why we call this a propagation step. We call this a propagation step. This right here is a propagation step. When one free radical reacts, it created another free radical."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "That's why we call this a propagation step. We call this a propagation step. This right here is a propagation step. When one free radical reacts, it created another free radical. What's that free radical likely to do? You might be tempted to say, hey, it's going to just react with that other chlorine. Think about it."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "When one free radical reacts, it created another free radical. What's that free radical likely to do? You might be tempted to say, hey, it's going to just react with that other chlorine. Think about it. These molecules, there's a gazillion of them in this solution. The odds that this guy is going to react exactly with that other free radical is actually very low, especially early on in the reaction where most of the molecules are still either methane or molecular chlorine. This guy is much more likely to bump into another molecular chlorine than he is to bump into one of these original free radicals that formed."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Think about it. These molecules, there's a gazillion of them in this solution. The odds that this guy is going to react exactly with that other free radical is actually very low, especially early on in the reaction where most of the molecules are still either methane or molecular chlorine. This guy is much more likely to bump into another molecular chlorine than he is to bump into one of these original free radicals that formed. If he bumps into another molecular chlorine in just the right way, let me draw another molecular chlorine. That's another molecular chlorine. Each of these, 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This guy is much more likely to bump into another molecular chlorine than he is to bump into one of these original free radicals that formed. If he bumps into another molecular chlorine in just the right way, let me draw another molecular chlorine. That's another molecular chlorine. Each of these, 1, 2, 3, 4, 5, 6, 7. There is a bond here. If they bump in just the right way, this chlorine electron might get contributed. This free unpaired electron will be contributed."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Each of these, 1, 2, 3, 4, 5, 6, 7. There is a bond here. If they bump in just the right way, this chlorine electron might get contributed. This free unpaired electron will be contributed. This CH3, I guess we could call it, this carbon free radical, or this methyl free radical, will then form a bond with this chlorine. What's everything going to look like after that? After that happens, this is now bonded to a chlorine."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This free unpaired electron will be contributed. This CH3, I guess we could call it, this carbon free radical, or this methyl free radical, will then form a bond with this chlorine. What's everything going to look like after that? After that happens, this is now bonded to a chlorine. It's now chloromethane. Let me draw it. It's carbon, hydrogen, hydrogen, hydrogen."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "After that happens, this is now bonded to a chlorine. It's now chloromethane. Let me draw it. It's carbon, hydrogen, hydrogen, hydrogen. Now it's bonded to a chlorine. Let me draw the electrons so we can keep track of everything. We have that magenta electron right over there."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "It's carbon, hydrogen, hydrogen, hydrogen. Now it's bonded to a chlorine. Let me draw the electrons so we can keep track of everything. We have that magenta electron right over there. Then we have the chlorine with its 1, 2, 3, 4, 5, 6, 7 valence electrons. They are now bonded. This is chloromethane."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We have that magenta electron right over there. Then we have the chlorine with its 1, 2, 3, 4, 5, 6, 7 valence electrons. They are now bonded. This is chloromethane. Now you have another free radical. This guy, and I should have drawn it there, this guy, the bond was broken. He gets back his electron."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This is chloromethane. Now you have another free radical. This guy, and I should have drawn it there, this guy, the bond was broken. He gets back his electron. He's sitting over here. He is now a free radical. This is another propagation step."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "He gets back his electron. He's sitting over here. He is now a free radical. This is another propagation step. We still have that original free radical guy sitting out over here. We keep forming more and more free radicals as this happens. Eventually, we're going to start running out of methane."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This is another propagation step. We still have that original free radical guy sitting out over here. We keep forming more and more free radicals as this happens. Eventually, we're going to start running out of methane. We're going to start running out of the molecular chlorines. They're going to be less likely to react. You're actually going to have more free radicals around."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Eventually, we're going to start running out of methane. We're going to start running out of the molecular chlorines. They're going to be less likely to react. You're actually going to have more free radicals around. Once the concentration of free radicals gets high enough, then you might start to see them reacting with each other. When the concentration of free radicals get high enough, you might see, instead of this step happening, this will happen a long time until most of the free radicals or most of the non-free radicals disappear. Once we have a soup of mainly free radicals, you'll see things like this."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "You're actually going to have more free radicals around. Once the concentration of free radicals gets high enough, then you might start to see them reacting with each other. When the concentration of free radicals get high enough, you might see, instead of this step happening, this will happen a long time until most of the free radicals or most of the non-free radicals disappear. Once we have a soup of mainly free radicals, you'll see things like this. You'll see the methyl free radical. Let me draw it like this. You'll see him maybe reacting with another methyl free radical."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Once we have a soup of mainly free radicals, you'll see things like this. You'll see the methyl free radical. Let me draw it like this. You'll see him maybe reacting with another methyl free radical. With another methyl free radical. Where they both contribute an electron to form a bond. Then, once the bond forms, you have ethane."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "You'll see him maybe reacting with another methyl free radical. With another methyl free radical. Where they both contribute an electron to form a bond. Then, once the bond forms, you have ethane. I could just write it CH3H3C. You might have something like this. This type of a step where two free radicals cancel each other out, this is a termination step."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Then, once the bond forms, you have ethane. I could just write it CH3H3C. You might have something like this. This type of a step where two free radicals cancel each other out, this is a termination step. Because it's starting to lower the concentration of free radicals in the solution. But this is only once the concentration of free radicals becomes really high. You might also see some of the chlorines cancel out with each other again."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "This type of a step where two free radicals cancel each other out, this is a termination step. Because it's starting to lower the concentration of free radicals in the solution. But this is only once the concentration of free radicals becomes really high. You might also see some of the chlorines cancel out with each other again. So a chlorine free radical and another chlorine free radical. I'll only draw the unpaired electron. They can bond with each other and form molecular chlorine again."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "You might also see some of the chlorines cancel out with each other again. So a chlorine free radical and another chlorine free radical. I'll only draw the unpaired electron. They can bond with each other and form molecular chlorine again. That again is a termination step. Or you could see something like the methyl free radical. Let me just for shorthand, I'll write it like this."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "They can bond with each other and form molecular chlorine again. That again is a termination step. Or you could see something like the methyl free radical. Let me just for shorthand, I'll write it like this. H3C, the methyl free radical and a chlorine free radical. Might also just straight up react and form chloromethane. And form H3CCl."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "Let me just for shorthand, I'll write it like this. H3C, the methyl free radical and a chlorine free radical. Might also just straight up react and form chloromethane. And form H3CCl. So this will all happen once the concentration of free radicals gets really high. Now another thing that might happen once this reaction proceeds. We have a lot of the propagation steps."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And form H3CCl. So this will all happen once the concentration of free radicals gets really high. Now another thing that might happen once this reaction proceeds. We have a lot of the propagation steps. Is that you might have a situation where you already have a chloro. You already have a chloromethane. So it looks like this."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We have a lot of the propagation steps. Is that you might have a situation where you already have a chloro. You already have a chloromethane. So it looks like this. You already have a chloromethane. And once you have enough of these, it then becomes more likely that some free radical chlorine. That some free radical chlorine might be able to react with this thing."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "So it looks like this. You already have a chloromethane. And once you have enough of these, it then becomes more likely that some free radical chlorine. That some free radical chlorine might be able to react with this thing. So it might actually add another chlorine to this molecule. And the way it would do it, this chlorine over here. I'm just drawing the free electron pairs."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "That some free radical chlorine might be able to react with this thing. So it might actually add another chlorine to this molecule. And the way it would do it, this chlorine over here. I'm just drawing the free electron pairs. It would form a bond with this hydrogen right over there. They would both contribute their electrons. And then the carbon would take back its electron."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "I'm just drawing the free electron pairs. It would form a bond with this hydrogen right over there. They would both contribute their electrons. And then the carbon would take back its electron. Notice all the half arrows. And then we'd be left with the hydrogen and the chlorine would have bonded. And now this guy is going to be a free radical."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And then the carbon would take back its electron. Notice all the half arrows. And then we'd be left with the hydrogen and the chlorine would have bonded. And now this guy is going to be a free radical. But he's going to be a chlorinated free radical. So it's going to look like this. He has a free electron over there."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "And now this guy is going to be a free radical. But he's going to be a chlorinated free radical. So it's going to look like this. He has a free electron over there. Hydrogen. And then he might be able to react with another chlorine molecule. He contributes an electron."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "He has a free electron over there. Hydrogen. And then he might be able to react with another chlorine molecule. He contributes an electron. Maybe this guy contributes an electron. This guy, I don't want to draw a full arrow. He contributes an electron to a bond."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "He contributes an electron. Maybe this guy contributes an electron. This guy, I don't want to draw a full arrow. He contributes an electron to a bond. And then this guy takes his electron back and becomes a free radical. And then we're left with what? We're left with a doubly chlorinated methane."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "He contributes an electron to a bond. And then this guy takes his electron back and becomes a free radical. And then we're left with what? We're left with a doubly chlorinated methane. So then we have Cl, Cl, and then a hydrogen and a hydrogen. And this could actually keep happening. As the concentration of these get higher, then it becomes more likely that this can react with another chlorine."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We're left with a doubly chlorinated methane. So then we have Cl, Cl, and then a hydrogen and a hydrogen. And this could actually keep happening. As the concentration of these get higher, then it becomes more likely that this can react with another chlorine. And of course this chlorine over here becomes another free radical. But the general idea here that I wanted to show you is that once a free radical reaction starts, and the first step requires some energy to break this chlorine-chlorine bond, but once it happens, these guys are highly reactive. We'll start reacting with other things."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "As the concentration of these get higher, then it becomes more likely that this can react with another chlorine. And of course this chlorine over here becomes another free radical. But the general idea here that I wanted to show you is that once a free radical reaction starts, and the first step requires some energy to break this chlorine-chlorine bond, but once it happens, these guys are highly reactive. We'll start reacting with other things. And as they react with other things, it causes more and more free radicals. So it starts this chain reaction. And actually, all in all, this required energy to occur."}, {"video_title": "Free radical reactions Substitution and elimination reactions Organic chemistry Khan Academy.mp3", "Sentence": "We'll start reacting with other things. And as they react with other things, it causes more and more free radicals. So it starts this chain reaction. And actually, all in all, this required energy to occur. This step right here, this propagation step, it requires a little bit of energy, but it's almost neutral. It requires energy to break this bond, but it creates energy when this bond is formed. It still requires a little net energy."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I want to do a quick primer on refraction. And our focus here is going to be on seismic waves. But the principles, how things refract when they go from a fast to a slow medium or slow to a fast medium, it's actually the same as you would see when you're studying light waves or actually any type of wave. So let's think about it a little bit. So let's say I have a slow medium right over here. And let's say I have a fast medium right over here. Let's say, just so that we can travel through both solid and liquid, let's think about maybe P waves."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's think about it a little bit. So let's say I have a slow medium right over here. And let's say I have a fast medium right over here. Let's say, just so that we can travel through both solid and liquid, let's think about maybe P waves. And a slow medium could be maybe some type of liquid. And our fast medium could be some type of solid. So let me draw the boundary right over here."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say, just so that we can travel through both solid and liquid, let's think about maybe P waves. And a slow medium could be maybe some type of liquid. And our fast medium could be some type of solid. So let me draw the boundary right over here. And let's say, so if I have something that comes out P wave, let's say it's going through the water and it's going right perpendicular to the boundary, it will then just continue to travel in the faster medium in the same direction. If it's going right, if it goes right at the boundary, it'll just travel faster in the faster medium. And that's because that faster medium is going to be more dense."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me draw the boundary right over here. And let's say, so if I have something that comes out P wave, let's say it's going through the water and it's going right perpendicular to the boundary, it will then just continue to travel in the faster medium in the same direction. If it's going right, if it goes right at the boundary, it'll just travel faster in the faster medium. And that's because that faster medium is going to be more dense. And the molecules are going to bump into each other faster. In the same amount of time, more molecules, kind of the chain reaction, is going to be able to travel further because they're more closely packed and they rebound faster than it would in the slow medium. So that's obviously no refraction is going on."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's because that faster medium is going to be more dense. And the molecules are going to bump into each other faster. In the same amount of time, more molecules, kind of the chain reaction, is going to be able to travel further because they're more closely packed and they rebound faster than it would in the slow medium. So that's obviously no refraction is going on. It has not been deflected. And just as a bit of a reminder, in general, refraction is when a wave gets deflected. Reflection is when it bounces back."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's obviously no refraction is going on. It has not been deflected. And just as a bit of a reminder, in general, refraction is when a wave gets deflected. Reflection is when it bounces back. Refraction is when it gets deflected a little bit. Let me just make that clear. So if I have some type of boundary here and I have a wave that bounces off, that's reflection."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Reflection is when it bounces back. Refraction is when it gets deflected a little bit. Let me just make that clear. So if I have some type of boundary here and I have a wave that bounces off, that's reflection. But if the wave goes through the boundary and just gets bent a little bit, its direction changes, that is refraction. And that's what we're talking about. So clearly, so far, this P wave has not been refracted."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if I have some type of boundary here and I have a wave that bounces off, that's reflection. But if the wave goes through the boundary and just gets bent a little bit, its direction changes, that is refraction. And that's what we're talking about. So clearly, so far, this P wave has not been refracted. But if this P wave comes in at an angle, so let's make this P wave come in at an angle, what's going to happen is, and the way you should think about it, it's the easiest way to think about which direction will be refracted, or at least the way I think about it, is literally I imagine some type of vehicle with wheels on it. So this is the top view of my vehicle. So if I have some type of vehicle, and the wheels will be able to move slowly in this medium."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So clearly, so far, this P wave has not been refracted. But if this P wave comes in at an angle, so let's make this P wave come in at an angle, what's going to happen is, and the way you should think about it, it's the easiest way to think about which direction will be refracted, or at least the way I think about it, is literally I imagine some type of vehicle with wheels on it. So this is the top view of my vehicle. So if I have some type of vehicle, and the wheels will be able to move slowly in this medium. You can view it as on mud so it doesn't get good traction. And then the fast medium, maybe it's a road, so it gets good traction and it can move faster. So what's going to happen when the vehicle gets to the boundary?"}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if I have some type of vehicle, and the wheels will be able to move slowly in this medium. You can view it as on mud so it doesn't get good traction. And then the fast medium, maybe it's a road, so it gets good traction and it can move faster. So what's going to happen when the vehicle gets to the boundary? Well, this bottom right wheel is going to go on the fast medium before any of the other wheels do. So it's going to get the traction first. These wheels on the left side of the vehicle, these wheels right here, these are still going to be stuck in the mud."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what's going to happen when the vehicle gets to the boundary? Well, this bottom right wheel is going to go on the fast medium before any of the other wheels do. So it's going to get the traction first. These wheels on the left side of the vehicle, these wheels right here, these are still going to be stuck in the mud. So what's going to happen is, this wheel right over here is moving faster, so it's essentially going to be able to turn the vehicle. These guys are still stuck in the mud. And so you fast forward a little bit, the direction of the vehicle will change."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These wheels on the left side of the vehicle, these wheels right here, these are still going to be stuck in the mud. So what's going to happen is, this wheel right over here is moving faster, so it's essentially going to be able to turn the vehicle. These guys are still stuck in the mud. And so you fast forward a little bit, the direction of the vehicle will change. And so the vehicle will now move in a direction something like this. The same thing would happen in a wave. If the P wave is approaching the boundary like this, and something analogous to this is happening at the molecular level."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you fast forward a little bit, the direction of the vehicle will change. And so the vehicle will now move in a direction something like this. The same thing would happen in a wave. If the P wave is approaching the boundary like this, and something analogous to this is happening at the molecular level. You can kind of view it as even billiard balls, and maybe they're kind of hitting each other. I won't go into that, because that can kind of get a little confusing, depending on the different cases and the different boundaries. But this is the easiest way to think about in which direction it will refract, and hopefully it makes a little bit of intuitive sense."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If the P wave is approaching the boundary like this, and something analogous to this is happening at the molecular level. You can kind of view it as even billiard balls, and maybe they're kind of hitting each other. I won't go into that, because that can kind of get a little confusing, depending on the different cases and the different boundaries. But this is the easiest way to think about in which direction it will refract, and hopefully it makes a little bit of intuitive sense. And so when you go from a slow to a fast medium, our P wave, its angle would accentuate in that direction. If you went from the fast medium to the slow medium, once again, you can just go through the same thought experiment. So let's say you have our wave coming in like that."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is the easiest way to think about in which direction it will refract, and hopefully it makes a little bit of intuitive sense. And so when you go from a slow to a fast medium, our P wave, its angle would accentuate in that direction. If you went from the fast medium to the slow medium, once again, you can just go through the same thought experiment. So let's say you have our wave coming in like that. Draw the car. Visualize the car here. Visualize the car right here."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say you have our wave coming in like that. Draw the car. Visualize the car here. Visualize the car right here. And you say, well, look, this tire is going to get stuck in the mud, because now we're going from, it was on the road, now this top left, this top tire right over here is getting stuck in the mud first, so it's going to be moving slower. So these tires are going to be able to move faster, so the vehicle is going to turn. So you'll be refracted in a direction like that when you're going from the fast to the slow medium."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Visualize the car right here. And you say, well, look, this tire is going to get stuck in the mud, because now we're going from, it was on the road, now this top left, this top tire right over here is getting stuck in the mud first, so it's going to be moving slower. So these tires are going to be able to move faster, so the vehicle is going to turn. So you'll be refracted in a direction like that when you're going from the fast to the slow medium. So that's just a primer on refraction generally. Now let's think about what would happen when sound waves are traveling through the Earth. And this will help inform us of, essentially, how do we figure out what the actual structure of the Earth is?"}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you'll be refracted in a direction like that when you're going from the fast to the slow medium. So that's just a primer on refraction generally. Now let's think about what would happen when sound waves are traveling through the Earth. And this will help inform us of, essentially, how do we figure out what the actual structure of the Earth is? So if the Earth was just made up of some uniform material, and you had an earthquake right here on Earth, maybe a little bit below the surface. So it's happening in the crust, but a little bit below the surface of the Earth. If Earth was of uniform density, if it was all the same material, how would those, let's just think about the P waves, because P waves can travel in anything."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this will help inform us of, essentially, how do we figure out what the actual structure of the Earth is? So if the Earth was just made up of some uniform material, and you had an earthquake right here on Earth, maybe a little bit below the surface. So it's happening in the crust, but a little bit below the surface of the Earth. If Earth was of uniform density, if it was all the same material, how would those, let's just think about the P waves, because P waves can travel in anything. Let's think about how those P waves would travel. Well, they would just go in straight lines. There's nothing that would refract the P waves."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If Earth was of uniform density, if it was all the same material, how would those, let's just think about the P waves, because P waves can travel in anything. Let's think about how those P waves would travel. Well, they would just go in straight lines. There's nothing that would refract the P waves. It would just go in straight lines, radially outward, from where the earthquake occurred. Now, at a first approximation, we know that as we go deeper and deeper into Earth, there's more and more rock above that. The weight of that rock is kind of compressing the rock below it."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's nothing that would refract the P waves. It would just go in straight lines, radially outward, from where the earthquake occurred. Now, at a first approximation, we know that as we go deeper and deeper into Earth, there's more and more rock above that. The weight of that rock is kind of compressing the rock below it. So you get higher and higher pressures, and higher and higher densities. So this is a uniform Earth. So this is a uniform."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The weight of that rock is kind of compressing the rock below it. So you get higher and higher pressures, and higher and higher densities. So this is a uniform Earth. So this is a uniform. But let's imagine an Earth that's made up of uniform material, that's all solid, a completely solid Earth, but one where the density is constantly increasing as you go down. So let's just think about it in, before we go into the continuous case, because we're talking about the density as you go deeper, it's just getting continuously more dense. Let's think about the discrete case, where we have the least dense layer."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is a uniform. But let's imagine an Earth that's made up of uniform material, that's all solid, a completely solid Earth, but one where the density is constantly increasing as you go down. So let's just think about it in, before we go into the continuous case, because we're talking about the density as you go deeper, it's just getting continuously more dense. Let's think about the discrete case, where we have the least dense layer. So let me draw it right over here. Let's say this is the surface of the Earth, and this is least dense. Then let's say you have another layer over here that is more dense."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's think about the discrete case, where we have the least dense layer. So let me draw it right over here. Let's say this is the surface of the Earth, and this is least dense. Then let's say you have another layer over here that is more dense. So this is more dense. Let's say you have another layer that's even more dense. So you have another layer over here that's even more dense."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then let's say you have another layer over here that is more dense. So this is more dense. Let's say you have another layer that's even more dense. So you have another layer over here that's even more dense. And then let's do one more layer. Let's do this layer here, that this is the densest layer. So in general, your P wave, your seismic wave, is going to travel faster and denser material."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have another layer over here that's even more dense. And then let's do one more layer. Let's do this layer here, that this is the densest layer. So in general, your P wave, your seismic wave, is going to travel faster and denser material. So it's going to travel the fastest here, then here, then here. It's going to travel the slowest in this least dense material. So if you're coming in at an angle, let's think about what's going to happen."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in general, your P wave, your seismic wave, is going to travel faster and denser material. So it's going to travel the fastest here, then here, then here. It's going to travel the slowest in this least dense material. So if you're coming in at an angle, let's think about what's going to happen. So let's say you have your P wave coming in at an angle like this. So it's going straight through the least dense material. What's going to happen when it goes into the slightly shallower angle?"}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you're coming in at an angle, let's think about what's going to happen. So let's say you have your P wave coming in at an angle like this. So it's going straight through the least dense material. What's going to happen when it goes into the slightly shallower angle? So let's say it's like that. What's going to happen when it goes into the more dense material? So once again, let's imagine our little car."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What's going to happen when it goes into the slightly shallower angle? So let's say it's like that. What's going to happen when it goes into the more dense material? So once again, let's imagine our little car. So this tire is going to be able to go faster before the tires on the other side. So the car is going to be deflected to the left, to the down left. So now it's going to travel like this."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So once again, let's imagine our little car. So this tire is going to be able to go faster before the tires on the other side. So the car is going to be deflected to the left, to the down left. So now it's going to travel like this. So it's now going to travel something like this. Now what's going to happen at this boundary? Once again, imagine the car."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So now it's going to travel like this. So it's now going to travel something like this. Now what's going to happen at this boundary? Once again, imagine the car. This tire right here is going to be able to travel faster before the other tire, so it'll be deflected even more in that direction. Then we go into the densest material. Once again, the tire is on kind of the bottom side."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once again, imagine the car. This tire right here is going to be able to travel faster before the other tire, so it'll be deflected even more in that direction. Then we go into the densest material. Once again, the tire is on kind of the bottom side. When we look at it this way, we're going to be able to move faster before the other tire, so we're going to get deflected even more. So you see as you go from least dense material to more dense material, you're kind of curving outward. So if this was continuous, if you had a continuous structure where as you go down, it just gets more and more dense."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once again, the tire is on kind of the bottom side. When we look at it this way, we're going to be able to move faster before the other tire, so we're going to get deflected even more. So you see as you go from least dense material to more dense material, you're kind of curving outward. So if this was continuous, if you had a continuous structure where as you go down, it just gets more and more dense. As you go, so this is less dense, and then it just continuously gets more dense. So this is the most dense down here. How would the refraction look?"}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if this was continuous, if you had a continuous structure where as you go down, it just gets more and more dense. As you go, so this is less dense, and then it just continuously gets more dense. So this is the most dense down here. How would the refraction look? Well, then it would just be a continuous curve. It would look like this. Your P wave would constantly be refracted out like that."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "How would the refraction look? Well, then it would just be a continuous curve. It would look like this. Your P wave would constantly be refracted out like that. It would curve outwards. So if Earth, so this was the simplest example, where Earth is uniform. And that's pretty easy to dismiss, that obviously things will get denser because of more pressure down."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Your P wave would constantly be refracted out like that. It would curve outwards. So if Earth, so this was the simplest example, where Earth is uniform. And that's pretty easy to dismiss, that obviously things will get denser because of more pressure down. So let's say we assume another thing. We have a uniform Earth in terms of composition, but let's say it gets denser. So denser at the center."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's pretty easy to dismiss, that obviously things will get denser because of more pressure down. So let's say we assume another thing. We have a uniform Earth in terms of composition, but let's say it gets denser. So denser at the center. Then how would the P waves travel? Then how would the P waves travel? Or how would any seismic waves travel?"}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So denser at the center. Then how would the P waves travel? Then how would the P waves travel? Or how would any seismic waves travel? Well, then if you have your earthquake right over here, the ones that are going straight down still would go straight down, because we know that we won't get refracted if we're kind of going perpendicular to the change in medium or the change in boundaries. But things that are coming at a slight angle, as they get deeper they're going to get deflected more and more and more, and they're going to be refracted outward, just like we saw in this example here. If they go in this angle, they're going to be refracted outward like that."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or how would any seismic waves travel? Well, then if you have your earthquake right over here, the ones that are going straight down still would go straight down, because we know that we won't get refracted if we're kind of going perpendicular to the change in medium or the change in boundaries. But things that are coming at a slight angle, as they get deeper they're going to get deflected more and more and more, and they're going to be refracted outward, just like we saw in this example here. If they go in this angle, they're going to be refracted outward like that. If they go here they're going to be refracted outward like that. They're going to be refracted outward like that. If you're here, you're going to be refracted outward like that."}, {"video_title": "Refraction of seismic waves Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If they go in this angle, they're going to be refracted outward like that. If they go here they're going to be refracted outward like that. They're going to be refracted outward like that. If you're here, you're going to be refracted outward like that. If you're here, you're going to be refracted outward like this. Now what we're going to do in the next few videos is use what we just learned about refraction in the case of seismic waves, and hopefully we learned it in this video, and how it would refract as we're going through ever increasing denser material. We're going to use that information to essentially try to figure out the composition of the Earth based on what we've actually observed."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the lower the pKa value, the stronger the acid. pKa values are used a lot in organic chemistry, so it's really important to become familiar with them. For example, if our acid is HCl, the pKa of this proton is approximately negative seven. So if a base comes along, some generic base, and takes this proton, these electrons are left behind on the chlorine, so let's use a different color here. So the electrons in magenta come off onto the chlorine to form the chloride anion. So the chloride anion is the conjugate base to HCl. The lower the pKa value, the stronger the acid."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if a base comes along, some generic base, and takes this proton, these electrons are left behind on the chlorine, so let's use a different color here. So the electrons in magenta come off onto the chlorine to form the chloride anion. So the chloride anion is the conjugate base to HCl. The lower the pKa value, the stronger the acid. And out of all the acids I have on this pKa table, HCl has the lowest pKa value. So this is the strongest acid out of the ones on this pKa table. And from general chemistry, the stronger the acid, the weaker the conjugate base."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The lower the pKa value, the stronger the acid. And out of all the acids I have on this pKa table, HCl has the lowest pKa value. So this is the strongest acid out of the ones on this pKa table. And from general chemistry, the stronger the acid, the weaker the conjugate base. Because for an acid to be strong, the conjugate base must be weak to resist reprotonation. If it's easy to lose a proton, it should be really hard to regain it. Otherwise, you wouldn't have a strong acid."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And from general chemistry, the stronger the acid, the weaker the conjugate base. Because for an acid to be strong, the conjugate base must be weak to resist reprotonation. If it's easy to lose a proton, it should be really hard to regain it. Otherwise, you wouldn't have a strong acid. So HCl has the lowest pKa value on this table, so it is the strongest acid. And the chloride anion must therefore be the weakest base out of all of the conjugate bases on the right side of our table. Next, let's look at this acid."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Otherwise, you wouldn't have a strong acid. So HCl has the lowest pKa value on this table, so it is the strongest acid. And the chloride anion must therefore be the weakest base out of all of the conjugate bases on the right side of our table. Next, let's look at this acid. So this is the acidic proton. So if a base came along, some generic base, and took this proton and left these electrons behind on the oxygen, we will use this color again. So the electrons in magenta come off onto the oxygen to give us acetone as our conjugate base."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at this acid. So this is the acidic proton. So if a base came along, some generic base, and took this proton and left these electrons behind on the oxygen, we will use this color again. So the electrons in magenta come off onto the oxygen to give us acetone as our conjugate base. The approximate pKa for this proton is negative three. And I've seen lots of different pKa values for this proton. I've seen negative six, I've seen negative seven."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta come off onto the oxygen to give us acetone as our conjugate base. The approximate pKa for this proton is negative three. And I've seen lots of different pKa values for this proton. I've seen negative six, I've seen negative seven. So use whatever pKa values you are given in your course. So pKa tables don't always match exactly. And this is one that I've seen lots of different values for."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I've seen negative six, I've seen negative seven. So use whatever pKa values you are given in your course. So pKa tables don't always match exactly. And this is one that I've seen lots of different values for. Next we have hydronium, H3O+. So if a base comes along and takes a proton from hydronium, the pKa of that proton is approximately negative two. So these electrons would come off onto the oxygen."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And this is one that I've seen lots of different values for. Next we have hydronium, H3O+. So if a base comes along and takes a proton from hydronium, the pKa of that proton is approximately negative two. So these electrons would come off onto the oxygen. So the electrons in magenta come off onto the oxygen. And those would be these electrons right here to form water as our conjugate base. So H3O+, has a higher pKa value than HCl."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons would come off onto the oxygen. So the electrons in magenta come off onto the oxygen. And those would be these electrons right here to form water as our conjugate base. So H3O+, has a higher pKa value than HCl. Therefore H3O+, is not as strong of an acid. Or we could say it the other way around. HCl has a lower value for the pKa."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So H3O+, has a higher pKa value than HCl. Therefore H3O+, is not as strong of an acid. Or we could say it the other way around. HCl has a lower value for the pKa. Therefore HCl is a stronger acid than H3O+. All right, let's look at some more examples for acids and pKa values. So next let's look at acetic acid."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "HCl has a lower value for the pKa. Therefore HCl is a stronger acid than H3O+. All right, let's look at some more examples for acids and pKa values. So next let's look at acetic acid. So this compound right here. Here's the acidic proton on acetic acid. And that proton has a pKa value of approximately five."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So next let's look at acetic acid. So this compound right here. Here's the acidic proton on acetic acid. And that proton has a pKa value of approximately five. So if a base takes that proton, these electrons are left behind on the oxygen, which give that oxygen a negative one formal charge. So the acetate anion is the conjugate base to acetic acid. Next, our next compound, we have these two protons right here."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And that proton has a pKa value of approximately five. So if a base takes that proton, these electrons are left behind on the oxygen, which give that oxygen a negative one formal charge. So the acetate anion is the conjugate base to acetic acid. Next, our next compound, we have these two protons right here. And let's say this proton, it doesn't really matter which one you choose, has a pKa value of approximately nine. So if a base takes that proton, then the electrons in magenta here are left behind on a carbon. Let me go ahead and circle the carbon."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Next, our next compound, we have these two protons right here. And let's say this proton, it doesn't really matter which one you choose, has a pKa value of approximately nine. So if a base takes that proton, then the electrons in magenta here are left behind on a carbon. Let me go ahead and circle the carbon. So this carbon would get a negative one formal charge. Our next compound is phenol. And the acidic proton on phenol is this one right here."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and circle the carbon. So this carbon would get a negative one formal charge. Our next compound is phenol. And the acidic proton on phenol is this one right here. So if a base takes that proton, these electrons in magenta are left behind on the oxygen, which give the oxygen a negative one formal charge. So this is our conjugate base. So the pKa for that proton is approximately 10."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the acidic proton on phenol is this one right here. So if a base takes that proton, these electrons in magenta are left behind on the oxygen, which give the oxygen a negative one formal charge. So this is our conjugate base. So the pKa for that proton is approximately 10. All right, let's get some more space down here. Let's look at some more. All right, next we have water."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the pKa for that proton is approximately 10. All right, let's get some more space down here. Let's look at some more. All right, next we have water. All right, so what's the pKa for this proton on water? It's approximately 15.7. So the conjugate base to water would be hydroxide."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, next we have water. All right, so what's the pKa for this proton on water? It's approximately 15.7. So the conjugate base to water would be hydroxide. Next we have ethanol. So this is the acidic proton on ethanol. The pKa value is approximately 16."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the conjugate base to water would be hydroxide. Next we have ethanol. So this is the acidic proton on ethanol. The pKa value is approximately 16. So the electrons in magenta here would end up on the oxygen, giving the oxygen a negative one formal charge. So this is the ethoxide anion. So how much more acidic is phenol than ethanol?"}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The pKa value is approximately 16. So the electrons in magenta here would end up on the oxygen, giving the oxygen a negative one formal charge. So this is the ethoxide anion. So how much more acidic is phenol than ethanol? Remember, the lower the pKa value, the stronger the acid. Phenol has a pKa value of 10 for this proton, whereas ethanol has a pKa value of 16, approximately, for this proton. So how much more acidic is this compound compared to this compound?"}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So how much more acidic is phenol than ethanol? Remember, the lower the pKa value, the stronger the acid. Phenol has a pKa value of 10 for this proton, whereas ethanol has a pKa value of 16, approximately, for this proton. So how much more acidic is this compound compared to this compound? Well, remember the definition for pKa. It's the negative log of the Ka. And each pKa unit is an order of magnitude."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So how much more acidic is this compound compared to this compound? Well, remember the definition for pKa. It's the negative log of the Ka. And each pKa unit is an order of magnitude. So how many units difference do we have between these pKa values? This one's 10 and this one's 16. So that's a difference of six."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And each pKa unit is an order of magnitude. So how many units difference do we have between these pKa values? This one's 10 and this one's 16. So that's a difference of six. But really, each unit is an order of magnitude. So this proton on phenol is 10 to the sixth times more acidic. So this compound is one million times more acidic than ethanol."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's a difference of six. But really, each unit is an order of magnitude. So this proton on phenol is 10 to the sixth times more acidic. So this compound is one million times more acidic than ethanol. So that's really how to think about acid strength when you're looking at a pKa table. And it's so much easier to work with pKa values, which is why they're used so often in organic chemistry. All right, next, let's look at another alcohol."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this compound is one million times more acidic than ethanol. So that's really how to think about acid strength when you're looking at a pKa table. And it's so much easier to work with pKa values, which is why they're used so often in organic chemistry. All right, next, let's look at another alcohol. So this proton has a pKa value of approximately 17, and this would be the conjugate base. This proton on this next alcohol, the pKa is approximately 18. So we get this as our conjugate base."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, next, let's look at another alcohol. So this proton has a pKa value of approximately 17, and this would be the conjugate base. This proton on this next alcohol, the pKa is approximately 18. So we get this as our conjugate base. And we'll talk about the reasons why. For example, we have these alcohols here. Later, we'll talk about the reasons why they have different pKa values."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we get this as our conjugate base. And we'll talk about the reasons why. For example, we have these alcohols here. Later, we'll talk about the reasons why they have different pKa values. All right, let's look at some more. So the pKa value for this proton is approximately 19. So if the base comes along and takes that proton that I just circled in red, these electrons in magenta would be these electrons, and they're on this carbon now, which gives that carbon a negative one formal charge."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Later, we'll talk about the reasons why they have different pKa values. All right, let's look at some more. So the pKa value for this proton is approximately 19. So if the base comes along and takes that proton that I just circled in red, these electrons in magenta would be these electrons, and they're on this carbon now, which gives that carbon a negative one formal charge. All right, next we have acetylene. So if we deprotonate acetylene, this pKa has a value of approximately 25. So the electrons in magenta here are left behind on this carbon to give us the conjugate base."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if the base comes along and takes that proton that I just circled in red, these electrons in magenta would be these electrons, and they're on this carbon now, which gives that carbon a negative one formal charge. All right, next we have acetylene. So if we deprotonate acetylene, this pKa has a value of approximately 25. So the electrons in magenta here are left behind on this carbon to give us the conjugate base. All right, for ammonia, for ammonia, if a base took that proton, these electrons would come off onto the nitrogen, so that would be these electrons here, giving us our conjugate base with a pKa value of approximately 36. And this is another compound you see a lot of different pKa values for. I've seen 33, I've seen 38."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta here are left behind on this carbon to give us the conjugate base. All right, for ammonia, for ammonia, if a base took that proton, these electrons would come off onto the nitrogen, so that would be these electrons here, giving us our conjugate base with a pKa value of approximately 36. And this is another compound you see a lot of different pKa values for. I've seen 33, I've seen 38. So again, use whatever pKa values you are given in your class. All right, let's finish this off here. So our last few compounds."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I've seen 33, I've seen 38. So again, use whatever pKa values you are given in your class. All right, let's finish this off here. So our last few compounds. So what's the pKa for this proton? It's approximately 44. So the electrons in magenta come off onto your carbon, and this would be your conjugate base."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So our last few compounds. So what's the pKa for this proton? It's approximately 44. So the electrons in magenta come off onto your carbon, and this would be your conjugate base. And our last example, ethane, is the weakest acid out of all the acids I have listed on this pKa table. So the pKa for this proton, approximately 50. And the electrons in magenta here would be left behind on this carbon."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta come off onto your carbon, and this would be your conjugate base. And our last example, ethane, is the weakest acid out of all the acids I have listed on this pKa table. So the pKa for this proton, approximately 50. And the electrons in magenta here would be left behind on this carbon. So if this is the weakest acid, if ethane is the weakest acid, then this must be on the right here, this must be our strongest base. So you can also use a pKa table to think about the strengths of bases. So if you had some acid here, let me go ahead and draw one, so HA right here."}, {"video_title": "Using a pKa table Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons in magenta here would be left behind on this carbon. So if this is the weakest acid, if ethane is the weakest acid, then this must be on the right here, this must be our strongest base. So you can also use a pKa table to think about the strengths of bases. So if you had some acid here, let me go ahead and draw one, so HA right here. So the electrons in magenta would take this proton, and these electrons would be left behind. And so you would reprotonate, you would form your ethane. So if this functioned as a base and took a proton, then you would make your ethane here."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We've done a lot of electrophilic aromatic substitution reactions. Let's look at the possibility of a nucleophilic aromatic substitution. And so if we started here with bromobenzene and we added a nucleophile, something like the hydroxide anion right here, so negative 1 formal charge, it could function as a nucleophile. And it would, of course, attack the carbon that is bonded to our halogen here. And so when the nucleophile attacks here, if we're thinking about an SN2 type mechanism, it's a concerted mechanism where these electrons should kick off onto the bromine. So if this happened, we would get our benzene ring. And we would get now the OH has substituted in for the bromine."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it would, of course, attack the carbon that is bonded to our halogen here. And so when the nucleophile attacks here, if we're thinking about an SN2 type mechanism, it's a concerted mechanism where these electrons should kick off onto the bromine. So if this happened, we would get our benzene ring. And we would get now the OH has substituted in for the bromine. And the bromine has left in the form of an anion, so the bromide anion here, which has a negative 1 formal charge and is relatively stable on its own. So it's a decent leaving group. So the problem with this is that when our nucleophile is attacking this carbon right here, this is an sp2 hybridized carbon, which is part of this benzene ring, of course."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we would get now the OH has substituted in for the bromine. And the bromine has left in the form of an anion, so the bromide anion here, which has a negative 1 formal charge and is relatively stable on its own. So it's a decent leaving group. So the problem with this is that when our nucleophile is attacking this carbon right here, this is an sp2 hybridized carbon, which is part of this benzene ring, of course. And the benzene ring is going to get in the way of the nucleophile attack via an SN2 type mechanism. And so because of that ring, because you're working with an sp2 hybridized carbon, the nucleophile can't attack in the proper orientation. And so an SN2 mechanism is not possible."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the problem with this is that when our nucleophile is attacking this carbon right here, this is an sp2 hybridized carbon, which is part of this benzene ring, of course. And the benzene ring is going to get in the way of the nucleophile attack via an SN2 type mechanism. And so because of that ring, because you're working with an sp2 hybridized carbon, the nucleophile can't attack in the proper orientation. And so an SN2 mechanism is not possible. So SN2 does not occur at an sp2 hybridized carbon. And so this reaction doesn't proceed this way. What about an SN1 type mechanism?"}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so an SN2 mechanism is not possible. So SN2 does not occur at an sp2 hybridized carbon. And so this reaction doesn't proceed this way. What about an SN1 type mechanism? So if we thought about an SN1 type mechanism, we know the first step in that is dissociation. So these electrons in here are coming off onto the bromine. And so we can go ahead and draw our benzene ring."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "What about an SN1 type mechanism? So if we thought about an SN1 type mechanism, we know the first step in that is dissociation. So these electrons in here are coming off onto the bromine. And so we can go ahead and draw our benzene ring. And since we took a bond away from the carbon that's bonded to the bromine, that would get a plus 1 formal charge like that. And so we form a carbocation. The problem is that this is a very unstable carbocation."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we can go ahead and draw our benzene ring. And since we took a bond away from the carbon that's bonded to the bromine, that would get a plus 1 formal charge like that. And so we form a carbocation. The problem is that this is a very unstable carbocation. We can't really draw any resonance structures for it. And so since it's an unstable carbocation, it's not likely to form. And so an SN1 type mechanism is highly unlikely."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The problem is that this is a very unstable carbocation. We can't really draw any resonance structures for it. And so since it's an unstable carbocation, it's not likely to form. And so an SN1 type mechanism is highly unlikely. It is actually possible if you have an incredible leaving group. But for our purposes, we'll say it's extremely unlikely. So SN2 is out."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so an SN1 type mechanism is highly unlikely. It is actually possible if you have an incredible leaving group. But for our purposes, we'll say it's extremely unlikely. So SN2 is out. SN1 is out. And so you might think that you can't do a nucleophilic aromatic substitution. But as a matter of fact, you can."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So SN2 is out. SN1 is out. And so you might think that you can't do a nucleophilic aromatic substitution. But as a matter of fact, you can. And let's take a look at the criteria in order to do so. So your ring must have an electron withdrawing group, so withdrawing some electron density from the ring. We have that here."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But as a matter of fact, you can. And let's take a look at the criteria in order to do so. So your ring must have an electron withdrawing group, so withdrawing some electron density from the ring. We have that here. Of course, this nitro group is electron withdrawing. The ring must have a leaving group. And we just saw how our halogen here can act as a leaving group."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have that here. Of course, this nitro group is electron withdrawing. The ring must have a leaving group. And we just saw how our halogen here can act as a leaving group. The leaving group is ortho or para to the electron withdrawing group. Well, in this case, those two groups are both ortho to each other. So this molecule could undergo nucleophilic aromatic substitution."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we just saw how our halogen here can act as a leaving group. The leaving group is ortho or para to the electron withdrawing group. Well, in this case, those two groups are both ortho to each other. So this molecule could undergo nucleophilic aromatic substitution. And so let's think about, once again, our hydroxide functioning as the nucleophile. So we have our negatively charged hydroxide anion functioning as a nucleophile, attacking the carbon, once again, that's bonded to our halogen. This time, we're going to move some electrons around."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this molecule could undergo nucleophilic aromatic substitution. And so let's think about, once again, our hydroxide functioning as the nucleophile. So we have our negatively charged hydroxide anion functioning as a nucleophile, attacking the carbon, once again, that's bonded to our halogen. This time, we're going to move some electrons around. So these pi electrons are going to move into here to form a double bond. And these electrons are going to move out onto the oxygen. So let's go ahead and show the result of all those electrons moving."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This time, we're going to move some electrons around. So these pi electrons are going to move into here to form a double bond. And these electrons are going to move out onto the oxygen. So let's go ahead and show the result of all those electrons moving. So we have our ring. We have our pi electrons here. We have our bromine still attached to the ring."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the result of all those electrons moving. So we have our ring. We have our pi electrons here. We have our bromine still attached to the ring. So let's go ahead and put those lone pairs of electrons on it. And now we have an OH attached to our ring, too, like that. So let's put those lone pairs of electrons on."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our bromine still attached to the ring. So let's go ahead and put those lone pairs of electrons on it. And now we have an OH attached to our ring, too, like that. So let's put those lone pairs of electrons on. Now our nitrogen is double bonded to our ring, like that. The nitrogen is still bonded to an oxygen on the right. The oxygen on the right still has a negative 1 formal charge."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's put those lone pairs of electrons on. Now our nitrogen is double bonded to our ring, like that. The nitrogen is still bonded to an oxygen on the right. The oxygen on the right still has a negative 1 formal charge. And now the nitrogen is bonded to an oxygen on the left, also with a negative 1 formal charge, like that. The nitrogen itself has a plus 1 formal charge. So let's see if we can show the movement of all of those electrons here."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen on the right still has a negative 1 formal charge. And now the nitrogen is bonded to an oxygen on the left, also with a negative 1 formal charge, like that. The nitrogen itself has a plus 1 formal charge. So let's see if we can show the movement of all of those electrons here. And so if I show these electrons in magenta on our hydroxide anion, you could think about them as being this bond now. And these pi electrons in here, like that, these you could think about moving out to here to form a pi bond with the nitrogen. And then finally, these electrons right in here moved out onto this oxygen on the left to give it a negative 1 formal charge."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's see if we can show the movement of all of those electrons here. And so if I show these electrons in magenta on our hydroxide anion, you could think about them as being this bond now. And these pi electrons in here, like that, these you could think about moving out to here to form a pi bond with the nitrogen. And then finally, these electrons right in here moved out onto this oxygen on the left to give it a negative 1 formal charge. And so the first step in this mechanism is the addition of the nucleophile. So let me go ahead and write that. So it's the addition step, where we add our nucleophile to the ring."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, these electrons right in here moved out onto this oxygen on the left to give it a negative 1 formal charge. And so the first step in this mechanism is the addition of the nucleophile. So let me go ahead and write that. So it's the addition step, where we add our nucleophile to the ring. And when you do that, you're, of course, adding electrons to the ring. And that negative charge that you're adding actually ends up on this oxygen here. So this oxygen has a negative 1 formal charge."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's the addition step, where we add our nucleophile to the ring. And when you do that, you're, of course, adding electrons to the ring. And that negative charge that you're adding actually ends up on this oxygen here. So this oxygen has a negative 1 formal charge. It's able to stabilize that, since it's very electronegative. And so the presence of your electron withdrawing group withdraws some electron density from the ring. It allows some of that electron density to be temporarily stored on your electron withdrawing group."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this oxygen has a negative 1 formal charge. It's able to stabilize that, since it's very electronegative. And so the presence of your electron withdrawing group withdraws some electron density from the ring. It allows some of that electron density to be temporarily stored on your electron withdrawing group. And that stabilizes this intermediate here. You can actually draw some other resonance structures. But this is the one that we're most concerned with, the one showing the negative charge on the oxygen."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It allows some of that electron density to be temporarily stored on your electron withdrawing group. And that stabilizes this intermediate here. You can actually draw some other resonance structures. But this is the one that we're most concerned with, the one showing the negative charge on the oxygen. So once again, this electron withdrawing group is able to stabilize that negative charge that was added to the ring. And it's only able to stabilize it, because it is ortho to the leaving group. And so in our next step, we're going to show the leaving group leaving."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But this is the one that we're most concerned with, the one showing the negative charge on the oxygen. So once again, this electron withdrawing group is able to stabilize that negative charge that was added to the ring. And it's only able to stabilize it, because it is ortho to the leaving group. And so in our next step, we're going to show the leaving group leaving. So this is the elimination step, so the elimination of the leaving group. And again, that nitro group was temporarily holding on to some of that negative charge in those electrons. And they're going to move back into here, which would push these electrons back into here."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so in our next step, we're going to show the leaving group leaving. So this is the elimination step, so the elimination of the leaving group. And again, that nitro group was temporarily holding on to some of that negative charge in those electrons. And they're going to move back into here, which would push these electrons back into here. And then these electrons would kick off onto your bromine. And so we can go ahead and show the final product now, where we have our benzene ring. And the bromine has left in the form of the bromide anion."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And they're going to move back into here, which would push these electrons back into here. And then these electrons would kick off onto your bromine. And so we can go ahead and show the final product now, where we have our benzene ring. And the bromine has left in the form of the bromide anion. So we have an OH attached to our ring. And now we have our nitro group back, the way it looked before, with a negative 1 formal charge on this oxygen. And this oxygen over here double bonded to the nitrogen, nitrogen having a plus 1 formal charge."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the bromine has left in the form of the bromide anion. So we have an OH attached to our ring. And now we have our nitro group back, the way it looked before, with a negative 1 formal charge on this oxygen. And this oxygen over here double bonded to the nitrogen, nitrogen having a plus 1 formal charge. So the electrons in green over here on this structure, so these electrons moved back into here to form this bond. And the electrons in blue moved back into here to reform your ring this way. And we're done."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this oxygen over here double bonded to the nitrogen, nitrogen having a plus 1 formal charge. So the electrons in green over here on this structure, so these electrons moved back into here to form this bond. And the electrons in blue moved back into here to reform your ring this way. And we're done. We've seen that the OH has substituted in for the halogen here. So this is called the addition-elimination mechanism. And you can see why."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're done. We've seen that the OH has substituted in for the halogen here. So this is called the addition-elimination mechanism. And you can see why. First you add your nucleophile. And then that electron density is temporarily stored in the electron withdrawing group. And it comes off again to eliminate your halogen like that."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you can see why. First you add your nucleophile. And then that electron density is temporarily stored in the electron withdrawing group. And it comes off again to eliminate your halogen like that. All right, so that was an example of a situation where the two groups are ortho to each other. Let's do one where the two groups are para to each other. So let's look at this next example here."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it comes off again to eliminate your halogen like that. All right, so that was an example of a situation where the two groups are ortho to each other. Let's do one where the two groups are para to each other. So let's look at this next example here. And so once again, we have an electron withdrawing group, our nitro group. We have a leaving group, our halogen. And so therefore, we can perform nucleophilic aromatic substitution."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this next example here. And so once again, we have an electron withdrawing group, our nitro group. We have a leaving group, our halogen. And so therefore, we can perform nucleophilic aromatic substitution. And so I'm going to use the same nucleophiles before. I'm going to use hydroxide. So we have a negative 1 formal charge."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, we can perform nucleophilic aromatic substitution. And so I'm going to use the same nucleophiles before. I'm going to use hydroxide. So we have a negative 1 formal charge. And our nucleophile is going to attack, of course, the carbon that is bonded to our leaving group, our halogen, which would push these electrons over to here, which would push these electrons over to here. And then that pushes these electrons off onto our oxygen once again. So let's go ahead and show, again, all of those electrons after they have moved here."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a negative 1 formal charge. And our nucleophile is going to attack, of course, the carbon that is bonded to our leaving group, our halogen, which would push these electrons over to here, which would push these electrons over to here. And then that pushes these electrons off onto our oxygen once again. So let's go ahead and show, again, all of those electrons after they have moved here. So we have our bromine still attached to our ring. Our OH is now attached to our ring like that. So it would have two lone pairs of electrons on it."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show, again, all of those electrons after they have moved here. So we have our bromine still attached to our ring. Our OH is now attached to our ring like that. So it would have two lone pairs of electrons on it. And we have some pi electrons here, pi electrons here. Nitrogen is once again double bonded to our ring now. It's bonded to this oxygen on the left, which has a negative 1 formal charge."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it would have two lone pairs of electrons on it. And we have some pi electrons here, pi electrons here. Nitrogen is once again double bonded to our ring now. It's bonded to this oxygen on the left, which has a negative 1 formal charge. And the oxygen on the right now has a negative 1 formal charge like that, with this nitrogen here being positively charged. So two negatively charged oxygens and one positively charged nitrogen. And so once again, this is the addition step, the addition of the nucleophile to the ring."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's bonded to this oxygen on the left, which has a negative 1 formal charge. And the oxygen on the right now has a negative 1 formal charge like that, with this nitrogen here being positively charged. So two negatively charged oxygens and one positively charged nitrogen. And so once again, this is the addition step, the addition of the nucleophile to the ring. And now that negative charge, when we added some electrons to our ring, that negative charge is partially residing on this oxygen here. And so our electron withdrawing group stabilizes this intermediate. The next step, of course, is elimination."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, this is the addition step, the addition of the nucleophile to the ring. And now that negative charge, when we added some electrons to our ring, that negative charge is partially residing on this oxygen here. And so our electron withdrawing group stabilizes this intermediate. The next step, of course, is elimination. And so if these electrons move back into here, that would push these electrons back into here, which pushes these electrons over to here. And then these electrons come off onto our leaving group. And so the bromide anion leaves."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The next step, of course, is elimination. And so if these electrons move back into here, that would push these electrons back into here, which pushes these electrons over to here. And then these electrons come off onto our leaving group. And so the bromide anion leaves. And we now have our OH has substituted in for it. So we have our ring. We now have our OH on there."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the bromide anion leaves. And we now have our OH has substituted in for it. So we have our ring. We now have our OH on there. And we have, of course, our nitro group. So over here, double bonded to this oxygen. This nitrogen is a single bond to this oxygen."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We now have our OH on there. And we have, of course, our nitro group. So over here, double bonded to this oxygen. This nitrogen is a single bond to this oxygen. And negative 1 formal charge and a plus 1 formal charge like that. And so this is an example of the addition elimination mechanism where your electron withdrawing group and your leaving group are para to each other. If you try to do this where the two groups are meta to each other, you'll see that it doesn't work."}, {"video_title": "Nucleophilic aromatic substitution I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This nitrogen is a single bond to this oxygen. And negative 1 formal charge and a plus 1 formal charge like that. And so this is an example of the addition elimination mechanism where your electron withdrawing group and your leaving group are para to each other. If you try to do this where the two groups are meta to each other, you'll see that it doesn't work. You can't show that electron density out onto this oxygen. You can't show your electrons moving around to do that. So it's only when those groups are either ortho or para to each other."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "On the left we have one compound, on the right we have its mirror image. We're going to assign an RS to each of our enantiomers. So let's start with step one. In step one we prioritize the four groups attached to our chiral center. And we do that according to atomic number. Let's start with the enantiomer on the left. We know that this carbon is our chiral center and we have four different things attached to this carbon."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "In step one we prioritize the four groups attached to our chiral center. And we do that according to atomic number. Let's start with the enantiomer on the left. We know that this carbon is our chiral center and we have four different things attached to this carbon. We go over here to the right to a very shortened version of our periodic table. And we can see that bromine has the highest atomic number out of these four atoms. So bromine gets highest priority."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that this carbon is our chiral center and we have four different things attached to this carbon. We go over here to the right to a very shortened version of our periodic table. And we can see that bromine has the highest atomic number out of these four atoms. So bromine gets highest priority. We're going to give the bromine a number one. Chlorine has the next highest atomic number at 17. So chlorine gets second highest priority."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So bromine gets highest priority. We're going to give the bromine a number one. Chlorine has the next highest atomic number at 17. So chlorine gets second highest priority. Fluorine has the next highest with a nine. So fluorine gets a three. And finally hydrogen is the lowest priority group, the lowest atomic number of one."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So chlorine gets second highest priority. Fluorine has the next highest with a nine. So fluorine gets a three. And finally hydrogen is the lowest priority group, the lowest atomic number of one. So hydrogen gets a four. So step one is done. Step two, orient the groups so that the lowest priority group is projecting or pointing away from you."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And finally hydrogen is the lowest priority group, the lowest atomic number of one. So hydrogen gets a four. So step one is done. Step two, orient the groups so that the lowest priority group is projecting or pointing away from you. Our lowest priority group is hydrogen and it's already going away from us. The hydrogen is on a dash here. So step two is pretty much done."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Step two, orient the groups so that the lowest priority group is projecting or pointing away from you. Our lowest priority group is hydrogen and it's already going away from us. The hydrogen is on a dash here. So step two is pretty much done. Step three, determine if the sequence one, two, three is clockwise or counterclockwise. So we're going to ignore our hydrogen here. So I'll just kind of rub it out here."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So step two is pretty much done. Step three, determine if the sequence one, two, three is clockwise or counterclockwise. So we're going to ignore our hydrogen here. So I'll just kind of rub it out here. And look at what's happening with one, two, and three. So one, two, and three are going around in this direction. That is clockwise."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So I'll just kind of rub it out here. And look at what's happening with one, two, and three. So one, two, and three are going around in this direction. That is clockwise. Therefore this is the R enantiomer. So I'm going to write R bromo chlorofluoromethane. Let's do the same thing for its mirror image."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That is clockwise. Therefore this is the R enantiomer. So I'm going to write R bromo chlorofluoromethane. Let's do the same thing for its mirror image. So this compound on the right. We know that this carbon is our chiral center. We already know how to assign priority."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's do the same thing for its mirror image. So this compound on the right. We know that this carbon is our chiral center. We already know how to assign priority. Bromine gets a number one. Chlorine gets a number two. Fluorine gets a number three."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We already know how to assign priority. Bromine gets a number one. Chlorine gets a number two. Fluorine gets a number three. And hydrogen gets a number four. So step one is done. Step two, point the lowest priority group away from you."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Fluorine gets a number three. And hydrogen gets a number four. So step one is done. Step two, point the lowest priority group away from you. Well that's already happening here. So step two is done. Step three, determine if the sequence one, two, three is clockwise or counterclockwise."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Step two, point the lowest priority group away from you. Well that's already happening here. So step two is done. Step three, determine if the sequence one, two, three is clockwise or counterclockwise. So let's ignore this group going away from us, the lowest priority group. Let's go around one, two, and three. So going around in a circle, one, two, and three in this direction."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Step three, determine if the sequence one, two, three is clockwise or counterclockwise. So let's ignore this group going away from us, the lowest priority group. Let's go around one, two, and three. So going around in a circle, one, two, and three in this direction. We know that is counterclockwise. So this must be the S enantiomer. So this would be S bromo chlorofluoromethane."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So going around in a circle, one, two, and three in this direction. We know that is counterclockwise. So this must be the S enantiomer. So this would be S bromo chlorofluoromethane. So step three is done. So that's how you assign a configuration to a chiral center. Here we have another pair of enantiomers."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this would be S bromo chlorofluoromethane. So step three is done. So that's how you assign a configuration to a chiral center. Here we have another pair of enantiomers. So this alcohol on the left, and its mirror image on the right. Let's start with the one on the left. We know from earlier videos that this is the chiral center."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here we have another pair of enantiomers. So this alcohol on the left, and its mirror image on the right. Let's start with the one on the left. We know from earlier videos that this is the chiral center. And our goal is to assign a configuration to this chiral center. Let's go ahead and redraw the molecule. So that carbon is our chiral center."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know from earlier videos that this is the chiral center. And our goal is to assign a configuration to this chiral center. Let's go ahead and redraw the molecule. So that carbon is our chiral center. Attached to that carbon is our OH. So I'll draw the OH on a wedge. We know that hydrogen is there going away from us."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon is our chiral center. Attached to that carbon is our OH. So I'll draw the OH on a wedge. We know that hydrogen is there going away from us. So that's a dash. So even though it's not drawn in, we already know it's there. To the right we have a methyl group."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that hydrogen is there going away from us. So that's a dash. So even though it's not drawn in, we already know it's there. To the right we have a methyl group. So a CH3. So let's draw in a CH3. And then finally to the left we have an ethyl group."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "To the right we have a methyl group. So a CH3. So let's draw in a CH3. And then finally to the left we have an ethyl group. So that would be a CH2. And then CH3. All right, let's go back to our chiral center."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then finally to the left we have an ethyl group. So that would be a CH2. And then CH3. All right, let's go back to our chiral center. So here is our carbon. That's our chiral center. Let's look at the atoms that are directly bonded to this carbon."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's go back to our chiral center. So here is our carbon. That's our chiral center. Let's look at the atoms that are directly bonded to this carbon. So there's an oxygen directly bonded to the carbon. There's a hydrogen. And then we have two carbons."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at the atoms that are directly bonded to this carbon. So there's an oxygen directly bonded to the carbon. There's a hydrogen. And then we have two carbons. So let's assign priorities. If we look over here at our very shortened version of the periodic table, we know that oxygen has the highest atomic number out of those atoms. So oxygen gets highest priority."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we have two carbons. So let's assign priorities. If we look over here at our very shortened version of the periodic table, we know that oxygen has the highest atomic number out of those atoms. So oxygen gets highest priority. The OH group gets the highest priority. So this gets a number one. Hydrogen has the lowest atomic number."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen gets highest priority. The OH group gets the highest priority. So this gets a number one. Hydrogen has the lowest atomic number. So the hydrogen is the lowest priority group. So we say that's group number four. Finally we have two carbons."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Hydrogen has the lowest atomic number. So the hydrogen is the lowest priority group. So we say that's group number four. Finally we have two carbons. And two carbons would of course be a tie. Carbon has atomic number of six. So we have to find a way to break the tie."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Finally we have two carbons. And two carbons would of course be a tie. Carbon has atomic number of six. So we have to find a way to break the tie. So the way to do a tie breaker is to look at the atoms that are directly bonded to those carbons. So we'll start with the carbon on the right. The carbon on the right is directly bonded to three hydrogens."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have to find a way to break the tie. So the way to do a tie breaker is to look at the atoms that are directly bonded to those carbons. So we'll start with the carbon on the right. The carbon on the right is directly bonded to three hydrogens. One, two, three. So let's go ahead and write that out. So hydrogen, hydrogen, hydrogen."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the right is directly bonded to three hydrogens. One, two, three. So let's go ahead and write that out. So hydrogen, hydrogen, hydrogen. The carbon over here is directly bonded to a carbon, a hydrogen, and a hydrogen. And we're gonna put those atoms in order of decreasing atomic number. So this carbon has a higher atomic number than these hydrogens."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So hydrogen, hydrogen, hydrogen. The carbon over here is directly bonded to a carbon, a hydrogen, and a hydrogen. And we're gonna put those atoms in order of decreasing atomic number. So this carbon has a higher atomic number than these hydrogens. So we write carbon, hydrogen, hydrogen. Next we compare these lists. So on the left we have carbon, hydrogen, hydrogen."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon has a higher atomic number than these hydrogens. So we write carbon, hydrogen, hydrogen. Next we compare these lists. So on the left we have carbon, hydrogen, hydrogen. On the right we have hydrogen, hydrogen, hydrogen. And we look for the first point of difference. Well that's carbon versus this hydrogen here."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So on the left we have carbon, hydrogen, hydrogen. On the right we have hydrogen, hydrogen, hydrogen. And we look for the first point of difference. Well that's carbon versus this hydrogen here. Carbon has a higher atomic number than this hydrogen. So the carbon wins. So this group gets a higher priority."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well that's carbon versus this hydrogen here. Carbon has a higher atomic number than this hydrogen. So the carbon wins. So this group gets a higher priority. So the ethyl group has a higher priority than the methyl group. So the ethyl group must be the second highest priority. So this gets a number two."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this group gets a higher priority. So the ethyl group has a higher priority than the methyl group. So the ethyl group must be the second highest priority. So this gets a number two. And the methyl group must get a number three. So now we've assigned priority to all of our groups. Let's go down here and let's write it in."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this gets a number two. And the methyl group must get a number three. So now we've assigned priority to all of our groups. Let's go down here and let's write it in. The OH got highest priority, so that gets a number one. The ethyl group got second highest priority, so that gets a number two. The methyl group got third highest priority, so that's a number three."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's go down here and let's write it in. The OH got highest priority, so that gets a number one. The ethyl group got second highest priority, so that gets a number two. The methyl group got third highest priority, so that's a number three. And the hydrogen, our lowest priority group, is pointing away from us. So that takes care of step two. Step three is to see what the sequence is doing."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The methyl group got third highest priority, so that's a number three. And the hydrogen, our lowest priority group, is pointing away from us. So that takes care of step two. Step three is to see what the sequence is doing. So if we go around in a circle, one, two, three, we're going around this way. We're going around clockwise. And we know if we're going around clockwise, that must be the R-enantiomer."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Step three is to see what the sequence is doing. So if we go around in a circle, one, two, three, we're going around this way. We're going around clockwise. And we know if we're going around clockwise, that must be the R-enantiomer. So this is R-2-butanol. If that's R-2-butanol, the mirror image must be S-2-butanol. So let's go ahead and let's double check and make sure that's true."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we know if we're going around clockwise, that must be the R-enantiomer. So this is R-2-butanol. If that's R-2-butanol, the mirror image must be S-2-butanol. So let's go ahead and let's double check and make sure that's true. So this is our chiral center. This is our chiral center. We know the OH is highest priority, so that gets a number one."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and let's double check and make sure that's true. So this is our chiral center. This is our chiral center. We know the OH is highest priority, so that gets a number one. Let me go ahead and change colors for this. So this one gets a number one. We know that our ethyl group gets a number two."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know the OH is highest priority, so that gets a number one. Let me go ahead and change colors for this. So this one gets a number one. We know that our ethyl group gets a number two. We know our methyl group gets a number three. And we also know there's a hydrogen going away from us in space. So our lowest priority group is projecting away."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that our ethyl group gets a number two. We know our methyl group gets a number three. And we also know there's a hydrogen going away from us in space. So our lowest priority group is projecting away. So all we have to do now is look at what's happening with one, two, and three. And one, two, and three are going around this way. And that, of course, is counterclockwise."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So our lowest priority group is projecting away. So all we have to do now is look at what's happening with one, two, and three. And one, two, and three are going around this way. And that, of course, is counterclockwise. And counterclockwise is S. So this is S-2-butanol. We just saw that this is S-2-butanol. But what if you were given the dot structure on the right and asked to assign a configuration to the chiral center?"}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And that, of course, is counterclockwise. And counterclockwise is S. So this is S-2-butanol. We just saw that this is S-2-butanol. But what if you were given the dot structure on the right and asked to assign a configuration to the chiral center? So here is the chiral center. If the OH this time is going away from us, that means that the hydrogen is coming out at us in space. So we've already seen how to assign priority."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But what if you were given the dot structure on the right and asked to assign a configuration to the chiral center? So here is the chiral center. If the OH this time is going away from us, that means that the hydrogen is coming out at us in space. So we've already seen how to assign priority. The OH gets a number one. The ethyl gets a number two. And the methyl gets a number three."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we've already seen how to assign priority. The OH gets a number one. The ethyl gets a number two. And the methyl gets a number three. And the hydrogen gets a number four. When you go to step two, step two says to orient the group so the lowest priority group is projecting away from you. But that's not what we have here."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And the methyl gets a number three. And the hydrogen gets a number four. When you go to step two, step two says to orient the group so the lowest priority group is projecting away from you. But that's not what we have here. Here we have our lowest priority group coming out at us in space because this is a wedge. So one thing you could do would be to take this compound and in your head mentally rotate it so the hydrogen is pointing away from you in space. And when you do that, you'll see that it's the same as the one on the left."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But that's not what we have here. Here we have our lowest priority group coming out at us in space because this is a wedge. So one thing you could do would be to take this compound and in your head mentally rotate it so the hydrogen is pointing away from you in space. And when you do that, you'll see that it's the same as the one on the left. So this is just two different ways to represent the same enantiomer. So in the video on drawing enantiomers, I actually showed you the video where I rotated this compound to prove that these two drawings represent the same compound. So now your hydrogen is going away from you in space."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And when you do that, you'll see that it's the same as the one on the left. So this is just two different ways to represent the same enantiomer. So in the video on drawing enantiomers, I actually showed you the video where I rotated this compound to prove that these two drawings represent the same compound. So now your hydrogen is going away from you in space. And that's how we got S because this was a number one, this was a number two, and this was a number three. So we went around this way and we saw that is counterclockwise and so we got S. But what if you didn't want to rotate the molecule in your head? Sometimes it's pretty difficult to do."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So now your hydrogen is going away from you in space. And that's how we got S because this was a number one, this was a number two, and this was a number three. So we went around this way and we saw that is counterclockwise and so we got S. But what if you didn't want to rotate the molecule in your head? Sometimes it's pretty difficult to do. For this one, it's not too bad, but it can get a little bit tricky. So there is a trick that you can use. There is a trick that you can do."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes it's pretty difficult to do. For this one, it's not too bad, but it can get a little bit tricky. So there is a trick that you can use. There is a trick that you can do. And you can just start with the drawing on the right. You can go ahead and number your groups in terms of priority. And then you can go ahead and just ignore the hydrogen for the time being, even though the hydrogen is coming out at us in space here, if you ignore it and look at one, two, three."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There is a trick that you can do. And you can just start with the drawing on the right. You can go ahead and number your groups in terms of priority. And then you can go ahead and just ignore the hydrogen for the time being, even though the hydrogen is coming out at us in space here, if you ignore it and look at one, two, three. So one, two, three is going around this way. And that would be clockwise. Now we know that clockwise is R. So I'm going to write it looks R here."}, {"video_title": "R,S system Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then you can go ahead and just ignore the hydrogen for the time being, even though the hydrogen is coming out at us in space here, if you ignore it and look at one, two, three. So one, two, three is going around this way. And that would be clockwise. Now we know that clockwise is R. So I'm going to write it looks R here. It looks R. But since the hydrogen is coming out at us in space, we can take the opposite of how it looks. So if it looks R and the hydrogen is coming out in space, you know it's actually the S enantiomer. So that's a little trick that you can do instead of rotating the molecule in your head."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I start with my alkene on the left and I add a halogen to it and some water, you can see that an OH and a halogen are added anti to each other. So anti, or on opposite sides of where the double bond used to be. The mechanism for this reaction starts off the exact same way the halogenation reaction did. And so we have our halogen approaching our alkene. And we saw in that video that the halogen is usually, of course, nonpolar because those two atoms have the exact same electronegativity. So these blue electrons in here are pulled with equal force to either halogen. So it's a nonpolar molecule."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so we have our halogen approaching our alkene. And we saw in that video that the halogen is usually, of course, nonpolar because those two atoms have the exact same electronegativity. So these blue electrons in here are pulled with equal force to either halogen. So it's a nonpolar molecule. However, if the pi electrons in my alkene here get too close to the electrons in blue, we saw how that could induce a temporary dipole on the halogen molecule. So those electrons in blue are repelled by the electrons in magenta and pushed closer to the top halogen, which gives the top halogen a partial negative charge and leaves the bottom halogen with a partial positive charge. The bottom halogen is now an electrophile."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's a nonpolar molecule. However, if the pi electrons in my alkene here get too close to the electrons in blue, we saw how that could induce a temporary dipole on the halogen molecule. So those electrons in blue are repelled by the electrons in magenta and pushed closer to the top halogen, which gives the top halogen a partial negative charge and leaves the bottom halogen with a partial positive charge. The bottom halogen is now an electrophile. So it wants electrons. It's going to get electrons from those pi electrons here, which are going to move out and nucleophilic attack that partially positively charged halogen atom. And then this lone pair of electrons is going to form a bond with this carbon."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The bottom halogen is now an electrophile. So it wants electrons. It's going to get electrons from those pi electrons here, which are going to move out and nucleophilic attack that partially positively charged halogen atom. And then this lone pair of electrons is going to form a bond with this carbon. At the same time, these blue electrons move out onto the halogen. So when we draw the result of all those electrons moving around, we're going to form a bond between the carbon on the right and the halogen. And we use the magenta electrons to show that."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then this lone pair of electrons is going to form a bond with this carbon. At the same time, these blue electrons move out onto the halogen. So when we draw the result of all those electrons moving around, we're going to form a bond between the carbon on the right and the halogen. And we use the magenta electrons to show that. So there's now a bond there. And we used red electrons before to show these electrons in here, forming a bond with the carbon on the left. That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we use the magenta electrons to show that. So there's now a bond there. And we used red electrons before to show these electrons in here, forming a bond with the carbon on the left. That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge. We called this our cyclic halonium ion in an earlier video. And if I think about that cyclic halonium ion, I think about the halogen being very electronegative. It's going to attract, I'll say, the electrons in magenta again, just to be consistent, closer towards it."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That halogen had two lone pairs of electrons still on it, like that, which gives that halogen a plus 1 formal charge. We called this our cyclic halonium ion in an earlier video. And if I think about that cyclic halonium ion, I think about the halogen being very electronegative. It's going to attract, I'll say, the electrons in magenta again, just to be consistent, closer towards it. So it's going to take away a little bit of electron density from this carbon right down here. So I'm going to say this carbon is partially positive. And it's going to have some partial carbocationic character."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to attract, I'll say, the electrons in magenta again, just to be consistent, closer towards it. So it's going to take away a little bit of electron density from this carbon right down here. So I'm going to say this carbon is partially positive. And it's going to have some partial carbocationic character. So in the next step of the mechanism, water's going to come along. And water's going to function as a nucleophile. So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And it's going to have some partial carbocationic character. So in the next step of the mechanism, water's going to come along. And water's going to function as a nucleophile. So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here. And so when that lone pair of electrons on oxygen attacks this carbon, that's going to kick the electrons in magenta off onto your halogen. And so let's go ahead and draw the product. We're going to have on the left carbon, this halogen now used to have two lone pairs of electrons."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So one of the lone pairs of electrons on water is going to nucleophilic attack our electrophile, which is this carbon right here. And so when that lone pair of electrons on oxygen attacks this carbon, that's going to kick the electrons in magenta off onto your halogen. And so let's go ahead and draw the product. We're going to have on the left carbon, this halogen now used to have two lone pairs of electrons. It picked up the ones in magenta. So now it looks like that. And on the right, we still have the carbon on the right bonded to two other things, except now it's now bonded to what used to be our water molecule."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We're going to have on the left carbon, this halogen now used to have two lone pairs of electrons. It picked up the ones in magenta. So now it looks like that. And on the right, we still have the carbon on the right bonded to two other things, except now it's now bonded to what used to be our water molecule. So the oxygen is now bonded to the carbon. And there's still one lone pair of electrons on that oxygen, giving it a plus 1 formal charge. So let's go ahead and highlight these electrons here in blue."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And on the right, we still have the carbon on the right bonded to two other things, except now it's now bonded to what used to be our water molecule. So the oxygen is now bonded to the carbon. And there's still one lone pair of electrons on that oxygen, giving it a plus 1 formal charge. So let's go ahead and highlight these electrons here in blue. Those electrons in blue are the ones that formed this bond between the carbon and the oxygen. And so we're almost done. The last step of the mechanism would just be an acid-base reaction."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and highlight these electrons here in blue. Those electrons in blue are the ones that formed this bond between the carbon and the oxygen. And so we're almost done. The last step of the mechanism would just be an acid-base reaction. So another water molecule comes along. And one of the lone pairs of electrons on the water molecule is going to function as a base and take this proton, leaving these two electrons behind on the oxygen. And we are finally done."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The last step of the mechanism would just be an acid-base reaction. So another water molecule comes along. And one of the lone pairs of electrons on the water molecule is going to function as a base and take this proton, leaving these two electrons behind on the oxygen. And we are finally done. We have formed our halohydrin. So I have my halogen on one side. And then I now have my OH on the opposite side like that."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we are finally done. We have formed our halohydrin. So I have my halogen on one side. And then I now have my OH on the opposite side like that. Let's go ahead and do an example so we can examine the stereochemistry a little bit more here. So if I start with an alkene. So I'm going to start with this alkene right here."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then I now have my OH on the opposite side like that. Let's go ahead and do an example so we can examine the stereochemistry a little bit more here. So if I start with an alkene. So I'm going to start with this alkene right here. And to this alkene, we are going to add bromine and water. And we're going to think about doing this two different ways here. So we'll start with the way on the left, so Br2 and H2O."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to start with this alkene right here. And to this alkene, we are going to add bromine and water. And we're going to think about doing this two different ways here. So we'll start with the way on the left, so Br2 and H2O. And then we'll come back and we'll go ahead and do this on the right, so Br2 and H2O. So on the left side, I'm going to think about the formation of that bromonium ion here. So I'm going to, once again, look at this molecule a little bit from above."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start with the way on the left, so Br2 and H2O. And then we'll come back and we'll go ahead and do this on the right, so Br2 and H2O. So on the left side, I'm going to think about the formation of that bromonium ion here. So I'm going to, once again, look at this molecule a little bit from above. So looking down. And I'm going to say the bromonium ion is going to form this way. So the bromine is going to form on top here."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to, once again, look at this molecule a little bit from above. So looking down. And I'm going to say the bromonium ion is going to form this way. So the bromine is going to form on top here. And there's going to be two lone pairs of electrons on that bromine. It's going to have a plus 1 formal charge. And if I look at this carbon right here, that's this carbon."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the bromine is going to form on top here. And there's going to be two lone pairs of electrons on that bromine. It's going to have a plus 1 formal charge. And if I look at this carbon right here, that's this carbon. So I'm going to say that my methyl group is now going down in space. So now I have a methyl group down in space like that. So with the addition of my bromonium ion, that would be my intermediate."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And if I look at this carbon right here, that's this carbon. So I'm going to say that my methyl group is now going down in space. So now I have a methyl group down in space like that. So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here's H2O. And I think about which carbon will the oxygen attack. So I have two options."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So with the addition of my bromonium ion, that would be my intermediate. And so now, when I think about water coming along and acting as a nucleophile, so here's H2O. And I think about which carbon will the oxygen attack. So I have two options. This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the oxygen is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I have two options. This oxygen could attack the carbon on the left, or it could attack the carbon on the right. It's been proven that the oxygen is going to attack the most substituted carbon. So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation or a partial carbocation character. So you could think about it as being like a partial secondary carbocation on the left here, if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red, if I think about that being a carbocation, that would be bonded to one, two, three other carbons."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at the carbon on the left, and if I think about what sort of carbocation would that be, the carbon on the left is bonded to two other carbons. So this would be similar to a secondary carbocation or a partial carbocation character. So you could think about it as being like a partial secondary carbocation on the left here, if this was a partial positive. Or on the right, if I think about this carbon right here, the one in red, if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Or on the right, if I think about this carbon right here, the one in red, if I think about that being a carbocation, that would be bonded to one, two, three other carbons. So it's like a tertiary carbocation. And we know that tertiary carbocations are more stable than secondary. So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character. And that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So even though this isn't a full carbocation, this carbon in red exhibits some partial carbocation character. And that is where our water is going to attack. So the nucleophile is going to attack the electrophile. And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons in here would kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. So what would we have here?"}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And it's more stable for it to attack this one on the right, since it has partial carbocation character similar to a tertiary carbocation. And if it attacks that carbon on the right, these electrons in here would kick off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. So what would we have here? Let's go ahead and draw our ring. And the bromine is going to swing over to the carbon on the left. It's now going to have three lone pairs of electrons around it like that."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what would we have here? Let's go ahead and draw our ring. And the bromine is going to swing over to the carbon on the left. It's now going to have three lone pairs of electrons around it like that. And the methyl group that was down relative to the plane is going to be pushed up when that water nucleophilic attacks. So now the methyl group is up. And this oxygen is now going to be bonded to this carbon."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's now going to have three lone pairs of electrons around it like that. And the methyl group that was down relative to the plane is going to be pushed up when that water nucleophilic attacks. So now the methyl group is up. And this oxygen is now going to be bonded to this carbon. And so we still have our two hydrogens attached to it like that. And there's a lone pair of electrons on this oxygen, giving it a plus 1 formal charge. So once again, let's go ahead and highlight those electrons."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this oxygen is now going to be bonded to this carbon. And so we still have our two hydrogens attached to it like that. And there's a lone pair of electrons on this oxygen, giving it a plus 1 formal charge. So once again, let's go ahead and highlight those electrons. So I'll draw them in blue here. These electrons right here, those are the ones that formed this bond. So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So once again, let's go ahead and highlight those electrons. So I'll draw them in blue here. These electrons right here, those are the ones that formed this bond. So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here. So on the left, I showed the bromine adding from the top. If I think about the alkene portion of my starting material, if I think about the alkene portion, so I'm talking about right here, well, there's a chance that the bromonium ion could form from below that plane as well. So let's go ahead and draw the result of that over here on the right."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go back and let's think about the formation of that cyclic bromonium ion in a different way here. So on the left, I showed the bromine adding from the top. If I think about the alkene portion of my starting material, if I think about the alkene portion, so I'm talking about right here, well, there's a chance that the bromonium ion could form from below that plane as well. So let's go ahead and draw the result of that over here on the right. So if I'm showing another bromonium ion, that is possible. And this time, the bromine is going to add from below the plane like that. So it's going to get two lone pairs of electrons and have a plus 1 formal charge, just like usual."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the result of that over here on the right. So if I'm showing another bromonium ion, that is possible. And this time, the bromine is going to add from below the plane like that. So it's going to get two lone pairs of electrons and have a plus 1 formal charge, just like usual. And I will say that this is the carbon that has the methyl on it. So this is the carbon that's going to have the methyl. So let's go ahead and draw that in as well."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to get two lone pairs of electrons and have a plus 1 formal charge, just like usual. And I will say that this is the carbon that has the methyl on it. So this is the carbon that's going to have the methyl. So let's go ahead and draw that in as well. So let's see, I'll put it like that. So that's my CH3. And this time, when I think about where will water attack, so let's go ahead and think about water as my nucleophile."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that in as well. So let's see, I'll put it like that. So that's my CH3. And this time, when I think about where will water attack, so let's go ahead and think about water as my nucleophile. It's the same idea. If I compare the carbons on either side, so I compare this carbon with this carbon, it's the carbon on the right that's going to be the most stable partial carbocation. So I'll draw a partial carbocation here."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this time, when I think about where will water attack, so let's go ahead and think about water as my nucleophile. It's the same idea. If I compare the carbons on either side, so I compare this carbon with this carbon, it's the carbon on the right that's going to be the most stable partial carbocation. So I'll draw a partial carbocation here. It would be the most stable one. So that is where my nucleophile is going to attack. So I can think about this lone pair of electrons on oxygen going to attack right here, which would kick these electrons off onto the bromine."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll draw a partial carbocation here. It would be the most stable one. So that is where my nucleophile is going to attack. So I can think about this lone pair of electrons on oxygen going to attack right here, which would kick these electrons off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. So let's see what that would look like. So I have my ring like that."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can think about this lone pair of electrons on oxygen going to attack right here, which would kick these electrons off onto the bromine. So let's go ahead and draw the results of that nucleophilic attack. So let's see what that would look like. So I have my ring like that. And I now am going to have my oxygen bonded here. It still has two hydrogens attached to it. It has a lone pair of electrons on it, which give it a plus 1 formal charge."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I have my ring like that. And I now am going to have my oxygen bonded here. It still has two hydrogens attached to it. It has a lone pair of electrons on it, which give it a plus 1 formal charge. And when the oxygen attacked, that is going to push down this methyl group. So this methyl group is going to be pushed down relative to the plane of the ring. So now we have a methyl group down at this carbon."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It has a lone pair of electrons on it, which give it a plus 1 formal charge. And when the oxygen attacked, that is going to push down this methyl group. So this methyl group is going to be pushed down relative to the plane of the ring. So now we have a methyl group down at this carbon. And the bromine is going to swing over to the carbon on the left. So the bromine is going to swing over to here. And so that's the position of my bromine now."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now we have a methyl group down at this carbon. And the bromine is going to swing over to the carbon on the left. So the bromine is going to swing over to here. And so that's the position of my bromine now. And so in the last step, now I have these two guys right here. They're both going to lose a proton in the next step. So it's an acid-base reaction."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so that's the position of my bromine now. And so in the last step, now I have these two guys right here. They're both going to lose a proton in the next step. So it's an acid-base reaction. And we could show water coming along. So for the molecule on the left, water comes along. And water is going to take this proton."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's an acid-base reaction. And we could show water coming along. So for the molecule on the left, water comes along. And water is going to take this proton. These electrons are going to kick off onto that oxygen. And we can draw that product. So if we were to draw that product, we would look down this way."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And water is going to take this proton. These electrons are going to kick off onto that oxygen. And we can draw that product. So if we were to draw that product, we would look down this way. And we would treat this as being the top carbon here. So there's a methyl group going up at that carbon. So I can say that this is going to, after it loses a proton, that carbon in blue is going to have a methyl group up relative to the plane of the ring."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if we were to draw that product, we would look down this way. And we would treat this as being the top carbon here. So there's a methyl group going up at that carbon. So I can say that this is going to, after it loses a proton, that carbon in blue is going to have a methyl group up relative to the plane of the ring. And it's going to have an OH group down. So this OH is going to be down relative to the plane. So I can go ahead and put OH going away from me in space."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can say that this is going to, after it loses a proton, that carbon in blue is going to have a methyl group up relative to the plane of the ring. And it's going to have an OH group down. So this OH is going to be down relative to the plane. So I can go ahead and put OH going away from me in space. And then this bromine over here, this bromine is going to be coming out at me. So I can go ahead and show a wedge for that bromine. So that's one of our possible products."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and put OH going away from me in space. And then this bromine over here, this bromine is going to be coming out at me. So I can go ahead and show a wedge for that bromine. So that's one of our possible products. Over here on the right, if this is what happens on the right, I do the same thing. I put my eye right here. And I stare down with this carbon being the top carbon."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's one of our possible products. Over here on the right, if this is what happens on the right, I do the same thing. I put my eye right here. And I stare down with this carbon being the top carbon. So I can go ahead and draw my cyclohexane ring. And I can see that this time my OH will be coming out at me after it loses a proton. So if I wanted to, I could go ahead and draw water in here and show the last step of the mechanism, lone pair of electrons taking this proton, leaving these electrons behind, giving me an OH coming out at me in space."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I stare down with this carbon being the top carbon. So I can go ahead and draw my cyclohexane ring. And I can see that this time my OH will be coming out at me after it loses a proton. So if I wanted to, I could go ahead and draw water in here and show the last step of the mechanism, lone pair of electrons taking this proton, leaving these electrons behind, giving me an OH coming out at me in space. So there would be an OH coming out at me. And then this methyl group would therefore be going away from me in space. So I can go ahead and show that methyl group as a dash."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I wanted to, I could go ahead and draw water in here and show the last step of the mechanism, lone pair of electrons taking this proton, leaving these electrons behind, giving me an OH coming out at me in space. So there would be an OH coming out at me. And then this methyl group would therefore be going away from me in space. So I can go ahead and show that methyl group as a dash. And then finally, this bromine over here would be going away from me. So that would be a dash on my ring like that. So bromine going away from me."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and show that methyl group as a dash. And then finally, this bromine over here would be going away from me. So that would be a dash on my ring like that. So bromine going away from me. When I finally get to my two products, I can analyze them. And I can see that they are enantiomers to each other. These two guys are enantiomers to each other."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So bromine going away from me. When I finally get to my two products, I can analyze them. And I can see that they are enantiomers to each other. These two guys are enantiomers to each other. So they're different molecules. And we can look at the absolute configurations really fast. I can see that I have a carbon coming out at me on a wedge and an oxygen going away from me."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "These two guys are enantiomers to each other. So they're different molecules. And we can look at the absolute configurations really fast. I can see that I have a carbon coming out at me on a wedge and an oxygen going away from me. And it's been reversed over here on the right. This time the oxygen is coming out at me on a wedge. And the carbon is going away from me on the dash."}, {"video_title": "Halohydrin formation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I can see that I have a carbon coming out at me on a wedge and an oxygen going away from me. And it's been reversed over here on the right. This time the oxygen is coming out at me on a wedge. And the carbon is going away from me on the dash. When I look at the bromine, it's coming out at me on the left. And the bromine's going away from me on the right. So I can see that I have different absolute configurations at both chirality centers."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we got a value of a little bit over 3,000 wave numbers. However, that wave number depends on the hybridization state of this carbon. So let's look at some examples here. So if we look at an example where the carbon is SP hybridized, so we know this is an SP hybridized carbon because we have a triple bond here. So we're talking about a carbon-hydrogen bond where the carbon is SP hybridized. The signal for this carbon-hydrogen bond stretch shows up about 3,300 wave numbers. If we look at this next example here, so now we have a carbon that has a double bond to it, so it must be SP2 hybridized."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we look at an example where the carbon is SP hybridized, so we know this is an SP hybridized carbon because we have a triple bond here. So we're talking about a carbon-hydrogen bond where the carbon is SP hybridized. The signal for this carbon-hydrogen bond stretch shows up about 3,300 wave numbers. If we look at this next example here, so now we have a carbon that has a double bond to it, so it must be SP2 hybridized. So we're talking about a carbon-hydrogen bond where the carbon is SP2 hybridized. The signal for this carbon-hydrogen bond stretch shows up about 3,100 wave numbers. And then finally, if we look at a situation where we have only single bonds to this carbon, we're talking about an SP3 hybridized carbon here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If we look at this next example here, so now we have a carbon that has a double bond to it, so it must be SP2 hybridized. So we're talking about a carbon-hydrogen bond where the carbon is SP2 hybridized. The signal for this carbon-hydrogen bond stretch shows up about 3,100 wave numbers. And then finally, if we look at a situation where we have only single bonds to this carbon, we're talking about an SP3 hybridized carbon here. So we're talking about a carbon-hydrogen bond where the carbon is SP3 hybridized. The signal for this carbon-hydrogen bond stretch shows up about 2,900 wave numbers. And so how do we explain these different wave numbers?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And then finally, if we look at a situation where we have only single bonds to this carbon, we're talking about an SP3 hybridized carbon here. So we're talking about a carbon-hydrogen bond where the carbon is SP3 hybridized. The signal for this carbon-hydrogen bond stretch shows up about 2,900 wave numbers. And so how do we explain these different wave numbers? Because they're all carbon-hydrogen bonds. Well, we need to think about the hybridization. So let's do that."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so how do we explain these different wave numbers? Because they're all carbon-hydrogen bonds. Well, we need to think about the hybridization. So let's do that. So if we look at the SP hybridized carbon, remember that means that this carbon has two SP hybrid orbitals. And an SP hybrid orbital has the most S character out of all these orbitals that we've discussed here. So it actually 50% S character, if you remember that from the videos on hybridization."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's do that. So if we look at the SP hybridized carbon, remember that means that this carbon has two SP hybrid orbitals. And an SP hybrid orbital has the most S character out of all these orbitals that we've discussed here. So it actually 50% S character, if you remember that from the videos on hybridization. So 50% S character for an SP hybridized orbital. For an SP2 hybridized orbital, it's about 33% S character. And finally, for an SP3 hybridized orbital, it's about 25% S character."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So it actually 50% S character, if you remember that from the videos on hybridization. So 50% S character for an SP hybridized orbital. For an SP2 hybridized orbital, it's about 33% S character. And finally, for an SP3 hybridized orbital, it's about 25% S character. And so going back to the SP hybridized carbon, so the SP hybrid orbital is 50% S character. That means, remember what that means, the electron density is closest to the nucleus. So if that's the case, then we're talking about this bond right here, this bond being the shortest bond, because the electron density is closest to the nucleus, the more S character you have."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And finally, for an SP3 hybridized orbital, it's about 25% S character. And so going back to the SP hybridized carbon, so the SP hybrid orbital is 50% S character. That means, remember what that means, the electron density is closest to the nucleus. So if that's the case, then we're talking about this bond right here, this bond being the shortest bond, because the electron density is closest to the nucleus, the more S character you have. And if this is the shortest bond, it must be the strongest bond out of these three that we're talking about. So this carbon-hydrogen bond, where the carbon is SP hybridized, is stronger than the carbon-hydrogen bond where the carbon is SP2 hybridized. This bond, though, has more S character than this one, so this bond is stronger than this one."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if that's the case, then we're talking about this bond right here, this bond being the shortest bond, because the electron density is closest to the nucleus, the more S character you have. And if this is the shortest bond, it must be the strongest bond out of these three that we're talking about. So this carbon-hydrogen bond, where the carbon is SP hybridized, is stronger than the carbon-hydrogen bond where the carbon is SP2 hybridized. This bond, though, has more S character than this one, so this bond is stronger than this one. And so this order of bond strength explains the wave numbers, because if you remember from the previous video, the bond strength affects the force constant or the spring constant, K. So as you increase in bond strength, you increase K. And we saw that increasing K increases the frequency or the wave number. So this increases the frequency of bond vibration, or increases the wave number where you would find the signal on your IR spectrum. And so since this is the strongest bond, this is the highest value for the wave number, so we're gonna find this signal, right, more to the left on our IR spectrum when we're looking at it."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This bond, though, has more S character than this one, so this bond is stronger than this one. And so this order of bond strength explains the wave numbers, because if you remember from the previous video, the bond strength affects the force constant or the spring constant, K. So as you increase in bond strength, you increase K. And we saw that increasing K increases the frequency or the wave number. So this increases the frequency of bond vibration, or increases the wave number where you would find the signal on your IR spectrum. And so since this is the strongest bond, this is the highest value for the wave number, so we're gonna find this signal, right, more to the left on our IR spectrum when we're looking at it. All right, now that we understand this idea, so the hybridization, we can look at some IR spectra for hydrocarbons, and we can analyze them. So let's do that. First, let's compare alkanes and alkynes."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so since this is the strongest bond, this is the highest value for the wave number, so we're gonna find this signal, right, more to the left on our IR spectrum when we're looking at it. All right, now that we understand this idea, so the hybridization, we can look at some IR spectra for hydrocarbons, and we can analyze them. So let's do that. First, let's compare alkanes and alkynes. So let's go down here and let's look at some spectra. So let me just pull down here, and we can look at two IR spectra. The first one is for this molecule, which is decane, so hopefully I have the right number of carbons drawn there."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "First, let's compare alkanes and alkynes. So let's go down here and let's look at some spectra. So let me just pull down here, and we can look at two IR spectra. The first one is for this molecule, which is decane, so hopefully I have the right number of carbons drawn there. The second one is for one octyne, so a triple bond in this molecule. Let's compare these two to think about the differences. All right, one of the things that's sometimes helpful to do is to draw a line, draw a line around 3,000."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The first one is for this molecule, which is decane, so hopefully I have the right number of carbons drawn there. The second one is for one octyne, so a triple bond in this molecule. Let's compare these two to think about the differences. All right, one of the things that's sometimes helpful to do is to draw a line, draw a line around 3,000. So let me draw a line around 3,000. I'm gonna try to draw it for both here, too. So I'm gonna draw a line around 3,000 for both."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, one of the things that's sometimes helpful to do is to draw a line, draw a line around 3,000. So let me draw a line around 3,000. I'm gonna try to draw it for both here, too. So I'm gonna draw a line around 3,000 for both. So we can compare these two spectra. All right, we know that a carbon-hydrogen, let me go ahead and write this in here, a carbon-hydrogen bond where the carbon is sp3 hybridized, so that signal for that stretch shows up under 3,000, and so that's why it's helpful to draw a line around 3,000. So that's what we're talking about when we're talking about this complicated-looking thing in here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna draw a line around 3,000 for both. So we can compare these two spectra. All right, we know that a carbon-hydrogen, let me go ahead and write this in here, a carbon-hydrogen bond where the carbon is sp3 hybridized, so that signal for that stretch shows up under 3,000, and so that's why it's helpful to draw a line around 3,000. So that's what we're talking about when we're talking about this complicated-looking thing in here. So it's not really worth your time to analyze this in great detail, and of course my drawing isn't perfect to begin with, but think about under 3,000, right? That's where you expect to find your signal for your carbon-hydrogen bond. We're talking about an sp3 hybridized carbon."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's what we're talking about when we're talking about this complicated-looking thing in here. So it's not really worth your time to analyze this in great detail, and of course my drawing isn't perfect to begin with, but think about under 3,000, right? That's where you expect to find your signal for your carbon-hydrogen bond. We're talking about an sp3 hybridized carbon. And those are the only types of carbons that we have in decade. And so if we think about the diagnostic region versus the fingerprint region, right? So if I draw a line here to separate those two regions, in the diagnostic region, all we have is this, right?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're talking about an sp3 hybridized carbon. And those are the only types of carbons that we have in decade. And so if we think about the diagnostic region versus the fingerprint region, right? So if I draw a line here to separate those two regions, in the diagnostic region, all we have is this, right? So all we're thinking about is carbon-hydrogen, the carbon is sp3 hybridized. So a very simple spectrum to analyze. We move on to one-octyne, so now we're looking at this one down here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if I draw a line here to separate those two regions, in the diagnostic region, all we have is this, right? So all we're thinking about is carbon-hydrogen, the carbon is sp3 hybridized. So a very simple spectrum to analyze. We move on to one-octyne, so now we're looking at this one down here. So we see that same kind of thing, right? Because obviously we have carbon-hydrogen sp3 hybridized, right, also in this molecule. And so this isn't gonna really help you too much when you're analyzing spectra, but it's useful to know what you're looking at, drawing a line at 3,000, and thinking about that's what that represents."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We move on to one-octyne, so now we're looking at this one down here. So we see that same kind of thing, right? Because obviously we have carbon-hydrogen sp3 hybridized, right, also in this molecule. And so this isn't gonna really help you too much when you're analyzing spectra, but it's useful to know what you're looking at, drawing a line at 3,000, and thinking about that's what that represents. All right, so once you draw that line at 3,000, it allows you to see some differences. All right, so for example, right, this signal right here, if we drop down, it's pretty close to 3,300, right? So this would be 3,100, 3,200, so 3,300."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this isn't gonna really help you too much when you're analyzing spectra, but it's useful to know what you're looking at, drawing a line at 3,000, and thinking about that's what that represents. All right, so once you draw that line at 3,000, it allows you to see some differences. All right, so for example, right, this signal right here, if we drop down, it's pretty close to 3,300, right? So this would be 3,100, 3,200, so 3,300. So approximately 3,300 wave numbers. And we know what that signal represents, right? We can go back up to here, and we can look at about 3,300 is where we would expect to see this carbon-hydrogen bond stretch where that carbon is sp hybridized."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this would be 3,100, 3,200, so 3,300. So approximately 3,300 wave numbers. And we know what that signal represents, right? We can go back up to here, and we can look at about 3,300 is where we would expect to see this carbon-hydrogen bond stretch where that carbon is sp hybridized. So that's what we're looking at there. So let's go down, look at the dot structure, and see if we can figure out what it means on the dot structure. All right, so this signal, right, this signal must be a carbon-hydrogen bond where the carbon is sp hybridized."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We can go back up to here, and we can look at about 3,300 is where we would expect to see this carbon-hydrogen bond stretch where that carbon is sp hybridized. So that's what we're looking at there. So let's go down, look at the dot structure, and see if we can figure out what it means on the dot structure. All right, so this signal, right, this signal must be a carbon-hydrogen bond where the carbon is sp hybridized. And that would be right here, right? So we have a carbon here, and then we have a hydrogen right here. We also have a carbon right here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so this signal, right, this signal must be a carbon-hydrogen bond where the carbon is sp hybridized. And that would be right here, right? So we have a carbon here, and then we have a hydrogen right here. We also have a carbon right here. So that gives you your eight carbons. And so this bond, let me go ahead and highlight it here. So this bond right here, this carbon-hydrogen bond where this carbon is sp hybridized, that's this signal, that's this signal on our spectrum."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We also have a carbon right here. So that gives you your eight carbons. And so this bond, let me go ahead and highlight it here. So this bond right here, this carbon-hydrogen bond where this carbon is sp hybridized, that's this signal, that's this signal on our spectrum. And so once again, it's useful, it's useful for analyzing here. All right, we have something else that shows up in the diagnostic region for this alkyne, and it's this signal right here. So if we drop down, right, what's the wave number where this signal appears?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this bond right here, this carbon-hydrogen bond where this carbon is sp hybridized, that's this signal, that's this signal on our spectrum. And so once again, it's useful, it's useful for analyzing here. All right, we have something else that shows up in the diagnostic region for this alkyne, and it's this signal right here. So if we drop down, right, what's the wave number where this signal appears? The wave number is about 2,100, so approximately 2,100, maybe a little bit higher than that. And that's the carbon-triple bond stretch, right? So that's the carbon-carbon-triple bond stretch that we talked about in an earlier video, right?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if we drop down, right, what's the wave number where this signal appears? The wave number is about 2,100, so approximately 2,100, maybe a little bit higher than that. And that's the carbon-triple bond stretch, right? So that's the carbon-carbon-triple bond stretch that we talked about in an earlier video, right? So that's approximately the triple bond region when you're looking at your spectrum. And of course, obviously we have one. This is an alkyne here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So that's the carbon-carbon-triple bond stretch that we talked about in an earlier video, right? So that's approximately the triple bond region when you're looking at your spectrum. And of course, obviously we have one. This is an alkyne here. And so hopefully this just shows you the differences. And once again, your fingerprint region over here is unique for each of these molecules here. So this shows you the differences and helps you to think about how to analyze your IR spectrum."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is an alkyne here. And so hopefully this just shows you the differences. And once again, your fingerprint region over here is unique for each of these molecules here. So this shows you the differences and helps you to think about how to analyze your IR spectrum. Let's look at two more. All right, so let's, actually, let's look at one more here, and let's compare these two. So now we have a spectrum for an alkyne, right?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this shows you the differences and helps you to think about how to analyze your IR spectrum. Let's look at two more. All right, so let's, actually, let's look at one more here, and let's compare these two. So now we have a spectrum for an alkyne, right? So here we have one hexene, and let's see if we can analyze this one. So we're gonna do the same thing. We're gonna draw a line around 1,500, right?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So now we have a spectrum for an alkyne, right? So here we have one hexene, and let's see if we can analyze this one. So we're gonna do the same thing. We're gonna draw a line around 1,500, right? So draw a line around 1,500. We're gonna draw a line around 3,000, and let's analyze this one, right? So we know what this one's talking about."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We're gonna draw a line around 1,500, right? So draw a line around 1,500. We're gonna draw a line around 3,000, and let's analyze this one, right? So we know what this one's talking about. We know this is talking about a carbon-hydrogen where the carbon is sp3 hybridized. All right, but what's this signal right here? We drop down, and that's pretty close to 3,100, right?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we know what this one's talking about. We know this is talking about a carbon-hydrogen where the carbon is sp3 hybridized. All right, but what's this signal right here? We drop down, and that's pretty close to 3,100, right? So that signal is approximately 3,100. And so we know what that must represent, right? That's a carbon-hydrogen bond where the carbon is sp2 hybridized."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "We drop down, and that's pretty close to 3,100, right? So that signal is approximately 3,100. And so we know what that must represent, right? That's a carbon-hydrogen bond where the carbon is sp2 hybridized. So that stretch occurs at this frequency or this wave number. And so we know an sp2 hybridized carbon must be present, and obviously here with this double bond, right, we know we have an sp2 hybridized carbon. Notice a difference between this one and the one we just talked about."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's a carbon-hydrogen bond where the carbon is sp2 hybridized. So that stretch occurs at this frequency or this wave number. And so we know an sp2 hybridized carbon must be present, and obviously here with this double bond, right, we know we have an sp2 hybridized carbon. Notice a difference between this one and the one we just talked about. So let me go ahead and highlight this here. So this signal shows up around 3,100, right? So I draw a line up here, and we saw this signal at 3,300."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Notice a difference between this one and the one we just talked about. So let me go ahead and highlight this here. So this signal shows up around 3,100, right? So I draw a line up here, and we saw this signal at 3,300. So it's useful to think about, it helps you distinguish on your IR spectra if you draw a line there and think about where the wave number is. What wave number does this signal appear? All right, we also have something else showing up in this one."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I draw a line up here, and we saw this signal at 3,300. So it's useful to think about, it helps you distinguish on your IR spectra if you draw a line there and think about where the wave number is. What wave number does this signal appear? All right, we also have something else showing up in this one. Let me go ahead and draw this one in here. So what is this? What is this guy right here?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, we also have something else showing up in this one. Let me go ahead and draw this one in here. So what is this? What is this guy right here? That looks like a pretty obvious signal. So we can drop down here and check out approximately where does that show up? What's that wave number?"}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What is this guy right here? That looks like a pretty obvious signal. So we can drop down here and check out approximately where does that show up? What's that wave number? All right, well, if this is 1,500, this is 1,600, this is 1,700. That's pretty much in between. It's approximately 1,650."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "What's that wave number? All right, well, if this is 1,500, this is 1,600, this is 1,700. That's pretty much in between. It's approximately 1,650. And in an earlier video, we said that was in the double bond region. So that's the carbon-carbon double bond stretch signal right in here. And obviously, there's a double bond in our molecule."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's approximately 1,650. And in an earlier video, we said that was in the double bond region. So that's the carbon-carbon double bond stretch signal right in here. And obviously, there's a double bond in our molecule. So again, comparing these two, remember we talked about the fact that a triple bond vibrates faster than a double bond. And so the signal is different. The triple bond vibrates faster, so it has a higher wave number."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And obviously, there's a double bond in our molecule. So again, comparing these two, remember we talked about the fact that a triple bond vibrates faster than a double bond. And so the signal is different. The triple bond vibrates faster, so it has a higher wave number. The double bond doesn't vibrate as fast, so it has a lower wave number. So these are important things to think about. Finally, let's compare this alkene to an arene here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The triple bond vibrates faster, so it has a higher wave number. The double bond doesn't vibrate as fast, so it has a lower wave number. So these are important things to think about. Finally, let's compare this alkene to an arene here. So let's look at toluene right here. All right, so if we do the same thing, if we draw a line around 3,000, so somewhere around 3,000, and we know this is below 3,000, so we know that this must be talking about carbon-hydrogen where the carbon is sp3 hybridized. That's a carbon-hydrogen bond stretch where the carbon's sp3 hybridized."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Finally, let's compare this alkene to an arene here. So let's look at toluene right here. All right, so if we do the same thing, if we draw a line around 3,000, so somewhere around 3,000, and we know this is below 3,000, so we know that this must be talking about carbon-hydrogen where the carbon is sp3 hybridized. That's a carbon-hydrogen bond stretch where the carbon's sp3 hybridized. Well, this carbon right here on toluene, so this is toluene, that carbon's sp3 hybridized, so that makes sense. We have this one little peak here, this one little signal that's a little bit higher than 3,000, so it's pretty close to 3,100. And we know that's approximately where we would find a carbon-hydrogen bond stretch where the carbon's sp2 hybridized, so somewhere around 3,100 here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "That's a carbon-hydrogen bond stretch where the carbon's sp3 hybridized. Well, this carbon right here on toluene, so this is toluene, that carbon's sp3 hybridized, so that makes sense. We have this one little peak here, this one little signal that's a little bit higher than 3,000, so it's pretty close to 3,100. And we know that's approximately where we would find a carbon-hydrogen bond stretch where the carbon's sp2 hybridized, so somewhere around 3,100 here. And so this makes this, at first you might think, oh, how do we tell this apart? So this is very similar. This is very similar to this situation."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we know that's approximately where we would find a carbon-hydrogen bond stretch where the carbon's sp2 hybridized, so somewhere around 3,100 here. And so this makes this, at first you might think, oh, how do we tell this apart? So this is very similar. This is very similar to this situation. So this looks very similar to this, and so it can be tricky sometimes if you're just glancing at that part of your spectrum. Let's think about the double bond region, too. So the double bond region right up here, so this is 1,600, that's for this one."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This is very similar to this situation. So this looks very similar to this, and so it can be tricky sometimes if you're just glancing at that part of your spectrum. Let's think about the double bond region, too. So the double bond region right up here, so this is 1,600, that's for this one. I'm gonna draw a line down here, and let's try to connect the 1,600s right here like that. And let's think about what happened. So here's the signal for the carbon-carbon double bond here, and when you're talking about an aromatic double bond stretch, so carbon-carbon aromatic, that usually shows up lower than 1,600."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the double bond region right up here, so this is 1,600, that's for this one. I'm gonna draw a line down here, and let's try to connect the 1,600s right here like that. And let's think about what happened. So here's the signal for the carbon-carbon double bond here, and when you're talking about an aromatic double bond stretch, so carbon-carbon aromatic, that usually shows up lower than 1,600. So here it looks like we have two signals, so it depends on what kind of compound you're dealing with, but we can see that this usually shows up lower. So this is somewhere usually around 1,600 to 1,450, but we're talking about the carbon-carbon double bond stretching here. And there's some other subtle things that clue you in, like this right here."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's the signal for the carbon-carbon double bond here, and when you're talking about an aromatic double bond stretch, so carbon-carbon aromatic, that usually shows up lower than 1,600. So here it looks like we have two signals, so it depends on what kind of compound you're dealing with, but we can see that this usually shows up lower. So this is somewhere usually around 1,600 to 1,450, but we're talking about the carbon-carbon double bond stretching here. And there's some other subtle things that clue you in, like this right here. So we don't really see that on this one, on this spectrum. And then these down here, we're missing those here too. And it'd be way too much to get into in this video to talk about what these mean, and it's a little bit more than we've talked about so far, so that'll have to be a different video."}, {"video_title": "IR spectra for hydrocarbons Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And there's some other subtle things that clue you in, like this right here. So we don't really see that on this one, on this spectrum. And then these down here, we're missing those here too. And it'd be way too much to get into in this video to talk about what these mean, and it's a little bit more than we've talked about so far, so that'll have to be a different video. But there are subtle differences, and the easiest one is to think about the fact that this aromatic carbon-carbon double bond shows up at a lower wave number than the one that we talked about right here. So just look at the sheer multitude of signals, and sometimes that clues you into the fact that you're dealing with this benzene ring here for a toluene. All right, so that sums up just a quick intro to looking at IR spectra for hydrocarbons."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Notice I have some hydrogens in green here, and that's just to help us visualize rotation around the carbon-carbon bond. So I'm going to rotate the front carbon and keep the back carbon stationary. So now we get the staggered conformation of ethane. I'm going to keep rotating 60 degrees. So I'm going to rotate again, and now we have an eclipse conformation. I rotate again, and now we have a staggered conformation of ethane. I rotate another 60 degrees, and we get an eclipse conformation."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to keep rotating 60 degrees. So I'm going to rotate again, and now we have an eclipse conformation. I rotate again, and now we have a staggered conformation of ethane. I rotate another 60 degrees, and we get an eclipse conformation. I rotate again, and we get a staggered, and you get the idea. One more rotation, and we get back to an eclipse conformation. Here we have a graph of the potential energies of the conformations that we just saw in the video."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "I rotate another 60 degrees, and we get an eclipse conformation. I rotate again, and we get a staggered, and you get the idea. One more rotation, and we get back to an eclipse conformation. Here we have a graph of the potential energies of the conformations that we just saw in the video. So we started out right here with this eclipsed conformation of ethane, so we are at this particular potential energy. As we rotate to get to this staggered conformation, you can see there's a decrease in potential energy. So this staggered conformation has a lower potential energy than the eclipse conformation."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Here we have a graph of the potential energies of the conformations that we just saw in the video. So we started out right here with this eclipsed conformation of ethane, so we are at this particular potential energy. As we rotate to get to this staggered conformation, you can see there's a decrease in potential energy. So this staggered conformation has a lower potential energy than the eclipse conformation. We rotate again, we get back up here to this eclipse conformation. Notice that takes energy. So it takes energy to go from this staggered conformation to this eclipse conformation."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So this staggered conformation has a lower potential energy than the eclipse conformation. We rotate again, we get back up here to this eclipse conformation. Notice that takes energy. So it takes energy to go from this staggered conformation to this eclipse conformation. Going from the eclipse to the staggered, that's a decrease in energy. Going from the staggered up to this eclipse, that's an increase in energy. Going from the eclipse down to the staggered again is a decrease, and finally back to the original eclipse conformation would be an increase in energy."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So it takes energy to go from this staggered conformation to this eclipse conformation. Going from the eclipse to the staggered, that's a decrease in energy. Going from the staggered up to this eclipse, that's an increase in energy. Going from the eclipse down to the staggered again is a decrease, and finally back to the original eclipse conformation would be an increase in energy. Notice that all of our eclipsed conformations here have the same potential energy. If I draw a line, this is all the same potential energy. Whoops, I didn't draw a very nice line there, but you get the idea."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Going from the eclipse down to the staggered again is a decrease, and finally back to the original eclipse conformation would be an increase in energy. Notice that all of our eclipsed conformations here have the same potential energy. If I draw a line, this is all the same potential energy. Whoops, I didn't draw a very nice line there, but you get the idea. Therefore, we say that these are degenerate in terms of energy. Same thing for the staggered conformations. All of these staggered conformations, if I draw a line in here, have the same potential energy."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "Whoops, I didn't draw a very nice line there, but you get the idea. Therefore, we say that these are degenerate in terms of energy. Same thing for the staggered conformations. All of these staggered conformations, if I draw a line in here, have the same potential energy. So the staggered conformations are lower in energy than the eclipse conformations. And actually the difference is 12 kilojoules per mole. So if I write that in here, so 12 kilojoules per mole, that's talking about the difference in energy between the eclipse conformations and the staggered conformations."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "All of these staggered conformations, if I draw a line in here, have the same potential energy. So the staggered conformations are lower in energy than the eclipse conformations. And actually the difference is 12 kilojoules per mole. So if I write that in here, so 12 kilojoules per mole, that's talking about the difference in energy between the eclipse conformations and the staggered conformations. Lower in energy means more stable. And the easy way of thinking about that is imagine these things as hills. And if I have a boulder or a rock or something down here at the bottom of the hill, and I'm comparing that boulder or rock up here to a boulder at the top of the hill, in physics you can set your potential energy equal to zero at the ground."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So if I write that in here, so 12 kilojoules per mole, that's talking about the difference in energy between the eclipse conformations and the staggered conformations. Lower in energy means more stable. And the easy way of thinking about that is imagine these things as hills. And if I have a boulder or a rock or something down here at the bottom of the hill, and I'm comparing that boulder or rock up here to a boulder at the top of the hill, in physics you can set your potential energy equal to zero at the ground. So let's say that this is ground level. I say my potential energy, U, is equal to zero joules at this point. And so the boulder in this valley here, let's say it's 10 joules."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And if I have a boulder or a rock or something down here at the bottom of the hill, and I'm comparing that boulder or rock up here to a boulder at the top of the hill, in physics you can set your potential energy equal to zero at the ground. So let's say that this is ground level. I say my potential energy, U, is equal to zero joules at this point. And so the boulder in this valley here, let's say it's 10 joules. Let's say this is 10 joules here at this point. And then it would take energy to push this boulder up this hill to this point. And let's say the final potential energy of the boulder at this point would be 22 joules."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And so the boulder in this valley here, let's say it's 10 joules. Let's say this is 10 joules here at this point. And then it would take energy to push this boulder up this hill to this point. And let's say the final potential energy of the boulder at this point would be 22 joules. It takes energy in order to do that. And this final position is less stable. And this is the higher potential energy."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And let's say the final potential energy of the boulder at this point would be 22 joules. It takes energy in order to do that. And this final position is less stable. And this is the higher potential energy. So higher in potential energy means less stable. Lower in potential energy means more stable. So why do we have a difference in energy between the staggered and the eclipse conformation?"}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And this is the higher potential energy. So higher in potential energy means less stable. Lower in potential energy means more stable. So why do we have a difference in energy between the staggered and the eclipse conformation? Well, this is called torsional strain. So this difference in energy, this 12 kilojoules per mole, is called torsional strain. And the source of torsional strain has been a topic of debate."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So why do we have a difference in energy between the staggered and the eclipse conformation? Well, this is called torsional strain. So this difference in energy, this 12 kilojoules per mole, is called torsional strain. And the source of torsional strain has been a topic of debate. One of the current theories has to do with molecular orbital theory. I'm going to go with one of the older ones, which talks about the electron pair repulsions. And the electron pair repulsions are greatest when the bonds are eclipsed."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And the source of torsional strain has been a topic of debate. One of the current theories has to do with molecular orbital theory. I'm going to go with one of the older ones, which talks about the electron pair repulsions. And the electron pair repulsions are greatest when the bonds are eclipsed. So if you think about the electrons in this bond being close to the electrons in this bond, and you have that same thing over here and over here. So in space, these electron pairs, these bonding electron pairs, are closer together in the eclipse conformation than they are in the staggered. If I go down here to the staggered, you can see, if you're thinking about these electron pairs, they're further away than they are in the eclipse conformation."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And the electron pair repulsions are greatest when the bonds are eclipsed. So if you think about the electrons in this bond being close to the electrons in this bond, and you have that same thing over here and over here. So in space, these electron pairs, these bonding electron pairs, are closer together in the eclipse conformation than they are in the staggered. If I go down here to the staggered, you can see, if you're thinking about these electron pairs, they're further away than they are in the eclipse conformation. And we know that electrons will repel. So electron pair repulsions are greatest when the bonds are eclipsed. And therefore, that's higher energy."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "If I go down here to the staggered, you can see, if you're thinking about these electron pairs, they're further away than they are in the eclipse conformation. And we know that electrons will repel. So electron pair repulsions are greatest when the bonds are eclipsed. And therefore, that's higher energy. And the electron pairs are further away from each other when you're talking about staggered, therefore, more stable. The total energy cost between the two conformations is 12 kilojoules per mole. And we have three pairs of eclipsed hydrogens."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, that's higher energy. And the electron pairs are further away from each other when you're talking about staggered, therefore, more stable. The total energy cost between the two conformations is 12 kilojoules per mole. And we have three pairs of eclipsed hydrogens. If we go back up to here, here's one pair of eclipsed hydrogens. Here's another pair. And here's another pair."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And we have three pairs of eclipsed hydrogens. If we go back up to here, here's one pair of eclipsed hydrogens. Here's another pair. And here's another pair. So if the total energy is 12 kilojoules per mole, and I have three pairs of eclipsed hydrogens, we could say that the energy cost for each pair of eclipsed hydrogens is 4 kilojoules per mole. So this would be 4 kilojoules per mole for this pair of eclipsed hydrogens, 4 kilojoules per mole for this one, and 4 kilojoules per mole for this one, adding up for a total of 12. So our total energy cost is 12."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "And here's another pair. So if the total energy is 12 kilojoules per mole, and I have three pairs of eclipsed hydrogens, we could say that the energy cost for each pair of eclipsed hydrogens is 4 kilojoules per mole. So this would be 4 kilojoules per mole for this pair of eclipsed hydrogens, 4 kilojoules per mole for this one, and 4 kilojoules per mole for this one, adding up for a total of 12. So our total energy cost is 12. And now we can think about two hydrogens eclipsing each other as having an energy cost of 4 kilojoules per mole. We've just seen that the staggered conformation of ethane is more stable than the eclipsed conformation of ethane. And if you want to turn a staggered conformation into an eclipsed conformation, you would need energy."}, {"video_title": "Conformational analysis of ethane Organic chemistry Khan Academy.mp3", "Sentence": "So our total energy cost is 12. And now we can think about two hydrogens eclipsing each other as having an energy cost of 4 kilojoules per mole. We've just seen that the staggered conformation of ethane is more stable than the eclipsed conformation of ethane. And if you want to turn a staggered conformation into an eclipsed conformation, you would need energy. And at room temperature, there's enough energy for the staggered conformation to turn into the eclipsed. Equilibrium is reached between the two conformations. And at room temperature, approximately 99% of ethane molecules have an approximately staggered conformation, whereas only about 1% have an eclipsed conformation."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, I promised you that I would show a concrete example of electrophilic aromatic substitution. So let's do that right here. So let's say we have some benzene. And it's a solution with some molecular bromine. So I'll draw the molecular bromine like this. So it's one bromine right there. It has one, two, three, four, five, six, seven valence electrons, and it's bonded to another bromine that has one, two, three, four, five, six, seven electrons."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it's a solution with some molecular bromine. So I'll draw the molecular bromine like this. So it's one bromine right there. It has one, two, three, four, five, six, seven valence electrons, and it's bonded to another bromine that has one, two, three, four, five, six, seven electrons. And these two guys in the middle are bonded to each other, they can kind of act as an electron pair or make both bromines feel like they have eight electrons. This guy thinks he has a magenta one, that guy thinks he has the blue one. And let's say we have some iron bromide in our mix as well, and what we're going to see is that this is going to catalyze the reaction."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It has one, two, three, four, five, six, seven valence electrons, and it's bonded to another bromine that has one, two, three, four, five, six, seven electrons. And these two guys in the middle are bonded to each other, they can kind of act as an electron pair or make both bromines feel like they have eight electrons. This guy thinks he has a magenta one, that guy thinks he has the blue one. And let's say we have some iron bromide in our mix as well, and what we're going to see is that this is going to catalyze the reaction. So the iron bromide, we have some iron and it's bonded to three bromines just like that. So we haven't seen iron bromide much, but the way I think about it is that bromine molecules are much more electronegative than the iron. So even though these look like fair covalent bonds, if we thought about oxidation, these guys are going to be hogging the electrons."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let's say we have some iron bromide in our mix as well, and what we're going to see is that this is going to catalyze the reaction. So the iron bromide, we have some iron and it's bonded to three bromines just like that. So we haven't seen iron bromide much, but the way I think about it is that bromine molecules are much more electronegative than the iron. So even though these look like fair covalent bonds, if we thought about oxidation, these guys are going to be hogging the electrons. This actually would have an oxidation number of three. And in reality, they are hogging the electrons. They are more electronegative."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So even though these look like fair covalent bonds, if we thought about oxidation, these guys are going to be hogging the electrons. This actually would have an oxidation number of three. And in reality, they are hogging the electrons. They are more electronegative. So the way I think about it is that this iron will have a slightly positive charge, because the electrons are being hogged away from it. So it wouldn't mind to gain an electron, or it might want to accept an electron. It might want to act as a Lewis acid."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "They are more electronegative. So the way I think about it is that this iron will have a slightly positive charge, because the electrons are being hogged away from it. So it wouldn't mind to gain an electron, or it might want to accept an electron. It might want to act as a Lewis acid. Remember, a Lewis acid will accept an electron. So this might want to act as a Lewis acid. So who can it nab electrons from?"}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It might want to act as a Lewis acid. Remember, a Lewis acid will accept an electron. So this might want to act as a Lewis acid. So who can it nab electrons from? Well, maybe it can nab an electron from this bromine right here. And I'm not saying that this is always going to occur, but under the right circumstances, if they bump into each other with the right energies, it can happen. So this electron, let me do it in a different color that I haven't used yet, this green color."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So who can it nab electrons from? Well, maybe it can nab an electron from this bromine right here. And I'm not saying that this is always going to occur, but under the right circumstances, if they bump into each other with the right energies, it can happen. So this electron, let me do it in a different color that I haven't used yet, this green color. Let's say this electron right here gets nabbed by that iron. Then what do we have? Well, then we have a situation."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this electron, let me do it in a different color that I haven't used yet, this green color. Let's say this electron right here gets nabbed by that iron. Then what do we have? Well, then we have a situation. We have this bromine, the blue one, with one, two, three, four, five, six, seven valence electrons. We have the magenta bromine with one, two, three, four, five, now it only has the sixth valence electron right here. The seventh got nabbed by the iron."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Well, then we have a situation. We have this bromine, the blue one, with one, two, three, four, five, six, seven valence electrons. We have the magenta bromine with one, two, three, four, five, now it only has the sixth valence electron right here. The seventh got nabbed by the iron. So the iron has the seventh valence electron, and then you have the rest of the molecule. So then you have your iron, and it's attached, of course, to the three bromines just like that. And then our bonds, these guys were bonded, they still are bonded."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The seventh got nabbed by the iron. So the iron has the seventh valence electron, and then you have the rest of the molecule. So then you have your iron, and it's attached, of course, to the three bromines just like that. And then our bonds, these guys were bonded, they still are bonded. And now these guys are bonded. These were in a pair, this electron jumped over to the iron, now we have another bond. But because this bromine lost an electron, it was neutral, it lost an electron, it now has a positive charge."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then our bonds, these guys were bonded, they still are bonded. And now these guys are bonded. These were in a pair, this electron jumped over to the iron, now we have another bond. But because this bromine lost an electron, it was neutral, it lost an electron, it now has a positive charge. And the iron, now that it gained this electron, now has a negative charge. So let's think about what's going to happen now. Now we're going to bring the benzene into the mix."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But because this bromine lost an electron, it was neutral, it lost an electron, it now has a positive charge. And the iron, now that it gained this electron, now has a negative charge. So let's think about what's going to happen now. Now we're going to bring the benzene into the mix. So let me redraw the benzene just like that. And we have this double bond, that double bond, and then, just to make things clear, let me draw this double bond with the two electrons on either end. So we have the orange electron, you have your green electron right over there, and I'll draw the double bond as being green."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now we're going to bring the benzene into the mix. So let me redraw the benzene just like that. And we have this double bond, that double bond, and then, just to make things clear, let me draw this double bond with the two electrons on either end. So we have the orange electron, you have your green electron right over there, and I'll draw the double bond as being green. Now let's think about this molecule right here. We have a bromine with a positive charge. Bromines are really, really, really electronegative."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have the orange electron, you have your green electron right over there, and I'll draw the double bond as being green. Now let's think about this molecule right here. We have a bromine with a positive charge. Bromines are really, really, really electronegative. You might see them with a negative charge. With a positive charge, it really wants to grab an electron. And in the right circumstances, you can imagine where it really wants to grab that electron right there."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Bromines are really, really, really electronegative. You might see them with a negative charge. With a positive charge, it really wants to grab an electron. And in the right circumstances, you can imagine where it really wants to grab that electron right there. So maybe if there's just some way it could pull this electron, but the only way it could pull this electron is maybe if this, because if it just took that electron, then this bromine would have a positive charge, which isn't cool. So this bromine maybe would want to pull an electron. If this guy gets an electron, then this guy can get an electron."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And in the right circumstances, you can imagine where it really wants to grab that electron right there. So maybe if there's just some way it could pull this electron, but the only way it could pull this electron is maybe if this, because if it just took that electron, then this bromine would have a positive charge, which isn't cool. So this bromine maybe would want to pull an electron. If this guy gets an electron, then this guy can get an electron. So you can imagine this thing as a whole really, really wants to grab an electron and might be very good at doing it. So this is our strong electrophile. So what actually will happen in the bromination of this benzene ring, and let me draw some hydrogens here just to make things clear."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If this guy gets an electron, then this guy can get an electron. So you can imagine this thing as a whole really, really wants to grab an electron and might be very good at doing it. So this is our strong electrophile. So what actually will happen in the bromination of this benzene ring, and let me draw some hydrogens here just to make things clear. We already have hydrogens on all of these carbons. It's sometimes important to visualize this when we're doing electrophilic aromatic substitutions. So we already have a hydrogen on all of these molecules."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So what actually will happen in the bromination of this benzene ring, and let me draw some hydrogens here just to make things clear. We already have hydrogens on all of these carbons. It's sometimes important to visualize this when we're doing electrophilic aromatic substitutions. So we already have a hydrogen on all of these molecules. So maybe this is so electrophilic, it can actually break the aromatic ring, nab this electron right there. So maybe this electron right there goes to the bromine. Maybe I should even do it this way, just so you make it clear, it kind of replaces that one."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we already have a hydrogen on all of these molecules. So maybe this is so electrophilic, it can actually break the aromatic ring, nab this electron right there. So maybe this electron right there goes to the bromine. Maybe I should even do it this way, just so you make it clear, it kind of replaces that one. Although, obviously, the electrons are a bit fungible. So maybe it goes over there. And then when it goes to this, let me make it clear, it's going to the blue bromine."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Maybe I should even do it this way, just so you make it clear, it kind of replaces that one. Although, obviously, the electrons are a bit fungible. So maybe it goes over there. And then when it goes to this, let me make it clear, it's going to the blue bromine. So this electron right here goes to the blue bromine. If the blue bromine gets an electron, then it can let go of this blue electron. So this blue electron can then go to this bromine right over here."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then when it goes to this, let me make it clear, it's going to the blue bromine. So this electron right here goes to the blue bromine. If the blue bromine gets an electron, then it can let go of this blue electron. So this blue electron can then go to this bromine right over here. And then what does our situation look like? Well, if we have that, then let me draw our benzene ring first, let me draw the benzene ring. This double bond, that double bond, let me draw all of the hydrogens, one hydrogen, I want to do them in purple."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this blue electron can then go to this bromine right over here. And then what does our situation look like? Well, if we have that, then let me draw our benzene ring first, let me draw the benzene ring. This double bond, that double bond, let me draw all of the hydrogens, one hydrogen, I want to do them in purple. So we have one hydrogen, two hydrogens, three hydrogens, four hydrogens, five hydrogens, and six hydrogens. This orange electron is still with this carbon right here. But that electron got nabbed by this bromine."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This double bond, that double bond, let me draw all of the hydrogens, one hydrogen, I want to do them in purple. So we have one hydrogen, two hydrogens, three hydrogens, four hydrogens, five hydrogens, and six hydrogens. This orange electron is still with this carbon right here. But that electron got nabbed by this bromine. So that electron got nabbed by this bromine right over here. I've kind of flipped it around, and now it has its other six valence electrons. One, two, three, four, five, six."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But that electron got nabbed by this bromine. So that electron got nabbed by this bromine right over here. I've kind of flipped it around, and now it has its other six valence electrons. One, two, three, four, five, six. The electron got taken away from this carbon. So now that carbon will have a positive charge. But we saw in the last video, it's actually resonance stabilized."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six. The electron got taken away from this carbon. So now that carbon will have a positive charge. But we saw in the last video, it's actually resonance stabilized. That electron could jump there, that electron can jump there. So it's not as stable as a nice aromatic benzene ring like this, but it's not a ridiculous carbocation. It's stable enough for it to exist for some small amount of time while we kind of hit our transition state."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But we saw in the last video, it's actually resonance stabilized. That electron could jump there, that electron can jump there. So it's not as stable as a nice aromatic benzene ring like this, but it's not a ridiculous carbocation. It's stable enough for it to exist for some small amount of time while we kind of hit our transition state. And then this molecule over here, what's it going to look like? Well, this bromine had a positive charge. Now, it gained an electron."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's stable enough for it to exist for some small amount of time while we kind of hit our transition state. And then this molecule over here, what's it going to look like? Well, this bromine had a positive charge. Now, it gained an electron. Let me draw it. So you have your bromine. It gained an electron."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, it gained an electron. Let me draw it. So you have your bromine. It gained an electron. So now it is neutral again. So let's see, it had the one, two, three, four, five, six. Now it gained this blue electron."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It gained an electron. So now it is neutral again. So let's see, it had the one, two, three, four, five, six. Now it gained this blue electron. So now it's seven valence electrons, back to being neutral. It's bonded to the iron bromide. Let me draw the bromines."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now it gained this blue electron. So now it's seven valence electrons, back to being neutral. It's bonded to the iron bromide. Let me draw the bromines. One, two, three. And so we've given this blue bromine to the benzene ring, but it's not happy here. It doesn't want to break its aromaticity."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw the bromines. One, two, three. And so we've given this blue bromine to the benzene ring, but it's not happy here. It doesn't want to break its aromaticity. It wants an electron back. So how can it gain an electron back? Well, this thing, actually let me make it very clear, this thing had a negative charge right here."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't want to break its aromaticity. It wants an electron back. So how can it gain an electron back? Well, this thing, actually let me make it very clear, this thing had a negative charge right here. So you can imagine, say, we had this electron right here. So maybe this thing right over here can now act as an actual base. It can nab a proton off of the benzene ring, just like we saw in the last video."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Well, this thing, actually let me make it very clear, this thing had a negative charge right here. So you can imagine, say, we had this electron right here. So maybe this thing right over here can now act as an actual base. It can nab a proton off of the benzene ring, just like we saw in the last video. This is now the base. This whole complex, to some degree, acted as a strong electrophile. Now that we got rid of one of these bromines, this thing might want to grab a proton now, since it is positive, and will act as our base."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It can nab a proton off of the benzene ring, just like we saw in the last video. This is now the base. This whole complex, to some degree, acted as a strong electrophile. Now that we got rid of one of these bromines, this thing might want to grab a proton now, since it is positive, and will act as our base. And we'll nab one of these protons. If it nabs the proton, then the leftover electron is still there. That electron is still there."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now that we got rid of one of these bromines, this thing might want to grab a proton now, since it is positive, and will act as our base. And we'll nab one of these protons. If it nabs the proton, then the leftover electron is still there. That electron is still there. Let me do that in a different color. That electron is still there. And then that electron can be given to that original carbocation, and we'll have a nice aromatic ring again."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That electron is still there. Let me do that in a different color. That electron is still there. And then that electron can be given to that original carbocation, and we'll have a nice aromatic ring again. So how would that look? So you could imagine a situation where this electron, this green electron right here, gets given to the hydrogen nucleus. If it's given to that hydrogen nucleus, then this red electron right here can then be given to the carbocation."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then that electron can be given to that original carbocation, and we'll have a nice aromatic ring again. So how would that look? So you could imagine a situation where this electron, this green electron right here, gets given to the hydrogen nucleus. If it's given to that hydrogen nucleus, then this red electron right here can then be given to the carbocation. And then what are we left with? What are we left with? Well, we have our, let me draw what we started with."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "If it's given to that hydrogen nucleus, then this red electron right here can then be given to the carbocation. And then what are we left with? What are we left with? Well, we have our, let me draw what we started with. So we have our ring. We have this double bond and that double bond right there. Now let me draw all of our hydrogens."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have our, let me draw what we started with. So we have our ring. We have this double bond and that double bond right there. Now let me draw all of our hydrogens. We have this hydrogen, that hydrogen, this one over here, this one over here, and we have this one over here. Now this hydrogen just now got nabbed. So this hydrogen over here got nabbed."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now let me draw all of our hydrogens. We have this hydrogen, that hydrogen, this one over here, this one over here, and we have this one over here. Now this hydrogen just now got nabbed. So this hydrogen over here got nabbed. It got given, actually the hydrogen nucleus got given this green electron, which is paired with this magenta one right over here. So it is now bonded to the bromine. We now have hydrogen bromide."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen over here got nabbed. It got given, actually the hydrogen nucleus got given this green electron, which is paired with this magenta one right over here. So it is now bonded to the bromine. We now have hydrogen bromide. And this had one, two, three, four, five, six other valence electrons. I'll keep the colors consistent. And this is now bonded."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We now have hydrogen bromide. And this had one, two, three, four, five, six other valence electrons. I'll keep the colors consistent. And this is now bonded. Now this electron went away from the iron, so the iron will now lose its negative charge, the iron bromide. So now it is back in its original form. So we have our iron bonded to three bromines."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this is now bonded. Now this electron went away from the iron, so the iron will now lose its negative charge, the iron bromide. So now it is back in its original form. So we have our iron bonded to three bromines. Just like that, it has lost its negative charge. And this electron right here has now gone to the carbocation. So that electron right there has now gone to the carbocation."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our iron bonded to three bromines. Just like that, it has lost its negative charge. And this electron right here has now gone to the carbocation. So that electron right there has now gone to the carbocation. It is right there. So this bond, you can imagine it now being this double bond. So it was magenta, so I'll draw it in magenta just there."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that electron right there has now gone to the carbocation. It is right there. So this bond, you can imagine it now being this double bond. So it was magenta, so I'll draw it in magenta just there. And we can't forget the whole point of this whole video was the bromination. So now we now have this bromo group there. So we have this orange electron is right over here."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it was magenta, so I'll draw it in magenta just there. And we can't forget the whole point of this whole video was the bromination. So now we now have this bromo group there. So we have this orange electron is right over here. And it is bonded to that bromine. Just like that. And we have brominated the benzene ring using, and just let's be clear, we had a negative charge and a positive charge."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have this orange electron is right over here. And it is bonded to that bromine. Just like that. And we have brominated the benzene ring using, and just let's be clear, we had a negative charge and a positive charge. Now it's canceled out. The thing with the negative charge gave an electron to the positive charge. It is now neutral."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have brominated the benzene ring using, and just let's be clear, we had a negative charge and a positive charge. Now it's canceled out. The thing with the negative charge gave an electron to the positive charge. It is now neutral. That is now neutral. Now let's be clear on what happened here. We want to just map it to what we saw before."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It is now neutral. That is now neutral. Now let's be clear on what happened here. We want to just map it to what we saw before. We had a benzene ring. Right here we have no strong electrophile or strong base yet, it had to become a strong electrophile. When this thing gave an electron to the iron bromine it became this larger molecule."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We want to just map it to what we saw before. We had a benzene ring. Right here we have no strong electrophile or strong base yet, it had to become a strong electrophile. When this thing gave an electron to the iron bromine it became this larger molecule. Now this whole thing can act as a pretty strong electrophile. This grabbed an electron. It broke the aromaticity of the benzene ring, but just long enough for this bromine to form."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "When this thing gave an electron to the iron bromine it became this larger molecule. Now this whole thing can act as a pretty strong electrophile. This grabbed an electron. It broke the aromaticity of the benzene ring, but just long enough for this bromine to form. But once this bromine, it gained an electron and bonded to the benzene, and then gave up an electron to this other bromine that really wanted to get an electron because it had a positive charge. And then once it got it, now this whole thing acted as the base. So we kind of have the same molecule changing up a little bit, acting as an electrophile or acting as a base."}, {"video_title": "Bromination of benzene Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It broke the aromaticity of the benzene ring, but just long enough for this bromine to form. But once this bromine, it gained an electron and bonded to the benzene, and then gave up an electron to this other bromine that really wanted to get an electron because it had a positive charge. And then once it got it, now this whole thing acted as the base. So we kind of have the same molecule changing up a little bit, acting as an electrophile or acting as a base. And then once it acts as a base, this bromine, this magenta bromine, nabs the proton, allows this electron to go back to that carbocation, and then we're left with the iron bromide again. So this thing really didn't change through the whole reaction. That's why we can call it a catalyst."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In the previous video, we looked at the mechanism for the Birch reduction. In this video, we're going to see what happens with the Birch reduction when you have a substituted benzene ring. So if you have a benzene ring with a substituent on it, and you add sodium, liquid ammonia, and an alcohol, the substituent is going to affect which carbons are reduced. So for example, if you have an electron withdrawing group on your ring, the carbon that's bonded to your y substituent here is reduced. If you have an electron donating group on your ring, the carbon that's bonded to that substituent is not reduced. Let's see if we can learn why by going through a mechanism for each one of these examples. And we'll start with an electron withdrawing group."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So for example, if you have an electron withdrawing group on your ring, the carbon that's bonded to your y substituent here is reduced. If you have an electron donating group on your ring, the carbon that's bonded to that substituent is not reduced. Let's see if we can learn why by going through a mechanism for each one of these examples. And we'll start with an electron withdrawing group. And so here we have an electron withdrawing group that is an ester on our benzene ring. And we're going to start with our sodium, which we know has one valence electron. I'm going to go ahead and just do the mechanism."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with an electron withdrawing group. And so here we have an electron withdrawing group that is an ester on our benzene ring. And we're going to start with our sodium, which we know has one valence electron. I'm going to go ahead and just do the mechanism. And we'll talk about details as to why it happens that way when I'm done here. So this lone electron on sodium is going to be donated to this carbon. And we have a bond here, which consists of two electrons."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and just do the mechanism. And we'll talk about details as to why it happens that way when I'm done here. So this lone electron on sodium is going to be donated to this carbon. And we have a bond here, which consists of two electrons. One of those electrons is also going to move out onto that carbon. And one of those electrons is going to move into here. With our other bond, similar idea."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have a bond here, which consists of two electrons. One of those electrons is also going to move out onto that carbon. And one of those electrons is going to move into here. With our other bond, similar idea. One of the electrons is going to move into here. And one of them is going to move off onto this carbon. So let's go ahead and show the movement of all of those electrons."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "With our other bond, similar idea. One of the electrons is going to move into here. And one of them is going to move off onto this carbon. So let's go ahead and show the movement of all of those electrons. So once again, we have our ester group up here. We added an electron to this carbon from the sodium. We got another electron from that bond."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the movement of all of those electrons. So once again, we have our ester group up here. We added an electron to this carbon from the sodium. We got another electron from that bond. So that gives this carbon a negative 1 formal charge. And over here, we formed a pi bond. There was already a pi bond over here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We got another electron from that bond. So that gives this carbon a negative 1 formal charge. And over here, we formed a pi bond. There was already a pi bond over here. And then, of course, we get one electron on that carbon. So this is our radical anion. And if you had a hard time following those electrons, please go back and watch the previous video and take some time and study the mechanism really well."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "There was already a pi bond over here. And then, of course, we get one electron on that carbon. So this is our radical anion. And if you had a hard time following those electrons, please go back and watch the previous video and take some time and study the mechanism really well. So in the next step of the mechanism, we know that's the protonation step. So we have an alcohol comes along here. And we're going to protonate our anion."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if you had a hard time following those electrons, please go back and watch the previous video and take some time and study the mechanism really well. So in the next step of the mechanism, we know that's the protonation step. So we have an alcohol comes along here. And we're going to protonate our anion. So these electrons here pick up a proton and kick those electrons off onto the oxygen. So let's go ahead and draw what we have now. So pi electrons here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to protonate our anion. So these electrons here pick up a proton and kick those electrons off onto the oxygen. So let's go ahead and draw what we have now. So pi electrons here. Up here, we have our electron withdrawing group, our ester. And we had an electron on this carbon. There was already a hydrogen bonded to this carbon."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So pi electrons here. Up here, we have our electron withdrawing group, our ester. And we had an electron on this carbon. There was already a hydrogen bonded to this carbon. And those electrons picked up a proton. And so we now have two hydrogens bonded to that carbon. In the third step of our mechanism, we started off with an electron."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "There was already a hydrogen bonded to this carbon. And those electrons picked up a proton. And so we now have two hydrogens bonded to that carbon. In the third step of our mechanism, we started off with an electron. And then we went to a protonation. And so we know the third step is electron again. And so you can think about another sodium coming along with its valence electron."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "In the third step of our mechanism, we started off with an electron. And then we went to a protonation. And so we know the third step is electron again. And so you can think about another sodium coming along with its valence electron. And of course, it's going to donate its valence electron to the carbon that already has an electron on it, so to this carbon right here. So let's go ahead and draw the result of that. So I have my ring."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so you can think about another sodium coming along with its valence electron. And of course, it's going to donate its valence electron to the carbon that already has an electron on it, so to this carbon right here. So let's go ahead and draw the result of that. So I have my ring. I have my pi electrons in my ring. I have my hydrogens already bonded to it. Up here, I have my ester group bonded to my ring."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I have my ring. I have my pi electrons in my ring. I have my hydrogens already bonded to it. Up here, I have my ester group bonded to my ring. I had one electron on this carbon. And the sodium just donated another one. So that gives me a negative 1 formal charge."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Up here, I have my ester group bonded to my ring. I had one electron on this carbon. And the sodium just donated another one. So that gives me a negative 1 formal charge. And so this is our anion right here. In the final step for our mechanism, we know that's another protonation step. So I go ahead and show my anion picking up a proton from this alcohol here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that gives me a negative 1 formal charge. And so this is our anion right here. In the final step for our mechanism, we know that's another protonation step. So I go ahead and show my anion picking up a proton from this alcohol here. And so we can go ahead and show and draw the final product. So I have my ring. I have these pi electrons."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I go ahead and show my anion picking up a proton from this alcohol here. And so we can go ahead and show and draw the final product. So I have my ring. I have these pi electrons. I have these hydrogens on this carbon. And I have my ester group coming off of my ring. And I added on a proton to that carbon."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I have these pi electrons. I have these hydrogens on this carbon. And I have my ester group coming off of my ring. And I added on a proton to that carbon. So that's the final product. And you can see, of course, that the carbon that is bonded to our electron withdrawing group was reduced. It gained a proton here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I added on a proton to that carbon. So that's the final product. And you can see, of course, that the carbon that is bonded to our electron withdrawing group was reduced. It gained a proton here. And the question is why. Why do we get this as our product? And a way to think about that is to go back here to this structure, so this anion step."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It gained a proton here. And the question is why. Why do we get this as our product? And a way to think about that is to go back here to this structure, so this anion step. And I can draw a resonance structure for this anion because of the presence of that electron withdrawing group. So let me go ahead and put in my resonance brackets and my resonance arrow here. And this negative charge, I'm going to go ahead and use red here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And a way to think about that is to go back here to this structure, so this anion step. And I can draw a resonance structure for this anion because of the presence of that electron withdrawing group. So let me go ahead and put in my resonance brackets and my resonance arrow here. And this negative charge, I'm going to go ahead and use red here. So these electrons right here can participate in resonance due to the presence of our electron withdrawing group. And we know that this oxygen has two lone pairs of electrons on it. So if these electrons moved in here to form a bond, that would push these pi electrons off onto the oxygen."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And this negative charge, I'm going to go ahead and use red here. So these electrons right here can participate in resonance due to the presence of our electron withdrawing group. And we know that this oxygen has two lone pairs of electrons on it. So if these electrons moved in here to form a bond, that would push these pi electrons off onto the oxygen. So let's go ahead and draw our resonance structure. So we have our ring. We have our pi electrons."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if these electrons moved in here to form a bond, that would push these pi electrons off onto the oxygen. So let's go ahead and draw our resonance structure. So we have our ring. We have our pi electrons. Now the carbon on our ring is double bonded to the carbon out here. And that carbon is bonded to an oxygen. That oxygen now has three lone pairs of electrons, which gives it a negative 1 formal charge."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have our pi electrons. Now the carbon on our ring is double bonded to the carbon out here. And that carbon is bonded to an oxygen. That oxygen now has three lone pairs of electrons, which gives it a negative 1 formal charge. And we still have this part over here on our molecule and these hydrogens, of course. And so let me show those electrons in red there. So those electrons in red moved in here to form a pi bond."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That oxygen now has three lone pairs of electrons, which gives it a negative 1 formal charge. And we still have this part over here on our molecule and these hydrogens, of course. And so let me show those electrons in red there. So those electrons in red moved in here to form a pi bond. And so there's resonance stabilization of electron density at the ipso position. So the ipso position is the position that's right next to our substituent. So it's the carbon on our ring that's bonded to our substituent."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So those electrons in red moved in here to form a pi bond. And so there's resonance stabilization of electron density at the ipso position. So the ipso position is the position that's right next to our substituent. So it's the carbon on our ring that's bonded to our substituent. And because of the presence of our electron withdrawing group, electron density at that position is stabilized. And actually, if you go back here to the radical anion, these electrons are actually delocalized. And so the presence of that electron withdrawing group stabilizes electron density at the ipso position and also the para position."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's the carbon on our ring that's bonded to our substituent. And because of the presence of our electron withdrawing group, electron density at that position is stabilized. And actually, if you go back here to the radical anion, these electrons are actually delocalized. And so the presence of that electron withdrawing group stabilizes electron density at the ipso position and also the para position. And so that's why you see the reduction happening at those carbons. So now that we have an idea about why we get this as the product when you get an electron withdrawing group present on your ring, let's go ahead and do one for an electron donating group. And we'll try to compare."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the presence of that electron withdrawing group stabilizes electron density at the ipso position and also the para position. And so that's why you see the reduction happening at those carbons. So now that we have an idea about why we get this as the product when you get an electron withdrawing group present on your ring, let's go ahead and do one for an electron donating group. And we'll try to compare. So the electron donating group here is a methoxy group. So here's our methoxy group, which we know the oxygen has two lone pairs of electrons. And it's those lone pairs of electrons."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we'll try to compare. So the electron donating group here is a methoxy group. So here's our methoxy group, which we know the oxygen has two lone pairs of electrons. And it's those lone pairs of electrons. It's one of those lone pairs that can contribute and increase the electron density in your ring. And so therefore, the methoxy group is an electron donating group, as we saw in some of the other videos here. So we'll start the mechanism."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And it's those lone pairs of electrons. It's one of those lone pairs that can contribute and increase the electron density in your ring. And so therefore, the methoxy group is an electron donating group, as we saw in some of the other videos here. So we'll start the mechanism. And once again, let me just go ahead and do the mechanism. We'll talk about why it occurs when I'm done here. So if the sodium donates its valence electron to the ring, and it's going to donate it to this carbon right here on our ring."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start the mechanism. And once again, let me just go ahead and do the mechanism. We'll talk about why it occurs when I'm done here. So if the sodium donates its valence electron to the ring, and it's going to donate it to this carbon right here on our ring. For this bond, one of the electrons is going to move out onto that same carbon that the sodium donated electron to. And then the other electron is going to move into here. So for this bond, one of the electrons is going to move into here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if the sodium donates its valence electron to the ring, and it's going to donate it to this carbon right here on our ring. For this bond, one of the electrons is going to move out onto that same carbon that the sodium donated electron to. And then the other electron is going to move into here. So for this bond, one of the electrons is going to move into here. And one of the electrons is going to move out onto this carbon. So let's go ahead and draw the result of all of those electrons moving. So once again, we have our methoxy group up here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So for this bond, one of the electrons is going to move into here. And one of the electrons is going to move out onto this carbon. So let's go ahead and draw the result of all of those electrons moving. So once again, we have our methoxy group up here. So this carbon got an electron from sodium. This carbon got an electron from the bond. And so now it has a negative 1 formal charge on it."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we have our methoxy group up here. So this carbon got an electron from sodium. This carbon got an electron from the bond. And so now it has a negative 1 formal charge on it. We formed a pi bond over here. And we also have an unpaired electron over here. So this is our radical anion."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so now it has a negative 1 formal charge on it. We formed a pi bond over here. And we also have an unpaired electron over here. So this is our radical anion. Next, of course, is protonation. So we can go ahead and show our alcohol over here. So our carbanion functions as a base, picks up a proton."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this is our radical anion. Next, of course, is protonation. So we can go ahead and show our alcohol over here. So our carbanion functions as a base, picks up a proton. And let's go ahead and show the result after our ring is protonated here. So this is protonation in the ortho position. So if I think about this carbon right here, protonation occurs ortho to our methoxy group."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So our carbanion functions as a base, picks up a proton. And let's go ahead and show the result after our ring is protonated here. So this is protonation in the ortho position. So if I think about this carbon right here, protonation occurs ortho to our methoxy group. So there was already a hydrogen on that ring. And we protonated. So now there are two hydrogens on that ring."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I think about this carbon right here, protonation occurs ortho to our methoxy group. So there was already a hydrogen on that ring. And we protonated. So now there are two hydrogens on that ring. Pi bond over here. We still have our unpaired electron. So third step of our mechanism, of course, would be another electron, which we get from sodium."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So now there are two hydrogens on that ring. Pi bond over here. We still have our unpaired electron. So third step of our mechanism, of course, would be another electron, which we get from sodium. So sodium comes along and donates a valence electron right here. So let's go ahead and draw the result of that. So we have our ring."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So third step of our mechanism, of course, would be another electron, which we get from sodium. So sodium comes along and donates a valence electron right here. So let's go ahead and draw the result of that. So we have our ring. We have our electron donating group. We have pi electrons here and here. We had these hydrogens."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have our ring. We have our electron donating group. We have pi electrons here and here. We had these hydrogens. And at this carbon, we had one electron. We got another one from another sodium. So that gives us a negative 1 formal charge."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We had these hydrogens. And at this carbon, we had one electron. We got another one from another sodium. So that gives us a negative 1 formal charge. And so that's our anion. So the next step would be protonation. So this time, protonation is occurring meta to our electron donating group."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that gives us a negative 1 formal charge. And so that's our anion. So the next step would be protonation. So this time, protonation is occurring meta to our electron donating group. So if I think about this carbon right here, now we're talking about meta protonation compared to the position of our electron donating group. And so these electrons are going to pick up a proton from our alcohol, kicks these electrons in here off. And so now we can go ahead and show the product."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this time, protonation is occurring meta to our electron donating group. So if I think about this carbon right here, now we're talking about meta protonation compared to the position of our electron donating group. And so these electrons are going to pick up a proton from our alcohol, kicks these electrons in here off. And so now we can go ahead and show the product. So once again, our methoxy group, our pi electrons, we had hydrogens on that carbon. There was already a hydrogen on this carbon. It picked up a proton."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so now we can go ahead and show the product. So once again, our methoxy group, our pi electrons, we had hydrogens on that carbon. There was already a hydrogen on this carbon. It picked up a proton. And so now there are two hydrogens. And so now we're done. This is our product."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It picked up a proton. And so now there are two hydrogens. And so now we're done. This is our product. And so of course, the question is, why is this a little bit different from the electron withdrawing group? Why does the electron donating group work this way? So once again, we can look at the carbon that's attached to our methoxy group, our electron donating group."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This is our product. And so of course, the question is, why is this a little bit different from the electron withdrawing group? Why does the electron donating group work this way? So once again, we can look at the carbon that's attached to our methoxy group, our electron donating group. That carbon was not reduced. And the way to think about it is, or one way of thinking about it is, in this mechanism, you can see we have a lot of electron density in the ortho and the meta position. So we have a lot of electron density in the ortho and the meta position."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we can look at the carbon that's attached to our methoxy group, our electron donating group. That carbon was not reduced. And the way to think about it is, or one way of thinking about it is, in this mechanism, you can see we have a lot of electron density in the ortho and the meta position. So we have a lot of electron density in the ortho and the meta position. And so those are the carbons that are reduced. And it makes sense that we don't reduce the carbon that's attached to our electron donating group because, of course, our electron donating group would donate some electron density to that carbon. So let me go ahead and draw a little picture here so you can understand a little bit better about what I'm talking about."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a lot of electron density in the ortho and the meta position. And so those are the carbons that are reduced. And it makes sense that we don't reduce the carbon that's attached to our electron donating group because, of course, our electron donating group would donate some electron density to that carbon. So let me go ahead and draw a little picture here so you can understand a little bit better about what I'm talking about. If I had my electron donating group, I can move these electrons into here. And so there's a little bit more electron density. And so if we had a negative 1 formal charge or some electron density right at this carbon, these electrons here would, of course, destabilize an anion or electron density at this position, at the ipso position."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw a little picture here so you can understand a little bit better about what I'm talking about. If I had my electron donating group, I can move these electrons into here. And so there's a little bit more electron density. And so if we had a negative 1 formal charge or some electron density right at this carbon, these electrons here would, of course, destabilize an anion or electron density at this position, at the ipso position. And so you're not going to see reduction at the ipso and the para position. So that's what we saw when we had the electron withdrawing group. An electron withdrawing group stabilized electron density at the ipso and the para position here."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if we had a negative 1 formal charge or some electron density right at this carbon, these electrons here would, of course, destabilize an anion or electron density at this position, at the ipso position. And so you're not going to see reduction at the ipso and the para position. So that's what we saw when we had the electron withdrawing group. An electron withdrawing group stabilized electron density at the ipso and the para position here. But that's not the case for an electron donating group. It would destabilize any electron density at those positions. And so that's why you think about an ortho and a meta."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "An electron withdrawing group stabilized electron density at the ipso and the para position here. But that's not the case for an electron donating group. It would destabilize any electron density at those positions. And so that's why you think about an ortho and a meta. It's more out of the way. And so it doesn't destabilize it as much. And of course, you could think about this ortho and this meta."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that's why you think about an ortho and a meta. It's more out of the way. And so it doesn't destabilize it as much. And of course, you could think about this ortho and this meta. It doesn't really matter because of the symmetry for this example. But that's one way to think about it. Maybe it's not the best explanation."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, you could think about this ortho and this meta. It doesn't really matter because of the symmetry for this example. But that's one way to think about it. Maybe it's not the best explanation. But it's a good way of thinking about it here. And at one time, there was some disagreement over the order of protonation. So it turns out that you do have an ortho protonation first and then followed by a meta protonation."}, {"video_title": "Birch reduction II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Maybe it's not the best explanation. But it's a good way of thinking about it here. And at one time, there was some disagreement over the order of protonation. So it turns out that you do have an ortho protonation first and then followed by a meta protonation. But some people used to think that a meta protonation was first followed by an ortho. But I think that studies have shown that ortho was first and now meta. So this just gives you a little bit of insight into what electron withdrawing groups and what electron donating groups, what effect they can have, what effect those directors can have on the Birch reduction."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and do that. So we can see that the longest carbon chain for this example has five carbons, which we know is pentane. The second step is to number the chain to give the substituents the lowest number possible. So I could number this chain from the left, or I could number it from the right. If I number it from the left, I give that substituent a number of two. If I numbered it from the right, that substituent would have a number of four. So numbering it from the left for this example is the correct way to approach it."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So I could number this chain from the left, or I could number it from the right. If I number it from the left, I give that substituent a number of two. If I numbered it from the right, that substituent would have a number of four. So numbering it from the left for this example is the correct way to approach it. Identify and name your substituents is step three. Well, I know that this is a methyl group, and that methyl group is found on carbon two. So I would write 2-methylpentane, and that would be the correct IUPAC name for this dot structure."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So numbering it from the left for this example is the correct way to approach it. Identify and name your substituents is step three. Well, I know that this is a methyl group, and that methyl group is found on carbon two. So I would write 2-methylpentane, and that would be the correct IUPAC name for this dot structure. I don't have to worry about step four because I have only one substituent. So when we see multiple substituents, we'll have to think about using the alphabet rule. So 2-methylpentane is the correct IUPAC name for this molecule."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So I would write 2-methylpentane, and that would be the correct IUPAC name for this dot structure. I don't have to worry about step four because I have only one substituent. So when we see multiple substituents, we'll have to think about using the alphabet rule. So 2-methylpentane is the correct IUPAC name for this molecule. And from that name, you should be able to draw the dot structure. Let's do another one here. So let's follow our steps."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So 2-methylpentane is the correct IUPAC name for this molecule. And from that name, you should be able to draw the dot structure. Let's do another one here. So let's follow our steps. Find the longest carbon chain and name it. So I have one, two, three, four, five, six. So this would be hexane for my longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow our steps. Find the longest carbon chain and name it. So I have one, two, three, four, five, six. So this would be hexane for my longest carbon chain. Number to give your substituents the lowest number possible. So once again, I have a choice of numbering from the left or from the right. In this case, numbering from the left would give my substituents the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So this would be hexane for my longest carbon chain. Number to give your substituents the lowest number possible. So once again, I have a choice of numbering from the left or from the right. In this case, numbering from the left would give my substituents the lowest number possible. What are my substituents? Well, I have two methyl groups this time. So I have to use a prefix."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "In this case, numbering from the left would give my substituents the lowest number possible. What are my substituents? Well, I have two methyl groups this time. So I have to use a prefix. I have two methyl groups, so I'm going to call this dimethyl. So dimethylhexane. And those methyl groups are coming off of carbon two."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So I have to use a prefix. I have two methyl groups, so I'm going to call this dimethyl. So dimethylhexane. And those methyl groups are coming off of carbon two. So I have to write 2, 2-dimethylhexane. So when you have prefixes, you would use di for two. You would use tri for three."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And those methyl groups are coming off of carbon two. So I have to write 2, 2-dimethylhexane. So when you have prefixes, you would use di for two. You would use tri for three. You would use tetra for four. You would use penta for five and hexa for six. All right, so let's do another nomenclature example."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "You would use tri for three. You would use tetra for four. You would use penta for five and hexa for six. All right, so let's do another nomenclature example. Let's look at this molecule. Find the longest carbon chain. So let's see, that would be 1, 2, 3, 4, 5, 6, 7, and name it."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's do another nomenclature example. Let's look at this molecule. Find the longest carbon chain. So let's see, that would be 1, 2, 3, 4, 5, 6, 7, and name it. So 7 would be heptane. So I have heptane here. Number to give your substituents the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So let's see, that would be 1, 2, 3, 4, 5, 6, 7, and name it. So 7 would be heptane. So I have heptane here. Number to give your substituents the lowest number possible. My options are to start from the left or to start from the right. This time, it makes more sense to start from the right because that gives my first substituent a number of 2, which would be lower than if I started from the left. So 1, 2, 3, 4, 5, 6, and 7."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Number to give your substituents the lowest number possible. My options are to start from the left or to start from the right. This time, it makes more sense to start from the right because that gives my first substituent a number of 2, which would be lower than if I started from the left. So 1, 2, 3, 4, 5, 6, and 7. So that is heptane. Identify your substituents and name them is the third step. So that's a methyl group coming off of carbon 2, and that is an ethyl group coming off of carbon 4."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6, and 7. So that is heptane. Identify your substituents and name them is the third step. So that's a methyl group coming off of carbon 2, and that is an ethyl group coming off of carbon 4. Which one comes first, the ethyl group or the methyl group? Step four says it's the alphabet. You arrange them alphabetically."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So that's a methyl group coming off of carbon 2, and that is an ethyl group coming off of carbon 4. Which one comes first, the ethyl group or the methyl group? Step four says it's the alphabet. You arrange them alphabetically. So E comes before M. So you're going to put ethyl before methyl. So you have an ethyl group coming off of carbon 4, so 4-ethyl, and a methyl group coming off of carbon 2, so 2-methyl, if I can squeeze it in here. So I have 4-ethyl-2-methylheptane is the official IUPAC name for this molecule."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "You arrange them alphabetically. So E comes before M. So you're going to put ethyl before methyl. So you have an ethyl group coming off of carbon 4, so 4-ethyl, and a methyl group coming off of carbon 2, so 2-methyl, if I can squeeze it in here. So I have 4-ethyl-2-methylheptane is the official IUPAC name for this molecule. What about something like this? So this is a little bit more complicated. Let's look at my longest carbon chain and how many carbons are in my longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So I have 4-ethyl-2-methylheptane is the official IUPAC name for this molecule. What about something like this? So this is a little bit more complicated. Let's look at my longest carbon chain and how many carbons are in my longest carbon chain. That would be 8. This is the exact same dot structure. So both of these are going to be octane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at my longest carbon chain and how many carbons are in my longest carbon chain. That would be 8. This is the exact same dot structure. So both of these are going to be octane. Number your carbon chain to give the substituents the lowest number possible. Well, if I number from the left, I would have my substituent with a number of 2. If I number from the right, I would have a substituent with a number of 2."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So both of these are going to be octane. Number your carbon chain to give the substituents the lowest number possible. Well, if I number from the left, I would have my substituent with a number of 2. If I number from the right, I would have a substituent with a number of 2. So it's not immediately obvious to me which would be the correct way to start from. So let me go ahead and just number all of them. 1, 2, 3, 4, 5, 6, 7, 8."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "If I number from the right, I would have a substituent with a number of 2. So it's not immediately obvious to me which would be the correct way to start from. So let me go ahead and just number all of them. 1, 2, 3, 4, 5, 6, 7, 8. And if I go this way, 1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 8. So let's go ahead and name both of these. And let's see what we get."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7, 8. And if I go this way, 1, 2, 3, 1, 2, 3, 4, 5, 6, 7, 8. So let's go ahead and name both of these. And let's see what we get. Well, what kind of substituents do I have on the left? I have, let's see, how many methyl groups? A total of 5 methyl groups."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And let's see what we get. Well, what kind of substituents do I have on the left? I have, let's see, how many methyl groups? A total of 5 methyl groups. So it would be pentamethyl. So pentamethyl octane. And obviously, they both would be pentamethyl octane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "A total of 5 methyl groups. So it would be pentamethyl. So pentamethyl octane. And obviously, they both would be pentamethyl octane. It's the same dot structure here. So pentamethyl octane. Where are those methyl groups?"}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And obviously, they both would be pentamethyl octane. It's the same dot structure here. So pentamethyl octane. Where are those methyl groups? Well, on the left, they are on 2, 3, 3, 7, 7. So 2, 3, 3, 7, 7 pentamethyl octane. But on the right, my methyl groups are on 2, 2, 6, 6, 7."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Where are those methyl groups? Well, on the left, they are on 2, 3, 3, 7, 7. So 2, 3, 3, 7, 7 pentamethyl octane. But on the right, my methyl groups are on 2, 2, 6, 6, 7. So let me go ahead and put that down. 2, 2, 6, 6, 7 pentamethyl octane. So I want to give my substituents the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "But on the right, my methyl groups are on 2, 2, 6, 6, 7. So let me go ahead and put that down. 2, 2, 6, 6, 7 pentamethyl octane. So I want to give my substituents the lowest number possible. But I've already seen that that first number, 2 versus 2, is a tie. So that doesn't quite work. What I need to do is go to the next number."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So I want to give my substituents the lowest number possible. But I've already seen that that first number, 2 versus 2, is a tie. So that doesn't quite work. What I need to do is go to the next number. So I have a 3 over here. And I have a 2 over here. And you always want to do the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "What I need to do is go to the next number. So I have a 3 over here. And I have a 2 over here. And you always want to do the lowest number possible. So since 2 is a lower number than 3, this is actually the correct IUPAC name for this molecule, 2, 2, 6, 6, 7 pentamethyl octane. This is called the first point of difference rule. You want to give your substituents the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And you always want to do the lowest number possible. So since 2 is a lower number than 3, this is actually the correct IUPAC name for this molecule, 2, 2, 6, 6, 7 pentamethyl octane. This is called the first point of difference rule. You want to give your substituents the lowest number possible. So if there's a tie with the first number, you go to the second number and compare those numbers. And if there's a tie with the second number, you go on to the third number and so on. So in this case, the first point of difference came with the second number."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "You want to give your substituents the lowest number possible. So if there's a tie with the first number, you go to the second number and compare those numbers. And if there's a tie with the second number, you go on to the third number and so on. So in this case, the first point of difference came with the second number. Let's look at cycloalkanes. So here I have a cycloalkane. And you name it the same way you would name a straight chain alkane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So in this case, the first point of difference came with the second number. Let's look at cycloalkanes. So here I have a cycloalkane. And you name it the same way you would name a straight chain alkane. You first find the longest number of carbons. In this case, it's in a ring. So we have a total of six carbons in a ring, which we know is cyclohexane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And you name it the same way you would name a straight chain alkane. You first find the longest number of carbons. In this case, it's in a ring. So we have a total of six carbons in a ring, which we know is cyclohexane. So that's my parent name. So cyclohexane is my parent name. Number the ring to give the substituents the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So we have a total of six carbons in a ring, which we know is cyclohexane. So that's my parent name. So cyclohexane is my parent name. Number the ring to give the substituents the lowest number possible. Well, here's my substituent. I know that's an ethyl group. I want to number my ring to give that substituent the lowest number possible."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Number the ring to give the substituents the lowest number possible. Well, here's my substituent. I know that's an ethyl group. I want to number my ring to give that substituent the lowest number possible. So obviously, that must be number one. So which way do I go around my ring? Well, it doesn't matter for this example, because it'd be the exact same, since I have only one substituent."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "I want to number my ring to give that substituent the lowest number possible. So obviously, that must be number one. So which way do I go around my ring? Well, it doesn't matter for this example, because it'd be the exact same, since I have only one substituent. So six carbons all the way around, with an ethyl group coming off of carbon one. So I could call this one ethyl cyclohexane. Or you could even leave off the one and just say ethyl cyclohexane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Well, it doesn't matter for this example, because it'd be the exact same, since I have only one substituent. So six carbons all the way around, with an ethyl group coming off of carbon one. So I could call this one ethyl cyclohexane. Or you could even leave off the one and just say ethyl cyclohexane. It's implied that the ethyl group is coming off of carbon one here. So that's how to name a cycloalkane. Let's look at a more complicated cycloalkane here."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Or you could even leave off the one and just say ethyl cyclohexane. It's implied that the ethyl group is coming off of carbon one here. So that's how to name a cycloalkane. Let's look at a more complicated cycloalkane here. So now I have, again, I can see that it's cyclohexane. That's my base name here. So this is going to be cyclohexane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at a more complicated cycloalkane here. So now I have, again, I can see that it's cyclohexane. That's my base name here. So this is going to be cyclohexane. So I can go ahead and put that down. Next, I have to number to give my substituents the lowest number possible. So which one of these two carbons is going to be number one?"}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to be cyclohexane. So I can go ahead and put that down. Next, I have to number to give my substituents the lowest number possible. So which one of these two carbons is going to be number one? I could make this carbon number one, or I could make this carbon number one. Well, first point of difference rule. If I make the top carbon number one, that's going to give me two number ones."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So which one of these two carbons is going to be number one? I could make this carbon number one, or I could make this carbon number one. Well, first point of difference rule. If I make the top carbon number one, that's going to give me two number ones. Versus if I make this carbon down here number one, I have only one substituent coming off of carbon one. So the first point of difference rule says this could not be carbon number one. So carbon number one has to be this top carbon up here like that."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "If I make the top carbon number one, that's going to give me two number ones. Versus if I make this carbon down here number one, I have only one substituent coming off of carbon one. So the first point of difference rule says this could not be carbon number one. So carbon number one has to be this top carbon up here like that. So how do I number my ring? Well, I could go this way. I could go one, two, three, four, five, six."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So carbon number one has to be this top carbon up here like that. So how do I number my ring? Well, I could go this way. I could go one, two, three, four, five, six. Or I could go the other way. I could say that's number one, two, three, four, five, six. Which way is correct?"}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "I could go one, two, three, four, five, six. Or I could go the other way. I could say that's number one, two, three, four, five, six. Which way is correct? Well, if I think about my substituents, this would be an ethyl group coming off of carbon three, whereas this would be an ethyl group coming off of carbon five. So the one on the left is the correct way to number your ring, because it gives your substituents the lowest number possible. So now we're ready to go ahead and name it."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Which way is correct? Well, if I think about my substituents, this would be an ethyl group coming off of carbon three, whereas this would be an ethyl group coming off of carbon five. So the one on the left is the correct way to number your ring, because it gives your substituents the lowest number possible. So now we're ready to go ahead and name it. Let's look at all of my substituents on this first example here. I have two methyl groups coming off carbon one, so that would be dimethyl. I have an ethyl group coming off of carbon three."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So now we're ready to go ahead and name it. Let's look at all of my substituents on this first example here. I have two methyl groups coming off carbon one, so that would be dimethyl. I have an ethyl group coming off of carbon three. So when I think about the alphabet rule, I'm going to put the ethyl group first. So I have 3-ethyl-1-1-dimethylcyclohexane as the official name. Now, sometimes students get confused with the alphabet rule."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "I have an ethyl group coming off of carbon three. So when I think about the alphabet rule, I'm going to put the ethyl group first. So I have 3-ethyl-1-1-dimethylcyclohexane as the official name. Now, sometimes students get confused with the alphabet rule. Think about the parent name. Think about M versus E. E comes before M in the alphabet. Prefixes don't matter."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Now, sometimes students get confused with the alphabet rule. Think about the parent name. Think about M versus E. E comes before M in the alphabet. Prefixes don't matter. So don't think about alphabetizing something with the D for di. Think about ethyl versus methyl, and that will give you 3-ethyl-1-1-dimethylcyclohexane for as the correct IUPAC name for this molecule. Let's look at another example here."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Prefixes don't matter. So don't think about alphabetizing something with the D for di. Think about ethyl versus methyl, and that will give you 3-ethyl-1-1-dimethylcyclohexane for as the correct IUPAC name for this molecule. Let's look at another example here. So what do I do? What do I do for this one? Well, first thing, identify the longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at another example here. So what do I do? What do I do for this one? Well, first thing, identify the longest carbon chain. In this case, what kind of cycloalkane is it? Well, five carbons, so it's going to be cyclopentane. So cyclopentane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Well, first thing, identify the longest carbon chain. In this case, what kind of cycloalkane is it? Well, five carbons, so it's going to be cyclopentane. So cyclopentane. So they're both going to be cyclopentane here, since it's the same molecule. So cyclopentane. Well, which one gets a number one?"}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So cyclopentane. So they're both going to be cyclopentane here, since it's the same molecule. So cyclopentane. Well, which one gets a number one? I could make this a number one and this a number two. Or I could make this a number one and this a number two. So let's just go ahead and name them both."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Well, which one gets a number one? I could make this a number one and this a number two. Or I could make this a number one and this a number two. So let's just go ahead and name them both. And then let's see which one is the correct name. On the left, I'd have 1-ethyl-2-methylcyclopentane. And on the right, I would have 2-ethyl-1-methylcyclopentane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So let's just go ahead and name them both. And then let's see which one is the correct name. On the left, I'd have 1-ethyl-2-methylcyclopentane. And on the right, I would have 2-ethyl-1-methylcyclopentane. And the question is, which is the correct IUPAC name? First point of difference rule doesn't really work, because I have one that's one, then I have one that's two. So there is no first point of difference for this."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And on the right, I would have 2-ethyl-1-methylcyclopentane. And the question is, which is the correct IUPAC name? First point of difference rule doesn't really work, because I have one that's one, then I have one that's two. So there is no first point of difference for this. So if the first point of difference rules fails, then you go to the alphabet rule. And you say to yourself, all right, so which one is going to get a number one? Is it the ethyl that's going to get a number one?"}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So there is no first point of difference for this. So if the first point of difference rules fails, then you go to the alphabet rule. And you say to yourself, all right, so which one is going to get a number one? Is it the ethyl that's going to get a number one? Or is it the methyl that's going to get a number one? If first point of difference rule fails, you go to the alphabet rule. So E comes before M. So the ethyl group gets priority."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Is it the ethyl that's going to get a number one? Or is it the methyl that's going to get a number one? If first point of difference rule fails, you go to the alphabet rule. So E comes before M. So the ethyl group gets priority. So ethyl is number one. And therefore, this is the correct IUPAC name for the molecule like that. All right, so let's look at one more here."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So E comes before M. So the ethyl group gets priority. So ethyl is number one. And therefore, this is the correct IUPAC name for the molecule like that. All right, so let's look at one more here. So how do I name this guy? So it's kind of funny looking. Well, let's see how many carbons are in my longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "All right, so let's look at one more here. So how do I name this guy? So it's kind of funny looking. Well, let's see how many carbons are in my longest carbon chain. So there's one, two, three, four, five, six, and seven. So there are seven carbons in that chain. How many carbons are there in the cyclope portion?"}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "Well, let's see how many carbons are in my longest carbon chain. So there's one, two, three, four, five, six, and seven. So there are seven carbons in that chain. How many carbons are there in the cyclope portion? Well, there's one, two, and three. So I have more carbons in my straight chain alkane than I do in my cycloalkane. And therefore, I'm going to name this as an alkane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "How many carbons are there in the cyclope portion? Well, there's one, two, and three. So I have more carbons in my straight chain alkane than I do in my cycloalkane. And therefore, I'm going to name this as an alkane. So there are seven carbons. So it's going to be a heptane. And I have a cycloalkane as an alkyl group now."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, I'm going to name this as an alkane. So there are seven carbons. So it's going to be a heptane. And I have a cycloalkane as an alkyl group now. So if this wasn't an alkyl group, I'd call it cyclopropane. I drop the ane ending because it's now an alkyl group. So it becomes cyclopropyl."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "And I have a cycloalkane as an alkyl group now. So if this wasn't an alkyl group, I'd call it cyclopropane. I drop the ane ending because it's now an alkyl group. So it becomes cyclopropyl. So I have a cyclopropyl group coming off of my heptane, coming off of, let's see, which carbon is that? Let's go and number it. One, two, three, four, five, six, seven."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "So it becomes cyclopropyl. So I have a cyclopropyl group coming off of my heptane, coming off of, let's see, which carbon is that? Let's go and number it. One, two, three, four, five, six, seven. So I have a cyclopropyl group coming off of carbon four. So it's four cyclopropyl heptane. So when you have more carbons in your chain than you do in your cycloalkane, you name it as a cycloalkyl group and then your alkane."}, {"video_title": "Alkane and cycloalkane nomenclature II Organic chemistry Khan Academy.mp3", "Sentence": "One, two, three, four, five, six, seven. So I have a cyclopropyl group coming off of carbon four. So it's four cyclopropyl heptane. So when you have more carbons in your chain than you do in your cycloalkane, you name it as a cycloalkyl group and then your alkane. Let's remind ourselves an example up here where we had six carbons in our cycloalkane and only two carbons in our alkyl. In that case, you name it as an alkyl cyclohexane. And if there's a tie, let's say, if there's a tie, the tie goes to the cycloalkane."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's assign a formal charge to carbon in the methane molecule. Remember that each bond consists of two electrons, so I'm gonna draw in these electrons in these bonds because it's gonna make it easier for us to assign a formal charge to carbon. So to find a formal charge for carbon, the formal charge is equal to the number of valence electrons in the free atom, or the number of valence electrons that carbon is supposed to have. We know that carbon is supposed to have four valence electrons, so I could write a four here, and from that, we subtract the number of valence electrons that carbon actually has in the drawing. Remember when we drew our dot structures, we knew that each bond here came from one valence electron from hydrogen and one valence electron from carbon, so I could give that one valence electron back to hydrogen and one valence electron to carbon, and so we're going to divide up all of our bonds that way, give one valence electron to hydrogen and the other valence electron to carbon because that makes it easier for us to see that carbon has four valence electrons in our drawing. Let me highlight them here, one, two, three, four. We're going to subtract four from four, so four minus four is equal to zero, so carbon has a formal charge of zero in methane."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We know that carbon is supposed to have four valence electrons, so I could write a four here, and from that, we subtract the number of valence electrons that carbon actually has in the drawing. Remember when we drew our dot structures, we knew that each bond here came from one valence electron from hydrogen and one valence electron from carbon, so I could give that one valence electron back to hydrogen and one valence electron to carbon, and so we're going to divide up all of our bonds that way, give one valence electron to hydrogen and the other valence electron to carbon because that makes it easier for us to see that carbon has four valence electrons in our drawing. Let me highlight them here, one, two, three, four. We're going to subtract four from four, so four minus four is equal to zero, so carbon has a formal charge of zero in methane. Let's do another example, this one down here. You can see it's different because this time we have three bonds, so let me draw in the electrons in those bonds, and let's find the formal charge on carbon. The formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that we subtract the number of valence electrons that carbon actually has in our drawing."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We're going to subtract four from four, so four minus four is equal to zero, so carbon has a formal charge of zero in methane. Let's do another example, this one down here. You can see it's different because this time we have three bonds, so let me draw in the electrons in those bonds, and let's find the formal charge on carbon. The formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that we subtract the number of valence electrons that carbon actually has in our drawing. We divide up the electrons in our bonds just like we did before, and we can see that carbon has only three electrons around it this time, so I'll highlight those, one, two, and three. Four minus three is equal to plus one, so carbon has a formal charge of plus one. Carbon's supposed to have four valence electrons."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that we subtract the number of valence electrons that carbon actually has in our drawing. We divide up the electrons in our bonds just like we did before, and we can see that carbon has only three electrons around it this time, so I'll highlight those, one, two, and three. Four minus three is equal to plus one, so carbon has a formal charge of plus one. Carbon's supposed to have four valence electrons. It has only three around it, so it lost one of its electrons, which gives it a formal charge of plus one. Let me go ahead and redraw that. Over here on the right we have carbon with three bonds to hydrogen, and this carbon has a plus one formal charge, so we can represent that by putting a plus charge here next to the carbon."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Carbon's supposed to have four valence electrons. It has only three around it, so it lost one of its electrons, which gives it a formal charge of plus one. Let me go ahead and redraw that. Over here on the right we have carbon with three bonds to hydrogen, and this carbon has a plus one formal charge, so we can represent that by putting a plus charge here next to the carbon. Notice that carbon does not have an octet of electrons around it. It has only six electrons around it, and that's actually okay. Carbon can never exceed an octet, but it's okay for carbon to have less than eight electrons."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Over here on the right we have carbon with three bonds to hydrogen, and this carbon has a plus one formal charge, so we can represent that by putting a plus charge here next to the carbon. Notice that carbon does not have an octet of electrons around it. It has only six electrons around it, and that's actually okay. Carbon can never exceed an octet, but it's okay for carbon to have less than eight electrons. This is a carbon with a plus one formal charge. It's a positively charged carbon. We call those carbocations."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Carbon can never exceed an octet, but it's okay for carbon to have less than eight electrons. This is a carbon with a plus one formal charge. It's a positively charged carbon. We call those carbocations. Let me write that down here. This is a carbocation, and carbocations come up a lot in organic chemistry mechanisms, so it's really important to understand them. Let's think about the pattern that we see here."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We call those carbocations. Let me write that down here. This is a carbocation, and carbocations come up a lot in organic chemistry mechanisms, so it's really important to understand them. Let's think about the pattern that we see here. We have three single bonds around this carbon. Let me go ahead and highlight them here. Here's one, two, and three."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about the pattern that we see here. We have three single bonds around this carbon. Let me go ahead and highlight them here. Here's one, two, and three. We have three single bonds around our carbon, and we have zero lone pairs of electrons around that carbon, so three bonds plus zero lone pairs of electrons will give you a positively charged carbon, will give you a carbocation. What is the hybridization of this positively charged carbon? Well, there's one, two, three sigma bonds and zero lone pairs of electrons."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here's one, two, and three. We have three single bonds around our carbon, and we have zero lone pairs of electrons around that carbon, so three bonds plus zero lone pairs of electrons will give you a positively charged carbon, will give you a carbocation. What is the hybridization of this positively charged carbon? Well, there's one, two, three sigma bonds and zero lone pairs of electrons. From the videos on hybridization, you should know this carbon is sp2 hybridized, and therefore, this carbon will have trigonal planar geometry around it. Again, that's important when you do your organic chemistry mechanisms, so carbocations are extremely important to understand. Let's look at some other examples of carbocations and analyze them a little bit, too."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Well, there's one, two, three sigma bonds and zero lone pairs of electrons. From the videos on hybridization, you should know this carbon is sp2 hybridized, and therefore, this carbon will have trigonal planar geometry around it. Again, that's important when you do your organic chemistry mechanisms, so carbocations are extremely important to understand. Let's look at some other examples of carbocations and analyze them a little bit, too. Let's start with the carbocation on the far left. The carbon with the plus one formal charge is this one in the center here. What is this carbon in red bonded to?"}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at some other examples of carbocations and analyze them a little bit, too. Let's start with the carbocation on the far left. The carbon with the plus one formal charge is this one in the center here. What is this carbon in red bonded to? Well, the carbon in red is bonded to a CH3 group up here, which we call a methyl group in organic chemistry. The carbon in red is bonded to another CH3 group here and another CH3 group here. The carbon in red already has three single bonds and zero lone pairs of electrons, and so the carbon in red is a plus one formal charge."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "What is this carbon in red bonded to? Well, the carbon in red is bonded to a CH3 group up here, which we call a methyl group in organic chemistry. The carbon in red is bonded to another CH3 group here and another CH3 group here. The carbon in red already has three single bonds and zero lone pairs of electrons, and so the carbon in red is a plus one formal charge. Let's look at this carbocation right here, and let's highlight the carbon with the plus one formal charge. It's this one. This carbon in red is bonded to a CH3 group on the left and a CH3 group on the right, so we only have two bonds here."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "The carbon in red already has three single bonds and zero lone pairs of electrons, and so the carbon in red is a plus one formal charge. Let's look at this carbocation right here, and let's highlight the carbon with the plus one formal charge. It's this one. This carbon in red is bonded to a CH3 group on the left and a CH3 group on the right, so we only have two bonds here. We only have two bonds at this point, but we know in order for that carbon in red to have a plus one formal charge, we need three bonds, like the example on the left. The example on the left, we have three bonds here to that carbon. Where is the last bond?"}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "This carbon in red is bonded to a CH3 group on the left and a CH3 group on the right, so we only have two bonds here. We only have two bonds at this point, but we know in order for that carbon in red to have a plus one formal charge, we need three bonds, like the example on the left. The example on the left, we have three bonds here to that carbon. Where is the last bond? The last bond, of course, must be to a hydrogen, so we draw it in here like that, so the carbon in red is bonded to a hydrogen. Usually, you leave off your hydrogens when you make these drawings, but it's important to understand what's actually there. Move on to the last example."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Where is the last bond? The last bond, of course, must be to a hydrogen, so we draw it in here like that, so the carbon in red is bonded to a hydrogen. Usually, you leave off your hydrogens when you make these drawings, but it's important to understand what's actually there. Move on to the last example. This time, the positive one formal charge is on this carbon in red, and that carbon in red is directly bonded to one other carbon, so that's one bond, but we know we need a total of three bonds, so the carbon in red must have two more bonds, and those two other bonds must be to hydrogen, so we draw in there's one bond to hydrogen and there's another bond to hydrogen, so it's important to recognize these patterns. Let's do another formal charge. Let's assign formal charge to another carbon."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Move on to the last example. This time, the positive one formal charge is on this carbon in red, and that carbon in red is directly bonded to one other carbon, so that's one bond, but we know we need a total of three bonds, so the carbon in red must have two more bonds, and those two other bonds must be to hydrogen, so we draw in there's one bond to hydrogen and there's another bond to hydrogen, so it's important to recognize these patterns. Let's do another formal charge. Let's assign formal charge to another carbon. Let's put in our electrons and our bonds. We put those in, and our goal is to find the formal charge on carbon, so the formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that, we subtract the number of valence electrons that carbon actually has in our drawing. We divide up these electrons here and these bonds, and this time, carbon has a lone pair of electrons on it, so how many electrons are around carbon in our drawing?"}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's assign formal charge to another carbon. Let's put in our electrons and our bonds. We put those in, and our goal is to find the formal charge on carbon, so the formal charge on carbon is equal to the number of valence electrons that carbon is supposed to have, which we know is four, and from that, we subtract the number of valence electrons that carbon actually has in our drawing. We divide up these electrons here and these bonds, and this time, carbon has a lone pair of electrons on it, so how many electrons are around carbon in our drawing? This time, there's one, two, three, and then two more from this lone pair, so four and five, so four minus five gives us a formal charge of negative one, so carbon is supposed to have four valence electrons. Here, it has five, so it's like it's gained an extra electron, which gives it a negative one formal charge. Let me go ahead and redraw that."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We divide up these electrons here and these bonds, and this time, carbon has a lone pair of electrons on it, so how many electrons are around carbon in our drawing? This time, there's one, two, three, and then two more from this lone pair, so four and five, so four minus five gives us a formal charge of negative one, so carbon is supposed to have four valence electrons. Here, it has five, so it's like it's gained an extra electron, which gives it a negative one formal charge. Let me go ahead and redraw that. We have carbon with three bonds to hydrogen and one lone pair of electrons on this carbon, a negative one formal charge, so we can represent that here with our negative sign next to that carbon. A carbon with a negative charge is called a carbanion, so this is a carbanion, and let's analyze the pattern that we have for our carbanion. We have one, two, three bonds."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and redraw that. We have carbon with three bonds to hydrogen and one lone pair of electrons on this carbon, a negative one formal charge, so we can represent that here with our negative sign next to that carbon. A carbon with a negative charge is called a carbanion, so this is a carbanion, and let's analyze the pattern that we have for our carbanion. We have one, two, three bonds. Let me write that down. We have three bonds, and this time, we have one lone pair of electrons, so we have one lone pair, so three bonds plus one, three single bonds plus one lone pair of electrons for a carbon will give us a negative one formal charge on that carbon. We will have a carbanion."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have one, two, three bonds. Let me write that down. We have three bonds, and this time, we have one lone pair of electrons, so we have one lone pair, so three bonds plus one, three single bonds plus one lone pair of electrons for a carbon will give us a negative one formal charge on that carbon. We will have a carbanion. These also come up in mechanisms in organic chemistry, so let's analyze some carbanions. Down here, let's start with the carbanion on the left, and the negative one formal charge is on this carbon, which I just marked in red, so we should have three bonds and one lone pair of electrons on that carbon. Let's analyze it."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "We will have a carbanion. These also come up in mechanisms in organic chemistry, so let's analyze some carbanions. Down here, let's start with the carbanion on the left, and the negative one formal charge is on this carbon, which I just marked in red, so we should have three bonds and one lone pair of electrons on that carbon. Let's analyze it. The carbon in red is bonded to a CH3, a CH3, and a CH3, so that takes care of our three bonds. Then, of course, here is the one lone pair of electrons. Let's move on to the next example."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's analyze it. The carbon in red is bonded to a CH3, a CH3, and a CH3, so that takes care of our three bonds. Then, of course, here is the one lone pair of electrons. Let's move on to the next example. The carbon with a negative one formal charge is this carbon that I just marked in red. The carbon in red is directly bonded to a carbon here and directly bonded to a carbon here, so that's two bonds. I need a total of three bonds because I already have a lone pair of electrons."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's move on to the next example. The carbon with a negative one formal charge is this carbon that I just marked in red. The carbon in red is directly bonded to a carbon here and directly bonded to a carbon here, so that's two bonds. I need a total of three bonds because I already have a lone pair of electrons. Here's my lone pair of electrons on that carbon. I need one more bond, and that bond, of course, must be to a hydrogen, so I can draw in a hydrogen here. Again, that hydrogen is usually left off when you're drawing dot structures, but it's important to realize that that hydrogen is actually there."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "I need a total of three bonds because I already have a lone pair of electrons. Here's my lone pair of electrons on that carbon. I need one more bond, and that bond, of course, must be to a hydrogen, so I can draw in a hydrogen here. Again, that hydrogen is usually left off when you're drawing dot structures, but it's important to realize that that hydrogen is actually there. Finally, one more example. The negative one formal charge is on this carbon, and that carbon is directly bonded to one other carbon. It already has a lone pair of electrons, so so far we have one bond and one lone pair."}, {"video_title": "Formal charge on carbon Resonance and acid-base chemistry Organic chemistry Khan Academy.mp3", "Sentence": "Again, that hydrogen is usually left off when you're drawing dot structures, but it's important to realize that that hydrogen is actually there. Finally, one more example. The negative one formal charge is on this carbon, and that carbon is directly bonded to one other carbon. It already has a lone pair of electrons, so so far we have one bond and one lone pair. We need a total of three bonds, so we need two more bonds on that carbon in red. Those last two bonds, of course, must be to two hydrogens. It's important to be able to assign formal charge and to do the math."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Terminal alkynes can function as weak acids if you react them with a very strong base. So something like sodium amide. So this NH2 minus over here came from Na plus NH2 minus, so sodium amide, which is a very strong base. And so if the amide anion functions as a base, a lone pair of electrons in this nitrogen is going to take this proton right here. This is the acidic proton on terminal alkynes. And that leaves these electrons in here to kick off onto your carbon. So if I take NH2 minus and that picks up an H plus, well, that would form NH3."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so if the amide anion functions as a base, a lone pair of electrons in this nitrogen is going to take this proton right here. This is the acidic proton on terminal alkynes. And that leaves these electrons in here to kick off onto your carbon. So if I take NH2 minus and that picks up an H plus, well, that would form NH3. So now I have nitrogen with three hydrogens attached to it and one lone pair of electrons. So when sodium amide functions as a base, it forms ammonia as its product. What is our other product?"}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if I take NH2 minus and that picks up an H plus, well, that would form NH3. So now I have nitrogen with three hydrogens attached to it and one lone pair of electrons. So when sodium amide functions as a base, it forms ammonia as its product. What is our other product? So we had our carbon triple bonded to another carbon with an R group here. And then lone pair of electrons on this carbon. So these electrons right here were the electrons in this bond."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What is our other product? So we had our carbon triple bonded to another carbon with an R group here. And then lone pair of electrons on this carbon. So these electrons right here were the electrons in this bond. So those electrons in there moved onto that carbon. That gives this carbon a negative 1 formal charge like that. So we form a carb anion here, also called an alkanide anion for this portion."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here were the electrons in this bond. So those electrons in there moved onto that carbon. That gives this carbon a negative 1 formal charge like that. So we form a carb anion here, also called an alkanide anion for this portion. So this is an alkanide anion, a carb anion here. It's a relatively stable conjugate base because the electrons, these two electrons here, are housed in an sp hybridized orbital, which has a lot of s character to it. So it's relatively small."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we form a carb anion here, also called an alkanide anion for this portion. So this is an alkanide anion, a carb anion here. It's a relatively stable conjugate base because the electrons, these two electrons here, are housed in an sp hybridized orbital, which has a lot of s character to it. So it's relatively small. So those negatively charged electrons are held a little bit more closely to the positively charged nucleus of this carbon here. So that somewhat stabilizes the conjugate base, which is the reason why a terminal alkyne can function as an acid. So once we've formed our alkanide anion, we can use that alkanide anion to do an alkylation reaction."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's relatively small. So those negatively charged electrons are held a little bit more closely to the positively charged nucleus of this carbon here. So that somewhat stabilizes the conjugate base, which is the reason why a terminal alkyne can function as an acid. So once we've formed our alkanide anion, we can use that alkanide anion to do an alkylation reaction. So let's go ahead and redraw that alkanide anion here. So I'm going to draw this portion, this R group over here. And then we have our carbon triple bonded to another carbon."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So once we've formed our alkanide anion, we can use that alkanide anion to do an alkylation reaction. So let's go ahead and redraw that alkanide anion here. So I'm going to draw this portion, this R group over here. And then we have our carbon triple bonded to another carbon. A negative charge on this carbon on the right here. And this can now function as a nucleophile. So a negatively charged anion can function as a nucleophile."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our carbon triple bonded to another carbon. A negative charge on this carbon on the right here. And this can now function as a nucleophile. So a negatively charged anion can function as a nucleophile. And if we react this alkanide anion with an alkyl halide, let's go ahead and draw an alkyl halide here. So I'm going to put hydrogen there. And I'll put my halogen over here on the right."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So a negatively charged anion can function as a nucleophile. And if we react this alkanide anion with an alkyl halide, let's go ahead and draw an alkyl halide here. So I'm going to put hydrogen there. And I'll put my halogen over here on the right. So put my lone pairs of electrons. Let's draw in one R group right here, and then a hydrogen over here. So here is my alkyl halide."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'll put my halogen over here on the right. So put my lone pairs of electrons. Let's draw in one R group right here, and then a hydrogen over here. So here is my alkyl halide. And if you react a strong nucleophile with an alkyl halide that is not very sterically hindered, this is a primary alkyl halide right here, you're going to get an SN2 reaction. So think about this being an SN2 reaction. We have an alkyl halide, which has a polarized bond between the carbon and the halogen."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here is my alkyl halide. And if you react a strong nucleophile with an alkyl halide that is not very sterically hindered, this is a primary alkyl halide right here, you're going to get an SN2 reaction. So think about this being an SN2 reaction. We have an alkyl halide, which has a polarized bond between the carbon and the halogen. So the halogen is more electronegative. It's going to pull the electrons and the bond between it and carbon closer to itself. So this halogen ends up being partially negative."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have an alkyl halide, which has a polarized bond between the carbon and the halogen. So the halogen is more electronegative. It's going to pull the electrons and the bond between it and carbon closer to itself. So this halogen ends up being partially negative. This carbon, therefore, will be partially positive, like that. So we have an electrophile. This carbon right here is partial positive charge."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this halogen ends up being partially negative. This carbon, therefore, will be partially positive, like that. So we have an electrophile. This carbon right here is partial positive charge. It wants electrons. Of course, our nucleophile has those electrons. So the lone pair of electrons on our carbon can attack our electrophile."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here is partial positive charge. It wants electrons. Of course, our nucleophile has those electrons. So the lone pair of electrons on our carbon can attack our electrophile. So nucleophile attacks electrophile. And an SN2 mechanism, remember, is a concerted mechanism, meaning the nucleophile attacks the electrophile at the same time your leaving group is leaving here. So these electrons are going to kick off onto your halogen."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the lone pair of electrons on our carbon can attack our electrophile. So nucleophile attacks electrophile. And an SN2 mechanism, remember, is a concerted mechanism, meaning the nucleophile attacks the electrophile at the same time your leaving group is leaving here. So these electrons are going to kick off onto your halogen. So let's go ahead and draw the product. So now we would have an R group, carbon triple bonded to another carbon. And now this is bonded to yet another carbon."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons are going to kick off onto your halogen. So let's go ahead and draw the product. So now we would have an R group, carbon triple bonded to another carbon. And now this is bonded to yet another carbon. So we formed a carbon-carbon bond here in this reaction. And then this hydrogen is up here. This R group is still here coming out at us."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now this is bonded to yet another carbon. So we formed a carbon-carbon bond here in this reaction. And then this hydrogen is up here. This R group is still here coming out at us. And then the hydrogen going away from us like that in space. And then we have our halogen over here with now four lone pairs of electrons and negative 1 formal charge. It is stable on its own as an anion like that."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This R group is still here coming out at us. And then the hydrogen going away from us like that in space. And then we have our halogen over here with now four lone pairs of electrons and negative 1 formal charge. It is stable on its own as an anion like that. So this is an alkylation reaction. We put an alkyl group onto our alkyne. So our alkyl group consisted of this carbon and whatever this R group is here."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It is stable on its own as an anion like that. So this is an alkylation reaction. We put an alkyl group onto our alkyne. So our alkyl group consisted of this carbon and whatever this R group is here. And we formed a new carbon-carbon bond. So we alkylated our alkyne. Let's look at an example of an acid-base reaction followed by an alkylation."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So our alkyl group consisted of this carbon and whatever this R group is here. And we formed a new carbon-carbon bond. So we alkylated our alkyne. Let's look at an example of an acid-base reaction followed by an alkylation. So let's start with acetylene, the simplest alkyne. So we have carbon triple bonded to another carbon, and then two hydrogens on either side. And if we react that with sodium amide here."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an example of an acid-base reaction followed by an alkylation. So let's start with acetylene, the simplest alkyne. So we have carbon triple bonded to another carbon, and then two hydrogens on either side. And if we react that with sodium amide here. So we know sodium amide being a strong base. If we use one molar equivalent, it's going to take off one of these acidic protons. So let's say it's the proton on the right here."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And if we react that with sodium amide here. So we know sodium amide being a strong base. If we use one molar equivalent, it's going to take off one of these acidic protons. So let's say it's the proton on the right here. And so we're going to lose that proton. So we're going to leave those two electrons behind on this carbon, making this carbon negatively charged. And then the positively charged sodium ion is going to interact with that negatively charged carbanion like that."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say it's the proton on the right here. And so we're going to lose that proton. So we're going to leave those two electrons behind on this carbon, making this carbon negatively charged. And then the positively charged sodium ion is going to interact with that negatively charged carbanion like that. So that's the first reaction, formation of your alkanide anion. And then if you want to do an alkylation, it's a separate reaction. You take this, and let's react it with ethyl bromide, so CH3CH2Br."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then the positively charged sodium ion is going to interact with that negatively charged carbanion like that. So that's the first reaction, formation of your alkanide anion. And then if you want to do an alkylation, it's a separate reaction. You take this, and let's react it with ethyl bromide, so CH3CH2Br. If you think about what's going to happen, the lone pair of electrons on the carbon is going to attack this carbon, the one that's bonded to your halogen like that. The halogen is going to leave, and you're going to put this alkyl group onto your alkyne. So you're going to end up with an ethyl group on your alkyne."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "You take this, and let's react it with ethyl bromide, so CH3CH2Br. If you think about what's going to happen, the lone pair of electrons on the carbon is going to attack this carbon, the one that's bonded to your halogen like that. The halogen is going to leave, and you're going to put this alkyl group onto your alkyne. So you're going to end up with an ethyl group on your alkyne. So let's go ahead and draw that. So we have hydrogen, and then carbon triple bonded to another carbon. And then we just have to put our alkyl group on there, so a CH2CH3 like that."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you're going to end up with an ethyl group on your alkyne. So let's go ahead and draw that. So we have hydrogen, and then carbon triple bonded to another carbon. And then we just have to put our alkyl group on there, so a CH2CH3 like that. So we've alkylated our alkyne. This is a very useful reaction for organic synthesis. So let's take the molecule we just made, and let's make something else with it."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we just have to put our alkyl group on there, so a CH2CH3 like that. So we've alkylated our alkyne. This is a very useful reaction for organic synthesis. So let's take the molecule we just made, and let's make something else with it. So if I took this, let me go ahead and redraw it over here. So if I took this alkyne, so we just form this. And let's react it for two steps."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's take the molecule we just made, and let's make something else with it. So if I took this, let me go ahead and redraw it over here. So if I took this alkyne, so we just form this. And let's react it for two steps. Let's first react it with our base again. So let's use sodium amide right here. And in our second step, we'll react with a primary alkyl halide."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's react it for two steps. Let's first react it with our base again. So let's use sodium amide right here. And in our second step, we'll react with a primary alkyl halide. So let's go ahead and draw a primary alkyl halide here. So that is our molecule. So we think to ourselves, what happens?"}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And in our second step, we'll react with a primary alkyl halide. So let's go ahead and draw a primary alkyl halide here. So that is our molecule. So we think to ourselves, what happens? I have a strong base. I still have an acidic proton left on my alkyne, so the proton over here on the left. So that's what the base is going to do."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we think to ourselves, what happens? I have a strong base. I still have an acidic proton left on my alkyne, so the proton over here on the left. So that's what the base is going to do. The base is going to take that proton, forming a negatively charged carbanion, an alkanide anion. And then that anion is going to be our nucleophile for an SN2 reaction. So when you're thinking about it, these electrons in here that are going to be on that carbon, giving a negative 1 formal charge, are going to come all the way over here and attack this carbon and attach all of this alkyl group to our carbon."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that's what the base is going to do. The base is going to take that proton, forming a negatively charged carbanion, an alkanide anion. And then that anion is going to be our nucleophile for an SN2 reaction. So when you're thinking about it, these electrons in here that are going to be on that carbon, giving a negative 1 formal charge, are going to come all the way over here and attack this carbon and attach all of this alkyl group to our carbon. So let's go ahead and draw the products of that. We're going to have our benzene ring. So let's go ahead and draw our benzene ring here."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when you're thinking about it, these electrons in here that are going to be on that carbon, giving a negative 1 formal charge, are going to come all the way over here and attack this carbon and attach all of this alkyl group to our carbon. So let's go ahead and draw the products of that. We're going to have our benzene ring. So let's go ahead and draw our benzene ring here. So let's put in our electrons going around my benzene ring. And then on that benzene ring is a CH2. So that CH2, the CH2 is the red one that we marked right here."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw our benzene ring here. So let's put in our electrons going around my benzene ring. And then on that benzene ring is a CH2. So that CH2, the CH2 is the red one that we marked right here. And this is the alkyl group that gets put onto your alkyne. So let's just go ahead and finish drawing our alkyne here. So we have now our triple bond, carbon triple bonded to another carbon."}, {"video_title": "Alkyne acidity and alkylation Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that CH2, the CH2 is the red one that we marked right here. And this is the alkyl group that gets put onto your alkyne. So let's just go ahead and finish drawing our alkyne here. So we have now our triple bond, carbon triple bonded to another carbon. And then our ethyl group. So CH2, CH3. So you'll see in later videos how we use the acidity of a terminal alkynes to alkylate when we do a few different synthesis problems."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're all familiar with what a meter looks like. The average adult male is a little under 2 meters. If you were to divide a meter into a thousand units, you would get a millimeter. I think we probably know what a millimeter is if you've ever looked at a meter stick. It's the smallest measurement on that meter stick. So it's already pretty hard to look at. If you were to divide each of those millimeters into a thousand sections, you'd get a micrometer."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I think we probably know what a millimeter is if you've ever looked at a meter stick. It's the smallest measurement on that meter stick. So it's already pretty hard to look at. If you were to divide each of those millimeters into a thousand sections, you'd get a micrometer. Or another way to think about a micrometer is it's one millionth of a meter. So this is kind of beyond what we're capable of really perceiving. If you were to take each of those micrometers and divide them into a thousand sections, you would get a nanometer."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to divide each of those millimeters into a thousand sections, you'd get a micrometer. Or another way to think about a micrometer is it's one millionth of a meter. So this is kind of beyond what we're capable of really perceiving. If you were to take each of those micrometers and divide them into a thousand sections, you would get a nanometer. So now we're at one billionth of a meter. You divide that by a thousand. You get a picometer."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you were to take each of those micrometers and divide them into a thousand sections, you would get a nanometer. So now we're at one billionth of a meter. You divide that by a thousand. You get a picometer. So a picometer is one thousand billionth of a meter. Or you could say a trillionth of a meter. You divide one of those by a thousand and you would get a femtometer."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You get a picometer. So a picometer is one thousand billionth of a meter. Or you could say a trillionth of a meter. You divide one of those by a thousand and you would get a femtometer. So these are unimaginably small things. Now once you're familiar with the units, let's explore what types of things we can expect to find at these different scales. I'll start over here, and I've written them on the left as well, but it's more compelling when you see the pictures."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You divide one of those by a thousand and you would get a femtometer. So these are unimaginably small things. Now once you're familiar with the units, let's explore what types of things we can expect to find at these different scales. I'll start over here, and I've written them on the left as well, but it's more compelling when you see the pictures. We'll start over here with the b. I've arbitrarily picked something of this scale. There's many, many, many, almost an infinite number of things I could have picked at this scale. But the average b is about two centimeters long."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll start over here, and I've written them on the left as well, but it's more compelling when you see the pictures. We'll start over here with the b. I've arbitrarily picked something of this scale. There's many, many, many, almost an infinite number of things I could have picked at this scale. But the average b is about two centimeters long. This b right over here. It's about, give or take, one hundredth the length of the average adult human being. But once again, a honey bee, not too exciting, although it is pretty exciting to see it zoomed in like this."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the average b is about two centimeters long. This b right over here. It's about, give or take, one hundredth the length of the average adult human being. But once again, a honey bee, not too exciting, although it is pretty exciting to see it zoomed in like this. But a honey bee is something that we can relate to. We've all seen honey bees. Now, what I want to do is zoom in or look at something that's fifty times smaller than a honey bee."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But once again, a honey bee, not too exciting, although it is pretty exciting to see it zoomed in like this. But a honey bee is something that we can relate to. We've all seen honey bees. Now, what I want to do is zoom in or look at something that's fifty times smaller than a honey bee. So something that if I were to kind of show how big it is relative to this honey bee, it would look something like this. I'm doing it very rough. And that is a dust mite."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, what I want to do is zoom in or look at something that's fifty times smaller than a honey bee. So something that if I were to kind of show how big it is relative to this honey bee, it would look something like this. I'm doing it very rough. And that is a dust mite. And this right here, these are both pictures of dust mites. Now, dust mites look like these strange and alien creatures. But what's amazing about them is that they are everywhere."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that is a dust mite. And this right here, these are both pictures of dust mites. Now, dust mites look like these strange and alien creatures. But what's amazing about them is that they are everywhere. They're all around us. You probably have many of them lying on your skin or wherever right now, which is kind of a creepy idea. But we're talking about scale here."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's amazing about them is that they are everywhere. They're all around us. You probably have many of them lying on your skin or wherever right now, which is kind of a creepy idea. But we're talking about scale here. And the average dust mite, so we were talking about centimeters before. Now we'll talk about millimeters. The average dust mite is less than half of a millimeter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we're talking about scale here. And the average dust mite, so we were talking about centimeters before. Now we'll talk about millimeters. The average dust mite is less than half of a millimeter. Or if you want to talk in micrometers, it's about 400 micrometers long. So this length right over here is about 400 micrometers, so about 150th the length. Remember, this huge thing that I'm showing right here, this is a honey bee."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The average dust mite is less than half of a millimeter. Or if you want to talk in micrometers, it's about 400 micrometers long. So this length right over here is about 400 micrometers, so about 150th the length. Remember, this huge thing that I'm showing right here, this is a honey bee. It's about 150th the length of a honey bee. Or maybe to put it in other terms that you might be familiar with, this is a zoomed in picture of human hair. And you might say, oh my god, this person has horrible hair."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, this huge thing that I'm showing right here, this is a honey bee. It's about 150th the length of a honey bee. Or maybe to put it in other terms that you might be familiar with, this is a zoomed in picture of human hair. And you might say, oh my god, this person has horrible hair. But no, if you were to look at your own hair under an electron microscope, you'd be lucky if it looked this good. This person actually I've seen pictures of more damaged hair than this. This is probably smooth and silky hair right here."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you might say, oh my god, this person has horrible hair. But no, if you were to look at your own hair under an electron microscope, you'd be lucky if it looked this good. This person actually I've seen pictures of more damaged hair than this. This is probably smooth and silky hair right here. But the diameter of human hair, and this is on average. It depends on whose hair you're talking about. The diameter of human hair is about 100 micrometers thick."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is probably smooth and silky hair right here. But the diameter of human hair, and this is on average. It depends on whose hair you're talking about. The diameter of human hair is about 100 micrometers thick. That's the diameter. So it's about a fourth the length of a dust mite. Or if I were to draw some human hair relative to this honey bee, it would look something like this."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The diameter of human hair is about 100 micrometers thick. That's the diameter. So it's about a fourth the length of a dust mite. Or if I were to draw some human hair relative to this honey bee, it would look something like this. And I'm drawing the whole hair, so its width would be the width of this thing that I just drew. Remember, we're looking at a honey bee here. It looks like some type of giant, but it is a honey bee."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or if I were to draw some human hair relative to this honey bee, it would look something like this. And I'm drawing the whole hair, so its width would be the width of this thing that I just drew. Remember, we're looking at a honey bee here. It looks like some type of giant, but it is a honey bee. Let's zoom in even more. So we started with the honey bee. We zoomed in by 50 to get the dust mite."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It looks like some type of giant, but it is a honey bee. Let's zoom in even more. So we started with the honey bee. We zoomed in by 50 to get the dust mite. We zoomed in by another factor of 4 to get the width of human hair. If we zoom in, we're in the micrometer range now. If we zoom in by roughly another factor of 10, we get to the scale of cells."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We zoomed in by 50 to get the dust mite. We zoomed in by another factor of 4 to get the width of human hair. If we zoom in, we're in the micrometer range now. If we zoom in by roughly another factor of 10, we get to the scale of cells. And this right here is a red blood cell. I think this is a white blood cell right over here. About 6 to 8 micrometers."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we zoom in by roughly another factor of 10, we get to the scale of cells. And this right here is a red blood cell. I think this is a white blood cell right over here. About 6 to 8 micrometers. So once again, if I were to draw a cell relative to this human hair, it would probably look something like this. Something on a similar scale that we can still kind of relate to is the width of spider silk. It's about 3 to 8 micrometers."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "About 6 to 8 micrometers. So once again, if I were to draw a cell relative to this human hair, it would probably look something like this. Something on a similar scale that we can still kind of relate to is the width of spider silk. It's about 3 to 8 micrometers. So if I were to draw some spider silk on the same diagram, it would look something like this. This is an actual image of spider silk. So once again, something that we can kind of perceive."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's about 3 to 8 micrometers. So if I were to draw some spider silk on the same diagram, it would look something like this. This is an actual image of spider silk. So once again, something that we can kind of perceive. You can bump into it. You can touch spider silk. You can see it if the sun is reflecting just right or if it has a little bit of moisture on it."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So once again, something that we can kind of perceive. You can bump into it. You can touch spider silk. You can see it if the sun is reflecting just right or if it has a little bit of moisture on it. But it's about the thinnest thing that humans can perceive. And this is in the ones of micrometer range. At that same range, you start to have some of your larger bacteria."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can see it if the sun is reflecting just right or if it has a little bit of moisture on it. But it's about the thinnest thing that humans can perceive. And this is in the ones of micrometer range. At that same range, you start to have some of your larger bacteria. Bacteria can be anywhere from, and I'm speaking very roughly, 1 to 10 micrometers. So in general, they're smaller than cells. Most bacteria are smaller than most cells."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "At that same range, you start to have some of your larger bacteria. Bacteria can be anywhere from, and I'm speaking very roughly, 1 to 10 micrometers. So in general, they're smaller than cells. Most bacteria are smaller than most cells. And just to figure out where we sit on our scale, have it over here. So we started off, I want to keep reminding ourselves, humans, you divide by 100, you get to the B. So each of these slashes right here are dividing by 10."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Most bacteria are smaller than most cells. And just to figure out where we sit on our scale, have it over here. So we started off, I want to keep reminding ourselves, humans, you divide by 100, you get to the B. So each of these slashes right here are dividing by 10. So this is divide by 10. Divide by 10 again, you're divided in size by 100. Divide by 10 again, you get to millimeter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So each of these slashes right here are dividing by 10. So this is divide by 10. Divide by 10 again, you're divided in size by 100. Divide by 10 again, you get to millimeter. You've divided by 1,000. Divide by 10 again, you are doing tenths of millimeters, which is about the size of the human hair. You divide again by 10, you're going into tens of micrometers."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Divide by 10 again, you get to millimeter. You've divided by 1,000. Divide by 10 again, you are doing tenths of millimeters, which is about the size of the human hair. You divide again by 10, you're going into tens of micrometers. By 10 again, you get into the micrometer range. So now we're talking about cells, we're talking about bacteria. Now things are going to get really crazy."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You divide again by 10, you're going into tens of micrometers. By 10 again, you get into the micrometer range. So now we're talking about cells, we're talking about bacteria. Now things are going to get really crazy. This was in the ones of micrometer range. Now we're going to start getting into the hundreds of nanometer range. Just to get a sense of things."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now things are going to get really crazy. This was in the ones of micrometer range. Now we're going to start getting into the hundreds of nanometer range. Just to get a sense of things. So remember, a nanometer is a thousandth of a micrometer. Or 100 nanometers would be a tenth of a micrometer. And this picture right here, this big, enormous planet or asteroid looking thing, this is a white blood cell."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just to get a sense of things. So remember, a nanometer is a thousandth of a micrometer. Or 100 nanometers would be a tenth of a micrometer. And this picture right here, this big, enormous planet or asteroid looking thing, this is a white blood cell. The enormous blue thing in this picture. And so if I were to zoom out, it might look something like this right over here. But what's really fascinating about this picture for multiple reasons are these little green things that are emerging, that are essentially reproducing, emerging from the surface of this white blood cell."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this picture right here, this big, enormous planet or asteroid looking thing, this is a white blood cell. The enormous blue thing in this picture. And so if I were to zoom out, it might look something like this right over here. But what's really fascinating about this picture for multiple reasons are these little green things that are emerging, that are essentially reproducing, emerging from the surface of this white blood cell. And these things right here, these are AIDS viruses. So now if we zoom in, roughly another factor of about 100 to 1000 from the size of a cell, you're now getting to the size of a virus. And all of the genetic material necessary to replicate that virus is right inside each of these little capsids, right inside each of these little green containers."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's really fascinating about this picture for multiple reasons are these little green things that are emerging, that are essentially reproducing, emerging from the surface of this white blood cell. And these things right here, these are AIDS viruses. So now if we zoom in, roughly another factor of about 100 to 1000 from the size of a cell, you're now getting to the size of a virus. And all of the genetic material necessary to replicate that virus is right inside each of these little capsids, right inside each of these little green containers. So now going back to our scale, we are down to the scale of a virus. So we're in the hundreds of nanometer range. If we divide by 10 and then divide by 10, you get to the nanometer range."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And all of the genetic material necessary to replicate that virus is right inside each of these little capsids, right inside each of these little green containers. So now going back to our scale, we are down to the scale of a virus. So we're in the hundreds of nanometer range. If we divide by 10 and then divide by 10, you get to the nanometer range. And right in the ones of nanometer range, you get to the width of the double helix of a DNA molecule. So this right here is, if you were to zoom in, and this is an artist's depiction of it, obviously this is not a picture, so to speak, of a DNA molecule. But the width of this double helix is about 2 nanometers, or another way to think about it, about 160th the diameter of one of these viral capsids."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we divide by 10 and then divide by 10, you get to the nanometer range. And right in the ones of nanometer range, you get to the width of the double helix of a DNA molecule. So this right here is, if you were to zoom in, and this is an artist's depiction of it, obviously this is not a picture, so to speak, of a DNA molecule. But the width of this double helix is about 2 nanometers, or another way to think about it, about 160th the diameter of one of these viral capsids. It would have to be, because it's going to have to get all wound up and fit into one of these viral capsids. And DNA, just to make it clear, this is just the width of DNA. It's much, much, much, much, much, much longer, and we can talk about that in future videos."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the width of this double helix is about 2 nanometers, or another way to think about it, about 160th the diameter of one of these viral capsids. It would have to be, because it's going to have to get all wound up and fit into one of these viral capsids. And DNA, just to make it clear, this is just the width of DNA. It's much, much, much, much, much, much longer, and we can talk about that in future videos. So once again, we're at a very, very small scale. If you want to think of it in terms of meters, we're at 2 billionths of a meter. You could put 500 million of these side by side to get to a meter."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's much, much, much, much, much, much longer, and we can talk about that in future videos. So once again, we're at a very, very small scale. If you want to think of it in terms of meters, we're at 2 billionths of a meter. You could put 500 million of these side by side to get to a meter. Or you could even think of it this way. This is 2 millionths of a millimeter. So once again, super small."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You could put 500 million of these side by side to get to a meter. Or you could even think of it this way. This is 2 millionths of a millimeter. So once again, super small. You could put these side by side, one DNA, another DNA. If you made them touch, you could put 500,000 next to each other in a millimeter. So this is an unbelievably small amount of space."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So once again, super small. You could put these side by side, one DNA, another DNA. If you made them touch, you could put 500,000 next to each other in a millimeter. So this is an unbelievably small amount of space. It's going to fit into another unit that's not kind of in the conventional prefix followed by meters, and this is an angstrom. And 10 angstroms equal 1 nanometer. So the width of this DNA double helix would be 2 nanometers or 20 angstroms."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is an unbelievably small amount of space. It's going to fit into another unit that's not kind of in the conventional prefix followed by meters, and this is an angstrom. And 10 angstroms equal 1 nanometer. So the width of this DNA double helix would be 2 nanometers or 20 angstroms. Now, if we were to divide again by 10, you get to something that's 2 angstroms or 0.2 nanometers wide, and that is a water molecule. Maybe instead of using red, I should have used blue or something. But this right here is the oxygen, and it is bonded to the two hydrogens right over here."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the width of this DNA double helix would be 2 nanometers or 20 angstroms. Now, if we were to divide again by 10, you get to something that's 2 angstroms or 0.2 nanometers wide, and that is a water molecule. Maybe instead of using red, I should have used blue or something. But this right here is the oxygen, and it is bonded to the two hydrogens right over here. So this is beyond, frankly, human perception, or even really stuff that we can conceptualize, not to even speak of perception. We're still imagining how small we're dealing with right over here. Remember, we're dealing with less than a fifth of a billionth of a meter, or a fifth of a millionth of a millimeter, something that I really can't fathom."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this right here is the oxygen, and it is bonded to the two hydrogens right over here. So this is beyond, frankly, human perception, or even really stuff that we can conceptualize, not to even speak of perception. We're still imagining how small we're dealing with right over here. Remember, we're dealing with less than a fifth of a billionth of a meter, or a fifth of a millionth of a millimeter, something that I really can't fathom. But we're going to get even smaller than that. If we were to zoom in on one of these hydrogen atoms, and now things start to get kind of abstract, and we start dealing in the quantum realm, and it's hard to define where one thing ends and one thing begins, and what is real and what is not real, and all of that silliness. But if we try our best to do it, if we were to zoom in and we were to put some boundary on a hydrogen atom, because the electrons actually could jump around anywhere, but if we set some boundary of where the electrons are most likely to be found, the diameter of a hydrogen atom is roughly one angstrom, which makes sense from this diagram, too."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, we're dealing with less than a fifth of a billionth of a meter, or a fifth of a millionth of a millimeter, something that I really can't fathom. But we're going to get even smaller than that. If we were to zoom in on one of these hydrogen atoms, and now things start to get kind of abstract, and we start dealing in the quantum realm, and it's hard to define where one thing ends and one thing begins, and what is real and what is not real, and all of that silliness. But if we try our best to do it, if we were to zoom in and we were to put some boundary on a hydrogen atom, because the electrons actually could jump around anywhere, but if we set some boundary of where the electrons are most likely to be found, the diameter of a hydrogen atom is roughly one angstrom, which makes sense from this diagram, too. It's about half of the diameter of this water molecule. What's extra crazy is, one, this atom is super, super, duper small, something that we can't, you know, this is one ten billionth of a meter, or one ten millionth of a millimeter, so something we really, really can't fathom. But what's crazier than that is that it's mostly free space."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if we try our best to do it, if we were to zoom in and we were to put some boundary on a hydrogen atom, because the electrons actually could jump around anywhere, but if we set some boundary of where the electrons are most likely to be found, the diameter of a hydrogen atom is roughly one angstrom, which makes sense from this diagram, too. It's about half of the diameter of this water molecule. What's extra crazy is, one, this atom is super, super, duper small, something that we can't, you know, this is one ten billionth of a meter, or one ten millionth of a millimeter, so something we really, really can't fathom. But what's crazier than that is that it's mostly free space. We've gotten this small. We're trying to get to these fundamental units, and this thing right here is mostly free space. And that's because if you look at an electron, and when we say radius here, it's really hard to define where it starts and ends, and you have to do some things related to the charge, and we're not even thinking about quantum effects and all of that."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's crazier than that is that it's mostly free space. We've gotten this small. We're trying to get to these fundamental units, and this thing right here is mostly free space. And that's because if you look at an electron, and when we say radius here, it's really hard to define where it starts and ends, and you have to do some things related to the charge, and we're not even thinking about quantum effects and all of that. An electron has a radius of 3 times 10 to the negative 5th angstroms, and the nucleus of a hydrogen atom, which is really just a proton, has a radius a little bit, and, you know, don't even worry about this number right here. The general idea is it's the same order of magnitude. It's about one ten thousandth of an angstrom."}, {"video_title": "Scale of the small Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that's because if you look at an electron, and when we say radius here, it's really hard to define where it starts and ends, and you have to do some things related to the charge, and we're not even thinking about quantum effects and all of that. An electron has a radius of 3 times 10 to the negative 5th angstroms, and the nucleus of a hydrogen atom, which is really just a proton, has a radius a little bit, and, you know, don't even worry about this number right here. The general idea is it's the same order of magnitude. It's about one ten thousandth of an angstrom. And just to give a sense of what it's like, if you view the entire atomic radius to be about an angstrom, kind of just have a conception for scale of the atom and how much free space there is in an atom, if we even want to think what is free space. Imagine a nucleus being maybe a marble at the center of a football stadium, of a domed football stadium, and imagine an electron being a honeybee just randomly jumping around random parts of that entire volume inside of that football stadium. And obviously it's a quantum honeybee, so it can jump around from spot to spot, and it's not easy to predict where it's going to go next and all of the rest."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "We've already seen the mechanism for this acid-base reaction in the video on organic acid-base mechanisms, but I want to run through it really quickly again. So on the left we have acetic acid, which is going to function as our acid, and then we have hydroxide over here, which will be our base. Our base is going to take the acidic proton on acetic acid, leaving these electrons behind on the oxygen to give us the acetate anion here. And if we protonate OH minus, then we would form H2O. So those would be our two products. Let's think about the reverse reaction. So the acetate anion would function as a base and take this proton on water, leaving these electrons behind."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "And if we protonate OH minus, then we would form H2O. So those would be our two products. Let's think about the reverse reaction. So the acetate anion would function as a base and take this proton on water, leaving these electrons behind. That would give us back acetic acid and hydroxide. So for the reverse reaction, the acetate anion here is functioning as a base, and water is functioning as an acid. So it's donating a proton."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So the acetate anion would function as a base and take this proton on water, leaving these electrons behind. That would give us back acetic acid and hydroxide. So for the reverse reaction, the acetate anion here is functioning as a base, and water is functioning as an acid. So it's donating a proton. What if our goal was to find the equilibrium constant for the forward reaction? So what is the equilibrium constant for the forward reaction? Which would be, the stuff on the left would be the reactants, and the stuff on the right would of course be the products."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So it's donating a proton. What if our goal was to find the equilibrium constant for the forward reaction? So what is the equilibrium constant for the forward reaction? Which would be, the stuff on the left would be the reactants, and the stuff on the right would of course be the products. So we could figure that out just using pKa values. So if we know the pKa values for the two acids in our reaction, we can figure out the equilibrium constant for that reaction. So we need to know the pKa of the acid on the left."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "Which would be, the stuff on the left would be the reactants, and the stuff on the right would of course be the products. So we could figure that out just using pKa values. So if we know the pKa values for the two acids in our reaction, we can figure out the equilibrium constant for that reaction. So we need to know the pKa of the acid on the left. So we already know that acetic acid is the acid on the left side here. And acetic acid has a pKa, this proton right here has a pKa of approximately five. On the right, what's functioning as an acid?"}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So we need to know the pKa of the acid on the left. So we already know that acetic acid is the acid on the left side here. And acetic acid has a pKa, this proton right here has a pKa of approximately five. On the right, what's functioning as an acid? That's of course water. So what is the pKa of this proton right here on water? That's approximately 16."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "On the right, what's functioning as an acid? That's of course water. So what is the pKa of this proton right here on water? That's approximately 16. So let's plug that in to our equation. So the pKeq for the forward reaction is equal to the pKa of the acid on the left, which would be approximately five, minus the pKa of the acid on the right, which is approximately 16. So five minus 16 gives us a pKeq equal to negative 11."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "That's approximately 16. So let's plug that in to our equation. So the pKeq for the forward reaction is equal to the pKa of the acid on the left, which would be approximately five, minus the pKa of the acid on the right, which is approximately 16. So five minus 16 gives us a pKeq equal to negative 11. So how do we go from the pKeq to the Keq? Well, we can do that because we know from general chemistry, pKeq is equal to the negative log of Keq. So to solve for Keq, first we need to put the negative sign on the left, so we have negative pKeq is equal to the log of Keq."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So five minus 16 gives us a pKeq equal to negative 11. So how do we go from the pKeq to the Keq? Well, we can do that because we know from general chemistry, pKeq is equal to the negative log of Keq. So to solve for Keq, first we need to put the negative sign on the left, so we have negative pKeq is equal to the log of Keq. And how do I get rid of that log? I have to take 10 to both sides. If I take 10 to both sides, that gets rid of our log, so we know that Keq is equal to 10 to the negative pKeq."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So to solve for Keq, first we need to put the negative sign on the left, so we have negative pKeq is equal to the log of Keq. And how do I get rid of that log? I have to take 10 to both sides. If I take 10 to both sides, that gets rid of our log, so we know that Keq is equal to 10 to the negative pKeq. So we can just take our answer here for pKeq, and we just need to plug that in to our equation here. So we get that the Keq, so let me go ahead and put that right here, Keq is equal to 10 to the negative, so that negative sign that I just wrote here is this negative sign, 10 to the negative pKeq, which was negative 11. So 10 to the negative, negative 11, which of course is 10 to the 11th."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "If I take 10 to both sides, that gets rid of our log, so we know that Keq is equal to 10 to the negative pKeq. So we can just take our answer here for pKeq, and we just need to plug that in to our equation here. So we get that the Keq, so let me go ahead and put that right here, Keq is equal to 10 to the negative, so that negative sign that I just wrote here is this negative sign, 10 to the negative pKeq, which was negative 11. So 10 to the negative, negative 11, which of course is 10 to the 11th. So the equilibrium constant for the forward reaction is equal to 10 to the 11th. And we know what that means from general chemistry. We know that when K is much greater than one like this, at equilibrium we have way more products than we do reactants."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So 10 to the negative, negative 11, which of course is 10 to the 11th. So the equilibrium constant for the forward reaction is equal to 10 to the 11th. And we know what that means from general chemistry. We know that when K is much greater than one like this, at equilibrium we have way more products than we do reactants. So the equilibrium lies to the right. The equilibrium lies to the right, and we have a large amount of our products compared to our reactants. If you wanted to do that a faster way, if you just wanted to figure out which direction, which direction the equilibrium lies, look at your pKa values, let's go back up here, and you can see on the left it's five, and on the right it's 16."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "We know that when K is much greater than one like this, at equilibrium we have way more products than we do reactants. So the equilibrium lies to the right. The equilibrium lies to the right, and we have a large amount of our products compared to our reactants. If you wanted to do that a faster way, if you just wanted to figure out which direction, which direction the equilibrium lies, look at your pKa values, let's go back up here, and you can see on the left it's five, and on the right it's 16. We figured out that the equilibrium lies to the right, so therefore the equilibrium lies to the side that has the acid with the higher pKa value. So the equilibrium favors the weaker acid. So that's the short way of figuring out the position of equilibrium using pKa values."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "If you wanted to do that a faster way, if you just wanted to figure out which direction, which direction the equilibrium lies, look at your pKa values, let's go back up here, and you can see on the left it's five, and on the right it's 16. We figured out that the equilibrium lies to the right, so therefore the equilibrium lies to the side that has the acid with the higher pKa value. So the equilibrium favors the weaker acid. So that's the short way of figuring out the position of equilibrium using pKa values. Here's another organic acid-base mechanism that we've seen before. Acetone on the left functions as a base and takes a proton from H3O+, which is hydronium, leaving these electrons behind on the oxygen. So hydronium functions as an acid."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So that's the short way of figuring out the position of equilibrium using pKa values. Here's another organic acid-base mechanism that we've seen before. Acetone on the left functions as a base and takes a proton from H3O+, which is hydronium, leaving these electrons behind on the oxygen. So hydronium functions as an acid. If you protonate acetone, you would get this. So this must be the conjugate acid to acetone. And if you take away an H+, from hydronium, you are left with water."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So hydronium functions as an acid. If you protonate acetone, you would get this. So this must be the conjugate acid to acetone. And if you take away an H+, from hydronium, you are left with water. So water must be the conjugate base to H3O+. So for the reverse reaction, if water functions as a base, water's going to take this proton, leaving these electrons behind on the oxygen, giving us back acetone and forming hydronium, H3O+. To use our pKa values to predict the position of equilibrium, we need to find the pKa for the acid on the left, and from that we subtract the pKa for the acid on the right."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "And if you take away an H+, from hydronium, you are left with water. So water must be the conjugate base to H3O+. So for the reverse reaction, if water functions as a base, water's going to take this proton, leaving these electrons behind on the oxygen, giving us back acetone and forming hydronium, H3O+. To use our pKa values to predict the position of equilibrium, we need to find the pKa for the acid on the left, and from that we subtract the pKa for the acid on the right. The acid on the left is hydronium, and hydronium has a pKa of approximately negative two. The acid on the right is right here, and the pKa for this proton is approximately negative three. So to find the pKeq for the forward reaction, for the forward reaction, meaning the stuff on the left is the reactants and the stuff on the right is the products, the pKeq is equal to the pKa of the acid on the left, which would be negative two, minus the pKa of the acid on the right, which is negative three."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "To use our pKa values to predict the position of equilibrium, we need to find the pKa for the acid on the left, and from that we subtract the pKa for the acid on the right. The acid on the left is hydronium, and hydronium has a pKa of approximately negative two. The acid on the right is right here, and the pKa for this proton is approximately negative three. So to find the pKeq for the forward reaction, for the forward reaction, meaning the stuff on the left is the reactants and the stuff on the right is the products, the pKeq is equal to the pKa of the acid on the left, which would be negative two, minus the pKa of the acid on the right, which is negative three. So negative two minus negative three is equal to plus one. So the pKeq for the forward reaction is plus one. We know how to find the Keq because from the previous example, we saw that the Keq is equal to 10 to the negative pKeq."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "So to find the pKeq for the forward reaction, for the forward reaction, meaning the stuff on the left is the reactants and the stuff on the right is the products, the pKeq is equal to the pKa of the acid on the left, which would be negative two, minus the pKa of the acid on the right, which is negative three. So negative two minus negative three is equal to plus one. So the pKeq for the forward reaction is plus one. We know how to find the Keq because from the previous example, we saw that the Keq is equal to 10 to the negative pKeq. So we plug in our pKeq, which is one, into here, and so we get the Keq for the forward reaction is equal to 10 to the negative first. So K is less than one, and we know what it means when K is less than one. That means at equilibrium, we have more reactants than products."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "We know how to find the Keq because from the previous example, we saw that the Keq is equal to 10 to the negative pKeq. So we plug in our pKeq, which is one, into here, and so we get the Keq for the forward reaction is equal to 10 to the negative first. So K is less than one, and we know what it means when K is less than one. That means at equilibrium, we have more reactants than products. So the equilibrium lies to the left, and we have more reactants than we do products at equilibrium. So we could also do this using the shorter way. To figure out the position of equilibrium, we could look at our pKa values and say, all right, on the left we have negative two, on the right we have negative three, and we know that the equilibrium favors the acid with the higher pKa value, favors the formation of the acid with the higher pKa value."}, {"video_title": "Using pKa values to predict the position of equilibrium Organic chemistry Khan Academy.mp3", "Sentence": "That means at equilibrium, we have more reactants than products. So the equilibrium lies to the left, and we have more reactants than we do products at equilibrium. So we could also do this using the shorter way. To figure out the position of equilibrium, we could look at our pKa values and say, all right, on the left we have negative two, on the right we have negative three, and we know that the equilibrium favors the acid with the higher pKa value, favors the formation of the acid with the higher pKa value. It's a little bit tricky because we have two negative values for our pKa, but negative two is closer to zero than negative three. So negative two is the higher pKa, and our equilibrium favors the formation of the acid with the higher pKa. Our equilibrium favors the formation of the weaker acid, so the equilibrium lies to the left."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "Let's start with polar protic solvents. A polar protic solvent is a solvent that has at least one hydrogen connected to an electronegative atom. For example, if we look at water here, we can see we have a hydrogen directly connected to an electronegative atom, which is oxygen. So water is an example of a polar protic solvent. Next, we have methanol, which again has a hydrogen directly connected to an electronegative atom, an oxygen. And finally, acetic acid, which has the same thing here. Here's our hydrogen and here is our oxygen."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So water is an example of a polar protic solvent. Next, we have methanol, which again has a hydrogen directly connected to an electronegative atom, an oxygen. And finally, acetic acid, which has the same thing here. Here's our hydrogen and here is our oxygen. So these polar protic solvents favor an SN1 mechanism. So let me write that in here. So an SN1 mechanism is favored by a polar protic solvent."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "Here's our hydrogen and here is our oxygen. So these polar protic solvents favor an SN1 mechanism. So let me write that in here. So an SN1 mechanism is favored by a polar protic solvent. And let's look at why. So down here I have tert-butyl bromide, and for an SN1 mechanism, the first step here would be loss of a leaving group. So these electrons come off onto the bromine to form our bromide anion, and we're gonna form a carbocation as well."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So an SN1 mechanism is favored by a polar protic solvent. And let's look at why. So down here I have tert-butyl bromide, and for an SN1 mechanism, the first step here would be loss of a leaving group. So these electrons come off onto the bromine to form our bromide anion, and we're gonna form a carbocation as well. So let me draw in the carbocation first. So we have a carbon that is bonded to three methyl groups, and this is a planar carbocation, so I'm trying to show that. Our carbon has a plus one formal charge."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So these electrons come off onto the bromine to form our bromide anion, and we're gonna form a carbocation as well. So let me draw in the carbocation first. So we have a carbon that is bonded to three methyl groups, and this is a planar carbocation, so I'm trying to show that. Our carbon has a plus one formal charge. And we're also gonna have our bromine here, which we have three lone pairs of electrons. I'll put those in. And then we're gonna get one more lone pair of electrons on the bromine that came from this bond in here."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "Our carbon has a plus one formal charge. And we're also gonna have our bromine here, which we have three lone pairs of electrons. I'll put those in. And then we're gonna get one more lone pair of electrons on the bromine that came from this bond in here. So highlighting those electrons in magenta. Here are those electrons in magenta, and bromine has a negative one formal charge. It is the bromide anion."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "And then we're gonna get one more lone pair of electrons on the bromine that came from this bond in here. So highlighting those electrons in magenta. Here are those electrons in magenta, and bromine has a negative one formal charge. It is the bromide anion. So we have this carbocation and this anion in our SN1 mechanism, and we know this is the rate determining step of our SN1 mechanism, the loss of a leaving group. If we're using a polar protic solvent, such as water, water can stabilize both the cation and the anion. For example, for our carbocation, we know that carbon has a positive charge on it."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "It is the bromide anion. So we have this carbocation and this anion in our SN1 mechanism, and we know this is the rate determining step of our SN1 mechanism, the loss of a leaving group. If we're using a polar protic solvent, such as water, water can stabilize both the cation and the anion. For example, for our carbocation, we know that carbon has a positive charge on it. And if we look at water, we know that this oxygen here is a partial negative charge, since oxygen's more electronegative than hydrogen, and this hydrogen would have a partial positive charge. So the negative portion of this molecule, the oxygen, would interact with this positive charge on our carbocation. So let's go ahead and show a water molecule here."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "For example, for our carbocation, we know that carbon has a positive charge on it. And if we look at water, we know that this oxygen here is a partial negative charge, since oxygen's more electronegative than hydrogen, and this hydrogen would have a partial positive charge. So the negative portion of this molecule, the oxygen, would interact with this positive charge on our carbocation. So let's go ahead and show a water molecule here. And the partially negative oxygen, with these two lone pairs of electrons here on the oxygen, will help to stabilize our carbocation. And for our negative anion, for our bromide anion here, which is negatively charged, it'd be the other end of the water molecule. So if I draw in my water molecule right here, so two lone pairs of electrons on the oxygen, our partial positive hydrogens would interact and help to stabilize that anion."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So let's go ahead and show a water molecule here. And the partially negative oxygen, with these two lone pairs of electrons here on the oxygen, will help to stabilize our carbocation. And for our negative anion, for our bromide anion here, which is negatively charged, it'd be the other end of the water molecule. So if I draw in my water molecule right here, so two lone pairs of electrons on the oxygen, our partial positive hydrogens would interact and help to stabilize that anion. So polar protic solvents help to stabilize both the carbocation and the anion, and that solvation of both cations and anions helps the SN1 mechanism proceed. So that's why a polar protic solvent will favor an SN1 mechanism. Now let's look at polar aprotic solvents."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So if I draw in my water molecule right here, so two lone pairs of electrons on the oxygen, our partial positive hydrogens would interact and help to stabilize that anion. So polar protic solvents help to stabilize both the carbocation and the anion, and that solvation of both cations and anions helps the SN1 mechanism proceed. So that's why a polar protic solvent will favor an SN1 mechanism. Now let's look at polar aprotic solvents. So first let's look at dimethyl sulfoxide, so more commonly known as DMSO. So here's the D, M, S, and O. Oxygen is more electronegative than sulfur, so the oxygen's going to withdraw some electron density and become partially negative, and the sulfur would be partially positive."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "Now let's look at polar aprotic solvents. So first let's look at dimethyl sulfoxide, so more commonly known as DMSO. So here's the D, M, S, and O. Oxygen is more electronegative than sulfur, so the oxygen's going to withdraw some electron density and become partially negative, and the sulfur would be partially positive. A polar aprotic solvent does not have a hydrogen directly connected to an electronegative atom. So if we think about the hydrogens on DMSO, so let me just sketch them in here real fast, there'd be three on this carbon, and there'd be three on this carbon. So here we have hydrogens directly connected to a carbon, and of course carbon is not very electronegative."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "Oxygen is more electronegative than sulfur, so the oxygen's going to withdraw some electron density and become partially negative, and the sulfur would be partially positive. A polar aprotic solvent does not have a hydrogen directly connected to an electronegative atom. So if we think about the hydrogens on DMSO, so let me just sketch them in here real fast, there'd be three on this carbon, and there'd be three on this carbon. So here we have hydrogens directly connected to a carbon, and of course carbon is not very electronegative. So that's why this is a polar aprotic solvent. Next let's look at DMF. So DMF is the short way of writing this one here."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So here we have hydrogens directly connected to a carbon, and of course carbon is not very electronegative. So that's why this is a polar aprotic solvent. Next let's look at DMF. So DMF is the short way of writing this one here. Again, no hydrogen directly connected to an electronegative atom. This hydrogen is directly connected to this carbon, and then this carbon would have three hydrogens on it, and then this carbon would have three hydrogens on it. So DMF is a polar aprotic solvent."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So DMF is the short way of writing this one here. Again, no hydrogen directly connected to an electronegative atom. This hydrogen is directly connected to this carbon, and then this carbon would have three hydrogens on it, and then this carbon would have three hydrogens on it. So DMF is a polar aprotic solvent. And finally, let's look at this last one here. So the abbreviation would be HMPA. So let me write that down here."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So DMF is a polar aprotic solvent. And finally, let's look at this last one here. So the abbreviation would be HMPA. So let me write that down here. So HMPA. Again, no hydrogens directly connected to an electronegative atom. Polar aprotic solvents favor an SN2 mechanism."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So let me write that down here. So HMPA. Again, no hydrogens directly connected to an electronegative atom. Polar aprotic solvents favor an SN2 mechanism. So let's look at why. Down here I have an SN2 reaction. On the left we have this alkyl halide."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "Polar aprotic solvents favor an SN2 mechanism. So let's look at why. Down here I have an SN2 reaction. On the left we have this alkyl halide. Let's say we have sodium hydroxide. We could use DMSO as our solvent, so let me write that in here. So we're gonna use DMSO."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "On the left we have this alkyl halide. Let's say we have sodium hydroxide. We could use DMSO as our solvent, so let me write that in here. So we're gonna use DMSO. And we know in an SN2 mechanism, the nucleophile attacks our alkyl halide at the same time our leaving group leaves. So our nucleophile is the hydroxide ion that's going to attack this carbon, and these electrons are gonna come off onto the bromine to form our bromide anion. So our OH replaces our bromine, and we can see that over here in our product."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So we're gonna use DMSO. And we know in an SN2 mechanism, the nucleophile attacks our alkyl halide at the same time our leaving group leaves. So our nucleophile is the hydroxide ion that's going to attack this carbon, and these electrons are gonna come off onto the bromine to form our bromide anion. So our OH replaces our bromine, and we can see that over here in our product. In an SN2 mechanism, we need a strong nucleophile to attack our alkyl halide. And DMSO is gonna help us increase the effectiveness of our nucleophile, which is our hydroxide ion. So let's look at some pictures of how it helps us."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So our OH replaces our bromine, and we can see that over here in our product. In an SN2 mechanism, we need a strong nucleophile to attack our alkyl halide. And DMSO is gonna help us increase the effectiveness of our nucleophile, which is our hydroxide ion. So let's look at some pictures of how it helps us. So we have sodium hydroxide here. So first let's focus in on the sodium, our cation. So here's the sodium cation."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So let's look at some pictures of how it helps us. So we have sodium hydroxide here. So first let's focus in on the sodium, our cation. So here's the sodium cation. DMSO is a good solvator of cations, and that's because the oxygen has a partial negative charge, the sulfur has a partial positive charge, and these lone pairs of electrons on the oxygen help to stabilize the positive charge on our sodium. So same thing over here, partial negative, partial positive, and again, we are able to solvate our cation. So the fact that a polar aprotic solvent is a good solvator of a cation means we can separate this ion from our nucleophile, and that increases the effectiveness of the hydroxide ion."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So here's the sodium cation. DMSO is a good solvator of cations, and that's because the oxygen has a partial negative charge, the sulfur has a partial positive charge, and these lone pairs of electrons on the oxygen help to stabilize the positive charge on our sodium. So same thing over here, partial negative, partial positive, and again, we are able to solvate our cation. So the fact that a polar aprotic solvent is a good solvator of a cation means we can separate this ion from our nucleophile, and that increases the effectiveness of the hydroxide ion. The hydroxide ion itself is not solvated by a polar aprotic solvent. So you might think, okay, well, if the oxygen is partially negative and the sulfur is partially positive, the partially positive sulfur could interact with our negatively charged nucleophile. But remember, we have these bulky methyl groups here, and because of steric hindrance, that prevents our hydroxide ion from interacting with DMSO."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "So the fact that a polar aprotic solvent is a good solvator of a cation means we can separate this ion from our nucleophile, and that increases the effectiveness of the hydroxide ion. The hydroxide ion itself is not solvated by a polar aprotic solvent. So you might think, okay, well, if the oxygen is partially negative and the sulfur is partially positive, the partially positive sulfur could interact with our negatively charged nucleophile. But remember, we have these bulky methyl groups here, and because of steric hindrance, that prevents our hydroxide ion from interacting with DMSO. So the hydroxide ion is all by itself, which of course increases its effectiveness as a nucleophile. It's better able to attack the alkyl halide. If we had used something like water, we know that water is a polar protic solvent, with the oxygen being partially negative and the hydrogens being partially positive."}, {"video_title": "Sn1 and Sn2 solvent.mp3", "Sentence": "But remember, we have these bulky methyl groups here, and because of steric hindrance, that prevents our hydroxide ion from interacting with DMSO. So the hydroxide ion is all by itself, which of course increases its effectiveness as a nucleophile. It's better able to attack the alkyl halide. If we had used something like water, we know that water is a polar protic solvent, with the oxygen being partially negative and the hydrogens being partially positive. And a polar protic solvent would interact with our nucleophile, solvating it, and essentially decreasing the effectiveness of our nucleophile. So that's why polar protic solvents don't work as well if you want an SN2 mechanism. The polar aprotic solvent increases the effectiveness of our nucleophile, therefore favoring our SN2 mechanism."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And primary alkyl halides work the best. We'll talk about why in a minute. And what happens is you end up putting the R prime group from your alkyl halide onto what used to be your alcohol to form your ether, like that. So let's look at the mechanism for the Williamson ether synthesis, where you start with your alcohol. We know that alcohols can function as weak acids. So if you react an alcohol with a strong base, something like sodium hydride, so we know that the hydride portion of the molecule is going to function as a strong base. This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at the mechanism for the Williamson ether synthesis, where you start with your alcohol. We know that alcohols can function as weak acids. So if you react an alcohol with a strong base, something like sodium hydride, so we know that the hydride portion of the molecule is going to function as a strong base. This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen. So if we're drawing the product of that acid-base reaction, we now have an oxygen with three lone pairs of electrons around it, giving it a negative 1 formal charge. And we call that an alkoxide anion, which would interact with the positively charged sodium ion floating around. So there's some electrostatic or ionic interaction between those opposite charges."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen. So if we're drawing the product of that acid-base reaction, we now have an oxygen with three lone pairs of electrons around it, giving it a negative 1 formal charge. And we call that an alkoxide anion, which would interact with the positively charged sodium ion floating around. So there's some electrostatic or ionic interaction between those opposite charges. And here's where you introduce your alkyl halides. So if we draw our alkyl halide, it would look like this. And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative and our carbon is going to be partially positive."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So there's some electrostatic or ionic interaction between those opposite charges. And here's where you introduce your alkyl halides. So if we draw our alkyl halide, it would look like this. And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative and our carbon is going to be partially positive. Partially positive carbon means that that carbon wants electrons. It's going to function as an electrophile in the next step of the mechanism. And a lone pair of electrons in the oxygen is going to function as a nucleophile."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative and our carbon is going to be partially positive. Partially positive carbon means that that carbon wants electrons. It's going to function as an electrophile in the next step of the mechanism. And a lone pair of electrons in the oxygen is going to function as a nucleophile. So opposite charges attract. Lone pair of electrons on our nucleophile are going to attack our electrophile, our carbon. At the same time, the electrons in the bond between the carbon and the halogen are going to kick off onto the halogen like that."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And a lone pair of electrons in the oxygen is going to function as a nucleophile. So opposite charges attract. Lone pair of electrons on our nucleophile are going to attack our electrophile, our carbon. At the same time, the electrons in the bond between the carbon and the halogen are going to kick off onto the halogen like that. So this is an SN2 type mechanism. So this is an SN2 type mechanism, which is why a primary alkyl halide will work the best, because that has the decreased steric hindrance compared to other alkyl halides. So what will happen is, after the nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "At the same time, the electrons in the bond between the carbon and the halogen are going to kick off onto the halogen like that. So this is an SN2 type mechanism. So this is an SN2 type mechanism, which is why a primary alkyl halide will work the best, because that has the decreased steric hindrance compared to other alkyl halides. So what will happen is, after the nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether. So if we wanted to, we could just rewrite our ether like this to show it as we added on an R prime group like that. Let's look at an example of the Williamson ether synthesis. So if I start with a molecule over here on the left, and it's kind of an interesting looking molecule."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So what will happen is, after the nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether. So if we wanted to, we could just rewrite our ether like this to show it as we added on an R prime group like that. Let's look at an example of the Williamson ether synthesis. So if I start with a molecule over here on the left, and it's kind of an interesting looking molecule. It's called beta-naphthol. And so beta-naphthol has two rings together like this. And then there's an OH coming off of one of the rings like that."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So if I start with a molecule over here on the left, and it's kind of an interesting looking molecule. It's called beta-naphthol. And so beta-naphthol has two rings together like this. And then there's an OH coming off of one of the rings like that. So that's beta-naphthol. And in the first part, we're going to add potassium hydroxide. So in the first step, potassium hydroxide as our base."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And then there's an OH coming off of one of the rings like that. So that's beta-naphthol. And in the first part, we're going to add potassium hydroxide. So in the first step, potassium hydroxide as our base. Now, potassium hydroxide is not as strong of a base as sodium hydride is. But in this case, it's OK to use a little bit weaker base. So the lone pair of electrons on the hydroxide, we're going to take that proton, leaving these electrons behind on the oxygen."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So in the first step, potassium hydroxide as our base. Now, potassium hydroxide is not as strong of a base as sodium hydride is. But in this case, it's OK to use a little bit weaker base. So the lone pair of electrons on the hydroxide, we're going to take that proton, leaving these electrons behind on the oxygen. So when we draw the conjugate base to beta-naphthol, we can go ahead and show that. We're going to take off that proton, which is going to leave that oxygen there with three lone pairs of electrons. I'm giving it a negative 1 formal charge."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So the lone pair of electrons on the hydroxide, we're going to take that proton, leaving these electrons behind on the oxygen. So when we draw the conjugate base to beta-naphthol, we can go ahead and show that. We're going to take off that proton, which is going to leave that oxygen there with three lone pairs of electrons. I'm giving it a negative 1 formal charge. So this is our alkoxide anion. And this alkoxide anion is resonance stabilized. So a resonance stabilized conjugate base stabilizes the conjugate base, which makes beta-naphthol a little bit better acid than other alcohols that we will talk about."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "I'm giving it a negative 1 formal charge. So this is our alkoxide anion. And this alkoxide anion is resonance stabilized. So a resonance stabilized conjugate base stabilizes the conjugate base, which makes beta-naphthol a little bit better acid than other alcohols that we will talk about. So since beta-naphthol is a little bit more acidic, that's why it's OK for us to use a weaker base for this example. So potassium hydroxide is strong enough to take away the acidic proton on beta-naphthol because the conjugate base to beta-naphthol is resonance stabilized. So in the second step, once we have formed our alkoxide anion, this is where we add our alkyl halide."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So a resonance stabilized conjugate base stabilizes the conjugate base, which makes beta-naphthol a little bit better acid than other alcohols that we will talk about. So since beta-naphthol is a little bit more acidic, that's why it's OK for us to use a weaker base for this example. So potassium hydroxide is strong enough to take away the acidic proton on beta-naphthol because the conjugate base to beta-naphthol is resonance stabilized. So in the second step, once we have formed our alkoxide anion, this is where we add our alkyl halide. So if I add my alkyl halide in my second step, let's see if we can have enough room here. I'm going to use methyl iodide as our alkyl halide. So methyl iodine looks like that."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So in the second step, once we have formed our alkoxide anion, this is where we add our alkyl halide. So if I add my alkyl halide in my second step, let's see if we can have enough room here. I'm going to use methyl iodide as our alkyl halide. So methyl iodine looks like that. And once again, we know this carbon is going to be the electrophilic carbon. So nucleophile, electrophile. So a lone pair of electrons on the oxygen attacks the carbon, kicks these electrons off onto the iodine, and we form our product."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So methyl iodine looks like that. And once again, we know this carbon is going to be the electrophilic carbon. So nucleophile, electrophile. So a lone pair of electrons on the oxygen attacks the carbon, kicks these electrons off onto the iodine, and we form our product. So let's go ahead and draw the ether product that will result. So these rings are going to stay the same like that. And we now are going to have our oxygen attached to a methyl group, which came from the methyl iodine like that."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons on the oxygen attacks the carbon, kicks these electrons off onto the iodine, and we form our product. So let's go ahead and draw the ether product that will result. So these rings are going to stay the same like that. And we now are going to have our oxygen attached to a methyl group, which came from the methyl iodine like that. So we formed our product. This product is called Neuralin, which is a fixative used in perfume. So this has an interesting smell to it."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And we now are going to have our oxygen attached to a methyl group, which came from the methyl iodine like that. So we formed our product. This product is called Neuralin, which is a fixative used in perfume. So this has an interesting smell to it. So if you ever get a chance to do this Williamson ether synthesis, it's just interesting to see what Neuralin smells like, what it looks like, and to think about it as being a component of some perfumes. Let's think about synthesizing an ether. So if you were given a problem where the question says something like, OK, here is the ether that you want to synthesize."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So this has an interesting smell to it. So if you ever get a chance to do this Williamson ether synthesis, it's just interesting to see what Neuralin smells like, what it looks like, and to think about it as being a component of some perfumes. Let's think about synthesizing an ether. So if you were given a problem where the question says something like, OK, here is the ether that you want to synthesize. What would you need in order to do so? So you need to think about, OK, there's my ether. And I'm going to make it from some other things over here."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "So if you were given a problem where the question says something like, OK, here is the ether that you want to synthesize. What would you need in order to do so? So you need to think about, OK, there's my ether. And I'm going to make it from some other things over here. And if I analyze the alkyl groups attached to my ether, I have a methyl group over here and this would be like a cyclohexyl group over here. And one of those two groups I'm going to use for my alkyl halides. You want to use the group that's the least sterically hindered since it's an SN2 type mechanism."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to make it from some other things over here. And if I analyze the alkyl groups attached to my ether, I have a methyl group over here and this would be like a cyclohexyl group over here. And one of those two groups I'm going to use for my alkyl halides. You want to use the group that's the least sterically hindered since it's an SN2 type mechanism. So you want to go with the methyl group. So in your second step, you would need to add something like methyl iodide. That's the least sterically hindered."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "You want to use the group that's the least sterically hindered since it's an SN2 type mechanism. So you want to go with the methyl group. So in your second step, you would need to add something like methyl iodide. That's the least sterically hindered. So that's going to improve your yield on this reaction. So that's the second step. And in the first step, you'd have to add a strong base."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "That's the least sterically hindered. So that's going to improve your yield on this reaction. So that's the second step. And in the first step, you'd have to add a strong base. So we'll use a sodium hydride here. And your alcohol, therefore, must come from this. So this must be where your alcohol comes from."}, {"video_title": "Williamson ether synthesis Organic chemistry Khan Academy.mp3", "Sentence": "And in the first step, you'd have to add a strong base. So we'll use a sodium hydride here. And your alcohol, therefore, must come from this. So this must be where your alcohol comes from. So if I'm going to show my starting alcohol, it would have to look like this. So if I add that alcohol in the first step, sodium hydride, I take off that proton, form an alkoxide. That alkoxide nucleophilic attacks the methyl iodide to add the methyl group on and to form the ether on the right."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now that we know all of our reactions, let's see if we can put those reactions together to synthesize some simple organic compounds. And so our goal is to make this molecule from benzene. And one approach that you can use is the concept of retrosynthesis. So you try to think backwards. And you think to yourself, what can be an immediate precursor to this molecule? Well, to do that, we have to analyze the groups that are attached to our ring. And so if I look at this bromine up here, I know this bromine is an ortho para director because I know it has lone pairs of electrons around it."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you try to think backwards. And you think to yourself, what can be an immediate precursor to this molecule? Well, to do that, we have to analyze the groups that are attached to our ring. And so if I look at this bromine up here, I know this bromine is an ortho para director because I know it has lone pairs of electrons around it. So this bromine is an ortho para director. If I look at the nitro group, I know this is a meta director. I know it's meta because there's a plus 1 formal charge on that nitrogen."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so if I look at this bromine up here, I know this bromine is an ortho para director because I know it has lone pairs of electrons around it. So this bromine is an ortho para director. If I look at the nitro group, I know this is a meta director. I know it's meta because there's a plus 1 formal charge on that nitrogen. And then over here for this acyl group attached to our ring, I know this is also a meta director because this carbonyl carbon right here has partially positive like that. And so when we try to figure out which of these groups was added last, it makes sense that the bromine was added last because this bromine right here is meta to both our nitro group and our acyl group. And so we can go ahead and draw the precursor."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I know it's meta because there's a plus 1 formal charge on that nitrogen. And then over here for this acyl group attached to our ring, I know this is also a meta director because this carbonyl carbon right here has partially positive like that. And so when we try to figure out which of these groups was added last, it makes sense that the bromine was added last because this bromine right here is meta to both our nitro group and our acyl group. And so we can go ahead and draw the precursor. We go ahead and just take the bromine off. So we're left with a benzene ring. And we have an acyl group on our ring."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we can go ahead and draw the precursor. We go ahead and just take the bromine off. So we're left with a benzene ring. And we have an acyl group on our ring. And we also have a nitro group. So next, we just have to remember how to put a bromine on a benzene ring. And of course, it's a bromination reaction."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have an acyl group on our ring. And we also have a nitro group. So next, we just have to remember how to put a bromine on a benzene ring. And of course, it's a bromination reaction. So you would need some bromine and a catalyst, so something like iron bromide. So FeBr3 will work for that. Let's see if we can figure out the next precursor here."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, it's a bromination reaction. So you would need some bromine and a catalyst, so something like iron bromide. So FeBr3 will work for that. Let's see if we can figure out the next precursor here. So what could we do to make this molecule? Well, once again, we have two groups on here. We have a nitro group and we have an acyl group."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's see if we can figure out the next precursor here. So what could we do to make this molecule? Well, once again, we have two groups on here. We have a nitro group and we have an acyl group. So we could do a nitration to put the nitro group on. And we could do a Friedel-Crafts acylation to put this acyl group on our ring. So the question is, which one of these comes first?"}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We have a nitro group and we have an acyl group. So we could do a nitration to put the nitro group on. And we could do a Friedel-Crafts acylation to put this acyl group on our ring. So the question is, which one of these comes first? And it turns out that you can't really do a Friedel-Crafts alkylation or acylation with a moderate or strongly deactivating group already on your ring. And this nitro group here is strongly deactivating, which means we can't put the nitro group on first and then add our acyl group. The acyl group must come on before the nitro group, which means in this step, we're going to put on the nitro group."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the question is, which one of these comes first? And it turns out that you can't really do a Friedel-Crafts alkylation or acylation with a moderate or strongly deactivating group already on your ring. And this nitro group here is strongly deactivating, which means we can't put the nitro group on first and then add our acyl group. The acyl group must come on before the nitro group, which means in this step, we're going to put on the nitro group. So the immediate precursor to this molecule, we just take off our nitro group and we're left with our benzene ring and an acyl group attached to our benzene ring like that. And so we need to do a nitration, which requires, of course, concentrated nitric acid and also concentrated sulfuric acid like that. Now all we have to do is go from benzene to this molecule."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The acyl group must come on before the nitro group, which means in this step, we're going to put on the nitro group. So the immediate precursor to this molecule, we just take off our nitro group and we're left with our benzene ring and an acyl group attached to our benzene ring like that. And so we need to do a nitration, which requires, of course, concentrated nitric acid and also concentrated sulfuric acid like that. Now all we have to do is go from benzene to this molecule. And we know how to do that, of course. That's a Friedel-Crafts acylation reaction. And so when we think about what kind of acyl chloride we're going to use, just count the number of carbons here."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now all we have to do is go from benzene to this molecule. And we know how to do that, of course. That's a Friedel-Crafts acylation reaction. And so when we think about what kind of acyl chloride we're going to use, just count the number of carbons here. So one and then two. So we need a two-carbon acyl chloride. So we're going to draw here a two-carbon acyl chloride like that."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so when we think about what kind of acyl chloride we're going to use, just count the number of carbons here. So one and then two. So we need a two-carbon acyl chloride. So we're going to draw here a two-carbon acyl chloride like that. And then we need a catalyst, something like aluminum chloride will work for our catalyst. So our synthesis is complete. We'd start with a Friedel-Crafts acylation."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to draw here a two-carbon acyl chloride like that. And then we need a catalyst, something like aluminum chloride will work for our catalyst. So our synthesis is complete. We'd start with a Friedel-Crafts acylation. And the acyl group is a metadirector, which would direct the nitro group to the meta position. And then finally, we have two metadirectors, which we now brominate, which would direct the bromine to the final position. And we are complete."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We'd start with a Friedel-Crafts acylation. And the acyl group is a metadirector, which would direct the nitro group to the meta position. And then finally, we have two metadirectors, which we now brominate, which would direct the bromine to the final position. And we are complete. Let's do another problem here. And actually, it's the exact same groups that we just saw in the previous problem. But this target molecule looks a little bit different."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we are complete. Let's do another problem here. And actually, it's the exact same groups that we just saw in the previous problem. But this target molecule looks a little bit different. And so we're going to need to do the reactions that we did in the previous synthesis in a different order here. So once again, let's start by analyzing the groups. So once again, we know that this bromine is an ortho para director because of the lone pairs of electrons on it."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But this target molecule looks a little bit different. And so we're going to need to do the reactions that we did in the previous synthesis in a different order here. So once again, let's start by analyzing the groups. So once again, we know that this bromine is an ortho para director because of the lone pairs of electrons on it. We know the nitro group is a meta director because of the plus 1 formal charge. And our acyl group is a meta director because of the partial positive charge on our carbonyl carbon right here. And so when we look at those groups and we think about which of those reactions was done last, it makes sense that this nitration was done last."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we know that this bromine is an ortho para director because of the lone pairs of electrons on it. We know the nitro group is a meta director because of the plus 1 formal charge. And our acyl group is a meta director because of the partial positive charge on our carbonyl carbon right here. And so when we look at those groups and we think about which of those reactions was done last, it makes sense that this nitration was done last. And that's because this nitro group is meta to our acyl group because our acyl group is a meta director. And our bromine, more importantly, is an ortho para director. And of course, the nitro group is ortho to the bromine."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so when we look at those groups and we think about which of those reactions was done last, it makes sense that this nitration was done last. And that's because this nitro group is meta to our acyl group because our acyl group is a meta director. And our bromine, more importantly, is an ortho para director. And of course, the nitro group is ortho to the bromine. And so it makes sense the last reaction was a nitration reaction. So to draw the precursor to this, all we do is take off that nitro group. And we would have our benzene ring like this."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And of course, the nitro group is ortho to the bromine. And so it makes sense the last reaction was a nitration reaction. So to draw the precursor to this, all we do is take off that nitro group. And we would have our benzene ring like this. And we would have a bromine on our ring. And we would already have our acyl group on our ring like that. So our last reaction was a nitration reaction."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we would have our benzene ring like this. And we would have a bromine on our ring. And we would already have our acyl group on our ring like that. So our last reaction was a nitration reaction. So we need to add, once again, nitric acid and concentrated nitric acid and concentrated sulfuric acid for our nitration. So when we think about the precursor to this molecule, so once again, we have an ortho para director on our ring. And we have a meta director on our ring."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So our last reaction was a nitration reaction. So we need to add, once again, nitric acid and concentrated nitric acid and concentrated sulfuric acid for our nitration. So when we think about the precursor to this molecule, so once again, we have an ortho para director on our ring. And we have a meta director on our ring. And we have our groups, our bromine and our acyl group are para to each other, which means that the ortho para director directed the acyl group to the para position as the major product. And so we can think about doing a Friedel-Crafts acylation reaction here. So that means that we're taking off the acyl group."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have a meta director on our ring. And we have our groups, our bromine and our acyl group are para to each other, which means that the ortho para director directed the acyl group to the para position as the major product. And so we can think about doing a Friedel-Crafts acylation reaction here. So that means that we're taking off the acyl group. So we're left with bromobenzene to start with over here like that. So we have bromobenzene. And we're doing a Friedel-Crafts acylation."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that means that we're taking off the acyl group. So we're left with bromobenzene to start with over here like that. So we have bromobenzene. And we're doing a Friedel-Crafts acylation. And once again, we need two carbons on our acyl group. So let me just point that out, one and two. So we need a two-carbon acyl chloride."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're doing a Friedel-Crafts acylation. And once again, we need two carbons on our acyl group. So let me just point that out, one and two. So we need a two-carbon acyl chloride. So I go ahead and put on a two-carbon acyl chloride like that. Once again, our catalyst, something like aluminum chloride will work. And you might think to yourself that I know that the halogen, the bromine, is deactivating."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we need a two-carbon acyl chloride. So I go ahead and put on a two-carbon acyl chloride like that. Once again, our catalyst, something like aluminum chloride will work. And you might think to yourself that I know that the halogen, the bromine, is deactivating. So how can we do a Friedel-Crafts acylation with a deactivating group on there, even though it's an ortho para director? Well, remember, it's only weakly deactivating. And so you can't do an alkylation or acylation with a moderate or strongly deactivating group."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you might think to yourself that I know that the halogen, the bromine, is deactivating. So how can we do a Friedel-Crafts acylation with a deactivating group on there, even though it's an ortho para director? Well, remember, it's only weakly deactivating. And so you can't do an alkylation or acylation with a moderate or strongly deactivating group. And so it turns out, since this is weakly deactivating, you can still do this. And you'll get the para product right as your major product over here. So I'm sure you get a little bit of ortho as well."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so you can't do an alkylation or acylation with a moderate or strongly deactivating group. And so it turns out, since this is weakly deactivating, you can still do this. And you'll get the para product right as your major product over here. So I'm sure you get a little bit of ortho as well. So now all we have to do is go from benzene to bromobenzene. And of course, that's really simple. It's just a bromination reaction again."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I'm sure you get a little bit of ortho as well. So now all we have to do is go from benzene to bromobenzene. And of course, that's really simple. It's just a bromination reaction again. So we have our bromine, and then we have our catalyst, and then our synthesis is complete. So for this time, we start out with a bromination reaction to form bromobenzene. The bromine is an ortho para director."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It's just a bromination reaction again. So we have our bromine, and then we have our catalyst, and then our synthesis is complete. So for this time, we start out with a bromination reaction to form bromobenzene. The bromine is an ortho para director. This is an ortho para director. And so it's going to put this acyl group on our ring in the para position as our major product here. And then, of course, we nitrate it."}, {"video_title": "Synthesis of substituted benzene rings I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The bromine is an ortho para director. This is an ortho para director. And so it's going to put this acyl group on our ring in the para position as our major product here. And then, of course, we nitrate it. And we have an ortho para director and a meta director, which means the nitro group will end up in this position. And we're done. So that's how to think about synthesis problems."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "In the previous video, we started with the molecular formula C3H8O, and we looked at one of the possible Lewis dot structures that you can draw that has that molecular formula. From this Lewis dot structure, we looked at other ways to represent the same molecule. However, we didn't have time to talk about bond line structures, so let's start this video by taking this Lewis dot structure and turning it into a bond line structure. If you look at the drawing on the left, it implies that these three carbons are in a perfectly straight line. But the drawing on the right does a little bit better job of showing what the molecule looks like in reality. Those carbons are not in a perfectly straight line. You can see there's a bend to them like that."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If you look at the drawing on the left, it implies that these three carbons are in a perfectly straight line. But the drawing on the right does a little bit better job of showing what the molecule looks like in reality. Those carbons are not in a perfectly straight line. You can see there's a bend to them like that. So when you're drawing a bond line structure and you have a carbon chain, you want to show that carbon chain in a zigzag pattern. Next, let's think about the carbon-hydrogen bonds. If you were to draw every carbon-hydrogen bond in an organic chemistry class, it would take you forever."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "You can see there's a bend to them like that. So when you're drawing a bond line structure and you have a carbon chain, you want to show that carbon chain in a zigzag pattern. Next, let's think about the carbon-hydrogen bonds. If you were to draw every carbon-hydrogen bond in an organic chemistry class, it would take you forever. So we leave those out in bond line structures. Carbon is still bonded to these hydrogens, but we're going to ignore them for our bond line structure just to simplify things. So I'm gonna draw this around the carbon-hydrogen bonds, so we're going to ignore them for the time being."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "If you were to draw every carbon-hydrogen bond in an organic chemistry class, it would take you forever. So we leave those out in bond line structures. Carbon is still bonded to these hydrogens, but we're going to ignore them for our bond line structure just to simplify things. So I'm gonna draw this around the carbon-hydrogen bonds, so we're going to ignore them for the time being. And now we have our three carbons drawn like that. And this carbon is bonded to an oxygen, and this oxygen is bonded to a hydrogen. I'll put in lone pairs of electrons on that oxygen."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I'm gonna draw this around the carbon-hydrogen bonds, so we're going to ignore them for the time being. And now we have our three carbons drawn like that. And this carbon is bonded to an oxygen, and this oxygen is bonded to a hydrogen. I'll put in lone pairs of electrons on that oxygen. Next, we can simplify this even further. We can leave out those carbons. So the carbons are still there, I'm just talking about not drawing these C's in here because it can get kind of confusing."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I'll put in lone pairs of electrons on that oxygen. Next, we can simplify this even further. We can leave out those carbons. So the carbons are still there, I'm just talking about not drawing these C's in here because it can get kind of confusing. So we take out those C's, and I'll leave off the lone pairs of electrons on the oxygen. And we have our bond line structure. So this is our bond line structure."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the carbons are still there, I'm just talking about not drawing these C's in here because it can get kind of confusing. So we take out those C's, and I'll leave off the lone pairs of electrons on the oxygen. And we have our bond line structure. So this is our bond line structure. It contains the same information as our Lewis dot structure does, but it's obviously much easier to draw. It takes less time. It's gonna help you out throughout your course when you're looking at chemical reactions."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this is our bond line structure. It contains the same information as our Lewis dot structure does, but it's obviously much easier to draw. It takes less time. It's gonna help you out throughout your course when you're looking at chemical reactions. So let's focus in on some carbons here. So this carbon in red, that's this carbon. So the carbon's still there, we're just not drawing in the C. And let's look at our other carbons."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It's gonna help you out throughout your course when you're looking at chemical reactions. So let's focus in on some carbons here. So this carbon in red, that's this carbon. So the carbon's still there, we're just not drawing in the C. And let's look at our other carbons. So the carbon on the right is the one in magenta, so that's this carbon right here. And the carbon on the left is in blue, so that's this carbon. The carbon in blue is still bonded to three hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the carbon's still there, we're just not drawing in the C. And let's look at our other carbons. So the carbon on the right is the one in magenta, so that's this carbon right here. And the carbon on the left is in blue, so that's this carbon. The carbon in blue is still bonded to three hydrogens. This carbon in blue is still bonded to three hydrogens. We just leave them off in our bond line structure. The carbon on the right is still bonded to three hydrogens, but again, we leave those off when we're drawing a bond line structure."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The carbon in blue is still bonded to three hydrogens. This carbon in blue is still bonded to three hydrogens. We just leave them off in our bond line structure. The carbon on the right is still bonded to three hydrogens, but again, we leave those off when we're drawing a bond line structure. And the carbon in the middle, this red carbon here, is bonded to an OH. That's already shown in our bond line structure, and it's bonded to one more hydrogen. So there still is a hydrogen bonded to that carbon."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the right is still bonded to three hydrogens, but again, we leave those off when we're drawing a bond line structure. And the carbon in the middle, this red carbon here, is bonded to an OH. That's already shown in our bond line structure, and it's bonded to one more hydrogen. So there still is a hydrogen bonded to that carbon. We just leave them off to make things easier to see. So let me go ahead and erase what I just did here. So let's take off those hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So there still is a hydrogen bonded to that carbon. We just leave them off to make things easier to see. So let me go ahead and erase what I just did here. So let's take off those hydrogens. So those hydrogens are still there. We just know that they are there. So this just takes some practice to figure out what these bond line structures mean."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let's take off those hydrogens. So those hydrogens are still there. We just know that they are there. So this just takes some practice to figure out what these bond line structures mean. So let's do several examples of understanding bond line structures and the information that they contain. Let's start by analyzing this bond line structure. And we'll start with this carbon right here in magenta."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this just takes some practice to figure out what these bond line structures mean. So let's do several examples of understanding bond line structures and the information that they contain. Let's start by analyzing this bond line structure. And we'll start with this carbon right here in magenta. So that carbon in magenta, I'll represent it over here, is bonded to another carbon. I'll use light blue for that. So that's this carbon right here."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with this carbon right here in magenta. So that carbon in magenta, I'll represent it over here, is bonded to another carbon. I'll use light blue for that. So that's this carbon right here. So there's a bond between those two carbons. And let me draw in that bond. Now we have another carbon."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that's this carbon right here. So there's a bond between those two carbons. And let me draw in that bond. Now we have another carbon. I'll use red right here. So the carbon in red is up here. I'll show the bond between those two carbons."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Now we have another carbon. I'll use red right here. So the carbon in red is up here. I'll show the bond between those two carbons. And then let's use green for the next carbon. So we have a carbon right here in green. And finally, there's one more carbon to think about."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I'll show the bond between those two carbons. And then let's use green for the next carbon. So we have a carbon right here in green. And finally, there's one more carbon to think about. So let me, let's see, what color do we need to use here? Let's use dark blue. So we have one more carbon right here in dark blue."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And finally, there's one more carbon to think about. So let me, let's see, what color do we need to use here? Let's use dark blue. So we have one more carbon right here in dark blue. And I'll show that bond. So we have five carbons in this molecule. So five carbons."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we have one more carbon right here in dark blue. And I'll show that bond. So we have five carbons in this molecule. So five carbons. So let's write the molecular formula. So it would be C5. Next, let's figure out how many hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So five carbons. So let's write the molecular formula. So it would be C5. Next, let's figure out how many hydrogens. Now to do that, you need to remember that a neutral carbon atom forms four bonds. So let's see how many bonds we already have. We'll start with the carbon in magenta."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's figure out how many hydrogens. Now to do that, you need to remember that a neutral carbon atom forms four bonds. So let's see how many bonds we already have. We'll start with the carbon in magenta. The carbon in magenta already has one bond. And a neutral carbon atom forms four bonds. So if that carbon already has one bond, it needs three bonds to hydrogen."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We'll start with the carbon in magenta. The carbon in magenta already has one bond. And a neutral carbon atom forms four bonds. So if that carbon already has one bond, it needs three bonds to hydrogen. So one bond to hydrogen, two bonds to hydrogen, and three bonds to hydrogen. Next, let's go with this top carbon here. So the one in red."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So if that carbon already has one bond, it needs three bonds to hydrogen. So one bond to hydrogen, two bonds to hydrogen, and three bonds to hydrogen. Next, let's go with this top carbon here. So the one in red. This carbon already has one bond. So it needs a total of four. So it needs three more bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the one in red. This carbon already has one bond. So it needs a total of four. So it needs three more bonds. And those bonds are to hydrogen. So it's implied that those bonds are to hydrogen. Next, we'll go for the light blue carbon in here."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it needs three more bonds. And those bonds are to hydrogen. So it's implied that those bonds are to hydrogen. Next, we'll go for the light blue carbon in here. So how many bonds does this carbon already have? Well, here's one, here's two, and here's three. That carbon already has three bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, we'll go for the light blue carbon in here. So how many bonds does this carbon already have? Well, here's one, here's two, and here's three. That carbon already has three bonds. So it only needs one more. So we can draw in one hydrogen. We know that carbon is bonded to only one hydrogen."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That carbon already has three bonds. So it only needs one more. So we can draw in one hydrogen. We know that carbon is bonded to only one hydrogen. Next, we'll do the green carbon. So the green carbon right here already has two bonds. Here's one, and here's another one."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We know that carbon is bonded to only one hydrogen. Next, we'll do the green carbon. So the green carbon right here already has two bonds. Here's one, and here's another one. So if we think about a neutral carbon atom forming four bonds, that carbon needs two more bonds. And those bonds are to hydrogen. So we draw in those hydrogens there."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Here's one, and here's another one. So if we think about a neutral carbon atom forming four bonds, that carbon needs two more bonds. And those bonds are to hydrogen. So we draw in those hydrogens there. And finally, the carbon in dark blue. The carbon in dark blue already has one bond. So it needs three more."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we draw in those hydrogens there. And finally, the carbon in dark blue. The carbon in dark blue already has one bond. So it needs three more. So there's one, there's two, and there's three. So now we've drawn out the complete Lewis dot structure for this bond line structure over here. And how many total hydrogens do we have?"}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it needs three more. So there's one, there's two, and there's three. So now we've drawn out the complete Lewis dot structure for this bond line structure over here. And how many total hydrogens do we have? Well, if you count those up, you'll get 12. So the molecular formula is C5H12. Let's do another one."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And how many total hydrogens do we have? Well, if you count those up, you'll get 12. So the molecular formula is C5H12. Let's do another one. So let's look at this next bond line structure here. And let's focus in on our carbon. So we'll start with this carbon right here in magenta."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one. So let's look at this next bond line structure here. And let's focus in on our carbon. So we'll start with this carbon right here in magenta. So I'll draw that in right here. The carbon in magenta is bonded to two other carbons. So I'll make this top carbon here red."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we'll start with this carbon right here in magenta. So I'll draw that in right here. The carbon in magenta is bonded to two other carbons. So I'll make this top carbon here red. So there's a bond to the carbon in red. And there's a bond to this carbon here in light blue. So let's draw in those bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So I'll make this top carbon here red. So there's a bond to the carbon in red. And there's a bond to this carbon here in light blue. So let's draw in those bonds. So we draw in those bonds here. And let's just keep going with our carbons. Let's assign our carbons first, and we'll come back to our hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw in those bonds. So we draw in those bonds here. And let's just keep going with our carbons. Let's assign our carbons first, and we'll come back to our hydrogens. So next, let's make this carbon right here in green. So we have another bond of a carbon to a carbon. And then let's go with dark blue."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's assign our carbons first, and we'll come back to our hydrogens. So next, let's make this carbon right here in green. So we have another bond of a carbon to a carbon. And then let's go with dark blue. So we have dark blue over here for this carbon. So let's show that bond. And then we have another carbon over here."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then let's go with dark blue. So we have dark blue over here for this carbon. So let's show that bond. And then we have another carbon over here. So that carbon is right here. And we can show our last bond. The carbon in red is bonded to a chlorine."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then we have another carbon over here. So that carbon is right here. And we can show our last bond. The carbon in red is bonded to a chlorine. So let me go ahead and show that. So there's our chlorine. And now let's think about hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "The carbon in red is bonded to a chlorine. So let me go ahead and show that. So there's our chlorine. And now let's think about hydrogens. And let's start with the carbon in red. So this carbon in red, how many bonds does it already have? Well, one, two, and three."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And now let's think about hydrogens. And let's start with the carbon in red. So this carbon in red, how many bonds does it already have? Well, one, two, and three. It already has three bonds, so it needs one more. And so it's implied that that bond is to a hydrogen. So that carbon's bonded to one hydrogen."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, one, two, and three. It already has three bonds, so it needs one more. And so it's implied that that bond is to a hydrogen. So that carbon's bonded to one hydrogen. Next, let's do the carbon in magenta. So over here, how many bonds does that carbon in magenta already have? Well, here's one, and here's two."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon's bonded to one hydrogen. Next, let's do the carbon in magenta. So over here, how many bonds does that carbon in magenta already have? Well, here's one, and here's two. So it already has two. So that carbon needs two more. So we can draw in a hydrogen here and a hydrogen here."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, here's one, and here's two. So it already has two. So that carbon needs two more. So we can draw in a hydrogen here and a hydrogen here. It's the same situation for all of the carbons around our ring. If we go to this carbon here in light blue, it already has two bonds. So it needs two more bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we can draw in a hydrogen here and a hydrogen here. It's the same situation for all of the carbons around our ring. If we go to this carbon here in light blue, it already has two bonds. So it needs two more bonds. And that must mean two bonds to hydrogen. So we can go around the entire ring and add in two hydrogens to all of these carbons. So let me make sure I use the correct colors here."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it needs two more bonds. And that must mean two bonds to hydrogen. So we can go around the entire ring and add in two hydrogens to all of these carbons. So let me make sure I use the correct colors here. So in blue, and then we have this one over here. So what's the total molecular formula for this compound? Well, we have a total of six carbons, right?"}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me make sure I use the correct colors here. So in blue, and then we have this one over here. So what's the total molecular formula for this compound? Well, we have a total of six carbons, right? One, two, three, four, five, six. So C6, and how many total hydrogens? We have two on five carbons, and then we have another one here."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, we have a total of six carbons, right? One, two, three, four, five, six. So C6, and how many total hydrogens? We have two on five carbons, and then we have another one here. So two times five is 10, plus one is 11. So H11, and then we have a chlorine as well. So C6H11Cl would be the molecular formula for this compound."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We have two on five carbons, and then we have another one here. So two times five is 10, plus one is 11. So H11, and then we have a chlorine as well. So C6H11Cl would be the molecular formula for this compound. Let's look at two more examples, and we'll start with this carbon right here in magenta. So I'll draw in that carbon. That carbon in magenta is bonded to this carbon in blue, but notice there are two bonds between those two carbons."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So C6H11Cl would be the molecular formula for this compound. Let's look at two more examples, and we'll start with this carbon right here in magenta. So I'll draw in that carbon. That carbon in magenta is bonded to this carbon in blue, but notice there are two bonds between those two carbons. So the carbon in magenta is bonded to the carbon in blue, but there's a double bond between our carbons this time. And the carbon on the right here in red, there's a single bond between the carbon in blue and the carbon in red. So for the molecular formula, so far we know there are a total of three carbons in this compound."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That carbon in magenta is bonded to this carbon in blue, but notice there are two bonds between those two carbons. So the carbon in magenta is bonded to the carbon in blue, but there's a double bond between our carbons this time. And the carbon on the right here in red, there's a single bond between the carbon in blue and the carbon in red. So for the molecular formula, so far we know there are a total of three carbons in this compound. Next, we need to think about hydrogens. So we know a neutral carbon atom forms four bonds. So carbon forms four bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So for the molecular formula, so far we know there are a total of three carbons in this compound. Next, we need to think about hydrogens. So we know a neutral carbon atom forms four bonds. So carbon forms four bonds. How many bonds does the carbon in magenta already have? Well, here's one bond, and here's another bond. So the carbon in magenta needs two more bonds, and those bonds must be two hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So carbon forms four bonds. How many bonds does the carbon in magenta already have? Well, here's one bond, and here's another bond. So the carbon in magenta needs two more bonds, and those bonds must be two hydrogens. So let me draw in those carbon-hydrogen bonds like that. Next, we think about the carbon in blue. How many bonds does the carbon in blue already have?"}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the carbon in magenta needs two more bonds, and those bonds must be two hydrogens. So let me draw in those carbon-hydrogen bonds like that. Next, we think about the carbon in blue. How many bonds does the carbon in blue already have? Well, here's one, here's two, and here's three. So the carbon in blue already has three bonds. It needs one more."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "How many bonds does the carbon in blue already have? Well, here's one, here's two, and here's three. So the carbon in blue already has three bonds. It needs one more. So we show one carbon-hydrogen bond. And finally, the carbon in red already has one bond, so it needs three more. So we draw in three carbon-hydrogen bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It needs one more. So we show one carbon-hydrogen bond. And finally, the carbon in red already has one bond, so it needs three more. So we draw in three carbon-hydrogen bonds. So how many total hydrogens do we have? That would be six hydrogens. So the molecular formula is C3H6."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So we draw in three carbon-hydrogen bonds. So how many total hydrogens do we have? That would be six hydrogens. So the molecular formula is C3H6. We can start to think about hybridization states here, too, because if you look at this carbon and this carbon, you know both of those carbons are sp2 hybridized. And if those carbons are sp2 hybridized, we're talking about trigonal planar geometry around those atoms. And we try to show that in our dot structure as best we can, right?"}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the molecular formula is C3H6. We can start to think about hybridization states here, too, because if you look at this carbon and this carbon, you know both of those carbons are sp2 hybridized. And if those carbons are sp2 hybridized, we're talking about trigonal planar geometry around those atoms. And we try to show that in our dot structure as best we can, right? Approximately 120 degree bond angles around here. So hybridization can come into it as well. Next, let's look at this one right here, which has a triple bond."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we try to show that in our dot structure as best we can, right? Approximately 120 degree bond angles around here. So hybridization can come into it as well. Next, let's look at this one right here, which has a triple bond. And triple bonds often confuse students on bond line structures. So let's assign our carbons again. Let's start with this one right here in magenta."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, let's look at this one right here, which has a triple bond. And triple bonds often confuse students on bond line structures. So let's assign our carbons again. Let's start with this one right here in magenta. So let me draw in that carbon in magenta. The carbon in magenta is bonded to this carbon in blue, and there's a single bond between those two carbons. Next, there's a bond between the carbon in blue and this carbon right here in red."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with this one right here in magenta. So let me draw in that carbon in magenta. The carbon in magenta is bonded to this carbon in blue, and there's a single bond between those two carbons. Next, there's a bond between the carbon in blue and this carbon right here in red. So that carbon in red, there's a single bond between those. The carbon in red is bonded to one more carbon on the opposite side of our triple bond. So that carbon in blue is right there."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Next, there's a bond between the carbon in blue and this carbon right here in red. So that carbon in red, there's a single bond between those. The carbon in red is bonded to one more carbon on the opposite side of our triple bond. So that carbon in blue is right there. There's a triple bond between the carbon in red and the carbon in blue. So now we have our carbons drawn out. That's four carbons."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon in blue is right there. There's a triple bond between the carbon in red and the carbon in blue. So now we have our carbons drawn out. That's four carbons. So this would be C4 so far for the molecular formula. Next, we need to think about hydrogens. So the carbon in magenta already has one bond, so it needs three more bonds."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "That's four carbons. So this would be C4 so far for the molecular formula. Next, we need to think about hydrogens. So the carbon in magenta already has one bond, so it needs three more bonds. So we draw in three bonds of carbon to hydrogen. The carbon in blue here already has two bonds. There's one and there's two."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So the carbon in magenta already has one bond, so it needs three more bonds. So we draw in three bonds of carbon to hydrogen. The carbon in blue here already has two bonds. There's one and there's two. So the carbon in blue needs two more. What about the carbon in red? Well, the carbon in red has one bond, two, three, and four."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "There's one and there's two. So the carbon in blue needs two more. What about the carbon in red? Well, the carbon in red has one bond, two, three, and four. The carbon in red already has four bonds. So the carbon in red doesn't have any hydrogens on it at all. And finally, the carbon in blue."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, the carbon in red has one bond, two, three, and four. The carbon in red already has four bonds. So the carbon in red doesn't have any hydrogens on it at all. And finally, the carbon in blue. The carbon in blue has three bonds, one, two, three. So three bonds already, which means the carbon in blue needs one more bond and that bond is two hydrogen. So now we have all of our hydrogens."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And finally, the carbon in blue. The carbon in blue has three bonds, one, two, three. So three bonds already, which means the carbon in blue needs one more bond and that bond is two hydrogen. So now we have all of our hydrogens. That's a total of six hydrogens. So we can complete the molecular formula, C4H6. And once again, thinking about hybridization."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So now we have all of our hydrogens. That's a total of six hydrogens. So we can complete the molecular formula, C4H6. And once again, thinking about hybridization. This carbon and this carbon, they're both SP hybridized. And so we know the geometry is linear around those carbons. And so that's why we draw this as being a straight line on our bond line structures."}, {"video_title": "Bond-line structures Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And once again, thinking about hybridization. This carbon and this carbon, they're both SP hybridized. And so we know the geometry is linear around those carbons. And so that's why we draw this as being a straight line on our bond line structures. It's because of the geometry. We're trying to reflect the structure of the molecule the best that we can. So practice your bond line structures because they're extremely important for everything that you will do in organic chemistry."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So the geometry of the atoms around this carbon happens to be planar. And so actually, this entire molecule is planar. So you can think about all this in a plane here. And the bond angles are close to 120 degrees, so approximately 120 degree bond angles. And this carbon that I've underlined here is bonded to only three atoms, so a hydrogen, a hydrogen, and a carbon. And so we must need a different hybridization for each of the carbons present in the ethylene molecule. And so we're going to start with our electron configurations over here, the excited state."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And the bond angles are close to 120 degrees, so approximately 120 degree bond angles. And this carbon that I've underlined here is bonded to only three atoms, so a hydrogen, a hydrogen, and a carbon. And so we must need a different hybridization for each of the carbons present in the ethylene molecule. And so we're going to start with our electron configurations over here, the excited state. So we have carbons, four valence electrons represented. So one, two, three, and four. And in the video on sp3 hybridization, we took all four of these orbitals and combined them to make four sp3 hybrid orbitals."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so we're going to start with our electron configurations over here, the excited state. So we have carbons, four valence electrons represented. So one, two, three, and four. And in the video on sp3 hybridization, we took all four of these orbitals and combined them to make four sp3 hybrid orbitals. In this case, we only have a carbon bonded to three atoms, so we only need three of our orbitals. So we're going to promote the s orbital. So we're going to promote the s orbital up."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And in the video on sp3 hybridization, we took all four of these orbitals and combined them to make four sp3 hybrid orbitals. In this case, we only have a carbon bonded to three atoms, so we only need three of our orbitals. So we're going to promote the s orbital. So we're going to promote the s orbital up. And this time, we only need two of the p orbitals. So we're going to take one of the p's and then another one of the p's here. That is going to leave one of our p orbitals unhybridized."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we're going to promote the s orbital up. And this time, we only need two of the p orbitals. So we're going to take one of the p's and then another one of the p's here. That is going to leave one of our p orbitals unhybridized. So each one of these orbitals has one electron in it like that. And this is no longer an s orbital. This is an sp2 hybrid orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "That is going to leave one of our p orbitals unhybridized. So each one of these orbitals has one electron in it like that. And this is no longer an s orbital. This is an sp2 hybrid orbital. This is no longer a p orbital. This is an sp2 hybrid orbital. And same with this one, an sp2 hybrid orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "This is an sp2 hybrid orbital. This is no longer a p orbital. This is an sp2 hybrid orbital. And same with this one, an sp2 hybrid orbital. We call this sp2 hybridization. So let me go ahead and write this up here. Let's use a different color here."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And same with this one, an sp2 hybrid orbital. We call this sp2 hybridization. So let me go ahead and write this up here. Let's use a different color here. So this is sp2 hybridization because we're using one s orbital and two p orbitals to form our new hybrid orbitals. And so this carbon right here is sp2 hybridized. And the same with this carbon."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Let's use a different color here. So this is sp2 hybridization because we're using one s orbital and two p orbitals to form our new hybrid orbitals. And so this carbon right here is sp2 hybridized. And the same with this carbon. So notice that we left a p orbital untouched. So we have a p orbital unhybridized like that. In terms of the shape of our new hybrid orbital, let's go ahead and get some more space down here."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And the same with this carbon. So notice that we left a p orbital untouched. So we have a p orbital unhybridized like that. In terms of the shape of our new hybrid orbital, let's go ahead and get some more space down here. So we're taking one s orbital. We know s orbitals are shaped like spheres. We're taking two p orbitals."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "In terms of the shape of our new hybrid orbital, let's go ahead and get some more space down here. So we're taking one s orbital. We know s orbitals are shaped like spheres. We're taking two p orbitals. We know that a p orbital is shaped like a dumbbell. So we're going to take these orbitals and hybridize them to form three sp2 hybrid orbitals. And they have a bigger front lobe and a smaller back lobe here like that."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "We're taking two p orbitals. We know that a p orbital is shaped like a dumbbell. So we're going to take these orbitals and hybridize them to form three sp2 hybrid orbitals. And they have a bigger front lobe and a smaller back lobe here like that. And once again, when we draw the pictures, we're going to ignore the smaller back lobe. So this gives us our sp2 hybrid orbitals. In terms of what percentage character, we have three orbitals that we're taking here."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And they have a bigger front lobe and a smaller back lobe here like that. And once again, when we draw the pictures, we're going to ignore the smaller back lobe. So this gives us our sp2 hybrid orbitals. In terms of what percentage character, we have three orbitals that we're taking here. And one of them is an s orbital. So one out of three gives us 33% s character in our new hybrid sp2 orbital. And then we have two p orbitals."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "In terms of what percentage character, we have three orbitals that we're taking here. And one of them is an s orbital. So one out of three gives us 33% s character in our new hybrid sp2 orbital. And then we have two p orbitals. So two out of three gives us 67% p character. So 33% s character and 67% p character. So there's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then we have two p orbitals. So two out of three gives us 67% p character. So 33% s character and 67% p character. So there's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital. And since the electron density in an s orbital is closer to the nucleus, so we think about the electron density here being closer to the nucleus, that means that we could think about this lobe right here being a little bit shorter, with the electron density being closer to the nucleus. And that's going to have an effect on the length of the bonds that we're going to be forming. So let's go ahead and draw the picture of the ethylene molecule now."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So there's more s character in an sp2 hybrid orbital than an sp3 hybrid orbital. And since the electron density in an s orbital is closer to the nucleus, so we think about the electron density here being closer to the nucleus, that means that we could think about this lobe right here being a little bit shorter, with the electron density being closer to the nucleus. And that's going to have an effect on the length of the bonds that we're going to be forming. So let's go ahead and draw the picture of the ethylene molecule now. So we know that each of the carbons in ethylene, so just going back up here to emphasize the point, each of these carbons here is sp2 hybridized. That means each of those carbons is going to have three sp2 hybrid orbitals around it and one unhybridized p orbital. So let's go ahead and draw that."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the picture of the ethylene molecule now. So we know that each of the carbons in ethylene, so just going back up here to emphasize the point, each of these carbons here is sp2 hybridized. That means each of those carbons is going to have three sp2 hybrid orbitals around it and one unhybridized p orbital. So let's go ahead and draw that. So we have a carbon right here. And this is an sp2 hybridized orbital. So we're going to draw in, there's one sp2 hybrid orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that. So we have a carbon right here. And this is an sp2 hybridized orbital. So we're going to draw in, there's one sp2 hybrid orbital. Here's another sp2 hybrid orbital. And here's another one. And we go back up to here."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we're going to draw in, there's one sp2 hybrid orbital. Here's another sp2 hybrid orbital. And here's another one. And we go back up to here. And we can see that each one of those orbitals, let me go ahead and mark this, each one of those sp2 hybrid orbitals has one electron in it. So each one of these orbitals has one electron. I go back down here."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And we go back up to here. And we can see that each one of those orbitals, let me go ahead and mark this, each one of those sp2 hybrid orbitals has one electron in it. So each one of these orbitals has one electron. I go back down here. And I put in the one electron in each one of my orbitals like that. And I know that each of those carbons is going to have an unhybridized p orbital here. So an unhybridized p orbital with one electron, too."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "I go back down here. And I put in the one electron in each one of my orbitals like that. And I know that each of those carbons is going to have an unhybridized p orbital here. So an unhybridized p orbital with one electron, too. So let me go ahead and draw that in. So I'll go ahead and use a different color. So we have our unhybridized p orbital like that."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So an unhybridized p orbital with one electron, too. So let me go ahead and draw that in. So I'll go ahead and use a different color. So we have our unhybridized p orbital like that. And there's one electron in our unhybridized p orbital. So each of the carbons was sp2 hybridized. So let me go ahead and draw the dot structure right here again so we can take a look at it, the dot structure for ethylene."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have our unhybridized p orbital like that. And there's one electron in our unhybridized p orbital. So each of the carbons was sp2 hybridized. So let me go ahead and draw the dot structure right here again so we can take a look at it, the dot structure for ethylene. So let's do the other carbon now. So the carbon on the right is also sp2 hybridized. So we can go ahead and draw in an sp2 hybrid orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw the dot structure right here again so we can take a look at it, the dot structure for ethylene. So let's do the other carbon now. So the carbon on the right is also sp2 hybridized. So we can go ahead and draw in an sp2 hybrid orbital. And there's one electron in that orbital. And then here's another one with one electron. And then here's another one with one electron."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and draw in an sp2 hybrid orbital. And there's one electron in that orbital. And then here's another one with one electron. And then here's another one with one electron. And this carbon, being sp2 hybridized, also has an unhybridized p orbital with one electron. So we go ahead and draw in that p orbital with its one electron. We also have some hydrogens."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then here's another one with one electron. And this carbon, being sp2 hybridized, also has an unhybridized p orbital with one electron. So we go ahead and draw in that p orbital with its one electron. We also have some hydrogens. So we have some hydrogens to think about here. So each carbon is bonded to two hydrogens. So let me go ahead and put in the hydrogens."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "We also have some hydrogens. So we have some hydrogens to think about here. So each carbon is bonded to two hydrogens. So let me go ahead and put in the hydrogens. So hydrogen has a valence electron in an unhybridized s orbital. So I'm going ahead and putting in the s orbital and the one valence electron from hydrogen like this. When we take a look at what we've drawn here, we can see some head-on overlap of orbitals, which we know from our earlier video is called a sigma bond."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and put in the hydrogens. So hydrogen has a valence electron in an unhybridized s orbital. So I'm going ahead and putting in the s orbital and the one valence electron from hydrogen like this. When we take a look at what we've drawn here, we can see some head-on overlap of orbitals, which we know from our earlier video is called a sigma bond. So here's a head-on overlap of orbitals. So that's a sigma bond. Here's another head-on overlap of orbitals."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "When we take a look at what we've drawn here, we can see some head-on overlap of orbitals, which we know from our earlier video is called a sigma bond. So here's a head-on overlap of orbitals. So that's a sigma bond. Here's another head-on overlap of orbitals. The carbon-carbon bond, here's also a head-on overlap of orbitals. And then we have these two over here. And so we have a total of five sigma bonds in our molecule."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Here's another head-on overlap of orbitals. The carbon-carbon bond, here's also a head-on overlap of orbitals. And then we have these two over here. And so we have a total of five sigma bonds in our molecule. So let me go ahead and write that over here. So there are five sigma bonds. And if I'm trying to find those on my dot structure, that would be a sigma bond."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And so we have a total of five sigma bonds in our molecule. So let me go ahead and write that over here. So there are five sigma bonds. And if I'm trying to find those on my dot structure, that would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond. And then these over here."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And if I'm trying to find those on my dot structure, that would be a sigma bond. This would be a sigma bond. One of these two is a sigma bond. And then these over here. So a total of five sigma bonds. And then we have a new type of bonding. So these unhybridized p orbitals can overlap side by side."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And then these over here. So a total of five sigma bonds. And then we have a new type of bonding. So these unhybridized p orbitals can overlap side by side. So up here and down here, we get side by side overlap of our p orbitals. And this creates a pi bond. So a pi bond is side by side overlap."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So these unhybridized p orbitals can overlap side by side. So up here and down here, we get side by side overlap of our p orbitals. And this creates a pi bond. So a pi bond is side by side overlap. So there is overlap above and below this sigma bond here. And that's going to prevent free rotation. So when we're looking at the example of ethane, we had free rotation about the sigma bond that connected the two carbons."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So a pi bond is side by side overlap. So there is overlap above and below this sigma bond here. And that's going to prevent free rotation. So when we're looking at the example of ethane, we had free rotation about the sigma bond that connected the two carbons. But because of this pi bond here, this pi bond is going to prevent rotation. So we don't get different conformations of the ethylene molecule. So no free rotation due to the pi bonds."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So when we're looking at the example of ethane, we had free rotation about the sigma bond that connected the two carbons. But because of this pi bond here, this pi bond is going to prevent rotation. So we don't get different conformations of the ethylene molecule. So no free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds. So I'm just going to say it's this one right here. So if you have a double bond, one of those bonds is a sigma bond."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So no free rotation due to the pi bonds. When you're looking at the dot structure, one of these bonds is the pi bonds. So I'm just going to say it's this one right here. So if you have a double bond, one of those bonds is a sigma bond. And one of those bonds is a pi bond. So we have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, so let me go ahead and use a different color for that."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So if you have a double bond, one of those bonds is a sigma bond. And one of those bonds is a pi bond. So we have a total of one pi bond in the ethylene molecule. If you're thinking about the distance between the two carbons, so let me go ahead and use a different color for that. The distance between this carbon and this carbon. So it turns out to be approximately 1.34 angstroms, which is shorter than the distance between the two carbons in the ethane molecule. Remember, for ethane, the distance was approximately 1.54 angstroms."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "If you're thinking about the distance between the two carbons, so let me go ahead and use a different color for that. The distance between this carbon and this carbon. So it turns out to be approximately 1.34 angstroms, which is shorter than the distance between the two carbons in the ethane molecule. Remember, for ethane, the distance was approximately 1.54 angstroms. And so a double bond is shorter than a single bond. And one way to think about that is the increased s character. So this increased s character means the electron density is closer to the nucleus."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Remember, for ethane, the distance was approximately 1.54 angstroms. And so a double bond is shorter than a single bond. And one way to think about that is the increased s character. So this increased s character means the electron density is closer to the nucleus. And so that's going to make this lobe a little bit shorter than before. And that's going to decrease the distance between these two carbon atoms here. So 1.34 angstroms."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So this increased s character means the electron density is closer to the nucleus. And so that's going to make this lobe a little bit shorter than before. And that's going to decrease the distance between these two carbon atoms here. So 1.34 angstroms. All right, let's look at the dot structure again and see how we can analyze this using the concept of steric numbers. So let me go ahead and redraw the dot structure. So we have our carbon-carbon double bond here and our hydrogens like that."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So 1.34 angstroms. All right, let's look at the dot structure again and see how we can analyze this using the concept of steric numbers. So let me go ahead and redraw the dot structure. So we have our carbon-carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number, remember to find the hybridization, we can use this concept. Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. So if my goal was to find the steric number for this carbon, I count up my number of sigma bonds."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So we have our carbon-carbon double bond here and our hydrogens like that. If you're approaching this situation using steric number, remember to find the hybridization, we can use this concept. Steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. So if my goal was to find the steric number for this carbon, I count up my number of sigma bonds. So that's 1, 2. And then I know in a double bond, one of those is sigma and one of those is pi. So one of those is a sigma bond."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So if my goal was to find the steric number for this carbon, I count up my number of sigma bonds. So that's 1, 2. And then I know in a double bond, one of those is sigma and one of those is pi. So one of those is a sigma bond. So I have a total of three sigma bonds. I have zero lone pairs of electrons around that carbon. So 3 plus 0 gives me a steric number of 3."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So one of those is a sigma bond. So I have a total of three sigma bonds. I have zero lone pairs of electrons around that carbon. So 3 plus 0 gives me a steric number of 3. So I need three hybrid orbitals. And we've just seen in this video that three sp2 hybrid orbitals form if you're dealing with sp2 hybridization. So if you get a steric number of 3, you're going to think about sp2 hybridization, one s orbital and two p orbitals hybridizing."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So 3 plus 0 gives me a steric number of 3. So I need three hybrid orbitals. And we've just seen in this video that three sp2 hybrid orbitals form if you're dealing with sp2 hybridization. So if you get a steric number of 3, you're going to think about sp2 hybridization, one s orbital and two p orbitals hybridizing. So that carbon is sp2 hybridized. And of course, this one is too. So both of them are sp2 hybridized."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So if you get a steric number of 3, you're going to think about sp2 hybridization, one s orbital and two p orbitals hybridizing. So that carbon is sp2 hybridized. And of course, this one is too. So both of them are sp2 hybridized. Let's do another example. Let's do boron trifluoride, so BF3. If you want to draw the dot structure for BF3, you would have boron."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So both of them are sp2 hybridized. Let's do another example. Let's do boron trifluoride, so BF3. If you want to draw the dot structure for BF3, you would have boron. And then you would surround it with your fluorines here. And you would have an octet of electrons around each fluorine. So I go ahead and put those in on my dot structure."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "If you want to draw the dot structure for BF3, you would have boron. And then you would surround it with your fluorines here. And you would have an octet of electrons around each fluorine. So I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here, so what is the hybridization state of this boron? Let's use the concept of steric number. So once again, let's use steric number."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So I go ahead and put those in on my dot structure. If your goal is to figure out the hybridization of this boron here, so what is the hybridization state of this boron? Let's use the concept of steric number. So once again, let's use steric number. So we'll find the hybridization of this boron. Steric number is equal to number of sigma bonds. So that's 1, 2, 3, so three sigma bonds plus lone pairs of electrons."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So once again, let's use steric number. So we'll find the hybridization of this boron. Steric number is equal to number of sigma bonds. So that's 1, 2, 3, so three sigma bonds plus lone pairs of electrons. That's 0. So a steric number of 3 tells us this boron is sp2 hybridized. So this boron is going to have three sp2 hybrid orbitals and one p orbital, one unhybridized p orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So that's 1, 2, 3, so three sigma bonds plus lone pairs of electrons. That's 0. So a steric number of 3 tells us this boron is sp2 hybridized. So this boron is going to have three sp2 hybrid orbitals and one p orbital, one unhybridized p orbital. So let's go ahead and draw that. So we have a boron here bonded to three fluorines. And it also is going to have an unhybridized p orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So this boron is going to have three sp2 hybrid orbitals and one p orbital, one unhybridized p orbital. So let's go ahead and draw that. So we have a boron here bonded to three fluorines. And it also is going to have an unhybridized p orbital. Now remember, when you are dealing with boron, it has one less valence electron than carbon. Carbon had four valence electrons. Boron has only three."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "And it also is going to have an unhybridized p orbital. Now remember, when you are dealing with boron, it has one less valence electron than carbon. Carbon had four valence electrons. Boron has only three. So when you're thinking about the sp2 hybrid orbitals that you create, sp2 hybrid orbital, sp2, sp2, and then one unhybridized p orbital right here, boron only has three valence electrons. So let's go ahead and put in those valence electrons, 1, 2, and 3. So it doesn't have any electrons in its unhybridized p orbital."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "Boron has only three. So when you're thinking about the sp2 hybrid orbitals that you create, sp2 hybrid orbital, sp2, sp2, and then one unhybridized p orbital right here, boron only has three valence electrons. So let's go ahead and put in those valence electrons, 1, 2, and 3. So it doesn't have any electrons in its unhybridized p orbital. And so over here, when you're looking at the picture, this has an empty orbital. And so boron can accept a pair of electrons. So if you're thinking about its chemical behavior, one of the things that BF3 can do, the boron can accept an electron pair and function as a Lewis acid."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So it doesn't have any electrons in its unhybridized p orbital. And so over here, when you're looking at the picture, this has an empty orbital. And so boron can accept a pair of electrons. So if you're thinking about its chemical behavior, one of the things that BF3 can do, the boron can accept an electron pair and function as a Lewis acid. And so that's one way in thinking about how hybridization allows you to think about the structure and how something might react. So this boron turns out to be sp2 hybridized. So this boron here is sp2 hybridized."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So if you're thinking about its chemical behavior, one of the things that BF3 can do, the boron can accept an electron pair and function as a Lewis acid. And so that's one way in thinking about how hybridization allows you to think about the structure and how something might react. So this boron turns out to be sp2 hybridized. So this boron here is sp2 hybridized. And so we can also talk about the geometry of the molecule. It's planar. So around this boron, it's planar."}, {"video_title": "sp\u00b2 hybridization AP Chemistry Khan Academy.mp3", "Sentence": "So this boron here is sp2 hybridized. And so we can also talk about the geometry of the molecule. It's planar. So around this boron, it's planar. And so therefore, your bond angles are 120 degrees. So if you had boron right here, and you're thinking about a circle, a circle is 360 degrees. So if you divide 360 by 3, you get 120 degrees for all of these bond angles."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What I want to do in this video is kind of introduce you to the idea of, one, how carbon-14 comes about and how it gets into all living things. And then, either later in this video or in future videos, we'll talk about how it's actually used to date things. How we use it to actually figure out that that bone is 12,000 years old, that that person died 18,000 years ago, whatever it might be. So let me draw the Earth. So let me just draw the surface of the Earth like that. It's just a little section of the surface of the Earth. And then we have the atmosphere of the Earth."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me draw the Earth. So let me just draw the surface of the Earth like that. It's just a little section of the surface of the Earth. And then we have the atmosphere of the Earth. I'll draw that in yellow. So then you have the Earth's atmosphere right over here. Let me write that down."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then we have the atmosphere of the Earth. I'll draw that in yellow. So then you have the Earth's atmosphere right over here. Let me write that down. Atmosphere. And 78% the most abundant element in our atmosphere is nitrogen. It is 78% nitrogen."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me write that down. Atmosphere. And 78% the most abundant element in our atmosphere is nitrogen. It is 78% nitrogen. And if I write nitrogen, its symbol is just N. And it has 7 protons and it also has 7 neutrons. So it has an atomic mass of roughly 14. And this is the most typical isotope of nitrogen."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is 78% nitrogen. And if I write nitrogen, its symbol is just N. And it has 7 protons and it also has 7 neutrons. So it has an atomic mass of roughly 14. And this is the most typical isotope of nitrogen. And we talk about the word isotope in the chemistry playlist. An isotope, the protons define what element it is. But this number up here can change depending on the number of neutrons you have."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is the most typical isotope of nitrogen. And we talk about the word isotope in the chemistry playlist. An isotope, the protons define what element it is. But this number up here can change depending on the number of neutrons you have. So the different versions of a given element, those are each called isotopes. I just view them in my head as versions of an element. So anyway, we have our atmosphere."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this number up here can change depending on the number of neutrons you have. So the different versions of a given element, those are each called isotopes. I just view them in my head as versions of an element. So anyway, we have our atmosphere. And then coming from our sun, we have what's commonly called cosmic rays. But they're actually not rays. They're cosmic particles."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So anyway, we have our atmosphere. And then coming from our sun, we have what's commonly called cosmic rays. But they're actually not rays. They're cosmic particles. They're mainly, you can view them as just single protons, which is the same thing as a hydrogen nucleus. They can also be alpha particles, which is the same thing as a helium nucleus. And there's even a few electrons."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're cosmic particles. They're mainly, you can view them as just single protons, which is the same thing as a hydrogen nucleus. They can also be alpha particles, which is the same thing as a helium nucleus. And there's even a few electrons. And they're going to come in and they're going to bump into things in our atmosphere. And they're actually going to form neutrons. So they're actually going to form neutrons."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And there's even a few electrons. And they're going to come in and they're going to bump into things in our atmosphere. And they're actually going to form neutrons. So they're actually going to form neutrons. And we'll show a neutron with a lowercase n and a 1 for its mass number. And we don't write anything because it has no protons down here. Like we had for nitrogen, we had 7 protons."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they're actually going to form neutrons. And we'll show a neutron with a lowercase n and a 1 for its mass number. And we don't write anything because it has no protons down here. Like we had for nitrogen, we had 7 protons. So it's not really an element. It is a subatomic particle. But you have these neutrons form."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Like we had for nitrogen, we had 7 protons. So it's not really an element. It is a subatomic particle. But you have these neutrons form. And every now and then, and let's just be clear, this isn't like a typical reaction. But every now and then, one of those neutrons will bump into one of the nitrogen-14s in just the right way. So that it bumps off one of the protons in the nitrogen."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But you have these neutrons form. And every now and then, and let's just be clear, this isn't like a typical reaction. But every now and then, one of those neutrons will bump into one of the nitrogen-14s in just the right way. So that it bumps off one of the protons in the nitrogen. And essentially replaces that proton with itself. So let me make it clear. So it bumps off one of the protons."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that it bumps off one of the protons in the nitrogen. And essentially replaces that proton with itself. So let me make it clear. So it bumps off one of the protons. So instead of 7 protons, we now have 6 protons. But this number 14 doesn't go down to 13 because it replaces it with itself. So this still stays at 14."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it bumps off one of the protons. So instead of 7 protons, we now have 6 protons. But this number 14 doesn't go down to 13 because it replaces it with itself. So this still stays at 14. And now since it now only has 6 protons, this is no longer nitrogen by definition. This is now carbon. And that proton that was bumped off just kind of gets emitted."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this still stays at 14. And now since it now only has 6 protons, this is no longer nitrogen by definition. This is now carbon. And that proton that was bumped off just kind of gets emitted. So then let me just do that in another color. And a proton that's just flying around, you could call that hydrogen-1. And it can gain an electron some ways."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that proton that was bumped off just kind of gets emitted. So then let me just do that in another color. And a proton that's just flying around, you could call that hydrogen-1. And it can gain an electron some ways. If it doesn't gain an electron, it's just a hydrogen ion, a positive ion. Either way, or a hydrogen nucleus. But this process, and once again, it's not a typical process."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it can gain an electron some ways. If it doesn't gain an electron, it's just a hydrogen ion, a positive ion. Either way, or a hydrogen nucleus. But this process, and once again, it's not a typical process. But it happens every now and then. This is how carbon-14 forms. So this right here is carbon-14."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this process, and once again, it's not a typical process. But it happens every now and then. This is how carbon-14 forms. So this right here is carbon-14. You can essentially view it as a nitrogen-14, and one of the protons is replaced with a neutron. And what's interesting about this is this is constantly being formed in our atmosphere. Not in huge quantities, but in reasonable quantities."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right here is carbon-14. You can essentially view it as a nitrogen-14, and one of the protons is replaced with a neutron. And what's interesting about this is this is constantly being formed in our atmosphere. Not in huge quantities, but in reasonable quantities. So let me write this down. Constantly being formed. Constant formation."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Not in huge quantities, but in reasonable quantities. So let me write this down. Constantly being formed. Constant formation. And what happens is, and let me be very clear. Let's look at the periodic table over here. Typical carbon."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Constant formation. And what happens is, and let me be very clear. Let's look at the periodic table over here. Typical carbon. So carbon by definition, if you have 6 protons, carbon by definition has 6 protons. But the typical isotope, the most common isotope of carbon, is carbon-12. So carbon-12 is the most common."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Typical carbon. So carbon by definition, if you have 6 protons, carbon by definition has 6 protons. But the typical isotope, the most common isotope of carbon, is carbon-12. So carbon-12 is the most common. So most of the carbon in your body is carbon-12. But what's interesting is that a small fraction of carbon-14 forms, and then this carbon-14 can then also combine with oxygen to form carbon dioxide. And then that carbon dioxide gets absorbed into the rest of the atmosphere, into our oceans."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So carbon-12 is the most common. So most of the carbon in your body is carbon-12. But what's interesting is that a small fraction of carbon-14 forms, and then this carbon-14 can then also combine with oxygen to form carbon dioxide. And then that carbon dioxide gets absorbed into the rest of the atmosphere, into our oceans. It can be fixed by plants. When people talk about carbon fixation, they're really talking about using mainly light energy from the sun to take gaseous carbon and turn it into actual kind of organic tissue. And so this carbon-14 makes its way, it's constantly being formed, it makes its way into oceans."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then that carbon dioxide gets absorbed into the rest of the atmosphere, into our oceans. It can be fixed by plants. When people talk about carbon fixation, they're really talking about using mainly light energy from the sun to take gaseous carbon and turn it into actual kind of organic tissue. And so this carbon-14 makes its way, it's constantly being formed, it makes its way into oceans. It's already in the air, but it completely mixes through the whole atmosphere. Oceans and the air. And then it makes its way into plants."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this carbon-14 makes its way, it's constantly being formed, it makes its way into oceans. It's already in the air, but it completely mixes through the whole atmosphere. Oceans and the air. And then it makes its way into plants. And then it makes its way, and plants are really just made out of that fixed carbon, that carbon that was taken in gaseous form and put into, I guess you could say, into kind of solid form, put into living form. That's what wood pretty much is. It gets put into plants, and then it gets put into the things that eat the plants."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then it makes its way into plants. And then it makes its way, and plants are really just made out of that fixed carbon, that carbon that was taken in gaseous form and put into, I guess you could say, into kind of solid form, put into living form. That's what wood pretty much is. It gets put into plants, and then it gets put into the things that eat the plants. So that could be us. Now why is this even interesting? I've just explained a mechanism where some of our body, even though carbon-12 is the most common isotope, some of our body, while we're living, gets made up of this carbon-14 thing."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It gets put into plants, and then it gets put into the things that eat the plants. So that could be us. Now why is this even interesting? I've just explained a mechanism where some of our body, even though carbon-12 is the most common isotope, some of our body, while we're living, gets made up of this carbon-14 thing. Well, the interesting thing is the only time you can take in this carbon-14 is while you're alive, while you're eating new things. Because as soon as you die and you get buried under the ground, there's no way for the carbon-14 to become part of your tissue anymore because you're not eating anything with the new carbon-14. And what's interesting here is once you die, you're not going to get any new carbon-14, and that carbon-14 that you did have at your death is going to decay via beta decay, and we learned about this, back into nitrogen-14."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I've just explained a mechanism where some of our body, even though carbon-12 is the most common isotope, some of our body, while we're living, gets made up of this carbon-14 thing. Well, the interesting thing is the only time you can take in this carbon-14 is while you're alive, while you're eating new things. Because as soon as you die and you get buried under the ground, there's no way for the carbon-14 to become part of your tissue anymore because you're not eating anything with the new carbon-14. And what's interesting here is once you die, you're not going to get any new carbon-14, and that carbon-14 that you did have at your death is going to decay via beta decay, and we learned about this, back into nitrogen-14. So this process reverses. So it'll decay back into nitrogen-14, and beta decay, you emit an electron and an electron antineutrino. I won't go into the details of that."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And what's interesting here is once you die, you're not going to get any new carbon-14, and that carbon-14 that you did have at your death is going to decay via beta decay, and we learned about this, back into nitrogen-14. So this process reverses. So it'll decay back into nitrogen-14, and beta decay, you emit an electron and an electron antineutrino. I won't go into the details of that. But essentially what you have happening here is you have one of the neutrons is turning into a proton and emitting this stuff in the process. Now why is this interesting? So I just said while you're living, you have kind of straight-up carbon-14, and carbon-14 is constantly doing this decay thing."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I won't go into the details of that. But essentially what you have happening here is you have one of the neutrons is turning into a proton and emitting this stuff in the process. Now why is this interesting? So I just said while you're living, you have kind of straight-up carbon-14, and carbon-14 is constantly doing this decay thing. But what's interesting is as soon as you die and you're not ingesting any more plants or breathing from the atmosphere, if you are a plant, or fixing from the atmosphere, and this even applies to plants, once a plant dies, it's no longer taking in carbon dioxide from the atmosphere and turning it into new tissue. The carbon-14 in that tissue gets frozen, and this carbon-14 does this decay at a specific rate. And then you can use that rate to actually determine how long ago that thing must have died."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I just said while you're living, you have kind of straight-up carbon-14, and carbon-14 is constantly doing this decay thing. But what's interesting is as soon as you die and you're not ingesting any more plants or breathing from the atmosphere, if you are a plant, or fixing from the atmosphere, and this even applies to plants, once a plant dies, it's no longer taking in carbon dioxide from the atmosphere and turning it into new tissue. The carbon-14 in that tissue gets frozen, and this carbon-14 does this decay at a specific rate. And then you can use that rate to actually determine how long ago that thing must have died. So the rate at which this happens, so the rate of carbon-14 decay is essentially half disappears, half gone in roughly 5,730 years. And this is actually called a half-life. And we talk about it in other videos."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then you can use that rate to actually determine how long ago that thing must have died. So the rate at which this happens, so the rate of carbon-14 decay is essentially half disappears, half gone in roughly 5,730 years. And this is actually called a half-life. And we talk about it in other videos. This is called a half-life. And I want to be clear here. You don't know which half of it's gone."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we talk about it in other videos. This is called a half-life. And I want to be clear here. You don't know which half of it's gone. It's a probabilistic thing. You can't just say, oh, all of the carbon-14s on the left are going to decay and all the carbon-14s on the right aren't going to decay in that 5,730 years. What it's essentially saying is any given carbon-14 atom has a 50% chance of decaying into nitrogen-14 in 5,730 years."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You don't know which half of it's gone. It's a probabilistic thing. You can't just say, oh, all of the carbon-14s on the left are going to decay and all the carbon-14s on the right aren't going to decay in that 5,730 years. What it's essentially saying is any given carbon-14 atom has a 50% chance of decaying into nitrogen-14 in 5,730 years. So over the course of 5,730 years, roughly half of them will have decayed. Now why is that interesting? Well, if you know that all living things have a certain proportion of carbon-14 in their tissue as kind of part of what makes them up, and then if you were to find some bone, let's just say find some bone right here."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What it's essentially saying is any given carbon-14 atom has a 50% chance of decaying into nitrogen-14 in 5,730 years. So over the course of 5,730 years, roughly half of them will have decayed. Now why is that interesting? Well, if you know that all living things have a certain proportion of carbon-14 in their tissue as kind of part of what makes them up, and then if you were to find some bone, let's just say find some bone right here. You dig it up on some type of archaeology dig. And you say, hey, that bone has one half the carbon-14 of all the living things that you see right now. So it would be a pretty reasonable estimate to say, well, that thing must be 5,730 years old."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, if you know that all living things have a certain proportion of carbon-14 in their tissue as kind of part of what makes them up, and then if you were to find some bone, let's just say find some bone right here. You dig it up on some type of archaeology dig. And you say, hey, that bone has one half the carbon-14 of all the living things that you see right now. So it would be a pretty reasonable estimate to say, well, that thing must be 5,730 years old. Even better, maybe you dig a little deeper and you find another bone. Maybe you find another bone, maybe a couple of feet even deeper. And you say, wow, this thing right over here has 1 4th the carbon-14."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it would be a pretty reasonable estimate to say, well, that thing must be 5,730 years old. Even better, maybe you dig a little deeper and you find another bone. Maybe you find another bone, maybe a couple of feet even deeper. And you say, wow, this thing right over here has 1 4th the carbon-14. This has 1 4th the carbon-14 that I would expect to find in something living. So how old is this? Well, if it only has 1 4th the carbon-14, it must have gone through two half-lives."}, {"video_title": "Carbon 14 dating 1 Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you say, wow, this thing right over here has 1 4th the carbon-14. This has 1 4th the carbon-14 that I would expect to find in something living. So how old is this? Well, if it only has 1 4th the carbon-14, it must have gone through two half-lives. After one half-life, it would have one half the carbon. And then after another half-life, half of that also turns into nitrogen-14. And so this would involve two half-lives, which is the same thing as 2 times 5,730 years."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video, I've mentioned how I've been asked to make predictions for the year 2060. In the last one, I focused on education. That's obviously what I'm working on now, so I have some opinions about that. But what I want to do in this video is do maybe slightly more broad and wild predictions. And my one prediction I'll make is I'm probably completely not going to predict the real reality of 2060. And probably the really big things to predict I will completely miss. But with that out of the way, it is fun to predict things, so let's give a shot."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what I want to do in this video is do maybe slightly more broad and wild predictions. And my one prediction I'll make is I'm probably completely not going to predict the real reality of 2060. And probably the really big things to predict I will completely miss. But with that out of the way, it is fun to predict things, so let's give a shot. So the first area that I will predict is what's going to happen in the field of medicine. In particular, I think that the human lifespan is going to increase dramatically. So I'll be conservative and say that the lifespan, the average human lifespan, is going to be, especially in the developed world, and once again, hopefully by 2060, most of the world is developed, the average human lifespan is going to be over 100 years old."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But with that out of the way, it is fun to predict things, so let's give a shot. So the first area that I will predict is what's going to happen in the field of medicine. In particular, I think that the human lifespan is going to increase dramatically. So I'll be conservative and say that the lifespan, the average human lifespan, is going to be, especially in the developed world, and once again, hopefully by 2060, most of the world is developed, the average human lifespan is going to be over 100 years old. And I won't say whether that's going to be a good or a bad thing. There's arguments either way about the proper way to live and what happens to global populations. But if the world is, for the most part, developed and educated, you actually will probably have a lower rate of reproduction, and so you might actually have some space in the world for older people."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I'll be conservative and say that the lifespan, the average human lifespan, is going to be, especially in the developed world, and once again, hopefully by 2060, most of the world is developed, the average human lifespan is going to be over 100 years old. And I won't say whether that's going to be a good or a bad thing. There's arguments either way about the proper way to live and what happens to global populations. But if the world is, for the most part, developed and educated, you actually will probably have a lower rate of reproduction, and so you might actually have some space in the world for older people. But this, I think, is, there's a very strong chance of this happening because we are starting to understand the molecular basis of aging. It's not a given thing that because of some form of wear and tear, things have to die after 70 years or 80 years or 90 years, and we're starting to understand the mechanisms and how to maybe improve the repair mechanisms or how to augment it in some way. So I definitely think this is going to happen."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if the world is, for the most part, developed and educated, you actually will probably have a lower rate of reproduction, and so you might actually have some space in the world for older people. But this, I think, is, there's a very strong chance of this happening because we are starting to understand the molecular basis of aging. It's not a given thing that because of some form of wear and tear, things have to die after 70 years or 80 years or 90 years, and we're starting to understand the mechanisms and how to maybe improve the repair mechanisms or how to augment it in some way. So I definitely think this is going to happen. I don't know whether it's going to be a positive or negative, but it's likely to happen. The next thing, and this is kind of closely related, this is still biological, is you're going to have a close integration between the digital and the biological. So digital and biological."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I definitely think this is going to happen. I don't know whether it's going to be a positive or negative, but it's likely to happen. The next thing, and this is kind of closely related, this is still biological, is you're going to have a close integration between the digital and the biological. So digital and biological. And once again, I won't make any statement of whether this is a good or a bad thing, but it seems like it's just an extrapolation of the direction we're already going in, digital and biological integration. So already, you're getting more and more in your handheld devices. Imagine when your screen is now no longer in your palm, but it's being projected onto your retina from some little thing."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So digital and biological. And once again, I won't make any statement of whether this is a good or a bad thing, but it seems like it's just an extrapolation of the direction we're already going in, digital and biological integration. So already, you're getting more and more in your handheld devices. Imagine when your screen is now no longer in your palm, but it's being projected onto your retina from some little thing. And then eventually you have a direct connection with your retina, and your brain can directly access areas of memory through some biological and digital interface. So I definitely think this is going to happen. This is a big deal, because this is starting to, and I think it's already happening with a lot of what you see around technology, it's really going to reshape what the individual human experience is going to be."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Imagine when your screen is now no longer in your palm, but it's being projected onto your retina from some little thing. And then eventually you have a direct connection with your retina, and your brain can directly access areas of memory through some biological and digital interface. So I definitely think this is going to happen. This is a big deal, because this is starting to, and I think it's already happening with a lot of what you see around technology, it's really going to reshape what the individual human experience is going to be. We already see people kind of living in virtual realities and playing these immersive games and spending all of their time on social networking platforms. What happens when they're literally plugged in all the time, when almost their cell phone is in the brain? I don't know."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is a big deal, because this is starting to, and I think it's already happening with a lot of what you see around technology, it's really going to reshape what the individual human experience is going to be. We already see people kind of living in virtual realities and playing these immersive games and spending all of their time on social networking platforms. What happens when they're literally plugged in all the time, when almost their cell phone is in the brain? I don't know. I'm not going to comment whether it's a good or a bad thing, but it does look like a trend that's going to keep on going through the next 50 years. Now, if we take that even further, we're talking about a digital and biological integration, but if you go at the extreme, and actually it's probably both of these top two things combined in some way, is that we are learning, and once again, not making a comment whether it's good or bad, it's just an extrapolation of what we're already seeing, we are seeing more and more ability to understand our genome, to molecularly target things, to manipulate actual biology. And so what you have is that you can actually have, let me write this, manipulation of biology."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't know. I'm not going to comment whether it's a good or a bad thing, but it does look like a trend that's going to keep on going through the next 50 years. Now, if we take that even further, we're talking about a digital and biological integration, but if you go at the extreme, and actually it's probably both of these top two things combined in some way, is that we are learning, and once again, not making a comment whether it's good or bad, it's just an extrapolation of what we're already seeing, we are seeing more and more ability to understand our genome, to molecularly target things, to manipulate actual biology. And so what you have is that you can actually have, let me write this, manipulation of biology. And why this is on some level creepy, it could be creepy or it could be exciting, depending on how it all plays out, but it could do things like augment intelligence, augment human intelligence, which would, if you think about all the progress of society and all the things that are already accelerating, imagine how society will change if intelligence itself is augmented. As someone with a limited intelligence, I can't even imagine what will happen as soon as you do this. And obviously, the more you augment intelligence, the more that you can learn how to augment intelligence and increase lifespan and do digital and biological integration even more."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so what you have is that you can actually have, let me write this, manipulation of biology. And why this is on some level creepy, it could be creepy or it could be exciting, depending on how it all plays out, but it could do things like augment intelligence, augment human intelligence, which would, if you think about all the progress of society and all the things that are already accelerating, imagine how society will change if intelligence itself is augmented. As someone with a limited intelligence, I can't even imagine what will happen as soon as you do this. And obviously, the more you augment intelligence, the more that you can learn how to augment intelligence and increase lifespan and do digital and biological integration even more. So these things I see as some form of a trend. We'll see how it all plays out, and hopefully it plays out in the feel-good care bear version versus some type of crazy society and we all turn into the Borg in some way. Now the other trend, and these are the things that just popped into my brain today when I pressed record, but the other trend that I think is interesting is some of the things that we've taken for granted in terms of how nations interact with each other, and nation states in particular interact with each other."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And obviously, the more you augment intelligence, the more that you can learn how to augment intelligence and increase lifespan and do digital and biological integration even more. So these things I see as some form of a trend. We'll see how it all plays out, and hopefully it plays out in the feel-good care bear version versus some type of crazy society and we all turn into the Borg in some way. Now the other trend, and these are the things that just popped into my brain today when I pressed record, but the other trend that I think is interesting is some of the things that we've taken for granted in terms of how nations interact with each other, and nation states in particular interact with each other. And just as a kind of an overview, so there's a notion called a nation state. And in everyday language, the word nation and the word state almost means the same thing, but they kind of mean different things if you want to be a little bit more formal about it. That's why people use the word nation state."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now the other trend, and these are the things that just popped into my brain today when I pressed record, but the other trend that I think is interesting is some of the things that we've taken for granted in terms of how nations interact with each other, and nation states in particular interact with each other. And just as a kind of an overview, so there's a notion called a nation state. And in everyday language, the word nation and the word state almost means the same thing, but they kind of mean different things if you want to be a little bit more formal about it. That's why people use the word nation state. That's like, to a lot of people, it doesn't mean saying like state, state. The difference between a nation and a state is these are a group of people that feel some type of common identity. It could be a language, it could be a culture, it could be a value system."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's why people use the word nation state. That's like, to a lot of people, it doesn't mean saying like state, state. The difference between a nation and a state is these are a group of people that feel some type of common identity. It could be a language, it could be a culture, it could be a value system. So this is some kind of common identity. And it often is somehow associated with geography, but it does not have to be associated with geography. It could even be a religion or it could be whatever else."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It could be a language, it could be a culture, it could be a value system. So this is some kind of common identity. And it often is somehow associated with geography, but it does not have to be associated with geography. It could even be a religion or it could be whatever else. That's what a nation is. A state is a formal governance structure that makes the laws and has the institutions to make society function. Now, a nation state is what most of us live in today because it both has an identity and some type of formal institutions."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It could even be a religion or it could be whatever else. That's what a nation is. A state is a formal governance structure that makes the laws and has the institutions to make society function. Now, a nation state is what most of us live in today because it both has an identity and some type of formal institutions. So a very pure nation state would be someplace like Japan where there's relatively uniform in terms of ethnicity and religion and in terms of culture. And you have that same group of people are governing themselves. In a place like the United States, ethnicity, religion, that's diverse, but what gives identity is a notion of shared values and a notion of maybe a common history or whatever else or a certain kind of world view."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, a nation state is what most of us live in today because it both has an identity and some type of formal institutions. So a very pure nation state would be someplace like Japan where there's relatively uniform in terms of ethnicity and religion and in terms of culture. And you have that same group of people are governing themselves. In a place like the United States, ethnicity, religion, that's diverse, but what gives identity is a notion of shared values and a notion of maybe a common history or whatever else or a certain kind of world view. And obviously there's a formal state structure. Now, what I think is going to be interesting here, and I actually have no idea how all of this is going to play out, but when you see things like some of the revolutions in the Middle East due to things like people being able to communicate irrespective of the traditional media, I think there's going to be some interesting questions on what happens to the nation state, especially nation states that are able to secure their power by kind of having a bottleneck on access to information. And all of that is, I think, going to change in a very dramatic way as you have more and more integration between people, cross-border communications, when people realize that the people on the other side of the border really aren't that different than themselves."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In a place like the United States, ethnicity, religion, that's diverse, but what gives identity is a notion of shared values and a notion of maybe a common history or whatever else or a certain kind of world view. And obviously there's a formal state structure. Now, what I think is going to be interesting here, and I actually have no idea how all of this is going to play out, but when you see things like some of the revolutions in the Middle East due to things like people being able to communicate irrespective of the traditional media, I think there's going to be some interesting questions on what happens to the nation state, especially nation states that are able to secure their power by kind of having a bottleneck on access to information. And all of that is, I think, going to change in a very dramatic way as you have more and more integration between people, cross-border communications, when people realize that the people on the other side of the border really aren't that different than themselves. And another interesting thing, even the notion of democracy, and once again, I don't know how it's going to play out, but all notions of representational democracy that we have today are somewhat based on geography. They're somewhat based on geography, and that's because when the major representational democracies came about, that was the best way to represent each other. Hey, let me pick some representatives from our county or from our region, and they'll go elect other representatives, and they'll go to the national government."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And all of that is, I think, going to change in a very dramatic way as you have more and more integration between people, cross-border communications, when people realize that the people on the other side of the border really aren't that different than themselves. And another interesting thing, even the notion of democracy, and once again, I don't know how it's going to play out, but all notions of representational democracy that we have today are somewhat based on geography. They're somewhat based on geography, and that's because when the major representational democracies came about, that was the best way to represent each other. Hey, let me pick some representatives from our county or from our region, and they'll go elect other representatives, and they'll go to the national government. But now that we have this instantaneous communication with people, you might be able to have different types of a representational democracy, or maybe you could even have more direct democracies. Who knows? Because of things like communication and technology and whatever else."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Hey, let me pick some representatives from our county or from our region, and they'll go elect other representatives, and they'll go to the national government. But now that we have this instantaneous communication with people, you might be able to have different types of a representational democracy, or maybe you could even have more direct democracies. Who knows? Because of things like communication and technology and whatever else. And then the other way that I think nation-states, or the way they fundamentally interact is going to change is actually in things like warfare. And once again, already seeing this trend. In particular, I think developed countries are not going to have humans on the front line."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because of things like communication and technology and whatever else. And then the other way that I think nation-states, or the way they fundamentally interact is going to change is actually in things like warfare. And once again, already seeing this trend. In particular, I think developed countries are not going to have humans on the front line. No humans on the front line. And depending on your point of view, this could be a very good thing, or it could be a kind of a scary thing. Because if you have no humans on the front line, and you're already seeing things like this with drones, predator drones, and you see these kind of robot bomb detectors and things like that."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In particular, I think developed countries are not going to have humans on the front line. No humans on the front line. And depending on your point of view, this could be a very good thing, or it could be a kind of a scary thing. Because if you have no humans on the front line, and you're already seeing things like this with drones, predator drones, and you see these kind of robot bomb detectors and things like that. And it doesn't even have to be these big things. There's already a DARPA-funded project to work on miniature insects that could be used as some form of reconnaissance. Or you could imagine eventually they could have these little things on them that could knock someone out, or who knows what they do."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because if you have no humans on the front line, and you're already seeing things like this with drones, predator drones, and you see these kind of robot bomb detectors and things like that. And it doesn't even have to be these big things. There's already a DARPA-funded project to work on miniature insects that could be used as some form of reconnaissance. Or you could imagine eventually they could have these little things on them that could knock someone out, or who knows what they do. The exciting thing is that all of a sudden a human won't be there to get shot, and so hopefully military casualties would go down. The scary thing here is if you don't have humans on the front line, nations might be willing to enter into war, especially developed nations. They might not take it as seriously, and so it might be something that they do a little bit more when they're in the mood."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or you could imagine eventually they could have these little things on them that could knock someone out, or who knows what they do. The exciting thing is that all of a sudden a human won't be there to get shot, and so hopefully military casualties would go down. The scary thing here is if you don't have humans on the front line, nations might be willing to enter into war, especially developed nations. They might not take it as seriously, and so it might be something that they do a little bit more when they're in the mood. And it would actually create a huge disparity between developed and developing nations when this happens. And you already see that to a certain degree. A developed nation, they don't have to put as many humans in harm's way, and it's purely driven by their wealth to have capital that can go, these robots and these drones, and their technological innovation, while in the developing countries they actually would have to use human beings."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They might not take it as seriously, and so it might be something that they do a little bit more when they're in the mood. And it would actually create a huge disparity between developed and developing nations when this happens. And you already see that to a certain degree. A developed nation, they don't have to put as many humans in harm's way, and it's purely driven by their wealth to have capital that can go, these robots and these drones, and their technological innovation, while in the developing countries they actually would have to use human beings. So their costs would be much, much, much, much higher. So it's an interesting question, and once again, who knows how this plays out, whether it's a good or bad thing in the long term. I think a similar thing with this is, I think you're going to see more and more non-lethal weapons, which once again, it's very similar to this."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "A developed nation, they don't have to put as many humans in harm's way, and it's purely driven by their wealth to have capital that can go, these robots and these drones, and their technological innovation, while in the developing countries they actually would have to use human beings. So their costs would be much, much, much, much higher. So it's an interesting question, and once again, who knows how this plays out, whether it's a good or bad thing in the long term. I think a similar thing with this is, I think you're going to see more and more non-lethal weapons, which once again, it's very similar to this. It sounds like it's a good thing. If there's a gun that instead of having to kill someone, it incapacitates them in some way, or it stuns them in some way, kind of the classic Star Trek, put your phasers on, stun things. I guess the scarier version of non-lethal weapons is the threshold for using it becomes much lower."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I think a similar thing with this is, I think you're going to see more and more non-lethal weapons, which once again, it's very similar to this. It sounds like it's a good thing. If there's a gun that instead of having to kill someone, it incapacitates them in some way, or it stuns them in some way, kind of the classic Star Trek, put your phasers on, stun things. I guess the scarier version of non-lethal weapons is the threshold for using it becomes much lower. So if a government wants to subject its citizens or subject another group of citizens, it can literally just stun them, or it can make them pass out, or it can control them in some way. So this could be a little bit scarier. So who knows how all of this plays out."}, {"video_title": "Random predictions for 2060 Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I guess the scarier version of non-lethal weapons is the threshold for using it becomes much lower. So if a government wants to subject its citizens or subject another group of citizens, it can literally just stun them, or it can make them pass out, or it can control them in some way. So this could be a little bit scarier. So who knows how all of this plays out. So those are my predictions or things to think about over the next 50 years. These are just the things that happened to jump into my brain today, probably based on some of the science fiction books I've been reading or whatever else. But take them with a huge grain of salt."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Collective learners. And to understand that, let's think about how even our closest relative in the animal kingdom, the chimpanzee, might communicate. So you might have one chimpanzee, and over the course of his or her lifetime, they're able to learn a bunch of cool experiences, and they're even able to learn to use tools, manipulate tools, and who knows, maybe even make tools, maybe even get a twig someplace and take off the leaves and then use that to go get ants out of a hole or whatever else. So they're able to learn all of this stuff over a lifetime. Now, unfortunate for chimpanzees, well, what is fortunate for chimpanzees is they do teach some of these things that they've learned to other members of their group, often their offspring, but what's unfortunate for chimpanzees is they don't have a great way to communicate with each other. So for most chimpanzees, the way that they're able to teach is essentially by kind of showing, not showing and telling, just showing. And so because this is such a unprecise or not exact and such an inefficient way of communication, they're really, all of the nuances of what this chimpanzee might be able to accumulate over his or her lifetime aren't able to be conveyed to the next generation or to the other chimpanzees around."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they're able to learn all of this stuff over a lifetime. Now, unfortunate for chimpanzees, well, what is fortunate for chimpanzees is they do teach some of these things that they've learned to other members of their group, often their offspring, but what's unfortunate for chimpanzees is they don't have a great way to communicate with each other. So for most chimpanzees, the way that they're able to teach is essentially by kind of showing, not showing and telling, just showing. And so because this is such a unprecise or not exact and such an inefficient way of communication, they're really, all of the nuances of what this chimpanzee might be able to accumulate over his or her lifetime aren't able to be conveyed to the next generation or to the other chimpanzees around. So you have tremendous energy loss. And in particular, all that can be conveyed are maybe the specific movements or what you might be able to kind of observe in the present, all of the other things that maybe the chimpanzee is learning about the times of year where this is appropriate. And maybe they can convey some of that by showing them at the right times of year."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so because this is such a unprecise or not exact and such an inefficient way of communication, they're really, all of the nuances of what this chimpanzee might be able to accumulate over his or her lifetime aren't able to be conveyed to the next generation or to the other chimpanzees around. So you have tremendous energy loss. And in particular, all that can be conveyed are maybe the specific movements or what you might be able to kind of observe in the present, all of the other things that maybe the chimpanzee is learning about the times of year where this is appropriate. And maybe they can convey some of that by showing them at the right times of year. But other nuanced aspects of it or particular ways to hold something or twist something can only be shown, it can't be described in a very precise way. So you have all of this loss of experience, this loss of information. And then over the course of these animals' life, they may be able to learn the same amount again."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And maybe they can convey some of that by showing them at the right times of year. But other nuanced aspects of it or particular ways to hold something or twist something can only be shown, it can't be described in a very precise way. So you have all of this loss of experience, this loss of information. And then over the course of these animals' life, they may be able to learn the same amount again. They're able to learn maybe the same amount again. But then when they need to communicate it, they have the exact same problem. It's hard to communicate it with what they have at their disposal, which is really just showing the other chimpanzees what they've done."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then over the course of these animals' life, they may be able to learn the same amount again. They're able to learn maybe the same amount again. But then when they need to communicate it, they have the exact same problem. It's hard to communicate it with what they have at their disposal, which is really just showing the other chimpanzees what they've done. And so once again, you have a loss of information. And what you have in this type of circumstances, generation after generation, even though there is learning over the course of an individual chimpanzee's life, and even though they can communicate to some of that to each other, that form of communication is so, it loses so much information and so much nuance that you never have an overall accumulation of knowledge and wisdom in this chimpanzee population. Now, humans, on the other hand, have something called symbolic language."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's hard to communicate it with what they have at their disposal, which is really just showing the other chimpanzees what they've done. And so once again, you have a loss of information. And what you have in this type of circumstances, generation after generation, even though there is learning over the course of an individual chimpanzee's life, and even though they can communicate to some of that to each other, that form of communication is so, it loses so much information and so much nuance that you never have an overall accumulation of knowledge and wisdom in this chimpanzee population. Now, humans, on the other hand, have something called symbolic language. And I'll talk about this in a second. But for now, it's safe to say that human language is far more precise and far more efficient than just being able to show someone something. Imagine if you had to learn how to do something without being able to communicate verbally, if you just had to look at someone else's actions."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, humans, on the other hand, have something called symbolic language. And I'll talk about this in a second. But for now, it's safe to say that human language is far more precise and far more efficient than just being able to show someone something. Imagine if you had to learn how to do something without being able to communicate verbally, if you just had to look at someone else's actions. And then you'd have a good idea of how difficult it is for chimpanzees to teach each other. But in the case of human beings, we have this thing called symbolic language that's a very precise, a very efficient way of communicating. So from one human being to another, you can actually communicate a good deal, maybe not every single nuance and every single experience, but a good chunk of it."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Imagine if you had to learn how to do something without being able to communicate verbally, if you just had to look at someone else's actions. And then you'd have a good idea of how difficult it is for chimpanzees to teach each other. But in the case of human beings, we have this thing called symbolic language that's a very precise, a very efficient way of communicating. So from one human being to another, you can actually communicate a good deal, maybe not every single nuance and every single experience, but a good chunk of it. So right here, I'm drawing about that much of it to the next, to some other human being. Maybe it is the offspring, maybe it is another member of the tribe or the group, whatever it is. And then this human being might come up with some other innovations."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So from one human being to another, you can actually communicate a good deal, maybe not every single nuance and every single experience, but a good chunk of it. So right here, I'm drawing about that much of it to the next, to some other human being. Maybe it is the offspring, maybe it is another member of the tribe or the group, whatever it is. And then this human being might come up with some other innovations. They're able to build off of all of this learning from that previous generation or from that other human being that's around, and they're able to come up with their own nuances and their own innovations. And this one right over here might come up with his or her own nuances and innovations. And because they have a good communication mechanism, this one could even communicate to that one what he's learned or what she's learned and communicate a good chunk of that."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then this human being might come up with some other innovations. They're able to build off of all of this learning from that previous generation or from that other human being that's around, and they're able to come up with their own nuances and their own innovations. And this one right over here might come up with his or her own nuances and innovations. And because they have a good communication mechanism, this one could even communicate to that one what he's learned or what she's learned and communicate a good chunk of that. Maybe not all of it, but maybe a reasonable bit. They can describe exactly how they do something, the times of years, when it's good to do it, when it's not good to do it, how to plan for the future, what's the history of this new learning. And so what you have going on here is because of this strong communication mechanism, so strong, precise, efficient communication, what you have is a human group, or eventually a human civilization, is able to have a collective memory."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And because they have a good communication mechanism, this one could even communicate to that one what he's learned or what she's learned and communicate a good chunk of that. Maybe not all of it, but maybe a reasonable bit. They can describe exactly how they do something, the times of years, when it's good to do it, when it's not good to do it, how to plan for the future, what's the history of this new learning. And so what you have going on here is because of this strong communication mechanism, so strong, precise, efficient communication, what you have is a human group, or eventually a human civilization, is able to have a collective memory. In the case of the chimpanzees, they're every generation, every chimpanzee is having to relearn the things that the other chimpanzees might have already done in previous generations. They're not able to really move forward or build on those in significant ways. In humans, as information is learned and experience gained, a good bit of that is able to be passed on to other humans."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so what you have going on here is because of this strong communication mechanism, so strong, precise, efficient communication, what you have is a human group, or eventually a human civilization, is able to have a collective memory. In the case of the chimpanzees, they're every generation, every chimpanzee is having to relearn the things that the other chimpanzees might have already done in previous generations. They're not able to really move forward or build on those in significant ways. In humans, as information is learned and experience gained, a good bit of that is able to be passed on to other humans. And so this might be passed on. So all of this might be passed on, or a good chunk of this could be passed on to the next generation. And I'm not even talking about written language yet."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In humans, as information is learned and experience gained, a good bit of that is able to be passed on to other humans. And so this might be passed on. So all of this might be passed on, or a good chunk of this could be passed on to the next generation. And I'm not even talking about written language yet. This could still just be oral communication, which is still a very strong, precise, efficient means of communication. Written communication takes it to another level. But then this person over here, maybe she comes up with other innovations."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm not even talking about written language yet. This could still just be oral communication, which is still a very strong, precise, efficient means of communication. Written communication takes it to another level. But then this person over here, maybe she comes up with other innovations. And at some point you might say, well look, if everyone keeps having innovations and they keep learning what everyone learned in previous generations, maybe this will tap out the total amount of memory that a human being even has. And there's actually a case that maybe this is one of the reasons why humans even have larger memories, because there is all of this collective knowledge to gain from one generation to another, from one human being to another. But there are some limits to this."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But then this person over here, maybe she comes up with other innovations. And at some point you might say, well look, if everyone keeps having innovations and they keep learning what everyone learned in previous generations, maybe this will tap out the total amount of memory that a human being even has. And there's actually a case that maybe this is one of the reasons why humans even have larger memories, because there is all of this collective knowledge to gain from one generation to another, from one human being to another. But there are some limits to this. And this is the other element where this collective aspect of collective memory and collective learning becomes really powerful. A human being, because of the strong communication mechanism is not just limited to the knowledge and the experience in their memory. They are able to tap into, so this human being right over here does not have this skill set."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But there are some limits to this. And this is the other element where this collective aspect of collective memory and collective learning becomes really powerful. A human being, because of the strong communication mechanism is not just limited to the knowledge and the experience in their memory. They are able to tap into, so this human being right over here does not have this skill set. And that skill set maybe gets passed on to another human being. So let me copy and paste that. So let's say you copy, let's say you paste that."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They are able to tap into, so this human being right over here does not have this skill set. And that skill set maybe gets passed on to another human being. So let me copy and paste that. So let's say you copy, let's say you paste that. This other human being that's maybe living at the same time, and when that becomes relevant, when that becomes relevant, they could actually tap into it. And maybe they could learn it from that human being, or maybe it's in a different part of society and this human being can build certain tools or build certain things using this information, using that knowledge right over there. And then this human being doesn't need to know that information."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say you copy, let's say you paste that. This other human being that's maybe living at the same time, and when that becomes relevant, when that becomes relevant, they could actually tap into it. And maybe they could learn it from that human being, or maybe it's in a different part of society and this human being can build certain tools or build certain things using this information, using that knowledge right over there. And then this human being doesn't need to know that information. They can just leverage the output of that information to then build on top of it. So what it allows human beings to do is not only convey information and build on information from generation to generation, human to human, it allows all of the human brains collectively at any given point of time to be one collective memory bank that can be used to develop or innovate in specific domains and adapt to specific parts of the ecosystem or to teach each other. So all of a sudden, this is really unique as far as we can tell in the animal kingdom."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then this human being doesn't need to know that information. They can just leverage the output of that information to then build on top of it. So what it allows human beings to do is not only convey information and build on information from generation to generation, human to human, it allows all of the human brains collectively at any given point of time to be one collective memory bank that can be used to develop or innovate in specific domains and adapt to specific parts of the ecosystem or to teach each other. So all of a sudden, this is really unique as far as we can tell in the animal kingdom. All of a sudden, it's not all about the brain or the memory of one individual member of a species. It now becomes about the brain or the memory of the entire civilization or the entire group of the species. And just as an example of that, there's, as far as I know, there's no human being who knows how to do everything that all human beings know how to do."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So all of a sudden, this is really unique as far as we can tell in the animal kingdom. All of a sudden, it's not all about the brain or the memory of one individual member of a species. It now becomes about the brain or the memory of the entire civilization or the entire group of the species. And just as an example of that, there's, as far as I know, there's no human being who knows how to do everything that all human beings know how to do. I could imagine that there is a chimpanzee that knows how to do everything that any other chimpanzee knows how to do. There are no humans that can be a fighter pilot, a doctor, a gymnast, a lawyer, understands philosophy, speaks 20 different languages. As far as I know, that human being does not exist."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just as an example of that, there's, as far as I know, there's no human being who knows how to do everything that all human beings know how to do. I could imagine that there is a chimpanzee that knows how to do everything that any other chimpanzee knows how to do. There are no humans that can be a fighter pilot, a doctor, a gymnast, a lawyer, understands philosophy, speaks 20 different languages. As far as I know, that human being does not exist. And that's okay because they can tap into the experiences, the abilities of other human beings to build up their civilization. None of us, as far as we know, knows how to do everything that we need to actually build our civilization. But the information is in our collective memory to actually do it."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As far as I know, that human being does not exist. And that's okay because they can tap into the experiences, the abilities of other human beings to build up their civilization. None of us, as far as we know, knows how to do everything that we need to actually build our civilization. But the information is in our collective memory to actually do it. Now, the next thing you might say is, okay, I started with this premise that we have a strong, precise, efficient means of communication and that other animals don't. But don't other animals actually have some form of language? So for example, don't some, you know, for example, even monkeys, when they screech, when they are in danger, that's a form of communication, maybe a form of language."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the information is in our collective memory to actually do it. Now, the next thing you might say is, okay, I started with this premise that we have a strong, precise, efficient means of communication and that other animals don't. But don't other animals actually have some form of language? So for example, don't some, you know, for example, even monkeys, when they screech, when they are in danger, that's a form of communication, maybe a form of language. Maybe certain animals, birds, monkeys, maybe they have a song that they sing that can convey certain things. Maybe it's when they are looking for a mate. Isn't that a form of communication?"}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So for example, don't some, you know, for example, even monkeys, when they screech, when they are in danger, that's a form of communication, maybe a form of language. Maybe certain animals, birds, monkeys, maybe they have a song that they sing that can convey certain things. Maybe it's when they are looking for a mate. Isn't that a form of communication? And these are, these are a form of communication and a form of language. But these don't really come in play in terms of the teaching learning. You don't see one chimpanzee making screeching sounds or learning sounds."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Isn't that a form of communication? And these are, these are a form of communication and a form of language. But these don't really come in play in terms of the teaching learning. You don't see one chimpanzee making screeching sounds or learning sounds. They might do a little bit just to warn, maybe as a warning, but there's no deep nuance or deep precision that's being able to convey by these one-off sounds or even one-off gestures. And what's particularly powerful about human language is that it is a symbolic language. It is a symbolic language."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You don't see one chimpanzee making screeching sounds or learning sounds. They might do a little bit just to warn, maybe as a warning, but there's no deep nuance or deep precision that's being able to convey by these one-off sounds or even one-off gestures. And what's particularly powerful about human language is that it is a symbolic language. It is a symbolic language. And when I say it's a symbolic language, I'm even saying it in a broader sense than even just written symbols. I'm talking about even the sounds themselves. So let's go to a time where we did not even have writings."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is a symbolic language. And when I say it's a symbolic language, I'm even saying it in a broader sense than even just written symbols. I'm talking about even the sounds themselves. So let's go to a time where we did not even have writings. And when we talk about symbolic languages, let's think about a non-symbolic language. So in a non-symbolic language, you might have some sound, let's call it sound one, and it has some meaning. Let's call it meaning one."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's go to a time where we did not even have writings. And when we talk about symbolic languages, let's think about a non-symbolic language. So in a non-symbolic language, you might have some sound, let's call it sound one, and it has some meaning. Let's call it meaning one. Meaning one. So this might be a certain type of scream. It means that a predator is approaching."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's call it meaning one. Meaning one. So this might be a certain type of scream. It means that a predator is approaching. Then you might have something like a sound two or gesture two, and then it has some other meaning. It has meaning two. It might be a certain type of song, which means that I am in the mood to reproduce or whatever else."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It means that a predator is approaching. Then you might have something like a sound two or gesture two, and then it has some other meaning. It has meaning two. It might be a certain type of song, which means that I am in the mood to reproduce or whatever else. You might have gesture three. Gesture three. That has some direct meaning."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It might be a certain type of song, which means that I am in the mood to reproduce or whatever else. You might have gesture three. Gesture three. That has some direct meaning. It might mean that I have found food or something like that. So meaning three. What humans have, they can do this, where particular sounds have particular meanings."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That has some direct meaning. It might mean that I have found food or something like that. So meaning three. What humans have, they can do this, where particular sounds have particular meanings. So for example, in humans, you could have sound one. It refers to meaning one. I'll just refer it to meaning one."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What humans have, they can do this, where particular sounds have particular meanings. So for example, in humans, you could have sound one. It refers to meaning one. I'll just refer it to meaning one. You could have sound two that refers to meaning two. You could have sound three that refers to meaning three. So these are just direct representations, but what is really powerful about symbolic languages is that these oral symbols can be combined according to set rules or grammars to have an infinite number of meanings."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll just refer it to meaning one. You could have sound two that refers to meaning two. You could have sound three that refers to meaning three. So these are just direct representations, but what is really powerful about symbolic languages is that these oral symbols can be combined according to set rules or grammars to have an infinite number of meanings. So this is what really makes human language transcend other languages and really makes it this robust, precise communication mechanism is you could have combinations. Sound one, sound two, sound three will now have another meaning, meaning four. Then you could maybe have a combination where you have sound three, sound one, and sound two might have meaning five."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So these are just direct representations, but what is really powerful about symbolic languages is that these oral symbols can be combined according to set rules or grammars to have an infinite number of meanings. So this is what really makes human language transcend other languages and really makes it this robust, precise communication mechanism is you could have combinations. Sound one, sound two, sound three will now have another meaning, meaning four. Then you could maybe have a combination where you have sound three, sound one, and sound two might have meaning five. And if you have tens of thousands of sounds, when really our oral words are those sounds in a given language, then all of a sudden you can have infinite meanings by putting them in different combinations. And if you think this is a little bit abstract, imagine that sound one is the sound me saying the word dog. And I'm not even gonna write it down because I wanna imagine a world even before written communication."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Then you could maybe have a combination where you have sound three, sound one, and sound two might have meaning five. And if you have tens of thousands of sounds, when really our oral words are those sounds in a given language, then all of a sudden you can have infinite meanings by putting them in different combinations. And if you think this is a little bit abstract, imagine that sound one is the sound me saying the word dog. And I'm not even gonna write it down because I wanna imagine a world even before written communication. So sound one is the sound dog. Sound two is the sound eats. And sound three is the sound man."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'm not even gonna write it down because I wanna imagine a world even before written communication. So sound one is the sound dog. Sound two is the sound eats. And sound three is the sound man. So literally, sound one, if you heard dog, you'd think, okay, I'd visualize a dog of some type. And even there, you'd have some visualization of a dog, and we all have one maybe. Sound two, if you heard eats, you'd say, okay, I imagine eating in some way."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And sound three is the sound man. So literally, sound one, if you heard dog, you'd think, okay, I'd visualize a dog of some type. And even there, you'd have some visualization of a dog, and we all have one maybe. Sound two, if you heard eats, you'd say, okay, I imagine eating in some way. And sound three, man, you have some visualization of it. And if it was a non-symbolic language, that's all you could get out of those three sounds. But now in a symbolic language, we can combine those."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Sound two, if you heard eats, you'd say, okay, I imagine eating in some way. And sound three, man, you have some visualization of it. And if it was a non-symbolic language, that's all you could get out of those three sounds. But now in a symbolic language, we can combine those. We could say dog eats man. So once again, we just reused the three sounds, the three symbols, but now they're referring to a whole new, a much more complex meaning than just referring to certain objects or certain actions. Or you could have man eats dog."}, {"video_title": "Collective learning Life on earth and in the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But now in a symbolic language, we can combine those. We could say dog eats man. So once again, we just reused the three sounds, the three symbols, but now they're referring to a whole new, a much more complex meaning than just referring to certain objects or certain actions. Or you could have man eats dog. And it's not pleasant, but I guess in a desperate situation. But once again, it is another meaning that we can get out of the same sounds. And what these symbolic languages do, besides giving you an infinite number of meanings, they're allowed to give you more nuance and really refer to things that are abstract, including and maybe most importantly, things like the present, the future, the past, kind of hypothetical things that really are necessary in order to really communicate or optimally communicate all of the experiences or the learnings from one entity to another."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And angle strain is the increase in energy that's associated with a bond angle that deviates from the ideal bond angle of 109.5 degrees, and this number should sound familiar to you. This was the bond angle for a carbon of tetrahedral geometry. So if you go through and you analyze these, the bond angle in here for this triangle must be 60 degrees, and 60 degrees is a long ways off from 109.5 degrees, meaning cyclopropane has significant angle strain. For cyclobutane, this angle would be 90 degrees, and 90 degrees is still a ways off from 109.5 degrees, so cyclobutane also has a large amount of angle strain, although not as much as cyclopropane. For cyclopentane, this bond angle is 108 degrees, and 108 degrees is pretty close to 109.5 degrees. Closer than it would be for cyclohexane, this bond angle is 120 degrees. And so the theory was cyclopentane is the most stable out of the cycloalkanes because this bond angle is closest to 109.5 degrees."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "For cyclobutane, this angle would be 90 degrees, and 90 degrees is still a ways off from 109.5 degrees, so cyclobutane also has a large amount of angle strain, although not as much as cyclopropane. For cyclopentane, this bond angle is 108 degrees, and 108 degrees is pretty close to 109.5 degrees. Closer than it would be for cyclohexane, this bond angle is 120 degrees. And so the theory was cyclopentane is the most stable out of the cycloalkanes because this bond angle is closest to 109.5 degrees. However, that conclusion doesn't hold up if you look at the heats of combustion of the cycloalkanes. And first we'll start with cyclopropane. So if you define the heat of combustion as the negative change in the enthalpy, cyclopropane gives off 2,091 kilojoules for every one mole of cyclopropane that is combusted."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And so the theory was cyclopentane is the most stable out of the cycloalkanes because this bond angle is closest to 109.5 degrees. However, that conclusion doesn't hold up if you look at the heats of combustion of the cycloalkanes. And first we'll start with cyclopropane. So if you define the heat of combustion as the negative change in the enthalpy, cyclopropane gives off 2,091 kilojoules for every one mole of cyclopropane that is combusted. And if you count the number of CH2 groups on cyclopropane, let's go back up here and let's count them up. So here's one, two, and three on the drawing, so that's why there's a three here. Now you can't analyze the cycloalkanes in terms of just the heats of combustion, so if we look at those, we can see that they increase, 2,091 to 2,721 to 3,291, and then 3,920."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if you define the heat of combustion as the negative change in the enthalpy, cyclopropane gives off 2,091 kilojoules for every one mole of cyclopropane that is combusted. And if you count the number of CH2 groups on cyclopropane, let's go back up here and let's count them up. So here's one, two, and three on the drawing, so that's why there's a three here. Now you can't analyze the cycloalkanes in terms of just the heats of combustion, so if we look at those, we can see that they increase, 2,091 to 2,721 to 3,291, and then 3,920. But that's what we expect to happen because as we go from cyclopropane to cyclobutane, pentane, and hexane, we're increasing in number of carbons. And we already know from the earlier video on heats of combustion, if you increase the amount of carbons that you have, you would expect an increase in the heats of combustion. So you can't really compare the cycloalkanes directly in terms of just the heats of combustion."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "Now you can't analyze the cycloalkanes in terms of just the heats of combustion, so if we look at those, we can see that they increase, 2,091 to 2,721 to 3,291, and then 3,920. But that's what we expect to happen because as we go from cyclopropane to cyclobutane, pentane, and hexane, we're increasing in number of carbons. And we already know from the earlier video on heats of combustion, if you increase the amount of carbons that you have, you would expect an increase in the heats of combustion. So you can't really compare the cycloalkanes directly in terms of just the heats of combustion. You have to compare them in terms of their heats of combustion divided by the number of CH2 groups. And that gives you a better idea of the stability. So if you take 2,091, which is the heat of combustion of cyclopropane, and divide that by the number of CH2 groups, which is three, you get approximately 697."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So you can't really compare the cycloalkanes directly in terms of just the heats of combustion. You have to compare them in terms of their heats of combustion divided by the number of CH2 groups. And that gives you a better idea of the stability. So if you take 2,091, which is the heat of combustion of cyclopropane, and divide that by the number of CH2 groups, which is three, you get approximately 697. So again, this is the heat of combustion divided by the number of CH2 groups in kilojoules per mole. And this is a much better way to compare the stability of the cycloalkanes. Notice cyclopropane has the highest value here, 697."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if you take 2,091, which is the heat of combustion of cyclopropane, and divide that by the number of CH2 groups, which is three, you get approximately 697. So again, this is the heat of combustion divided by the number of CH2 groups in kilojoules per mole. And this is a much better way to compare the stability of the cycloalkanes. Notice cyclopropane has the highest value here, 697. Cyclobutane goes down to 680. Cyclopentane is 658. And cyclohexane is approximately 653."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "Notice cyclopropane has the highest value here, 697. Cyclobutane goes down to 680. Cyclopentane is 658. And cyclohexane is approximately 653. If you remember back to the video on heats of combustion, this number here, 653 kilojoules per mole, is approximately the same value we got for a straight-chain alkane when you added on a CH2 group. So each additional CH2 group increased the heat of combustion by approximately 653 or 654 kilojoules per mole. And that tells us that cyclohexane is pretty much strain-free."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And cyclohexane is approximately 653. If you remember back to the video on heats of combustion, this number here, 653 kilojoules per mole, is approximately the same value we got for a straight-chain alkane when you added on a CH2 group. So each additional CH2 group increased the heat of combustion by approximately 653 or 654 kilojoules per mole. And that tells us that cyclohexane is pretty much strain-free. Cyclohexane is about as stable as an open-chain alkane. And so we know that this idea of cyclohexane being flat must not be true. So cyclohexane isn't flat, as we'll see in later videos."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And that tells us that cyclohexane is pretty much strain-free. Cyclohexane is about as stable as an open-chain alkane. And so we know that this idea of cyclohexane being flat must not be true. So cyclohexane isn't flat, as we'll see in later videos. So cyclohexane is the most stable out of these cycloalkanes. Cyclopentane is a little bit higher in energy and therefore a little bit more unstable. And cyclobutane even higher than that."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So cyclohexane isn't flat, as we'll see in later videos. So cyclohexane is the most stable out of these cycloalkanes. Cyclopentane is a little bit higher in energy and therefore a little bit more unstable. And cyclobutane even higher than that. And finally, cyclopropane at 697 for a heat of combustion per CH2 group. This is the most unstable. This is the highest heat of combustion."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And cyclobutane even higher than that. And finally, cyclopropane at 697 for a heat of combustion per CH2 group. This is the most unstable. This is the highest heat of combustion. This is the highest in energy. And so let's analyze why cyclopropane has such a relatively high heat of combustion per CH2 group. Here we have a model of the cyclopropane molecule."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "This is the highest heat of combustion. This is the highest in energy. And so let's analyze why cyclopropane has such a relatively high heat of combustion per CH2 group. Here we have a model of the cyclopropane molecule. And if I turn it to the side, you can see that all three carbon atoms are in the same plane. So cyclopropane is planar. You can also see that these bonds are bent."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "Here we have a model of the cyclopropane molecule. And if I turn it to the side, you can see that all three carbon atoms are in the same plane. So cyclopropane is planar. You can also see that these bonds are bent. The significant angle strain means the orbitals don't overlap very well, which leads to these bent bonds. You can see the plastic is even bending in the model set. There's another source of strain associated with cyclopropane."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "You can also see that these bonds are bent. The significant angle strain means the orbitals don't overlap very well, which leads to these bent bonds. You can see the plastic is even bending in the model set. There's another source of strain associated with cyclopropane. And we can see it if we look down one of the carbon-carbon bonds. So the front hydrogens are eclipsing the hydrogens in the back. And cyclopropane is locked into this eclipse conformation."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "There's another source of strain associated with cyclopropane. And we can see it if we look down one of the carbon-carbon bonds. So the front hydrogens are eclipsing the hydrogens in the back. And cyclopropane is locked into this eclipse conformation. So all this increased strain means that a three-membered ring is very reactive and highly susceptible to ring-opening reactions. So I'm gonna take the ring here. I'm gonna break it and open up the ring so we can see that decreased the strain."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And cyclopropane is locked into this eclipse conformation. So all this increased strain means that a three-membered ring is very reactive and highly susceptible to ring-opening reactions. So I'm gonna take the ring here. I'm gonna break it and open up the ring so we can see that decreased the strain. Those bonds even look straight now. Here we have the cyclobutane molecule. And you can see there is some angle strain here, although not as much as in cyclopropane."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "I'm gonna break it and open up the ring so we can see that decreased the strain. Those bonds even look straight now. Here we have the cyclobutane molecule. And you can see there is some angle strain here, although not as much as in cyclopropane. If we turn it to the side, you can also see some torsional strain. The hydrogens in the front are eclipsing the hydrogens in the back. To relieve this torsional strain, cyclobutane can adopt a non-planar conformation."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And you can see there is some angle strain here, although not as much as in cyclopropane. If we turn it to the side, you can also see some torsional strain. The hydrogens in the front are eclipsing the hydrogens in the back. To relieve this torsional strain, cyclobutane can adopt a non-planar conformation. This is called the Puckard conformation. And if you turn it to the side here, so you're staring down one of the carbon-carbon bonds, you can see how that's relieved some of the torsional strain. So the hydrogens in front are no longer eclipsing the hydrogens in the back."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "To relieve this torsional strain, cyclobutane can adopt a non-planar conformation. This is called the Puckard conformation. And if you turn it to the side here, so you're staring down one of the carbon-carbon bonds, you can see how that's relieved some of the torsional strain. So the hydrogens in front are no longer eclipsing the hydrogens in the back. And finally we have the cyclopentane molecule, which has much less angle strain than cyclopropane or cyclobutane. But if you turn it to the side, you can see that the planar conformation is destabilized by torsional strain. So we have some eclipsed hydrogens there."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So the hydrogens in front are no longer eclipsing the hydrogens in the back. And finally we have the cyclopentane molecule, which has much less angle strain than cyclopropane or cyclobutane. But if you turn it to the side, you can see that the planar conformation is destabilized by torsional strain. So we have some eclipsed hydrogens there. Some of that torsional strain can be relieved in a non-planar conformation. So one of the non-planar conformations will be to rotate the carbon up like that, and that's called the envelope conformation. And you can see that four carbons are in the same plane."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So we have some eclipsed hydrogens there. Some of that torsional strain can be relieved in a non-planar conformation. So one of the non-planar conformations will be to rotate the carbon up like that, and that's called the envelope conformation. And you can see that four carbons are in the same plane. So this carbon right here, this one, the one in the back, and this one in the back are in the same plane. And this fifth one here is up out of the plane. So this looks a little bit like an envelope, and so that's why it's called the envelope conformation."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "And you can see that four carbons are in the same plane. So this carbon right here, this one, the one in the back, and this one in the back are in the same plane. And this fifth one here is up out of the plane. So this looks a little bit like an envelope, and so that's why it's called the envelope conformation. Now that we understand the stability of cycloalkanes, let's do a quick problem. On the left we have ethyl cyclopropane. On the right we have methyl cyclobutane."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So this looks a little bit like an envelope, and so that's why it's called the envelope conformation. Now that we understand the stability of cycloalkanes, let's do a quick problem. On the left we have ethyl cyclopropane. On the right we have methyl cyclobutane. They're isomers of each other. They both have the molecular formula C5H10. The first question is, which isomer is more stable?"}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "On the right we have methyl cyclobutane. They're isomers of each other. They both have the molecular formula C5H10. The first question is, which isomer is more stable? Well, we're comparing a three-membered ring to a four-membered ring. And we know that cyclopropane is higher energy. There's more strain associated with it."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "The first question is, which isomer is more stable? Well, we're comparing a three-membered ring to a four-membered ring. And we know that cyclopropane is higher energy. There's more strain associated with it. So ethyl cyclopropane must be the less stable isomer. So this one is less stable, which makes methyl cyclobutane the more stable isomer. So there's not as much strain in methyl cyclobutane."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "There's more strain associated with it. So ethyl cyclopropane must be the less stable isomer. So this one is less stable, which makes methyl cyclobutane the more stable isomer. So there's not as much strain in methyl cyclobutane. So we've answered our first question. Our second question is, which one has the higher heat of combustion? Well, we know that the more stable compound has a lower heat of combustion."}, {"video_title": "Stability of cycloalkanes Organic chemistry Khan Academy.mp3", "Sentence": "So there's not as much strain in methyl cyclobutane. So we've answered our first question. Our second question is, which one has the higher heat of combustion? Well, we know that the more stable compound has a lower heat of combustion. We know that from the heat of combustion video. That must mean methyl cyclobutane has the lower heat of combustion because this one is more stable, which means that ethyl cyclopropane must have the higher heat of combustion. So the higher heat of combustion."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say I have something like this. And hopefully you're reasonably familiar with what this represents. And I'll review it a little bit. So let's say I have something like this. Let's say that that is the molecular structure right there. And so the first question you should be asking is, how many carbons are there there? And some of you might say, wait, how is that even a molecule?"}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say I have something like this. Let's say that that is the molecular structure right there. And so the first question you should be asking is, how many carbons are there there? And some of you might say, wait, how is that even a molecule? And just as a review, the endpoint of every line represents a carbon. So that's a carbon, that's a carbon, that's a carbon, that's a carbon, that's a carbon, that's a carbon. And we have one, two, three, four, five, six carbons."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And some of you might say, wait, how is that even a molecule? And just as a review, the endpoint of every line represents a carbon. So that's a carbon, that's a carbon, that's a carbon, that's a carbon, that's a carbon, that's a carbon. And we have one, two, three, four, five, six carbons. And we have no double bonds. So if you have all of that information, you're ready to name this molecule. And before I actually name it, let me just kind of give you all of the different prefixes."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And we have one, two, three, four, five, six carbons. And we have no double bonds. So if you have all of that information, you're ready to name this molecule. And before I actually name it, let me just kind of give you all of the different prefixes. So if you have one carbon, the prefix is meth. If you have two carbons, the prefix is eth. And it's good to memorize at least up to about 10."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And before I actually name it, let me just kind of give you all of the different prefixes. So if you have one carbon, the prefix is meth. If you have two carbons, the prefix is eth. And it's good to memorize at least up to about 10. And actually, it kind of repeats after that. If you have three carbons, the prefix is prop. Prop like propane."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And it's good to memorize at least up to about 10. And actually, it kind of repeats after that. If you have three carbons, the prefix is prop. Prop like propane. And you've heard of ethane and methane. So you'll see all of that soon. Four, you're talking about bute."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Prop like propane. And you've heard of ethane and methane. So you'll see all of that soon. Four, you're talking about bute. Five, and after five, it kind of becomes the traditional prefixes that we associate with a lot of these numbers. So at five, it's pent like pentagon. Six, it's hex like hexagon."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Four, you're talking about bute. Five, and after five, it kind of becomes the traditional prefixes that we associate with a lot of these numbers. So at five, it's pent like pentagon. Six, it's hex like hexagon. Seven is hept. Eight is oct like octagon. Nine is non."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Six, it's hex like hexagon. Seven is hept. Eight is oct like octagon. Nine is non. Ten is dec. And then after that, it kind of starts to have a pattern here. You're not going to really deal with things much beyond the teens. But I'll just write them down here just out of interest."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Nine is non. Ten is dec. And then after that, it kind of starts to have a pattern here. You're not going to really deal with things much beyond the teens. But I'll just write them down here just out of interest. 11 is un-dec. And for those of you who know French, I'm not one of them, but I know that one in French is un or in Spanish uno. So it's kind of one and 10, 11. 12 is do-dec. Do or dos if you're speaking Spanish."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "But I'll just write them down here just out of interest. 11 is un-dec. And for those of you who know French, I'm not one of them, but I know that one in French is un or in Spanish uno. So it's kind of one and 10, 11. 12 is do-dec. Do or dos if you're speaking Spanish. For two, dec for 10. Two and 10, that's 12. 13, you could imagine what it's going to be."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "12 is do-dec. Do or dos if you're speaking Spanish. For two, dec for 10. Two and 10, that's 12. 13, you could imagine what it's going to be. It is tri-dec. 14 is tetra-dec. A tetrapod is something with four legs. And then after that, it becomes very systematic. At 15 is pentadec."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "13, you could imagine what it's going to be. It is tri-dec. 14 is tetra-dec. A tetrapod is something with four legs. And then after that, it becomes very systematic. At 15 is pentadec. Notice pent, pent, five and 10. 16 is hexadec. 17 is heptadec."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "At 15 is pentadec. Notice pent, pent, five and 10. 16 is hexadec. 17 is heptadec. So it just goes on and on and on. I don't think I have to go away. It's hexadec, heptadec."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "17 is heptadec. So it just goes on and on and on. I don't think I have to go away. It's hexadec, heptadec. Octadec is 18. 19 is nonadec. And then 20 is actually iso."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "It's hexadec, heptadec. Octadec is 18. 19 is nonadec. And then 20 is actually iso. But we won't even go into that. This will probably serve our purposes. I mean, I could go up to 16 is hexadec."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And then 20 is actually iso. But we won't even go into that. This will probably serve our purposes. I mean, I could go up to 16 is hexadec. So this is just how many carbons are in our longest chain. What I drew here is just one chain. So we can immediately try to name it."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "I mean, I could go up to 16 is hexadec. So this is just how many carbons are in our longest chain. What I drew here is just one chain. So we can immediately try to name it. How many carbons do we have here? Well, we have one, two, three, four, five, six carbons. So we'll be dealing with hex as a prefix."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So we can immediately try to name it. How many carbons do we have here? Well, we have one, two, three, four, five, six carbons. So we'll be dealing with hex as a prefix. And then to get, I guess, the postfix on this prefix, or kind of the root, you look to see if there are any double bonds here. And there are no double bonds here. And if we have no double bonds in this carbon chain, we're dealing with an alkane."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So we'll be dealing with hex as a prefix. And then to get, I guess, the postfix on this prefix, or kind of the root, you look to see if there are any double bonds here. And there are no double bonds here. And if we have no double bonds in this carbon chain, we're dealing with an alkane. This is called an alkane, which is a general term for all of the chains of carbons that have no double bonds on them, or no triple bonds, all single bonds. So in this situation, you take hex for 6. So this is hex."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And if we have no double bonds in this carbon chain, we're dealing with an alkane. This is called an alkane, which is a general term for all of the chains of carbons that have no double bonds on them, or no triple bonds, all single bonds. So in this situation, you take hex for 6. So this is hex. And then, because it's an alkane, it gets the ane from alkane. So this is hexane. Let's do another one."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So this is hex. And then, because it's an alkane, it gets the ane from alkane. So this is hexane. Let's do another one. Let's say I have this thing right here. I'll draw, let's make it even longer. So let's say I have that thing right there."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Let's do another one. Let's say I have this thing right here. I'll draw, let's make it even longer. So let's say I have that thing right there. So how many carbons do we have? We have 1, 2, 3, 4, 5, 6, 7 carbons. They're all single bonds."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say I have that thing right there. So how many carbons do we have? We have 1, 2, 3, 4, 5, 6, 7 carbons. They're all single bonds. So it's an alkane. So this will be 7 carbons. It is heptane, because we have all single bonds."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "They're all single bonds. So it's an alkane. So this will be 7 carbons. It is heptane, because we have all single bonds. Now, if things form a chain, or if things form a ring, I should say, we put the prefix cyclo in front of it. So if I have, let me show you what I'm talking about. So if I just have 5 carbons, 1, 2, 3, 4, 5, OK, so that's 5 right there."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "It is heptane, because we have all single bonds. Now, if things form a chain, or if things form a ring, I should say, we put the prefix cyclo in front of it. So if I have, let me show you what I'm talking about. So if I just have 5 carbons, 1, 2, 3, 4, 5, OK, so that's 5 right there. I have 1, 2, 3, 4, 5 carbons in a chain. If I just have 5 carbons in a chain like this, this would be pentane. But if I have 5 carbons, then they form a ring."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if I just have 5 carbons, 1, 2, 3, 4, 5, OK, so that's 5 right there. I have 1, 2, 3, 4, 5 carbons in a chain. If I just have 5 carbons in a chain like this, this would be pentane. But if I have 5 carbons, then they form a ring. So let me draw it. So it's 1, 2, 3, 4, 5 carbons, and it forms a ring. Let me make the drawing a little bit better."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "But if I have 5 carbons, then they form a ring. So let me draw it. So it's 1, 2, 3, 4, 5 carbons, and it forms a ring. Let me make the drawing a little bit better. So really, I'm just drawing a pentagon. But notice, this has 5 carbons on it. I could draw the carbons here."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Let me make the drawing a little bit better. So really, I'm just drawing a pentagon. But notice, this has 5 carbons on it. I could draw the carbons here. Carbon, carbon, carbon, carbon, carbon. And just as a review, what you don't see is the hydrogens they're bonded to. Each of these guys have 2 bonds."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "I could draw the carbons here. Carbon, carbon, carbon, carbon, carbon. And just as a review, what you don't see is the hydrogens they're bonded to. Each of these guys have 2 bonds. So they must have 2 bonds with something else, and those are going to be with hydrogens. And I'll draw it here just as a bit of a review. But you notice, very quickly, the drawing gets extremely messy when you draw the 2 hydrogens on each of these carbons."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Each of these guys have 2 bonds. So they must have 2 bonds with something else, and those are going to be with hydrogens. And I'll draw it here just as a bit of a review. But you notice, very quickly, the drawing gets extremely messy when you draw the 2 hydrogens on each of these carbons. So it's a little bit over. Maybe I shouldn't be doing that. But there you go."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "But you notice, very quickly, the drawing gets extremely messy when you draw the 2 hydrogens on each of these carbons. So it's a little bit over. Maybe I shouldn't be doing that. But there you go. So it becomes very messy when you draw the hydrogens. So it's better to just assume that they're there. If we don't draw all 4 bonds of the carbon, the other 2 bonds are going to be with hydrogen."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "But there you go. So it becomes very messy when you draw the hydrogens. So it's better to just assume that they're there. If we don't draw all 4 bonds of the carbon, the other 2 bonds are going to be with hydrogen. So here, you might say, OK, this is an alkane, because I don't have any double bonds here. All of these are single bonds with the carbon. I have 5 carbons."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "If we don't draw all 4 bonds of the carbon, the other 2 bonds are going to be with hydrogen. So here, you might say, OK, this is an alkane, because I don't have any double bonds here. All of these are single bonds with the carbon. I have 5 carbons. So you might say this is pentane. But you have to think about one more thing. It's in a ring."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "I have 5 carbons. So you might say this is pentane. But you have to think about one more thing. It's in a ring. So we add the prefix cyclo to it. So this is, because it's a ring, we write cyclopentane. So let me just break that apart."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "It's in a ring. So we add the prefix cyclo to it. So this is, because it's a ring, we write cyclopentane. So let me just break that apart. This tells us that we're dealing with a ring. You see that this is a ring right there. This tells us that we're dealing with 5 carbons."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So let me just break that apart. This tells us that we're dealing with a ring. You see that this is a ring right there. This tells us that we're dealing with 5 carbons. And then this tells us, right here, the ane part, that tells us that they are all single bonds. All carbon-carbon single bonds. No double or triple bonds."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "This tells us that we're dealing with 5 carbons. And then this tells us, right here, the ane part, that tells us that they are all single bonds. All carbon-carbon single bonds. No double or triple bonds. All single bonds. So let's go the other way. Let's start with the word, and let's see if we can figure out what the actual structure would look like."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "No double or triple bonds. All single bonds. So let's go the other way. Let's start with the word, and let's see if we can figure out what the actual structure would look like. So let's say I have cyclonane. So what is this telling me? This tells me I'm dealing with a ring."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Let's start with the word, and let's see if we can figure out what the actual structure would look like. So let's say I have cyclonane. So what is this telling me? This tells me I'm dealing with a ring. That is a ring. It's going to have a ring structure. It's going to have 9 carbons, 9 C's."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "This tells me I'm dealing with a ring. That is a ring. It's going to have a ring structure. It's going to have 9 carbons, 9 C's. And then it's an alkane. So they're all going to be single bonds. So if I want to draw it, I want to draw 9 carbons in a ring."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to have 9 carbons, 9 C's. And then it's an alkane. So they're all going to be single bonds. So if I want to draw it, I want to draw 9 carbons in a ring. It's not a trivial thing to draw. Let me try my best. Let's see."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So if I want to draw it, I want to draw 9 carbons in a ring. It's not a trivial thing to draw. Let me try my best. Let's see. That's 1, 2, 3, 4, 5, 6, 7, 8. Let's see. Let me draw it."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Let's see. That's 1, 2, 3, 4, 5, 6, 7, 8. Let's see. Let me draw it. Let me try a better shot at it. Let's see, I have 1, 2, 3, 4, 5, 6, 7, 8, and then 9. And then you can connect the last two."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw it. Let me try a better shot at it. Let's see, I have 1, 2, 3, 4, 5, 6, 7, 8, and then 9. And then you can connect the last two. So let me make sure that this is. And obviously I could have drawn it better than that, but hopefully you can see all the edges here. So I have 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And then you can connect the last two. So let me make sure that this is. And obviously I could have drawn it better than that, but hopefully you can see all the edges here. So I have 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. So that is, it's in a cycle, it's in a ring. I have 9 carbons, and they're all single bonded. So this is cyclononane, although there's probably better ways to draw that ring right there."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So I have 1, 2, 3, 4, 5, 6, 7, 8, 9 carbons. So that is, it's in a cycle, it's in a ring. I have 9 carbons, and they're all single bonded. So this is cyclononane, although there's probably better ways to draw that ring right there. So if someone were to tell you octane, and that word might feel familiar to you from the gas station. They are literally talking about the molecule octane. And now you know, or at least you have a sense of what they're talking about."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "So this is cyclononane, although there's probably better ways to draw that ring right there. So if someone were to tell you octane, and that word might feel familiar to you from the gas station. They are literally talking about the molecule octane. And now you know, or at least you have a sense of what they're talking about. The oct tells you that you have 8 carbons. There's no cyclo in front of it, so it's not a cycle, it's just going to be a chain. And then the ane part tells you that they're all single bonds."}, {"video_title": "Naming simple alkanes Organic chemistry Khan Academy.mp3", "Sentence": "And now you know, or at least you have a sense of what they're talking about. The oct tells you that you have 8 carbons. There's no cyclo in front of it, so it's not a cycle, it's just going to be a chain. And then the ane part tells you that they're all single bonds. So it's just going to be 8 carbons in a chain. 1, 2, 3, 4, 5, 6, 7, 8. It's just going to look like that."}, {"video_title": "Quasar correction Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I want to make a quick correction to the video on quasars. In that video, I said, and I mistakenly said, that the Cretan disk that's really forming or releasing the energy of the quasar, that's releasing energy predominantly in the X-ray part of the electromagnetic spectrum. And that was incorrect. Most quasars are actually emitting electromagnetic radiation across the spectrum, all the way from X-rays. As high frequency as X-rays, all the way down to infrared. And some quasars even release super high frequency gamma rays, and they'll release low frequency electromagnetic waves all the way down to radio waves. So I just wanted to make that correction."}, {"video_title": "Quasar correction Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Most quasars are actually emitting electromagnetic radiation across the spectrum, all the way from X-rays. As high frequency as X-rays, all the way down to infrared. And some quasars even release super high frequency gamma rays, and they'll release low frequency electromagnetic waves all the way down to radio waves. So I just wanted to make that correction. It's not predominantly in the X-ray part of the spectrum. It's across the spectrum right over here. It's this entire range of the spectrum, and sometimes even a wider range."}, {"video_title": "Quasar correction Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I just wanted to make that correction. It's not predominantly in the X-ray part of the spectrum. It's across the spectrum right over here. It's this entire range of the spectrum, and sometimes even a wider range. Now, the other thing I want to clarify is this is the range of the spectrum that's being emitted. But we have to remember that most, or actually all of these quasars are quite far away. The closest is 780 million light years away."}, {"video_title": "Quasar correction Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's this entire range of the spectrum, and sometimes even a wider range. Now, the other thing I want to clarify is this is the range of the spectrum that's being emitted. But we have to remember that most, or actually all of these quasars are quite far away. The closest is 780 million light years away. Many of them are many, many billions of light years away. And so they're moving away from us at a very fast speed. Or they're getting redshifted because the universe is expanding so fast relative to us at that point."}, {"video_title": "Quasar correction Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The closest is 780 million light years away. Many of them are many, many billions of light years away. And so they're moving away from us at a very fast speed. Or they're getting redshifted because the universe is expanding so fast relative to us at that point. Or that coordinate is moving so fast away from our coordinate. And so even though this is the spectrum that's being emitted, it's all going to be redshifted. It's all going to be redshifted down."}, {"video_title": "Quasar correction Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Or they're getting redshifted because the universe is expanding so fast relative to us at that point. Or that coordinate is moving so fast away from our coordinate. And so even though this is the spectrum that's being emitted, it's all going to be redshifted. It's all going to be redshifted down. And so we are going to observe things at a much lower frequency, maybe around the radio part of the frequency. So everything will be redshifted down. And that's why these were originally called quasi-stellar radio sources."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Alkanes all have the general molecular formula of CnH2n plus 2, where n is the number of carbons that are in your molecule. For example, if you have one carbon in your molecule, you plug one into your formula there, and you would get C1. And you would get 1 times 2 plus 2, which is, of course, 4. So CH4, which is methane. And we've seen the dot structure for methane many times. So carbon with four hydrogens around it like that. So methane is the simplest alkane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So CH4, which is methane. And we've seen the dot structure for methane many times. So carbon with four hydrogens around it like that. So methane is the simplest alkane. What about two carbons? We'll plug it into your formula. That would be C2."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So methane is the simplest alkane. What about two carbons? We'll plug it into your formula. That would be C2. And that would be 2 times 2, which is 4, plus 2, which is 6. So C2H6 is the molecular formula for a two-carbon alkane, which we call ethane. So this would be the dot structure for ethane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "That would be C2. And that would be 2 times 2, which is 4, plus 2, which is 6. So C2H6 is the molecular formula for a two-carbon alkane, which we call ethane. So this would be the dot structure for ethane. Carbon bonded to another carbon with six hydrogens around it like that. So there are millions upon millions of organic compounds. And you have to have some sort of logical way to name them."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So this would be the dot structure for ethane. Carbon bonded to another carbon with six hydrogens around it like that. So there are millions upon millions of organic compounds. And you have to have some sort of logical way to name them. So one carbon is methane. Two carbons is ethane. And these names were determined by what's called the IUPAC nomenclature system."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And you have to have some sort of logical way to name them. So one carbon is methane. Two carbons is ethane. And these names were determined by what's called the IUPAC nomenclature system. IUPAC stands for International Union of Pure and Applied Chemistry, which was a bunch of chemists getting together and saying, we're going to name all of these molecules in a very systematic way. So for the rest of this course, we're going to focus on IUPAC nomenclature for naming organic molecules. And I've kind of summarized IUPAC nomenclature up in this little table here."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And these names were determined by what's called the IUPAC nomenclature system. IUPAC stands for International Union of Pure and Applied Chemistry, which was a bunch of chemists getting together and saying, we're going to name all of these molecules in a very systematic way. So for the rest of this course, we're going to focus on IUPAC nomenclature for naming organic molecules. And I've kind of summarized IUPAC nomenclature up in this little table here. If you have one carbon, your parent name is meth. And if you're working with an alkane, your ending is anane. So therefore, a one-carbon alkane is called methane, as we have already seen."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And I've kind of summarized IUPAC nomenclature up in this little table here. If you have one carbon, your parent name is meth. And if you're working with an alkane, your ending is anane. So therefore, a one-carbon alkane is called methane, as we have already seen. A two-carbon alkane, the root is eth. And so that would be ethane. And so of course, your ending would change depending on what functional group you're working with."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So therefore, a one-carbon alkane is called methane, as we have already seen. A two-carbon alkane, the root is eth. And so that would be ethane. And so of course, your ending would change depending on what functional group you're working with. But here, we're just working with alkanes. Three carbons, the root is prop. So if I were to draw a three-carbon alkane like that, that would be the dot structure for propane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And so of course, your ending would change depending on what functional group you're working with. But here, we're just working with alkanes. Three carbons, the root is prop. So if I were to draw a three-carbon alkane like that, that would be the dot structure for propane. Four carbons is bute. So if I were to draw a four-carbon alkane like that, that would be butane. And of course, five carbons would be pentane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So if I were to draw a three-carbon alkane like that, that would be the dot structure for propane. Four carbons is bute. So if I were to draw a four-carbon alkane like that, that would be butane. And of course, five carbons would be pentane. So I could go ahead and draw pentane. And we could keep going here. But I think you get the idea."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And of course, five carbons would be pentane. So I could go ahead and draw pentane. And we could keep going here. But I think you get the idea. These are all straight-chain alkanes, meaning it's just one line of carbons, one carbon right after the other. Six carbons would be hex or hexane. Seven carbons, hept or heptane, eight octane, nine nonane, 10 is decane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "But I think you get the idea. These are all straight-chain alkanes, meaning it's just one line of carbons, one carbon right after the other. Six carbons would be hex or hexane. Seven carbons, hept or heptane, eight octane, nine nonane, 10 is decane. And we can see the rest of them here. Undec for 11 carbons, dodec, tridec, tetradec, pentadec, and icos for 20. So this holds true not only for alkanes, but for other functional groups."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Seven carbons, hept or heptane, eight octane, nine nonane, 10 is decane. And we can see the rest of them here. Undec for 11 carbons, dodec, tridec, tetradec, pentadec, and icos for 20. So this holds true not only for alkanes, but for other functional groups. So it's important to memorize all of these parent names here. So so far, we've talked about straight-chain alkanes. What about if you get a branched-chain alkane like this?"}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So this holds true not only for alkanes, but for other functional groups. So it's important to memorize all of these parent names here. So so far, we've talked about straight-chain alkanes. What about if you get a branched-chain alkane like this? So you can see it's no longer just one carbon, one after another in a straight chain. You can see there's something coming off of that carbon chain. So first, let's find the longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "What about if you get a branched-chain alkane like this? So you can see it's no longer just one carbon, one after another in a straight chain. You can see there's something coming off of that carbon chain. So first, let's find the longest carbon chain. So this is a skill that you have to develop when you're doing IUPAC nomenclature. So if I start over here, I find the longest carbon chain. And if I go like this, you can see the carbon chain that I'm going for like that."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So first, let's find the longest carbon chain. So this is a skill that you have to develop when you're doing IUPAC nomenclature. So if I start over here, I find the longest carbon chain. And if I go like this, you can see the carbon chain that I'm going for like that. How many carbons are there in that chain? Well, there was one right here, and then two, and then three, and then four, and then five. So there are five carbons in this chain."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And if I go like this, you can see the carbon chain that I'm going for like that. How many carbons are there in that chain? Well, there was one right here, and then two, and then three, and then four, and then five. So there are five carbons in this chain. And when we go back up here to our IUPAC nomenclature table, we see that five carbons will have pent, and it will be pentane. So the parent name will be pentane for this example. So for this molecule, you're going to name this pentane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So there are five carbons in this chain. And when we go back up here to our IUPAC nomenclature table, we see that five carbons will have pent, and it will be pentane. So the parent name will be pentane for this example. So for this molecule, you're going to name this pentane. And what is coming off of my straight chain alkane? I have something coming off of it right here. I have a one carbon CH3 group branching off of my pentane molecule."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So for this molecule, you're going to name this pentane. And what is coming off of my straight chain alkane? I have something coming off of it right here. I have a one carbon CH3 group branching off of my pentane molecule. This is called a substituent. So a substituent is something coming off of your parent chain. It's a group or groups that is connected to your parent chain, so a group connected to your parent chain."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "I have a one carbon CH3 group branching off of my pentane molecule. This is called a substituent. So a substituent is something coming off of your parent chain. It's a group or groups that is connected to your parent chain, so a group connected to your parent chain. So how are we going to name this substituent? Well, it's one carbon, and this is what's called an alkyl group. So an alkyl group."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "It's a group or groups that is connected to your parent chain, so a group connected to your parent chain. So how are we going to name this substituent? Well, it's one carbon, and this is what's called an alkyl group. So an alkyl group. And notice it's not an alkane. It's an alkyl. So the ending for an alkyl group will be Yl."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So an alkyl group. And notice it's not an alkane. It's an alkyl. So the ending for an alkyl group will be Yl. But we're still going to use our parent name to name alkyl groups. So if there's one carbon, I go back up here to my IUPAC table here, and I say, well, one carbon in organic chemistry has the parent name of meth. And this is an alkyl group, which has a Yl ending."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So the ending for an alkyl group will be Yl. But we're still going to use our parent name to name alkyl groups. So if there's one carbon, I go back up here to my IUPAC table here, and I say, well, one carbon in organic chemistry has the parent name of meth. And this is an alkyl group, which has a Yl ending. So I have meth plus Yl. So this is called a methyl group, which we've said several times already in these videos. So we have a methyl group attached to my parent chain here, which is called pentane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And this is an alkyl group, which has a Yl ending. So I have meth plus Yl. So this is called a methyl group, which we've said several times already in these videos. So we have a methyl group attached to my parent chain here, which is called pentane. So there's a methyl group coming off of pentane in the second position. We will get into details about IUPAC nomenclature in the next video. So for right now, let's just be able to identify substituents coming off of our parent chain."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So we have a methyl group attached to my parent chain here, which is called pentane. So there's a methyl group coming off of pentane in the second position. We will get into details about IUPAC nomenclature in the next video. So for right now, let's just be able to identify substituents coming off of our parent chain. And the first step when you're doing IUPAC nomenclature is always to find the longest carbon chain possible. So for this molecule, it is a 5-carbon chain. Let's look at a more challenging example."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So for right now, let's just be able to identify substituents coming off of our parent chain. And the first step when you're doing IUPAC nomenclature is always to find the longest carbon chain possible. So for this molecule, it is a 5-carbon chain. Let's look at a more challenging example. Here are three dot structures for the exact same molecule. And let's see if we can find some carbon chains for this molecule here. So let's find the longest carbon chain we possibly can."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at a more challenging example. Here are three dot structures for the exact same molecule. And let's see if we can find some carbon chains for this molecule here. So let's find the longest carbon chain we possibly can. So I'm going to start with the top left one. And I'm going to look at this portion of the molecule here. And I'm going to try to find my longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So let's find the longest carbon chain we possibly can. So I'm going to start with the top left one. And I'm going to look at this portion of the molecule here. And I'm going to try to find my longest carbon chain. So that might be my first guess here. So how many carbons in that chain? Well, this would be 1, 2, 3, 4, 5, 6, and 7."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And I'm going to try to find my longest carbon chain. So that might be my first guess here. So how many carbons in that chain? Well, this would be 1, 2, 3, 4, 5, 6, and 7. So if you were to name this as a parent chain, this would be heptane, since I have seven carbons on it like that. What about the substituents coming off of heptane? Well, I have a 1-carbon alkyl group coming off of the second position."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Well, this would be 1, 2, 3, 4, 5, 6, and 7. So if you were to name this as a parent chain, this would be heptane, since I have seven carbons on it like that. What about the substituents coming off of heptane? Well, I have a 1-carbon alkyl group coming off of the second position. So 1 carbon we've seen. That would be called a methyl group. I have a 2-carbon substituent coming off of the third carbon."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Well, I have a 1-carbon alkyl group coming off of the second position. So 1 carbon we've seen. That would be called a methyl group. I have a 2-carbon substituent coming off of the third carbon. So let's go back up here and refresh our memory. What would be the parent name for 2 carbons for organic chemistry? So 2 carbons would be eth."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "I have a 2-carbon substituent coming off of the third carbon. So let's go back up here and refresh our memory. What would be the parent name for 2 carbons for organic chemistry? So 2 carbons would be eth. And since this is an alkyl group, it would be called an ethyl group. So we have an ethyl group coming off of carbon 3. So this is a methyl group right here."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So 2 carbons would be eth. And since this is an alkyl group, it would be called an ethyl group. So we have an ethyl group coming off of carbon 3. So this is a methyl group right here. And then this is an ethyl group. So we have a methyl group coming off of carbon 2. And we have an ethyl group coming off of carbon 3."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So this is a methyl group right here. And then this is an ethyl group. So we have a methyl group coming off of carbon 2. And we have an ethyl group coming off of carbon 3. And coming off of carbon 4 is yet another ethyl group like that. Let's say we chose a different way to find our longest carbon chain. Let's say we started down here."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And we have an ethyl group coming off of carbon 3. And coming off of carbon 4 is yet another ethyl group like that. Let's say we chose a different way to find our longest carbon chain. Let's say we started down here. So let's say we said, oh, that looks like that might be the longest carbon chain to me at first glance. So let's see what we have. Let's see how many carbons we have if we said this is our longest carbon chain."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we started down here. So let's say we said, oh, that looks like that might be the longest carbon chain to me at first glance. So let's see what we have. Let's see how many carbons we have if we said this is our longest carbon chain. So let's number them. Let's call this carbon 1. Let's call this carbon 2."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Let's see how many carbons we have if we said this is our longest carbon chain. So let's number them. Let's call this carbon 1. Let's call this carbon 2. 3, 4, 5, 6, and 7. So once again, this would be called heptane. What sort of substituents do we have coming off this molecule?"}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Let's call this carbon 2. 3, 4, 5, 6, and 7. So once again, this would be called heptane. What sort of substituents do we have coming off this molecule? We have a methyl group coming off of carbon 2. We have an ethyl group coming off of carbon 3. And we have another ethyl group coming off of carbon 4."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "What sort of substituents do we have coming off this molecule? We have a methyl group coming off of carbon 2. We have an ethyl group coming off of carbon 3. And we have another ethyl group coming off of carbon 4. So that's the exact same situation we had for the first example here. So these are the same thing. So it doesn't really matter which one of those you chose."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And we have another ethyl group coming off of carbon 4. So that's the exact same situation we had for the first example here. So these are the same thing. So it doesn't really matter which one of those you chose. You would be naming it the exact same name. Let's compare those two to the molecule, to the example down here. Again, it's the same molecule."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So it doesn't really matter which one of those you chose. You would be naming it the exact same name. Let's compare those two to the molecule, to the example down here. Again, it's the same molecule. But let's say you chose a different path. Let's say you chose down here. So you said, oh, this looks like it's the longest carbon chain to me."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Again, it's the same molecule. But let's say you chose a different path. Let's say you chose down here. So you said, oh, this looks like it's the longest carbon chain to me. So you go like this. And you say, all right, that's my longest carbon chain. How many carbons are in that?"}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So you said, oh, this looks like it's the longest carbon chain to me. So you go like this. And you say, all right, that's my longest carbon chain. How many carbons are in that? Well, this would be 1, 2, 3, 4, 5, 6, and 7. So what sort of substituents do we have in this molecule? Well, coming off of carbon 4, we can see there is an ethyl group."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "How many carbons are in that? Well, this would be 1, 2, 3, 4, 5, 6, and 7. So what sort of substituents do we have in this molecule? Well, coming off of carbon 4, we can see there is an ethyl group. Coming off of carbon 3, we can see this looks kind of complicated. It's not really a straight chain. This is much more complex substituent, which we'll get to naming in a future video."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "Well, coming off of carbon 4, we can see there is an ethyl group. Coming off of carbon 3, we can see this looks kind of complicated. It's not really a straight chain. This is much more complex substituent, which we'll get to naming in a future video. So for this molecule, we have a total of two substituents. For the top molecule, we have an example of three substituents. So the question is, which one of these will be the correct way to name my molecule according to IUPAC nomenclature?"}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "This is much more complex substituent, which we'll get to naming in a future video. So for this molecule, we have a total of two substituents. For the top molecule, we have an example of three substituents. So the question is, which one of these will be the correct way to name my molecule according to IUPAC nomenclature? So I have two chains of equal length. Both of these chains are 7 carbons. So how do I break that tie?"}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So the question is, which one of these will be the correct way to name my molecule according to IUPAC nomenclature? So I have two chains of equal length. Both of these chains are 7 carbons. So how do I break that tie? IUPAC rules state you choose the parent chain with the greatest number of substituents. So the top one has three substituents. The bottom example has two substituents."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So how do I break that tie? IUPAC rules state you choose the parent chain with the greatest number of substituents. So the top one has three substituents. The bottom example has two substituents. So if you were to name this molecule using IUPAC nomenclature, you would choose the top way of naming it, which again, we'll get to in more detail in the next few videos here. So let's look at cycloalkanes now. So we've just done straight chain alkanes."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "The bottom example has two substituents. So if you were to name this molecule using IUPAC nomenclature, you would choose the top way of naming it, which again, we'll get to in more detail in the next few videos here. So let's look at cycloalkanes now. So we've just done straight chain alkanes. We've looked at branch chain alkanes. Let's look at cycloalkanes. So this is a pretty funny dot structure here."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So we've just done straight chain alkanes. We've looked at branch chain alkanes. Let's look at cycloalkanes. So this is a pretty funny dot structure here. Let's see how many carbons are in this triangle. Well, of course, there's one, two, and three carbons. So if I were to draw what this molecule looks like, if I were to draw all the atoms involved, well, there would be three carbons like that."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So this is a pretty funny dot structure here. Let's see how many carbons are in this triangle. Well, of course, there's one, two, and three carbons. So if I were to draw what this molecule looks like, if I were to draw all the atoms involved, well, there would be three carbons like that. And to complete the octets around carbon, there would have to be two hydrogens on each carbon like that. So that's a cycloalkane. So it's carbon forming rings now."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So if I were to draw what this molecule looks like, if I were to draw all the atoms involved, well, there would be three carbons like that. And to complete the octets around carbon, there would have to be two hydrogens on each carbon like that. So that's a cycloalkane. So it's carbon forming rings now. So the molecular formula for this molecule, there would be three carbons and a total of six hydrogens. And we can see that the pattern for a cycloalkane would therefore have to be, if you have n carbons, you must have 2n hydrogens for a cycloalkane. This cycloalkane has three carbons."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So it's carbon forming rings now. So the molecular formula for this molecule, there would be three carbons and a total of six hydrogens. And we can see that the pattern for a cycloalkane would therefore have to be, if you have n carbons, you must have 2n hydrogens for a cycloalkane. This cycloalkane has three carbons. So we go back up here to our IUPAC nomenclature table. And we say that three carbons should be prop. And since it's an alkane, it'd be propane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "This cycloalkane has three carbons. So we go back up here to our IUPAC nomenclature table. And we say that three carbons should be prop. And since it's an alkane, it'd be propane. But it's a cycloalkane. So we would actually call this cyclopropane. So this molecule is called cyclopropane, like that."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And since it's an alkane, it'd be propane. But it's a cycloalkane. So we would actually call this cyclopropane. So this molecule is called cyclopropane, like that. The next molecule looks like a square. How many carbons are in it? There are four."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So this molecule is called cyclopropane, like that. The next molecule looks like a square. How many carbons are in it? There are four. So that would be butane. But since it's a cyclic molecule, it would be called cyclobutane. And let's do two more examples."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "There are four. So that would be butane. But since it's a cyclic molecule, it would be called cyclobutane. And let's do two more examples. So the next one's a pentagon. So that's five carbons. So it would be pentane."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "And let's do two more examples. So the next one's a pentagon. So that's five carbons. So it would be pentane. So it'd be cyclopentane. So here we have cyclopentane. And probably the most important cycloalkane would, of course, be six carbons."}, {"video_title": "Alkane and cycloalkane nomenclature I Organic chemistry Khan Academy.mp3", "Sentence": "So it would be pentane. So it'd be cyclopentane. So here we have cyclopentane. And probably the most important cycloalkane would, of course, be six carbons. And six carbons would be cyclohexane. So this guy down here would be named as cyclohexane. So that's an introduction to alkanes and cycloalkanes."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "We're going to use this screenshot to redraw these two molecules. We start with the molecule on the left. There's a carbon here, and the carbon is bonded to a hydrogen. The hydrogen's going straight up, and that bond is in the plane of the page. We draw our carbon, and we draw the hydrogen going straight up, and we show that this bond is in the plane of the page. I decided to use chlorine as being yellow, and so therefore, this bond is also in the plane of the page. I'm going to go ahead and draw in my chlorine like that."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "The hydrogen's going straight up, and that bond is in the plane of the page. We draw our carbon, and we draw the hydrogen going straight up, and we show that this bond is in the plane of the page. I decided to use chlorine as being yellow, and so therefore, this bond is also in the plane of the page. I'm going to go ahead and draw in my chlorine like that. I made bromine red, and this bromine's coming out at us in space, so we use a wedge to show the bromine coming out at us. I decided to make fluorine green, and this fluorine's going away from us in space, and so we can show that with a dash. The fluorine's going away."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and draw in my chlorine like that. I made bromine red, and this bromine's coming out at us in space, so we use a wedge to show the bromine coming out at us. I decided to make fluorine green, and this fluorine's going away from us in space, and so we can show that with a dash. The fluorine's going away. This molecule on the right, we already saw in the previous video, this molecule on the right is the mirror image to the one on the left, but you can't superimpose the molecule on the right with the one on the left. Therefore, it's a different molecule. Let's go ahead and draw it."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "The fluorine's going away. This molecule on the right, we already saw in the previous video, this molecule on the right is the mirror image to the one on the left, but you can't superimpose the molecule on the right with the one on the left. Therefore, it's a different molecule. Let's go ahead and draw it. Once again, it has a carbon in the center bonded to a hydrogen that's going up, so I can draw that in there. It's also bonded to a chlorine with the bond in the plane of the page. This time, the chlorine is going to the left."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw it. Once again, it has a carbon in the center bonded to a hydrogen that's going up, so I can draw that in there. It's also bonded to a chlorine with the bond in the plane of the page. This time, the chlorine is going to the left. The bromine is still coming out at us in space, so I draw in the bromine, and then finally, this fluorine is going away from us, so I can go ahead and draw in the fluorine going away from us in space. Let's use these images here to talk about three definitions. Let me just move down here, and let's look at these three different definitions."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "This time, the chlorine is going to the left. The bromine is still coming out at us in space, so I draw in the bromine, and then finally, this fluorine is going away from us, so I can go ahead and draw in the fluorine going away from us in space. Let's use these images here to talk about three definitions. Let me just move down here, and let's look at these three different definitions. We'll start with stereoisomers. Stereoisomers are isomers that differ in the three-dimensional arrangement of atoms. Let's think about what the word isomer means again."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Let me just move down here, and let's look at these three different definitions. We'll start with stereoisomers. Stereoisomers are isomers that differ in the three-dimensional arrangement of atoms. Let's think about what the word isomer means again. Isomer means same parts. These two different molecules are composed of the same parts. Each of these molecules contains one carbon, one hydrogen, one fluorine, one bromine, and one chlorine."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about what the word isomer means again. Isomer means same parts. These two different molecules are composed of the same parts. Each of these molecules contains one carbon, one hydrogen, one fluorine, one bromine, and one chlorine. In terms of what kind of isomer are they, we've talked about structural isomers before, or structural or constitutional isomers, but we can't classify these as being structural isomers. Let me go ahead and draw one more dot structure. This time, I'm going to leave out the stereochemistry."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Each of these molecules contains one carbon, one hydrogen, one fluorine, one bromine, and one chlorine. In terms of what kind of isomer are they, we've talked about structural isomers before, or structural or constitutional isomers, but we can't classify these as being structural isomers. Let me go ahead and draw one more dot structure. This time, I'm going to leave out the stereochemistry. I'm just going to show a carbon bonded to a hydrogen, bonded to a fluorine, bonded to a bromine, bonded to a chlorine. I've left out the stereochemistry. You could see that this dot structure I just drew could represent either of these two dot structures that has the stereochemistry shown."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "This time, I'm going to leave out the stereochemistry. I'm just going to show a carbon bonded to a hydrogen, bonded to a fluorine, bonded to a bromine, bonded to a chlorine. I've left out the stereochemistry. You could see that this dot structure I just drew could represent either of these two dot structures that has the stereochemistry shown. They're all connected in the same way. They all have a carbon directly bonded to a hydrogen, a fluorine, a bromine, and a chlorine. You can't say that these two isomers are structural isomers of each other."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "You could see that this dot structure I just drew could represent either of these two dot structures that has the stereochemistry shown. They're all connected in the same way. They all have a carbon directly bonded to a hydrogen, a fluorine, a bromine, and a chlorine. You can't say that these two isomers are structural isomers of each other. You have to say they are stereoisomers. They differ in the three-dimensional arrangement of atoms around that central carbon. These are stereoisomers."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "You can't say that these two isomers are structural isomers of each other. You have to say they are stereoisomers. They differ in the three-dimensional arrangement of atoms around that central carbon. These are stereoisomers. Our next definition is enantiomers. Enantiomers are stereoisomers that are non-superimposable mirror images. Once again, we saw in the previous video that this molecule on the right is the mirror image to the one on the left, but when we tried to superimpose the one on the right on the one on the left, we couldn't do so."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "These are stereoisomers. Our next definition is enantiomers. Enantiomers are stereoisomers that are non-superimposable mirror images. Once again, we saw in the previous video that this molecule on the right is the mirror image to the one on the left, but when we tried to superimpose the one on the right on the one on the left, we couldn't do so. They're different molecules. They are enantiomers of each other, which is Greek for opposite. Finally, our last definition here is a chiral center, or a chirality center, or a stereogenic center, or whatever term you'd want to use there."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Once again, we saw in the previous video that this molecule on the right is the mirror image to the one on the left, but when we tried to superimpose the one on the right on the one on the left, we couldn't do so. They're different molecules. They are enantiomers of each other, which is Greek for opposite. Finally, our last definition here is a chiral center, or a chirality center, or a stereogenic center, or whatever term you'd want to use there. It has a tetrahedral, it's a tetrahedral carbon, so I think that's sp3 hybridized. If I look at this carbon here, and this dot structure, this is a tetrahedral arrangement of atoms, tetrahedral geometry. It has four different groups attached to that carbon, in this case, four different atoms, so a hydrogen, a fluorine, a bromine, a chlorine."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Finally, our last definition here is a chiral center, or a chirality center, or a stereogenic center, or whatever term you'd want to use there. It has a tetrahedral, it's a tetrahedral carbon, so I think that's sp3 hybridized. If I look at this carbon here, and this dot structure, this is a tetrahedral arrangement of atoms, tetrahedral geometry. It has four different groups attached to that carbon, in this case, four different atoms, so a hydrogen, a fluorine, a bromine, a chlorine. Anytime you have this tetrahedral carbon that has four different groups attached to it, you create a chiral center. This carbon right here is a chiral center, or a chirality center. If you're starting without stereochemistry, if you start with this dot structure right here, and you identify that you have one chiral center present in this dot structure, we've just seen one chiral center means two possible stereoisomers."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "It has four different groups attached to that carbon, in this case, four different atoms, so a hydrogen, a fluorine, a bromine, a chlorine. Anytime you have this tetrahedral carbon that has four different groups attached to it, you create a chiral center. This carbon right here is a chiral center, or a chirality center. If you're starting without stereochemistry, if you start with this dot structure right here, and you identify that you have one chiral center present in this dot structure, we've just seen one chiral center means two possible stereoisomers. We have two possible stereoisomers. We could draw, we could write out a little formula here. Two to the n, where n is the number of chiral centers."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "If you're starting without stereochemistry, if you start with this dot structure right here, and you identify that you have one chiral center present in this dot structure, we've just seen one chiral center means two possible stereoisomers. We have two possible stereoisomers. We could draw, we could write out a little formula here. Two to the n, where n is the number of chiral centers. Let me go ahead and write this. This is the number of chiral centers, or chirality centers. Two to whatever power that is."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Two to the n, where n is the number of chiral centers. Let me go ahead and write this. This is the number of chiral centers, or chirality centers. Two to whatever power that is. In this case, for this dot structure, we had one chiral center. We're going to say two to the first power. This is equal to two, of course."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "Two to whatever power that is. In this case, for this dot structure, we had one chiral center. We're going to say two to the first power. This is equal to two, of course. This number tells us how many stereoisomers we have. We've already talked about that. One chiral center gives us two stereoisomers."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "This is equal to two, of course. This number tells us how many stereoisomers we have. We've already talked about that. One chiral center gives us two stereoisomers. These two stereoisomers that we drew have our non-superimposable mirror images. These are non-superimposable mirror images. These two stereoisomers."}, {"video_title": "Stereoisomers, enantiomers, and chirality centers Organic chemistry Khan Academy.mp3", "Sentence": "One chiral center gives us two stereoisomers. These two stereoisomers that we drew have our non-superimposable mirror images. These are non-superimposable mirror images. These two stereoisomers. They're a special type of stereoisomer that we call enantiomers. We'll talk much more about number of stereoisomers in a later video. In the next video, we're going to go in more detail about chiral centers and chirality centers and how to identify the number of chiral centers in a molecule."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if we start with a carboxylic acid and add thionyl chloride, we can form our acyl chloride, and we would also form a sulfur dioxide, an HCl, in this reaction. So let's look at the structure of thionyl chloride. So here we have the dot structure right here. And we could draw a resonance structure for this. So we could show these electrons in here moving off onto the oxygen. So let's go ahead and draw what we would form from that. Our oxygen would now have three lone pairs of electrons on it, giving it a negative 1 formal charge."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And we could draw a resonance structure for this. So we could show these electrons in here moving off onto the oxygen. So let's go ahead and draw what we would form from that. Our oxygen would now have three lone pairs of electrons on it, giving it a negative 1 formal charge. Our sulfur would still be bonded to these chlorines here. It would still have a lone pair of electrons, and it would get a plus 1 formal charge like that. So this is a major contributor to the overall structure."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "Our oxygen would now have three lone pairs of electrons on it, giving it a negative 1 formal charge. Our sulfur would still be bonded to these chlorines here. It would still have a lone pair of electrons, and it would get a plus 1 formal charge like that. So this is a major contributor to the overall structure. Oxygen is more electronegative than sulfur. And if you think about this pi bond in here, there's ineffective overlap of those p orbitals. And that's because sulfur and oxygen are in different periods on the periodic table."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So this is a major contributor to the overall structure. Oxygen is more electronegative than sulfur. And if you think about this pi bond in here, there's ineffective overlap of those p orbitals. And that's because sulfur and oxygen are in different periods on the periodic table. So sulfur is in the third period, so it has a larger p orbital than oxygen. Oxygen's in the second period. And so you get ineffective overlap of these orbitals here."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And that's because sulfur and oxygen are in different periods on the periodic table. So sulfur is in the third period, so it has a larger p orbital than oxygen. Oxygen's in the second period. And so you get ineffective overlap of these orbitals here. And so that's another reason why this is going to contribute to the overall structure. Plus you have these chlorines here withdrawing some electron density from the sulfur. So chlorine is more electronegative than the sulfur."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And so you get ineffective overlap of these orbitals here. And so that's another reason why this is going to contribute to the overall structure. Plus you have these chlorines here withdrawing some electron density from the sulfur. So chlorine is more electronegative than the sulfur. So the end result of all this is going to make this sulfur very electrophilic right here. And so therefore, our carboxylic acid is able to act as a nucleophile. So if these electrons move into here, these electrons can attack our sulfur."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So chlorine is more electronegative than the sulfur. So the end result of all this is going to make this sulfur very electrophilic right here. And so therefore, our carboxylic acid is able to act as a nucleophile. So if these electrons move into here, these electrons can attack our sulfur. So the nucleophile attacks our electrophile. And then these electrons kick off onto the oxygen. So let's go ahead and show that."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if these electrons move into here, these electrons can attack our sulfur. So the nucleophile attacks our electrophile. And then these electrons kick off onto the oxygen. So let's go ahead and show that. We'd have our R group bonded to our carbon. And then that's bonded to an oxygen. The oxygen has two lone pairs of electrons on it."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. We'd have our R group bonded to our carbon. And then that's bonded to an oxygen. The oxygen has two lone pairs of electrons on it. The oxygen formed a bond with the sulfur. And now the sulfur is bonded to this oxygen, which gets a negative 1 formal charge. The sulfur is bonded to two chlorines."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen has two lone pairs of electrons on it. The oxygen formed a bond with the sulfur. And now the sulfur is bonded to this oxygen, which gets a negative 1 formal charge. The sulfur is bonded to two chlorines. So we draw in our chlorines here with all the lone pairs of electrons. And there's still a lone pair of electrons on our sulfur. We now have a double bond between the carbon and this oxygen."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "The sulfur is bonded to two chlorines. So we draw in our chlorines here with all the lone pairs of electrons. And there's still a lone pair of electrons on our sulfur. We now have a double bond between the carbon and this oxygen. And one lone pair of electrons on this oxygen gives it a plus 1 formal charge. So following some of our electrons, if these electrons in magenta move in here, they'd form our double bond. And then we could think about these electrons in blue forming the bond between the oxygen and the sulfur."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "We now have a double bond between the carbon and this oxygen. And one lone pair of electrons on this oxygen gives it a plus 1 formal charge. So following some of our electrons, if these electrons in magenta move in here, they'd form our double bond. And then we could think about these electrons in blue forming the bond between the oxygen and the sulfur. And then finally, we could think about these electrons in here in green moving off onto our oxygen like that. And so for the next step, we could think about these electrons moving in here to form our double bond between oxygen and sulfur. And now we kick off the chloride anion as a leaving group."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then we could think about these electrons in blue forming the bond between the oxygen and the sulfur. And then finally, we could think about these electrons in here in green moving off onto our oxygen like that. And so for the next step, we could think about these electrons moving in here to form our double bond between oxygen and sulfur. And now we kick off the chloride anion as a leaving group. And we know the chloride anion is an excellent leaving group because it's stable on its own. So when we draw the result of that, we would have our R group. We would have our carbon."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And now we kick off the chloride anion as a leaving group. And we know the chloride anion is an excellent leaving group because it's stable on its own. So when we draw the result of that, we would have our R group. We would have our carbon. We would have that carbon double bonded to our oxygen here with a lone pair of electrons plus 1 formal charge. And then we would have our oxygen with two lone pairs of electrons. And our sulfur is now double bonded to this oxygen."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "We would have our carbon. We would have that carbon double bonded to our oxygen here with a lone pair of electrons plus 1 formal charge. And then we would have our oxygen with two lone pairs of electrons. And our sulfur is now double bonded to this oxygen. There's still a lone pair of electrons on that sulfur, but only one chlorine bonded to the sulfur now. So we lost the chloride anion. So let's go ahead and show some of those electrons."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And our sulfur is now double bonded to this oxygen. There's still a lone pair of electrons on that sulfur, but only one chlorine bonded to the sulfur now. So we lost the chloride anion. So let's go ahead and show some of those electrons. So let's say these electrons in red here move in to reform the double bond between oxygen and sulfur. And then we have some electrons kick off onto the chlorine. So we have the chloride anion that forms."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show some of those electrons. So let's say these electrons in red here move in to reform the double bond between oxygen and sulfur. And then we have some electrons kick off onto the chlorine. So we have the chloride anion that forms. Let's go ahead and show that as well. So these electrons in here come off onto chlorine. Then we have the chloride anion."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So we have the chloride anion that forms. Let's go ahead and show that as well. So these electrons in here come off onto chlorine. Then we have the chloride anion. So let's draw in the chloride anion. Let's get some more space down here. So we also have the chloride anion that forms."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "Then we have the chloride anion. So let's draw in the chloride anion. Let's get some more space down here. So we also have the chloride anion that forms. So we draw that in here like that. So a negative 1 formal charge. So at this stage, we need to consider whether the chloride anion is going to function as a base or a nucleophile."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So we also have the chloride anion that forms. So we draw that in here like that. So a negative 1 formal charge. So at this stage, we need to consider whether the chloride anion is going to function as a base or a nucleophile. And it could do both. So let's first think about the chloride anion functioning as a base. And so if it functions as a base, it could take this proton, leaving these electrons behind on the oxygen."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So at this stage, we need to consider whether the chloride anion is going to function as a base or a nucleophile. And it could do both. So let's first think about the chloride anion functioning as a base. And so if it functions as a base, it could take this proton, leaving these electrons behind on the oxygen. So let's go ahead and draw what we would form. We would form our carbonyl with two lone pairs of electrons on the oxygen. And then we would have this oxygen here with two lone pairs."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And so if it functions as a base, it could take this proton, leaving these electrons behind on the oxygen. So let's go ahead and draw what we would form. We would form our carbonyl with two lone pairs of electrons on the oxygen. And then we would have this oxygen here with two lone pairs. And then this sulfur would still be double bonded, one lone pair, and then the chlorine like that. So we're saying that in this acid-base reaction, these electrons in here move back on here to form the carbonyl. And so you could consider this to be the intermediate for this mechanism."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have this oxygen here with two lone pairs. And then this sulfur would still be double bonded, one lone pair, and then the chlorine like that. So we're saying that in this acid-base reaction, these electrons in here move back on here to form the carbonyl. And so you could consider this to be the intermediate for this mechanism. And you'll see some versions of this mechanism take this intermediate and continue on to form your product. I'm going to show the chloride anion functioning as a nucleophile over here for this. So it's an acid-base reaction."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And so you could consider this to be the intermediate for this mechanism. And you'll see some versions of this mechanism take this intermediate and continue on to form your product. I'm going to show the chloride anion functioning as a nucleophile over here for this. So it's an acid-base reaction. So it's possible to, once again, protonate your carbonyl. That's going to activate it. So this carbon right here becomes more electrophilic."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So it's an acid-base reaction. So it's possible to, once again, protonate your carbonyl. That's going to activate it. So this carbon right here becomes more electrophilic. And that means that the chloride anion can function as a nucleophile and attack our electrophile, since the chloride anion isn't a great nucleophile on its own. So if the chloride anion attacks here, that would push these electrons off onto the oxygen. And we could go ahead and draw what we would form."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here becomes more electrophilic. And that means that the chloride anion can function as a nucleophile and attack our electrophile, since the chloride anion isn't a great nucleophile on its own. So if the chloride anion attacks here, that would push these electrons off onto the oxygen. And we could go ahead and draw what we would form. So we would have our R group. We would have our carbon. We would have this oxygen with two lone pairs of electrons bonded to hydrogen here."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And we could go ahead and draw what we would form. So we would have our R group. We would have our carbon. We would have this oxygen with two lone pairs of electrons bonded to hydrogen here. So let's show those electrons in blue. So if these electrons in blue kick off onto the oxygen, we could say that those are these electrons. And then we could also say that these electrons here in green on the chloride anion are going to form a bond between the chlorine and the carbon."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "We would have this oxygen with two lone pairs of electrons bonded to hydrogen here. So let's show those electrons in blue. So if these electrons in blue kick off onto the oxygen, we could say that those are these electrons. And then we could also say that these electrons here in green on the chloride anion are going to form a bond between the chlorine and the carbon. So we can go ahead and draw in the bond between the chlorine and the carbon like that. And then we still have our oxygen right here, bonded to our sulfur, double bonded to this oxygen. And then we have our lone pair of electrons here."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then we could also say that these electrons here in green on the chloride anion are going to form a bond between the chlorine and the carbon. So we can go ahead and draw in the bond between the chlorine and the carbon like that. And then we still have our oxygen right here, bonded to our sulfur, double bonded to this oxygen. And then we have our lone pair of electrons here. And then we have our chlorine. So what all this does is make a much better leaving group than what we started off with. So you could think about this as our leaving group."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our lone pair of electrons here. And then we have our chlorine. So what all this does is make a much better leaving group than what we started off with. So you could think about this as our leaving group. This is our leaving group in the next step of the mechanism. And if we go back up here, that's a much better leaving group than the OH that we started out with. And so if that's going to be our leaving group, we could show these electrons in here moving in to form our double bond, which would kick these electrons off onto our oxygen."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about this as our leaving group. This is our leaving group in the next step of the mechanism. And if we go back up here, that's a much better leaving group than the OH that we started out with. And so if that's going to be our leaving group, we could show these electrons in here moving in to form our double bond, which would kick these electrons off onto our oxygen. So let's go ahead and show what we would make from that. We would now form our R group, our carbon. We would double bond it to an oxygen like this."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And so if that's going to be our leaving group, we could show these electrons in here moving in to form our double bond, which would kick these electrons off onto our oxygen. So let's go ahead and show what we would make from that. We would now form our R group, our carbon. We would double bond it to an oxygen like this. Lone pair of electrons plus one formal charge. And then we'd have our chlorine over here with all of its electrons too. So let's go ahead and show where those electrons came from."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "We would double bond it to an oxygen like this. Lone pair of electrons plus one formal charge. And then we'd have our chlorine over here with all of its electrons too. So let's go ahead and show where those electrons came from. So if these electrons move in here, that would reform our carbonyl. So I can go ahead and draw that in. And then we still have this bond in green here just to clarify."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show where those electrons came from. So if these electrons move in here, that would reform our carbonyl. So I can go ahead and draw that in. And then we still have this bond in green here just to clarify. And then we would have some electrons in, let's make them magenta. So these electrons in here are going to kick off onto the oxygen. So we can go ahead and draw that."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then we still have this bond in green here just to clarify. And then we would have some electrons in, let's make them magenta. So these electrons in here are going to kick off onto the oxygen. So we can go ahead and draw that. We would have our oxygen. So we would have three lone pairs of electrons. One of them would be the magenta ones."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So we can go ahead and draw that. We would have our oxygen. So we would have three lone pairs of electrons. One of them would be the magenta ones. So we can go ahead and do that. That gives our oxygen a negative 1 formal charge because it's bonded to the sulfur here. The sulfur is still double bonded to this oxygen."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "One of them would be the magenta ones. So we can go ahead and do that. That gives our oxygen a negative 1 formal charge because it's bonded to the sulfur here. The sulfur is still double bonded to this oxygen. And once again, still a lone pair of electrons and a chlorine over here like that. The reason why this is a good leaving group is because we can push these electrons into here, push these electrons off onto here, and we can form sulfur dioxide as a gas. So if we go ahead and draw the result of that, we would now have this oxygen double bonded to the sulfur, double bonded to this oxygen."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "The sulfur is still double bonded to this oxygen. And once again, still a lone pair of electrons and a chlorine over here like that. The reason why this is a good leaving group is because we can push these electrons into here, push these electrons off onto here, and we can form sulfur dioxide as a gas. So if we go ahead and draw the result of that, we would now have this oxygen double bonded to the sulfur, double bonded to this oxygen. So draw in our lone pairs of electrons here. And we still have a lone pair of electrons on our sulfur. So if these electrons in red here move in to form this bond, we've now formed sulfur dioxide."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if we go ahead and draw the result of that, we would now have this oxygen double bonded to the sulfur, double bonded to this oxygen. So draw in our lone pairs of electrons here. And we still have a lone pair of electrons on our sulfur. So if these electrons in red here move in to form this bond, we've now formed sulfur dioxide. And we also formed the chloride anion. So once again, follow some electrons. If these electrons move off onto chlorine, we have the chloride anion."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if these electrons in red here move in to form this bond, we've now formed sulfur dioxide. And we also formed the chloride anion. So once again, follow some electrons. If these electrons move off onto chlorine, we have the chloride anion. So now we have our chloride anion, which could function as a base and in the last step of our mechanism, deprotonate. So take this proton, leave these electrons behind. And finally, we formed our acyl chloride."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "If these electrons move off onto chlorine, we have the chloride anion. So now we have our chloride anion, which could function as a base and in the last step of our mechanism, deprotonate. So take this proton, leave these electrons behind. And finally, we formed our acyl chloride. So we have our carbon double bonded to an oxygen with two lone pairs of electrons. And we have our chlorine over here like that. So if we follow those final electrons, these electrons in blue, move off onto your carbonyl."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And finally, we formed our acyl chloride. So we have our carbon double bonded to an oxygen with two lone pairs of electrons. And we have our chlorine over here like that. So if we follow those final electrons, these electrons in blue, move off onto your carbonyl. And so you would also form HCl by this. So you form HCl, which is also a gas. So the formation of these gases, the formation of sulfur dioxide and HCl, drives the reaction to completion here."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if we follow those final electrons, these electrons in blue, move off onto your carbonyl. And so you would also form HCl by this. So you form HCl, which is also a gas. So the formation of these gases, the formation of sulfur dioxide and HCl, drives the reaction to completion here. So that was a long mechanism. And once again, you'll see some slight variations on this. So if I go back up to here, you could show this step, the nucleophile attacking the electrophile, and directly form here if you wanted to."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So the formation of these gases, the formation of sulfur dioxide and HCl, drives the reaction to completion here. So that was a long mechanism. And once again, you'll see some slight variations on this. So if I go back up to here, you could show this step, the nucleophile attacking the electrophile, and directly form here if you wanted to. So as your nucleophile attacks your sulfur, these electrons could kick off onto chlorine to form the chloride anion at this step. And then you would get this at this point. And then once again, I talked about the possibility of using this as your intermediate."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "So if I go back up to here, you could show this step, the nucleophile attacking the electrophile, and directly form here if you wanted to. So as your nucleophile attacks your sulfur, these electrons could kick off onto chlorine to form the chloride anion at this step. And then you would get this at this point. And then once again, I talked about the possibility of using this as your intermediate. So depending on which textbook you look in, you might see some slight variations on this mechanism. And it is a bit of a long one here. All right, let's look at two other ways to make an acyl chloride."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "And then once again, I talked about the possibility of using this as your intermediate. So depending on which textbook you look in, you might see some slight variations on this mechanism. And it is a bit of a long one here. All right, let's look at two other ways to make an acyl chloride. So starting with a carboxylic acid, you could add phosphorus pentachloride or phosphorus trichloride. And both of those will give you an acyl chloride as well. And the mechanism is pretty similar."}, {"video_title": "Preparation of acyl (acid) chlorides Organic chemistry Khan Academy.mp3", "Sentence": "All right, let's look at two other ways to make an acyl chloride. So starting with a carboxylic acid, you could add phosphorus pentachloride or phosphorus trichloride. And both of those will give you an acyl chloride as well. And the mechanism is pretty similar. And also, we could think about instead of phosphorus trichloride, we could add PBr3, so phosphorus tribromide. And instead of putting a chlorine here, that would put a bromine. And so we could form an acyl bromide this way."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So let's look at this SN1 reaction. On the left is our alkyl halide, ethanol is our solvent, and on the right is our product. The first step should be loss of a leaving group. So these electrons come off onto the iodine to form the iodide anion. We're taking a bond away from this carbon in red, so the carbon in red is gonna get a plus one formal charge. So let's draw this in. We're gonna form a carbocation, and the carbon in red is this carbon."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So these electrons come off onto the iodine to form the iodide anion. We're taking a bond away from this carbon in red, so the carbon in red is gonna get a plus one formal charge. So let's draw this in. We're gonna form a carbocation, and the carbon in red is this carbon. So this carbon gets a plus one formal charge. If we look at this carbocation, it's secondary. The carbon in red is directly bonded to two other carbons, so this is a secondary carbocation."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "We're gonna form a carbocation, and the carbon in red is this carbon. So this carbon gets a plus one formal charge. If we look at this carbocation, it's secondary. The carbon in red is directly bonded to two other carbons, so this is a secondary carbocation. And if you think about the video on carbocations and rearrangements, we can have a rearrangement here to form a more stable carbocation. A methyl shift makes sense, so this methyl group is gonna shift over to this carbon. And let's draw what we would make now."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "The carbon in red is directly bonded to two other carbons, so this is a secondary carbocation. And if you think about the video on carbocations and rearrangements, we can have a rearrangement here to form a more stable carbocation. A methyl shift makes sense, so this methyl group is gonna shift over to this carbon. And let's draw what we would make now. So we would have a methyl group now that's moved over. So the carbon in red is this carbon. And we're taking a bond away from the carbon that I just circled in green."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "And let's draw what we would make now. So we would have a methyl group now that's moved over. So the carbon in red is this carbon. And we're taking a bond away from the carbon that I just circled in green. So the carbon I circled in green is this carbon, and that carbon gets a plus one formal charge now. So this is our carbocation. What kind of carbocation is it?"}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "And we're taking a bond away from the carbon that I just circled in green. So the carbon I circled in green is this carbon, and that carbon gets a plus one formal charge now. So this is our carbocation. What kind of carbocation is it? Well, if we look at the carbon that I circled in green, it's directly bonded to one, two, three other carbons, so it is a tertiary carbocation, which we know is more stable than a secondary carbocation. So a methyl shift increases the stability, going from a secondary to a tertiary carbocation. This carbocation is our electrophile, and now it's time for our nucleophile to attack our electrophile."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "What kind of carbocation is it? Well, if we look at the carbon that I circled in green, it's directly bonded to one, two, three other carbons, so it is a tertiary carbocation, which we know is more stable than a secondary carbocation. So a methyl shift increases the stability, going from a secondary to a tertiary carbocation. This carbocation is our electrophile, and now it's time for our nucleophile to attack our electrophile. Our nucleophile is our solvent, ethanol. So this is a solvolysis reaction. So if I draw in an ethanol molecule here and put in two lone pairs of electrons on the oxygen, so one of those lone pairs is gonna form a bond with the carbon that I circled in green."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "This carbocation is our electrophile, and now it's time for our nucleophile to attack our electrophile. Our nucleophile is our solvent, ethanol. So this is a solvolysis reaction. So if I draw in an ethanol molecule here and put in two lone pairs of electrons on the oxygen, so one of those lone pairs is gonna form a bond with the carbon that I circled in green. So the nucleophile attacks the electrophile, and we form a bond between the oxygen and the carbon. So if I sketch this in here, now we would have a bond to this carbon, this oxygen is still bonded to a hydrogen, still bonded to the ethyl, and let's say that the lone pair of electrons that form the bond, let's make them these electrons in magenta. So those electrons in magenta on the oxygen form this bond."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So if I draw in an ethanol molecule here and put in two lone pairs of electrons on the oxygen, so one of those lone pairs is gonna form a bond with the carbon that I circled in green. So the nucleophile attacks the electrophile, and we form a bond between the oxygen and the carbon. So if I sketch this in here, now we would have a bond to this carbon, this oxygen is still bonded to a hydrogen, still bonded to the ethyl, and let's say that the lone pair of electrons that form the bond, let's make them these electrons in magenta. So those electrons in magenta on the oxygen form this bond. We still have a lone pair of electrons left on the oxygen, so here they are, so that's, here they are right here, and that's a plus one formal charge on the oxygen. If we compare this to our final product, notice all we have to do is deprotonate. So the last step of the mechanism is just a proton transfer, an acid-base reaction."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So those electrons in magenta on the oxygen form this bond. We still have a lone pair of electrons left on the oxygen, so here they are, so that's, here they are right here, and that's a plus one formal charge on the oxygen. If we compare this to our final product, notice all we have to do is deprotonate. So the last step of the mechanism is just a proton transfer, an acid-base reaction. Another molecule of ethanol could come along and serve as the base. So if I draw in my lone pairs of electrons on the oxygen, one of those lone pairs could pick up this proton, leave these electrons behind on this oxygen, and finally give us our product. So you could draw your product a few different ways, but that is our SN1 mechanism with a carbocation rearrangement."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So the last step of the mechanism is just a proton transfer, an acid-base reaction. Another molecule of ethanol could come along and serve as the base. So if I draw in my lone pairs of electrons on the oxygen, one of those lone pairs could pick up this proton, leave these electrons behind on this oxygen, and finally give us our product. So you could draw your product a few different ways, but that is our SN1 mechanism with a carbocation rearrangement. Let's do another carbocation rearrangement in an SN1 mechanism, but this time we're starting with this alcohol. In the previous example, the first step was loss of a leaving group, but if we showed the electrons going onto the oxygen now to form the hydroxide ion, that doesn't work because the hydroxide ion is a bad leaving group, so we actually have to protonate the alcohol first to form a good leaving group. So the first step of this mechanism is a proton transfer."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So you could draw your product a few different ways, but that is our SN1 mechanism with a carbocation rearrangement. Let's do another carbocation rearrangement in an SN1 mechanism, but this time we're starting with this alcohol. In the previous example, the first step was loss of a leaving group, but if we showed the electrons going onto the oxygen now to form the hydroxide ion, that doesn't work because the hydroxide ion is a bad leaving group, so we actually have to protonate the alcohol first to form a good leaving group. So the first step of this mechanism is a proton transfer. The alcohol's gonna function as a base, and H3O plus is gonna donate a proton. So the hydronium ion is gonna act as our acid, so that's H3O plus, and our alcohol acts as our base. A lone pair of electrons on the oxygen pick up a proton from hydronium."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So the first step of this mechanism is a proton transfer. The alcohol's gonna function as a base, and H3O plus is gonna donate a proton. So the hydronium ion is gonna act as our acid, so that's H3O plus, and our alcohol acts as our base. A lone pair of electrons on the oxygen pick up a proton from hydronium. So our first step is a proton transfer, and that gives us this oxygen here with a plus one formal charge. So still one lone pair of electrons, a plus one formal charge on that oxygen, and let's say that this lone pair of electrons picked up this proton to form this bond. Now we are ready for loss of a leaving group because when these electrons come off onto this oxygen now, that gives us water, and water is a good leaving group."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "A lone pair of electrons on the oxygen pick up a proton from hydronium. So our first step is a proton transfer, and that gives us this oxygen here with a plus one formal charge. So still one lone pair of electrons, a plus one formal charge on that oxygen, and let's say that this lone pair of electrons picked up this proton to form this bond. Now we are ready for loss of a leaving group because when these electrons come off onto this oxygen now, that gives us water, and water is a good leaving group. So first step is proton transfer. Second step is loss of a leaving group, and we're taking a bond away from this carbon in red, so that carbon in red gets a plus one formal charge. So let me draw this in here."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "Now we are ready for loss of a leaving group because when these electrons come off onto this oxygen now, that gives us water, and water is a good leaving group. So first step is proton transfer. Second step is loss of a leaving group, and we're taking a bond away from this carbon in red, so that carbon in red gets a plus one formal charge. So let me draw this in here. So the carbon in red is this carbon. So let me write a plus one formal charge on this carbon. What kind of a carbocation is that?"}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So let me draw this in here. So the carbon in red is this carbon. So let me write a plus one formal charge on this carbon. What kind of a carbocation is that? The carbon in red is directly bonded to two other carbons, so this is a secondary carbocation, and we think about the possibility of a rearrangement. So in the carbocation and rearrangements video, well, another shift that we did was a hydride shift. So on this carbon in magenta, there's still a hydrogen, and a hydride shift would give us a more stable carbocation."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "What kind of a carbocation is that? The carbon in red is directly bonded to two other carbons, so this is a secondary carbocation, and we think about the possibility of a rearrangement. So in the carbocation and rearrangements video, well, another shift that we did was a hydride shift. So on this carbon in magenta, there's still a hydrogen, and a hydride shift would give us a more stable carbocation. Remember, a hydride shift, think about this hydrogen and these two electrons here shifting over to this carbon in red. So let's draw in what we would have now. So now, if you think about that carbon in red, we already had a hydrogen bonded to it."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So on this carbon in magenta, there's still a hydrogen, and a hydride shift would give us a more stable carbocation. Remember, a hydride shift, think about this hydrogen and these two electrons here shifting over to this carbon in red. So let's draw in what we would have now. So now, if you think about that carbon in red, we already had a hydrogen bonded to it. So let me go ahead and highlight this carbon as being the carbon in red. We already had a hydrogen bonded to it, in the example in the carbocation on the left, I should say. So let me draw in that original hydrogen."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So now, if you think about that carbon in red, we already had a hydrogen bonded to it. So let me go ahead and highlight this carbon as being the carbon in red. We already had a hydrogen bonded to it, in the example in the carbocation on the left, I should say. So let me draw in that original hydrogen. With the hydride shift, we're adding in another hydrogen here to this carbon in red. So there's no more formal charge on the carbon in red. The formal charge moves to this carbon, the one I circled in green."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So let me draw in that original hydrogen. With the hydride shift, we're adding in another hydrogen here to this carbon in red. So there's no more formal charge on the carbon in red. The formal charge moves to this carbon, the one I circled in green. It's losing a bond, so that carbon that I circled in green now gets a plus one formal charge. What kind of a carbocation did we form? Well, the carbon in green is directly bonded to one, two, three other carbons."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "The formal charge moves to this carbon, the one I circled in green. It's losing a bond, so that carbon that I circled in green now gets a plus one formal charge. What kind of a carbocation did we form? Well, the carbon in green is directly bonded to one, two, three other carbons. So this is a tertiary carbocation. A hydride shift gives us a more stable carbocation. Now we formed our electrophile, and our nucleophile will come along in the next step, and our nucleophile should be methanol."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "Well, the carbon in green is directly bonded to one, two, three other carbons. So this is a tertiary carbocation. A hydride shift gives us a more stable carbocation. Now we formed our electrophile, and our nucleophile will come along in the next step, and our nucleophile should be methanol. So let me draw in a molecule of methanol here. So put in a hydrogen, two lone pairs of electrons on the oxygen. So our nucleophile will attack our electrophile."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "Now we formed our electrophile, and our nucleophile will come along in the next step, and our nucleophile should be methanol. So let me draw in a molecule of methanol here. So put in a hydrogen, two lone pairs of electrons on the oxygen. So our nucleophile will attack our electrophile. So a lone pair of electrons on this oxygen are gonna form a bond with this carbon that I circled in green. So let's draw the result of our nucleophilic attack. So now we're forming a bond between the carbon in green and this oxygen here."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So our nucleophile will attack our electrophile. So a lone pair of electrons on this oxygen are gonna form a bond with this carbon that I circled in green. So let's draw the result of our nucleophilic attack. So now we're forming a bond between the carbon in green and this oxygen here. So let me sketch in the rest of this. And we still have a methyl on this oxygen. We still have a hydrogen on this oxygen."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So now we're forming a bond between the carbon in green and this oxygen here. So let me sketch in the rest of this. And we still have a methyl on this oxygen. We still have a hydrogen on this oxygen. And let me highlight those electrons. So our electrons, let's make them blue. So these electrons in blue here on this oxygen form a bond between the oxygen and that carbon."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "We still have a hydrogen on this oxygen. And let me highlight those electrons. So our electrons, let's make them blue. So these electrons in blue here on this oxygen form a bond between the oxygen and that carbon. We still have a lone pair of electrons on this oxygen, which gives the oxygen a plus one formal charge. And also notice what I did with this group right here. So these two carbons, I just drew them going down this way."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So these electrons in blue here on this oxygen form a bond between the oxygen and that carbon. We still have a lone pair of electrons on this oxygen, which gives the oxygen a plus one formal charge. And also notice what I did with this group right here. So these two carbons, I just drew them going down this way. There's free rotation around this bond right here. So you can draw it however you want. I just drew it to look a little bit more like our product here."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "So these two carbons, I just drew them going down this way. There's free rotation around this bond right here. So you can draw it however you want. I just drew it to look a little bit more like our product here. When we compare our two, the last step must be a proton transfer. All we have to do is take a proton away and we have our final product. So I could have, I have a couple of choices here as my base."}, {"video_title": "Sn1 mechanism carbocation rearrangement.mp3", "Sentence": "I just drew it to look a little bit more like our product here. When we compare our two, the last step must be a proton transfer. All we have to do is take a proton away and we have our final product. So I could have, I have a couple of choices here as my base. I could use either the water molecule that we lost back in this step, or I could use another molecule of methanol as our base. It doesn't really matter what we use here. I'll just use the water molecule."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And when chemists first made cyclooctatetraene, which is this molecule right here, they assumed it would react like benzene, because it looks like it has alternating single double bonds and it has a ring. And so they just assumed that it would behave like benzene. It turns out cyclooctatetraene does not react like benzene. So benzene did not react the way cyclohexene did up here. So cyclohexene will give us a mix of enantiomers when the bromine adds across double bond. Benzene does not do that. Cyclooctatetraene does, as a matter of fact."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So benzene did not react the way cyclohexene did up here. So cyclohexene will give us a mix of enantiomers when the bromine adds across double bond. Benzene does not do that. Cyclooctatetraene does, as a matter of fact. So it will give you a mixture of enantiomers, and the bromine will add across one of those double bonds. It turns out cyclooctatetraene isn't even conjugated. So it looks like it has alternating single and double bonds in this dot structure, but it behaves like isolated double bonds, like four different isolated double bonds."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Cyclooctatetraene does, as a matter of fact. So it will give you a mixture of enantiomers, and the bromine will add across one of those double bonds. It turns out cyclooctatetraene isn't even conjugated. So it looks like it has alternating single and double bonds in this dot structure, but it behaves like isolated double bonds, like four different isolated double bonds. And it's not aromatic. So let's see if we can analyze the reasons for this reaction here. And so we're going to just real quickly review the criteria to determine if a compound is aromatic or not."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it looks like it has alternating single and double bonds in this dot structure, but it behaves like isolated double bonds, like four different isolated double bonds. And it's not aromatic. So let's see if we can analyze the reasons for this reaction here. And so we're going to just real quickly review the criteria to determine if a compound is aromatic or not. So a compound or ion is aromatic if it contains a ring of continuously overlapping p orbitals and also has 4n plus 2, where n is an integer, pi electrons in the ring, which is Huckel's rule. So here is my cyclooctatetraene molecule. And if I count my pi electrons, I can see there are a total of 8 pi electrons in this molecule."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we're going to just real quickly review the criteria to determine if a compound is aromatic or not. So a compound or ion is aromatic if it contains a ring of continuously overlapping p orbitals and also has 4n plus 2, where n is an integer, pi electrons in the ring, which is Huckel's rule. So here is my cyclooctatetraene molecule. And if I count my pi electrons, I can see there are a total of 8 pi electrons in this molecule. So 8 pi electrons. And each carbon in cyclooctatetraene is sp2 hybridized. So each carbon has a free p orbital."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And if I count my pi electrons, I can see there are a total of 8 pi electrons in this molecule. So 8 pi electrons. And each carbon in cyclooctatetraene is sp2 hybridized. So each carbon has a free p orbital. Now, the fact that cyclooctatetraene has 8 atoms in the ring means that there's a little bit of angle strain in this molecule. So one possible conformation for the molecule to adopt is what's called a tub conformation. So you can see these two carbons can swing up on either side to adopt a tub conformation."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So each carbon has a free p orbital. Now, the fact that cyclooctatetraene has 8 atoms in the ring means that there's a little bit of angle strain in this molecule. So one possible conformation for the molecule to adopt is what's called a tub conformation. So you can see these two carbons can swing up on either side to adopt a tub conformation. Now, if each carbon in the tub conformation is sp2 hybridized, I could draw a p orbital on each carbon in the tub conformation here. So when I do that, I would get this as a picture. Now, the problem is it's a little difficult for the p orbitals to overlap."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So you can see these two carbons can swing up on either side to adopt a tub conformation. Now, if each carbon in the tub conformation is sp2 hybridized, I could draw a p orbital on each carbon in the tub conformation here. So when I do that, I would get this as a picture. Now, the problem is it's a little difficult for the p orbitals to overlap. So it might be easy for these guys to overlap right here. But because of the tub conformation, it'd be hard to get overlap of these orbitals. And so it turns out that this is the reason why the molecule acts like it's not conjugated."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, the problem is it's a little difficult for the p orbitals to overlap. So it might be easy for these guys to overlap right here. But because of the tub conformation, it'd be hard to get overlap of these orbitals. And so it turns out that this is the reason why the molecule acts like it's not conjugated. It's because it does have p orbitals, but they can't really overlap in the tub conformation. And so this violates the first criteria for a compound to be aromatic. And therefore, we say that cyclooctatetraene is non-aromatic."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it turns out that this is the reason why the molecule acts like it's not conjugated. It's because it does have p orbitals, but they can't really overlap in the tub conformation. And so this violates the first criteria for a compound to be aromatic. And therefore, we say that cyclooctatetraene is non-aromatic. So it's non-aromatic because it violates the first criteria. It does not have a ring of continuously overlapping p orbitals. The p orbitals don't really overlap very well in the tub conformation."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And therefore, we say that cyclooctatetraene is non-aromatic. So it's non-aromatic because it violates the first criteria. It does not have a ring of continuously overlapping p orbitals. The p orbitals don't really overlap very well in the tub conformation. What if cyclooctatetraene adopted a planar conformation? So we're just going to pretend like cyclooctatetraene adopts a planar conformation down here. And once again, each carbon is sp2 hybridized."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The p orbitals don't really overlap very well in the tub conformation. What if cyclooctatetraene adopted a planar conformation? So we're just going to pretend like cyclooctatetraene adopts a planar conformation down here. And once again, each carbon is sp2 hybridized. So each carbon is going to get a p orbital like that. And so we have a total of eight carbons, so eight p orbitals, so eight atomic orbitals. So let me go ahead and write that here."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And once again, each carbon is sp2 hybridized. So each carbon is going to get a p orbital like that. And so we have a total of eight carbons, so eight p orbitals, so eight atomic orbitals. So let me go ahead and write that here. So we have a total of eight atomic orbitals for cyclooctatetraene. And according to MO theory, those eight atomic orbitals are going to give us eight molecular orbitals. So the atomic orbitals are going to cease to exist and give us eight molecular orbitals."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write that here. So we have a total of eight atomic orbitals for cyclooctatetraene. And according to MO theory, those eight atomic orbitals are going to give us eight molecular orbitals. So the atomic orbitals are going to cease to exist and give us eight molecular orbitals. Drawing those eight molecular orbitals would be way too much for this video. So we're not going to actually draw them, but we are going to show where they are in terms of energy using our frost circle. And so this is what we saw in the last video, so how to draw a frost circle."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the atomic orbitals are going to cease to exist and give us eight molecular orbitals. Drawing those eight molecular orbitals would be way too much for this video. So we're not going to actually draw them, but we are going to show where they are in terms of energy using our frost circle. And so this is what we saw in the last video, so how to draw a frost circle. I'm going to go ahead and put a line through the center of my frost circle to help me draw the polygon in here. So what kind of a polygon do I draw? Well, I have an eight-membered ring, so I'm going to draw an eight-sided polygon in here."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this is what we saw in the last video, so how to draw a frost circle. I'm going to go ahead and put a line through the center of my frost circle to help me draw the polygon in here. So what kind of a polygon do I draw? Well, I have an eight-membered ring, so I'm going to draw an eight-sided polygon in here. I'm going to start from the bottom here. So I'm going to attempt to draw an eight-sided figure in here. So it would be something like this."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Well, I have an eight-membered ring, so I'm going to draw an eight-sided polygon in here. I'm going to start from the bottom here. So I'm going to attempt to draw an eight-sided figure in here. So it would be something like this. Now remember, when you are drawing your frost circle and you're inscribing your polygon, the important thing is where the polygon intersects with your circle. And every point where the polygon intersects with your circle represents a molecular orbital. And so I can see that I would have a total of eight molecular orbitals, because I have eight points of intersection between my polygon and between my circle."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it would be something like this. Now remember, when you are drawing your frost circle and you're inscribing your polygon, the important thing is where the polygon intersects with your circle. And every point where the polygon intersects with your circle represents a molecular orbital. And so I can see that I would have a total of eight molecular orbitals, because I have eight points of intersection between my polygon and between my circle. And again, the nice thing about a frost circle is it shows you the relative energies of your eight molecular orbitals. So I would have three bonding molecular orbitals, which would be the ones down here. So these three points of intersection will give you the relative energies."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so I can see that I would have a total of eight molecular orbitals, because I have eight points of intersection between my polygon and between my circle. And again, the nice thing about a frost circle is it shows you the relative energies of your eight molecular orbitals. So I would have three bonding molecular orbitals, which would be the ones down here. So these three points of intersection will give you the relative energies. And so I have three bonding molecular orbitals. And then up here at the top, I also have three molecular orbitals. But these are my anti-bonding molecular orbitals."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these three points of intersection will give you the relative energies. And so I have three bonding molecular orbitals. And then up here at the top, I also have three molecular orbitals. But these are my anti-bonding molecular orbitals. Those are higher in energy. And my two points of intersection that are right on the center line here represent two non-bonding molecular orbitals like that. So when I fill my molecular orbitals, again, it's analogous to electron configurations."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But these are my anti-bonding molecular orbitals. Those are higher in energy. And my two points of intersection that are right on the center line here represent two non-bonding molecular orbitals like that. So when I fill my molecular orbitals, again, it's analogous to electron configurations. I have a total of eight pi electrons that I need to worry about for a planar cyclooctatetraene molecule. And so I can go ahead and start to fill in my eight pi electrons like that. So that takes care of six of them."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So when I fill my molecular orbitals, again, it's analogous to electron configurations. I have a total of eight pi electrons that I need to worry about for a planar cyclooctatetraene molecule. And so I can go ahead and start to fill in my eight pi electrons like that. So that takes care of six of them. And I have two more. And since this is analogous to electron configurations, I'm going to follow Hund's rule and not pair up my electrons in an orbital here. So that represents my eight pi electrons."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that takes care of six of them. And I have two more. And since this is analogous to electron configurations, I'm going to follow Hund's rule and not pair up my electrons in an orbital here. So that represents my eight pi electrons. And since I have unpaired electrons, I have two unpaired electrons, that predicts a very unstable molecule if it were to adopt a planar conformation. And if I think about it in terms of Huckel's rule, I know it doesn't follow Huckel's rule. Huckel's rule is 4n plus 2, where n is an integer."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that represents my eight pi electrons. And since I have unpaired electrons, I have two unpaired electrons, that predicts a very unstable molecule if it were to adopt a planar conformation. And if I think about it in terms of Huckel's rule, I know it doesn't follow Huckel's rule. Huckel's rule is 4n plus 2, where n is an integer. And the 2 comes from the fact that I have this orbital down here. And I do have 4n right here. But I don't have 4n where n is an integer for these two electrons up here."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Huckel's rule is 4n plus 2, where n is an integer. And the 2 comes from the fact that I have this orbital down here. And I do have 4n right here. But I don't have 4n where n is an integer for these two electrons up here. And so this is where it breaks down. And so I have a total of eight pi electrons, which does not follow Huckel's rule. And so because the number of electrons is incorrect, this molecule is definitely not going to adopt a planar conformation."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But I don't have 4n where n is an integer for these two electrons up here. And so this is where it breaks down. And so I have a total of eight pi electrons, which does not follow Huckel's rule. And so because the number of electrons is incorrect, this molecule is definitely not going to adopt a planar conformation. And so cyclooctatetraene has a tub conformation and not planar. It is not aromatic. It is considered to be non-aromatic because of the violation of the first criteria."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so because the number of electrons is incorrect, this molecule is definitely not going to adopt a planar conformation. And so cyclooctatetraene has a tub conformation and not planar. It is not aromatic. It is considered to be non-aromatic because of the violation of the first criteria. But it is possible to react cyclooctatetraene. It's possible to oxidize it. And so let's see what happens when we do that."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It is considered to be non-aromatic because of the violation of the first criteria. But it is possible to react cyclooctatetraene. It's possible to oxidize it. And so let's see what happens when we do that. So if we take cyclooctatetraene and we oxidize it, so it's going to lose some electrons. I'm going to say that these pi electrons are going to stay. And we're going to lose the pi electrons on the left."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so let's see what happens when we do that. So if we take cyclooctatetraene and we oxidize it, so it's going to lose some electrons. I'm going to say that these pi electrons are going to stay. And we're going to lose the pi electrons on the left. So if I take away a bond from these two carbons that used to have that double bond there, they're going to be positively charged. And so I could draw a resonance structure for this. I could move these electrons over here."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to lose the pi electrons on the left. So if I take away a bond from these two carbons that used to have that double bond there, they're going to be positively charged. And so I could draw a resonance structure for this. I could move these electrons over here. And so if I go ahead and show that resonance structure, then this carbon still has a plus 1 charge. And the other positive charge moves over here to this carbon like that. And you could continue drawing resonance structures for this molecule."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "I could move these electrons over here. And so if I go ahead and show that resonance structure, then this carbon still has a plus 1 charge. And the other positive charge moves over here to this carbon like that. And you could continue drawing resonance structures for this molecule. And I'm not going to do that. I just want to show you that the positive charges are spread out throughout the entire ion here. And so one way to represent that would be to just show the electrons are spread out throughout this entire ion."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And you could continue drawing resonance structures for this molecule. And I'm not going to do that. I just want to show you that the positive charges are spread out throughout the entire ion here. And so one way to represent that would be to just show the electrons are spread out throughout this entire ion. And the whole thing has a 2 plus charge like that. And so when I analyze this dication that I got from cyclooctatetraene, I realize that all of the carbons are sp2 hybridized. So if I look at these, my carbocations, those are sp2 hybridized."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so one way to represent that would be to just show the electrons are spread out throughout this entire ion. And the whole thing has a 2 plus charge like that. And so when I analyze this dication that I got from cyclooctatetraene, I realize that all of the carbons are sp2 hybridized. So if I look at these, my carbocations, those are sp2 hybridized. Everything with a double bond on it is sp2 hybridized. And so I have eight sp2 hybridized carbons. And so each one of those sp2 hybridized carbons has a p orbital."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I look at these, my carbocations, those are sp2 hybridized. Everything with a double bond on it is sp2 hybridized. And so I have eight sp2 hybridized carbons. And so each one of those sp2 hybridized carbons has a p orbital. And I also have six pi electrons. I have two, four, and six pi electrons. So six pi electrons follows Huckel's rule."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so each one of those sp2 hybridized carbons has a p orbital. And I also have six pi electrons. I have two, four, and six pi electrons. So six pi electrons follows Huckel's rule. And so this ion turns out to be planar. So it actually does look like this up here. And so there's opportunity for those p orbitals to overlap side by side."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So six pi electrons follows Huckel's rule. And so this ion turns out to be planar. So it actually does look like this up here. And so there's opportunity for those p orbitals to overlap side by side. And so this ion actually fulfills the first criteria. It contains a ring of continuously overlapping p orbitals. And when you analyze the second criteria, it has six pi electrons."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so there's opportunity for those p orbitals to overlap side by side. And so this ion actually fulfills the first criteria. It contains a ring of continuously overlapping p orbitals. And when you analyze the second criteria, it has six pi electrons. And so it would fill the bonding molecular orbitals. It no longer has these two electrons up here. So it has six pi electrons."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And when you analyze the second criteria, it has six pi electrons. And so it would fill the bonding molecular orbitals. It no longer has these two electrons up here. So it has six pi electrons. It fulfills Huckel's rule. So there's a total of six pi electrons for this dication. And so this ion is said to be aromatic."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it has six pi electrons. It fulfills Huckel's rule. So there's a total of six pi electrons for this dication. And so this ion is said to be aromatic. So this is an aromatic ion. It fulfills both of the criteria for it to be aromatic. So it's extra stable."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this ion is said to be aromatic. So this is an aromatic ion. It fulfills both of the criteria for it to be aromatic. So it's extra stable. And that's the reason why it turns out to be planar, because it has eight carbons, like cyclooctatetraene. And so because of that number, it has some strain that it has to overcome. So it would like to turn into a tub to overcome some of that strain."}, {"video_title": "Aromatic stability II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's extra stable. And that's the reason why it turns out to be planar, because it has eight carbons, like cyclooctatetraene. And so because of that number, it has some strain that it has to overcome. So it would like to turn into a tub to overcome some of that strain. But the fact that it's a planar dication means that there must be some sort of extra stability in that ion that's forcing it to be planar. And that extra stability is due to the fact that it is aromatic. And so this ion has been proved to be planar."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So here is the dot structure for benzene, C6H6, and we can draw a resonance structure for this. So we could take these electrons right here, move them over to here. That would mean too many bonds to this carbon, so we'd have to take these electrons and push them to here, which would mean too many bonds to this carbon, and so finally we take these pi electrons and move them over to here. So we draw our resonance brackets, and go ahead and draw our other resonance structure for benzene. So the electrons moved over to here, to here, and then finally to here. And so let's follow those electrons, so let's make the ones on the top left here red. So these electrons in red, I'm showing them move to over here, and let's make these electrons over here green."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So we draw our resonance brackets, and go ahead and draw our other resonance structure for benzene. So the electrons moved over to here, to here, and then finally to here. And so let's follow those electrons, so let's make the ones on the top left here red. So these electrons in red, I'm showing them move to over here, and let's make these electrons over here green. So the electrons in green move down to here, and then finally we'll use blue. So these electrons in blue, I showed them moving over to here. And remember that the actual benzene molecule is a hybrid of these two resonance structures."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in red, I'm showing them move to over here, and let's make these electrons over here green. So the electrons in green move down to here, and then finally we'll use blue. So these electrons in blue, I showed them moving over to here. And remember that the actual benzene molecule is a hybrid of these two resonance structures. So if you're just drawing on a sheet of paper, you could use one or the other, but remember that it's actually the hybrid, because our dot structures are just not perfect ways to represent molecules or ions. And so those are our resonance structures for benzene. Next let's look at the phenoxide anion."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "And remember that the actual benzene molecule is a hybrid of these two resonance structures. So if you're just drawing on a sheet of paper, you could use one or the other, but remember that it's actually the hybrid, because our dot structures are just not perfect ways to represent molecules or ions. And so those are our resonance structures for benzene. Next let's look at the phenoxide anion. So here's the phenoxide anion down here, and I'm gonna try to color code the electrons. Let me go ahead and make these electrons in here red, and let's make these right here green, and then let's make these blue. So just like we did with the benzene ring up above."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "Next let's look at the phenoxide anion. So here's the phenoxide anion down here, and I'm gonna try to color code the electrons. Let me go ahead and make these electrons in here red, and let's make these right here green, and then let's make these blue. So just like we did with the benzene ring up above. And I could start off for the resonance structures for the phenoxide anion by doing the other resonance structure just like we did for benzene, like that. But I'm gonna save that for the end. So let's think about what we would do first."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So just like we did with the benzene ring up above. And I could start off for the resonance structures for the phenoxide anion by doing the other resonance structure just like we did for benzene, like that. But I'm gonna save that for the end. So let's think about what we would do first. Well, we know one of our patterns is a lone pair next to a pi bond, and that's what we have here. So we can think about a lone pair of electrons on this oxygen, so I'll make it magenta. That lone pair is next to the pi bond, the one in red."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about what we would do first. Well, we know one of our patterns is a lone pair next to a pi bond, and that's what we have here. So we can think about a lone pair of electrons on this oxygen, so I'll make it magenta. That lone pair is next to the pi bond, the one in red. And so we can go ahead and draw a resonance structure. And we take these electrons in magenta and move them into here. That would mean too many bonds to this carbon."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "That lone pair is next to the pi bond, the one in red. And so we can go ahead and draw a resonance structure. And we take these electrons in magenta and move them into here. That would mean too many bonds to this carbon. And so we take the electrons in red, and we push them off onto this carbon. So let's go ahead and draw our resonance structure. So we have our ring here, and we have now a double bond between the oxygen and the carbon, only two lone pairs of electrons on this oxygen now."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "That would mean too many bonds to this carbon. And so we take the electrons in red, and we push them off onto this carbon. So let's go ahead and draw our resonance structure. So we have our ring here, and we have now a double bond between the oxygen and the carbon, only two lone pairs of electrons on this oxygen now. The electrons in magenta moved in here to form a pi bond, and the electrons in red moved off onto this carbon right here. That's gonna give that carbon a negative one formal charge. So let's go ahead and draw a negative one formal charge here."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So we have our ring here, and we have now a double bond between the oxygen and the carbon, only two lone pairs of electrons on this oxygen now. The electrons in magenta moved in here to form a pi bond, and the electrons in red moved off onto this carbon right here. That's gonna give that carbon a negative one formal charge. So let's go ahead and draw a negative one formal charge here. The electrons in blue have not moved, and the electrons in green I haven't showed moving yet either, so we put those in there like that. All right, next we have the exact same pattern that we did before. We have a lone pair of electrons next to a pi bond."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw a negative one formal charge here. The electrons in blue have not moved, and the electrons in green I haven't showed moving yet either, so we put those in there like that. All right, next we have the exact same pattern that we did before. We have a lone pair of electrons next to a pi bond. So the lone pair of electrons are the electrons in red right here, next to a pi bond, the electrons in blue. So let's go ahead and draw another resonance structure. We could take these electrons in red, push them into here."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "We have a lone pair of electrons next to a pi bond. So the lone pair of electrons are the electrons in red right here, next to a pi bond, the electrons in blue. So let's go ahead and draw another resonance structure. We could take these electrons in red, push them into here. That would mean too many bonds to this carbon, so we have to take these electrons in blue and push them off onto this carbon. So let's draw that resonance structure. So once again, we have our carbon double bond to an oxygen up here."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "We could take these electrons in red, push them into here. That would mean too many bonds to this carbon, so we have to take these electrons in blue and push them off onto this carbon. So let's draw that resonance structure. So once again, we have our carbon double bond to an oxygen up here. We said that these electrons were the ones in magenta, and the electrons in red move in here to form a pi bond. The electrons in blue move off onto this carbon, and that gives this carbon a negative one formal charge. So this carbon has a negative one formal charge."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So once again, we have our carbon double bond to an oxygen up here. We said that these electrons were the ones in magenta, and the electrons in red move in here to form a pi bond. The electrons in blue move off onto this carbon, and that gives this carbon a negative one formal charge. So this carbon has a negative one formal charge. So this one right here, the one that has the blue electrons on it. All right, we still have our electrons in green over here, and we have the exact same pattern. We have a lone pair next to a pi bond, so the lone pair are the ones in blue, and this time the pi bond are the electrons in green here, so we can draw yet another resonance structure."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon has a negative one formal charge. So this one right here, the one that has the blue electrons on it. All right, we still have our electrons in green over here, and we have the exact same pattern. We have a lone pair next to a pi bond, so the lone pair are the ones in blue, and this time the pi bond are the electrons in green here, so we can draw yet another resonance structure. We could take the electrons in blue, move them into here. That would mean too many bonds to this carbon, so we have to take the electrons in green and push them off onto that carbon. And let's draw that resonance structure."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "We have a lone pair next to a pi bond, so the lone pair are the ones in blue, and this time the pi bond are the electrons in green here, so we can draw yet another resonance structure. We could take the electrons in blue, move them into here. That would mean too many bonds to this carbon, so we have to take the electrons in green and push them off onto that carbon. And let's draw that resonance structure. So once again, we have our ring. So we draw our ring in here. We have this double bond up here, so put in lone pairs of electrons on the oxygen, and these electrons were the ones in magenta, and we go around the ring, so we had our electrons in red right here."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "And let's draw that resonance structure. So once again, we have our ring. So we draw our ring in here. We have this double bond up here, so put in lone pairs of electrons on the oxygen, and these electrons were the ones in magenta, and we go around the ring, so we had our electrons in red right here. The electrons in blue moved into here, and finally the electrons in green moved off onto this carbon, so this carbon right here in green. Therefore, that carbon gets a negative one formal charge now. And thinking about it again, we once again have a lone pair of electrons next to a pi bond, so we have the electrons in green, so those electrons could move into here, and then that would mean too many bonds to this carbon, so we can take these electrons in magenta and push them off onto the oxygen."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "We have this double bond up here, so put in lone pairs of electrons on the oxygen, and these electrons were the ones in magenta, and we go around the ring, so we had our electrons in red right here. The electrons in blue moved into here, and finally the electrons in green moved off onto this carbon, so this carbon right here in green. Therefore, that carbon gets a negative one formal charge now. And thinking about it again, we once again have a lone pair of electrons next to a pi bond, so we have the electrons in green, so those electrons could move into here, and then that would mean too many bonds to this carbon, so we can take these electrons in magenta and push them off onto the oxygen. So let's go ahead and draw our last resonance structure here. So we have now a single bond to this top oxygen, and three lone pairs of electrons, giving that top oxygen a negative one formal charge. So the electrons in magenta, let's say that those electrons are these electrons right here."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "And thinking about it again, we once again have a lone pair of electrons next to a pi bond, so we have the electrons in green, so those electrons could move into here, and then that would mean too many bonds to this carbon, so we can take these electrons in magenta and push them off onto the oxygen. So let's go ahead and draw our last resonance structure here. So we have now a single bond to this top oxygen, and three lone pairs of electrons, giving that top oxygen a negative one formal charge. So the electrons in magenta, let's say that those electrons are these electrons right here. Going around our ring, we had electrons in red, we had electrons in blue right here, and then finally the electrons in green moved into here. So we have five total resonance structures for the phenoxide anion. I can go ahead and put brackets around all five of these."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "So the electrons in magenta, let's say that those electrons are these electrons right here. Going around our ring, we had electrons in red, we had electrons in blue right here, and then finally the electrons in green moved into here. So we have five total resonance structures for the phenoxide anion. I can go ahead and put brackets around all five of these. And since we're talking about resonance structures in the benzene ring, we can think about going back and forth between these two as a final thought here. So we could take these electrons in red, push them into here, which would take these electrons in green over to here, which would take those electrons in blue over to here, and then that would give us the one that we started with as well. So it's also important to think about the hybrid."}, {"video_title": "Resonance structures for benzene and the phenoxide anion Organic chemistry Khan Academy.mp3", "Sentence": "I can go ahead and put brackets around all five of these. And since we're talking about resonance structures in the benzene ring, we can think about going back and forth between these two as a final thought here. So we could take these electrons in red, push them into here, which would take these electrons in green over to here, which would take those electrons in blue over to here, and then that would give us the one that we started with as well. So it's also important to think about the hybrid. So the hybrid has the negative charge delocalized. So the negative charge is delocalized over, we can see in this resonance structure and this one, the negative charge is on the oxygen, and in this one it's on this carbon, and this one it's on this carbon, and this one it's on this carbon. So the negative charge is delocalized over oxygen and three carbons when you're thinking about the resonance hybrid."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "You're also going to form water in this reaction, and this reaction is at equilibrium, and so there are several things that you can do to shift the equilibrium to the right and to make more of your acetal products. One thing would be to remove the water as it forms. So if you decrease the concentration of this product, your equilibrium is going to shift to make more of it, and so therefore we're going to form more acetal. So another thing you could do to shift the equilibrium to the right would be to increase the concentration of one of your reactants. So you could increase the concentration of an aldehyde, and then that would, of course, once again, shift the equilibrium to the right and form more of your acetal products. So several things that you can do in the lab to increase your yield. This is done in an acidic environment, and so there are a couple different proton sources you can use."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So another thing you could do to shift the equilibrium to the right would be to increase the concentration of one of your reactants. So you could increase the concentration of an aldehyde, and then that would, of course, once again, shift the equilibrium to the right and form more of your acetal products. So several things that you can do in the lab to increase your yield. This is done in an acidic environment, and so there are a couple different proton sources you can use. You could use something like sulfuric acid, H2SO4, or you could use something like toluene sulfonic acid, so TSOH are two of the more common catalysts used to form your acetal. Let's look at a reaction here, and then we're going to do the mechanism for this reaction. You'll see it's a bit of a long mechanism."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This is done in an acidic environment, and so there are a couple different proton sources you can use. You could use something like sulfuric acid, H2SO4, or you could use something like toluene sulfonic acid, so TSOH are two of the more common catalysts used to form your acetal. Let's look at a reaction here, and then we're going to do the mechanism for this reaction. You'll see it's a bit of a long mechanism. I think it's a little bit easier to understand if you do it for an actual reaction here. So we have a cyclohexanone reacting with an excess of ethanol and using sulfuric acid as our catalyst. And so just looking at this general pattern up here for predicting the structure of your acetal, we can find this portion of the molecule and think about adding that to our ring."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "You'll see it's a bit of a long mechanism. I think it's a little bit easier to understand if you do it for an actual reaction here. So we have a cyclohexanone reacting with an excess of ethanol and using sulfuric acid as our catalyst. And so just looking at this general pattern up here for predicting the structure of your acetal, we can find this portion of the molecule and think about adding that to our ring. And so we have our ring here, and then we would have our oxygen, and then our R group, and then our oxygen, and then our R group like that. So that's the product. Kind of a funny looking molecule, but that is the acetal that we would make."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so just looking at this general pattern up here for predicting the structure of your acetal, we can find this portion of the molecule and think about adding that to our ring. And so we have our ring here, and then we would have our oxygen, and then our R group, and then our oxygen, and then our R group like that. So that's the product. Kind of a funny looking molecule, but that is the acetal that we would make. So let's think about a mechanism for this reaction. So if you have ethanol and sulfuric acid, one of the things that could happen is protonation of your ethanol. So let's go ahead and show a protonated ion."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Kind of a funny looking molecule, but that is the acetal that we would make. So let's think about a mechanism for this reaction. So if you have ethanol and sulfuric acid, one of the things that could happen is protonation of your ethanol. So let's go ahead and show a protonated ion. So this is one of the possible things that could happen first. So we have a protonated like that, and then we're going to show that functioning as an acid and reacting with cyclohexanone. So here we have cyclohexanone."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show a protonated ion. So this is one of the possible things that could happen first. So we have a protonated like that, and then we're going to show that functioning as an acid and reacting with cyclohexanone. So here we have cyclohexanone. And a lone pair of electrons in cyclohexanone are gonna pick up a proton. So a proton from somewhere, and this could be the acid over here on the left. So these electrons move over here to form ethanol, and we protonate our carbonyl."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So here we have cyclohexanone. And a lone pair of electrons in cyclohexanone are gonna pick up a proton. So a proton from somewhere, and this could be the acid over here on the left. So these electrons move over here to form ethanol, and we protonate our carbonyl. So let's go ahead and show that. So step one would be protonation of your carbonyl, and that is favored because that makes your carbon attached to your oxygen more electrophilic. So let's go ahead and show that."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons move over here to form ethanol, and we protonate our carbonyl. So let's go ahead and show that. So step one would be protonation of your carbonyl, and that is favored because that makes your carbon attached to your oxygen more electrophilic. So let's go ahead and show that. So we would have a proton now, right, bonded to our oxygen. Still one lone pair of electrons on our oxygen, so let's show those electrons. So these electrons in magenta, right, took this proton, and that forms this bond, which gives this oxygen a plus one formal charge like that."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. So we would have a proton now, right, bonded to our oxygen. Still one lone pair of electrons on our oxygen, so let's show those electrons. So these electrons in magenta, right, took this proton, and that forms this bond, which gives this oxygen a plus one formal charge like that. And we know that because of a resonance structure we could draw for this. That makes this carbon more electrophilic, right? So that carbon is going to function as an electrophile, and therefore a nucleophile can react with it."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in magenta, right, took this proton, and that forms this bond, which gives this oxygen a plus one formal charge like that. And we know that because of a resonance structure we could draw for this. That makes this carbon more electrophilic, right? So that carbon is going to function as an electrophile, and therefore a nucleophile can react with it. And we have a nucleophile present, of course. That would be ethanol. So a molecule of ethanol comes along, functions as a nucleophile, a lone pair of electrons attacks our electrophile, kicks these pi electrons off onto this oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon is going to function as an electrophile, and therefore a nucleophile can react with it. And we have a nucleophile present, of course. That would be ethanol. So a molecule of ethanol comes along, functions as a nucleophile, a lone pair of electrons attacks our electrophile, kicks these pi electrons off onto this oxygen. So that would be the second step, nucleophilic attack. So let's go ahead and write these out. So we had step one, right, protonation of our carbonyl."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So a molecule of ethanol comes along, functions as a nucleophile, a lone pair of electrons attacks our electrophile, kicks these pi electrons off onto this oxygen. So that would be the second step, nucleophilic attack. So let's go ahead and write these out. So we had step one, right, protonation of our carbonyl. All right, so step two, nucleophilic attack. And so when the nucleophile attacks, we would have this oxygen over here would now have two lone pairs of electrons around it. So let's show those."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we had step one, right, protonation of our carbonyl. All right, so step two, nucleophilic attack. And so when the nucleophile attacks, we would have this oxygen over here would now have two lone pairs of electrons around it. So let's show those. All right, so let's make them blue here. So these electrons, right, moved out onto our oxygen like that. And we just formed a bond between the oxygen on our ethanol and this carbon."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's show those. All right, so let's make them blue here. So these electrons, right, moved out onto our oxygen like that. And we just formed a bond between the oxygen on our ethanol and this carbon. So we have a bond here like that. And this still had a hydrogen attached to it, an ethyl group, and a plus one formal charge like that. All right, so next, let's get a little bit of room down here."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And we just formed a bond between the oxygen on our ethanol and this carbon. So we have a bond here like that. And this still had a hydrogen attached to it, an ethyl group, and a plus one formal charge like that. All right, so next, let's get a little bit of room down here. The third step would be deprotonation. So let me go ahead and write that. So step three, we deprotonate."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so next, let's get a little bit of room down here. The third step would be deprotonation. So let me go ahead and write that. So step three, we deprotonate. So another molecule of ethanol could come along and function as a base. And a lone pair of electrons on ethanol could take this proton, which leaves these electrons behind on our oxygen. So let's go ahead and show that."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So step three, we deprotonate. So another molecule of ethanol could come along and function as a base. And a lone pair of electrons on ethanol could take this proton, which leaves these electrons behind on our oxygen. So let's go ahead and show that. So next, we would have once again our ring. We would have an OH over here on the left. Let's go ahead and put in those electrons."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. So next, we would have once again our ring. We would have an OH over here on the left. Let's go ahead and put in those electrons. And then over here on the right, we would have this time two lone pairs of electrons on our oxygen. So let me go ahead and use green for those. These electrons right in here moved off onto our oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and put in those electrons. And then over here on the right, we would have this time two lone pairs of electrons on our oxygen. So let me go ahead and use green for those. These electrons right in here moved off onto our oxygen. And so if you look at that structure closely, that's a hemiacetal. All right, so deprotonation yields our hemiacetal here, which is the intermediate in our, which is an intermediate in our reaction. All right, so next step."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "These electrons right in here moved off onto our oxygen. And so if you look at that structure closely, that's a hemiacetal. All right, so deprotonation yields our hemiacetal here, which is the intermediate in our, which is an intermediate in our reaction. All right, so next step. Next step here is protonation. So let me go ahead and mark this as being step four. We're going to protonate this OH over here on the left."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so next step. Next step here is protonation. So let me go ahead and mark this as being step four. We're going to protonate this OH over here on the left. And so one of the possibilities would be once again a protonated ethanol over here functioning as an acid. So let's go ahead and draw that. All right, so plus one formal charge on this oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We're going to protonate this OH over here on the left. And so one of the possibilities would be once again a protonated ethanol over here functioning as an acid. So let's go ahead and draw that. All right, so plus one formal charge on this oxygen. And a lone pair of electrons picks up a proton, leaving these electrons behind. And so let's go ahead and show that. So we protonate the OH."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "All right, so plus one formal charge on this oxygen. And a lone pair of electrons picks up a proton, leaving these electrons behind. And so let's go ahead and show that. So we protonate the OH. And the reason why protonating the OH would be good is that would give us water as a leaving group. So let's once again show those electrons. Let's use magenta again."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we protonate the OH. And the reason why protonating the OH would be good is that would give us water as a leaving group. So let's once again show those electrons. Let's use magenta again. So these electrons right here picked up a proton. All right, and let's show these electrons as being that bond now. And then over here on the right, we have once again our oxygen and ethyl."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's use magenta again. So these electrons right here picked up a proton. All right, and let's show these electrons as being that bond now. And then over here on the right, we have once again our oxygen and ethyl. And then we have two lone pairs of electrons. And then let's keep this lone pair green right here. And then since we protonated the OH, we get a plus one formal charge on this oxygen here."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then over here on the right, we have once again our oxygen and ethyl. And then we have two lone pairs of electrons. And then let's keep this lone pair green right here. And then since we protonated the OH, we get a plus one formal charge on this oxygen here. And if you look closely, let me use red for this. If you look closely over here, you can kind of see water hiding. We know water is an excellent leaving group."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then since we protonated the OH, we get a plus one formal charge on this oxygen here. And if you look closely, let me use red for this. If you look closely over here, you can kind of see water hiding. We know water is an excellent leaving group. So if these electrons in green moved in here to reform the double bond, then that would kick these electrons off onto the oxygen. And then we would have water. So this is the dehydration portion."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We know water is an excellent leaving group. So if these electrons in green moved in here to reform the double bond, then that would kick these electrons off onto the oxygen. And then we would have water. So this is the dehydration portion. So we're going to form water. So let me go ahead and mark this as being the next step. So in the next step, when those electrons kick in there, so this would be step five."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is the dehydration portion. So we're going to form water. So let me go ahead and mark this as being the next step. So in the next step, when those electrons kick in there, so this would be step five. So step five, we're going to lose H2O, so the dehydration step. And we would be left with, once again, our ring. And this time, a double bond to this oxygen with an ethyl coming off of that oxygen like this."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So in the next step, when those electrons kick in there, so this would be step five. So step five, we're going to lose H2O, so the dehydration step. And we would be left with, once again, our ring. And this time, a double bond to this oxygen with an ethyl coming off of that oxygen like this. So let's go ahead and make sure we still have a lone pair of electrons on this oxygen, a plus one formal charge, and the electrons in green. So these electrons in here moved in here to give us our double bond once again. And once again, we have a plus one formal charge in the oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And this time, a double bond to this oxygen with an ethyl coming off of that oxygen like this. So let's go ahead and make sure we still have a lone pair of electrons on this oxygen, a plus one formal charge, and the electrons in green. So these electrons in here moved in here to give us our double bond once again. And once again, we have a plus one formal charge in the oxygen. So if you drew a resonance structure for this, you would actually have this carbon as being very electrophilic. So once again, we're going to get a nucleophile attacking our electrophile in the next step. So this would be step six."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And once again, we have a plus one formal charge in the oxygen. So if you drew a resonance structure for this, you would actually have this carbon as being very electrophilic. So once again, we're going to get a nucleophile attacking our electrophile in the next step. So this would be step six. So step six would be a nucleophilic attack. So in step six, a nucleophile comes along. Once again, ethanol is our nucleophile."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this would be step six. So step six would be a nucleophilic attack. So in step six, a nucleophile comes along. Once again, ethanol is our nucleophile. So here is ethanol. So let's go ahead and show ethanol right here with lone pairs of electrons. And one of these lone pairs of electrons, of course, would attack our electrophile."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Once again, ethanol is our nucleophile. So here is ethanol. So let's go ahead and show ethanol right here with lone pairs of electrons. And one of these lone pairs of electrons, of course, would attack our electrophile. So nucleophile attacks electrophile. And that would push these electrons in here off onto this oxygen. So let's go ahead and draw what we have next."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And one of these lone pairs of electrons, of course, would attack our electrophile. So nucleophile attacks electrophile. And that would push these electrons in here off onto this oxygen. So let's go ahead and draw what we have next. So we now have an oxygen with still a hydrogen on it, an ethyl right here, a lone pair of electrons, a plus one formal charge on this oxygen. So let's highlight those electrons. So in magenta here, these electrons formed a bond."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw what we have next. So we now have an oxygen with still a hydrogen on it, an ethyl right here, a lone pair of electrons, a plus one formal charge on this oxygen. So let's highlight those electrons. So in magenta here, these electrons formed a bond. So that oxygen is now bonded to that carbon. And then over here on the right, we have an oxygen with an ethyl group. And now there are two lone pairs of electrons on this oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So in magenta here, these electrons formed a bond. So that oxygen is now bonded to that carbon. And then over here on the right, we have an oxygen with an ethyl group. And now there are two lone pairs of electrons on this oxygen. So we are almost there, right? Last step. So step seven would be a deprotonation step."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And now there are two lone pairs of electrons on this oxygen. So we are almost there, right? Last step. So step seven would be a deprotonation step. So in step seven here, all we do is take that proton off, and we would form our acetal product. So once again, we could have a molecule of ethanol come along and function as a base. And so a lone pair of electrons take this proton, leaving these electrons behind on the oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So step seven would be a deprotonation step. So in step seven here, all we do is take that proton off, and we would form our acetal product. So once again, we could have a molecule of ethanol come along and function as a base. And so a lone pair of electrons take this proton, leaving these electrons behind on the oxygen. And then finally, we are able to draw our acetal product. So we would have, let's go ahead and make this a little bit more angled. So on the left, we would have our oxygen with an ethyl."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so a lone pair of electrons take this proton, leaving these electrons behind on the oxygen. And then finally, we are able to draw our acetal product. So we would have, let's go ahead and make this a little bit more angled. So on the left, we would have our oxygen with an ethyl. And then this carbon is also bonded to another oxygen with an ethyl coming off of it like that. And so let's go ahead and show those final electrons here on our oxygen like this. And once again, highlight these electrons came off onto our oxygen."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So on the left, we would have our oxygen with an ethyl. And then this carbon is also bonded to another oxygen with an ethyl coming off of it like that. And so let's go ahead and show those final electrons here on our oxygen like this. And once again, highlight these electrons came off onto our oxygen. So we've formed our acetal product. So a very long mechanism, and it's a little bit challenging. I think it was just easier to go through an actual reaction for this."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And once again, highlight these electrons came off onto our oxygen. So we've formed our acetal product. So a very long mechanism, and it's a little bit challenging. I think it was just easier to go through an actual reaction for this. So that's a long one. Let's do two quick problems to think about the acetal product here. So let's look at this next reaction."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "I think it was just easier to go through an actual reaction for this. So that's a long one. Let's do two quick problems to think about the acetal product here. So let's look at this next reaction. So here we have acetaldehyde. And then here we have butanol. This time we're going to use toluenesulfonic acid as our acid catalyst."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this next reaction. So here we have acetaldehyde. And then here we have butanol. This time we're going to use toluenesulfonic acid as our acid catalyst. And one of the things you could do is increase the concentration of one of your reactants. And if you increase the concentration of acid outside, you can actually drive this reaction to completion. If you think about the structure of the product, we know that we're going to be adding on this portion of our alcohol to this carbon."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This time we're going to use toluenesulfonic acid as our acid catalyst. And one of the things you could do is increase the concentration of one of your reactants. And if you increase the concentration of acid outside, you can actually drive this reaction to completion. If you think about the structure of the product, we know that we're going to be adding on this portion of our alcohol to this carbon. And that's going to happen twice. So therefore, we need to make sure we have two carbons. And those are our two carbons."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "If you think about the structure of the product, we know that we're going to be adding on this portion of our alcohol to this carbon. And that's going to happen twice. So therefore, we need to make sure we have two carbons. And those are our two carbons. And then we have that carbon bonded to an oxygen. We have four carbons in our product, so 1, 2, 3, 4. And then we know that it's going to be bonded to another oxygen and still 1, 2, 3, 4."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And those are our two carbons. And then we have that carbon bonded to an oxygen. We have four carbons in our product, so 1, 2, 3, 4. And then we know that it's going to be bonded to another oxygen and still 1, 2, 3, 4. So that would be our acetal product. So let's highlight some carbons here so we can follow along. So this carbon right here would be this carbon on the right."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we know that it's going to be bonded to another oxygen and still 1, 2, 3, 4. So that would be our acetal product. So let's highlight some carbons here so we can follow along. So this carbon right here would be this carbon on the right. This carbon that used to be our carbonyl carbon is going to be right here. And then let's switch colors for the butanol molecule. So oxygen right here would be this one and this one."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here would be this carbon on the right. This carbon that used to be our carbonyl carbon is going to be right here. And then let's switch colors for the butanol molecule. So oxygen right here would be this one and this one. And then we have 1, 2, 3, 4. So we have 1, 2, 3, 4, 1, 2, 3, and 4. So counting your carbons is one of the techniques you can use to figure out your final acetal product."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So oxygen right here would be this one and this one. And then we have 1, 2, 3, 4. So we have 1, 2, 3, 4, 1, 2, 3, and 4. So counting your carbons is one of the techniques you can use to figure out your final acetal product. Let's do one more reaction here. So this would be a ketone. So we have a four-carbon ketone, so butanone, reacting it with ethylene glycol."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So counting your carbons is one of the techniques you can use to figure out your final acetal product. Let's do one more reaction here. So this would be a ketone. So we have a four-carbon ketone, so butanone, reacting it with ethylene glycol. And once again, we use toluene sulfonic acid as our catalyst. And this one's a little bit different because we can see we have a diol as one of our reactants. Up here, we just had butanol, only one OH."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have a four-carbon ketone, so butanone, reacting it with ethylene glycol. And once again, we use toluene sulfonic acid as our catalyst. And this one's a little bit different because we can see we have a diol as one of our reactants. Up here, we just had butanol, only one OH. But this one has two on it. So trying to figure out the product here, sometimes it helps just to run through the mechanism really quickly. And so the toluene sulfonic acid is going to help us to protonate our carbonyl."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Up here, we just had butanol, only one OH. But this one has two on it. So trying to figure out the product here, sometimes it helps just to run through the mechanism really quickly. And so the toluene sulfonic acid is going to help us to protonate our carbonyl. And then we have our nucleophile attack. So one of these OHs is going to attack here. And so without going through all the steps in the mechanism again, that was obviously a pretty complicated mechanism."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so the toluene sulfonic acid is going to help us to protonate our carbonyl. And then we have our nucleophile attack. So one of these OHs is going to attack here. And so without going through all the steps in the mechanism again, that was obviously a pretty complicated mechanism. I'll jump to one of the later steps of the mechanism where we have already lost water, so minus H2O. So we've already gotten past the dehydration step. And then that would give us this as our intermediate."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so without going through all the steps in the mechanism again, that was obviously a pretty complicated mechanism. I'll jump to one of the later steps of the mechanism where we have already lost water, so minus H2O. So we've already gotten past the dehydration step. And then that would give us this as our intermediate. So there's actually going to be a plus 1 formal charge on this oxygen. And then we have these two carbons over here, and then our other OH on this side. So let's go ahead and color coordinate some of our atoms once again."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then that would give us this as our intermediate. So there's actually going to be a plus 1 formal charge on this oxygen. And then we have these two carbons over here, and then our other OH on this side. So let's go ahead and color coordinate some of our atoms once again. So this oxygen has already bonded. We've already lost water. So that oxygen is this oxygen right here."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and color coordinate some of our atoms once again. So this oxygen has already bonded. We've already lost water. So that oxygen is this oxygen right here. And then we still have another OH on this molecule. And that's this one over here like that. So when we get to this step, we're actually going to get an intramolecular nucleophilic attack."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So that oxygen is this oxygen right here. And then we still have another OH on this molecule. And that's this one over here like that. So when we get to this step, we're actually going to get an intramolecular nucleophilic attack. So these electrons are going to attack this carbon and kick these electrons off onto this oxygen. And so when you think about the final product, you're actually going to get a cyclic product here, a cyclic acetal. So we would have our four carbons."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So when we get to this step, we're actually going to get an intramolecular nucleophilic attack. So these electrons are going to attack this carbon and kick these electrons off onto this oxygen. And so when you think about the final product, you're actually going to get a cyclic product here, a cyclic acetal. So we would have our four carbons. And then we would have this oxygen, and then two carbons, and then this oxygen. And they're both bonded to this carbon right here. And so once again, let's highlight some of those carbons."}, {"video_title": "Formation of acetals Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our four carbons. And then we would have this oxygen, and then two carbons, and then this oxygen. And they're both bonded to this carbon right here. And so once again, let's highlight some of those carbons. So this carbon right here and this carbon right here are this carbon and this carbon, and in our final product like that. And so this is a cyclic acetal that we have formed, so a little bit trickier than the previous reaction. So in the next video, we'll see a use of cyclic acetals as a protecting group."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is the dominant place or the most likely place to find mountains or volcanoes on the surface of the earth. But that's not the only place that mountains or volcanoes can form. And probably the biggest example of volcanic activity, or the most popular one, this might be a slightly American, Amerocentric point of view. But the most often cited example of volcanic activity away from a plate boundary is Hawaii. So this right here, these are the Hawaiian islands. This is the big island of Hawaii. And it is experiencing an active volcano."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the most often cited example of volcanic activity away from a plate boundary is Hawaii. So this right here, these are the Hawaiian islands. This is the big island of Hawaii. And it is experiencing an active volcano. Lava or magma is flowing from underneath the ground. And once it surfaces, we call it lava. And that lava is actively making the island bigger."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it is experiencing an active volcano. Lava or magma is flowing from underneath the ground. And once it surfaces, we call it lava. And that lava is actively making the island bigger. So where is that volcanic activity coming from? And then how can we think about that volcanic activity or that kind of heat rising from below the surface of the earth to explain some of the geological features we see around Hawaii? So what we think is happening, and once again, this is all theory right here, is that Hawaii is sitting on top of a hotspot."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that lava is actively making the island bigger. So where is that volcanic activity coming from? And then how can we think about that volcanic activity or that kind of heat rising from below the surface of the earth to explain some of the geological features we see around Hawaii? So what we think is happening, and once again, this is all theory right here, is that Hawaii is sitting on top of a hotspot. In particular, the big island of Hawaii is sitting on top of the hotspot right now. And this hotspot, there's different theories on how it might emerge. But we think that at the mantle core boundary, and I don't know in this diagram whether they intended this white area to be the core, but let's just say that this is the outer core down here."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what we think is happening, and once again, this is all theory right here, is that Hawaii is sitting on top of a hotspot. In particular, the big island of Hawaii is sitting on top of the hotspot right now. And this hotspot, there's different theories on how it might emerge. But we think that at the mantle core boundary, and I don't know in this diagram whether they intended this white area to be the core, but let's just say that this is the outer core down here. So let's just say that this is the outer core for the sake of explaining things. It's possible that plumes of very hot material can kind of, just based on the fluid dynamics of what's happening at that mantle outer core boundary, that plumes of really hot material can kind of rise up. Let me do this in a darker color."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we think that at the mantle core boundary, and I don't know in this diagram whether they intended this white area to be the core, but let's just say that this is the outer core down here. So let's just say that this is the outer core for the sake of explaining things. It's possible that plumes of very hot material can kind of, just based on the fluid dynamics of what's happening at that mantle outer core boundary, that plumes of really hot material can kind of rise up. Let me do this in a darker color. Can rise up from the outer core. They would rise up from the outer core and then create hotspots underneath the moving lithospheric plates. Now, we don't know for sure whether the hotspots are being created by these mantle plumes, this material formed or heated up at the outer core mantle boundary."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me do this in a darker color. Can rise up from the outer core. They would rise up from the outer core and then create hotspots underneath the moving lithospheric plates. Now, we don't know for sure whether the hotspots are being created by these mantle plumes, this material formed or heated up at the outer core mantle boundary. But what we do feel pretty confident about is that there is this hotspot here. And it's independent of any of those convection patterns that we saw. I shouldn't say independent."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, we don't know for sure whether the hotspots are being created by these mantle plumes, this material formed or heated up at the outer core mantle boundary. But what we do feel pretty confident about is that there is this hotspot here. And it's independent of any of those convection patterns that we saw. I shouldn't say independent. It's obviously all related because we have all this fluidic motion going on in the mantle. But it's separate on some degree from all of those convection patterns that we talked about that would actually cause the plates to move. And to a large degree, or the way we think about it right now, this is stationary."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I shouldn't say independent. It's obviously all related because we have all this fluidic motion going on in the mantle. But it's separate on some degree from all of those convection patterns that we talked about that would actually cause the plates to move. And to a large degree, or the way we think about it right now, this is stationary. This hotspot is stationary relative to the plates. And the reason why we feel pretty good about thinking that it's stationary relative to the plates is we see this notion right here. If you look at the volcanic rock in Kauai, which is one of the older inhabited Hawaiian islands, the oldest rock that we've observed there is 5.5 million years old."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And to a large degree, or the way we think about it right now, this is stationary. This hotspot is stationary relative to the plates. And the reason why we feel pretty good about thinking that it's stationary relative to the plates is we see this notion right here. If you look at the volcanic rock in Kauai, which is one of the older inhabited Hawaiian islands, the oldest rock that we've observed there is 5.5 million years old. And it's all volcanic rock. Now, the oldest rock we've observed on the big island is about 700,000 years old. We also know that the Pacific plate, you could look at this diagram right over here, is moving in this general direction."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you look at the volcanic rock in Kauai, which is one of the older inhabited Hawaiian islands, the oldest rock that we've observed there is 5.5 million years old. And it's all volcanic rock. Now, the oldest rock we've observed on the big island is about 700,000 years old. We also know that the Pacific plate, you could look at this diagram right over here, is moving in this general direction. In fact, we know it from GPS measurements. It's moving exactly in the direction that the Hawaiian islands are kind of distributed in. So frankly, the only good explanation for why we see this pattern, why we see newer land here, and then as we go further and further up the Hawaiian island chain, we see older and older land."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We also know that the Pacific plate, you could look at this diagram right over here, is moving in this general direction. In fact, we know it from GPS measurements. It's moving exactly in the direction that the Hawaiian islands are kind of distributed in. So frankly, the only good explanation for why we see this pattern, why we see newer land here, and then as we go further and further up the Hawaiian island chain, we see older and older land. And actually, if we keep going, we have the Leeward Islands over here. And as we keep measuring the rock on the Leeward Islands, they get older and older as you go to the northwest. And then if you even look at what's below the ocean, this is the big island of Hawaii."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So frankly, the only good explanation for why we see this pattern, why we see newer land here, and then as we go further and further up the Hawaiian island chain, we see older and older land. And actually, if we keep going, we have the Leeward Islands over here. And as we keep measuring the rock on the Leeward Islands, they get older and older as you go to the northwest. And then if you even look at what's below the ocean, this is the big island of Hawaii. These are the main Hawaiian islands. These are the Leeward Islands. But you see, even beyond that, submerged under the Pacific Ocean, you continue to see a chain of islands."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then if you even look at what's below the ocean, this is the big island of Hawaii. These are the main Hawaiian islands. These are the Leeward Islands. But you see, even beyond that, submerged under the Pacific Ocean, you continue to see a chain of islands. So the explanation for what's happening here is that you have a stationary hotspot that is right now underneath the big island of Hawaii. And I just want to be clear. The big island is called the island of Hawaii."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But you see, even beyond that, submerged under the Pacific Ocean, you continue to see a chain of islands. So the explanation for what's happening here is that you have a stationary hotspot that is right now underneath the big island of Hawaii. And I just want to be clear. The big island is called the island of Hawaii. It is one of the islands in the state of Hawaii. So I don't want to cause you confusion. I'll just call it the big island from here on out."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The big island is called the island of Hawaii. It is one of the islands in the state of Hawaii. So I don't want to cause you confusion. I'll just call it the big island from here on out. So the hotspot is right under the big island. But if you were to rewind 5 million years ago, the entire Pacific plate was probably on the order of about 150, 200 miles, however far Kauai is from the big island. It was probably shifted that much to the southeast if you go back 5 million years ago."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I'll just call it the big island from here on out. So the hotspot is right under the big island. But if you were to rewind 5 million years ago, the entire Pacific plate was probably on the order of about 150, 200 miles, however far Kauai is from the big island. It was probably shifted that much to the southeast if you go back 5 million years ago. So 5 million years ago, when all of this was shifted down and to the right, then Kauai was on top of the hotspot. And so this is how each of these islands are formed. If you rewind a ton of years, then maybe this area over here on the Pacific plate was over the hotspot."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It was probably shifted that much to the southeast if you go back 5 million years ago. So 5 million years ago, when all of this was shifted down and to the right, then Kauai was on top of the hotspot. And so this is how each of these islands are formed. If you rewind a ton of years, then maybe this area over here on the Pacific plate was over the hotspot. An island formed there. Then the Pacific plate kept moving to the northwest. And new islands, new volcanoes kept forming."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If you rewind a ton of years, then maybe this area over here on the Pacific plate was over the hotspot. An island formed there. Then the Pacific plate kept moving to the northwest. And new islands, new volcanoes kept forming. Those volcanoes would release lava that would keep piling up, eventually go above the surface of the water and form this whole chain of islands. And as the whole Pacific plate kept moving to the northwest, it kept forming new islands. Now the one question you might ask is, well, how come the big island is bigger?"}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And new islands, new volcanoes kept forming. Those volcanoes would release lava that would keep piling up, eventually go above the surface of the water and form this whole chain of islands. And as the whole Pacific plate kept moving to the northwest, it kept forming new islands. Now the one question you might ask is, well, how come the big island is bigger? Has the plate kind of paused over there? Is it spending more time over the hotspot so that more lava can kind of form there to form this? Essentially, it's an underwater mountain that's now also above the water."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now the one question you might ask is, well, how come the big island is bigger? Has the plate kind of paused over there? Is it spending more time over the hotspot so that more lava can kind of form there to form this? Essentially, it's an underwater mountain that's now also above the water. And actually, if you go from the base of the Pacific Ocean to the top of the big island of Hawaii, it's actually 50% higher than Mount Everest. So you could really just view it as a big mountain. But the question is, this looks so much bigger than Kauai, and they keep getting smaller as you keep going to the northwest."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Essentially, it's an underwater mountain that's now also above the water. And actually, if you go from the base of the Pacific Ocean to the top of the big island of Hawaii, it's actually 50% higher than Mount Everest. So you could really just view it as a big mountain. But the question is, this looks so much bigger than Kauai, and they keep getting smaller as you keep going to the northwest. Is somehow the Pacific plate slowing? Is it spending more time here? And the answer is, it's probably not slowing."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the question is, this looks so much bigger than Kauai, and they keep getting smaller as you keep going to the northwest. Is somehow the Pacific plate slowing? Is it spending more time here? And the answer is, it's probably not slowing. What's happening is, at one time, Kauai was also probably a relatively large island. If you rewind maybe 5 million years ago, Kauai also might have been about that big. But over 5 million years, it's just experienced a ton of erosion."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the answer is, it's probably not slowing. What's happening is, at one time, Kauai was also probably a relatively large island. If you rewind maybe 5 million years ago, Kauai also might have been about that big. But over 5 million years, it's just experienced a ton of erosion. Remember, once it moved over the hotspot, new land wasn't being created. It's in the middle of the Pacific Ocean. It's experiencing weather."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But over 5 million years, it's just experienced a ton of erosion. Remember, once it moved over the hotspot, new land wasn't being created. It's in the middle of the Pacific Ocean. It's experiencing weather. 5 million years is a long period of time. And so it just got eroded over that time. So the older the island is, the more eroded it's going to be, and the smaller it's going to be."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's experiencing weather. 5 million years is a long period of time. And so it just got eroded over that time. So the older the island is, the more eroded it's going to be, and the smaller it's going to be. So if you go to these underwater mountains up here that don't even surface above the ocean, at one time, they might have surfaced. But due to the ocean and weather and whatnot, they've just been eroded over time to become smaller and smaller kind of remnants of volcanoes. So anyway, I thought you would find that entertaining of how the Hawaiian Islands actually got formed and how we can actually have these hotspots and this volcanic activity and actually even earthquake activity outside of actually plate boundaries."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the older the island is, the more eroded it's going to be, and the smaller it's going to be. So if you go to these underwater mountains up here that don't even surface above the ocean, at one time, they might have surfaced. But due to the ocean and weather and whatnot, they've just been eroded over time to become smaller and smaller kind of remnants of volcanoes. So anyway, I thought you would find that entertaining of how the Hawaiian Islands actually got formed and how we can actually have these hotspots and this volcanic activity and actually even earthquake activity outside of actually plate boundaries. Actually, while we're looking at this diagram, we talked about the trenches at plate boundaries. You can actually see it here because this shows the depth. And the really dark, dark, dark, dark blue is really deep parts of the ocean."}, {"video_title": "Hawaiian islands formation Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So anyway, I thought you would find that entertaining of how the Hawaiian Islands actually got formed and how we can actually have these hotspots and this volcanic activity and actually even earthquake activity outside of actually plate boundaries. Actually, while we're looking at this diagram, we talked about the trenches at plate boundaries. You can actually see it here because this shows the depth. And the really dark, dark, dark, dark blue is really deep parts of the ocean. So this right here is the Mariana Trench. And you can see over here the Pacific Plate just getting abducted, or not abducted, getting subducted into other plates underneath and forms these trenches here. Anyway, hopefully you found that entertaining."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then that energy begins to offset the actual gravitational force so the whole star, what's now a star, does not collapse on itself. And once we're there, we're now in the main sequence of a star. What I want to do in this video is to take off from that starting point and think about what happens in the star next. So in the main sequence, we have the core of the star. So this is the star's core. And you have hydrogen fusing into helium. And it's releasing just a ton of energy."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in the main sequence, we have the core of the star. So this is the star's core. And you have hydrogen fusing into helium. And it's releasing just a ton of energy. And that energy is what keeps the core from imploding. It's kind of the outward force to offset the gravitational force that wants to implode everything, that wants to crush everything. And so you have the core of a star, a star like the sun."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's releasing just a ton of energy. And that energy is what keeps the core from imploding. It's kind of the outward force to offset the gravitational force that wants to implode everything, that wants to crush everything. And so you have the core of a star, a star like the sun. And that energy then heats up all of the other gas on the outside of the core to create that really bright object that we see as a star, or in our case, in our sun's case, the sun. Now, as the hydrogen is fusing into helium, you can imagine that more and more helium is forming in the core. So I'll do the helium as green."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so you have the core of a star, a star like the sun. And that energy then heats up all of the other gas on the outside of the core to create that really bright object that we see as a star, or in our case, in our sun's case, the sun. Now, as the hydrogen is fusing into helium, you can imagine that more and more helium is forming in the core. So I'll do the helium as green. So more and more helium forms in the core. The closer you get to the center, the higher the pressures will be and the faster that this fusion, this ignition, will happen. In fact, the bigger the mass of the star, the more the pressure, the faster the fusion occurs."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I'll do the helium as green. So more and more helium forms in the core. The closer you get to the center, the higher the pressures will be and the faster that this fusion, this ignition, will happen. In fact, the bigger the mass of the star, the more the pressure, the faster the fusion occurs. And so you have this helium building up inside of the core as this hydrogen in the core gets fused. Now, what's going to happen there? Helium is a more dense atom."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, the bigger the mass of the star, the more the pressure, the faster the fusion occurs. And so you have this helium building up inside of the core as this hydrogen in the core gets fused. Now, what's going to happen there? Helium is a more dense atom. It's packing more mass in a smaller space. So as more and more of this hydrogen here turns into helium, what you're going to have is the core itself is going to shrink. So let me draw a smaller core here."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Helium is a more dense atom. It's packing more mass in a smaller space. So as more and more of this hydrogen here turns into helium, what you're going to have is the core itself is going to shrink. So let me draw a smaller core here. So the core itself is going to shrink. And now it has a lot more helium in it. And let's just stick it to the extreme point where it's all helium, where it's depleted."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me draw a smaller core here. So the core itself is going to shrink. And now it has a lot more helium in it. And let's just stick it to the extreme point where it's all helium, where it's depleted. But it's been shrinking. It's much denser. That same amount of mass that was in this sphere is now in a denser sphere, in a helium sphere."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's just stick it to the extreme point where it's all helium, where it's depleted. But it's been shrinking. It's much denser. That same amount of mass that was in this sphere is now in a denser sphere, in a helium sphere. So it's going to have just as much attraction to it, gravitational attraction. But things can get even closer to it. And we know that the closer you are to a mass, the stronger the pull of gravity."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That same amount of mass that was in this sphere is now in a denser sphere, in a helium sphere. So it's going to have just as much attraction to it, gravitational attraction. But things can get even closer to it. And we know that the closer you are to a mass, the stronger the pull of gravity. So then instead of having just the hydrogen fusion occurring at the core, you're now going to have hydrogen fusion in a shell around the core. So now you're going to have hydrogen fusing in a shell around the core. Let me just be clear."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we know that the closer you are to a mass, the stronger the pull of gravity. So then instead of having just the hydrogen fusion occurring at the core, you're now going to have hydrogen fusion in a shell around the core. So now you're going to have hydrogen fusing in a shell around the core. Let me just be clear. This isn't just happens all of a sudden. It is a gradual process. As we have more and more helium in the core, the core gets denser and denser and denser."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me just be clear. This isn't just happens all of a sudden. It is a gradual process. As we have more and more helium in the core, the core gets denser and denser and denser. And so the pressures become even larger and larger near the core. Because you're able to get closer to a more massive core since it is now more dense. And as that pressure near the core increases even more and more, the fusion reaction happens faster and faster and faster."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "As we have more and more helium in the core, the core gets denser and denser and denser. And so the pressures become even larger and larger near the core. Because you're able to get closer to a more massive core since it is now more dense. And as that pressure near the core increases even more and more, the fusion reaction happens faster and faster and faster. Until you get to this point. So here, let me be clear. You have a helium core."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as that pressure near the core increases even more and more, the fusion reaction happens faster and faster and faster. Until you get to this point. So here, let me be clear. You have a helium core. All of the hydrogen in the core has been used up. And then you have the hydrogen right outside of the core is now under enormous pressure. It's actually under more pressure than it was when it was just a pure hydrogen core."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have a helium core. All of the hydrogen in the core has been used up. And then you have the hydrogen right outside of the core is now under enormous pressure. It's actually under more pressure than it was when it was just a pure hydrogen core. Because there's so much mass on the outside here trying to, I guess you could say, exerting downward or gravitational force. Down trying to get to that even denser helium core. Because everything is able to get closer in."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's actually under more pressure than it was when it was just a pure hydrogen core. Because there's so much mass on the outside here trying to, I guess you could say, exerting downward or gravitational force. Down trying to get to that even denser helium core. Because everything is able to get closer in. And so now you have fusion occurring even faster. And it's occurring over a larger radius. So this faster fusion over a larger radius is then going to expel, the force is now going to expel, the energy that's released from this fusion is now going to expel these outer layers of the star even further."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because everything is able to get closer in. And so now you have fusion occurring even faster. And it's occurring over a larger radius. So this faster fusion over a larger radius is then going to expel, the force is now going to expel, the energy that's released from this fusion is now going to expel these outer layers of the star even further. So the whole time, this gradual process, as the hydrogen turns into helium or fuses into helium in the core, what the hydrogen right outside of the core, right outside of that area, starts to burn faster and faster. It starts to, or I shouldn't say burn, it starts to fuse faster and faster. And over a larger and larger radius."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this faster fusion over a larger radius is then going to expel, the force is now going to expel, the energy that's released from this fusion is now going to expel these outer layers of the star even further. So the whole time, this gradual process, as the hydrogen turns into helium or fuses into helium in the core, what the hydrogen right outside of the core, right outside of that area, starts to burn faster and faster. It starts to, or I shouldn't say burn, it starts to fuse faster and faster. And over a larger and larger radius. So the unintuitive thing is the fusion is happening faster over a larger radius. And the reason that is is because you have even a denser core that is causing even more gravitational pressure. And as that's happening, the star is getting brighter."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And over a larger and larger radius. So the unintuitive thing is the fusion is happening faster over a larger radius. And the reason that is is because you have even a denser core that is causing even more gravitational pressure. And as that's happening, the star is getting brighter. And it's also the fusion reactions, since they're happening in a more intense way and over a larger radius, are able to expel the material of the star even larger. So the radius of the star itself is getting bigger and bigger and bigger. So if this star would look like this, if this star, maybe let me draw it in white, this star looked like this."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as that's happening, the star is getting brighter. And it's also the fusion reactions, since they're happening in a more intense way and over a larger radius, are able to expel the material of the star even larger. So the radius of the star itself is getting bigger and bigger and bigger. So if this star would look like this, if this star, maybe let me draw it in white, this star looked like this. That's not white. This star looked like, what's happening to my color changer? There you go."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if this star would look like this, if this star, maybe let me draw it in white, this star looked like this. That's not white. This star looked like, what's happening to my color changer? There you go. OK, this star looked like this right over here. Now this star over here, since a faster fusion reaction is happening over a larger radius, is going to be far larger, and I'm not even drawing it to scale. In the case of our sun, when it gets to this point, it's going to be 100 times the diameter."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There you go. OK, this star looked like this right over here. Now this star over here, since a faster fusion reaction is happening over a larger radius, is going to be far larger, and I'm not even drawing it to scale. In the case of our sun, when it gets to this point, it's going to be 100 times the diameter. And at this point, it is a red giant. And the reason why it's redder than this one over here is that even though the fusion is happening more furiously, that energy is being dissipated over a larger surface area. So the actual surface temperature of the red giant at this point is actually going to be cooler."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the case of our sun, when it gets to this point, it's going to be 100 times the diameter. And at this point, it is a red giant. And the reason why it's redder than this one over here is that even though the fusion is happening more furiously, that energy is being dissipated over a larger surface area. So the actual surface temperature of the red giant at this point is actually going to be cooler. So it's going to emit a light at a larger wavelength, a redder wavelength than this thing over here. This thing, the core, was not burning as furiously as this thing over here, but that energy was being dissipated over a smaller volume. So this had a higher surface temperature."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the actual surface temperature of the red giant at this point is actually going to be cooler. So it's going to emit a light at a larger wavelength, a redder wavelength than this thing over here. This thing, the core, was not burning as furiously as this thing over here, but that energy was being dissipated over a smaller volume. So this had a higher surface temperature. This over here, the core is no longer burning. The core is now helium that's not burning. It's getting denser and denser as the helium packs in on it itself, but the hydrogen fusion over here is occurring more intensely."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this had a higher surface temperature. This over here, the core is no longer burning. The core is now helium that's not burning. It's getting denser and denser as the helium packs in on it itself, but the hydrogen fusion over here is occurring more intensely. It's occurring in a hotter way, but the surface here is less hot because it's just a larger surface area. So the increased heat is more than mitigated by how large the star has become. Now this is going to keep happening, keep happening, and the pressures keep intensifying because more and more helium is getting produced."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's getting denser and denser as the helium packs in on it itself, but the hydrogen fusion over here is occurring more intensely. It's occurring in a hotter way, but the surface here is less hot because it's just a larger surface area. So the increased heat is more than mitigated by how large the star has become. Now this is going to keep happening, keep happening, and the pressures keep intensifying because more and more helium is getting produced. And this core keeps collapsing, and the temperature here keeps going up. So we said that the first ignition, the first fusion occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now this is going to keep happening, keep happening, and the pressures keep intensifying because more and more helium is getting produced. And this core keeps collapsing, and the temperature here keeps going up. So we said that the first ignition, the first fusion occurs at around 10 million Kelvin. This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches 100 million. But let's just talk about the case in which it does."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This thing will keep heating up until it gets to 100 million Kelvin. And now I'm talking about a star that's about as massive as the sun. Some stars will never even be massive enough to condense the core so that its temperature reaches 100 million. But let's just talk about the case in which it does. So eventually you'll get to a point, so we're still sitting in the red giant phase. So we're this huge star over here. We have this helium core, and that helium core keeps getting condensed and condensed and condensed."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But let's just talk about the case in which it does. So eventually you'll get to a point, so we're still sitting in the red giant phase. So we're this huge star over here. We have this helium core, and that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger to expand."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We have this helium core, and that helium core keeps getting condensed and condensed and condensed. And then we have a shell of hydrogen that keeps fusing into helium around it. So this is our hydrogen shell. Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger to expand. But when the temperature gets sufficiently hot, and now I think you're going to get a sense of how heavier and heavier elements form in the universe. And all of the heavy elements that you see around us, including the ones that are in you, were formed this way from initially hydrogen. When it gets hot enough at 100 million Kelvin in this core, because of such enormous pressures, then the helium itself will start to fuse."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Hydrogen fusion is occurring in this yellow shell over here that's causing the radius of the star to get bigger and bigger to expand. But when the temperature gets sufficiently hot, and now I think you're going to get a sense of how heavier and heavier elements form in the universe. And all of the heavy elements that you see around us, including the ones that are in you, were formed this way from initially hydrogen. When it gets hot enough at 100 million Kelvin in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation, you have helium and you had hydrogen, and all sorts of combinations will form, but in general, the helium is mainly going to fuse into carbon and oxygen. And it will form into other things, and it becomes much more complicated, but I don't want to go into all of the details."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "When it gets hot enough at 100 million Kelvin in this core, because of such enormous pressures, then the helium itself will start to fuse. So then we're going to have a core in here where the helium itself will start to fuse. And now we're talking about a situation, you have helium and you had hydrogen, and all sorts of combinations will form, but in general, the helium is mainly going to fuse into carbon and oxygen. And it will form into other things, and it becomes much more complicated, but I don't want to go into all of the details. But let me just show you a periodic table. I didn't have this in the last one. I somehow lost it."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it will form into other things, and it becomes much more complicated, but I don't want to go into all of the details. But let me just show you a periodic table. I didn't have this in the last one. I somehow lost it. But we see hydrogen here has one proton. It actually has no neutrons. It was getting fused in the main sequence into helium."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I somehow lost it. But we see hydrogen here has one proton. It actually has no neutrons. It was getting fused in the main sequence into helium. Two protons, two neutrons. We would need four of these to get one of those, because this actually has an atomic mass of 4, if we're talking about helium-4. And then the helium, once we get to 100 million Kelvin, can start being fused."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It was getting fused in the main sequence into helium. Two protons, two neutrons. We would need four of these to get one of those, because this actually has an atomic mass of 4, if we're talking about helium-4. And then the helium, once we get to 100 million Kelvin, can start being fused. If you get roughly three of them, and there's all these other things that are coming and leaving the reactions, you can get to a carbon. You get four of them, at least as the starting raw material, you get to an oxygen. So we're starting to fuse heavier and heavier elements."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then the helium, once we get to 100 million Kelvin, can start being fused. If you get roughly three of them, and there's all these other things that are coming and leaving the reactions, you can get to a carbon. You get four of them, at least as the starting raw material, you get to an oxygen. So we're starting to fuse heavier and heavier elements. So what happens here is this helium is fusing into carbon and oxygen, so you start building a carbon and oxygen core. So I'm going to leave you there. I realize I'm already past my self-imposed limit of 10 minutes."}, {"video_title": "Becoming a red giant Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So we're starting to fuse heavier and heavier elements. So what happens here is this helium is fusing into carbon and oxygen, so you start building a carbon and oxygen core. So I'm going to leave you there. I realize I'm already past my self-imposed limit of 10 minutes. But what I want you to think about is what is likely to happen. What is likely to happen here if this star will never have the mass to begin to fuse this carbon and oxygen? If it does have the mass, if it is a super massive star, it eventually will be able to raise even this carbon and oxygen core to 600 million Kelvin and begin to fuse that into even heavier elements."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so if you think about the oxidation state of this carbon, if you assigned one, that's this carbon over here on your alcohol, and if you assign that oxidation state here, you'll see there's been a decrease in the oxidation state, so there's been a reduction. So lithium aluminum hydride is one way to reduce a carboxylic acid. You could also accomplish this with borane, and borane is actually more chemoselective, and so we'll talk about that at the end of the video. For right now, let's focus in on the possible mechanism of lithium aluminum hydride reacting with our carboxylic acid. So let's go ahead and redraw our carboxylic acid, so I'm going to go ahead and put in our carbonyl, and then we know the acidic proton on our carboxylic acid is the one on the oxygen. Lithium aluminum hydride can be a strong base, so I'm going to go ahead and draw in, so aluminum with four bonds to hydrogen, giving it a negative one formal charge, and then we have our lithium Li plus like that. So a hydride we know is a hydrogen with two electrons, and we know that's a strong base, so you could think about these two electrons here taking this proton and leaving these two electrons behind on your oxygen."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "For right now, let's focus in on the possible mechanism of lithium aluminum hydride reacting with our carboxylic acid. So let's go ahead and redraw our carboxylic acid, so I'm going to go ahead and put in our carbonyl, and then we know the acidic proton on our carboxylic acid is the one on the oxygen. Lithium aluminum hydride can be a strong base, so I'm going to go ahead and draw in, so aluminum with four bonds to hydrogen, giving it a negative one formal charge, and then we have our lithium Li plus like that. So a hydride we know is a hydrogen with two electrons, and we know that's a strong base, so you could think about these two electrons here taking this proton and leaving these two electrons behind on your oxygen. So an acid-base reaction is probably the first step of this mechanism, so if you take a proton from a carboxylic acid, you're left with the conjugate base, which is the carboxylate anion, so a negative one formal charge on this oxygen, and we could follow those electrons, so these electrons in magenta move on to this oxygen to form our carboxylate anion. Lithium is present, so it's probably going to bond with that oxygen, and we would also form hydrogen gas, so we would form H2, so let's show those electrons. So these electrons in blue could pick up this proton, so that forms H2, hydrogen gas, and then we took a bond away from the aluminum, so the aluminum is now only bonded to three hydrogens, and that takes away its formal charge, so formal charge of zero now on the aluminum."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So a hydride we know is a hydrogen with two electrons, and we know that's a strong base, so you could think about these two electrons here taking this proton and leaving these two electrons behind on your oxygen. So an acid-base reaction is probably the first step of this mechanism, so if you take a proton from a carboxylic acid, you're left with the conjugate base, which is the carboxylate anion, so a negative one formal charge on this oxygen, and we could follow those electrons, so these electrons in magenta move on to this oxygen to form our carboxylate anion. Lithium is present, so it's probably going to bond with that oxygen, and we would also form hydrogen gas, so we would form H2, so let's show those electrons. So these electrons in blue could pick up this proton, so that forms H2, hydrogen gas, and then we took a bond away from the aluminum, so the aluminum is now only bonded to three hydrogens, and that takes away its formal charge, so formal charge of zero now on the aluminum. So now that we've formed our carboxylate anion, that's going to react with AlH3, so let's go ahead and draw in our carboxylate anion here, so we have our carbonyl, we have our R group, and we have our oxygen with three lone pairs of electrons, so negative one formal charge. So next, our AlH3 comes along, so this is just one of the possibilities for the mechanism, and we're going to form a bond between the oxygen here and the aluminum, and we're going to form a bond between the carbon and the hydrogen. So if we think about these electrons in red right here, this carbon is partially positive because the oxygen is withdrawing some electron density from it, so these electrons in here could move in to form a bond, and at the same time, these electrons in blue here could move out to form a bond between the oxygen and the aluminum, so let's go ahead and show the result of that."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in blue could pick up this proton, so that forms H2, hydrogen gas, and then we took a bond away from the aluminum, so the aluminum is now only bonded to three hydrogens, and that takes away its formal charge, so formal charge of zero now on the aluminum. So now that we've formed our carboxylate anion, that's going to react with AlH3, so let's go ahead and draw in our carboxylate anion here, so we have our carbonyl, we have our R group, and we have our oxygen with three lone pairs of electrons, so negative one formal charge. So next, our AlH3 comes along, so this is just one of the possibilities for the mechanism, and we're going to form a bond between the oxygen here and the aluminum, and we're going to form a bond between the carbon and the hydrogen. So if we think about these electrons in red right here, this carbon is partially positive because the oxygen is withdrawing some electron density from it, so these electrons in here could move in to form a bond, and at the same time, these electrons in blue here could move out to form a bond between the oxygen and the aluminum, so let's go ahead and show the result of that. So we would now have our carbon, would be tetrahedral, so let's draw it like this. We would have a bond to this hydrogen right here, so the electrons in red move in here to form this bond, and then that carbon is still bonded to an oxygen with three lone pairs of electrons, so it still has a negative one formal charge like that. And then we would have our carbon bonded to this oxygen, this oxygen has two lone pairs of electrons on it, and we just formed a bond to the aluminum, so the electrons in blue form our bond here, and the aluminum is still bonded to two hydrogens, so we can go ahead and draw in those two hydrogens."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So if we think about these electrons in red right here, this carbon is partially positive because the oxygen is withdrawing some electron density from it, so these electrons in here could move in to form a bond, and at the same time, these electrons in blue here could move out to form a bond between the oxygen and the aluminum, so let's go ahead and show the result of that. So we would now have our carbon, would be tetrahedral, so let's draw it like this. We would have a bond to this hydrogen right here, so the electrons in red move in here to form this bond, and then that carbon is still bonded to an oxygen with three lone pairs of electrons, so it still has a negative one formal charge like that. And then we would have our carbon bonded to this oxygen, this oxygen has two lone pairs of electrons on it, and we just formed a bond to the aluminum, so the electrons in blue form our bond here, and the aluminum is still bonded to two hydrogens, so we can go ahead and draw in those two hydrogens. Alright, if we take these electrons and move them in here to form our double bond, we would have to push these electrons off onto our oxygen, and so we would have our oxygen bonded to the aluminum, and we have these two hydrogens here, and so the oxygen would now have three lone pairs of electrons giving it a negative one formal charge, and so the lithium is probably now going to bond to this oxygen, and we just reformed our carbonyl. So let's go ahead and show that. So we would form our carbonyl here like that, and then we have this hydrogen."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then we would have our carbon bonded to this oxygen, this oxygen has two lone pairs of electrons on it, and we just formed a bond to the aluminum, so the electrons in blue form our bond here, and the aluminum is still bonded to two hydrogens, so we can go ahead and draw in those two hydrogens. Alright, if we take these electrons and move them in here to form our double bond, we would have to push these electrons off onto our oxygen, and so we would have our oxygen bonded to the aluminum, and we have these two hydrogens here, and so the oxygen would now have three lone pairs of electrons giving it a negative one formal charge, and so the lithium is probably now going to bond to this oxygen, and we just reformed our carbonyl. So let's go ahead and show that. So we would form our carbonyl here like that, and then we have this hydrogen. So let's show some of those electrons. So if I say that these electrons in green here on this oxygen move in to form our carbonyl, and then we had the electrons in red, so this hydrogen is this hydrogen, and so we form an aldehyde. And so we have an aldehyde, and we have excess lithium aluminum hydride, and the lithium aluminum hydride is going to transfer a hydride to our aldehyde, so we can go ahead and reduce our aldehyde with another lithium aluminum hydride, so I'm going to draw that in."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we would form our carbonyl here like that, and then we have this hydrogen. So let's show some of those electrons. So if I say that these electrons in green here on this oxygen move in to form our carbonyl, and then we had the electrons in red, so this hydrogen is this hydrogen, and so we form an aldehyde. And so we have an aldehyde, and we have excess lithium aluminum hydride, and the lithium aluminum hydride is going to transfer a hydride to our aldehyde, so we can go ahead and reduce our aldehyde with another lithium aluminum hydride, so I'm going to draw that in. So we have lithium aluminum hydride, so a negative one formal charge on the aluminum. Our carbonyl is polarized, so partially negative oxygen, partially positive carbon right here. And so once again, we can think about these electrons, so these electrons right here from attacking this carbon, pushing these electrons in green off onto the oxygen."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so we have an aldehyde, and we have excess lithium aluminum hydride, and the lithium aluminum hydride is going to transfer a hydride to our aldehyde, so we can go ahead and reduce our aldehyde with another lithium aluminum hydride, so I'm going to draw that in. So we have lithium aluminum hydride, so a negative one formal charge on the aluminum. Our carbonyl is polarized, so partially negative oxygen, partially positive carbon right here. And so once again, we can think about these electrons, so these electrons right here from attacking this carbon, pushing these electrons in green off onto the oxygen. So let's get some more room. So to show what happens here, so this is what we've seen in earlier videos here. So now we would have our carbon bonded to an oxygen."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, we can think about these electrons, so these electrons right here from attacking this carbon, pushing these electrons in green off onto the oxygen. So let's get some more room. So to show what happens here, so this is what we've seen in earlier videos here. So now we would have our carbon bonded to an oxygen. This oxygen now has three lone pairs of electrons, so one of those lone pairs were the ones in green, so draw those in here like that, giving that a negative one formal charge. We're going to form a bond between carbon and hydrogen, so I'm going to show that. Let's use blue here."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So now we would have our carbon bonded to an oxygen. This oxygen now has three lone pairs of electrons, so one of those lone pairs were the ones in green, so draw those in here like that, giving that a negative one formal charge. We're going to form a bond between carbon and hydrogen, so I'm going to show that. Let's use blue here. So these electrons and this hydrogen are hydrides, so lithium aluminum hydride acts as a hydride transfer agent and transfers these two electrons and this hydrogen to our carbon, and then we still had this carbon in red here, sorry, this hydrogen bonded to the carbon in red like that, and then we had our R group here. So the final step, we would just protonate our alkoxide, so we could add something like dilute acid in our work up here, so if we add a dilute acid, H3O+, let me go ahead and draw that in, then our alkoxide could pick up a proton, leaving these electrons behind. So protonating our alkoxide would yield our alcohol as our product, so let me go ahead and draw those in."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Let's use blue here. So these electrons and this hydrogen are hydrides, so lithium aluminum hydride acts as a hydride transfer agent and transfers these two electrons and this hydrogen to our carbon, and then we still had this carbon in red here, sorry, this hydrogen bonded to the carbon in red like that, and then we had our R group here. So the final step, we would just protonate our alkoxide, so we could add something like dilute acid in our work up here, so if we add a dilute acid, H3O+, let me go ahead and draw that in, then our alkoxide could pick up a proton, leaving these electrons behind. So protonating our alkoxide would yield our alcohol as our product, so let me go ahead and draw those in. So we would have those two hydrogens, we have an OH, and then we would have an R group. And so that's one of the possibilities for the reduction of a carboxylic acid with lithium aluminum hydride, with the end result of transferring two hydrides. So this hydrogen and these electrons, and then also this hydrogen and these electrons, both came from lithium aluminum hydride."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So protonating our alkoxide would yield our alcohol as our product, so let me go ahead and draw those in. So we would have those two hydrogens, we have an OH, and then we would have an R group. And so that's one of the possibilities for the reduction of a carboxylic acid with lithium aluminum hydride, with the end result of transferring two hydrides. So this hydrogen and these electrons, and then also this hydrogen and these electrons, both came from lithium aluminum hydride. And the mechanism is definitely more complicated than the one I showed you, but this is a simple way to think about it. Alright, let's look at a practice problem. So if we had this compound over here on the left, and we added lithium aluminum hydride and a source of protons in our work up, we just talked about the fact that it would reduce a carboxylic acid, and we have a ketone present here as well, and we've seen in earlier videos that lithium aluminum hydride will reduce the ketone as well, and turn that into an alcohol on our work up."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen and these electrons, and then also this hydrogen and these electrons, both came from lithium aluminum hydride. And the mechanism is definitely more complicated than the one I showed you, but this is a simple way to think about it. Alright, let's look at a practice problem. So if we had this compound over here on the left, and we added lithium aluminum hydride and a source of protons in our work up, we just talked about the fact that it would reduce a carboxylic acid, and we have a ketone present here as well, and we've seen in earlier videos that lithium aluminum hydride will reduce the ketone as well, and turn that into an alcohol on our work up. So when we draw the product, we have our benzene ring, and we would turn that ketone into a secondary alcohol, as we've seen before. And then the carboxylic acid here would turn into a primary alcohol. So let's show some of these carbons."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So if we had this compound over here on the left, and we added lithium aluminum hydride and a source of protons in our work up, we just talked about the fact that it would reduce a carboxylic acid, and we have a ketone present here as well, and we've seen in earlier videos that lithium aluminum hydride will reduce the ketone as well, and turn that into an alcohol on our work up. So when we draw the product, we have our benzene ring, and we would turn that ketone into a secondary alcohol, as we've seen before. And then the carboxylic acid here would turn into a primary alcohol. So let's show some of these carbons. So this carbon right here is this carbon, and then let's do this carbon right here in red, is this carbon. So we reduced both functional groups using lithium aluminum hydride. If we did this reaction with borane, so BH3 instead, borane is actually chemoselective for the carboxylic acid group only, so it's only going to reduce this."}, {"video_title": "Reduction of carboxylic acids Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's show some of these carbons. So this carbon right here is this carbon, and then let's do this carbon right here in red, is this carbon. So we reduced both functional groups using lithium aluminum hydride. If we did this reaction with borane, so BH3 instead, borane is actually chemoselective for the carboxylic acid group only, so it's only going to reduce this. So if we draw the product using borane, we would have our benzene ring, and the borane wouldn't touch the ketone, so that is left here. It would reduce the carboxylic acid, so we would turn that into a primary alcohol. And so once again, this carbon in red is this carbon."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "Let's look at the regiochemistry of the E2 mechanism. So first we'll draw our products. We'll go through the mechanism, draw the products, and then we'll talk about why this reaction is regioselective. On the left is our alkyl halide, and here is our strong base, sodium ethoxide. So Na plus, and ethoxide has a negative charge. Since we're dealing with a strong base, we know we're going to do an E2 mechanism. The carbon that's directly bonded to the bromine would be the alpha carbon, and the carbons that are directly bonded to the alpha carbon are the beta carbons."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "On the left is our alkyl halide, and here is our strong base, sodium ethoxide. So Na plus, and ethoxide has a negative charge. Since we're dealing with a strong base, we know we're going to do an E2 mechanism. The carbon that's directly bonded to the bromine would be the alpha carbon, and the carbons that are directly bonded to the alpha carbon are the beta carbons. So I'm going to call this carbon beta one, I'm going to call this carbon beta two, and I'm going to call this carbon beta three. So sodium ethoxide is our strong base. It's going to take a proton from one of our beta carbons."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "The carbon that's directly bonded to the bromine would be the alpha carbon, and the carbons that are directly bonded to the alpha carbon are the beta carbons. So I'm going to call this carbon beta one, I'm going to call this carbon beta two, and I'm going to call this carbon beta three. So sodium ethoxide is our strong base. It's going to take a proton from one of our beta carbons. And let's think about beta one first. So I'm going to draw in a proton here on the beta one carbon, and ethoxide is a strong base, so I'm going to draw in the ethoxide anion. There's a negative one formal charge on the oxygen."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "It's going to take a proton from one of our beta carbons. And let's think about beta one first. So I'm going to draw in a proton here on the beta one carbon, and ethoxide is a strong base, so I'm going to draw in the ethoxide anion. There's a negative one formal charge on the oxygen. And the ethoxide anion is going to take this proton, and these electrons are going to move in here to form our double bond. At the same time, these electrons move off onto the bromine. So let's draw our product."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "There's a negative one formal charge on the oxygen. And the ethoxide anion is going to take this proton, and these electrons are going to move in here to form our double bond. At the same time, these electrons move off onto the bromine. So let's draw our product. If we take a proton from the beta one carbon, so it would look like that, and let's show those electrons. These electrons in here, and magenta moved in here to form our double bond. It would be the same result if we took a proton from beta two."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So let's draw our product. If we take a proton from the beta one carbon, so it would look like that, and let's show those electrons. These electrons in here, and magenta moved in here to form our double bond. It would be the same result if we took a proton from beta two. So we just took a proton from beta one. If we took a proton from beta two, we would get the same alkene. Let's move on to the beta three carbon."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "It would be the same result if we took a proton from beta two. So we just took a proton from beta one. If we took a proton from beta two, we would get the same alkene. Let's move on to the beta three carbon. So let's take a proton from our beta three carbon, and let's do that over here. So here's our proton, and let's draw in our ethoxide anion right here. So I'll put in my lone pairs of electrons on the oxygen, negative one formal charge."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "Let's move on to the beta three carbon. So let's take a proton from our beta three carbon, and let's do that over here. So here's our proton, and let's draw in our ethoxide anion right here. So I'll put in my lone pairs of electrons on the oxygen, negative one formal charge. So we're going to take this proton, and then these electrons would move into here, and then the electrons come off onto our leaving group to form the bromide anion. So the alkene that would form if we took a proton from our beta three carbon, let me draw it in here, so there's our double bond, we would form that. So our electrons in magenta came in here to form our double bond, and these would be the two products for this reaction."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So I'll put in my lone pairs of electrons on the oxygen, negative one formal charge. So we're going to take this proton, and then these electrons would move into here, and then the electrons come off onto our leaving group to form the bromide anion. So the alkene that would form if we took a proton from our beta three carbon, let me draw it in here, so there's our double bond, we would form that. So our electrons in magenta came in here to form our double bond, and these would be the two products for this reaction. It turns out that the isomer on the right is the major product. So this one is the major product, and the one on the left is the minor product. So we are talking about regiochemistry here, or think about the region of the molecule where the double bond forms."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So our electrons in magenta came in here to form our double bond, and these would be the two products for this reaction. It turns out that the isomer on the right is the major product. So this one is the major product, and the one on the left is the minor product. So we are talking about regiochemistry here, or think about the region of the molecule where the double bond forms. So this reaction is said to be regioselective, because one of the isomers is favored. Now let's talk about why. The major product on the right is a more stable alkene."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So we are talking about regiochemistry here, or think about the region of the molecule where the double bond forms. So this reaction is said to be regioselective, because one of the isomers is favored. Now let's talk about why. The major product on the right is a more stable alkene. It's more substituted. If we look at the carbons for our double bonds, we have one, two, three alkyl groups. This is a tri-substituted alkene, whereas our minor product on the left, this carbon has only two alkyl groups bonded to it."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "The major product on the right is a more stable alkene. It's more substituted. If we look at the carbons for our double bonds, we have one, two, three alkyl groups. This is a tri-substituted alkene, whereas our minor product on the left, this carbon has only two alkyl groups bonded to it. So this one is a di-substituted alkene. And we know from talking about stability of alkenes, the more substituted alkene is the more stable one. And we call this the Zaitsev product."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "This is a tri-substituted alkene, whereas our minor product on the left, this carbon has only two alkyl groups bonded to it. So this one is a di-substituted alkene. And we know from talking about stability of alkenes, the more substituted alkene is the more stable one. And we call this the Zaitsev product. So this would be the Zaitsev product, the more substituted alkene. And this is the major product. It's the most stable one."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "And we call this the Zaitsev product. So this would be the Zaitsev product, the more substituted alkene. And this is the major product. It's the most stable one. And so that's why this reaction is regioselective. Use sodium ethoxide, the major product is the more stable alkene. Now let's do this reaction using a different base."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "It's the most stable one. And so that's why this reaction is regioselective. Use sodium ethoxide, the major product is the more stable alkene. Now let's do this reaction using a different base. So we're starting with the same alkyl halides, and I actually left up the same products. But this time we're gonna go through thinking about potassium tert-butoxide as being our base. So there's a positive charge on potassium, a negative charge on the oxygen."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "Now let's do this reaction using a different base. So we're starting with the same alkyl halides, and I actually left up the same products. But this time we're gonna go through thinking about potassium tert-butoxide as being our base. So there's a positive charge on potassium, a negative charge on the oxygen. Potassium tert-butoxide is a sterically hindered base. So let me just add on to the drawing that I did in the previous example, where we used sodium ethoxide as the base. And I'm gonna change it to make tert-butoxide."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So there's a positive charge on potassium, a negative charge on the oxygen. Potassium tert-butoxide is a sterically hindered base. So let me just add on to the drawing that I did in the previous example, where we used sodium ethoxide as the base. And I'm gonna change it to make tert-butoxide. So let me draw in here. So now we have our tert-butoxide anion here functioning as a base. So if this takes a proton from the beta-1 or beta-2 positions, we form the disubstituted alkene."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "And I'm gonna change it to make tert-butoxide. So let me draw in here. So now we have our tert-butoxide anion here functioning as a base. So if this takes a proton from the beta-1 or beta-2 positions, we form the disubstituted alkene. And if we go over here, if we take a proton from the beta-3 position, let me add this stuff in here. So now we have our tert-butoxide anion. We're gonna form our trisubstituted alkene."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So if this takes a proton from the beta-1 or beta-2 positions, we form the disubstituted alkene. And if we go over here, if we take a proton from the beta-3 position, let me add this stuff in here. So now we have our tert-butoxide anion. We're gonna form our trisubstituted alkene. This time, the disubstituted alkene turns out to be the major product. So this one is the major product. And the trisubstituted alkene is the minor product."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "We're gonna form our trisubstituted alkene. This time, the disubstituted alkene turns out to be the major product. So this one is the major product. And the trisubstituted alkene is the minor product. And we can explain this by thinking about the fact that potassium tert-butoxide is a sterically hindered base. So on the left, we have these bulky methyl groups. But if we're taking a proton from beta-1 or beta-2, our base can be out to the side and relatively out of the way."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "And the trisubstituted alkene is the minor product. And we can explain this by thinking about the fact that potassium tert-butoxide is a sterically hindered base. So on the left, we have these bulky methyl groups. But if we're taking a proton from beta-1 or beta-2, our base can be out to the side and relatively out of the way. So it's easy for the base to take one of these protons. But on the right, as we get a little bit closer, at the beta-3 position, we have more steric hindrance here. So there's more steric hindrance that prevents the base from taking the proton as easily."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "But if we're taking a proton from beta-1 or beta-2, our base can be out to the side and relatively out of the way. So it's easy for the base to take one of these protons. But on the right, as we get a little bit closer, at the beta-3 position, we have more steric hindrance here. So there's more steric hindrance that prevents the base from taking the proton as easily. And so that's why the trisubstituted product turns out to be the minor product when you are using a sterically hindered base like potassium tert-butoxide. So in this case, the major product is the less substituted alkene. We call this the Hoffman product."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "So there's more steric hindrance that prevents the base from taking the proton as easily. And so that's why the trisubstituted product turns out to be the minor product when you are using a sterically hindered base like potassium tert-butoxide. So in this case, the major product is the less substituted alkene. We call this the Hoffman product. So let me write that in here. So this would be the Hoffman product. And this is the major product when a sterically hindered base is used."}, {"video_title": "E2 mechanism regioselectivity.mp3", "Sentence": "We call this the Hoffman product. So let me write that in here. So this would be the Hoffman product. And this is the major product when a sterically hindered base is used. So pay close attention to what base is used in an E2 mechanism. If you are using an unhindered strong base, something like sodium ethoxide, your major product is the Zaitsev product, the more substituted product. But if you're using a sterically hindered base, something like potassium tert-butoxide, the Hoffman product or the less substituted product is the major product."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So just to start off, I have here, this is the best depiction I could find where it didn't have copyrights. This is from NASA, of the Big Bang. And I've talked about it several times. The Big Bang occurred 13.7 billion years ago. And then if we go a little bit forward, actually a lot forward, we get to the formation of our actual solar system and the Earth. This is kind of the protoplanetary disk, or a depiction of a protoplanetary disk forming around our young sun. And so this right here is 4.5 billion years ago."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The Big Bang occurred 13.7 billion years ago. And then if we go a little bit forward, actually a lot forward, we get to the formation of our actual solar system and the Earth. This is kind of the protoplanetary disk, or a depiction of a protoplanetary disk forming around our young sun. And so this right here is 4.5 billion years ago. Now this over here, once again, these aren't pictures of them, these are just depictions, because no one was there with a camera. This is what we think the asteroid that killed the dinosaurs looked like when it was impacting Earth. And it killed the dinosaurs 65 million years ago."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this right here is 4.5 billion years ago. Now this over here, once again, these aren't pictures of them, these are just depictions, because no one was there with a camera. This is what we think the asteroid that killed the dinosaurs looked like when it was impacting Earth. And it killed the dinosaurs 65 million years ago. So until then, we had land dinosaurs, and then this, as far as the current theories go, got rid of them. Now we'll fast forward a little bit more. At about 3 million years ago, our ancestors look like this."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it killed the dinosaurs 65 million years ago. So until then, we had land dinosaurs, and then this, as far as the current theories go, got rid of them. Now we'll fast forward a little bit more. At about 3 million years ago, our ancestors look like this. This is Australopithecus afarensis. This is, I think, a depiction of Lucy. This is one of our, probably many of us share, I believe, the theory is that all of us have some DNA from her."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "At about 3 million years ago, our ancestors look like this. This is Australopithecus afarensis. This is, I think, a depiction of Lucy. This is one of our, probably many of us share, I believe, the theory is that all of us have some DNA from her. But this was 3 million years ago. And you fast forward some more, and you actually have the first modern humans appearing on the planet, people that looked and thought like you and me. This is 200,000 years ago."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is one of our, probably many of us share, I believe, the theory is that all of us have some DNA from her. But this was 3 million years ago. And you fast forward some more, and you actually have the first modern humans appearing on the planet, people that looked and thought like you and me. This is 200,000 years ago. That's right over here. Obviously, this drawing was done much later, but this is a depiction of a modern human. So 200,000 years ago."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is 200,000 years ago. That's right over here. Obviously, this drawing was done much later, but this is a depiction of a modern human. So 200,000 years ago. And then you fast forward even more. And I don't want to keep picking on Jesus. I did that when him getting on the jetliner, and I genuinely don't mean any offense to anyone."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So 200,000 years ago. And then you fast forward even more. And I don't want to keep picking on Jesus. I did that when him getting on the jetliner, and I genuinely don't mean any offense to anyone. I just keep picking Jesus because, frankly, our calendar is kind of, he's a good person that most people know about 2,000 years ago. And so when we associate a lot of modern history occurring after his birth. So I'll put this right here is obviously a painting of the birth of Jesus."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I did that when him getting on the jetliner, and I genuinely don't mean any offense to anyone. I just keep picking Jesus because, frankly, our calendar is kind of, he's a good person that most people know about 2,000 years ago. And so when we associate a lot of modern history occurring after his birth. So I'll put this right here is obviously a painting of the birth of Jesus. And this is 2,000 years ago. And then this might be a little bit American-centric, but the Declaration of Independence, it was a major event. Actually, even on a worldwide basis, it was the first secular democracy based on a kind of constitutional democracy that showed up on the planet."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So I'll put this right here is obviously a painting of the birth of Jesus. And this is 2,000 years ago. And then this might be a little bit American-centric, but the Declaration of Independence, it was a major event. Actually, even on a worldwide basis, it was the first secular democracy based on a kind of constitutional democracy that showed up on the planet. They had gotten rid of, they said, we don't want the King of England anymore. And this was about 234 years ago. And I always remember it because I was born almost on the 200th anniversary."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, even on a worldwide basis, it was the first secular democracy based on a kind of constitutional democracy that showed up on the planet. They had gotten rid of, they said, we don't want the King of England anymore. And this was about 234 years ago. And I always remember it because I was born almost on the 200th anniversary. So you just have to add my age to 200. So this is 234 years ago. So these are all events or periods of time that we've heard about and we've talked about, and people throw around these type of years."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I always remember it because I was born almost on the 200th anniversary. So you just have to add my age to 200. So this is 234 years ago. So these are all events or periods of time that we've heard about and we've talked about, and people throw around these type of years. But what I want to do in this video is relate it to time scales that we can comprehend. So instead of the Big Bang occurring 13.7 billion years ago, let's pretend like it occurred 10 years ago. Because most of us, especially if you're over the age of 10, can kind of understand what 10 years is."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So these are all events or periods of time that we've heard about and we've talked about, and people throw around these type of years. But what I want to do in this video is relate it to time scales that we can comprehend. So instead of the Big Bang occurring 13.7 billion years ago, let's pretend like it occurred 10 years ago. Because most of us, especially if you're over the age of 10, can kind of understand what 10 years is. It's a very, very long period of time, but something that's well within our lifetimes, well within our experience. So let's say the 13.7 billion, instead of saying the Big Bang occurred 13.7 billion years ago, let's pretend like it occurred 10 years ago. And if we pretend that it occurred 10 years ago, let's think about how many years or minutes or hours ago each of these events would have occurred."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Because most of us, especially if you're over the age of 10, can kind of understand what 10 years is. It's a very, very long period of time, but something that's well within our lifetimes, well within our experience. So let's say the 13.7 billion, instead of saying the Big Bang occurred 13.7 billion years ago, let's pretend like it occurred 10 years ago. And if we pretend that it occurred 10 years ago, let's think about how many years or minutes or hours ago each of these events would have occurred. So if the Big Bang, which is really 13.7 billion years, if it really had occurred 10 years ago and we scaled everything down, then the Earth would have been created about 3.3 years ago. So this would have been 3.3 years ago. So still nothing kind of amazing about this."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And if we pretend that it occurred 10 years ago, let's think about how many years or minutes or hours ago each of these events would have occurred. So if the Big Bang, which is really 13.7 billion years, if it really had occurred 10 years ago and we scaled everything down, then the Earth would have been created about 3.3 years ago. So this would have been 3.3 years ago. So still nothing kind of amazing about this. This is a significant fraction of the age of the universe. So not that mind-blowing just yet. But if we go all the way to when the dinosaurs were extinct, the last land dinosaurs, now the 65 million years, and this will give you an appreciation of the difference between million and billion, if the universe was only 10 years old, then the dinosaurs would have been extinct 17 days ago."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So still nothing kind of amazing about this. This is a significant fraction of the age of the universe. So not that mind-blowing just yet. But if we go all the way to when the dinosaurs were extinct, the last land dinosaurs, now the 65 million years, and this will give you an appreciation of the difference between million and billion, if the universe was only 10 years old, then the dinosaurs would have been extinct 17 days ago. Not even a month ago, the dinosaurs would have been extinct. So if the universe was created when I just graduated, well, I'm in my 30s now, so when I was 24, just last month, the dinosaurs would have gone extinct. And it gets even crazier."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if we go all the way to when the dinosaurs were extinct, the last land dinosaurs, now the 65 million years, and this will give you an appreciation of the difference between million and billion, if the universe was only 10 years old, then the dinosaurs would have been extinct 17 days ago. Not even a month ago, the dinosaurs would have been extinct. So if the universe was created when I just graduated, well, I'm in my 30s now, so when I was 24, just last month, the dinosaurs would have gone extinct. And it gets even crazier. 17 days ago, the dinosaurs would have extinct. Australopithecus afarensis would have walked on the Earth 19 hours ago yesterday. Yesterday, 19 hours ago, she would have been walking around on the planet."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it gets even crazier. 17 days ago, the dinosaurs would have extinct. Australopithecus afarensis would have walked on the Earth 19 hours ago yesterday. Yesterday, 19 hours ago, she would have been walking around on the planet. And modern humans wouldn't have shown up until 80 minutes ago. A little over an hour. There wasn't even a modern human."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Yesterday, 19 hours ago, she would have been walking around on the planet. And modern humans wouldn't have shown up until 80 minutes ago. A little over an hour. There wasn't even a modern human. Then the universe was 10 years. It didn't take until just very recently, the last hour, for us to see someone that looks something like us, looks and thinks something like us. Fast forward even more."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There wasn't even a modern human. Then the universe was 10 years. It didn't take until just very recently, the last hour, for us to see someone that looks something like us, looks and thinks something like us. Fast forward even more. The birth of Jesus. If the universe was 10 years old instead of 13.7 billion, and we scaled everything down, then the birth of Jesus would have been 46 seconds ago. And then if we fast forward all the way to the Declaration of Independence, this would have occurred 5 seconds ago."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Fast forward even more. The birth of Jesus. If the universe was 10 years old instead of 13.7 billion, and we scaled everything down, then the birth of Jesus would have been 46 seconds ago. And then if we fast forward all the way to the Declaration of Independence, this would have occurred 5 seconds ago. So this isn't quite as mind-blowing as the scale of the universe. But in my mind, this is still pretty amazing. All that's happened since 1776 on a global basis could have been encapsulated in 5 seconds if the age of the universe was 10 years."}, {"video_title": "Cosmological time scale 1 Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then if we fast forward all the way to the Declaration of Independence, this would have occurred 5 seconds ago. So this isn't quite as mind-blowing as the scale of the universe. But in my mind, this is still pretty amazing. All that's happened since 1776 on a global basis could have been encapsulated in 5 seconds if the age of the universe was 10 years. So hopefully that gives you a little bit of a perspective. In the next video, I'm going to compare, instead of condensing things in time, I'm going to compare this scale to kind of a distance scale. So we can kind of say, hey, if the universe was the number of pixels on my screen, how big would each of these things be?"}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So with a lower pKa value, acetic acid is more acidic than ethanol. We can explain why by looking at the conjugate bases. If ethanol donates this proton, the electrons in this bond, the electrons in magenta, are left behind on the oxygen. So let's draw in the conjugate base. This oxygen would have three lone pairs of electrons, and one of those lone pairs would be the electrons in magenta. That gives this oxygen a negative one formal charge. Let's draw the conjugate base for acetic acid."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw in the conjugate base. This oxygen would have three lone pairs of electrons, and one of those lone pairs would be the electrons in magenta. That gives this oxygen a negative one formal charge. Let's draw the conjugate base for acetic acid. If acetic acid donates this proton, then the electrons in magenta are left behind on the oxygen. So the conjugate base would have a carbon double bonded to an oxygen here with two lone pairs of electrons. And then on the right, we would have another oxygen, this one with three lone pairs."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw the conjugate base for acetic acid. If acetic acid donates this proton, then the electrons in magenta are left behind on the oxygen. So the conjugate base would have a carbon double bonded to an oxygen here with two lone pairs of electrons. And then on the right, we would have another oxygen, this one with three lone pairs. One of those lone pairs would be the electrons in magenta. So that gives this oxygen a negative one formal charge. Let's compare our two conjugate bases."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "And then on the right, we would have another oxygen, this one with three lone pairs. One of those lone pairs would be the electrons in magenta. So that gives this oxygen a negative one formal charge. Let's compare our two conjugate bases. Both of them have a negative charge on an oxygen, so there must be some other factor to stabilize a conjugate base. And that factor is resonance. So for the conjugate base on the right, we can take those electrons in magenta, we can take these electrons here, and we can move them in."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Let's compare our two conjugate bases. Both of them have a negative charge on an oxygen, so there must be some other factor to stabilize a conjugate base. And that factor is resonance. So for the conjugate base on the right, we can take those electrons in magenta, we can take these electrons here, and we can move them in. And then we can kick off these electrons onto the top oxygen. So let's draw the resulting resonance structure. So we would have for our top oxygen, we would now have three lone pairs of electrons around the top oxygen."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So for the conjugate base on the right, we can take those electrons in magenta, we can take these electrons here, and we can move them in. And then we can kick off these electrons onto the top oxygen. So let's draw the resulting resonance structure. So we would have for our top oxygen, we would now have three lone pairs of electrons around the top oxygen. Giving that top oxygen a negative one formal charge. This oxygen would now have only two lone pairs of electrons around it. And the electrons in magenta moved into here."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So we would have for our top oxygen, we would now have three lone pairs of electrons around the top oxygen. Giving that top oxygen a negative one formal charge. This oxygen would now have only two lone pairs of electrons around it. And the electrons in magenta moved into here. And these pi electrons, let me make them blue, moved off onto the top oxygen to give the top oxygen a negative one formal charge. So the negative charge on this oxygen is not localized to this one oxygen. It's spread out, it's delocalized."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons in magenta moved into here. And these pi electrons, let me make them blue, moved off onto the top oxygen to give the top oxygen a negative one formal charge. So the negative charge on this oxygen is not localized to this one oxygen. It's spread out, it's delocalized. So there's actually some negative charge on this oxygen too. So remember, resonance structures are not perfect. In reality, it's really a hybrid of our two resonance structures."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "It's spread out, it's delocalized. So there's actually some negative charge on this oxygen too. So remember, resonance structures are not perfect. In reality, it's really a hybrid of our two resonance structures. So the negative charge is spread out or delocalized over two oxygens. And when you spread out a negative charge, that has a stabilizing effect for the anion. So our conjugate base is stabilized by resonance."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "In reality, it's really a hybrid of our two resonance structures. So the negative charge is spread out or delocalized over two oxygens. And when you spread out a negative charge, that has a stabilizing effect for the anion. So our conjugate base is stabilized by resonance. And since our conjugate base is stabilized by resonance, that means acetic acid is more likely to donate to this proton. And that's why we see a lower pKa value. If we compare that to the conjugate base for ethanol, this is called the ethoxide anion, we can't draw a resonance structure."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So our conjugate base is stabilized by resonance. And since our conjugate base is stabilized by resonance, that means acetic acid is more likely to donate to this proton. And that's why we see a lower pKa value. If we compare that to the conjugate base for ethanol, this is called the ethoxide anion, we can't draw a resonance structure. We can't delocalize that negative charge. That negative charge is stuck on this one oxygen here. And that means this conjugate base is not as stable, it's not stabilized by resonance."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "If we compare that to the conjugate base for ethanol, this is called the ethoxide anion, we can't draw a resonance structure. We can't delocalize that negative charge. That negative charge is stuck on this one oxygen here. And that means this conjugate base is not as stable, it's not stabilized by resonance. So that means ethanol is not as likely to donate its proton. And that's why we see a higher pKa value for ethanol. So when you're trying to figure out the more acidic proton, draw the conjugate base and look for resonance."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "And that means this conjugate base is not as stable, it's not stabilized by resonance. So that means ethanol is not as likely to donate its proton. And that's why we see a higher pKa value for ethanol. So when you're trying to figure out the more acidic proton, draw the conjugate base and look for resonance. Here we have an organic compound. And our goal is to determine which is the more acidic proton. Is this the more acidic proton or is this the more acidic proton?"}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So when you're trying to figure out the more acidic proton, draw the conjugate base and look for resonance. Here we have an organic compound. And our goal is to determine which is the more acidic proton. Is this the more acidic proton or is this the more acidic proton? So let's say a base comes along and takes this top proton here. Well the electrons in this bond in magenta would be left behind on this carbon that I just circled in magenta. So let's draw that conjugate base."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Is this the more acidic proton or is this the more acidic proton? So let's say a base comes along and takes this top proton here. Well the electrons in this bond in magenta would be left behind on this carbon that I just circled in magenta. So let's draw that conjugate base. We would have our ring, we would have this carbon double bonded to this oxygen. This oxygen has two lone pairs of electrons on it. And the electrons in magenta would remain on this carbon right here in magenta, which gives that carbon a negative one formal charge."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw that conjugate base. We would have our ring, we would have this carbon double bonded to this oxygen. This oxygen has two lone pairs of electrons on it. And the electrons in magenta would remain on this carbon right here in magenta, which gives that carbon a negative one formal charge. Don't forget that carbon in magenta also has another hydrogen bonded to it. But I'm not drawing it in on the conjugate base just to make things easier to see. Let's say a base comes along and takes this proton."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "And the electrons in magenta would remain on this carbon right here in magenta, which gives that carbon a negative one formal charge. Don't forget that carbon in magenta also has another hydrogen bonded to it. But I'm not drawing it in on the conjugate base just to make things easier to see. Let's say a base comes along and takes this proton. So that would mean the electrons in dark blue would remain behind on this carbon in dark blue that I just circled. So let's draw that conjugate base. So let me draw in our ring here."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Let's say a base comes along and takes this proton. So that would mean the electrons in dark blue would remain behind on this carbon in dark blue that I just circled. So let's draw that conjugate base. So let me draw in our ring here. And then this oxygen would be, would have two lone pairs of electrons on it. The electrons in dark blue, the electrons in dark blue would end up on this carbon in dark blue, which gives that carbon in dark blue a negative one formal charge. Let me draw that in."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw in our ring here. And then this oxygen would be, would have two lone pairs of electrons on it. The electrons in dark blue, the electrons in dark blue would end up on this carbon in dark blue, which gives that carbon in dark blue a negative one formal charge. Let me draw that in. Again, remember this carbon also has a hydrogen bonded to it but I'm not gonna draw it in on the conjugate base because it makes it easier for us to see. Which of those two conjugate bases is the most stable? Well, the top conjugate base is stabilized by resonance."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw that in. Again, remember this carbon also has a hydrogen bonded to it but I'm not gonna draw it in on the conjugate base because it makes it easier for us to see. Which of those two conjugate bases is the most stable? Well, the top conjugate base is stabilized by resonance. You could take these electrons in magenta and you could move them in to form a double bond. So let's go ahead and show that. The electrons in magenta move in here and that would kick these electrons off onto the oxygen."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Well, the top conjugate base is stabilized by resonance. You could take these electrons in magenta and you could move them in to form a double bond. So let's go ahead and show that. The electrons in magenta move in here and that would kick these electrons off onto the oxygen. So we would have now, when we draw our ring, we'd have a double bond here and this oxygen would have three lone pairs of electrons around it, giving it a negative one formal charge. So the electrons in magenta moved in to here and the electrons in, let's make these light blue, electrons in light blue moved off onto the oxygen to give the oxygen a negative one formal charge. Now remember, oxygen is more electronegative than carbon."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "The electrons in magenta move in here and that would kick these electrons off onto the oxygen. So we would have now, when we draw our ring, we'd have a double bond here and this oxygen would have three lone pairs of electrons around it, giving it a negative one formal charge. So the electrons in magenta moved in to here and the electrons in, let's make these light blue, electrons in light blue moved off onto the oxygen to give the oxygen a negative one formal charge. Now remember, oxygen is more electronegative than carbon. So this oxygen here is better able to bear this negative charge than this carbon in magenta over here. So this on the right, this resonance structure actually contributes more to the overall hybrid. But our conjugate base is stabilized by resonance."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "Now remember, oxygen is more electronegative than carbon. So this oxygen here is better able to bear this negative charge than this carbon in magenta over here. So this on the right, this resonance structure actually contributes more to the overall hybrid. But our conjugate base is stabilized by resonance. That's in contrast to this conjugate base down here. We have a negative charge on this carbon but that negative charge is localized to that carbon in dark blue right here. We can't draw any resonance."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "But our conjugate base is stabilized by resonance. That's in contrast to this conjugate base down here. We have a negative charge on this carbon but that negative charge is localized to that carbon in dark blue right here. We can't draw any resonance. We can't show any resonance stabilization. Which means that this conjugate base is not as stable as what we drew up here, which did have some resonance stabilization. And that means that we found our answer."}, {"video_title": "Stabilization of a conjugate base resonance Organic chemistry Khan Academy.mp3", "Sentence": "We can't draw any resonance. We can't show any resonance stabilization. Which means that this conjugate base is not as stable as what we drew up here, which did have some resonance stabilization. And that means that we found our answer. The more acidic proton is the one that has the conjugate base that's resonance stabilized. So this proton that's next door to this carbon double bonded to the oxygen, that's on a carbon that's next door to this carbon double bonded to the oxygen, that is the acidic proton. And this is something that will come up later in the course."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "This is cyclopentadiene, and we have this bridging CH2 to think about. On the right is our dienophile, and we have these two aldehydes that are cis to each other. So in this video, we have stereochemistry for the diene and the dienophile, and we're also going to talk about the endo and exo products. But first, let's just think about our Diels-Alder reaction without any stereochemistry. So we're going to move our six pi electrons around. Take these pi electrons, and we're going to move them into here to form a bond between these two carbons. These pi electrons are going to form a bond between these two carbons."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "But first, let's just think about our Diels-Alder reaction without any stereochemistry. So we're going to move our six pi electrons around. Take these pi electrons, and we're going to move them into here to form a bond between these two carbons. These pi electrons are going to form a bond between these two carbons. And finally, these pi electrons are going to move down. So if we draw the product, again, not worrying about stereochemistry. We put our double bond in here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "These pi electrons are going to form a bond between these two carbons. And finally, these pi electrons are going to move down. So if we draw the product, again, not worrying about stereochemistry. We put our double bond in here. And these aldehydes, I'm going to abbreviate as CHO. So we have CHO, and then another aldehyde down here, so CHO. We can put in our bridging CH2, so I'll just draw that in."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We put our double bond in here. And these aldehydes, I'm going to abbreviate as CHO. So we have CHO, and then another aldehyde down here, so CHO. We can put in our bridging CH2, so I'll just draw that in. And let's follow our pi electrons. So our pi electrons in red formed a bond here. Our pi electrons in blue formed this bond."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We can put in our bridging CH2, so I'll just draw that in. And let's follow our pi electrons. So our pi electrons in red formed a bond here. Our pi electrons in blue formed this bond. And our pi electrons in magenta formed this bond. But notice, for our product, we haven't really shown any stereochemistry. So where are these aldehydes in space?"}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "Our pi electrons in blue formed this bond. And our pi electrons in magenta formed this bond. But notice, for our product, we haven't really shown any stereochemistry. So where are these aldehydes in space? How do they relate to this bridging CH2? Let's say the diene and the dienophile approach each other in this way. So we know that a bond forms here, so this bond in red."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So where are these aldehydes in space? How do they relate to this bridging CH2? Let's say the diene and the dienophile approach each other in this way. So we know that a bond forms here, so this bond in red. So that's a bond that forms between this carbon and this carbon. So that bond forms in here. And for the product, here you can see that bond."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So we know that a bond forms here, so this bond in red. So that's a bond that forms between this carbon and this carbon. So that bond forms in here. And for the product, here you can see that bond. And then we also have this bond in blue, which forms between this carbon and this carbon. So it's a bond that forms in here. And here is that bond on our product."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And for the product, here you can see that bond. And then we also have this bond in blue, which forms between this carbon and this carbon. So it's a bond that forms in here. And here is that bond on our product. We need to think about the stereochemistry of the diene first. So we look at our cyclopentadiene. We have this bridging CH2."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And here is that bond on our product. We need to think about the stereochemistry of the diene first. So we look at our cyclopentadiene. We have this bridging CH2. We know that inside substituents go up. So on our model set, here is that CH2. And here are the bonds that are leading to it."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We have this bridging CH2. We know that inside substituents go up. So on our model set, here is that CH2. And here are the bonds that are leading to it. And when we form our product, that CH2 goes up in space. Next, we need to think about the stereochem of our dienophile. So we have a double bond."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And here are the bonds that are leading to it. And when we form our product, that CH2 goes up in space. Next, we need to think about the stereochem of our dienophile. So we have a double bond. We have hydrogens on the left side. And if we draw a line here, we saw how to deal with this in earlier videos. The stuff on the left side goes down."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So we have a double bond. We have hydrogens on the left side. And if we draw a line here, we saw how to deal with this in earlier videos. The stuff on the left side goes down. So these hydrogens, if we follow them along, so here are the hydrogens. And we can see them in our final product. And if we're staring at the final product in this direction, those hydrogens are going away from us in space."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "The stuff on the left side goes down. So these hydrogens, if we follow them along, so here are the hydrogens. And we can see them in our final product. And if we're staring at the final product in this direction, those hydrogens are going away from us in space. So the hydrogens would be down. And these aldehydes here, so these aldehydes, I've made red in the model set. So here you can see the red, which is just standing in for our aldehyde."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And if we're staring at the final product in this direction, those hydrogens are going away from us in space. So the hydrogens would be down. And these aldehydes here, so these aldehydes, I've made red in the model set. So here you can see the red, which is just standing in for our aldehyde. And those aldehydes end up on the same side. And if we're staring down this way, here are the aldehydes. Those are actually coming out at us."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here you can see the red, which is just standing in for our aldehyde. And those aldehydes end up on the same side. And if we're staring down this way, here are the aldehydes. Those are actually coming out at us. So when we draw in our product, let's go ahead and start drawing this in here. We'd have a bicyclic compound. So we'd sketch that in."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "Those are actually coming out at us. So when we draw in our product, let's go ahead and start drawing this in here. We'd have a bicyclic compound. So we'd sketch that in. And we have our bridging CH2. So let's put that in. We have our double bond here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So we'd sketch that in. And we have our bridging CH2. So let's put that in. We have our double bond here. And then we have our aldehydes. Our aldehydes are up. So at this carbon, we need to draw in an aldehyde up, so CHO."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We have our double bond here. And then we have our aldehydes. Our aldehydes are up. So at this carbon, we need to draw in an aldehyde up, so CHO. We have a hydrogen that's down. So let's put in our hydrogen going down at this carbon. And then we have, at this carbon, an aldehyde going up, so CHO."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So at this carbon, we need to draw in an aldehyde up, so CHO. We have a hydrogen that's down. So let's put in our hydrogen going down at this carbon. And then we have, at this carbon, an aldehyde going up, so CHO. And then a hydrogen going down. So this is the exo product. So this is said to be the exo product, because we have our aldehydes up."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And then we have, at this carbon, an aldehyde going up, so CHO. And then a hydrogen going down. So this is the exo product. So this is said to be the exo product, because we have our aldehydes up. Our bridging CH2 is up. And then here is our double bond. So that's one of the possible products."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So this is said to be the exo product, because we have our aldehydes up. Our bridging CH2 is up. And then here is our double bond. So that's one of the possible products. What about if our dienophile approaches in this direction? So if we have our hydrogens here on our double bond, and we draw a line, we know the stuff on the right side of the double bond ends up. So here are the two hydrogens."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So that's one of the possible products. What about if our dienophile approaches in this direction? So if we have our hydrogens here on our double bond, and we draw a line, we know the stuff on the right side of the double bond ends up. So here are the two hydrogens. And when those two bonds form right in here, these hydrogens are up. If you're staring down this way, the stuff on the left side of the double bond now, our aldehydes are on the left side. So here are the aldehydes."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here are the two hydrogens. And when those two bonds form right in here, these hydrogens are up. If you're staring down this way, the stuff on the left side of the double bond now, our aldehydes are on the left side. So here are the aldehydes. And when our product is formed, we're looking this way. The aldehydes are going away from us in space. The aldehydes are down."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here are the aldehydes. And when our product is formed, we're looking this way. The aldehydes are going away from us in space. The aldehydes are down. Let's draw in this product. So we have our bicyclic. Let me go ahead and draw that in here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "The aldehydes are down. Let's draw in this product. So we have our bicyclic. Let me go ahead and draw that in here. So here's our bicyclic with our bridging CH2. Our double bond is in the back. And now when we're drawing these in, the hydrogens are up."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and draw that in here. So here's our bicyclic with our bridging CH2. Our double bond is in the back. And now when we're drawing these in, the hydrogens are up. So at this carbon, there's a hydrogen up. And then there's an aldehyde going down. So here is our CHO going down."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And now when we're drawing these in, the hydrogens are up. So at this carbon, there's a hydrogen up. And then there's an aldehyde going down. So here is our CHO going down. Same thing at this carbon. So a hydrogen's going up, and an aldehyde is going down. This is called the endo product."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here is our CHO going down. Same thing at this carbon. So a hydrogen's going up, and an aldehyde is going down. This is called the endo product. So let me write that in here. So this is endo. Our aldehydes are going down."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "This is called the endo product. So let me write that in here. So this is endo. Our aldehydes are going down. Our bridging CH2 is up. And here is our double bond. It turns out the endo product is the preferred product most of the time."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "Our aldehydes are going down. Our bridging CH2 is up. And here is our double bond. It turns out the endo product is the preferred product most of the time. And this is called the endo rule or the alder endo rule. And it's thought to be due to an interaction between the developing pi bond, which occurs between these two carbons, and the carbonyl groups, so the carbonyl groups for our aldehyde here. And even though I didn't put in the carbonyls, we know that these red spheres here symbolize those aldehydes."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "It turns out the endo product is the preferred product most of the time. And this is called the endo rule or the alder endo rule. And it's thought to be due to an interaction between the developing pi bond, which occurs between these two carbons, and the carbonyl groups, so the carbonyl groups for our aldehyde here. And even though I didn't put in the carbonyls, we know that these red spheres here symbolize those aldehydes. And so there's an interaction between the developing pi bond and those carbonyls, which stabilizes the dienophile approaching in this way. And that gives us our aldehydes down in space. That gives us our endo product."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And even though I didn't put in the carbonyls, we know that these red spheres here symbolize those aldehydes. And so there's an interaction between the developing pi bond and those carbonyls, which stabilizes the dienophile approaching in this way. And that gives us our aldehydes down in space. That gives us our endo product. So the endo product is preferred. With the exo product, these carbonyls, these aldehydes, would be way out here. That's too far away from the developing pi bonds."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "That gives us our endo product. So the endo product is preferred. With the exo product, these carbonyls, these aldehydes, would be way out here. That's too far away from the developing pi bonds. There's no interaction. So the exo product is not preferred. Let's do one more problem."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "That's too far away from the developing pi bonds. There's no interaction. So the exo product is not preferred. Let's do one more problem. On the left is our diene. On the right is our dienophile. And let's draw the product without stereochemistry again."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "Let's do one more problem. On the left is our diene. On the right is our dienophile. And let's draw the product without stereochemistry again. So we move our 6 pi electrons. So we form a bond between those two carbons. We form a bond between these two carbons."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And let's draw the product without stereochemistry again. So we move our 6 pi electrons. So we form a bond between those two carbons. We form a bond between these two carbons. And these pi electrons move down. So if we draw in our product, again, not worrying about stereochemistry, we'd have our aldehyde coming off of this carbon. We'd have a methyl group here and a methyl group here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We form a bond between these two carbons. And these pi electrons move down. So if we draw in our product, again, not worrying about stereochemistry, we'd have our aldehyde coming off of this carbon. We'd have a methyl group here and a methyl group here. Following our electrons, electrons in red, these pi electrons formed this bond. Our pi electrons in blue formed this bond. And our pi electrons in magenta moved down to here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We'd have a methyl group here and a methyl group here. Following our electrons, electrons in red, these pi electrons formed this bond. Our pi electrons in blue formed this bond. And our pi electrons in magenta moved down to here. So let's highlight those bonds on the model set, on the picture. So this bond in red formed between this carbon and this carbon. So formed in here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And our pi electrons in magenta moved down to here. So let's highlight those bonds on the model set, on the picture. So this bond in red formed between this carbon and this carbon. So formed in here. So there's that bond on our final product. The bond in blue formed between this carbon and this carbon. So here's that bond on our final product."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So formed in here. So there's that bond on our final product. The bond in blue formed between this carbon and this carbon. So here's that bond on our final product. But notice, if we draw the dienophile approaching this way, there is no stabilizing interaction between the developing pi bond and the carbonyl groups. So down here, I've flipped over the dienophile. So it has an endo approach to the diene."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here's that bond on our final product. But notice, if we draw the dienophile approaching this way, there is no stabilizing interaction between the developing pi bond and the carbonyl groups. So down here, I've flipped over the dienophile. So it has an endo approach to the diene. And now we think about our stereochemistry. So let's worry about the stereochemistry of our diene first. We know that inside substituents go up."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So it has an endo approach to the diene. And now we think about our stereochemistry. So let's worry about the stereochemistry of our diene first. We know that inside substituents go up. So these two hydrogens, here they are on the diene. And in our product, those two hydrogens are going up in space, if we're looking down in this direction. So those methyl groups, let me change colors here."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We know that inside substituents go up. So these two hydrogens, here they are on the diene. And in our product, those two hydrogens are going up in space, if we're looking down in this direction. So those methyl groups, let me change colors here. So these two methyl groups, which I've used yellow in the model set, those two methyl groups go down. So down relative to those hydrogens going up. So those methyl groups are going down if we're staring down this direction."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So those methyl groups, let me change colors here. So these two methyl groups, which I've used yellow in the model set, those two methyl groups go down. So down relative to those hydrogens going up. So those methyl groups are going down if we're staring down this direction. So let's draw that in for our product. We have our ring. So let's put in our double bond."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So those methyl groups are going down if we're staring down this direction. So let's draw that in for our product. We have our ring. So let's put in our double bond. And those methyl groups are going away from us. So we put those on a dash. So there's one methyl group, and there's the other one."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So let's put in our double bond. And those methyl groups are going away from us. So we put those on a dash. So there's one methyl group, and there's the other one. Next, let's think about the dienophile. So here is our dienophile. And if I draw a line here, we know the stuff to the left ends up down."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So there's one methyl group, and there's the other one. Next, let's think about the dienophile. So here is our dienophile. And if I draw a line here, we know the stuff to the left ends up down. So this aldehyde is going to be down in space. Let's follow that aldehyde through. So here is the aldehyde."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And if I draw a line here, we know the stuff to the left ends up down. So this aldehyde is going to be down in space. Let's follow that aldehyde through. So here is the aldehyde. We know that this endo approach gives us some stabilization between the developing pi bond and our carbonyl. So for our product, here is the aldehyde. And again, if we're looking down, this aldehyde is going away from us in space."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here is the aldehyde. We know that this endo approach gives us some stabilization between the developing pi bond and our carbonyl. So for our product, here is the aldehyde. And again, if we're looking down, this aldehyde is going away from us in space. So we'd have to put that aldehyde on a dash. So here is the aldehyde going away from us in space. What if the dienophile approached this way?"}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "And again, if we're looking down, this aldehyde is going away from us in space. So we'd have to put that aldehyde on a dash. So here is the aldehyde going away from us in space. What if the dienophile approached this way? Well, this is still an endo approach, because here would be our aldehyde, and our developing pi bond would be back here. So there's a stabilizing interaction. So here's the aldehyde in our product."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "What if the dienophile approached this way? Well, this is still an endo approach, because here would be our aldehyde, and our developing pi bond would be back here. So there's a stabilizing interaction. So here's the aldehyde in our product. And if we're staring down this way, this aldehyde is going away from us in space. So let's draw this product. We would have our ring."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So here's the aldehyde in our product. And if we're staring down this way, this aldehyde is going away from us in space. So let's draw this product. We would have our ring. We would have our double bond. And the aldehyde's going away from us at this carbon. So we put that aldehyde on a dash."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "We would have our ring. We would have our double bond. And the aldehyde's going away from us at this carbon. So we put that aldehyde on a dash. And our methyl groups are still going away from us. So this one's going down, and this one's going down. So we draw in our methyl groups."}, {"video_title": "Diels-Alder endo rule Organic chemistry Khan Academy.mp3", "Sentence": "So we put that aldehyde on a dash. And our methyl groups are still going away from us. So this one's going down, and this one's going down. So we draw in our methyl groups. What is the relationship between these two molecules? Well, they are enantiomers of each other. So you should get a pair of enantiomers."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So each of the six carbons in benzene, each of those six carbons has a double bond to it. So each of those six carbons is sp2 hybridized, which means that each of those carbons has a free p orbital. And because benzene is a planar molecule, those p orbitals can overlap side by side and allow for delocalization of the pi electrons in benzene. So if we count the number of pi electrons in benzene, we can see there are 2, 4, and 6. So 6 pi electrons fits Huckel's rule, which is the second criterion, which says that the ring has to have 4n plus 2 pi electrons. In this case, n is equal to 1. So 4 times 1 plus 2 gives us 6."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if we count the number of pi electrons in benzene, we can see there are 2, 4, and 6. So 6 pi electrons fits Huckel's rule, which is the second criterion, which says that the ring has to have 4n plus 2 pi electrons. In this case, n is equal to 1. So 4 times 1 plus 2 gives us 6. So 6 pi electrons for benzene. Just to remind you, n can be equal to 0, 1, 2, 3, or any other whole number. And that's called Huckel's rule."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So 4 times 1 plus 2 gives us 6. So 6 pi electrons for benzene. Just to remind you, n can be equal to 0, 1, 2, 3, or any other whole number. And that's called Huckel's rule. So you need Huckel's rule pi electrons in the ring for a compound to be aromatic. If we look at the pyridine molecule, pyridine is an analog to benzene. The only difference is that pyridine has a nitrogen in the ring instead of one of these carbons right here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that's called Huckel's rule. So you need Huckel's rule pi electrons in the ring for a compound to be aromatic. If we look at the pyridine molecule, pyridine is an analog to benzene. The only difference is that pyridine has a nitrogen in the ring instead of one of these carbons right here. So we say that pyridine is a heterocycle. A heterocycle is a cyclic compound that contains a heteroatom in the ring. A heteroatom is any atom other than carbon, so something like nitrogen, oxygen, or sulfur."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The only difference is that pyridine has a nitrogen in the ring instead of one of these carbons right here. So we say that pyridine is a heterocycle. A heterocycle is a cyclic compound that contains a heteroatom in the ring. A heteroatom is any atom other than carbon, so something like nitrogen, oxygen, or sulfur. And heterocycles can be aromatic too. So let's go ahead and analyze pyridine in a little bit more detail here. So here's the dot structure for pyridine."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "A heteroatom is any atom other than carbon, so something like nitrogen, oxygen, or sulfur. And heterocycles can be aromatic too. So let's go ahead and analyze pyridine in a little bit more detail here. So here's the dot structure for pyridine. And we'll start by looking at the carbons on pyridine. So pyridine has five carbons. Each of those carbons has a double bond to it."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So here's the dot structure for pyridine. And we'll start by looking at the carbons on pyridine. So pyridine has five carbons. Each of those carbons has a double bond to it. So each of those carbons is sp2 hybridized, meaning there's a free p orbital on each of those five carbons. I'm just going to sketch in those p orbitals on those five carbons like that. Let's analyze the nitrogen."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Each of those carbons has a double bond to it. So each of those carbons is sp2 hybridized, meaning there's a free p orbital on each of those five carbons. I'm just going to sketch in those p orbitals on those five carbons like that. Let's analyze the nitrogen. And let's figure out the hybridization of this nitrogen atom. The best way to do it is to figure out the steric number of this nitrogen atom. And so the steric number is equal to the number of sigma bonds plus number of lone pairs of electrons."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Let's analyze the nitrogen. And let's figure out the hybridization of this nitrogen atom. The best way to do it is to figure out the steric number of this nitrogen atom. And so the steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. So watch an earlier video if you'd like to see how to do steric number in more detail. So we could say this is a sigma bond. We could say this is a sigma bond."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the steric number is equal to the number of sigma bonds plus number of lone pairs of electrons. So watch an earlier video if you'd like to see how to do steric number in more detail. So we could say this is a sigma bond. We could say this is a sigma bond. And then we could say there's one pair of electrons on that nitrogen. So the steric number is equal to the number of sigma bonds, which is 2, plus number of lone pairs of electrons, which is 1. And so the steric number is equal to 3."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We could say this is a sigma bond. And then we could say there's one pair of electrons on that nitrogen. So the steric number is equal to the number of sigma bonds, which is 2, plus number of lone pairs of electrons, which is 1. And so the steric number is equal to 3. So this nitrogen must have three hybrid orbitals. And therefore, it's sp2 hybridized. So it has three sp2 hybridized orbitals."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the steric number is equal to 3. So this nitrogen must have three hybrid orbitals. And therefore, it's sp2 hybridized. So it has three sp2 hybridized orbitals. And therefore, one p orbital. So an unhybridized p orbital. So this nitrogen is also sp2 hybridized."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it has three sp2 hybridized orbitals. And therefore, one p orbital. So an unhybridized p orbital. So this nitrogen is also sp2 hybridized. And so we can go ahead and sketch in the p orbital like that. And if we look at the number of pi electrons in pyridine, there's 2, 4, and 6 pi electrons. So that fulfills Huckel's rule."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So this nitrogen is also sp2 hybridized. And so we can go ahead and sketch in the p orbital like that. And if we look at the number of pi electrons in pyridine, there's 2, 4, and 6 pi electrons. So that fulfills Huckel's rule. So there are 6 pi electrons. And we can see that the pyridine molecule is a ring of continuously overlapping p orbitals. These p orbitals can overlap side by side."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So that fulfills Huckel's rule. So there are 6 pi electrons. And we can see that the pyridine molecule is a ring of continuously overlapping p orbitals. These p orbitals can overlap side by side. And those 6 pi electrons can be delocalized throughout the ring. And so since this meets both of the criterion, pyridine is an aromatic molecule. So it has some extra stability associated with it."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "These p orbitals can overlap side by side. And those 6 pi electrons can be delocalized throughout the ring. And so since this meets both of the criterion, pyridine is an aromatic molecule. So it has some extra stability associated with it. Now, this lone pair of electrons on this nitrogen, that lone pair of electrons occupies an sp2 hybridized orbital. So we said that this nitrogen is sp2 hybridized, which means it has three sp2 hybrid orbitals. So one of those sp2 hybridized orbitals formed a bond with this carbon over here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it has some extra stability associated with it. Now, this lone pair of electrons on this nitrogen, that lone pair of electrons occupies an sp2 hybridized orbital. So we said that this nitrogen is sp2 hybridized, which means it has three sp2 hybrid orbitals. So one of those sp2 hybridized orbitals formed a bond with this carbon over here. One of them formed a bond with this carbon over here. And the last sp2 hybrid orbital actually contains that lone pair of electrons. So the lone pair of electrons on that nitrogen does not participate in resonance."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So one of those sp2 hybridized orbitals formed a bond with this carbon over here. One of them formed a bond with this carbon over here. And the last sp2 hybrid orbital actually contains that lone pair of electrons. So the lone pair of electrons on that nitrogen does not participate in resonance. That lone pair of electrons is localized to that nitrogen. And so any time you see a situation like pyridine, where you have a nitrogen with a lone pair of electrons and some electrons already participating in resonance, so those would be the electrons in magenta here, the electrons in magenta participate in resonance, so the electrons in blue cannot participate in resonance. They are localized to this nitrogen atom."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the lone pair of electrons on that nitrogen does not participate in resonance. That lone pair of electrons is localized to that nitrogen. And so any time you see a situation like pyridine, where you have a nitrogen with a lone pair of electrons and some electrons already participating in resonance, so those would be the electrons in magenta here, the electrons in magenta participate in resonance, so the electrons in blue cannot participate in resonance. They are localized to this nitrogen atom. So we've seen that pyridine is aromatic. Let's go ahead and do an example that's similar to pyridine. This is pyrimidine."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "They are localized to this nitrogen atom. So we've seen that pyridine is aromatic. Let's go ahead and do an example that's similar to pyridine. This is pyrimidine. So let's see if we can analyze the pyrimidine molecule. This is the same way that we analyzed the pyridine. So once again, if I start with my carbons here, each of these carbons is connected to a double bond."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "This is pyrimidine. So let's see if we can analyze the pyrimidine molecule. This is the same way that we analyzed the pyridine. So once again, if I start with my carbons here, each of these carbons is connected to a double bond. So I have four carbons, and therefore each carbon is sp2 hybridized. I can go ahead and sketch in a p orbital on each of my sp2 hybridized carbons like that. When I study the nitrogens in pyrimidine, I can see it's the exact same situation that we had in pyridine."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So once again, if I start with my carbons here, each of these carbons is connected to a double bond. So I have four carbons, and therefore each carbon is sp2 hybridized. I can go ahead and sketch in a p orbital on each of my sp2 hybridized carbons like that. When I study the nitrogens in pyrimidine, I can see it's the exact same situation that we had in pyridine. So I can see that there's a sigma bond here, a sigma bond here, and a lone pair of electrons here like that. And so I can see that this nitrogen is sp2 hybridized. And I can see that these pi electrons here are going to be participating in resonance."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "When I study the nitrogens in pyrimidine, I can see it's the exact same situation that we had in pyridine. So I can see that there's a sigma bond here, a sigma bond here, and a lone pair of electrons here like that. And so I can see that this nitrogen is sp2 hybridized. And I can see that these pi electrons here are going to be participating in resonance. So for that nitrogen, it's sp2 hybridized. It has a free p orbital. So I can go ahead and draw in the p orbital there."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I can see that these pi electrons here are going to be participating in resonance. So for that nitrogen, it's sp2 hybridized. It has a free p orbital. So I can go ahead and draw in the p orbital there. And I know that the lone pair of electrons in blue, since this nitrogen is sp2 hybridized, that lone pair of electrons is going to occupy an sp2 hybridized orbital. It's the exact same situation for this nitrogen. There's a sigma bond."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I can go ahead and draw in the p orbital there. And I know that the lone pair of electrons in blue, since this nitrogen is sp2 hybridized, that lone pair of electrons is going to occupy an sp2 hybridized orbital. It's the exact same situation for this nitrogen. There's a sigma bond. There's a sigma bond. We have a lone pair of electrons on that nitrogen. And then we also have some electrons already participating in resonance."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "There's a sigma bond. There's a sigma bond. We have a lone pair of electrons on that nitrogen. And then we also have some electrons already participating in resonance. And so this nitrogen is also sp2 hybridized. I can go ahead and draw a p orbital on that nitrogen, which means that that lone pair of electrons is not going to participate in resonance. That lone pair of electrons is going to occupy an sp2 hybridized orbital."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And then we also have some electrons already participating in resonance. And so this nitrogen is also sp2 hybridized. I can go ahead and draw a p orbital on that nitrogen, which means that that lone pair of electrons is not going to participate in resonance. That lone pair of electrons is going to occupy an sp2 hybridized orbital. It's going to be out to the side like that. And so for pyrimidine, once again, I have a total of 6 pi electrons. And those 6 pi electrons are going to be delocalized as the p orbitals overlap side by side in your ring."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That lone pair of electrons is going to occupy an sp2 hybridized orbital. It's going to be out to the side like that. And so for pyrimidine, once again, I have a total of 6 pi electrons. And those 6 pi electrons are going to be delocalized as the p orbitals overlap side by side in your ring. And so pyrimidine is also aromatic. It meets the criteria for it. And the lone pair of electrons on those nitrogens, those lone pairs are localized to those nitrogens."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And those 6 pi electrons are going to be delocalized as the p orbitals overlap side by side in your ring. And so pyrimidine is also aromatic. It meets the criteria for it. And the lone pair of electrons on those nitrogens, those lone pairs are localized to those nitrogens. The pyrimidine general structure is actually very important in biochemistry. So when you study biochemistry, you'll see how important it is. And here's an example."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And the lone pair of electrons on those nitrogens, those lone pairs are localized to those nitrogens. The pyrimidine general structure is actually very important in biochemistry. So when you study biochemistry, you'll see how important it is. And here's an example. This is the thymine molecule, which is, of course, found in DNA. And you'll always see in textbooks that thymine has a pyrimidine ring. But at first, it's not so obvious that a pyrimidine ring is present in thymine."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And here's an example. This is the thymine molecule, which is, of course, found in DNA. And you'll always see in textbooks that thymine has a pyrimidine ring. But at first, it's not so obvious that a pyrimidine ring is present in thymine. Because if I look at the nitrogens in thymine, we'll start with this nitrogen up top here. I can see that this nitrogen has, let's see, three sigma bonds to it and one lone pair of electrons. So three sigma bonds, so the steric number would be equal to three sigma bonds plus one lone pair of electrons."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But at first, it's not so obvious that a pyrimidine ring is present in thymine. Because if I look at the nitrogens in thymine, we'll start with this nitrogen up top here. I can see that this nitrogen has, let's see, three sigma bonds to it and one lone pair of electrons. So three sigma bonds, so the steric number would be equal to three sigma bonds plus one lone pair of electrons. The steric number should be equal to four, which implies four hybrid orbitals, which would mean that that nitrogen is sp3 hybridized. And if it's sp3 hybridized, you wouldn't have any p orbitals to participate for aromaticity. And so this must not be the case."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So three sigma bonds, so the steric number would be equal to three sigma bonds plus one lone pair of electrons. The steric number should be equal to four, which implies four hybrid orbitals, which would mean that that nitrogen is sp3 hybridized. And if it's sp3 hybridized, you wouldn't have any p orbitals to participate for aromaticity. And so this must not be the case. There must be a way to see a pyrimidine ring here. And the answer is because this nitrogen is actually not sp3 hybridized. It actually has a lone pair of electrons that are delocalized and not localized to this nitrogen, meaning we can draw a resonance structure for the thymine molecule."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this must not be the case. There must be a way to see a pyrimidine ring here. And the answer is because this nitrogen is actually not sp3 hybridized. It actually has a lone pair of electrons that are delocalized and not localized to this nitrogen, meaning we can draw a resonance structure for the thymine molecule. So this lone pair of electrons right here in this nitrogen are not localized to that nitrogen as we saw in the previous dot structures. Those electrons can move in here to form a pi bond between the nitrogen and that carbon. That would, of course, push these electrons in here off onto this oxygen."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "It actually has a lone pair of electrons that are delocalized and not localized to this nitrogen, meaning we can draw a resonance structure for the thymine molecule. So this lone pair of electrons right here in this nitrogen are not localized to that nitrogen as we saw in the previous dot structures. Those electrons can move in here to form a pi bond between the nitrogen and that carbon. That would, of course, push these electrons in here off onto this oxygen. So we can go ahead and draw a resonance structure. So let's go ahead and put the nitrogen in our ring. And let's go ahead and draw in the rest of our ring like that."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "That would, of course, push these electrons in here off onto this oxygen. So we can go ahead and draw a resonance structure. So let's go ahead and put the nitrogen in our ring. And let's go ahead and draw in the rest of our ring like that. And so that lone pair of electrons moved in to form a double bond between that nitrogen and that carbon. So that's our situation now. And for the top oxygen here, it had two lone pairs of electrons."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and draw in the rest of our ring like that. And so that lone pair of electrons moved in to form a double bond between that nitrogen and that carbon. So that's our situation now. And for the top oxygen here, it had two lone pairs of electrons. But it picked up one more lone pair of electrons, giving it a negative 1 formal charge. This nitrogen is still bonded to another hydrogen. And we can go ahead and draw in the rest of the molecule as well."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And for the top oxygen here, it had two lone pairs of electrons. But it picked up one more lone pair of electrons, giving it a negative 1 formal charge. This nitrogen is still bonded to another hydrogen. And we can go ahead and draw in the rest of the molecule as well. So there is our resonance structure for thymine. Now let's go ahead and analyze the nitrogen after we drew the resonance structure here. So if I want to figure out the steric number for this nitrogen now, I can see that there's a sigma bond here, there's a sigma bond here, and there's a sigma bond here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we can go ahead and draw in the rest of the molecule as well. So there is our resonance structure for thymine. Now let's go ahead and analyze the nitrogen after we drew the resonance structure here. So if I want to figure out the steric number for this nitrogen now, I can see that there's a sigma bond here, there's a sigma bond here, and there's a sigma bond here. And now no lone pairs of electrons around this nitrogen. So now the steric number would be equal to 3 plus 0, which is, of course, equal to 3. So the steric number of 3, we can say this nitrogen is sp2 hybridized."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So if I want to figure out the steric number for this nitrogen now, I can see that there's a sigma bond here, there's a sigma bond here, and there's a sigma bond here. And now no lone pairs of electrons around this nitrogen. So now the steric number would be equal to 3 plus 0, which is, of course, equal to 3. So the steric number of 3, we can say this nitrogen is sp2 hybridized. And so it has a p orbital now. And the lone pair of electrons that was on the nitrogen over here, I'm going to go ahead and put those in magenta. They're not localized to that nitrogen."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the steric number of 3, we can say this nitrogen is sp2 hybridized. And so it has a p orbital now. And the lone pair of electrons that was on the nitrogen over here, I'm going to go ahead and put those in magenta. They're not localized to that nitrogen. They're actually delocalized. And that lone pair of electrons can now participate in resonance. But we still don't have the exact pyrimidine structure here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "They're not localized to that nitrogen. They're actually delocalized. And that lone pair of electrons can now participate in resonance. But we still don't have the exact pyrimidine structure here. And so we can draw yet another resonance structure. So we can do the exact same thing with this lone pair of electrons down here on this nitrogen. So we can look at the other nitrogen now."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But we still don't have the exact pyrimidine structure here. And so we can draw yet another resonance structure. So we can do the exact same thing with this lone pair of electrons down here on this nitrogen. So we can look at the other nitrogen now. And we can do the exact same thing. So at first thought, it might look like that lone pair of electrons is localized to that nitrogen. But it's not."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we can look at the other nitrogen now. And we can do the exact same thing. So at first thought, it might look like that lone pair of electrons is localized to that nitrogen. But it's not. It's actually delocalized because of the resonance structure that we can draw. So pretty much the exact same thing we did before. So let's go ahead and draw in our ring here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But it's not. It's actually delocalized because of the resonance structure that we can draw. So pretty much the exact same thing we did before. So let's go ahead and draw in our ring here. So we have our two nitrogens in our ring. This nitrogen is bonded to a hydrogen right here. The lone pair of electrons now moves into here to form a double bond, a pi bond."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in our ring here. So we have our two nitrogens in our ring. This nitrogen is bonded to a hydrogen right here. The lone pair of electrons now moves into here to form a double bond, a pi bond. And this oxygen had two lone pairs of electrons. It picked up an extra lone pair, giving it a negative 1 formal charge like that. And also, we can draw in the rest of the molecule here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "The lone pair of electrons now moves into here to form a double bond, a pi bond. And this oxygen had two lone pairs of electrons. It picked up an extra lone pair, giving it a negative 1 formal charge like that. And also, we can draw in the rest of the molecule here. So double bond, methyl group. We have here our negatively charged oxygen up here. So negative 1 formal charge."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And also, we can draw in the rest of the molecule here. So double bond, methyl group. We have here our negatively charged oxygen up here. So negative 1 formal charge. And over here on our nitrogen, we have our hydrogen. And we have this in here. And I forgot to give a plus 1 formal charge on this nitrogen on this resonance structure right here."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So negative 1 formal charge. And over here on our nitrogen, we have our hydrogen. And we have this in here. And I forgot to give a plus 1 formal charge on this nitrogen on this resonance structure right here. So there's a plus 1 formal charge on this nitrogen. Obviously, it still has a plus 1 formal charge over here. And we have a plus 1 formal charge over here for this nitrogen."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And I forgot to give a plus 1 formal charge on this nitrogen on this resonance structure right here. So there's a plus 1 formal charge on this nitrogen. Obviously, it still has a plus 1 formal charge over here. And we have a plus 1 formal charge over here for this nitrogen. But if we focus in on both nitrogens, this nitrogen is now the exact same situation as this top nitrogen. They're both actually sp2 hybridized with a steric number of 2. So they have a p orbital."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we have a plus 1 formal charge over here for this nitrogen. But if we focus in on both nitrogens, this nitrogen is now the exact same situation as this top nitrogen. They're both actually sp2 hybridized with a steric number of 2. So they have a p orbital. Each one of them has a p orbital. And we now can see a little bit better that there actually are 6 pi electrons that can be delocalized throughout this ring. And now maybe it's a little bit more obvious that the thymine molecule contains the pyrimidine ring."}, {"video_title": "Aromatic heterocycles I Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So they have a p orbital. Each one of them has a p orbital. And we now can see a little bit better that there actually are 6 pi electrons that can be delocalized throughout this ring. And now maybe it's a little bit more obvious that the thymine molecule contains the pyrimidine ring. And therefore, it is aromatic and has some extra stability associated with it. So sometimes drawing resonance structures will allow you to see the possible aromaticity or extra stability in a molecule. So this is an example of a biological aromatic heterocycle, so a molecule found in biochemistry, which is obviously extremely important, that we can analyze using these simple concepts of aromaticity and organic chemistry."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "One way to do it would be to reflect this compound in the mirror. And if you look at this carbon skeleton, here we have our carbon skeleton with our OH group coming out at us in space. That's this model on the left. There's our carbon skeleton with our OH coming out at us in space. If we reflect this compound in the mirror, we'll see the enantiomer in the mirror. Our mirror image on the right is non-superimposable upon our model on the left. So let's just draw what we see."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's our carbon skeleton with our OH coming out at us in space. If we reflect this compound in the mirror, we'll see the enantiomer in the mirror. Our mirror image on the right is non-superimposable upon our model on the left. So let's just draw what we see. We see our carbon skeleton like this. All right, so let's draw that. So there is our carbon skeleton."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's just draw what we see. We see our carbon skeleton like this. All right, so let's draw that. So there is our carbon skeleton. And our OH group is coming out at us in space. So we could represent that with a wedge. So let's fill in our wedge here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there is our carbon skeleton. And our OH group is coming out at us in space. So we could represent that with a wedge. So let's fill in our wedge here. And let's draw our OH. And so this drawing on the right is the enantiomer to the drawing on the left. There's another way to represent the enantiomer on the right."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's fill in our wedge here. And let's draw our OH. And so this drawing on the right is the enantiomer to the drawing on the left. There's another way to represent the enantiomer on the right. And to do that, let's check out the video. So in the video, I imagine an axis going through this carbon. And then I rotate about this axis to give us another viewpoint of our other enantiomer."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's another way to represent the enantiomer on the right. And to do that, let's check out the video. So in the video, I imagine an axis going through this carbon. And then I rotate about this axis to give us another viewpoint of our other enantiomer. So here's a model of our enantiomer. And you can see our carbon skeleton with our OH coming out at us, attached to this carbon, and then the hydrogen going away from us in space. If we imagine rotating about an axis through this carbon, so let's go ahead and do that, we'll see another way to look at the enantiomer."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then I rotate about this axis to give us another viewpoint of our other enantiomer. So here's a model of our enantiomer. And you can see our carbon skeleton with our OH coming out at us, attached to this carbon, and then the hydrogen going away from us in space. If we imagine rotating about an axis through this carbon, so let's go ahead and do that, we'll see another way to look at the enantiomer. So now we have, for our carbon skeleton, you can see our carbon skeleton looks like this now. And then at this carbon, we have the OH going away from us in space, and the hydrogen's coming out at us. So here we have some pictures from the video to help us with our drawings."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If we imagine rotating about an axis through this carbon, so let's go ahead and do that, we'll see another way to look at the enantiomer. So now we have, for our carbon skeleton, you can see our carbon skeleton looks like this now. And then at this carbon, we have the OH going away from us in space, and the hydrogen's coming out at us. So here we have some pictures from the video to help us with our drawings. We can see that this picture is this compound. If you look at that carbon skeleton, and you can see the OH coming out at us in space. So in the video, we took this compound and we rotated it to give us this image on the right."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here we have some pictures from the video to help us with our drawings. We can see that this picture is this compound. If you look at that carbon skeleton, and you can see the OH coming out at us in space. So in the video, we took this compound and we rotated it to give us this image on the right. So these are just two images of the same compound. And this gives us another way to draw our enantiomer. This time our carbon skeleton's going like that, so let me go ahead and draw in our carbon skeleton."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So in the video, we took this compound and we rotated it to give us this image on the right. So these are just two images of the same compound. And this gives us another way to draw our enantiomer. This time our carbon skeleton's going like that, so let me go ahead and draw in our carbon skeleton. And our OH group is going away from us in space, so we have to put in the OH group with a dash like that. So these two drawings represent the same compound, the enantiomer that we were trying to draw. So there are two main ways to draw enantiomers, at least two ways that I like to use."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This time our carbon skeleton's going like that, so let me go ahead and draw in our carbon skeleton. And our OH group is going away from us in space, so we have to put in the OH group with a dash like that. So these two drawings represent the same compound, the enantiomer that we were trying to draw. So there are two main ways to draw enantiomers, at least two ways that I like to use. The first way is to reflect the compound in a mirror. And that's what we did first. We took this compound and we reflected it in the mirror, and we drew what we saw."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there are two main ways to draw enantiomers, at least two ways that I like to use. The first way is to reflect the compound in a mirror. And that's what we did first. We took this compound and we reflected it in the mirror, and we drew what we saw. And that gave us this drawing of the enantiomer. We've seen that this drawing on the right is the same thing as this drawing on the left here. And notice the difference between this drawing and our original compound."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We took this compound and we reflected it in the mirror, and we drew what we saw. And that gave us this drawing of the enantiomer. We've seen that this drawing on the right is the same thing as this drawing on the left here. And notice the difference between this drawing and our original compound. The carbon skeletons are the same if you look at these two. The only difference is we changed the wedge to a dash. So that's another very convenient way to draw an enantiomer."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And notice the difference between this drawing and our original compound. The carbon skeletons are the same if you look at these two. The only difference is we changed the wedge to a dash. So that's another very convenient way to draw an enantiomer. If you're starting with a wedge, change it to a dash. If you're starting with a dash, change it to a wedge. Now let's draw the enantiomer of this compound."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's another very convenient way to draw an enantiomer. If you're starting with a wedge, change it to a dash. If you're starting with a dash, change it to a wedge. Now let's draw the enantiomer of this compound. And the first method we'll use is the mirror method. So here's a simplified representation of our compound, so ignoring things like conformations of our ring. At this carbon, our bromine is going down in space, so that's this carbon, so you can see the bromine's going down."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now let's draw the enantiomer of this compound. And the first method we'll use is the mirror method. So here's a simplified representation of our compound, so ignoring things like conformations of our ring. At this carbon, our bromine is going down in space, so that's this carbon, so you can see the bromine's going down. And then at this carbon, bromine's going up in space. That's this carbon with our bromine going up. In the mirror, we can see the mirror image, or the enantiomer."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "At this carbon, our bromine is going down in space, so that's this carbon, so you can see the bromine's going down. And then at this carbon, bromine's going up in space. That's this carbon with our bromine going up. In the mirror, we can see the mirror image, or the enantiomer. So let's go ahead and draw our enantiomer here. So we draw our cyclohexane ring. And then at this carbon, we have our bromine going down in space."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "In the mirror, we can see the mirror image, or the enantiomer. So let's go ahead and draw our enantiomer here. So we draw our cyclohexane ring. And then at this carbon, we have our bromine going down in space. So let's go ahead and put in our bromine going down. And then at this carbon, we have our bromine going up in space. So that's a wedge."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then at this carbon, we have our bromine going down in space. So let's go ahead and put in our bromine going down. And then at this carbon, we have our bromine going up in space. So that's a wedge. So we draw in our bromine here. Let me fill in this wedge. And let's put in the bromine."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's a wedge. So we draw in our bromine here. Let me fill in this wedge. And let's put in the bromine. So this on the right is the enantiomer. And sometimes you don't have model sets or a mirror, but you can draw the mirror image by just using this drawing on the left. You can imagine a mirror right here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And let's put in the bromine. So this on the right is the enantiomer. And sometimes you don't have model sets or a mirror, but you can draw the mirror image by just using this drawing on the left. You can imagine a mirror right here. And just as you see up here, this bromine is reflected in the mirror. This bromine is reflected in the mirror, right? So these two, these bromines are reflecting each other."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can imagine a mirror right here. And just as you see up here, this bromine is reflected in the mirror. This bromine is reflected in the mirror, right? So these two, these bromines are reflecting each other. And then this carbon is opposite of this carbon, so that's this carbon and this one. And then this carbon is opposite of this one, so this carbon is opposite of that one. So just a few tricks to help you draw the mirror image."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these two, these bromines are reflecting each other. And then this carbon is opposite of this carbon, so that's this carbon and this one. And then this carbon is opposite of this one, so this carbon is opposite of that one. So just a few tricks to help you draw the mirror image. So there's another way to represent our enantiomer on the right. And let's go to the video to see the other way. So here we have our two enantiomers."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So just a few tricks to help you draw the mirror image. So there's another way to represent our enantiomer on the right. And let's go to the video to see the other way. So here we have our two enantiomers. So you can see they're mirror images of each other. If I rotate the enantiomer on the right, you can see it from a different viewpoint. Notice that both chiral centers have been inverted."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So here we have our two enantiomers. So you can see they're mirror images of each other. If I rotate the enantiomer on the right, you can see it from a different viewpoint. Notice that both chiral centers have been inverted. And just to prove that these are enantiomers of each other, let's try to superimpose one on top of the other. Notice how you can't do it. So these are non-superimposable mirror images of each other."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Notice that both chiral centers have been inverted. And just to prove that these are enantiomers of each other, let's try to superimpose one on top of the other. Notice how you can't do it. So these are non-superimposable mirror images of each other. So this picture on the right in the video shows you the relationship between our two enantiomers. So thinking about reflecting the molecule on the left in the mirror, we can see the enantiomer on the right. But we took this molecule on the right, this enantiomer, and I rotated it into this position."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So these are non-superimposable mirror images of each other. So this picture on the right in the video shows you the relationship between our two enantiomers. So thinking about reflecting the molecule on the left in the mirror, we can see the enantiomer on the right. But we took this molecule on the right, this enantiomer, and I rotated it into this position. So now let's draw the enantiomer from this perspective. We start out with our cyclohexane ring. And you can see at this carbon, our bromine is going down in space."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But we took this molecule on the right, this enantiomer, and I rotated it into this position. So now let's draw the enantiomer from this perspective. We start out with our cyclohexane ring. And you can see at this carbon, our bromine is going down in space. So that must be a dash. So there's my bromine. And then at this carbon, our bromine is coming up in space, coming out at us."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And you can see at this carbon, our bromine is going down in space. So that must be a dash. So there's my bromine. And then at this carbon, our bromine is coming up in space, coming out at us. So that's a wedge. So let me go ahead and draw in our wedge. And we'll put in our bromine."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then at this carbon, our bromine is coming up in space, coming out at us. So that's a wedge. So let me go ahead and draw in our wedge. And we'll put in our bromine. So this is just another way to represent our enantiomer, so on the right. So this drawing and this drawing are two different ways to represent the same molecule. So this is the enantiomer to the compound on the left."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we'll put in our bromine. So this is just another way to represent our enantiomer, so on the right. So this drawing and this drawing are two different ways to represent the same molecule. So this is the enantiomer to the compound on the left. So let's look at our original compound and compare this drawing on the right. Notice at this carbon, your bromine's coming out at you in space, whereas at this carbon, your bromine's going away from you in space. At this carbon, your bromine's going away from you in space."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So this is the enantiomer to the compound on the left. So let's look at our original compound and compare this drawing on the right. Notice at this carbon, your bromine's coming out at you in space, whereas at this carbon, your bromine's going away from you in space. At this carbon, your bromine's going away from you in space. And at this carbon, your bromine's coming out at you in space. So to draw our enantiomer, we just invert all chirality centers. So if you have a wedge, change it to a dash."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "At this carbon, your bromine's going away from you in space. And at this carbon, your bromine's coming out at you in space. So to draw our enantiomer, we just invert all chirality centers. So if you have a wedge, change it to a dash. If you have a dash, change it to a wedge. So this way is often easier. So just make sure to invert all your chiral centers to draw the enantiomer."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if you have a wedge, change it to a dash. If you have a dash, change it to a wedge. So this way is often easier. So just make sure to invert all your chiral centers to draw the enantiomer. For bicyclic compounds, it's easiest to use the mirror method. So if our goal is to draw the enantiomer of this compound, we can imagine a mirror right here. And we can use this picture above as a guide."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So just make sure to invert all your chiral centers to draw the enantiomer. For bicyclic compounds, it's easiest to use the mirror method. So if our goal is to draw the enantiomer of this compound, we can imagine a mirror right here. And we can use this picture above as a guide. You can see that the hydrogen is reflected in the mirror, and the chlorine is reflected in the mirror. So let's go ahead and draw those in. So we draw our hydrogen here, and then our bond going down to this carbon, and then a chlorine going straight down from here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And we can use this picture above as a guide. You can see that the hydrogen is reflected in the mirror, and the chlorine is reflected in the mirror. So let's go ahead and draw those in. So we draw our hydrogen here, and then our bond going down to this carbon, and then a chlorine going straight down from here. So notice how this hydrogen is reflected, and this chlorine is reflected. Next, let's think about reflecting this carbon. So we need to draw a line in this way right here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we draw our hydrogen here, and then our bond going down to this carbon, and then a chlorine going straight down from here. So notice how this hydrogen is reflected, and this chlorine is reflected. Next, let's think about reflecting this carbon. So we need to draw a line in this way right here. So now this carbon is reflected with that one. Let me highlight those carbons. So right here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we need to draw a line in this way right here. So now this carbon is reflected with that one. Let me highlight those carbons. So right here. I'm going to extend this line out a little bit so we can see where the horizontal is approximately, like that. So that's this carbon and this carbon. So that carbon's reflected in our mirror."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So right here. I'm going to extend this line out a little bit so we can see where the horizontal is approximately, like that. So that's this carbon and this carbon. So that carbon's reflected in our mirror. Next, let's draw a line up in space relative to that horizontal, like that. And that takes us to this carbon, which is reflecting this one in our mirror. Now let's worry about this one."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon's reflected in our mirror. Next, let's draw a line up in space relative to that horizontal, like that. And that takes us to this carbon, which is reflecting this one in our mirror. Now let's worry about this one. So this should go up in space. So let me draw a horizontal line down here just to help us with our drawing. So approximately horizontal at this point."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Now let's worry about this one. So this should go up in space. So let me draw a horizontal line down here just to help us with our drawing. So approximately horizontal at this point. We know we want to go up in space from this point. So let's do that. So that's this line."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So approximately horizontal at this point. We know we want to go up in space from this point. So let's do that. So that's this line. So this carbon is reflecting this carbon. Next, let's go ahead and reflect our top carbon here. So let's draw this line over to here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's this line. So this carbon is reflecting this carbon. Next, let's go ahead and reflect our top carbon here. So let's draw this line over to here. So you could think about this top carbon reflecting that one. So let's draw our lines over here. So from here and then back down to here."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw this line over to here. So you could think about this top carbon reflecting that one. So let's draw our lines over here. So from here and then back down to here. So this is more of a drawing exercise, really. And then we're going to go down a little bit. We'll leave that line broken so we can see that one's going behind."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So from here and then back down to here. So this is more of a drawing exercise, really. And then we're going to go down a little bit. We'll leave that line broken so we can see that one's going behind. And then we'll draw this to the carbon in the back. And then we know we have to draw down from this line to the carbon in the back. And then we can connect those lines."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We'll leave that line broken so we can see that one's going behind. And then we'll draw this to the carbon in the back. And then we know we have to draw down from this line to the carbon in the back. And then we can connect those lines. So finally, we've drawn our enantiomer. So on the right is the enantiomer to the compound on the left. And just to prove that this on the right is the enantiomer, let's look at a video where I try to superimpose the mirror image on the original compound."}, {"video_title": "Drawing enantiomers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And then we can connect those lines. So finally, we've drawn our enantiomer. So on the right is the enantiomer to the compound on the left. And just to prove that this on the right is the enantiomer, let's look at a video where I try to superimpose the mirror image on the original compound. Here we have our two enantiomers, so mirror images of each other that are non-superimposable. Just to prove they're non-superimposable, I'll rotate the enantiomer on the right and try to superimpose it on the one on the left. And you can see that they don't match up."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about, just as a visual aid, you could think about this proton spinning this way. A spinning proton is like a rotating sphere of charge and any moving charge creates a magnetic field. Therefore, you can say a proton is a tiny magnet, so like a bar magnet or a compass needle. So over here on the right, let's look at a compass needle, which has two poles. So we have the north pole, which I'll color in red here, and the south pole. So the compass needle is like a tiny bar magnet too, and so we could draw the magnetic field. So magnetic field lines go from the north pole to the south."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So over here on the right, let's look at a compass needle, which has two poles. So we have the north pole, which I'll color in red here, and the south pole. So the compass needle is like a tiny bar magnet too, and so we could draw the magnetic field. So magnetic field lines go from the north pole to the south. So I could draw in a magnetic field line here. So going from the north into the south. So going from the north to the south for our magnetic field line."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So magnetic field lines go from the north pole to the south. So I could draw in a magnetic field line here. So going from the north into the south. So going from the north to the south for our magnetic field line. We could also think about the magnetic dipole moment of the compass needle. The magnetic dipole moment is also called the magnetic moment, and it's a vector that points in the direction of the dipole's magnetic field. So we have two poles, north pole and a south pole, and the magnetic moment is going to point in this direction."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So going from the north to the south for our magnetic field line. We could also think about the magnetic dipole moment of the compass needle. The magnetic dipole moment is also called the magnetic moment, and it's a vector that points in the direction of the dipole's magnetic field. So we have two poles, north pole and a south pole, and the magnetic moment is going to point in this direction. So using this same idea, we can go back to the proton and think about it like a compass needle. So if it's spinning this way, it's going to have a north pole and a south pole. So let me go ahead and color the north pole red here."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we have two poles, north pole and a south pole, and the magnetic moment is going to point in this direction. So using this same idea, we can go back to the proton and think about it like a compass needle. So if it's spinning this way, it's going to have a north pole and a south pole. So let me go ahead and color the north pole red here. We could draw magnetic field lines. So we could draw a magnetic field line going from the north pole to the south pole. Let me go ahead and do one over here too like that."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and color the north pole red here. We could draw magnetic field lines. So we could draw a magnetic field line going from the north pole to the south pole. Let me go ahead and do one over here too like that. And therefore, we could also draw in the magnetic moment of the proton. So the magnetic moment points in the direction of our dipole's magnetic field. And so this is how we're going to think about a proton, like a tiny magnet with a magnetic moment."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and do one over here too like that. And therefore, we could also draw in the magnetic moment of the proton. So the magnetic moment points in the direction of our dipole's magnetic field. And so this is how we're going to think about a proton, like a tiny magnet with a magnetic moment. All right, let's go back to the idea of the compass needle because we know that a compass needle, if you put it into the Earth's magnetic field, the compass needle is going to point north. And so that's what I have down here. So the magnetic moment, the compass needle is pointing north like that."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this is how we're going to think about a proton, like a tiny magnet with a magnetic moment. All right, let's go back to the idea of the compass needle because we know that a compass needle, if you put it into the Earth's magnetic field, the compass needle is going to point north. And so that's what I have down here. So the magnetic moment, the compass needle is pointing north like that. And we know that opposite poles attract. So if this is the north pole of our little bar magnet, of our compass needle, this must be the magnetic south pole. And so I'm sure some of you are like, well, that's the geographic north pole."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the magnetic moment, the compass needle is pointing north like that. And we know that opposite poles attract. So if this is the north pole of our little bar magnet, of our compass needle, this must be the magnetic south pole. And so I'm sure some of you are like, well, that's the geographic north pole. And it is the geographic north pole. But if you're talking about magnets, it's actually the magnetic south pole because opposite poles attract. So if this is the south pole down here, this must be the magnetic north pole of the Earth."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so I'm sure some of you are like, well, that's the geographic north pole. And it is the geographic north pole. But if you're talking about magnets, it's actually the magnetic south pole because opposite poles attract. So if this is the south pole down here, this must be the magnetic north pole of the Earth. All right, so this is just what happens when you put a compass needle into the magnetic field of the Earth. And so if you wanted to make the compass needle point in the opposite direction, so here I have the compass needle pointed in this direction, you would have to put energy in. So here's my finger."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if this is the south pole down here, this must be the magnetic north pole of the Earth. All right, so this is just what happens when you put a compass needle into the magnetic field of the Earth. And so if you wanted to make the compass needle point in the opposite direction, so here I have the compass needle pointed in this direction, you would have to put energy in. So here's my finger. And so I had to rotate the compass needle. I had to put energy in in order to get the compass needle to point in this direction. And hopefully you can see this tiny little mark right here on the table that I left in."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's my finger. And so I had to rotate the compass needle. I had to put energy in in order to get the compass needle to point in this direction. And hopefully you can see this tiny little mark right here on the table that I left in. So you can see that I'm actually moving the compass needle with my finger. And so it took energy. And so this, having the compass needle point in this direction is higher in energy than this one."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And hopefully you can see this tiny little mark right here on the table that I left in. So you can see that I'm actually moving the compass needle with my finger. And so it took energy. And so this, having the compass needle point in this direction is higher in energy than this one. And so if I let go, right, if I just let go with my finger, the compass needle would automatically swing back and point in this direction again. So this is the lower energy state. And this is the higher energy state because I had to put energy in to make the compass needle point down."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this, having the compass needle point in this direction is higher in energy than this one. And so if I let go, right, if I just let go with my finger, the compass needle would automatically swing back and point in this direction again. So this is the lower energy state. And this is the higher energy state because I had to put energy in to make the compass needle point down. And so that's how we're gonna think about our proton. All right, so we could have a proton. And if we have an external magnetic field, let me go ahead and identify that."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And this is the higher energy state because I had to put energy in to make the compass needle point down. And so that's how we're gonna think about our proton. All right, so we could have a proton. And if we have an external magnetic field, let me go ahead and identify that. So this right here I'm saying is an external magnetic field that we're applying. So I'm gonna call this B naught. And if you put the proton in this external applied magnetic field, there's a quantized interaction between the magnetic moment of the proton and this external magnetic field."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if we have an external magnetic field, let me go ahead and identify that. So this right here I'm saying is an external magnetic field that we're applying. So I'm gonna call this B naught. And if you put the proton in this external applied magnetic field, there's a quantized interaction between the magnetic moment of the proton and this external magnetic field. And the magnetic moment of the proton either aligns with the external magnetic field or it aligns against the external magnetic field. So let me go ahead and draw that in. So here would be our magnetic moment aligning with the external magnetic field."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And if you put the proton in this external applied magnetic field, there's a quantized interaction between the magnetic moment of the proton and this external magnetic field. And the magnetic moment of the proton either aligns with the external magnetic field or it aligns against the external magnetic field. So let me go ahead and draw that in. So here would be our magnetic moment aligning with the external magnetic field. And then here would be the magnetic moment aligning against the external magnetic field. We could think about that relating to the spin of the proton, right? Because I said if it's spinning this way, this is the north pole and this is the south pole."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here would be our magnetic moment aligning with the external magnetic field. And then here would be the magnetic moment aligning against the external magnetic field. We could think about that relating to the spin of the proton, right? Because I said if it's spinning this way, this is the north pole and this is the south pole. So I could color in my north pole here red. And the magnetic moment was in this direction. And so for the other one, if the magnetic moment is now aligned against the applied magnetic field, the proton must be spinning in the opposite direction."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Because I said if it's spinning this way, this is the north pole and this is the south pole. So I could color in my north pole here red. And the magnetic moment was in this direction. And so for the other one, if the magnetic moment is now aligned against the applied magnetic field, the proton must be spinning in the opposite direction. So we could imagine, even though this isn't exactly what's happening, we could imagine the proton spinning this way, making this the north pole and this the south pole. So that's why this compass needle analogy helps so much because there's an energy difference between these two spin states. So when the magnetic moment is aligned with the magnetic field, this is the alpha spin state."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so for the other one, if the magnetic moment is now aligned against the applied magnetic field, the proton must be spinning in the opposite direction. So we could imagine, even though this isn't exactly what's happening, we could imagine the proton spinning this way, making this the north pole and this the south pole. So that's why this compass needle analogy helps so much because there's an energy difference between these two spin states. So when the magnetic moment is aligned with the magnetic field, this is the alpha spin state. And when the magnetic moment is aligned against the applied magnetic field, this is the beta spin state. And there's a difference in energy between these two spin states, just like there's a difference in energy between these states of the compass needle, right? So this one was higher in energy than this one."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So when the magnetic moment is aligned with the magnetic field, this is the alpha spin state. And when the magnetic moment is aligned against the applied magnetic field, this is the beta spin state. And there's a difference in energy between these two spin states, just like there's a difference in energy between these states of the compass needle, right? So this one was higher in energy than this one. And it's the exact same idea, or you could think about it as being the same, for our proton, right? So we have a difference in energy. So this spin state is higher in energy than this spin state."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this one was higher in energy than this one. And it's the exact same idea, or you could think about it as being the same, for our proton, right? So we have a difference in energy. So this spin state is higher in energy than this spin state. All right, let's go back to the analogy of the compass needle. If we were somehow able to increase the magnetic field of the Earth, it would take me more energy in order to make the compass point down, right? So I would have to put more energy in in order to change the direction of the compass needle."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this spin state is higher in energy than this spin state. All right, let's go back to the analogy of the compass needle. If we were somehow able to increase the magnetic field of the Earth, it would take me more energy in order to make the compass point down, right? So I would have to put more energy in in order to change the direction of the compass needle. Same idea with the proton. If you increase the applied magnetic field, so I'm now gonna draw a bigger magnetic field. So here's a bigger magnetic field, so a bigger B-naught."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I would have to put more energy in in order to change the direction of the compass needle. Same idea with the proton. If you increase the applied magnetic field, so I'm now gonna draw a bigger magnetic field. So here's a bigger magnetic field, so a bigger B-naught. So I've increased B-naught. I'm going to increase the energy difference between the two spin states. So I can draw a greater difference in energy between the alpha and the beta spin states."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here's a bigger magnetic field, so a bigger B-naught. So I've increased B-naught. I'm going to increase the energy difference between the two spin states. So I can draw a greater difference in energy between the alpha and the beta spin states. So now, this difference in energy, let me just go ahead and draw it in here. So this difference in energy, this difference in energy is greater than this difference in energy because we've applied a stronger magnetic field. And again, the compass analogy helps us understand that."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So I can draw a greater difference in energy between the alpha and the beta spin states. So now, this difference in energy, let me just go ahead and draw it in here. So this difference in energy, this difference in energy is greater than this difference in energy because we've applied a stronger magnetic field. And again, the compass analogy helps us understand that. All right, so now we've learned that we have these two different spin states. And it turns out a proton can absorb energy and flip from the lower spin state, the alpha spin state, to the beta spin state. So let's take a look at a diagram showing that."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And again, the compass analogy helps us understand that. All right, so now we've learned that we have these two different spin states. And it turns out a proton can absorb energy and flip from the lower spin state, the alpha spin state, to the beta spin state. So let's take a look at a diagram showing that. So here we go down here. All right, so if we apply, once again, if we apply an external magnetic field, there are two possible spin states for our proton, for our nucleus. So the nucleus could be in the alpha spin state or the beta spin state."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's take a look at a diagram showing that. So here we go down here. All right, so if we apply, once again, if we apply an external magnetic field, there are two possible spin states for our proton, for our nucleus. So the nucleus could be in the alpha spin state or the beta spin state. And let's say we have a proton or a nucleus in the alpha spin state. So there's a certain difference in energy between the alpha and the beta spin states. So there's a certain difference in energy."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the nucleus could be in the alpha spin state or the beta spin state. And let's say we have a proton or a nucleus in the alpha spin state. So there's a certain difference in energy between the alpha and the beta spin states. So there's a certain difference in energy. And the proton can absorb energy and flip to the higher energy spin state. So if we apply the right amount of energy, this proton can flip from the alpha spin state to the beta spin state. So let's draw that in here."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So there's a certain difference in energy. And the proton can absorb energy and flip to the higher energy spin state. So if we apply the right amount of energy, this proton can flip from the alpha spin state to the beta spin state. So let's draw that in here. This is alpha and this is beta. And when that happens, the nucleus is said to be in resonance with your applied magnetic field, and hence the term nuclear magnetic resonance. And so this energy difference, this energy difference between your two spin states corresponds to a frequency, because E is equal to h nu, where E is energy, right, and nu is the frequency."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw that in here. This is alpha and this is beta. And when that happens, the nucleus is said to be in resonance with your applied magnetic field, and hence the term nuclear magnetic resonance. And so this energy difference, this energy difference between your two spin states corresponds to a frequency, because E is equal to h nu, where E is energy, right, and nu is the frequency. And this frequency falls in the radio wave region of the electromagnetic spectrum. And so now we know enough to think about how an NMR works. And I should point out that I'm really only gonna talk about FT-NMR, right?"}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so this energy difference, this energy difference between your two spin states corresponds to a frequency, because E is equal to h nu, where E is energy, right, and nu is the frequency. And this frequency falls in the radio wave region of the electromagnetic spectrum. And so now we know enough to think about how an NMR works. And I should point out that I'm really only gonna talk about FT-NMR, right? Let me go ahead and rewrite that. So I'm only gonna talk about FT-NMR in this set of videos here in this tutorial. And in FT-NMR, you take a sample of your compound and you put it in an external magnetic field."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And I should point out that I'm really only gonna talk about FT-NMR, right? Let me go ahead and rewrite that. So I'm only gonna talk about FT-NMR in this set of videos here in this tutorial. And in FT-NMR, you take a sample of your compound and you put it in an external magnetic field. And the nuclei can either be in the alpha spin state or the beta spin state. There's a slight excess of nuclei in the alpha spin state. And so you hit the sample with a short pulse that contains a different range of frequencies, and that those excess nuclei can absorb the energy and flip from the alpha spin state to the beta spin state."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And in FT-NMR, you take a sample of your compound and you put it in an external magnetic field. And the nuclei can either be in the alpha spin state or the beta spin state. There's a slight excess of nuclei in the alpha spin state. And so you hit the sample with a short pulse that contains a different range of frequencies, and that those excess nuclei can absorb the energy and flip from the alpha spin state to the beta spin state. When the nuclei fall back down from the, fall from the beta spin state back down to the alpha spin state, so just like if I took my finger off the compass needle, the compass needle flips back to the lower energy state, the NMR machine can detect the energy that's given off and it gives us a signal on an NMR spectrum. And so down here, down here I'm showing you just a very simple NMR spectrum. And we get a signal, we get a signal, right?"}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so you hit the sample with a short pulse that contains a different range of frequencies, and that those excess nuclei can absorb the energy and flip from the alpha spin state to the beta spin state. When the nuclei fall back down from the, fall from the beta spin state back down to the alpha spin state, so just like if I took my finger off the compass needle, the compass needle flips back to the lower energy state, the NMR machine can detect the energy that's given off and it gives us a signal on an NMR spectrum. And so down here, down here I'm showing you just a very simple NMR spectrum. And we get a signal, we get a signal, right? So let me go ahead and draw that signal in here. So a signal looks like this. So like a peak right here."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we get a signal, we get a signal, right? So let me go ahead and draw that signal in here. So a signal looks like this. So like a peak right here. And this peak occurs at a certain frequency, right? So if you drop down to here, this represents a certain frequency. Over here, this is the intensity, so the number of absorptions."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So like a peak right here. And this peak occurs at a certain frequency, right? So if you drop down to here, this represents a certain frequency. Over here, this is the intensity, so the number of absorptions. So how high, or I'll talk about this in more detail later, your peak is here on your NMR spectrum. And so it's possible to get different signals at different frequencies. Let me go ahead and draw in another signal right down here like that."}, {"video_title": "Introduction to proton NMR Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Over here, this is the intensity, so the number of absorptions. So how high, or I'll talk about this in more detail later, your peak is here on your NMR spectrum. And so it's possible to get different signals at different frequencies. Let me go ahead and draw in another signal right down here like that. And so this signal's at this frequency and this signal is at this frequency. And if you have different frequencies, if you have different frequencies, right, you have different differences in energy here. And this is what helps us understand the structure of molecules."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "In the last video, I touched on the idea of a sigma bond. And that was a bond where, let me draw two nucleuses. And let me just draw one of the orbitals. Let's say that this is an sp3 hybridized orbital. And that's on this atom. And this is kind of its big lobe right there. And this guy has an sp3 hybridized orbital as well."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Let's say that this is an sp3 hybridized orbital. And that's on this atom. And this is kind of its big lobe right there. And this guy has an sp3 hybridized orbital as well. That's the small lobe. And then that's the big lobe like that. A sigma bond is one where there's an overlap kind of in the direction in which the lobes are pointed."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And this guy has an sp3 hybridized orbital as well. That's the small lobe. And then that's the big lobe like that. A sigma bond is one where there's an overlap kind of in the direction in which the lobes are pointed. And you might say, well, how can there be any other type of bond than that? Well, the other type of bond, so this right here, let me make this clear. This right here is a sigma bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "A sigma bond is one where there's an overlap kind of in the direction in which the lobes are pointed. And you might say, well, how can there be any other type of bond than that? Well, the other type of bond, so this right here, let me make this clear. This right here is a sigma bond. You say, well, what other kind of bond could there be where my two orbitals overlap kind of in the direction that they're pointing? And the other type of bond you could have, you can imagine if you have two p orbitals. So let me draw the nucleus of two atoms."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This right here is a sigma bond. You say, well, what other kind of bond could there be where my two orbitals overlap kind of in the direction that they're pointing? And the other type of bond you could have, you can imagine if you have two p orbitals. So let me draw the nucleus of two atoms. And I'll just draw one of each of their p orbitals. So let's say that that's the nucleus. And I'll just draw their p orbital."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw the nucleus of two atoms. And I'll just draw one of each of their p orbitals. So let's say that that's the nucleus. And I'll just draw their p orbital. So p orbital is just that dumbbell shape. Let me draw it a little bit closer together. So p orbital is that dumbbell shape."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And I'll just draw their p orbital. So p orbital is just that dumbbell shape. Let me draw it a little bit closer together. So p orbital is that dumbbell shape. So let me draw this guy's one of his p orbitals. I'm going to draw it a little bit bigger than that. And you'll see why in a second."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So p orbital is that dumbbell shape. So let me draw this guy's one of his p orbitals. I'm going to draw it a little bit bigger than that. And you'll see why in a second. So one of his p orbitals right there comes out like that. And then this guy over here also has a p orbital that is parallel to this p orbital. So it goes like that."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And you'll see why in a second. So one of his p orbitals right there comes out like that. And then this guy over here also has a p orbital that is parallel to this p orbital. So it goes like that. Well, let me draw that other one a little bit straighter. It goes, I want it to overlap more. So it goes like that."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it goes like that. Well, let me draw that other one a little bit straighter. It goes, I want it to overlap more. So it goes like that. I think you get the idea. So here, our two p orbitals are parallel to each other. You can imagine these are sp3 hybridized orbitals."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So it goes like that. I think you get the idea. So here, our two p orbitals are parallel to each other. You can imagine these are sp3 hybridized orbitals. They're pointing at each other. Here, they're parallel. p orbitals parallel to each other."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "You can imagine these are sp3 hybridized orbitals. They're pointing at each other. Here, they're parallel. p orbitals parallel to each other. And you see that they overlap on this kind of top lobe here and in this bottom lobe here. And this is a pi bond. And this is one pi bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "p orbitals parallel to each other. And you see that they overlap on this kind of top lobe here and in this bottom lobe here. And this is a pi bond. And this is one pi bond. So you could call it a pi, literally with the Greek letter pi. Pi bond. Sometimes you'll see this as just written as pi bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And this is one pi bond. So you could call it a pi, literally with the Greek letter pi. Pi bond. Sometimes you'll see this as just written as pi bond. And it's written, it's called a pi bond because it's the Greek letter for essentially p. And we're dealing with p orbitals overlapping. Now, sigma bonds, which are what form when you have a single bond, these are stronger than pi bonds. Pi bonds come into play once you start forming double or triple bonds on top of a sigma bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Sometimes you'll see this as just written as pi bond. And it's written, it's called a pi bond because it's the Greek letter for essentially p. And we're dealing with p orbitals overlapping. Now, sigma bonds, which are what form when you have a single bond, these are stronger than pi bonds. Pi bonds come into play once you start forming double or triple bonds on top of a sigma bond. To kind of get a better visualization of how that might work, let's think about ethene. So its molecular structure looks like this. So you have C double bonded to C. And then each of those guys have two hydrogens."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Pi bonds come into play once you start forming double or triple bonds on top of a sigma bond. To kind of get a better visualization of how that might work, let's think about ethene. So its molecular structure looks like this. So you have C double bonded to C. And then each of those guys have two hydrogens. So let me draw what it would look like, or our best visual or our best ability to kind of conceptualize what the orbitals around the carbon might look like. So let me draw. So first I'll draw the sp2 hybridized orbitals."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So you have C double bonded to C. And then each of those guys have two hydrogens. So let me draw what it would look like, or our best visual or our best ability to kind of conceptualize what the orbitals around the carbon might look like. So let me draw. So first I'll draw the sp2 hybridized orbitals. So let me just make it very clear what's going on here. So when we were dealing with methane, so methane, which is literally just a carbon bonded to four hydrogens. And if I actually wanted to draw it in a way that it kind of looks a little three dimensional with a tetrahedral structure, it might look like this."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So first I'll draw the sp2 hybridized orbitals. So let me just make it very clear what's going on here. So when we were dealing with methane, so methane, which is literally just a carbon bonded to four hydrogens. And if I actually wanted to draw it in a way that it kind of looks a little three dimensional with a tetrahedral structure, it might look like this. This hydrogen is pointing out a little bit. This hydrogen is kind of in the plane of the page. And then maybe that hydrogen is behind it."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And if I actually wanted to draw it in a way that it kind of looks a little three dimensional with a tetrahedral structure, it might look like this. This hydrogen is pointing out a little bit. This hydrogen is kind of in the plane of the page. And then maybe that hydrogen is behind it. And then you have one hydrogen popping up. That's methane. And we saw that these were all sp3 hybridized orbitals around the carbon."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then maybe that hydrogen is behind it. And then you have one hydrogen popping up. That's methane. And we saw that these were all sp3 hybridized orbitals around the carbon. And then they each formed sigma bonds with each of the hydrogens. We saw that in the last video. And when we drew its electron configuration, in order for this to happen, carbon's electron configuration when bonding in methane needed to look like this."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we saw that these were all sp3 hybridized orbitals around the carbon. And then they each formed sigma bonds with each of the hydrogens. We saw that in the last video. And when we drew its electron configuration, in order for this to happen, carbon's electron configuration when bonding in methane needed to look like this. It needed to look like 1s2. And then instead of having 2s2 and then 2p2, what you essentially have is, let me write it this way actually even better. In 1s you had two electrons."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And when we drew its electron configuration, in order for this to happen, carbon's electron configuration when bonding in methane needed to look like this. It needed to look like 1s2. And then instead of having 2s2 and then 2p2, what you essentially have is, let me write it this way actually even better. In 1s you had two electrons. And then instead of 2s you had two electrons and in each of the p's you had one, the s's and the p's all got mixed up. And you had a 2sp3 hybridized orbital, another 2sp3 hybridized orbital, another 2sp3 hybridized orbital, and then another 1sp3. Normally when carbon's sitting by itself, you'd expect a 2s here, and then you'd have a 2p in the x direction, a 2p in the y direction, and then a 2p in the z direction."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "In 1s you had two electrons. And then instead of 2s you had two electrons and in each of the p's you had one, the s's and the p's all got mixed up. And you had a 2sp3 hybridized orbital, another 2sp3 hybridized orbital, another 2sp3 hybridized orbital, and then another 1sp3. Normally when carbon's sitting by itself, you'd expect a 2s here, and then you'd have a 2p in the x direction, a 2p in the y direction, and then a 2p in the z direction. But we saw in the last video. They all get mixed up and they all have a 25% s character and a 75% p character when carbon bonds in methane. And then the electrons kind of separate out in that situation."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Normally when carbon's sitting by itself, you'd expect a 2s here, and then you'd have a 2p in the x direction, a 2p in the y direction, and then a 2p in the z direction. But we saw in the last video. They all get mixed up and they all have a 25% s character and a 75% p character when carbon bonds in methane. And then the electrons kind of separate out in that situation. When you're dealing with the carbons in ethene, remember eth is for two carbons and ene because we're dealing with an alkene. We have a double bond here. In this situation, the carbon's electron configuration when they bond in ethene looks more like this."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then the electrons kind of separate out in that situation. When you're dealing with the carbons in ethene, remember eth is for two carbons and ene because we're dealing with an alkene. We have a double bond here. In this situation, the carbon's electron configuration when they bond in ethene looks more like this. So you have your 1s and then you have your 1s orbital is still completely full. It has two electrons in it. But then in your 2 shell, you have, so let me just write, let me do this in a different color."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "In this situation, the carbon's electron configuration when they bond in ethene looks more like this. So you have your 1s and then you have your 1s orbital is still completely full. It has two electrons in it. But then in your 2 shell, you have, so let me just write, let me do this in a different color. So in our 2 shell, I'll show you what I mean in a second. I'm not writing the s or p's so far on purpose, but we're going to have four electrons just like we had before. We're still forming four bonds."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But then in your 2 shell, you have, so let me just write, let me do this in a different color. So in our 2 shell, I'll show you what I mean in a second. I'm not writing the s or p's so far on purpose, but we're going to have four electrons just like we had before. We're still forming four bonds. We're going to have kind of these four unpaired electrons. We're still forming one, two, three, four bonds with each of the carbon, so they're going to be separated out. But in this situation, instead of all of them being a mixture, kind of one part s, three parts p, the s mixes with two of the p orbitals."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "We're still forming four bonds. We're going to have kind of these four unpaired electrons. We're still forming one, two, three, four bonds with each of the carbon, so they're going to be separated out. But in this situation, instead of all of them being a mixture, kind of one part s, three parts p, the s mixes with two of the p orbitals. So what you have is 2sp2 orbitals. So you can imagine that the s orbital mixes with two of the p orbitals, so now it's one part s, two parts p. And then one of the p orbitals kind of stays by itself. And we need this p orbital to stay by itself because it is going to form, it is going to be what's responsible for the pi bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "But in this situation, instead of all of them being a mixture, kind of one part s, three parts p, the s mixes with two of the p orbitals. So what you have is 2sp2 orbitals. So you can imagine that the s orbital mixes with two of the p orbitals, so now it's one part s, two parts p. And then one of the p orbitals kind of stays by itself. And we need this p orbital to stay by itself because it is going to form, it is going to be what's responsible for the pi bond. And we're going to see that the pi bond does something very interesting to the molecule. It kind of makes it unrotatable around a bond axis, and you'll see what I mean in a second. So let me see if I can, in three dimensions, draw each of these carbons."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And we need this p orbital to stay by itself because it is going to form, it is going to be what's responsible for the pi bond. And we're going to see that the pi bond does something very interesting to the molecule. It kind of makes it unrotatable around a bond axis, and you'll see what I mean in a second. So let me see if I can, in three dimensions, draw each of these carbons. Let me do it in a different color. You have this carbon right there. So let's say that's the nucleus, although I'll put a c there so you know which carbon we're dealing with."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me see if I can, in three dimensions, draw each of these carbons. Let me do it in a different color. You have this carbon right there. So let's say that's the nucleus, although I'll put a c there so you know which carbon we're dealing with. And then I'll draw, you can assume that the 1s orbital, it's really small right around the carbon. And then you have these hybridized orbitals, the 2sp2 orbitals, and they're all going to be planar, kind of forming a triangle, or I guess maybe a p sign on some level, but I'll try to draw it in three dimensions here. So you have one, this is kind of coming out a little bit."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let's say that's the nucleus, although I'll put a c there so you know which carbon we're dealing with. And then I'll draw, you can assume that the 1s orbital, it's really small right around the carbon. And then you have these hybridized orbitals, the 2sp2 orbitals, and they're all going to be planar, kind of forming a triangle, or I guess maybe a p sign on some level, but I'll try to draw it in three dimensions here. So you have one, this is kind of coming out a little bit. Then you have one that's going in a little bit. And then you have, and they have another lobe a little bit on the other side, but I'm not going to draw them, it'll complicate it. They still have characteristics of p, so they'll have two lobes, but one is bigger than the other."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So you have one, this is kind of coming out a little bit. Then you have one that's going in a little bit. And then you have, and they have another lobe a little bit on the other side, but I'm not going to draw them, it'll complicate it. They still have characteristics of p, so they'll have two lobes, but one is bigger than the other. And then you have one that's maybe going in this side. So you can imagine that this is kind of a Mercedes sign if you drew a circle around it on its side. So that's this carbon right here."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "They still have characteristics of p, so they'll have two lobes, but one is bigger than the other. And then you have one that's maybe going in this side. So you can imagine that this is kind of a Mercedes sign if you drew a circle around it on its side. So that's this carbon right here. And of course, it has its hydrogens. So you have this hydrogen there, and so this hydrogen might be sitting right here. It just has one electron in its 1s orbital."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that's this carbon right here. And of course, it has its hydrogens. So you have this hydrogen there, and so this hydrogen might be sitting right here. It just has one electron in its 1s orbital. You have this hydrogen up here. It's sitting right over there. And now let's draw this carbon."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It just has one electron in its 1s orbital. You have this hydrogen up here. It's sitting right over there. And now let's draw this carbon. Now let's draw this carbon. This carbon will be sitting, I'm drawing it pretty close together, this carbon will be sitting right there. He has his 1s orbital, they have the exact same electron configuration, he has his 1s orbital right around him."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And now let's draw this carbon. Now let's draw this carbon. This carbon will be sitting, I'm drawing it pretty close together, this carbon will be sitting right there. He has his 1s orbital, they have the exact same electron configuration, he has his 1s orbital right around him. And then he has the exact same configuration. In either of these guys, we've so far only, or in this first guy, I've only drawn these first three. I haven't drawn this unhybridized p orbital yet."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "He has his 1s orbital, they have the exact same electron configuration, he has his 1s orbital right around him. And then he has the exact same configuration. In either of these guys, we've so far only, or in this first guy, I've only drawn these first three. I haven't drawn this unhybridized p orbital yet. So I'll do that in a second. But let me draw his bonds. So first of all, he has this, or you could imagine, that bond right there, which would be an sp2 hybridized bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I haven't drawn this unhybridized p orbital yet. So I'll do that in a second. But let me draw his bonds. So first of all, he has this, or you could imagine, that bond right there, which would be an sp2 hybridized bond. Let me do that in the same color as. So he has this bond right here, which would be an sp2 hybridized bond, just like that. And notice, this is a sigma bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So first of all, he has this, or you could imagine, that bond right there, which would be an sp2 hybridized bond. Let me do that in the same color as. So he has this bond right here, which would be an sp2 hybridized bond, just like that. And notice, this is a sigma bond. They overlap in kind of the direction that they're pointing in. That's the best way I could think about it. And then he's got these two hydrogens."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And notice, this is a sigma bond. They overlap in kind of the direction that they're pointing in. That's the best way I could think about it. And then he's got these two hydrogens. So one, he's got this guy in the back. And then there's one in the front. I'll draw it a little bigger so it's kind of pointing out at us."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And then he's got these two hydrogens. So one, he's got this guy in the back. And then there's one in the front. I'll draw it a little bigger so it's kind of pointing out at us. And then we have this hydrogen is sitting right over here. And these are also sigma bonds, just to be very clear about things. This is a s orbital overlapping with an sp2 orbital, but they're kind of overlapping in the direction that they're pointed, or kind of along the direction of each other, of the two atoms."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "I'll draw it a little bigger so it's kind of pointing out at us. And then we have this hydrogen is sitting right over here. And these are also sigma bonds, just to be very clear about things. This is a s orbital overlapping with an sp2 orbital, but they're kind of overlapping in the direction that they're pointed, or kind of along the direction of each other, of the two atoms. This is a sigma bond, and then we have this hydrogen in the back, which is also going to form a sigma bond. So everything I've drawn so far is a sigma bond. So that, maybe I don't want to make this picture too."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "This is a s orbital overlapping with an sp2 orbital, but they're kind of overlapping in the direction that they're pointed, or kind of along the direction of each other, of the two atoms. This is a sigma bond, and then we have this hydrogen in the back, which is also going to form a sigma bond. So everything I've drawn so far is a sigma bond. So that, maybe I don't want to make this picture too. So I could just put sigma bond there, sigma bond there, sigma bond there, sigma, sigma. So so far I've drawn this bond, this bond, this bond, this bond, and this bond. All of those sigma bonds."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So that, maybe I don't want to make this picture too. So I could just put sigma bond there, sigma bond there, sigma bond there, sigma, sigma. So so far I've drawn this bond, this bond, this bond, this bond, and this bond. All of those sigma bonds. So what happens to this last p orbital for each of these guys? Well, that's going to be kind of sticking out of the plane of the Mercedes sign, is the best way I can describe it. And let me see if I can do that in a color that I haven't done yet."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "All of those sigma bonds. So what happens to this last p orbital for each of these guys? Well, that's going to be kind of sticking out of the plane of the Mercedes sign, is the best way I can describe it. And let me see if I can do that in a color that I haven't done yet. Let me do this purple color. So you can imagine a pure p orbital, and I need to draw it even bigger than that, actually. A pure p orbital, it normally wouldn't be that big relative to things, but I have to make them overlap."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "And let me see if I can do that in a color that I haven't done yet. Let me do this purple color. So you can imagine a pure p orbital, and I need to draw it even bigger than that, actually. A pure p orbital, it normally wouldn't be that big relative to things, but I have to make them overlap. So it's a pure p orbital that's kind of going in, maybe you can imagine, the z-axis. That the other orbitals are kind of a Mercedes sign in the xy plane, and now you have the z-axis going straight up and down. Oh, and those bottom two have to overlap, so let me draw them bigger."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "A pure p orbital, it normally wouldn't be that big relative to things, but I have to make them overlap. So it's a pure p orbital that's kind of going in, maybe you can imagine, the z-axis. That the other orbitals are kind of a Mercedes sign in the xy plane, and now you have the z-axis going straight up and down. Oh, and those bottom two have to overlap, so let me draw them bigger. So it looks like that, and it looks like that, and they're going straight up and down. And notice, they are now overlapping. So this bond right here is this bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Oh, and those bottom two have to overlap, so let me draw them bigger. So it looks like that, and it looks like that, and they're going straight up and down. And notice, they are now overlapping. So this bond right here is this bond. I could have drawn them either way, but it's that second bond. And so what's happening now to the structure? So let me make it very clear."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So this bond right here is this bond. I could have drawn them either way, but it's that second bond. And so what's happening now to the structure? So let me make it very clear. This right here, that is a pi bond, and this right here is also. It's the same pi bond. It's this guy right here."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So let me make it very clear. This right here, that is a pi bond, and this right here is also. It's the same pi bond. It's this guy right here. It's the second bond in the double bond. But what's happening here? Well, first of all, by itself it would be a weaker bond, but because we already have a sigma bond, that's kind of a track that's making these molecules come close together, this pi bond will make them come even closer together."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "It's this guy right here. It's the second bond in the double bond. But what's happening here? Well, first of all, by itself it would be a weaker bond, but because we already have a sigma bond, that's kind of a track that's making these molecules come close together, this pi bond will make them come even closer together. So this distance right here is closer than if we were to just have a single sigma bond there. Now on top of that, the really interesting thing is if we just had a sigma bond here, both of these molecules could kind of rotate around the bond axis. They would be able to rotate around the bond axis if you just had one sigma bond there."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "Well, first of all, by itself it would be a weaker bond, but because we already have a sigma bond, that's kind of a track that's making these molecules come close together, this pi bond will make them come even closer together. So this distance right here is closer than if we were to just have a single sigma bond there. Now on top of that, the really interesting thing is if we just had a sigma bond here, both of these molecules could kind of rotate around the bond axis. They would be able to rotate around the bond axis if you just had one sigma bond there. But since we have these pi bonds that are parallel to each other and they're kind of overlapping and they're kind of locked into that configuration, you can no longer rotate. If one of these molecules rotates, the other one's going to rotate with it because these two guys are locked together. So what this pi bond does in this situation is it makes this carbon-carbon double bond, or it means that the double bonds are going to be rigid, that you can't have one molecule kind of flipping, swapping these two hydrogens without the other one having to flip with it."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "They would be able to rotate around the bond axis if you just had one sigma bond there. But since we have these pi bonds that are parallel to each other and they're kind of overlapping and they're kind of locked into that configuration, you can no longer rotate. If one of these molecules rotates, the other one's going to rotate with it because these two guys are locked together. So what this pi bond does in this situation is it makes this carbon-carbon double bond, or it means that the double bonds are going to be rigid, that you can't have one molecule kind of flipping, swapping these two hydrogens without the other one having to flip with it. So you wouldn't be able to kind of swap configurations of the hydrogens relative to the other side. That's what it causes. So hopefully that gives you a good understanding of the difference between sigma and pi bond."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So what this pi bond does in this situation is it makes this carbon-carbon double bond, or it means that the double bonds are going to be rigid, that you can't have one molecule kind of flipping, swapping these two hydrogens without the other one having to flip with it. So you wouldn't be able to kind of swap configurations of the hydrogens relative to the other side. That's what it causes. So hopefully that gives you a good understanding of the difference between sigma and pi bond. And if you're curious, when you're dealing with just to kind of make it clear, if we were dealing with ethine, so that we were just, this is the example of ethene, but ethine looks like this. You have a triple bond, and so you have each side bonding to one hydrogen. In this case, one of these, so the first bonds you can imagine, so these bonds are all sigma bonds."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So hopefully that gives you a good understanding of the difference between sigma and pi bond. And if you're curious, when you're dealing with just to kind of make it clear, if we were dealing with ethine, so that we were just, this is the example of ethene, but ethine looks like this. You have a triple bond, and so you have each side bonding to one hydrogen. In this case, one of these, so the first bonds you can imagine, so these bonds are all sigma bonds. They're actually sp hybridized. Your 2s orbital only mixes with one of the p's. So these are sp hybrid orbitals forming sigma bonds."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "In this case, one of these, so the first bonds you can imagine, so these bonds are all sigma bonds. They're actually sp hybridized. Your 2s orbital only mixes with one of the p's. So these are sp hybrid orbitals forming sigma bonds. So all of these right here. And then both of these are pi bonds. And if you had to imagine it, you could imagine another pi bond kind of coming out of the page, and another one here coming out of the page and into the page, and they too are overlapped."}, {"video_title": "Pi bonds and sp2 hybridized orbitals Structure and bonding Organic chemistry Khan Academy.mp3", "Sentence": "So these are sp hybrid orbitals forming sigma bonds. So all of these right here. And then both of these are pi bonds. And if you had to imagine it, you could imagine another pi bond kind of coming out of the page, and another one here coming out of the page and into the page, and they too are overlapped. And you just have one hydrogen pointing out in each direction. Maybe I'll make another video on that. So hopefully I didn't confuse you too much."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the blue proton has a signal with a chemical shift about 6.7 parts per million. So down here is a zoomed in view of the signal for the blue proton. Let's look at neighboring protons. So the blue proton is on this carbon, and we have a carbon next door right here with one proton, so there's one neighboring proton. Here's another carbon next door with one proton, so we have two neighboring protons. So let's try to apply the n plus one rule here. So if n is equal to two, we have two neighboring protons, we would expect n plus one peaks."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the blue proton is on this carbon, and we have a carbon next door right here with one proton, so there's one neighboring proton. Here's another carbon next door with one proton, so we have two neighboring protons. So let's try to apply the n plus one rule here. So if n is equal to two, we have two neighboring protons, we would expect n plus one peaks. So two plus one is equal to three. So a signal with three peaks or a triplet. But that's not what we see for the signal for the blue proton."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So if n is equal to two, we have two neighboring protons, we would expect n plus one peaks. So two plus one is equal to three. So a signal with three peaks or a triplet. But that's not what we see for the signal for the blue proton. We see one, two, three, four lines here. So the n plus one rule doesn't work in this case, and that's because the n plus one rule works when the neighboring protons are equivalent, and here the two neighboring protons are not equivalent. So we need a new way to explain the signal for the blue proton, and we're gonna use what's called a splitting tree."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "But that's not what we see for the signal for the blue proton. We see one, two, three, four lines here. So the n plus one rule doesn't work in this case, and that's because the n plus one rule works when the neighboring protons are equivalent, and here the two neighboring protons are not equivalent. So we need a new way to explain the signal for the blue proton, and we're gonna use what's called a splitting tree. So we're gonna start with the signal for the blue proton. So here's the signal for the blue proton. It's going to be split by the proton next to it, this proton in red, and the coupling constant between the red and the blue proton is 12 hertz."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we need a new way to explain the signal for the blue proton, and we're gonna use what's called a splitting tree. So we're gonna start with the signal for the blue proton. So here's the signal for the blue proton. It's going to be split by the proton next to it, this proton in red, and the coupling constant between the red and the blue proton is 12 hertz. So let's go ahead and show, let's show this signal is split into a doublet. Alright, so let me go ahead and draw in the blue lines here. So we get the signal split into a doublet, the coupling constant is 12 hertz, so this distance here represents 12 hertz."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "It's going to be split by the proton next to it, this proton in red, and the coupling constant between the red and the blue proton is 12 hertz. So let's go ahead and show, let's show this signal is split into a doublet. Alright, so let me go ahead and draw in the blue lines here. So we get the signal split into a doublet, the coupling constant is 12 hertz, so this distance here represents 12 hertz. So why can we think about it being split into a doublet? Well, you could think about a variation of the n plus one rule here. Alright, so we're talking about one neighboring proton, so n is equal to one, so one plus one is equal to two, so we split the signal into a doublet with two lines."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we get the signal split into a doublet, the coupling constant is 12 hertz, so this distance here represents 12 hertz. So why can we think about it being split into a doublet? Well, you could think about a variation of the n plus one rule here. Alright, so we're talking about one neighboring proton, so n is equal to one, so one plus one is equal to two, so we split the signal into a doublet with two lines. Alright, now let's think about what happens with the other proton, so this one. Okay, well that's going to take each line of the doublet that we just drew and split it into another doublet. So this line gets split into a doublet and this line gets split into a doublet."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Alright, so we're talking about one neighboring proton, so n is equal to one, so one plus one is equal to two, so we split the signal into a doublet with two lines. Alright, now let's think about what happens with the other proton, so this one. Okay, well that's going to take each line of the doublet that we just drew and split it into another doublet. So this line gets split into a doublet and this line gets split into a doublet. The coupling constant this time is six hertz, so this distance here represents six hertz and this distance represents six hertz. So why was each line split into two? Well, once again we can use a modification of the n plus one rule here."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this line gets split into a doublet and this line gets split into a doublet. The coupling constant this time is six hertz, so this distance here represents six hertz and this distance represents six hertz. So why was each line split into two? Well, once again we can use a modification of the n plus one rule here. Alright, so we're talking about one neighbor here, so n is equal to one, so one plus one is equal to two, so each line is split into two. Alright, so each line is split into a doublet for this line and for this line. This line is split into a doublet too."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, once again we can use a modification of the n plus one rule here. Alright, so we're talking about one neighbor here, so n is equal to one, so one plus one is equal to two, so each line is split into two. Alright, so each line is split into a doublet for this line and for this line. This line is split into a doublet too. So we get four lines for the signal of the blue proton. If we go over here we see those four lines, one, two, three, and four. And we call this a doublet of doublets or a double doublet."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "This line is split into a doublet too. So we get four lines for the signal of the blue proton. If we go over here we see those four lines, one, two, three, and four. And we call this a doublet of doublets or a double doublet. Alright, what would happen if the coupling constants were the same? So let's just pretend like they're both 12 hertz. Let's go ahead and draw what we would see for the signal."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we call this a doublet of doublets or a double doublet. Alright, what would happen if the coupling constants were the same? So let's just pretend like they're both 12 hertz. Let's go ahead and draw what we would see for the signal. So we have our blue proton's signal is split into a doublet by the red proton, so let's go ahead and draw in our doublet here. And let's say this distance represents 12 hertz, so 12 hertz here. And let's say the coupling constant over here was also 12 hertz, so instead of six hertz."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's go ahead and draw what we would see for the signal. So we have our blue proton's signal is split into a doublet by the red proton, so let's go ahead and draw in our doublet here. And let's say this distance represents 12 hertz, so 12 hertz here. And let's say the coupling constant over here was also 12 hertz, so instead of six hertz. So each line of the doublet we just drew is split into another doublet from our other neighbor, and this time I'm going to show a coupling constant of 12 hertz. So each line of the doublet in blue is split into another doublet. So the line in blue on the left is split into a doublet, and the line in blue on the right is split into a doublet, because of our one neighboring proton."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And let's say the coupling constant over here was also 12 hertz, so instead of six hertz. So each line of the doublet we just drew is split into another doublet from our other neighbor, and this time I'm going to show a coupling constant of 12 hertz. So each line of the doublet in blue is split into another doublet. So the line in blue on the left is split into a doublet, and the line in blue on the right is split into a doublet, because of our one neighboring proton. Well, I changed the coupling constants. Now I said to pretend like now we're dealing with 12 here. So the coupling constants are the same."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the line in blue on the left is split into a doublet, and the line in blue on the right is split into a doublet, because of our one neighboring proton. Well, I changed the coupling constants. Now I said to pretend like now we're dealing with 12 here. So the coupling constants are the same. And notice what this gives us. This gives us a triplet. So here is one line, and then we're going to get this peak, and then we're going to get this peak here."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the coupling constants are the same. And notice what this gives us. This gives us a triplet. So here is one line, and then we're going to get this peak, and then we're going to get this peak here. So if the coupling constants are the same, you get a triplet. You get what the n plus one rule predicted. And so that's just something to think about with the origin of the n plus one rule."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So here is one line, and then we're going to get this peak, and then we're going to get this peak here. So if the coupling constants are the same, you get a triplet. You get what the n plus one rule predicted. And so that's just something to think about with the origin of the n plus one rule. Let's do another example. So let's go down here, and let's look at this molecule. So we're going to focus in on this proton here in blue."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so that's just something to think about with the origin of the n plus one rule. Let's do another example. So let's go down here, and let's look at this molecule. So we're going to focus in on this proton here in blue. All right, and we have neighboring protons, right? So this neighboring proton, the coupling constant between those two is 12 hertz. And then over here, we have two neighboring protons, and the coupling constant between the magenta protons and the blue is seven hertz."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to focus in on this proton here in blue. All right, and we have neighboring protons, right? So this neighboring proton, the coupling constant between those two is 12 hertz. And then over here, we have two neighboring protons, and the coupling constant between the magenta protons and the blue is seven hertz. So let's think about the signal for the blue proton here. So here we have the signal for the blue proton, which isn't split. Let's think about what the red proton is going to do."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And then over here, we have two neighboring protons, and the coupling constant between the magenta protons and the blue is seven hertz. So let's think about the signal for the blue proton here. So here we have the signal for the blue proton, which isn't split. Let's think about what the red proton is going to do. We have one neighbor, right? So one plus one is equal to two. So we're going to split the signal for the blue proton in two."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about what the red proton is going to do. We have one neighbor, right? So one plus one is equal to two. So we're going to split the signal for the blue proton in two. So we get a doublet here. So let me go ahead and draw in the doublet. The coupling constant was 12 hertz."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to split the signal for the blue proton in two. So we get a doublet here. So let me go ahead and draw in the doublet. The coupling constant was 12 hertz. So this distance right here is 12 hertz. All right, now we have a situation where we're thinking about the magenta protons, and we have two of them, right? So n is equal to two."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "The coupling constant was 12 hertz. So this distance right here is 12 hertz. All right, now we have a situation where we're thinking about the magenta protons, and we have two of them, right? So n is equal to two. So two plus one is equal to three. So the magenta protons are going to split each line of the doublet that we just drew into a triplet, all right? So we're going to get a triplet here."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So n is equal to two. So two plus one is equal to three. So the magenta protons are going to split each line of the doublet that we just drew into a triplet, all right? So we're going to get a triplet here. Let me go ahead and draw that in. So for a triplet, and our coupling constant is seven hertz. So let me see if I can draw that here."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to get a triplet here. Let me go ahead and draw that in. So for a triplet, and our coupling constant is seven hertz. So let me see if I can draw that here. So that distance is supposed to represent seven hertz. So let me draw that in. All right, so this distance is seven hertz."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me see if I can draw that here. So that distance is supposed to represent seven hertz. So let me draw that in. All right, so this distance is seven hertz. This distance corresponds to seven hertz. And one line of our doublet is split into a triplet now. Same thing happens for the other line, right?"}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "All right, so this distance is seven hertz. This distance corresponds to seven hertz. And one line of our doublet is split into a triplet now. Same thing happens for the other line, right? Same thing happens for this line right here. We need to split that into a triplet because of these magenta protons, all right? So we're going to split that into a triplet."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Same thing happens for the other line, right? Same thing happens for this line right here. We need to split that into a triplet because of these magenta protons, all right? So we're going to split that into a triplet. Once again, we need to think about a distance of seven here, seven hertz. So we have on both sides, we have seven hertz. So let me go ahead and draw that in."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to split that into a triplet. Once again, we need to think about a distance of seven here, seven hertz. So we have on both sides, we have seven hertz. So let me go ahead and draw that in. So this is talking about seven hertz, and this is talking about seven hertz. So that one line is split into a triplet. So let's draw that in here, one, two, and three."}, {"video_title": "Complex splitting Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw that in. So this is talking about seven hertz, and this is talking about seven hertz. So that one line is split into a triplet. So let's draw that in here, one, two, and three. So finally, how many peaks would we expect for the signal for this blue proton? So one, two, three, four, five, six. So we would expect six peaks for the signal for this blue proton because of the neighboring protons which are in different types of environments."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And once again, I don't like the term that much, because it kind of implies some type of explosion. But what it really is is kind of an expansion of space, when space started to really start to expand from a singularity. But our best estimate of when this occurred is 13.7 billion years ago. And even though we're used to dealing with numbers in the billions, especially when we talk about large amounts of money and whatnot, this is an unbelievable amount of time. It seems like something that's tractable, but it really isn't. And in future videos, I'm actually going to talk about the timescale so we can really appreciate how long, or even start to appreciate, or appreciate that we can't appreciate how long 13.7 billion years is. And I also want to emphasize that this is the current best estimate, even in my lifetime."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And even though we're used to dealing with numbers in the billions, especially when we talk about large amounts of money and whatnot, this is an unbelievable amount of time. It seems like something that's tractable, but it really isn't. And in future videos, I'm actually going to talk about the timescale so we can really appreciate how long, or even start to appreciate, or appreciate that we can't appreciate how long 13.7 billion years is. And I also want to emphasize that this is the current best estimate, even in my lifetime. Even in my lifetime that I actually knew about the Big Bang and that I would pay attention to what the best estimate was, this number has been moving around. So I suspect that in the future, this number might become more accurate or might move around some. But this is our best guess."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I also want to emphasize that this is the current best estimate, even in my lifetime. Even in my lifetime that I actually knew about the Big Bang and that I would pay attention to what the best estimate was, this number has been moving around. So I suspect that in the future, this number might become more accurate or might move around some. But this is our best guess. Now with that said, I want to think about what this tells us about the size of the observable universe. So if all of the expansion started 13.7 billion years ago, all of everything we know in our three-dimensional universe was in a single point, the longest that any photon of light could be traveling that's reaching us right now. So our eye is right, so let's say that that is my eye right over here, that's my eyelashes, just like that."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is our best guess. Now with that said, I want to think about what this tells us about the size of the observable universe. So if all of the expansion started 13.7 billion years ago, all of everything we know in our three-dimensional universe was in a single point, the longest that any photon of light could be traveling that's reaching us right now. So our eye is right, so let's say that that is my eye right over here, that's my eyelashes, just like that. The longest that some photon of light is just getting to my eye, or maybe it's just getting to the lens of a telescope. The longest that that could have been traveling is 13.7 billion years. So it could be traveling 13.7 billion years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So our eye is right, so let's say that that is my eye right over here, that's my eyelashes, just like that. The longest that some photon of light is just getting to my eye, or maybe it's just getting to the lens of a telescope. The longest that that could have been traveling is 13.7 billion years. So it could be traveling 13.7 billion years. So when we looked at that depiction, this I think was two or three videos ago of the observable universe, I drew this circle. And when we see light coming from these remote objects, that light is getting to us right here. This is where we are."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it could be traveling 13.7 billion years. So when we looked at that depiction, this I think was two or three videos ago of the observable universe, I drew this circle. And when we see light coming from these remote objects, that light is getting to us right here. This is where we are. This is where, I guess, in the depiction the remote object was, but the light from that remote object is just now getting to us. And that light took 13.7 billion years to get to us. Now what I'm going to hesitate to do, because we're talking over such large distances, and we're talking on such large timescales, and timescales over which space itself is expanding, we're going to see in this video that you cannot say that this object over here, this is not necessarily 13.7 billion light years away."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is where we are. This is where, I guess, in the depiction the remote object was, but the light from that remote object is just now getting to us. And that light took 13.7 billion years to get to us. Now what I'm going to hesitate to do, because we're talking over such large distances, and we're talking on such large timescales, and timescales over which space itself is expanding, we're going to see in this video that you cannot say that this object over here, this is not necessarily 13.7 billion light years away. If we're talking about smaller timescales, or I guess smaller distances, you could say approximately that. The expansion of the universe itself would not make as much of a difference. And let me make it even more clear."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now what I'm going to hesitate to do, because we're talking over such large distances, and we're talking on such large timescales, and timescales over which space itself is expanding, we're going to see in this video that you cannot say that this object over here, this is not necessarily 13.7 billion light years away. If we're talking about smaller timescales, or I guess smaller distances, you could say approximately that. The expansion of the universe itself would not make as much of a difference. And let me make it even more clear. I'm talking about an object over there. We could even talk about that coordinate in space. And that coordinate in, and actually I should say that coordinate in space-time, because we're viewing it at a certain instant as well."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let me make it even more clear. I'm talking about an object over there. We could even talk about that coordinate in space. And that coordinate in, and actually I should say that coordinate in space-time, because we're viewing it at a certain instant as well. But that coordinate is not 13.7 billion light years away from our current coordinate. And there's a couple of reasons to think about it. First of all, think about it."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that coordinate in, and actually I should say that coordinate in space-time, because we're viewing it at a certain instant as well. But that coordinate is not 13.7 billion light years away from our current coordinate. And there's a couple of reasons to think about it. First of all, think about it. That light was emitted 13.7 billion years ago. When that light was emitted, we were much closer to that coordinate. This coordinate was much closer to that."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "First of all, think about it. That light was emitted 13.7 billion years ago. When that light was emitted, we were much closer to that coordinate. This coordinate was much closer to that. Where we are in the universe right now was much closer to that point of the universe. The other thing to think about is as this, let me actually draw it. So let's say that, let's go 300,000 years after that initial expansion of that singularity."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This coordinate was much closer to that. Where we are in the universe right now was much closer to that point of the universe. The other thing to think about is as this, let me actually draw it. So let's say that, let's go 300,000 years after that initial expansion of that singularity. So we're just 300,000 years into the universe's history right now. So this is roughly 300,000 years into the universe's life, I guess we could view it that way. And let's say at that point, well first of all, at that point, things haven't differentiated in a meaningful way yet right now."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say that, let's go 300,000 years after that initial expansion of that singularity. So we're just 300,000 years into the universe's history right now. So this is roughly 300,000 years into the universe's life, I guess we could view it that way. And let's say at that point, well first of all, at that point, things haven't differentiated in a meaningful way yet right now. I mean, we'll talk more about this when we talk about the cosmic microwave background radiation. But at this point of the universe, it was kind of this almost uniform white hot plasma of hydrogen. And we're going to talk about it was emitting microwave radiation."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And let's say at that point, well first of all, at that point, things haven't differentiated in a meaningful way yet right now. I mean, we'll talk more about this when we talk about the cosmic microwave background radiation. But at this point of the universe, it was kind of this almost uniform white hot plasma of hydrogen. And we're going to talk about it was emitting microwave radiation. And we'll talk more about that in a future video. But let's just think about two points in this early universe. So in this early universe, let's say you have that point."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we're going to talk about it was emitting microwave radiation. And we'll talk more about that in a future video. But let's just think about two points in this early universe. So in this early universe, let's say you have that point. And let's say you have the coordinate where we are right now. You have the coordinate where we are right now. In fact, I'll just make that roughly, I won't make it the center, just because I think it makes it easier to visualize if it's not the center."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So in this early universe, let's say you have that point. And let's say you have the coordinate where we are right now. You have the coordinate where we are right now. In fact, I'll just make that roughly, I won't make it the center, just because I think it makes it easier to visualize if it's not the center. And let's say at that very early stage in the universe, if you were able to just take some rulers instantaneously and measure that, you would measure this distance to be 30 million light years. And let's just say right at that point, this object over here, I'll do it in magenta, this object over here emits a photon, maybe in the microwave frequency range. And we'll see that that was the range that it was emitting in."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In fact, I'll just make that roughly, I won't make it the center, just because I think it makes it easier to visualize if it's not the center. And let's say at that very early stage in the universe, if you were able to just take some rulers instantaneously and measure that, you would measure this distance to be 30 million light years. And let's just say right at that point, this object over here, I'll do it in magenta, this object over here emits a photon, maybe in the microwave frequency range. And we'll see that that was the range that it was emitting in. But it emits a photon. So right, and that photon is traveling at the speed of light. It is light."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we'll see that that was the range that it was emitting in. But it emits a photon. So right, and that photon is traveling at the speed of light. It is light. And so that photon says, oh, you know what? I only got 30 million light years to travel. That's not too bad."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is light. And so that photon says, oh, you know what? I only got 30 million light years to travel. That's not too bad. I'm going to get there in 30 million years. And so I'm going to do it discrete. The math is more complicated than what I'm doing here."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's not too bad. I'm going to get there in 30 million years. And so I'm going to do it discrete. The math is more complicated than what I'm doing here. But I really just want to give you the idea of what's going on here. So let's just say, well, that photon says, in about 10 million years, I should be right about at that coordinate. I should be about 1 third of the distance."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The math is more complicated than what I'm doing here. But I really just want to give you the idea of what's going on here. So let's just say, well, that photon says, in about 10 million years, I should be right about at that coordinate. I should be about 1 third of the distance. But what happens over the course of those 10 million years? Well, over the course of those 10 million years, the universe has expanded some. The universe has expanded maybe a good deal."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I should be about 1 third of the distance. But what happens over the course of those 10 million years? Well, over the course of those 10 million years, the universe has expanded some. The universe has expanded maybe a good deal. So let me draw the expanded universe. So after 10 million years, the universe might look like this. Actually, it might even be bigger than that."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The universe has expanded maybe a good deal. So let me draw the expanded universe. So after 10 million years, the universe might look like this. Actually, it might even be bigger than that. Let me draw it like this. After 10 million years, the universe might have expanded a good bit. So this is 10 million years into the future."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, it might even be bigger than that. Let me draw it like this. After 10 million years, the universe might have expanded a good bit. So this is 10 million years into the future. Still, on a cosmological time scale, still almost at the kind of the infancy of the universe, because we're talking about 13.7 billion years. So let's say 10 million years go by. The universe has expanded."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is 10 million years into the future. Still, on a cosmological time scale, still almost at the kind of the infancy of the universe, because we're talking about 13.7 billion years. So let's say 10 million years go by. The universe has expanded. This coordinate, where we're sitting today at the present time, is now all the way over here. That coordinate where the photon was originally emitted is now going to be sitting right over here. And that photon, it said, OK, after 10 million light years, I'm going to get over there."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The universe has expanded. This coordinate, where we're sitting today at the present time, is now all the way over here. That coordinate where the photon was originally emitted is now going to be sitting right over here. And that photon, it said, OK, after 10 million light years, I'm going to get over there. And I'm approximating. And I'm doing it in a very discrete way. But I really just want to give you the idea."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And that photon, it said, OK, after 10 million light years, I'm going to get over there. And I'm approximating. And I'm doing it in a very discrete way. But I really just want to give you the idea. So that coordinate, where the photon roughly gets in 10 million light years, is about right over here. The whole universe has expanded. All the coordinates have gotten further away from each other."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I really just want to give you the idea. So that coordinate, where the photon roughly gets in 10 million light years, is about right over here. The whole universe has expanded. All the coordinates have gotten further away from each other. Now what just happened here? The universe has expanded. This distance that was 30 million light years, now, and I'm just making rough numbers here."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "All the coordinates have gotten further away from each other. Now what just happened here? The universe has expanded. This distance that was 30 million light years, now, and I'm just making rough numbers here. I don't know the actual numbers here. Now it is actually, this is really just for the sake of giving you the idea of why, well, giving you the intuition of what's going on. This distance now is no longer 30 million light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This distance that was 30 million light years, now, and I'm just making rough numbers here. I don't know the actual numbers here. Now it is actually, this is really just for the sake of giving you the idea of why, well, giving you the intuition of what's going on. This distance now is no longer 30 million light years. It could be, maybe it's 100 million. So this is now 100 million light years away from each other. The universe is expanding."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This distance now is no longer 30 million light years. It could be, maybe it's 100 million. So this is now 100 million light years away from each other. The universe is expanding. These coordinates, the space is actually spreading out. You can imagine it's kind of a trampoline or the surface of a balloon getting stretched thin. And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The universe is expanding. These coordinates, the space is actually spreading out. You can imagine it's kind of a trampoline or the surface of a balloon getting stretched thin. And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance. It has now gone, that distance now might be on the order of, maybe it's on the order of 30 million light years. And the math isn't exact here. I haven't done the math to figure it out."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so this coordinate where the light happens to be after 10 million years, it has been traveling for 10 million years, but it's gone a much larger distance. It has now gone, that distance now might be on the order of, maybe it's on the order of 30 million light years. And the math isn't exact here. I haven't done the math to figure it out. But the point here, so it's done 30 million light years, and I want to make it, actually I shouldn't even make it the same proportion, because the distance it's gone, the distance it has to go, because of the stretching, it's not going to be completely linear. At least when I'm thinking about it in my head, it shouldn't be, I think. But I'm not going to make a hard statement about that."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I haven't done the math to figure it out. But the point here, so it's done 30 million light years, and I want to make it, actually I shouldn't even make it the same proportion, because the distance it's gone, the distance it has to go, because of the stretching, it's not going to be completely linear. At least when I'm thinking about it in my head, it shouldn't be, I think. But I'm not going to make a hard statement about that. But the distance that it traversed, maybe this distance right here is now 20 million light years, because it got there, every time it moved some distance, the space that it had traversed is now stretched. So even though it's traveled for 10 million years, the space that it traversed is no longer just 10 million light years. It's now stretched to 20 million light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I'm not going to make a hard statement about that. But the distance that it traversed, maybe this distance right here is now 20 million light years, because it got there, every time it moved some distance, the space that it had traversed is now stretched. So even though it's traveled for 10 million years, the space that it traversed is no longer just 10 million light years. It's now stretched to 20 million light years. And the space that it has left to traverse is no longer only 20 million light years. It might now be 80 million light years. It is now 80 million light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's now stretched to 20 million light years. And the space that it has left to traverse is no longer only 20 million light years. It might now be 80 million light years. It is now 80 million light years. And so this photon might be getting frustrated. There's an optimistic way of viewing it. It's like, wow, I was able to cover 20 million light years in only 10 million years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It is now 80 million light years. And so this photon might be getting frustrated. There's an optimistic way of viewing it. It's like, wow, I was able to cover 20 million light years in only 10 million years. It looks like I'm moving faster than the speed of light. The reality is it's not, because the space coordinates themselves are spreading out. Those are getting thin."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's like, wow, I was able to cover 20 million light years in only 10 million years. It looks like I'm moving faster than the speed of light. The reality is it's not, because the space coordinates themselves are spreading out. Those are getting thin. So the photon is just moving at the speed of light. But the distance that it actually traversed in 10 million years is more than 10 million light years. It's 20 million light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Those are getting thin. So the photon is just moving at the speed of light. But the distance that it actually traversed in 10 million years is more than 10 million light years. It's 20 million light years. So you can't just multiply a rate times time on these cosmological scales here. Especially when the coordinates themselves, the distance coordinates are actually moving away from each other. But I think you see, or maybe you might see, where this is going."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's 20 million light years. So you can't just multiply a rate times time on these cosmological scales here. Especially when the coordinates themselves, the distance coordinates are actually moving away from each other. But I think you see, or maybe you might see, where this is going. Now this says, OK, this photon says, oh, well, you know, in another, let me write this, this is 80 million light years. In another 40 million light years, maybe I'm going to get over here. But the reality is over that next 40 million light years, sorry, over in 40 million years, I might get right over here, because this is 80 million light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But I think you see, or maybe you might see, where this is going. Now this says, OK, this photon says, oh, well, you know, in another, let me write this, this is 80 million light years. In another 40 million light years, maybe I'm going to get over here. But the reality is over that next 40 million light years, sorry, over in 40 million years, I might get right over here, because this is 80 million light years. But the reality is after 40 million years, so for another 40 million years go by, now all of a sudden the universe has expanded even more. I won't even draw the whole bubble, but the place where the photon was emitted from might be over here. And now our current position is over here, where the light got after 10 million years is now over here."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the reality is over that next 40 million light years, sorry, over in 40 million years, I might get right over here, because this is 80 million light years. But the reality is after 40 million years, so for another 40 million years go by, now all of a sudden the universe has expanded even more. I won't even draw the whole bubble, but the place where the photon was emitted from might be over here. And now our current position is over here, where the light got after 10 million years is now over here. And now where the light is after 40 million years is maybe it's over here. So now this distance, this distance between these two points, when we started it was 10 million light years, then it became 20 million light years. Maybe now it's on the order of, I don't know, maybe it's a billion light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And now our current position is over here, where the light got after 10 million years is now over here. And now where the light is after 40 million years is maybe it's over here. So now this distance, this distance between these two points, when we started it was 10 million light years, then it became 20 million light years. Maybe now it's on the order of, I don't know, maybe it's a billion light years. And maybe this distance over here, and I'm just making up these numbers, in fact that's probably too big for that point, maybe this is now 100 million light years. And now this distance maybe is, maybe this distance right here is, I don't know, 500 million light years. And maybe now the total distance between the two points is a billion light years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Maybe now it's on the order of, I don't know, maybe it's a billion light years. And maybe this distance over here, and I'm just making up these numbers, in fact that's probably too big for that point, maybe this is now 100 million light years. And now this distance maybe is, maybe this distance right here is, I don't know, 500 million light years. And maybe now the total distance between the two points is a billion light years. So as you can see, the photon might be getting frustrated as it covers more and more distance. It looks back and says, wow, you know, in only 50 million years I've been able to cover 600 million light years. That's pretty good."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And maybe now the total distance between the two points is a billion light years. So as you can see, the photon might be getting frustrated as it covers more and more distance. It looks back and says, wow, you know, in only 50 million years I've been able to cover 600 million light years. That's pretty good. But it's frustrated because what it thought was it only had to cover 30 million light years in distance. That keeps stretching out because space itself is stretching. So the reality, just going to the original idea, this photon that is just reaching us, that's been traveling for, let's say, 13.4 billion years."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's pretty good. But it's frustrated because what it thought was it only had to cover 30 million light years in distance. That keeps stretching out because space itself is stretching. So the reality, just going to the original idea, this photon that is just reaching us, that's been traveling for, let's say, 13.4 billion years. So it's reaching us just now. So let me just fast forward 13.4 billion years from this point now to get to the present day. So if I draw the whole visible universe right over here, this point right over here is going to be where it was emitted from, is right over there."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the reality, just going to the original idea, this photon that is just reaching us, that's been traveling for, let's say, 13.4 billion years. So it's reaching us just now. So let me just fast forward 13.4 billion years from this point now to get to the present day. So if I draw the whole visible universe right over here, this point right over here is going to be where it was emitted from, is right over there. We are sitting right over there. And actually, let me make something clear. If I'm drawing the whole observable universe, the center actually should be where we are because we can observe an equal distance."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if I draw the whole visible universe right over here, this point right over here is going to be where it was emitted from, is right over there. We are sitting right over there. And actually, let me make something clear. If I'm drawing the whole observable universe, the center actually should be where we are because we can observe an equal distance. If things aren't really strange, we can observe an equal distance in any direction. So actually, maybe we should put us at the center. So this was the entire observable universe."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If I'm drawing the whole observable universe, the center actually should be where we are because we can observe an equal distance. If things aren't really strange, we can observe an equal distance in any direction. So actually, maybe we should put us at the center. So this was the entire observable universe. And the photon was emitted from here 13.4 billion years ago. So 300,000 years after that initial Big Bang. And it's just getting to us."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this was the entire observable universe. And the photon was emitted from here 13.4 billion years ago. So 300,000 years after that initial Big Bang. And it's just getting to us. It is true that the photon has been traveling for 13.7 billion years. But what's kind of nutty about it is this object, since we've been expanding away from each other, this object is now, in our best estimates, this object is going to be about 46 billion light years away from us. And I want to make it very clear."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's just getting to us. It is true that the photon has been traveling for 13.7 billion years. But what's kind of nutty about it is this object, since we've been expanding away from each other, this object is now, in our best estimates, this object is going to be about 46 billion light years away from us. And I want to make it very clear. This object is now 46 billion light years away from us. When we just use light to observe it, it looks like, just based on light years, hey, this light's been traveling 13.7 billion years to reach us. That's our only way with light to think about the distance."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I want to make it very clear. This object is now 46 billion light years away from us. When we just use light to observe it, it looks like, just based on light years, hey, this light's been traveling 13.7 billion years to reach us. That's our only way with light to think about the distance. So maybe it's 13.4 billion light years away. But the reality is, if you had a ruler today, light year rulers, the space here has stretched so much that this is now 46 billion light years. And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like?"}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's our only way with light to think about the distance. So maybe it's 13.4 billion light years away. But the reality is, if you had a ruler today, light year rulers, the space here has stretched so much that this is now 46 billion light years. And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like? This thing that's actually 46 billion light years away, but the photon only took 13.7 billion years to reach us. What will this look like? Well, when we say look like, it's based on the photons that are reaching us right now."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And just to give you a hint of when we talk about the cosmic microwave background radiation, what will this point in space look like? This thing that's actually 46 billion light years away, but the photon only took 13.7 billion years to reach us. What will this look like? Well, when we say look like, it's based on the photons that are reaching us right now. Those photons left 13.4 billion years ago. So those photons are the photons being emitted from this primitive structure, from this white hot haze of hydrogen plasma. So what we're going to see is this white hot haze, this kind of white hot plasma."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, when we say look like, it's based on the photons that are reaching us right now. Those photons left 13.4 billion years ago. So those photons are the photons being emitted from this primitive structure, from this white hot haze of hydrogen plasma. So what we're going to see is this white hot haze, this kind of white hot plasma. White hot, undifferentiated, not differentiated into proper stable atoms, much less stars and galaxies. But we're going to see this white hot plasma. The reality today is that that point in space that's 46 billion years from now, it's probably differentiated into stable atoms and stars and planets and galaxies."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what we're going to see is this white hot haze, this kind of white hot plasma. White hot, undifferentiated, not differentiated into proper stable atoms, much less stars and galaxies. But we're going to see this white hot plasma. The reality today is that that point in space that's 46 billion years from now, it's probably differentiated into stable atoms and stars and planets and galaxies. And frankly, if that person, if there is a civilization there right now, and if they're sitting right there, and if they're observing photons being emitted from our coordinate, from our point in space right now, they're not going to see us. They're going to see us 13.4 billion years ago. They're going to see the super primitive state of our region of space, when it really was just a white hot plasma."}, {"video_title": "Radius of observable universe Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The reality today is that that point in space that's 46 billion years from now, it's probably differentiated into stable atoms and stars and planets and galaxies. And frankly, if that person, if there is a civilization there right now, and if they're sitting right there, and if they're observing photons being emitted from our coordinate, from our point in space right now, they're not going to see us. They're going to see us 13.4 billion years ago. They're going to see the super primitive state of our region of space, when it really was just a white hot plasma. And we're going to talk more about this in the next video, but think about it. Any photon that's coming from that period in time, so from any direction that's been traveling for 13.4 billion years, from any direction, it's going to come from that primitive state. Or it would have been emitted when the universe was in that primitive state, when it was just that white hot plasma, this undifferentiated mass."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this is an acid anhydride over here on the right. And we can form those from carboxylic acid. So if we start with the carboxylic acid, and our first step at a base, like sodium hydroxide, and our second step at an acyl chloride, then we'd form our acid anhydride as our product. And if you think about a mechanism, sodium hydroxide's a base. The hydroxide anion's gonna take this proton, right, leaving these electrons behind on the oxygen. And there are already two lone pairs of electrons on the oxygen to start with. So if we go ahead and draw the product, we would form a carboxylate anion."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And if you think about a mechanism, sodium hydroxide's a base. The hydroxide anion's gonna take this proton, right, leaving these electrons behind on the oxygen. And there are already two lone pairs of electrons on the oxygen to start with. So if we go ahead and draw the product, we would form a carboxylate anion. So this oxygen right here would have three lone pairs of electrons on it like that. And so if we follow some of those electrons, this would have a negative one formal charge. And if we put these electrons in magenta, those electrons come off onto our oxygen to form our carboxylate anion."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So if we go ahead and draw the product, we would form a carboxylate anion. So this oxygen right here would have three lone pairs of electrons on it like that. And so if we follow some of those electrons, this would have a negative one formal charge. And if we put these electrons in magenta, those electrons come off onto our oxygen to form our carboxylate anion. And that's gonna function as our nucleophile. So in the second step, we add our acyl chloride. I'm just gonna go ahead and draw in our acyl chloride."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And if we put these electrons in magenta, those electrons come off onto our oxygen to form our carboxylate anion. And that's gonna function as our nucleophile. So in the second step, we add our acyl chloride. I'm just gonna go ahead and draw in our acyl chloride. It's going to be our electrophile. And let's think about why here. We have the oxygen withdrawing electron density from this carbon because oxygen is more electronegative than carbon, so we have that."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "I'm just gonna go ahead and draw in our acyl chloride. It's going to be our electrophile. And let's think about why here. We have the oxygen withdrawing electron density from this carbon because oxygen is more electronegative than carbon, so we have that. And then we also have this chlorine doing it as well. Chlorine is more electronegative than carbon as well. So we have these two things withdrawing electron density."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We have the oxygen withdrawing electron density from this carbon because oxygen is more electronegative than carbon, so we have that. And then we also have this chlorine doing it as well. Chlorine is more electronegative than carbon as well. So we have these two things withdrawing electron density. And so this carbon is definitely electrophilic right here. And so we have a nucleophile that's going to attack our electrophile. So our nucleophile attacks our electrophile."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we have these two things withdrawing electron density. And so this carbon is definitely electrophilic right here. And so we have a nucleophile that's going to attack our electrophile. So our nucleophile attacks our electrophile. These electrons kick off onto our oxygen. And we can go ahead and show the result of our nucleophilic attack. So we would now have this carbon double bonded to this oxygen, and then this oxygen is now bonded to this carbon."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So our nucleophile attacks our electrophile. These electrons kick off onto our oxygen. And we can go ahead and show the result of our nucleophilic attack. So we would now have this carbon double bonded to this oxygen, and then this oxygen is now bonded to this carbon. And then we have an oxygen up here with three lone pairs of electrons, negative one formal charge. We still have this carbon bonded to a chlorine, and we still have an R prime group here like that. So following some electrons, let's go ahead and put in these lone pairs here."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we would now have this carbon double bonded to this oxygen, and then this oxygen is now bonded to this carbon. And then we have an oxygen up here with three lone pairs of electrons, negative one formal charge. We still have this carbon bonded to a chlorine, and we still have an R prime group here like that. So following some electrons, let's go ahead and put in these lone pairs here. The electrons in magenta formed the bond between the oxygen and the carbon. And then we could say that these electrons in here moved off onto our oxygen to give that oxygen a negative one formal charge. When we think about the next step, we know that the chloride anion is an excellent leaving group."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So following some electrons, let's go ahead and put in these lone pairs here. The electrons in magenta formed the bond between the oxygen and the carbon. And then we could say that these electrons in here moved off onto our oxygen to give that oxygen a negative one formal charge. When we think about the next step, we know that the chloride anion is an excellent leaving group. So if these electrons in blue move in here to reform our double bond, then these electrons would kick off onto the chlorine to form the chloride anion, which we know is a stable on its own as a leaving group. And so that forms our acid anhydride. So let's go ahead and draw in the final structure here."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "When we think about the next step, we know that the chloride anion is an excellent leaving group. So if these electrons in blue move in here to reform our double bond, then these electrons would kick off onto the chlorine to form the chloride anion, which we know is a stable on its own as a leaving group. And so that forms our acid anhydride. So let's go ahead and draw in the final structure here. So we would have our oxygen with lone pairs of electrons on our oxygen. We would have this bond. We reformed our carbonyl like that."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw in the final structure here. So we would have our oxygen with lone pairs of electrons on our oxygen. We would have this bond. We reformed our carbonyl like that. And then the chloride anion was our leaving group, so now we have an R prime group. And so we formed our acid anhydride. And so we could make these R groups the same, or we could make them different."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "We reformed our carbonyl like that. And then the chloride anion was our leaving group, so now we have an R prime group. And so we formed our acid anhydride. And so we could make these R groups the same, or we could make them different. And so this is a good way of forming a mixed anhydride as well as one that is symmetrical. So let's look at an example. So let's take acetic acid, and we're going to do two different reactions with acetic acid."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so we could make these R groups the same, or we could make them different. And so this is a good way of forming a mixed anhydride as well as one that is symmetrical. So let's look at an example. So let's take acetic acid, and we're going to do two different reactions with acetic acid. So the first thing we're going to do is add thionyl chloride to acetic acid. And we've seen that the addition of thionyl chloride converts a carboxylic acid into an acyl halide. So let's go ahead and show the conversion of that carboxylic acid into the corresponding acyl halides."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's take acetic acid, and we're going to do two different reactions with acetic acid. So the first thing we're going to do is add thionyl chloride to acetic acid. And we've seen that the addition of thionyl chloride converts a carboxylic acid into an acyl halide. So let's go ahead and show the conversion of that carboxylic acid into the corresponding acyl halides. And then we could take that acetic acid in a separate reaction. We could add sodium hydroxide. And the hydroxide takes this proton, which leave these electrons behind on the oxygen."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show the conversion of that carboxylic acid into the corresponding acyl halides. And then we could take that acetic acid in a separate reaction. We could add sodium hydroxide. And the hydroxide takes this proton, which leave these electrons behind on the oxygen. So let's scoot up a little bit so we can see the mechanism that we talked about before. And so this would form sodium acetate. So this would form sodium acetate, draw in our lone pairs of electrons on this oxygen, negative one formal charge, and a plus like that."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And the hydroxide takes this proton, which leave these electrons behind on the oxygen. So let's scoot up a little bit so we can see the mechanism that we talked about before. And so this would form sodium acetate. So this would form sodium acetate, draw in our lone pairs of electrons on this oxygen, negative one formal charge, and a plus like that. So this oxygen already had two lone pairs of electrons. And so now you have the situation that we talked about up here. Your carboxylate anion functions as a nucleophile, attacks your electrophilic carbon on your acyl chloride."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So this would form sodium acetate, draw in our lone pairs of electrons on this oxygen, negative one formal charge, and a plus like that. So this oxygen already had two lone pairs of electrons. And so now you have the situation that we talked about up here. Your carboxylate anion functions as a nucleophile, attacks your electrophilic carbon on your acyl chloride. So we could show these electrons attacking this carbon. These electrons kick off onto the oxygen. And then when those electrons move back in to reform your double bond, these electrons would kick off onto your chlorine."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "Your carboxylate anion functions as a nucleophile, attacks your electrophilic carbon on your acyl chloride. So we could show these electrons attacking this carbon. These electrons kick off onto the oxygen. And then when those electrons move back in to reform your double bond, these electrons would kick off onto your chlorine. And then so we can go ahead and draw our product. So just thinking about what happens in this mechanism, we can go ahead and draw our product, which would be a symmetrical anhydride. So we would have our oxygen right here, and then we would have our group over here."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then when those electrons move back in to reform your double bond, these electrons would kick off onto your chlorine. And then so we can go ahead and draw our product. So just thinking about what happens in this mechanism, we can go ahead and draw our product, which would be a symmetrical anhydride. So we would have our oxygen right here, and then we would have our group over here. So thinking about the R groups this time, so this R group is a methyl group. And then we could think about this oxygen being this oxygen. And then this portion of the acyl chloride is this portion for our final product, for our acid anhydride."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So we would have our oxygen right here, and then we would have our group over here. So thinking about the R groups this time, so this R group is a methyl group. And then we could think about this oxygen being this oxygen. And then this portion of the acyl chloride is this portion for our final product, for our acid anhydride. And so this is acetic anhydride. This right here is acetic anhydride, which is the one that's used most commonly in an undergraduate lab. And so this is a nice way of preparing acid anhydrides."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And then this portion of the acyl chloride is this portion for our final product, for our acid anhydride. And so this is acetic anhydride. This right here is acetic anhydride, which is the one that's used most commonly in an undergraduate lab. And so this is a nice way of preparing acid anhydrides. All right, let's look at another way to form acetic anhydride. You could start with two carboxylic acids, right? And this would be acetic acid and acetic acid, so the same carboxylic acid, and apply high heat."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And so this is a nice way of preparing acid anhydrides. All right, let's look at another way to form acetic anhydride. You could start with two carboxylic acids, right? And this would be acetic acid and acetic acid, so the same carboxylic acid, and apply high heat. And this time you think about a dehydration reaction. So you could think about losing an OH from one and a hydrogen from the other to form water. So you could think about losing water here in your dehydration reaction."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And this would be acetic acid and acetic acid, so the same carboxylic acid, and apply high heat. And this time you think about a dehydration reaction. So you could think about losing an OH from one and a hydrogen from the other to form water. So you could think about losing water here in your dehydration reaction. And then you can stick those portions of the molecule together. So you could take this portion, and then this portion, and put them together. And you can see that that is once again acetic anhydride."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So you could think about losing water here in your dehydration reaction. And then you can stick those portions of the molecule together. So you could take this portion, and then this portion, and put them together. And you can see that that is once again acetic anhydride. So let's go ahead and draw that. So we would form acetic anhydride here by dehydration. This way of doing it is not always the best way."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "And you can see that that is once again acetic anhydride. So let's go ahead and draw that. So we would form acetic anhydride here by dehydration. This way of doing it is not always the best way. It works for acetic acid, but it doesn't work for most carboxylic acids. All right, here's one more case where it can work if you have a situation like this. This is phthalic acid."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This way of doing it is not always the best way. It works for acetic acid, but it doesn't work for most carboxylic acids. All right, here's one more case where it can work if you have a situation like this. This is phthalic acid. So it's a dicarboxylic acid, and if you apply heat to it, you don't need as high of a heat as you need for the previous reaction. This heat is higher than this one. But you can once again form an anhydride."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "This is phthalic acid. So it's a dicarboxylic acid, and if you apply heat to it, you don't need as high of a heat as you need for the previous reaction. This heat is higher than this one. But you can once again form an anhydride. So if we think about a dehydration, losing OH from one and H from the other, and then we can go ahead and draw the product here. So we would form our benzene ring, and then we would form our anhydride like this. So once again, loss of water."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "But you can once again form an anhydride. So if we think about a dehydration, losing OH from one and H from the other, and then we can go ahead and draw the product here. So we would form our benzene ring, and then we would form our anhydride like this. So once again, loss of water. So the name of this acid anhydride would be phthalic anhydride because it's derived from phthalic acid. So this is a good way to form five or six-membered rings. In this case, we have a five-membered ring."}, {"video_title": "Preparation of acid anhydrides Carboxylic acids and derivatives Organic chemistry Khan Academy.mp3", "Sentence": "So once again, loss of water. So the name of this acid anhydride would be phthalic anhydride because it's derived from phthalic acid. So this is a good way to form five or six-membered rings. In this case, we have a five-membered ring. We have a carbon, an oxygen, a carbon, a carbon, and a carbon. So we have a five-membered ring this time. It also works for six-membered rings."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "The carbon bonded to that carbon would be a beta carbon, and we need a beta proton for our mechanism. The hydrogen and the halogen have to be on opposite sides of our bond here, if I draw in this line, they're on opposite sides, that's said to be anti. And these four atoms need to be nearly in the same plane. The hydrogen, the carbon, the carbon, and the halogen need to be nearly in the same plane, so that's planar. So we say this has to be antiperiplanar, so let me write this in here, so antiperiplanar, the anti meaning hydrogen and halogen on opposite sides, and the planar part meaning those four atoms are in the same plane. Our strong base comes along in our E2 mechanism, so a negative charge, and takes our beta proton. At the same time, these electrons move in here to form our double bond, and these electrons come off onto our leaving group."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "The hydrogen, the carbon, the carbon, and the halogen need to be nearly in the same plane, so that's planar. So we say this has to be antiperiplanar, so let me write this in here, so antiperiplanar, the anti meaning hydrogen and halogen on opposite sides, and the planar part meaning those four atoms are in the same plane. Our strong base comes along in our E2 mechanism, so a negative charge, and takes our beta proton. At the same time, these electrons move in here to form our double bond, and these electrons come off onto our leaving group. So let me show these electrons in here in blue, those electrons in blue move in to form our double bond, and so we form an alkene. If we're thinking about the stereoselectivity of an E2 mechanism, it can be helpful to look at this mechanism from a different point of view, for example, from a Newman projection. So here we have a Newman projection, and we can think about a base coming along in our Newman projection and taking our beta proton, so we're taking this proton here, and then these electrons would move in to here."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "At the same time, these electrons move in here to form our double bond, and these electrons come off onto our leaving group. So let me show these electrons in here in blue, those electrons in blue move in to form our double bond, and so we form an alkene. If we're thinking about the stereoselectivity of an E2 mechanism, it can be helpful to look at this mechanism from a different point of view, for example, from a Newman projection. So here we have a Newman projection, and we can think about a base coming along in our Newman projection and taking our beta proton, so we're taking this proton here, and then these electrons would move in to here. At the same time, these electrons come off onto our halogen, so it's a little bit easier to see that all four of those atoms are in the same plane if I draw a straight line through here. Or we could look at it from more of a saw horse projection, so here is our beta proton, and here is one of our carbons, here's another carbon, and here's our halogen, our strong base is going to take our beta proton, and these electrons would move in to here to form our double bond, and these electrons would come off onto our halogen, so that would give us our alkene. So let's take a look at an example where we're thinking about the E2 mechanism and stereoselectivity."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So here we have a Newman projection, and we can think about a base coming along in our Newman projection and taking our beta proton, so we're taking this proton here, and then these electrons would move in to here. At the same time, these electrons come off onto our halogen, so it's a little bit easier to see that all four of those atoms are in the same plane if I draw a straight line through here. Or we could look at it from more of a saw horse projection, so here is our beta proton, and here is one of our carbons, here's another carbon, and here's our halogen, our strong base is going to take our beta proton, and these electrons would move in to here to form our double bond, and these electrons would come off onto our halogen, so that would give us our alkene. So let's take a look at an example where we're thinking about the E2 mechanism and stereoselectivity. Let's say we have this alkyl halide, and we react it with sodium ethoxide, which is a strong base. The carbon bonded to the halogen is our alpha carbon, and the carbons bonded to the alpha carbon are the beta carbons, so I'll call this beta one, and then we have a beta carbon on the right, which I will call beta two. We're going to ignore beta two for right now, and just focus on beta one."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So let's take a look at an example where we're thinking about the E2 mechanism and stereoselectivity. Let's say we have this alkyl halide, and we react it with sodium ethoxide, which is a strong base. The carbon bonded to the halogen is our alpha carbon, and the carbons bonded to the alpha carbon are the beta carbons, so I'll call this beta one, and then we have a beta carbon on the right, which I will call beta two. We're going to ignore beta two for right now, and just focus on beta one. Beta one carbon has two hydrogens bonded to it, so I'll put one on a wedge and one on a dash, and let's highlight those. The one on the wedge I'm going to make green, and the one on the dash here I'm going to make white. And we're going to stare down that bond, so we're going to stare down this beta one alpha carbon bond, so if you put your eye right here, and you stare this way, and you stare down that bond, you're going to get a Newman projection."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "We're going to ignore beta two for right now, and just focus on beta one. Beta one carbon has two hydrogens bonded to it, so I'll put one on a wedge and one on a dash, and let's highlight those. The one on the wedge I'm going to make green, and the one on the dash here I'm going to make white. And we're going to stare down that bond, so we're going to stare down this beta one alpha carbon bond, so if you put your eye right here, and you stare this way, and you stare down that bond, you're going to get a Newman projection. And the hydrogen in green would be going up and to the right, so here's the hydrogen in green. The hydrogen in white would be going up and to the left, and then there'd be a methyl group going straight down, so we won't spend too much time on Newman projections since those are covered in earlier videos. On the back carbon, the bromine here would be down and to the right, and then there would also be, on this drawing, a hydrogen going away from us in space."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And we're going to stare down that bond, so we're going to stare down this beta one alpha carbon bond, so if you put your eye right here, and you stare this way, and you stare down that bond, you're going to get a Newman projection. And the hydrogen in green would be going up and to the right, so here's the hydrogen in green. The hydrogen in white would be going up and to the left, and then there'd be a methyl group going straight down, so we won't spend too much time on Newman projections since those are covered in earlier videos. On the back carbon, the bromine here would be down and to the right, and then there would also be, on this drawing, a hydrogen going away from us in space. On the Newman projection, that would be this hydrogen. And then finally, there's a methyl group going up in space, so here is our Newman projection staring down that bond. If you look at the hydrogen and the bromine, this hydrogen in white and the bromine, they're already antiperiplanar to each other, so if I draw in a line here, you can see that those four atoms are in the same plane, which is what we want."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "On the back carbon, the bromine here would be down and to the right, and then there would also be, on this drawing, a hydrogen going away from us in space. On the Newman projection, that would be this hydrogen. And then finally, there's a methyl group going up in space, so here is our Newman projection staring down that bond. If you look at the hydrogen and the bromine, this hydrogen in white and the bromine, they're already antiperiplanar to each other, so if I draw in a line here, you can see that those four atoms are in the same plane, which is what we want. However, to look at it a little bit easier, I've just turned this whole thing a little bit to the right, so this is really the same conformation. I haven't changed the conformation at all. I've just made that magenta line vertical."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "If you look at the hydrogen and the bromine, this hydrogen in white and the bromine, they're already antiperiplanar to each other, so if I draw in a line here, you can see that those four atoms are in the same plane, which is what we want. However, to look at it a little bit easier, I've just turned this whole thing a little bit to the right, so this is really the same conformation. I haven't changed the conformation at all. I've just made that magenta line vertical. So now we have our hydrogen in white is here, our bromine in red is here, and the hydrogen in green would be over here down and to the right. Notice that we have both of our methyl groups are anti to each other in both of these, right? So it's really the same Newman projection, just turned a little bit so we can see it better."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "I've just made that magenta line vertical. So now we have our hydrogen in white is here, our bromine in red is here, and the hydrogen in green would be over here down and to the right. Notice that we have both of our methyl groups are anti to each other in both of these, right? So it's really the same Newman projection, just turned a little bit so we can see it better. We know in our E2 mechanism, our strong base takes our beta proton. So our strong base would be the ethoxide anion. It's gonna take this beta proton."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So it's really the same Newman projection, just turned a little bit so we can see it better. We know in our E2 mechanism, our strong base takes our beta proton. So our strong base would be the ethoxide anion. It's gonna take this beta proton. These electrons move in here. These electrons come off onto our leaving group. And when the double bond forms, the two methyl groups end up on opposite sides of the double bond."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "It's gonna take this beta proton. These electrons move in here. These electrons come off onto our leaving group. And when the double bond forms, the two methyl groups end up on opposite sides of the double bond. So this actually gives us our trans product. So we would get our product as being trans right here. Maybe it's a little bit easier to see this mechanism from the Sawhorse projection."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And when the double bond forms, the two methyl groups end up on opposite sides of the double bond. So this actually gives us our trans product. So we would get our product as being trans right here. Maybe it's a little bit easier to see this mechanism from the Sawhorse projection. So on the right, so really, again, this is not a different conformation. This is just another way of looking at this molecule. Our alpha carbon would be here, because here's the bromine."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "Maybe it's a little bit easier to see this mechanism from the Sawhorse projection. So on the right, so really, again, this is not a different conformation. This is just another way of looking at this molecule. Our alpha carbon would be here, because here's the bromine. So this would be our beta one carbon. And our strong base would come along. And it's going to take this proton."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "Our alpha carbon would be here, because here's the bromine. So this would be our beta one carbon. And our strong base would come along. And it's going to take this proton. And these electrons would move in here to form our double bond at the same time that the bromine leaves. So this is our concerted mechanism. And when the double bond forms, these two methyl groups end up on opposite sides of the double bond."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And it's going to take this proton. And these electrons would move in here to form our double bond at the same time that the bromine leaves. So this is our concerted mechanism. And when the double bond forms, these two methyl groups end up on opposite sides of the double bond. So let's go to a video now where I show you all of this using a model set. And I think it's a little bit easier to see that you make the trans product. Also, we'll look at a different conformation and see the other product."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And when the double bond forms, these two methyl groups end up on opposite sides of the double bond. So let's go to a video now where I show you all of this using a model set. And I think it's a little bit easier to see that you make the trans product. Also, we'll look at a different conformation and see the other product. So taking the green proton. And I forgot to highlight this. This would be the white proton that we are taking."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "Also, we'll look at a different conformation and see the other product. So taking the green proton. And I forgot to highlight this. This would be the white proton that we are taking. And the green proton would be over here. So here's our alkyl halide. And you can see one hydrogen is green and one hydrogen is white."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "This would be the white proton that we are taking. And the green proton would be over here. So here's our alkyl halide. And you can see one hydrogen is green and one hydrogen is white. And we have the red for the bromine. And we're going to stare down this carbon-carbon bond. And notice I'm using these stretchy bonds to help us out with the E2 mechanism."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And you can see one hydrogen is green and one hydrogen is white. And we have the red for the bromine. And we're going to stare down this carbon-carbon bond. And notice I'm using these stretchy bonds to help us out with the E2 mechanism. So if we stare down that carbon-carbon bond, we'll see things from a Newman projection. And notice in this Newman projection, I'm just going to turn it a little bit so that we get our beta proton and our red bromine antiperiplanar and vertical. So next, think about these two methyl groups on the Newman projection as being anti to each other."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And notice I'm using these stretchy bonds to help us out with the E2 mechanism. So if we stare down that carbon-carbon bond, we'll see things from a Newman projection. And notice in this Newman projection, I'm just going to turn it a little bit so that we get our beta proton and our red bromine antiperiplanar and vertical. So next, think about these two methyl groups on the Newman projection as being anti to each other. So I'm going to turn it a little bit to see things from a sawhorse perspective. Think about taking this beta proton in the mechanism and pretend like that red bromine goes away. And you can see that now our two methyl groups are on opposite sides of the double bond that formed."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So next, think about these two methyl groups on the Newman projection as being anti to each other. So I'm going to turn it a little bit to see things from a sawhorse perspective. Think about taking this beta proton in the mechanism and pretend like that red bromine goes away. And you can see that now our two methyl groups are on opposite sides of the double bond that formed. So this is the trans product. If we go back to our Newman projection, we know there's free rotation around our single bond. So I can rotate to get a different conformation to put the green hydrogen antiperiplanar to the red bromine."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And you can see that now our two methyl groups are on opposite sides of the double bond that formed. So this is the trans product. If we go back to our Newman projection, we know there's free rotation around our single bond. So I can rotate to get a different conformation to put the green hydrogen antiperiplanar to the red bromine. And so from this Newman projection, you can see we have these two methyl groups that are now gauche to each other. So if I turn it to the side and I take this beta proton here, and again, pretend like the bromine goes away, here you can see that the two methyl groups are on the same side of the double bond that formed. So this is the cis product."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So I can rotate to get a different conformation to put the green hydrogen antiperiplanar to the red bromine. And so from this Newman projection, you can see we have these two methyl groups that are now gauche to each other. So if I turn it to the side and I take this beta proton here, and again, pretend like the bromine goes away, here you can see that the two methyl groups are on the same side of the double bond that formed. So this is the cis product. So here's what we saw in the video talking about this conformation. Our two methyl groups are gauche to each other here. And let me highlight our proton in green."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So this is the cis product. So here's what we saw in the video talking about this conformation. Our two methyl groups are gauche to each other here. And let me highlight our proton in green. So here's the proton in green. And the bromine is red. So if we're taking the proton in green, our strong base, which is our ethoxide anion, takes this proton, these electrons move into here, these electrons come off onto the bromine, and your two methyl groups end up on the same side."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And let me highlight our proton in green. So here's the proton in green. And the bromine is red. So if we're taking the proton in green, our strong base, which is our ethoxide anion, takes this proton, these electrons move into here, these electrons come off onto the bromine, and your two methyl groups end up on the same side. So this gives you the cis product. From the Sawhorse projection, if we take this proton, this proton is our green one, so let me highlight that. So this is the green proton."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So if we're taking the proton in green, our strong base, which is our ethoxide anion, takes this proton, these electrons move into here, these electrons come off onto the bromine, and your two methyl groups end up on the same side. So this gives you the cis product. From the Sawhorse projection, if we take this proton, this proton is our green one, so let me highlight that. So this is the green proton. And the bromine over here would be in red. So if we take the green proton, then these electrons move into here, the electrons come off onto our bromine, and the two methyl groups would be on the same side when the double bond forms. So this gives us our cis product."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So this is the green proton. And the bromine over here would be in red. So if we take the green proton, then these electrons move into here, the electrons come off onto our bromine, and the two methyl groups would be on the same side when the double bond forms. So this gives us our cis product. So let me go down here and draw it. So this would give us our cis product with our two methyl groups on the same side. And if we think about those two different conformations that have protons antiperiplanar, so this conformation had the white proton antiperiplanar, this conformation had the green proton antiperiplanar, the white one gave us the trans product, the green one gave us the cis product."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So this gives us our cis product. So let me go down here and draw it. So this would give us our cis product with our two methyl groups on the same side. And if we think about those two different conformations that have protons antiperiplanar, so this conformation had the white proton antiperiplanar, this conformation had the green proton antiperiplanar, the white one gave us the trans product, the green one gave us the cis product. We know that the first conformation is more stable. This conformation is more stable because our bulky methyl groups are anti to each other. Here, those two bulky methyl groups are gauche to each other and that's increased steric hindrance."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And if we think about those two different conformations that have protons antiperiplanar, so this conformation had the white proton antiperiplanar, this conformation had the green proton antiperiplanar, the white one gave us the trans product, the green one gave us the cis product. We know that the first conformation is more stable. This conformation is more stable because our bulky methyl groups are anti to each other. Here, those two bulky methyl groups are gauche to each other and that's increased steric hindrance. So this is not, this is the less stable conformation. And therefore, this product, our cis product, right, is not as stable as our trans product. To summarize what we saw for this alkyl halides, we know that our beta one carbon had two protons."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "Here, those two bulky methyl groups are gauche to each other and that's increased steric hindrance. So this is not, this is the less stable conformation. And therefore, this product, our cis product, right, is not as stable as our trans product. To summarize what we saw for this alkyl halides, we know that our beta one carbon had two protons. And we got two different products depending on which proton the base took. So we got the trans product and we got the cis product. The trans product turns out to be the major product for this reaction because this came from the more stable conformation."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "To summarize what we saw for this alkyl halides, we know that our beta one carbon had two protons. And we got two different products depending on which proton the base took. So we got the trans product and we got the cis product. The trans product turns out to be the major product for this reaction because this came from the more stable conformation. The cis product is the minor product for this reaction. So this one came from the less stable conformation. So this reaction is said to be stereoselective."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "The trans product turns out to be the major product for this reaction because this came from the more stable conformation. The cis product is the minor product for this reaction. So this one came from the less stable conformation. So this reaction is said to be stereoselective. We have these two stereoisomers and one stereoisomer is favored over the other. So the trans product is favored over the cis product and that's stereoselectivity. Now, we've ignored this beta two carbon until now."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So this reaction is said to be stereoselective. We have these two stereoisomers and one stereoisomer is favored over the other. So the trans product is favored over the cis product and that's stereoselectivity. Now, we've ignored this beta two carbon until now. So this beta two carbon has several protons and it's easy to get one in an antiperiplanar relationship with the bromine. I'll just draw one in really quickly here. So we think about a base taking our proton."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "Now, we've ignored this beta two carbon until now. So this beta two carbon has several protons and it's easy to get one in an antiperiplanar relationship with the bromine. I'll just draw one in really quickly here. So we think about a base taking our proton. So our base, our ethoxide anion takes this proton. These electrons move in here and these electrons come off onto the bromine. And so we're also gonna make another product."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "So we think about a base taking our proton. So our base, our ethoxide anion takes this proton. These electrons move in here and these electrons come off onto the bromine. And so we're also gonna make another product. So I'll put this in parentheses here. So we're gonna make this product. So our electrons in the magenta moved into here to form this double bond."}, {"video_title": "E2 mechanism stereoselectivity.mp3", "Sentence": "And so we're also gonna make another product. So I'll put this in parentheses here. So we're gonna make this product. So our electrons in the magenta moved into here to form this double bond. And so this is what we would get from our beta two carbon. So really we're focused on, in this video we're focused on our beta one carbon and thinking about stereoselectivity. But you can't forget that there are, there can be more than one beta carbon in a reaction."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we talked about how the geometry around our sp hybridized atoms must be linear. So this bond angle right here is 180 degrees, which makes this portion of the molecule with the triple bond and it's linear like that. And in that triple bond, we know that one of those bonds is a sigma bond and two of those bonds are pi bonds. So if you don't remember that stuff, you need to go back and watch the video on sp hybridization. So this particular alkyne right here has alkyl groups on either side of my triple bond. So it's a dye-substituted alkyne. And the more common term you'll see for that would be an internal alkyne."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if you don't remember that stuff, you need to go back and watch the video on sp hybridization. So this particular alkyne right here has alkyl groups on either side of my triple bond. So it's a dye-substituted alkyne. And the more common term you'll see for that would be an internal alkyne. So let's go ahead and write that. So it's an internal alkyne, which just means that the triple bond is found in the interior of the molecule, in the interior of the chain. The triple bond is not found on the end."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the more common term you'll see for that would be an internal alkyne. So let's go ahead and write that. So it's an internal alkyne, which just means that the triple bond is found in the interior of the molecule, in the interior of the chain. The triple bond is not found on the end. So that's an internal alkyne, which is different from a terminal alkyne. So let's go ahead and draw a terminal alkyne. And this is just where your triple bond is at the end of your carbon chain."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The triple bond is not found on the end. So that's an internal alkyne, which is different from a terminal alkyne. So let's go ahead and draw a terminal alkyne. And this is just where your triple bond is at the end of your carbon chain. So if I put a hydrogen on one side, now it's only monosubstituted. And the triple bond is on the end, or the terminal of my alkyne. So this is a terminal alkyne like this."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this is just where your triple bond is at the end of your carbon chain. So if I put a hydrogen on one side, now it's only monosubstituted. And the triple bond is on the end, or the terminal of my alkyne. So this is a terminal alkyne like this. So it's important to distinguish between these, because in future videos, we'll see how things like terminal alkynes have special properties. So let's talk about the nomenclature of alkynes now. And let's start with the simplest alkyne, a two-carbon alkyne."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is a terminal alkyne like this. So it's important to distinguish between these, because in future videos, we'll see how things like terminal alkynes have special properties. So let's talk about the nomenclature of alkynes now. And let's start with the simplest alkyne, a two-carbon alkyne. So a triple bond between two carbons like that. And let's just clear up that triple bond there a little bit. So this is our triple bond between our two carbons like that."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's start with the simplest alkyne, a two-carbon alkyne. So a triple bond between two carbons like that. And let's just clear up that triple bond there a little bit. So this is our triple bond between our two carbons like that. And I have hydrogens on either side. So when we talked about two carbons for nomenclature, our root was F. And since we're dealing with alkynes, our ending is going to be YNE here. So I stick the root together with the ending."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is our triple bond between our two carbons like that. And I have hydrogens on either side. So when we talked about two carbons for nomenclature, our root was F. And since we're dealing with alkynes, our ending is going to be YNE here. So I stick the root together with the ending. So I take F, and I add the ion to it. So it's called ethine. So ethine would be the IUPAC name for this molecule."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I stick the root together with the ending. So I take F, and I add the ion to it. So it's called ethine. So ethine would be the IUPAC name for this molecule. That's probably not the name that's used most frequently. This is the most famous alkyne, also called acetylene. So you've heard of acetylene torches before."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So ethine would be the IUPAC name for this molecule. That's probably not the name that's used most frequently. This is the most famous alkyne, also called acetylene. So you've heard of acetylene torches before. And you can do a very cool demonstration called underwater fireworks, where you use acetylene in there. So this is acetylene, which is the common name, or ethine, which is more the IUPAC name. And this is the simplest alkyne."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you've heard of acetylene torches before. And you can do a very cool demonstration called underwater fireworks, where you use acetylene in there. So this is acetylene, which is the common name, or ethine, which is more the IUPAC name. And this is the simplest alkyne. So let's name another one here. Let's get an actual chain, a carbon chain in here. And let's see how to name this one like that."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this is the simplest alkyne. So let's name another one here. Let's get an actual chain, a carbon chain in here. And let's see how to name this one like that. Well, the first thing you notice is the straight line that surrounds our triple bond there. And remember, that's because that portion of the molecule is linear, because those carbons are sp hybridized. So when you're drawing Lewis dot structures, you'll notice that almost everyone will draw that portion of the molecule straight to better reflect the actual geometry of the molecule."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's see how to name this one like that. Well, the first thing you notice is the straight line that surrounds our triple bond there. And remember, that's because that portion of the molecule is linear, because those carbons are sp hybridized. So when you're drawing Lewis dot structures, you'll notice that almost everyone will draw that portion of the molecule straight to better reflect the actual geometry of the molecule. So I need to figure out how many carbons are in this alkyne. So I find my longest carbon chain, including my triple bond. So let's see here."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when you're drawing Lewis dot structures, you'll notice that almost everyone will draw that portion of the molecule straight to better reflect the actual geometry of the molecule. So I need to figure out how many carbons are in this alkyne. So I find my longest carbon chain, including my triple bond. So let's see here. There's one carbon, two carbons, three carbon, four, and five. So a five-carbon alkyne. So remember, your root for five carbons would be pent."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's see here. There's one carbon, two carbons, three carbon, four, and five. So a five-carbon alkyne. So remember, your root for five carbons would be pent. So it would be pentane. If it was an alkyne, it would be pentine. So this is pentine."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So remember, your root for five carbons would be pent. So it would be pentane. If it was an alkyne, it would be pentine. So this is pentine. So I have that much so far. Next, I need to figure out how to number my chain. So I could number it from the left, or I could number it from the right."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is pentine. So I have that much so far. Next, I need to figure out how to number my chain. So I could number it from the left, or I could number it from the right. My goal is to give my triple bond the lowest number possible. So it's a lot like double bonds here. So I want to give my triple bond the lowest number possible."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I could number it from the left, or I could number it from the right. My goal is to give my triple bond the lowest number possible. So it's a lot like double bonds here. So I want to give my triple bond the lowest number possible. And that means, of course, I need to start from the right here. So if I start from the right, and I say this is carbon 1, then my triple bond starts at carbon 2. And then this would be carbon 3, carbon 4, and carbon 5, like that."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I want to give my triple bond the lowest number possible. And that means, of course, I need to start from the right here. So if I start from the right, and I say this is carbon 1, then my triple bond starts at carbon 2. And then this would be carbon 3, carbon 4, and carbon 5, like that. So the triple bond starts at carbon 2. So all I have to do is put a 2 in here and say 2-pentine. So in that respect, it's just like naming alkenes here."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then this would be carbon 3, carbon 4, and carbon 5, like that. So the triple bond starts at carbon 2. So all I have to do is put a 2 in here and say 2-pentine. So in that respect, it's just like naming alkenes here. All right, so that's 2-pentine. Let's do another one, so one that's a little bit more complicated than that. So let's look at this molecule here."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in that respect, it's just like naming alkenes here. All right, so that's 2-pentine. Let's do another one, so one that's a little bit more complicated than that. So let's look at this molecule here. So there's my triple bond. And then I have, let's see, I'll put in a chain and some methyl groups there. And so this is my molecule."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at this molecule here. So there's my triple bond. And then I have, let's see, I'll put in a chain and some methyl groups there. And so this is my molecule. All right, so once again, find the longest carbon chain that includes your triple bond. So that would be 1, 2, 3, 4, 5, 6, and 7. So 7 carbons would be hept, so it's heptine."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so this is my molecule. All right, so once again, find the longest carbon chain that includes your triple bond. So that would be 1, 2, 3, 4, 5, 6, and 7. So 7 carbons would be hept, so it's heptine. So let's go ahead and write heptine in here. So this molecule would be heptine, like that. And let's go ahead and number it."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So 7 carbons would be hept, so it's heptine. So let's go ahead and write heptine in here. So this molecule would be heptine, like that. And let's go ahead and number it. I want a number to give my triple bond the highest priority here. So in this case, if I start from the left or start from the right, there'd be a 2 for either that methyl group or for the start of my triple bond. The triple bond is going to win."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's go ahead and number it. I want a number to give my triple bond the highest priority here. So in this case, if I start from the left or start from the right, there'd be a 2 for either that methyl group or for the start of my triple bond. The triple bond is going to win. You want to give the triple bond the highest priority. And that's the reason you're going to start numbering from the left. So you're going to go 1, 2, 3, 4, 5, 6, and 7, like that."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The triple bond is going to win. You want to give the triple bond the highest priority. And that's the reason you're going to start numbering from the left. So you're going to go 1, 2, 3, 4, 5, 6, and 7, like that. And so now we can see that our triple bond starts at carbon 2. So I'm going to write 2-heptine here, like that. And what else do I have?"}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So you're going to go 1, 2, 3, 4, 5, 6, and 7, like that. And so now we can see that our triple bond starts at carbon 2. So I'm going to write 2-heptine here, like that. And what else do I have? Well, at carbon 5 and at carbon 6, I have methyl groups. So it'd be 5, 6-dimethyl. So I'll go ahead and put that in here, 5, 6-dimethyl, 2-heptine, like that."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And what else do I have? Well, at carbon 5 and at carbon 6, I have methyl groups. So it'd be 5, 6-dimethyl. So I'll go ahead and put that in here, 5, 6-dimethyl, 2-heptine, like that. And this is kind of the old school way of naming it. So old IUPAC recommendations would say to name it this way. And maybe most organic chemistry professors will still name this molecule this way."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'll go ahead and put that in here, 5, 6-dimethyl, 2-heptine, like that. And this is kind of the old school way of naming it. So old IUPAC recommendations would say to name it this way. And maybe most organic chemistry professors will still name this molecule this way. But there's a newer way that IUPAC recommends that you name this molecule. So it would still be 5, 6-dimethyl, except now you would go ahead and write hept. And then you would put the 2 in between the endings."}, {"video_title": "Alkyne nomenclature Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And maybe most organic chemistry professors will still name this molecule this way. But there's a newer way that IUPAC recommends that you name this molecule. So it would still be 5, 6-dimethyl, except now you would go ahead and write hept. And then you would put the 2 in between the endings. So it would be hept-2-ine, like that. So the first way is more old school. The second way is more the new way of doing it."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about how hydrogen bromide might react with this thing right here. Let's think about what this would be called. We have 1, 2, 3, 4, 5 carbons. It has a double bond right here. If we start numbering at this end, because that's where the double bond is, then this would be pent. This is pent. The double bond starts at the number 1 carbon, pent 1."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It has a double bond right here. If we start numbering at this end, because that's where the double bond is, then this would be pent. This is pent. The double bond starts at the number 1 carbon, pent 1. And it's obviously an alkene. It has a double bond. So it's pent 1 ene."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The double bond starts at the number 1 carbon, pent 1. And it's obviously an alkene. It has a double bond. So it's pent 1 ene. Sometimes this is called 1 pentene. Either way. So let's think about how these two characters might react with each other in some type of solvent."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's pent 1 ene. Sometimes this is called 1 pentene. Either way. So let's think about how these two characters might react with each other in some type of solvent. Usually when a solvent's not specified, it's usually water or alcohol. If this was water, then we would have a solution of hydrobromic acid. But let's not worry about that right now."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's think about how these two characters might react with each other in some type of solvent. Usually when a solvent's not specified, it's usually water or alcohol. If this was water, then we would have a solution of hydrobromic acid. But let's not worry about that right now. Let's just worry about how these two characters might react with each other. The first thing we might look at is this hydrogen bromide right there. As you get more experience there, you'll say, bromine is much more electronegative than hydrogen."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But let's not worry about that right now. Let's just worry about how these two characters might react with each other. The first thing we might look at is this hydrogen bromide right there. As you get more experience there, you'll say, bromine is much more electronegative than hydrogen. It likes to hog electrons much more. If that's a completely foreign concept to you, I'll do a little bit of a review. This is a periodic table."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "As you get more experience there, you'll say, bromine is much more electronegative than hydrogen. It likes to hog electrons much more. If that's a completely foreign concept to you, I'll do a little bit of a review. This is a periodic table. Electronegativity increases from the bottom left to the top right of the periodic table. Electronegativity. And really that's just a fancy way of saying, how much does an atom like to hog electrons?"}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is a periodic table. Electronegativity increases from the bottom left to the top right of the periodic table. Electronegativity. And really that's just a fancy way of saying, how much does an atom like to hog electrons? For example, fluorine loves to hog electrons, and calcium doesn't like to hog electrons so much. If we think about hydrogen bromide, hydrogen's way up here on the left side of the periodic table. Bromine's all the way to the right."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And really that's just a fancy way of saying, how much does an atom like to hog electrons? For example, fluorine loves to hog electrons, and calcium doesn't like to hog electrons so much. If we think about hydrogen bromide, hydrogen's way up here on the left side of the periodic table. Bromine's all the way to the right. Bromine is much more electronegative. In this situation, since bromine is more electronegative, it will hog the electrons in this bond. Since it's hogging the electrons in this bond, you'll have a partial negative charge on the bromine end, and you'll have a partial positive charge on the hydrogen end."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Bromine's all the way to the right. Bromine is much more electronegative. In this situation, since bromine is more electronegative, it will hog the electrons in this bond. Since it's hogging the electrons in this bond, you'll have a partial negative charge on the bromine end, and you'll have a partial positive charge on the hydrogen end. Whenever you do these reactions, it's actually useful to draw all of the valence electrons. Bromine right here has an atomic number of 35, which means it has 35 electrons in its neutral state. We don't have any charge here, so it's in a neutral state."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Since it's hogging the electrons in this bond, you'll have a partial negative charge on the bromine end, and you'll have a partial positive charge on the hydrogen end. Whenever you do these reactions, it's actually useful to draw all of the valence electrons. Bromine right here has an atomic number of 35, which means it has 35 electrons in its neutral state. We don't have any charge here, so it's in a neutral state. But you can look at its group. It's in group number 7 if you count from here. 1, 2, 3, 4, 5, 6, 7."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We don't have any charge here, so it's in a neutral state. But you can look at its group. It's in group number 7 if you count from here. 1, 2, 3, 4, 5, 6, 7. It has 7 valence electrons. Let me draw that. 7 electrons in its outermost shell."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3, 4, 5, 6, 7. It has 7 valence electrons. Let me draw that. 7 electrons in its outermost shell. Let me draw them right here. 1, 2, 3, 4, 5, 6. The 7th is in the bond with the hydrogen."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "7 electrons in its outermost shell. Let me draw them right here. 1, 2, 3, 4, 5, 6. The 7th is in the bond with the hydrogen. Hydrogen obviously has 1 electron. Hydrogen has 1 electron, and it's right there. These 2 guys are bonding with each other, and they both allow each other to pretend that they're part of a pair."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The 7th is in the bond with the hydrogen. Hydrogen obviously has 1 electron. Hydrogen has 1 electron, and it's right there. These 2 guys are bonding with each other, and they both allow each other to pretend that they're part of a pair. That's what gives it a lower energy level or makes it a little more stable. That's why the bond forms in the first place. Let's think about what might happen here."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "These 2 guys are bonding with each other, and they both allow each other to pretend that they're part of a pair. That's what gives it a lower energy level or makes it a little more stable. That's why the bond forms in the first place. Let's think about what might happen here. This guy is really electronegative, so maybe he wants to take this electron away, this green hydrogen electron. Maybe that happens. Let me just draw it out."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's think about what might happen here. This guy is really electronegative, so maybe he wants to take this electron away, this green hydrogen electron. Maybe that happens. Let me just draw it out. Maybe that happens. He takes that electron, that green electron away. It's just getting closer and closer."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me just draw it out. Maybe that happens. He takes that electron, that green electron away. It's just getting closer and closer. If there's only some other place that this hydrogen could get an electron from, then this guy could just go to the bromine, which is what bromine really wants to happen because it's so electronegative. Where can the hydrogen get an electron to replace the one that it's about to lose? We have a double bond here, and maybe one of these carbons lose an electron."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's just getting closer and closer. If there's only some other place that this hydrogen could get an electron from, then this guy could just go to the bromine, which is what bromine really wants to happen because it's so electronegative. Where can the hydrogen get an electron to replace the one that it's about to lose? We have a double bond here, and maybe one of these carbons lose an electron. In future videos, we'll talk more which one is more likely to give up the electron at this stage of the reaction. Just for simplicity, let's assume that this carbon right here gives up an electron. It's valence electrons."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We have a double bond here, and maybe one of these carbons lose an electron. In future videos, we'll talk more which one is more likely to give up the electron at this stage of the reaction. Just for simplicity, let's assume that this carbon right here gives up an electron. It's valence electrons. Just as a bit of review, here's carbon on the periodic table. It has six protons and six electrons in its stable state. Two are sitting in its first shell, and then the other four are its valence electrons."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's valence electrons. Just as a bit of review, here's carbon on the periodic table. It has six protons and six electrons in its stable state. Two are sitting in its first shell, and then the other four are its valence electrons. You can see that it's in group four. One, two, three, four. We can ignore these for now."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Two are sitting in its first shell, and then the other four are its valence electrons. You can see that it's in group four. One, two, three, four. We can ignore these for now. Usually, you're not going to be dealing with... We'll ignore those for now. It has four valence electrons. Most of what you're going to deal with, you just have to count the group numbers like this."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We can ignore these for now. Usually, you're not going to be dealing with... We'll ignore those for now. It has four valence electrons. Most of what you're going to deal with, you just have to count the group numbers like this. We won't worry too much about all of the metals and all of those right now. It has four valence electrons, and you see it right here. It has one, two, three, four valence electrons."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Most of what you're going to deal with, you just have to count the group numbers like this. We won't worry too much about all of the metals and all of those right now. It has four valence electrons, and you see it right here. It has one, two, three, four valence electrons. It has two in its one shell, so it actually has six, but you only draw the four out there. What we're saying is this green electron can go to the bromine as long as the hydrogen can take an electron maybe from this carbon right here. Let's draw that."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It has one, two, three, four valence electrons. It has two in its one shell, so it actually has six, but you only draw the four out there. What we're saying is this green electron can go to the bromine as long as the hydrogen can take an electron maybe from this carbon right here. Let's draw that. This electron right here is going to go to the hydrogen, and when that electron goes to the hydrogen, simultaneously that will allow this electron to go to bromine. Obviously, they won't be this far when it happens. There would be some type of collision that would have to occur in just the right way with just the right level of energy for this to occur."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's draw that. This electron right here is going to go to the hydrogen, and when that electron goes to the hydrogen, simultaneously that will allow this electron to go to bromine. Obviously, they won't be this far when it happens. There would be some type of collision that would have to occur in just the right way with just the right level of energy for this to occur. Maybe the hydrogen is getting really close to this part right at the right moment when this electron is being sucked away from the bromine. This has a partial positive charge attracted to the electron. That electron goes over there."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "There would be some type of collision that would have to occur in just the right way with just the right level of energy for this to occur. Maybe the hydrogen is getting really close to this part right at the right moment when this electron is being sucked away from the bromine. This has a partial positive charge attracted to the electron. That electron goes over there. It won't always happen, but this is a potential reaction mechanism. This whole step happens at once. This electron goes from this character to the hydrogen at the same time as hydrogen loses its electron to the bromine."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That electron goes over there. It won't always happen, but this is a potential reaction mechanism. This whole step happens at once. This electron goes from this character to the hydrogen at the same time as hydrogen loses its electron to the bromine. What's going to happen right after this step? Right after that happens, what will everything look like? The bromine will have gained the electron."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This electron goes from this character to the hydrogen at the same time as hydrogen loses its electron to the bromine. What's going to happen right after this step? Right after that happens, what will everything look like? The bromine will have gained the electron. It's now a bromide ion. Let me draw it like this. It had its original 7 valence electrons."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The bromine will have gained the electron. It's now a bromide ion. Let me draw it like this. It had its original 7 valence electrons. 1, 2, 3, 4, 5, 6, 7. Now it just stole an electron from the hydrogen. It was able to swipe an electron off the hydrogen."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It had its original 7 valence electrons. 1, 2, 3, 4, 5, 6, 7. Now it just stole an electron from the hydrogen. It was able to swipe an electron off the hydrogen. That's the electron it swiped off the hydrogen. Now what will this thing that was 1-pentene or pent-1-ene, what will this look like now? Let me draw it."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It was able to swipe an electron off the hydrogen. That's the electron it swiped off the hydrogen. Now what will this thing that was 1-pentene or pent-1-ene, what will this look like now? Let me draw it. We have a carbon, a hydrogen, and a hydrogen. Then you have a carbon, a hydrogen, and then you have the rest of the chain right over here. This double bond was broken."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw it. We have a carbon, a hydrogen, and a hydrogen. Then you have a carbon, a hydrogen, and then you have the rest of the chain right over here. This double bond was broken. This carbon lost an electron and went to the hydrogen. This bond right here now forms between this carbon and that hydrogen. Let me draw that bond."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This double bond was broken. This carbon lost an electron and went to the hydrogen. This bond right here now forms between this carbon and that hydrogen. Let me draw that bond. Actually, let me draw it. You have this electron right there. That electron will now be with the hydrogen."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw that bond. Actually, let me draw it. You have this electron right there. That electron will now be with the hydrogen. Let me draw the electron. Now we have that orange hydrogen. I'll try to keep the colors consistent so that we know where things came from."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That electron will now be with the hydrogen. Let me draw the electron. Now we have that orange hydrogen. I'll try to keep the colors consistent so that we know where things came from. Then we have this bond now to the hydrogen. This carbon now only has 3 valence electrons. 1, 2, 3."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'll try to keep the colors consistent so that we know where things came from. Then we have this bond now to the hydrogen. This carbon now only has 3 valence electrons. 1, 2, 3. It has 2 sitting in its first shell, so it actually has a total of 5 electrons. It has 6 protons, so it has a positive charge. This carbon right here has a positive charge."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "1, 2, 3. It has 2 sitting in its first shell, so it actually has a total of 5 electrons. It has 6 protons, so it has a positive charge. This carbon right here has a positive charge. Another way to think about it is it was completely neutral, and then it lost an electron. Now it will have a positive charge right over there. This is what we are left after that step of the reaction."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This carbon right here has a positive charge. Another way to think about it is it was completely neutral, and then it lost an electron. Now it will have a positive charge right over there. This is what we are left after that step of the reaction. As you can imagine, and of course we can't forget, bromine over here was neutral. It had 7 valence electrons, and that's when bromine is neutral, but now it has 8, so now this will have a negative charge. This will have a negative charge because it gained an electron."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This is what we are left after that step of the reaction. As you can imagine, and of course we can't forget, bromine over here was neutral. It had 7 valence electrons, and that's when bromine is neutral, but now it has 8, so now this will have a negative charge. This will have a negative charge because it gained an electron. In general, your total charge over here is 0, so our total charge should still be 0. We have a negative and we have a positive. They would cancel out, so our total charge is still 0."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This will have a negative charge because it gained an electron. In general, your total charge over here is 0, so our total charge should still be 0. We have a negative and we have a positive. They would cancel out, so our total charge is still 0. What's likely to happen for the next step of our reaction? We have this positive thing here. Maybe bromine just bumped just the right way to let go of this guy and steal his electron, but now you have this guy who's negative and this guy who's positive."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "They would cancel out, so our total charge is still 0. What's likely to happen for the next step of our reaction? We have this positive thing here. Maybe bromine just bumped just the right way to let go of this guy and steal his electron, but now you have this guy who's negative and this guy who's positive. Maybe they'll be attracted to each other. Maybe they'll just bump into each other the exact right way. If they bump in the exact right way, maybe this guy can swipe the electron from the bromide ion, from this negative ion right here."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Maybe bromine just bumped just the right way to let go of this guy and steal his electron, but now you have this guy who's negative and this guy who's positive. Maybe they'll be attracted to each other. Maybe they'll just bump into each other the exact right way. If they bump in the exact right way, maybe this guy can swipe the electron from the bromide ion, from this negative ion right here. You might say, isn't bromine more electronegative than carbon? It might be, but this guy is electron-rich. It's not just a regular bromine atom."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If they bump in the exact right way, maybe this guy can swipe the electron from the bromide ion, from this negative ion right here. You might say, isn't bromine more electronegative than carbon? It might be, but this guy is electron-rich. It's not just a regular bromine atom. This is a bromine plus an extra electron. He's already hogged an electron, so he's electron-rich. In this situation, he's negative, he's positive."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's not just a regular bromine atom. This is a bromine plus an extra electron. He's already hogged an electron, so he's electron-rich. In this situation, he's negative, he's positive. He can give this guy an electron. If they bump in just the right way, this electron can be swiped by this carbon right over here. This positive carbon, just to give you a little terminology, and we'll go over it in more detail in future videos, is actually called a carbocation."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "In this situation, he's negative, he's positive. He can give this guy an electron. If they bump in just the right way, this electron can be swiped by this carbon right over here. This positive carbon, just to give you a little terminology, and we'll go over it in more detail in future videos, is actually called a carbocation. It's a positive ion of carbon. That's where the word comes from. Anyway, if this electron gets swiped by this carbon, it'll then form a bond."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This positive carbon, just to give you a little terminology, and we'll go over it in more detail in future videos, is actually called a carbocation. It's a positive ion of carbon. That's where the word comes from. Anyway, if this electron gets swiped by this carbon, it'll then form a bond. Remember, this was the electron that was originally in a bond with this hydrogen. It's still going to be, you can imagine, paired up with this other purple or magenta electron right over there. If that happens, then we're going to be left with..."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Anyway, if this electron gets swiped by this carbon, it'll then form a bond. Remember, this was the electron that was originally in a bond with this hydrogen. It's still going to be, you can imagine, paired up with this other purple or magenta electron right over there. If that happens, then we're going to be left with... The next step is going to be... On this end of the molecule, we have C, carbon, hydrogen, hydrogen. Then we have this orange hydrogen that we stole from the hydrogen bromide. Then we have this carbon right here."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If that happens, then we're going to be left with... The next step is going to be... On this end of the molecule, we have C, carbon, hydrogen, hydrogen. Then we have this orange hydrogen that we stole from the hydrogen bromide. Then we have this carbon right here. It has a hydrogen. You have the rest of the chain, CH2, CH2, CH3. Now, since this guy stole an electron, a bond will form with the bromine."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Then we have this carbon right here. It has a hydrogen. You have the rest of the chain, CH2, CH2, CH3. Now, since this guy stole an electron, a bond will form with the bromine. Now, let me draw it. He's going to steal this. Let me draw it this way."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, since this guy stole an electron, a bond will form with the bromine. Now, let me draw it. He's going to steal this. Let me draw it this way. A bond will form. He's stealing this electron. Now this electron is with this carbon."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me draw it this way. A bond will form. He's stealing this electron. Now this electron is with this carbon. I can do it as one end of the bond. The other end of the bond will be that magenta electron right there on the bromine. Now the bromine lost an electron, so it's neutral."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now this electron is with this carbon. I can do it as one end of the bond. The other end of the bond will be that magenta electron right there on the bromine. Now the bromine lost an electron, so it's neutral. It once again has 1, 2, 3, 4, 5, 6, 7 valence electrons. Carbon is now neutral because it gained an electron. It once again has 1, 2, 3, 4 electrons."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now the bromine lost an electron, so it's neutral. It once again has 1, 2, 3, 4, 5, 6, 7 valence electrons. Carbon is now neutral because it gained an electron. It once again has 1, 2, 3, 4 electrons. Now everyone is happy again. In this video, we were able to get a mechanism. We can talk in the future about how likely it is to happen, how quickly it might happen."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It once again has 1, 2, 3, 4 electrons. Now everyone is happy again. In this video, we were able to get a mechanism. We can talk in the future about how likely it is to happen, how quickly it might happen. We were able to start with hydrogen bromide and 1-pentene or pentuanine and get to this thing just by pushing around electrons and just thinking about what is likely to happen based on what's electronegative and what might be able to gain or lose an electron. Just a bit of a review, what is this thing right here? This might be 1, 2, 3, 4, 5 carbon still, so it's going to be a pent."}, {"video_title": "Introduction to reaction mechanisms Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We can talk in the future about how likely it is to happen, how quickly it might happen. We were able to start with hydrogen bromide and 1-pentene or pentuanine and get to this thing just by pushing around electrons and just thinking about what is likely to happen based on what's electronegative and what might be able to gain or lose an electron. Just a bit of a review, what is this thing right here? This might be 1, 2, 3, 4, 5 carbon still, so it's going to be a pent. It's now an alkane, no double bond, so it's pentane. It has one group on it, so we'll start numbering closer to the group. It's 2-bromopentane."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's the eon in which either life first started to exist, or at least it first started to somewhat flourish. It's possible that maybe life first started to exist at the end of the Hadean Eon. And of course, this boundary is vague. And the Archean Eon is also the first eon where we still have rocks from that time. So we are able to find rocks that we can date to be roughly 3.8 billion years. Now, the other really interesting thing that happened in the Archean Eon, it really has pretty profound effects once we get into the Proterozoic Eon, is that you started to have cyanobacteria produce oxygen. And we said in the last video that they were producing oxygen, but most of that oxygen was being absorbed by iron in the oceans."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And the Archean Eon is also the first eon where we still have rocks from that time. So we are able to find rocks that we can date to be roughly 3.8 billion years. Now, the other really interesting thing that happened in the Archean Eon, it really has pretty profound effects once we get into the Proterozoic Eon, is that you started to have cyanobacteria produce oxygen. And we said in the last video that they were producing oxygen, but most of that oxygen was being absorbed by iron in the oceans. But what happens is we enter into the Proterozoic Eon. And Proterozoic Eon is right over here. I don't know if you can see that."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we said in the last video that they were producing oxygen, but most of that oxygen was being absorbed by iron in the oceans. But what happens is we enter into the Proterozoic Eon. And Proterozoic Eon is right over here. I don't know if you can see that. I'll rewrite it. Proterozoic. We're now in the Proterozoic Eon that starts up about 2.5 billion years ago."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I don't know if you can see that. I'll rewrite it. Proterozoic. We're now in the Proterozoic Eon that starts up about 2.5 billion years ago. And Proterozoic comes from the Greek for earlier life. And I'm not a Greek scholar, so any of you Greeks out there, forgive me if I'm not getting the translation exactly right. But what's really interesting about the Proterozoic Eon is that that oxygen that was being produced by the cyanobacteria at some point begins to saturate the iron and any other molecule that could have absorbed it before."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're now in the Proterozoic Eon that starts up about 2.5 billion years ago. And Proterozoic comes from the Greek for earlier life. And I'm not a Greek scholar, so any of you Greeks out there, forgive me if I'm not getting the translation exactly right. But what's really interesting about the Proterozoic Eon is that that oxygen that was being produced by the cyanobacteria at some point begins to saturate the iron and any other molecule that could have absorbed it before. And once it saturates, it starts to get released into the atmosphere. So the oxygen starts to get released and accumulate in the atmosphere. And we think this started to happen about 2.4 billion years ago."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But what's really interesting about the Proterozoic Eon is that that oxygen that was being produced by the cyanobacteria at some point begins to saturate the iron and any other molecule that could have absorbed it before. And once it saturates, it starts to get released into the atmosphere. So the oxygen starts to get released and accumulate in the atmosphere. And we think this started to happen about 2.4 billion years ago. So 2.4 billion years ago, oxygen begins to accumulate in the atmosphere. And of course, these dates, they might be moved around a few hundred million years if you get more and more data. But this is the current understanding of when things happen."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And we think this started to happen about 2.4 billion years ago. So 2.4 billion years ago, oxygen begins to accumulate in the atmosphere. And of course, these dates, they might be moved around a few hundred million years if you get more and more data. But this is the current understanding of when things happen. And maybe we'll look at the geological record or the fossil record, and we'll move these things around in the future. I could only imagine that 50 years from now or 100 years from now, if someone's still watching this video, a lot of this might say, hey, we found out later that oxygen started to accumulate in the atmosphere earlier or later or that eukaryotes occurred earlier or later. But this, as far as I can tell, is our best current understanding."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is the current understanding of when things happen. And maybe we'll look at the geological record or the fossil record, and we'll move these things around in the future. I could only imagine that 50 years from now or 100 years from now, if someone's still watching this video, a lot of this might say, hey, we found out later that oxygen started to accumulate in the atmosphere earlier or later or that eukaryotes occurred earlier or later. But this, as far as I can tell, is our best current understanding. But 2.4 billion years ago, oxygen begins to accumulate in the atmosphere. And what's interesting about this is once it accumulates and once it gets a critical amount of oxygen in the atmosphere, and I touched on this in the last video, about 2.3 billion years ago, we have something called the Great Oxygenation Event, sometimes called the Oxygen Catastrophe. And this is right here."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this, as far as I can tell, is our best current understanding. But 2.4 billion years ago, oxygen begins to accumulate in the atmosphere. And what's interesting about this is once it accumulates and once it gets a critical amount of oxygen in the atmosphere, and I touched on this in the last video, about 2.3 billion years ago, we have something called the Great Oxygenation Event, sometimes called the Oxygen Catastrophe. And this is right here. They mark it on this right here, 2.3 billion years ago or 2,300 million years ago. Atmosphere becomes oxygen rich. And it's not as oxygen rich as our current atmosphere, but it becomes oxygen rich enough that at least the environment becomes suitable for eukaryotic organisms or eukaryotic cells."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this is right here. They mark it on this right here, 2.3 billion years ago or 2,300 million years ago. Atmosphere becomes oxygen rich. And it's not as oxygen rich as our current atmosphere, but it becomes oxygen rich enough that at least the environment becomes suitable for eukaryotic organisms or eukaryotic cells. Now, the other interesting thing, and we might not care so much about it because we needed the oxygen, is that we think that this was actually the greatest extinction event in the history of Earth. That's why it's called the Oxygen Catastrophe. So this right over here, 2.3 billion years ago, I shouldn't giggle about it."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's not as oxygen rich as our current atmosphere, but it becomes oxygen rich enough that at least the environment becomes suitable for eukaryotic organisms or eukaryotic cells. Now, the other interesting thing, and we might not care so much about it because we needed the oxygen, is that we think that this was actually the greatest extinction event in the history of Earth. That's why it's called the Oxygen Catastrophe. So this right over here, 2.3 billion years ago, I shouldn't giggle about it. This is a serious matter. This is the greatest extinction event in Earth's history. And I'll do history with a capital H. In Earth's history of the Earth."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right over here, 2.3 billion years ago, I shouldn't giggle about it. This is a serious matter. This is the greatest extinction event in Earth's history. And I'll do history with a capital H. In Earth's history of the Earth. And that's because the cyanobacteria is producing all this oxygen, eventually saturates the iron, it accumulates in the air. Once it gets to enough concentration, it begins to actually suffocate. It's poisonous."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I'll do history with a capital H. In Earth's history of the Earth. And that's because the cyanobacteria is producing all this oxygen, eventually saturates the iron, it accumulates in the air. Once it gets to enough concentration, it begins to actually suffocate. It's poisonous. It's poisonous to most of the other organisms on the planet that were anaerobic, that did not need oxygen, that actually found oxygen poisonous. Now, but since we have oxygen, there's two interesting things that happened once that oxygen accumulated, other than causing this mass extinction event. Actually, three interesting things."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's poisonous. It's poisonous to most of the other organisms on the planet that were anaerobic, that did not need oxygen, that actually found oxygen poisonous. Now, but since we have oxygen, there's two interesting things that happened once that oxygen accumulated, other than causing this mass extinction event. Actually, three interesting things. Two of them are crucial to us eventually showing up on this planet. The first is that it became suitable now for eukaryotes to exist. Eukaryotic organisms, remember, these are organisms that have nuclear membranes around their DNA."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Actually, three interesting things. Two of them are crucial to us eventually showing up on this planet. The first is that it became suitable now for eukaryotes to exist. Eukaryotic organisms, remember, these are organisms that have nuclear membranes around their DNA. Most eukaryotes have other organelles, like mitochondria. They need oxygen. You can go to the biology playlist."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Eukaryotic organisms, remember, these are organisms that have nuclear membranes around their DNA. Most eukaryotes have other organelles, like mitochondria. They need oxygen. You can go to the biology playlist. We actually talk about respiration that occurs in the mitochondria. And that's obviously a process that needs oxygen. So one, now that oxygen is in the atmosphere, we're starting to have an environment where eukaryotes could at least exist."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You can go to the biology playlist. We actually talk about respiration that occurs in the mitochondria. And that's obviously a process that needs oxygen. So one, now that oxygen is in the atmosphere, we're starting to have an environment where eukaryotes could at least exist. And based on the fossil records, and when we look at how the DNA has changed over time, and we'll do multiple videos of that, we think that the first eukaryotes showed up about 2.2 billion years ago. Although there's some debate here. There's some evidence it might have been a little earlier, some evidence it might have been a little later."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So one, now that oxygen is in the atmosphere, we're starting to have an environment where eukaryotes could at least exist. And based on the fossil records, and when we look at how the DNA has changed over time, and we'll do multiple videos of that, we think that the first eukaryotes showed up about 2.2 billion years ago. Although there's some debate here. There's some evidence it might have been a little earlier, some evidence it might have been a little later. I'm sure that number will be refined. But about, give or take a few hundred millions of years, one prokaryote got engulfed by another prokaryote and said, hey, we do pretty well living together. The current theory is that mitochondria is actually descended from kind of an ancient prokaryotic cell, an ancient bacteria."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's some evidence it might have been a little earlier, some evidence it might have been a little later. I'm sure that number will be refined. But about, give or take a few hundred millions of years, one prokaryote got engulfed by another prokaryote and said, hey, we do pretty well living together. The current theory is that mitochondria is actually descended from kind of an ancient prokaryotic cell, an ancient bacteria. It actually has its own DNA. And actually, your mitochondrial DNA is passed down from your mother, and your mother's mother, and your mother's mother, so on and so forth. So it's kind of like another little animal living inside of a larger cell."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The current theory is that mitochondria is actually descended from kind of an ancient prokaryotic cell, an ancient bacteria. It actually has its own DNA. And actually, your mitochondrial DNA is passed down from your mother, and your mother's mother, and your mother's mother, so on and so forth. So it's kind of like another little animal living inside of a larger cell. And we are eukaryotes. We needed this to happen. The human body, we're not just one eukaryotic cell."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's kind of like another little animal living inside of a larger cell. And we are eukaryotes. We needed this to happen. The human body, we're not just one eukaryotic cell. We're made up of trillions. Estimates are 50 to 100 trillion eukaryotic cells. So these are our ancestors that had to come into being at that time."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The human body, we're not just one eukaryotic cell. We're made up of trillions. Estimates are 50 to 100 trillion eukaryotic cells. So these are our ancestors that had to come into being at that time. And once again, all of this is happening inside of the oceans. Now, the other interesting thing that happened. Remember, we're being bombarded with UV radiation from the sun."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So these are our ancestors that had to come into being at that time. And once again, all of this is happening inside of the oceans. Now, the other interesting thing that happened. Remember, we're being bombarded with UV radiation from the sun. So if you're on the land, let me draw the land and the ocean. So here is the ocean, and then here is the land, right over there in yellow, constantly being bombarded with UV radiation. And UV stands for ultraviolet, so I'm drawing it in purple."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Remember, we're being bombarded with UV radiation from the sun. So if you're on the land, let me draw the land and the ocean. So here is the ocean, and then here is the land, right over there in yellow, constantly being bombarded with UV radiation. And UV stands for ultraviolet, so I'm drawing it in purple. But it's even more violet than purple. So it's constantly being bombarded with ultraviolet radiation from the sun, which is very inhospitable to DNA and to life. And so the only life at this point could occur in the ocean, where it was protected to some degree from the ultraviolet radiation."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And UV stands for ultraviolet, so I'm drawing it in purple. But it's even more violet than purple. So it's constantly being bombarded with ultraviolet radiation from the sun, which is very inhospitable to DNA and to life. And so the only life at this point could occur in the ocean, where it was protected to some degree from the ultraviolet radiation. The land was just open to it. Anything on the land would have just gotten irradiated. Its DNA would get mutated."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so the only life at this point could occur in the ocean, where it was protected to some degree from the ultraviolet radiation. The land was just open to it. Anything on the land would have just gotten irradiated. Its DNA would get mutated. It just would not be able to live. So what happened, and what I guess has to happen, and the reason why we are able to live on land now is that we have an ozone layer. We have an ozone layer up in the upper atmosphere that helps absorb, that blocks most of the UV radiation from the sun."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Its DNA would get mutated. It just would not be able to live. So what happened, and what I guess has to happen, and the reason why we are able to live on land now is that we have an ozone layer. We have an ozone layer up in the upper atmosphere that helps absorb, that blocks most of the UV radiation from the sun. And now that oxygen began to accumulate, we have the oxygen catastrophe. Oxygen accumulates in the atmosphere. Some of that oxygen goes into the upper atmosphere."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We have an ozone layer up in the upper atmosphere that helps absorb, that blocks most of the UV radiation from the sun. And now that oxygen began to accumulate, we have the oxygen catastrophe. Oxygen accumulates in the atmosphere. Some of that oxygen goes into the upper atmosphere. So we're now in this time period right over here. It goes into the upper atmosphere. It actually reacts with the UV light to turn into ozone, which then can help actually block the UV light."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Some of that oxygen goes into the upper atmosphere. So we're now in this time period right over here. It goes into the upper atmosphere. It actually reacts with the UV light to turn into ozone, which then can help actually block the UV light. And I'll do another video maybe on the ozone oxygen cycle. So this oxygen production, it's crucial, one, to having an ozone layer so that eventually life can exist on the land. And it's also crucial because eukaryotic organisms need that oxygen."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It actually reacts with the UV light to turn into ozone, which then can help actually block the UV light. And I'll do another video maybe on the ozone oxygen cycle. So this oxygen production, it's crucial, one, to having an ozone layer so that eventually life can exist on the land. And it's also crucial because eukaryotic organisms need that oxygen. Now the third thing that happened, this is also a pretty significant event, we believe that that oxygen that started to accumulate in the atmosphere reacted with methane in the atmosphere. So it reacted with methane. And methane is a greenhouse gas."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it's also crucial because eukaryotic organisms need that oxygen. Now the third thing that happened, this is also a pretty significant event, we believe that that oxygen that started to accumulate in the atmosphere reacted with methane in the atmosphere. So it reacted with methane. And methane is a greenhouse gas. It helps retain heat in the atmosphere. And once it reacts with the oxygen and starts dropping out of the atmosphere as methane, we believe the Earth cooled down. And it entered its first, and some people believe its longest snowball period."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And methane is a greenhouse gas. It helps retain heat in the atmosphere. And once it reacts with the oxygen and starts dropping out of the atmosphere as methane, we believe the Earth cooled down. And it entered its first, and some people believe its longest snowball period. So that's what they talk about right here on this diagram. The first snowball Earth. It's sometimes called the Huronic glaciation."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And it entered its first, and some people believe its longest snowball period. So that's what they talk about right here on this diagram. The first snowball Earth. It's sometimes called the Huronic glaciation. And that happened because we weren't able to retain our heat, if that theory is correct. And so as the theory goes, the whole Earth essentially just iced over. So as we go through the Proterozoic eon, I guess the big markers of it is it's the first time that we now have an oxygen-rich atmosphere."}, {"video_title": "Ozone layer and eukaryotes show up in the Proterozoic eon Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's sometimes called the Huronic glaciation. And that happened because we weren't able to retain our heat, if that theory is correct. And so as the theory goes, the whole Earth essentially just iced over. So as we go through the Proterozoic eon, I guess the big markers of it is it's the first time that we now have an oxygen-rich atmosphere. It's the first time that eukaryotes can now come into existence. Because they now have oxygen to, I guess we could say, breathe. And the other big thing is now this is where the ozone forms."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's see how to form imines and enamines. We'll start with the formation of an imine. You start with an aldehyde or a ketone, and you add an imine to it, and you need an acid catalyst. And over here on the right, if your Y is equal to a hydrogen or an alkyl group, which we'll say R double prime, you will form an imine. So let's go ahead and write that. This would be an imine. And the Y could be other things, and we'll talk about those other things in the next video."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And over here on the right, if your Y is equal to a hydrogen or an alkyl group, which we'll say R double prime, you will form an imine. So let's go ahead and write that. This would be an imine. And the Y could be other things, and we'll talk about those other things in the next video. You also form water. And so if you, since this reaction is at equilibrium, if you wanted to shift the equilibrium to the right, you could remove water as it's formed, and the equilibrium will shift this way and give you more of your imine product. So let's look at the mechanism to form imines."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And the Y could be other things, and we'll talk about those other things in the next video. You also form water. And so if you, since this reaction is at equilibrium, if you wanted to shift the equilibrium to the right, you could remove water as it's formed, and the equilibrium will shift this way and give you more of your imine product. So let's look at the mechanism to form imines. And I've seen two different ways to start off the mechanism, and so I will present both ways, and you can choose which one you want to use. And so one way would be to think about an acid being present. So we'll say it's HA plus, a generic acid."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at the mechanism to form imines. And I've seen two different ways to start off the mechanism, and so I will present both ways, and you can choose which one you want to use. And so one way would be to think about an acid being present. So we'll say it's HA plus, a generic acid. And you could think about this as protonating your imine. So if you have an imine and protons, you protonate your imine to form this generic acid here. And then your carbonyl oxygen is going to be protonated."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we'll say it's HA plus, a generic acid. And you could think about this as protonating your imine. So if you have an imine and protons, you protonate your imine to form this generic acid here. And then your carbonyl oxygen is going to be protonated. A lone pair of electrons on the oxygen takes this proton, leaves these electrons behind. So let's go ahead and show the results of that. So now we have our oxygen with a plus one formal charge, and we still have our R prime group over here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then your carbonyl oxygen is going to be protonated. A lone pair of electrons on the oxygen takes this proton, leaves these electrons behind. So let's go ahead and show the results of that. So now we have our oxygen with a plus one formal charge, and we still have our R prime group over here. So let's show those electrons. So this lone pair of electrons here on the oxygen picks up this proton, and that forms this bond right here. We talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our oxygen with a plus one formal charge, and we still have our R prime group over here. So let's show those electrons. So this lone pair of electrons here on the oxygen picks up this proton, and that forms this bond right here. We talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic. And so we have a good electrophile. This carbon here is a good electrophile, and the imine can be a good nucleophile. So a lone pair of electrons in the nitrogen over here are going to attack our carbonyl carbon, and that's going to kick these electrons in here off onto the oxygen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We talked about in previous videos how protonation of the carbonyl oxygen makes the carbonyl carbon more electrophilic. And so we have a good electrophile. This carbon here is a good electrophile, and the imine can be a good nucleophile. So a lone pair of electrons in the nitrogen over here are going to attack our carbonyl carbon, and that's going to kick these electrons in here off onto the oxygen. So let's draw the results of that. So now we're going to have our nitrogen is forming a bond with our carbon. So let's show those electrons."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So a lone pair of electrons in the nitrogen over here are going to attack our carbonyl carbon, and that's going to kick these electrons in here off onto the oxygen. So let's draw the results of that. So now we're going to have our nitrogen is forming a bond with our carbon. So let's show those electrons. So let's make those blue. So this lone pair of electrons on the nitrogen forms a bond between this nitrogen and this carbon here. This nitrogen is also bonded to two hydrogens."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's show those electrons. So let's make those blue. So this lone pair of electrons on the nitrogen forms a bond between this nitrogen and this carbon here. This nitrogen is also bonded to two hydrogens. Let's draw those in. And still our Y group like that. That gives our nitrogen a plus one formal charge."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "This nitrogen is also bonded to two hydrogens. Let's draw those in. And still our Y group like that. That gives our nitrogen a plus one formal charge. And for our carbon, it's also bonded to this oxygen. This oxygen now has two lone pairs of electrons. So let's show the movement of those electrons."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "That gives our nitrogen a plus one formal charge. And for our carbon, it's also bonded to this oxygen. This oxygen now has two lone pairs of electrons. So let's show the movement of those electrons. So I'll use green here. So these electrons in green moved out onto our oxygen. The carbon is still bonded to our alkyl group, so R and R prime."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's show the movement of those electrons. So I'll use green here. So these electrons in green moved out onto our oxygen. The carbon is still bonded to our alkyl group, so R and R prime. So I'm saying we start with the ketone here. And so that's one way to start off your mechanism. And of course, there's another way to do it."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "The carbon is still bonded to our alkyl group, so R and R prime. So I'm saying we start with the ketone here. And so that's one way to start off your mechanism. And of course, there's another way to do it. So let's talk about just using our nitrogen as a nucleophile straight away. So the amine is a pretty good nucleophile. And so we could just show our nucleophile attacking our electrophile directly."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And of course, there's another way to do it. So let's talk about just using our nitrogen as a nucleophile straight away. So the amine is a pretty good nucleophile. And so we could just show our nucleophile attacking our electrophile directly. So partial negative oxygen, partial positive carbon. This is our electrophile. And then a lone pair of electrons on our nitrogen makes our amine a good nucleophile."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we could just show our nucleophile attacking our electrophile directly. So partial negative oxygen, partial positive carbon. This is our electrophile. And then a lone pair of electrons on our nitrogen makes our amine a good nucleophile. So it can just attack directly. Attack this carbon, push these electrons off onto your oxygen. So let's go ahead and draw the result of that."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then a lone pair of electrons on our nitrogen makes our amine a good nucleophile. So it can just attack directly. Attack this carbon, push these electrons off onto your oxygen. So let's go ahead and draw the result of that. We would have our carbon bonded to our nitrogen. Our nitrogen is bonded to two hydrogens and our Y group. And that would make our nitrogen a plus one formal charge."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw the result of that. We would have our carbon bonded to our nitrogen. Our nitrogen is bonded to two hydrogens and our Y group. And that would make our nitrogen a plus one formal charge. And then our oxygen, this time, would be a negative one formal charge. So negative one formal charge here. We have our alkyl groups down here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And that would make our nitrogen a plus one formal charge. And then our oxygen, this time, would be a negative one formal charge. So negative one formal charge here. We have our alkyl groups down here. So let's once again show our electrons. So if nitrogen attacks directly, this lone pair in blue forms this bond right here. And then we can say that these pi electrons in here kick off onto our oxygen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We have our alkyl groups down here. So let's once again show our electrons. So if nitrogen attacks directly, this lone pair in blue forms this bond right here. And then we can say that these pi electrons in here kick off onto our oxygen. So it doesn't really matter which lone pair you make them. So I'll just say it's that one right there. And then we can show this intermediate going to the one we already talked about."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we can say that these pi electrons in here kick off onto our oxygen. So it doesn't really matter which lone pair you make them. So I'll just say it's that one right there. And then we can show this intermediate going to the one we already talked about. If we just protonate our negative one formal charge on our oxygen here. So let's go ahead and show that. So if we have our generic acid, so HA plus, we could show a lone pair picking up this proton, leaving these electrons behind."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we can show this intermediate going to the one we already talked about. If we just protonate our negative one formal charge on our oxygen here. So let's go ahead and show that. So if we have our generic acid, so HA plus, we could show a lone pair picking up this proton, leaving these electrons behind. And so now that would give us this intermediate, the same one that we had before. So whichever way you would like to start off your mechanism. Once we've reached this intermediate, we can think about the base, the amine here, lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So if we have our generic acid, so HA plus, we could show a lone pair picking up this proton, leaving these electrons behind. And so now that would give us this intermediate, the same one that we had before. So whichever way you would like to start off your mechanism. Once we've reached this intermediate, we can think about the base, the amine here, lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogen. So let's show what we would make from that. So let's get a little more room down here. So we deprotonate."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Once we've reached this intermediate, we can think about the base, the amine here, lone pair of electrons picking up this proton and leaving these electrons behind on our nitrogen. So let's show what we would make from that. So let's get a little more room down here. So we deprotonate. And then we form this structure, a carbon bonded to a nitrogen. And then we still have our Y group. We still have a hydrogen on this nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we deprotonate. And then we form this structure, a carbon bonded to a nitrogen. And then we still have our Y group. We still have a hydrogen on this nitrogen. And we have a lone pair of electrons on this nitrogen. So that lone pair came from right in here. So you take that proton and leave those electrons in magenta behind on your nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We still have a hydrogen on this nitrogen. And we have a lone pair of electrons on this nitrogen. So that lone pair came from right in here. So you take that proton and leave those electrons in magenta behind on your nitrogen. So we also have an OH bonded to our carbon here. And then we also have our alkyl group, so R and R prime. So this intermediate is called a carbanolamine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So you take that proton and leave those electrons in magenta behind on your nitrogen. So we also have an OH bonded to our carbon here. And then we also have our alkyl group, so R and R prime. So this intermediate is called a carbanolamine. So let me go ahead and write that. So this is called a carbanolamine. And for our next step, we could protonate the OH group."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this intermediate is called a carbanolamine. So let me go ahead and write that. So this is called a carbanolamine. And for our next step, we could protonate the OH group. So we take our generic acid once again, so HA plus. And a lone pair of electrons on our oxygen could pick up this proton, leaving these electrons behind. So let's go ahead and show that."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And for our next step, we could protonate the OH group. So we take our generic acid once again, so HA plus. And a lone pair of electrons on our oxygen could pick up this proton, leaving these electrons behind. So let's go ahead and show that. So let's get, once again, some more space in here. So if we protonate our carbanolamine, we would have our carbon here, our nitrogen, lone pair of electrons on our nitrogen, our Y group, our hydrogen here, our alkyl groups. And then if we protonate the OH, we would form water as a good leaving group."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and show that. So let's get, once again, some more space in here. So if we protonate our carbanolamine, we would have our carbon here, our nitrogen, lone pair of electrons on our nitrogen, our Y group, our hydrogen here, our alkyl groups. And then if we protonate the OH, we would form water as a good leaving group. So right in here, this would be a plus 1 formal charge on our oxygen. And let's show those electrons. So in here, lone pair of electrons on the oxygen pick up this proton."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then if we protonate the OH, we would form water as a good leaving group. So right in here, this would be a plus 1 formal charge on our oxygen. And let's show those electrons. So in here, lone pair of electrons on the oxygen pick up this proton. And so forming this bond right here. And now you can see we have a good leaving group. So we have water as a good leaving group."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So in here, lone pair of electrons on the oxygen pick up this proton. And so forming this bond right here. And now you can see we have a good leaving group. So we have water as a good leaving group. So if these electrons came off onto the oxygen, these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen. So let's go ahead and show that. So our next step here, this is where we lose water."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we have water as a good leaving group. So if these electrons came off onto the oxygen, these electrons in the nitrogen could move in here to form a double bond between the carbon and the nitrogen. So let's go ahead and show that. So our next step here, this is where we lose water. So we're going to minus H2O. And let's go ahead and show our next structure here. We'd have our carbon double bonded to our nitrogen this time."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So our next step here, this is where we lose water. So we're going to minus H2O. And let's go ahead and show our next structure here. We'd have our carbon double bonded to our nitrogen this time. Our nitrogen is still bonded to our Y group and our hydrogen over here. That gives the nitrogen a plus 1 formal charge. And the carbon is still bonded to our alkyl groups."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We'd have our carbon double bonded to our nitrogen this time. Our nitrogen is still bonded to our Y group and our hydrogen over here. That gives the nitrogen a plus 1 formal charge. And the carbon is still bonded to our alkyl groups. And so let's show those electrons here on our nitrogen. So I'm going to go ahead and make those magenta. So these magenta electrons move in here to form our double bond between our carbon and our nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And the carbon is still bonded to our alkyl groups. And so let's show those electrons here on our nitrogen. So I'm going to go ahead and make those magenta. So these magenta electrons move in here to form our double bond between our carbon and our nitrogen. And this is an important structure. It's called an iminium ion. So let's go ahead and draw that."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So these magenta electrons move in here to form our double bond between our carbon and our nitrogen. And this is an important structure. It's called an iminium ion. So let's go ahead and draw that. So an iminium ion. And then we lose water, of course. So minus water at this stage."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's go ahead and draw that. So an iminium ion. And then we lose water, of course. So minus water at this stage. So we're almost to our final product. We would just have to deprotonate our iminium ion. And so we can do that with our amine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So minus water at this stage. So we're almost to our final product. We would just have to deprotonate our iminium ion. And so we can do that with our amine. We could take that proton and leave these electrons behind on the nitrogen. So that's going to give us our final product, where we have carbon double bonded to our nitrogen. And then we have nitrogen bonded to our y group here, lone pair of electrons on that nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we can do that with our amine. We could take that proton and leave these electrons behind on the nitrogen. So that's going to give us our final product, where we have carbon double bonded to our nitrogen. And then we have nitrogen bonded to our y group here, lone pair of electrons on that nitrogen. And then we have our alkyl groups. And so let's show those electrons here in blue. Moving in here off onto our nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we have nitrogen bonded to our y group here, lone pair of electrons on that nitrogen. And then we have our alkyl groups. And so let's show those electrons here in blue. Moving in here off onto our nitrogen. And then once again, if our y is a hydrogen or an alkyl group, we have formed an imine. So that's formation of an imine. Let's look at an example."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Moving in here off onto our nitrogen. And then once again, if our y is a hydrogen or an alkyl group, we have formed an imine. So that's formation of an imine. Let's look at an example. So here is our reaction. We're going to start with a ketone here, so cyclohexanone, and react it with a primary amine. So the nitrogen is bonded to one carbon."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "Let's look at an example. So here is our reaction. We're going to start with a ketone here, so cyclohexanone, and react it with a primary amine. So the nitrogen is bonded to one carbon. So this is a reaction with a primary amine. And then we're going to use sulfuric acid as our catalyst here. So to figure out the product of this reaction, it's kind of a long mechanism."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So the nitrogen is bonded to one carbon. So this is a reaction with a primary amine. And then we're going to use sulfuric acid as our catalyst here. So to figure out the product of this reaction, it's kind of a long mechanism. So it doesn't really make sense to run through the entire mechanism. But we could think about part of the mechanism here. We know that the nucleophile is going to be our amine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So to figure out the product of this reaction, it's kind of a long mechanism. So it doesn't really make sense to run through the entire mechanism. But we could think about part of the mechanism here. We know that the nucleophile is going to be our amine. And it's going to attack our carbonyl carbon here. So we know that in our mechanism, we lose one of these protons on our amine. So let's say it's that proton right there."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We know that the nucleophile is going to be our amine. And it's going to attack our carbonyl carbon here. So we know that in our mechanism, we lose one of these protons on our amine. So let's say it's that proton right there. And we know we're going to lose water, so minus H2O. And this is going to think about taking us to our iminium ion step. So we're going to have our ring."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's say it's that proton right there. And we know we're going to lose water, so minus H2O. And this is going to think about taking us to our iminium ion step. So we're going to have our ring. And so we've already lost water. And so now we're going to have a double bond to our nitrogen. And our nitrogen is going to have an ethyl group on it here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to have our ring. And so we've already lost water. And so now we're going to have a double bond to our nitrogen. And our nitrogen is going to have an ethyl group on it here. And then we still have one proton on our nitrogen. So we still have a hydrogen attached here at this point, which gives us a plus 1 formal charge on our nitrogen. So this is our iminium ion right here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And our nitrogen is going to have an ethyl group on it here. And then we still have one proton on our nitrogen. So we still have a hydrogen attached here at this point, which gives us a plus 1 formal charge on our nitrogen. So this is our iminium ion right here. And then in our last step, we know that our amine can come along and act as a base. So a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our amine product. So we go over here and we draw our product."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is our iminium ion right here. And then in our last step, we know that our amine can come along and act as a base. So a lone pair of electrons on the nitrogen could take this proton right here and leave these electrons behind on our nitrogen to give us our amine product. So we go over here and we draw our product. Double bond to our nitrogen. And then we have our ethyl group and lone pair of electrons on our nitrogen. So if I follow those electrons, so these electrons in here in blue ended up on our nitrogen to form our amine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So we go over here and we draw our product. Double bond to our nitrogen. And then we have our ethyl group and lone pair of electrons on our nitrogen. So if I follow those electrons, so these electrons in here in blue ended up on our nitrogen to form our amine. All right, so a few things that I want to point out here for the iminium ion. We still have a proton attached to it. So deprotonation for our last step will form an imine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So if I follow those electrons, so these electrons in here in blue ended up on our nitrogen to form our amine. All right, so a few things that I want to point out here for the iminium ion. We still have a proton attached to it. So deprotonation for our last step will form an imine. And this is, once again, done with a primary amine right here. And so that's a little bit different from what we're going to talk about next. We're going to talk about formation of an enamine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So deprotonation for our last step will form an imine. And this is, once again, done with a primary amine right here. And so that's a little bit different from what we're going to talk about next. We're going to talk about formation of an enamine. And formation of an enamine starts off the same way in terms of the mechanism. But it's this iminium ion step that changes a little bit. So let's look at reacting a ketone with a secondary amine this time."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We're going to talk about formation of an enamine. And formation of an enamine starts off the same way in terms of the mechanism. But it's this iminium ion step that changes a little bit. So let's look at reacting a ketone with a secondary amine this time. And so let's look at this one right here. So here we have a secondary amine. So our nitrogen bonded to two other carbons."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So let's look at reacting a ketone with a secondary amine this time. And so let's look at this one right here. So here we have a secondary amine. So our nitrogen bonded to two other carbons. So a secondary amine this time. And once again, we can kind of run through the mechanism, just thinking a little bit about what happens without going through each individual step. We know the nucleophile is going to attack here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So our nitrogen bonded to two other carbons. So a secondary amine this time. And once again, we can kind of run through the mechanism, just thinking a little bit about what happens without going through each individual step. We know the nucleophile is going to attack here. And we know we're going to lose a proton. So this is the only proton that we have left to lose. And so we lose that one."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We know the nucleophile is going to attack here. And we know we're going to lose a proton. So this is the only proton that we have left to lose. And so we lose that one. We're going to lose water in the mechanism to form our iminium ion. So let's go ahead and draw that out. We're going to have our ring."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we lose that one. We're going to lose water in the mechanism to form our iminium ion. So let's go ahead and draw that out. We're going to have our ring. We're going to have a double bond to this nitrogen. We've already lost the proton in red. So now we have two alkyl groups attached to our nitrogen and a plus 1 formal charge on our nitrogen here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "We're going to have our ring. We're going to have a double bond to this nitrogen. We've already lost the proton in red. So now we have two alkyl groups attached to our nitrogen and a plus 1 formal charge on our nitrogen here. And so in the previous reaction, let me go back up to here to the previous reaction again. Once again, we still had a proton on our iminium ion at this step. And so we're able to deprotonate here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So now we have two alkyl groups attached to our nitrogen and a plus 1 formal charge on our nitrogen here. And so in the previous reaction, let me go back up to here to the previous reaction again. Once again, we still had a proton on our iminium ion at this step. And so we're able to deprotonate here. But that's not the case with what we have for this ion. We don't have a proton on our nitrogen here. And so we can't deprotonate in the same place."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we're able to deprotonate here. But that's not the case with what we have for this ion. We don't have a proton on our nitrogen here. And so we can't deprotonate in the same place. We have to pick a carbon adjacent over here. So here's a proton over here on this carbon. And our base could come along."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And so we can't deprotonate in the same place. We have to pick a carbon adjacent over here. So here's a proton over here on this carbon. And our base could come along. So let's go ahead and draw out our amine base with a lone pair of electrons right here. And a lone pair could take this proton. That would push these electrons in here and then push these electrons off onto the nitrogen to give us our product."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And our base could come along. So let's go ahead and draw out our amine base with a lone pair of electrons right here. And a lone pair could take this proton. That would push these electrons in here and then push these electrons off onto the nitrogen to give us our product. So let me go ahead and draw this out here. So now we're going to have a double bond here. And then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "That would push these electrons in here and then push these electrons off onto the nitrogen to give us our product. So let me go ahead and draw this out here. So now we're going to have a double bond here. And then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen. So let's follow some electrons. So these electrons in here, once you deprotonate, move in here to form our double bond. And then these electrons in here move out onto our nitrogen like that to form a different product."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then we have our nitrogen with these ethyl groups and a lone pair of electrons on our nitrogen. So let's follow some electrons. So these electrons in here, once you deprotonate, move in here to form our double bond. And then these electrons in here move out onto our nitrogen like that to form a different product. So this is called an enamine. So let me go ahead and write this out here. An enamine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "And then these electrons in here move out onto our nitrogen like that to form a different product. So this is called an enamine. So let me go ahead and write this out here. An enamine. And the name comes from, en comes from the double bonds. So we form double bonds. That's the en portion."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "An enamine. And the name comes from, en comes from the double bonds. So we form double bonds. That's the en portion. And then the amine portion comes from this part up here. We have an amine. So this is an enamine."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "That's the en portion. And then the amine portion comes from this part up here. We have an amine. So this is an enamine. And so formation of an enamine happens because, once again, we don't have the same iminium ion that we had before because of the fact that we started with a secondary amine. So that's why it's important to recognize the kind of amine that you were reacting with your aldehyde or ketone. And if it's secondary, you're going to form an enamine here."}, {"video_title": "Formation of imines and enamines Aldehydes and ketones Organic chemistry Khan Academy.mp3", "Sentence": "So this is an enamine. And so formation of an enamine happens because, once again, we don't have the same iminium ion that we had before because of the fact that we started with a secondary amine. So that's why it's important to recognize the kind of amine that you were reacting with your aldehyde or ketone. And if it's secondary, you're going to form an enamine here. And enamines are useful synthetic intermediates. And so we'll talk much more about them in later videos. But once again, the mechanism is the same until you get to this last step here."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And someone brought up what I thought was a very good question. If the animals were the first, what did they eat? So I thought that that was, one, a good question. So it justified a whole video on clarifying exactly who was first on the land. So right here, this is a picture of algae on the coast. This is kind of algal scum right over here. So this right here is algae."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it justified a whole video on clarifying exactly who was first on the land. So right here, this is a picture of algae on the coast. This is kind of algal scum right over here. So this right here is algae. And just to be clear, sometimes cyanobacteria, which we talked about as the first photosynthetic organism, sometimes that's called blue-green algae. But that's really bacteria. Algae is considered to be eukaryotic."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right here is algae. And just to be clear, sometimes cyanobacteria, which we talked about as the first photosynthetic organism, sometimes that's called blue-green algae. But that's really bacteria. Algae is considered to be eukaryotic. And it just doesn't have the structures of modern plants. So this is algae right here. And our best estimate is that algae actually colonized kind of coastal rocks about 1.2 billion years ago."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Algae is considered to be eukaryotic. And it just doesn't have the structures of modern plants. So this is algae right here. And our best estimate is that algae actually colonized kind of coastal rocks about 1.2 billion years ago. So this is 1.2 billion years ago. G for giga, billion years ago. So if you wanted the first thing that even resembled or was close to plants or animals, and if you consider algae close to a plant, then this would be the winner of who got on land first."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And our best estimate is that algae actually colonized kind of coastal rocks about 1.2 billion years ago. So this is 1.2 billion years ago. G for giga, billion years ago. So if you wanted the first thing that even resembled or was close to plants or animals, and if you consider algae close to a plant, then this would be the winner of who got on land first. This is 1.2 billion years ago. Now, in the last video where I talk about animals colonizing the land first, they weren't animals that only existed on land. They would have been animals that probably spent most of their time in the ocean collecting food or whatever."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you wanted the first thing that even resembled or was close to plants or animals, and if you consider algae close to a plant, then this would be the winner of who got on land first. This is 1.2 billion years ago. Now, in the last video where I talk about animals colonizing the land first, they weren't animals that only existed on land. They would have been animals that probably spent most of their time in the ocean collecting food or whatever. And then they would show up on the land maybe to lay eggs. And if you think about it, back then, the land would have been a really good place to lay eggs because there wouldn't have been much else on the land. So you would have been protected from predators."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They would have been animals that probably spent most of their time in the ocean collecting food or whatever. And then they would show up on the land maybe to lay eggs. And if you think about it, back then, the land would have been a really good place to lay eggs because there wouldn't have been much else on the land. So you would have been protected from predators. So it might have been slug-like creatures like this. Some people talk about kind of spider-like creatures. But it still would have been at the coast."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you would have been protected from predators. So it might have been slug-like creatures like this. Some people talk about kind of spider-like creatures. But it still would have been at the coast. And these would have been creatures that would have spent a lot of time in the ocean and some time in their land. So this is what I was referring to as kind of animals colonizing the land before plants. And this would have happened about 530 million years ago."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But it still would have been at the coast. And these would have been creatures that would have spent a lot of time in the ocean and some time in their land. So this is what I was referring to as kind of animals colonizing the land before plants. And this would have happened about 530 million years ago. 530 million years ago. Now, the first living organisms to fully live on the land, their whole life is on the land, those would be plants. So it depends."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this would have happened about 530 million years ago. 530 million years ago. Now, the first living organisms to fully live on the land, their whole life is on the land, those would be plants. So it depends. If you think about things that live part of their life, you'd get the animals, things that live their whole life on the land. Then you go back to the plants. So this right here, this is what we think the first primitive plants would have looked like."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it depends. If you think about things that live part of their life, you'd get the animals, things that live their whole life on the land. Then you go back to the plants. So this right here, this is what we think the first primitive plants would have looked like. And the evidence, we actually don't have fossils from these plants themselves. We have fossils of their spores. But we think that the earliest fossils of their spores, which show that these existed, were about 475 million years ago."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right here, this is what we think the first primitive plants would have looked like. And the evidence, we actually don't have fossils from these plants themselves. We have fossils of their spores. But we think that the earliest fossils of their spores, which show that these existed, were about 475 million years ago. So this is 400. Let me do this in another color. This right over here is 475 million years ago."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But we think that the earliest fossils of their spores, which show that these existed, were about 475 million years ago. So this is 400. Let me do this in another color. This right over here is 475 million years ago. So 1.2 billion years ago, you have the algae. 530 million years ago, we have evidence of things kind of oozing out of the ocean and maybe laying their eggs or something. 475 million years ago, we have evidence of what we would kind of call really plants."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right over here is 475 million years ago. So 1.2 billion years ago, you have the algae. 530 million years ago, we have evidence of things kind of oozing out of the ocean and maybe laying their eggs or something. 475 million years ago, we have evidence of what we would kind of call really plants. But the evidence is really the fossils of their spores. And then the first evidence of real, I guess you could call them animals, that spend their entire life on the land. The oldest fossil we have, it was discovered in Scotland fairly recently in 2004."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "475 million years ago, we have evidence of what we would kind of call really plants. But the evidence is really the fossils of their spores. And then the first evidence of real, I guess you could call them animals, that spend their entire life on the land. The oldest fossil we have, it was discovered in Scotland fairly recently in 2004. And this is the fossil right over here. It was actually discovered by a bus driver, by Mike Newman. Mike Newman, who was a bus driver in Scotland."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The oldest fossil we have, it was discovered in Scotland fairly recently in 2004. And this is the fossil right over here. It was actually discovered by a bus driver, by Mike Newman. Mike Newman, who was a bus driver in Scotland. And they actually named the thing after him. It's called Pneumodesmus pneumoni. So they got the pneumoni from Mike Newman."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Mike Newman, who was a bus driver in Scotland. And they actually named the thing after him. It's called Pneumodesmus pneumoni. So they got the pneumoni from Mike Newman. And this fossil is 428 million years old. And right now, it's the oldest fossil we have of a true land animal. So if you think about true plants versus true land animals, things that spent their entire life on the land, the plants do win out."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they got the pneumoni from Mike Newman. And this fossil is 428 million years old. And right now, it's the oldest fossil we have of a true land animal. So if you think about true plants versus true land animals, things that spent their entire life on the land, the plants do win out. If you think about things that spend part of their time on the land, then the animals probably won out. If you view algae as plants, then the plants won out. So it depends where you want to draw the line."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So if you think about true plants versus true land animals, things that spent their entire life on the land, the plants do win out. If you think about things that spend part of their time on the land, then the animals probably won out. If you view algae as plants, then the plants won out. So it depends where you want to draw the line. And this first fossil, this is of a myriapod, which just means a lot of legs. Let me write over here. Myriapod."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it depends where you want to draw the line. And this first fossil, this is of a myriapod, which just means a lot of legs. Let me write over here. Myriapod. You probably know the word myriad. Myriad means a bunch of things, or a huge amount. So myriapod, a huge amount of legs."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Myriapod. You probably know the word myriad. Myriad means a bunch of things, or a huge amount. So myriapod, a huge amount of legs. And you might be familiar with the millipedes and centipedes. Those are myriapods. And so those first primitive myriapods, 428 million years ago, and they would have lived off of plants, and maybe other myriapods, and other slugs, and whatever other animals they might have found, might have looked something like that."}, {"video_title": "First living things on land clarification Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So myriapod, a huge amount of legs. And you might be familiar with the millipedes and centipedes. Those are myriapods. And so those first primitive myriapods, 428 million years ago, and they would have lived off of plants, and maybe other myriapods, and other slugs, and whatever other animals they might have found, might have looked something like that. So hopefully that gives a little bit of clarification over. It wasn't like you had dogs sitting on the land and they had nothing to eat. It was kind of a grayer in what you define a plant or an animal and who gets the bragging rights for being the first on the land."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so I'm using the term chirality center here, but you also might hear chiral center, or stereocenter, or stereogenic center, or asymmetric center, and they're all pretty much referring to the same concept. A tetrahedral carbon, an sp3 hybridized carbon that has four different groups attached to it. And so let's look for some chirality centers in these molecules, and we'll start with this alcohol here. And so I'm gonna go ahead and redraw this. I'm just gonna go ahead and draw out all of the atoms here. So we have four carbons. The carbon on the left has three hydrogens attached to it."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so I'm gonna go ahead and redraw this. I'm just gonna go ahead and draw out all of the atoms here. So we have four carbons. The carbon on the left has three hydrogens attached to it. So there's no way that's a chirality center. I need four different groups, and I have three of the same thing on that carbon. So that one is not one."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the left has three hydrogens attached to it. So there's no way that's a chirality center. I need four different groups, and I have three of the same thing on that carbon. So that one is not one. This next carbon here, we have an OH, and then we also have a hydrogen attached to it. So that's this carbon over here. And this is a chirality center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that one is not one. This next carbon here, we have an OH, and then we also have a hydrogen attached to it. So that's this carbon over here. And this is a chirality center. We have four different groups attached to this carbon. So I'm gonna go ahead and mark this carbon right here. This is a chirality center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And this is a chirality center. We have four different groups attached to this carbon. So I'm gonna go ahead and mark this carbon right here. This is a chirality center. There's a methyl group on this side. So that's one different group. There's an OH group."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a chirality center. There's a methyl group on this side. So that's one different group. There's an OH group. There's an ethyl group, and then there's also a hydrogen. So go ahead and draw in that hydrogen. So there are four different groups attached to that carbon, so that carbon is a chirality center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "There's an OH group. There's an ethyl group, and then there's also a hydrogen. So go ahead and draw in that hydrogen. So there are four different groups attached to that carbon, so that carbon is a chirality center. So this carbon right here. It's a chirality center. Let me go ahead and draw in the other hydrogens."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there are four different groups attached to that carbon, so that carbon is a chirality center. So this carbon right here. It's a chirality center. Let me go ahead and draw in the other hydrogens. So I have two hydrogens on this carbon, so that's not a chirality center. I have two of the same thing bonded to this carbon, and then finally this carbon over here with three hydrogens. Three of the same thing, so that carbon is not a chirality center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and draw in the other hydrogens. So I have two hydrogens on this carbon, so that's not a chirality center. I have two of the same thing bonded to this carbon, and then finally this carbon over here with three hydrogens. Three of the same thing, so that carbon is not a chirality center. So one chirality center in this alcohol. Alright, for our next example, let me go ahead and draw in the carbons. So we have three carbons, and the carbon on the left has three hydrogens bonded to it, right?"}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Three of the same thing, so that carbon is not a chirality center. So one chirality center in this alcohol. Alright, for our next example, let me go ahead and draw in the carbons. So we have three carbons, and the carbon on the left has three hydrogens bonded to it, right? So that carbon is not a chirality center. Same with the carbon on the right, three hydrogens, so there's no way it could be a chirality center. Let's focus in on this carbon right here."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So we have three carbons, and the carbon on the left has three hydrogens bonded to it, right? So that carbon is not a chirality center. Same with the carbon on the right, three hydrogens, so there's no way it could be a chirality center. Let's focus in on this carbon right here. So let's think about the hybridization of that carbon. We know from earlier videos, right, that's an sp2 hybridized carbon with trigonal planar geometry. And so immediately you know that it's not a chirality center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's focus in on this carbon right here. So let's think about the hybridization of that carbon. We know from earlier videos, right, that's an sp2 hybridized carbon with trigonal planar geometry. And so immediately you know that it's not a chirality center. That has to be sp3 hybridized giving you a tetrahedral geometry. And we don't have four different things bonded to that carbon, so none of these carbons are chirality centers. So there are zero chirality centers in this molecule."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so immediately you know that it's not a chirality center. That has to be sp3 hybridized giving you a tetrahedral geometry. And we don't have four different things bonded to that carbon, so none of these carbons are chirality centers. So there are zero chirality centers in this molecule. So that's acetone. Alright, let's do a ring example. So let's do this ring example right here."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there are zero chirality centers in this molecule. So that's acetone. Alright, let's do a ring example. So let's do this ring example right here. Alright, so let me go ahead and once again draw out, draw it on the molecule this time. So we have a hydrogen here, right, and we have two hydrogens on this carbon, right, two hydrogens on this carbon all the way around our ring. So let me go ahead and draw in all of these hydrogens."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let's do this ring example right here. Alright, so let me go ahead and once again draw out, draw it on the molecule this time. So we have a hydrogen here, right, and we have two hydrogens on this carbon, right, two hydrogens on this carbon all the way around our ring. So let me go ahead and draw in all of these hydrogens. Alright, and then we look for, we look for chirality centers or chiral centers. All of these carbons, let me highlight them in magenta, all of these carbons have two hydrogens bonded to them, so that's two of the same thing. So there's no way that's going to be, those are going to be chiral centers."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and draw in all of these hydrogens. Alright, and then we look for, we look for chirality centers or chiral centers. All of these carbons, let me highlight them in magenta, all of these carbons have two hydrogens bonded to them, so that's two of the same thing. So there's no way that's going to be, those are going to be chiral centers. What about this carbon right here? So it looks like it might be a chirality center, right? We have a chlorine bonded to it, a hydrogen, and then we have these things going in opposite directions."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So there's no way that's going to be, those are going to be chiral centers. What about this carbon right here? So it looks like it might be a chirality center, right? We have a chlorine bonded to it, a hydrogen, and then we have these things going in opposite directions. But in actuality, this is not a chiral center. And that's because there's a same path around this ring here. So the hydrogen is like one different group, the chlorine is another different group, so that's two different groups."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We have a chlorine bonded to it, a hydrogen, and then we have these things going in opposite directions. But in actuality, this is not a chiral center. And that's because there's a same path around this ring here. So the hydrogen is like one different group, the chlorine is another different group, so that's two different groups. But if you go around the ring this way, and you go around the ring this way, it's the same path both ways. You hit a CH2 and you hit a CH2. You hit a CH2 and you hit a CH2, and then you hit a CH2."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the hydrogen is like one different group, the chlorine is another different group, so that's two different groups. But if you go around the ring this way, and you go around the ring this way, it's the same path both ways. You hit a CH2 and you hit a CH2. You hit a CH2 and you hit a CH2, and then you hit a CH2. So it's the same path around the ring, so it's like two of the exact same groups bonded to that carbon. So another way of thinking about this is if I go ahead and focus on that carbon at the top there again, so I have a hydrogen here and a chlorine here, and I draw in a molecule like that, so I took out this last carbon down here, so I'm leaving out the last carbon. One way to think about it is that's two of the same things attached to this carbon."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You hit a CH2 and you hit a CH2, and then you hit a CH2. So it's the same path around the ring, so it's like two of the exact same groups bonded to that carbon. So another way of thinking about this is if I go ahead and focus on that carbon at the top there again, so I have a hydrogen here and a chlorine here, and I draw in a molecule like that, so I took out this last carbon down here, so I'm leaving out the last carbon. One way to think about it is that's two of the same things attached to this carbon. Here's an ethyl group and here's an ethyl group. So two of the same thing attached to that carbon, so this is not a chiral center. And so there are zero chirality centers for this molecule."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "One way to think about it is that's two of the same things attached to this carbon. Here's an ethyl group and here's an ethyl group. So two of the same thing attached to that carbon, so this is not a chiral center. And so there are zero chirality centers for this molecule. However, if we change things up, so let's look at this molecule now, we have a different path around the ring. So once again, we're focused in on this carbon, because we know we have a hydrogen bonded to that carbon here. So the paths around the ring, there's a different path around the ring."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so there are zero chirality centers for this molecule. However, if we change things up, so let's look at this molecule now, we have a different path around the ring. So once again, we're focused in on this carbon, because we know we have a hydrogen bonded to that carbon here. So the paths around the ring, there's a different path around the ring. Let me draw in the hydrogens once again. So there are two hydrogens on this carbon, only one hydrogen on this carbon. And so if you go to the left around the ring, you hit a CH2 right here."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So the paths around the ring, there's a different path around the ring. Let me draw in the hydrogens once again. So there are two hydrogens on this carbon, only one hydrogen on this carbon. And so if you go to the left around the ring, you hit a CH2 right here. If you go to the right around the ring, you hit a carbon bonded to only one hydrogen. And so it's a different path. And so it's like there are four different things attached to this carbon right here."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so if you go to the left around the ring, you hit a CH2 right here. If you go to the right around the ring, you hit a carbon bonded to only one hydrogen. And so it's a different path. And so it's like there are four different things attached to this carbon right here. So this carbon is a chirality center. If we look at some of these other carbons, let me use red for this. So if we look at this carbon right here, this carbon is sp2 hybridized."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so it's like there are four different things attached to this carbon right here. So this carbon is a chirality center. If we look at some of these other carbons, let me use red for this. So if we look at this carbon right here, this carbon is sp2 hybridized. There's a double bond there. So it's not a chirality center. Same with this carbon."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So if we look at this carbon right here, this carbon is sp2 hybridized. There's a double bond there. So it's not a chirality center. Same with this carbon. This carbon has two hydrogens bonded to it. This carbon has two hydrogens bonded to it. This carbon has two hydrogens bonded to it."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Same with this carbon. This carbon has two hydrogens bonded to it. This carbon has two hydrogens bonded to it. This carbon has two hydrogens bonded to it. And so there's only one chirality center in this molecule. All right, and let's finally do one more example, which just looks a little bit more challenging than the ones we've been doing. So it's a little bit scarier looking."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This carbon has two hydrogens bonded to it. And so there's only one chirality center in this molecule. All right, and let's finally do one more example, which just looks a little bit more challenging than the ones we've been doing. So it's a little bit scarier looking. This is the ibuprofen molecule. So let's go through this one by one. But let's start with this benzene ring here in the center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So it's a little bit scarier looking. This is the ibuprofen molecule. So let's go through this one by one. But let's start with this benzene ring here in the center. So let's just start with this carbon right here. So this carbon is sp2 hybridized. And as you go around the benzene ring, all of these carbons are sp2 hybridized."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "But let's start with this benzene ring here in the center. So let's just start with this carbon right here. So this carbon is sp2 hybridized. And as you go around the benzene ring, all of these carbons are sp2 hybridized. So that immediately means that these can't be chiral centers. You'd have tetrahedral geometry to be a chirality center. And so those carbons are all out."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And as you go around the benzene ring, all of these carbons are sp2 hybridized. So that immediately means that these can't be chiral centers. You'd have tetrahedral geometry to be a chirality center. And so those carbons are all out. Let's look at this carbon right here. I know that there are two hydrogens bonded to that carbon. So that carbon is out."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so those carbons are all out. Let's look at this carbon right here. I know that there are two hydrogens bonded to that carbon. So that carbon is out. Let's look at this carbon. Well, I know that this carbon is bonded to two methyl groups, which is two of the same thing. And so that carbon is out."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that carbon is out. Let's look at this carbon. Well, I know that this carbon is bonded to two methyl groups, which is two of the same thing. And so that carbon is out. Let's look at this carbon. I know there are three hydrogens bonded to that carbon. So that one's out."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "And so that carbon is out. Let's look at this carbon. I know there are three hydrogens bonded to that carbon. So that one's out. Same with this carbon. So these three are the same thing. So none of these carbons over here on the left side of the molecule are chirality centers."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that one's out. Same with this carbon. So these three are the same thing. So none of these carbons over here on the left side of the molecule are chirality centers. Let me use a different color for this carbon right here. So I know that there's only one hydrogen bonded to this carbon. I'll just put it in right here."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So none of these carbons over here on the left side of the molecule are chirality centers. Let me use a different color for this carbon right here. So I know that there's only one hydrogen bonded to this carbon. I'll just put it in right here. So let's think about the different groups. I have a methyl group right here. I have a hydrogen."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll just put it in right here. So let's think about the different groups. I have a methyl group right here. I have a hydrogen. I have this carboxylic acid over here. And then I have my benzene ring over here and the rest of the molecule. So that's four different groups attached to this carbon."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I have a hydrogen. I have this carboxylic acid over here. And then I have my benzene ring over here and the rest of the molecule. So that's four different groups attached to this carbon. And so this carbon that I've marked in blue here is a chirality center. So this is a chiral center. I'll go ahead and mark it."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's four different groups attached to this carbon. And so this carbon that I've marked in blue here is a chirality center. So this is a chiral center. I'll go ahead and mark it. Let's focus in on this carbon on the methyl group. There are three hydrogens. So that's not a chiral center."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll go ahead and mark it. Let's focus in on this carbon on the methyl group. There are three hydrogens. So that's not a chiral center. This carbon right here has a double bond to it. So it's sp2 hybridized. So that cannot be a chirality center either."}, {"video_title": "Identifying chirality centers Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "So that's not a chiral center. This carbon right here has a double bond to it. So it's sp2 hybridized. So that cannot be a chirality center either. And so we have only one chiral center in this molecule. And so this is a very important skill to practice, to look at the carbons, think about what's bonded to them. And if there are four different groups bonded to that carbon and it's a tetrahedral carbon sp3 hybridized, then we can call it a chirality center or a chiral center or whatever term your professor wants to use."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the benzene ring is going to function as a nucleophile. So all of the pi electrons in the benzene ring are negatively charged, of course, and they're attracted to positively charged things, like an electrophile. And so the benzene ring is going to function as a nucleophile. And some of these pi electrons are going to form a bond with our electrophile. So we get nucleophile attacking the electrophile, and the electrophile is going to add on to our ring. And I'm just going to say the electrophile adds on to this carbon right here, which means that this carbon has a plus 1 formal charge. And that will represent our positively charged sigma complex."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And some of these pi electrons are going to form a bond with our electrophile. So we get nucleophile attacking the electrophile, and the electrophile is going to add on to our ring. And I'm just going to say the electrophile adds on to this carbon right here, which means that this carbon has a plus 1 formal charge. And that will represent our positively charged sigma complex. So if you could imagine a substituent on our benzene ring that somehow increased the electron density in that ring, that would make that benzene ring even more nucleophilic. And that increased electron density would help to stabilize the positively charged sigma complex, which means the sigma complex is more likely to form. And so a substituent that increases electron density, we could call that an electron donating group."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And that will represent our positively charged sigma complex. So if you could imagine a substituent on our benzene ring that somehow increased the electron density in that ring, that would make that benzene ring even more nucleophilic. And that increased electron density would help to stabilize the positively charged sigma complex, which means the sigma complex is more likely to form. And so a substituent that increases electron density, we could call that an electron donating group. And an electron donating group would activate the ring towards electrophilic aromatic substitution, which means that the overall rate of the reaction would be faster than that compared to benzene. And so we call an electron donating group an activator. So let me go ahead and write an activator here."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so a substituent that increases electron density, we could call that an electron donating group. And an electron donating group would activate the ring towards electrophilic aromatic substitution, which means that the overall rate of the reaction would be faster than that compared to benzene. And so we call an electron donating group an activator. So let me go ahead and write an activator here. If we thought about the opposite situation, if we thought about a substituent on the ring that overall decreased the electron density in that ring, the ring would not be as nucleophilic. And you could think about the ring as being a little bit more positively charged. And so that would, of course, destabilize the positively charged sigma complex."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So let me go ahead and write an activator here. If we thought about the opposite situation, if we thought about a substituent on the ring that overall decreased the electron density in that ring, the ring would not be as nucleophilic. And you could think about the ring as being a little bit more positively charged. And so that would, of course, destabilize the positively charged sigma complex. And so a substituent that overall decreased the electron density in the ring, we could say it's an electron withdrawing group because it's withdrawing electron density from the ring. And that would, of course, deactivate the ring towards electrophilic aromatic substitution. And so we call this a deactivator."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so that would, of course, destabilize the positively charged sigma complex. And so a substituent that overall decreased the electron density in the ring, we could say it's an electron withdrawing group because it's withdrawing electron density from the ring. And that would, of course, deactivate the ring towards electrophilic aromatic substitution. And so we call this a deactivator. And so the reaction would be slower than that of benzene by itself. Let's see how the concepts of electron donating groups and electron withdrawing groups affect activating strength. And we'll start with strong activators here."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we call this a deactivator. And so the reaction would be slower than that of benzene by itself. Let's see how the concepts of electron donating groups and electron withdrawing groups affect activating strength. And we'll start with strong activators here. So first, we can look at the phenol molecule. And we can think about this carbon on our ring in a sigma bond to this oxygen right here. We know oxygen is more electronegative than carbon."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we'll start with strong activators here. So first, we can look at the phenol molecule. And we can think about this carbon on our ring in a sigma bond to this oxygen right here. We know oxygen is more electronegative than carbon. And so the oxygen can withdraw some electron density from the ring by an inductive effect. So oxygen being more electronegative, it can pull the electrons through that sigma bond closer to itself. And so it's withdrawing some electron density from the ring because of electronegativity."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "We know oxygen is more electronegative than carbon. And so the oxygen can withdraw some electron density from the ring by an inductive effect. So oxygen being more electronegative, it can pull the electrons through that sigma bond closer to itself. And so it's withdrawing some electron density from the ring because of electronegativity. And so we call this induction or the inductive effect. So there's some induction in this molecule. Now, since it's withdrawing some electron density, you might expect the OH group to be a deactivator."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it's withdrawing some electron density from the ring because of electronegativity. And so we call this induction or the inductive effect. So there's some induction in this molecule. Now, since it's withdrawing some electron density, you might expect the OH group to be a deactivator. But that's not what we observe. We observe the OH to be a strong activator. And so there must be another effect here to counteract this inductive effect."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Now, since it's withdrawing some electron density, you might expect the OH group to be a deactivator. But that's not what we observe. We observe the OH to be a strong activator. And so there must be another effect here to counteract this inductive effect. And of course, that effect is resonance. So let me go ahead and write resonance right over here. So we can draw a resonance."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so there must be another effect here to counteract this inductive effect. And of course, that effect is resonance. So let me go ahead and write resonance right over here. So we can draw a resonance. We can draw several resonance structures for the phenol molecule. But if you think about this lone pair of electrons on this oxygen, it's right next to our benzene ring. And so this lone pair of electrons can participate in resonance."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we can draw a resonance. We can draw several resonance structures for the phenol molecule. But if you think about this lone pair of electrons on this oxygen, it's right next to our benzene ring. And so this lone pair of electrons can participate in resonance. And move in here to form a pi bond, which would push these electrons in here off onto this carbon. So we can go ahead and draw a pi bond between our oxygen and our carbon now. There's still a lone pair of electrons on that oxygen."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so this lone pair of electrons can participate in resonance. And move in here to form a pi bond, which would push these electrons in here off onto this carbon. So we can go ahead and draw a pi bond between our oxygen and our carbon now. There's still a lone pair of electrons on that oxygen. It's still bonded to a hydrogen. So it has a plus 1 formal charge. And so we have these pi electrons here."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "There's still a lone pair of electrons on that oxygen. It's still bonded to a hydrogen. So it has a plus 1 formal charge. And so we have these pi electrons here. And we have these electrons move out onto this carbon, which gives that carbon a negative 1 formal charge. And we could keep drawing more resonance structures. But I'm going to stop there."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so we have these pi electrons here. And we have these electrons move out onto this carbon, which gives that carbon a negative 1 formal charge. And we could keep drawing more resonance structures. But I'm going to stop there. Because the point I'm trying to make is that we get some donation of electron density to the ring through a pi bond. So let me go ahead and highlight our pi bond here. So these electrons move in here to form a pi bond between that carbon and that oxygen."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "But I'm going to stop there. Because the point I'm trying to make is that we get some donation of electron density to the ring through a pi bond. So let me go ahead and highlight our pi bond here. So these electrons move in here to form a pi bond between that carbon and that oxygen. And so we get overlap of p orbitals between this carbon and this oxygen. So let me go ahead and sketch that really fast. So we have a carbon."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons move in here to form a pi bond between that carbon and that oxygen. And so we get overlap of p orbitals between this carbon and this oxygen. So let me go ahead and sketch that really fast. So we have a carbon. And we have an oxygen. And we're going to get overlap of p orbitals. Since carbon and oxygen are in the same period on the periodic table, their p orbitals are pretty much the same size."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So we have a carbon. And we have an oxygen. And we're going to get overlap of p orbitals. Since carbon and oxygen are in the same period on the periodic table, their p orbitals are pretty much the same size. And that means that you get good overlap. And therefore, the oxygen is able to donate some electron density to that ring there. So the lone pair of electrons on the oxygen is conjugated into the pi system of that ring."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "Since carbon and oxygen are in the same period on the periodic table, their p orbitals are pretty much the same size. And that means that you get good overlap. And therefore, the oxygen is able to donate some electron density to that ring there. So the lone pair of electrons on the oxygen is conjugated into the pi system of that ring. And so that's overall an electron donating effect. You're increasing the electron density in the ring. And so the resonance effect says that the OH group is an electron donating group, which would, of course, make it a strong activator."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So the lone pair of electrons on the oxygen is conjugated into the pi system of that ring. And so that's overall an electron donating effect. You're increasing the electron density in the ring. And so the resonance effect says that the OH group is an electron donating group, which would, of course, make it a strong activator. And that's, of course, what we observe experimentally. And so we can say that the resonance effect beats the inductive effect when you're talking about a strong activator here. So an atom that has a lone pair of electrons next to your benzene ring."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so the resonance effect says that the OH group is an electron donating group, which would, of course, make it a strong activator. And that's, of course, what we observe experimentally. And so we can say that the resonance effect beats the inductive effect when you're talking about a strong activator here. So an atom that has a lone pair of electrons next to your benzene ring. Now, the same idea holds true for aniline down here. So once again, thinking about nitrogen compared to this carbon, nitrogen is more electronegative. And so it's going to withdraw some electron density from the ring via the inductive effect through that sigma bond."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So an atom that has a lone pair of electrons next to your benzene ring. Now, the same idea holds true for aniline down here. So once again, thinking about nitrogen compared to this carbon, nitrogen is more electronegative. And so it's going to withdraw some electron density from the ring via the inductive effect through that sigma bond. So I can think about drawing an arrow showing the movement of electrons towards nitrogen. Now, nitrogen's not as electronegative as oxygen there. So it's not really withdrawing quite as much electron density via induction."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so it's going to withdraw some electron density from the ring via the inductive effect through that sigma bond. So I can think about drawing an arrow showing the movement of electrons towards nitrogen. Now, nitrogen's not as electronegative as oxygen there. So it's not really withdrawing quite as much electron density via induction. Since that nitrogen has a lone pair of electrons on it, it can also participate in resonance. And so just like the previous example, I could take this lone pair of electrons right next to the ring, move it into here to form a pi bond to push these electrons in here off onto that carbon. So I could go ahead and show one possible resonance structure here."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So it's not really withdrawing quite as much electron density via induction. Since that nitrogen has a lone pair of electrons on it, it can also participate in resonance. And so just like the previous example, I could take this lone pair of electrons right next to the ring, move it into here to form a pi bond to push these electrons in here off onto that carbon. So I could go ahead and show one possible resonance structure here. So I could show my pi bond between my carbon and my nitrogen. The nitrogen would now be positively charged. And we would have a lone pair of electrons on this carbon, which would make that carbon negatively charged."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "So I could go ahead and show one possible resonance structure here. So I could show my pi bond between my carbon and my nitrogen. The nitrogen would now be positively charged. And we would have a lone pair of electrons on this carbon, which would make that carbon negatively charged. Once again, you could draw more resonance structures. But we just don't have the time for that in this video. And so once again, this lone pair of electrons is actually conjugated into the pi system of the ring right there."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we would have a lone pair of electrons on this carbon, which would make that carbon negatively charged. Once again, you could draw more resonance structures. But we just don't have the time for that in this video. And so once again, this lone pair of electrons is actually conjugated into the pi system of the ring right there. And so that's increasing the electron density of the ring, which makes the ring more nucleophilic, stabilizes the positively charged sigma complex. And therefore, it's overall an electron donating group, which makes it an activator. And we could also think about the p orbitals of this carbon and this nitrogen."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so once again, this lone pair of electrons is actually conjugated into the pi system of the ring right there. And so that's increasing the electron density of the ring, which makes the ring more nucleophilic, stabilizes the positively charged sigma complex. And therefore, it's overall an electron donating group, which makes it an activator. And we could also think about the p orbitals of this carbon and this nitrogen. So they are also, of course, in the same period on the periodic table. So when you sketch in your p orbitals here, you can make them pretty much the same size. And so you get good overlap of those p orbitals."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And we could also think about the p orbitals of this carbon and this nitrogen. So they are also, of course, in the same period on the periodic table. So when you sketch in your p orbitals here, you can make them pretty much the same size. And so you get good overlap of those p orbitals. Now, aniline is actually even more reactive than the phenol example. And that's because of the nitrogen being less electronegative than oxygen. And so in the pi bond here, since this nitrogen, I'm going to go ahead and highlight our pi electrons."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so you get good overlap of those p orbitals. Now, aniline is actually even more reactive than the phenol example. And that's because of the nitrogen being less electronegative than oxygen. And so in the pi bond here, since this nitrogen, I'm going to go ahead and highlight our pi electrons. Let's use green this time. So these pi electrons right here, since nitrogen is less electronegative than oxygen, that means that those pi electrons are better able to be conjugated into the pi system of the ring, since nitrogen isn't pulling on them as much as the oxygen is. And so since you get a little bit more electron density donated to your ring because of the nitrogen being less electronegative, that means increased electron density makes this a better electron donating group, which activates the ring better towards electrophilic aromatic substitution."}, {"video_title": "Ortho-para directors II Aromatic Compounds Organic chemistry Khan Academy.mp3", "Sentence": "And so in the pi bond here, since this nitrogen, I'm going to go ahead and highlight our pi electrons. Let's use green this time. So these pi electrons right here, since nitrogen is less electronegative than oxygen, that means that those pi electrons are better able to be conjugated into the pi system of the ring, since nitrogen isn't pulling on them as much as the oxygen is. And so since you get a little bit more electron density donated to your ring because of the nitrogen being less electronegative, that means increased electron density makes this a better electron donating group, which activates the ring better towards electrophilic aromatic substitution. And so in the next video, we're going to use these same concepts of induction and resonance. And we're going to analyze a moderate activator, a weak activator. And we'll also look at an example of a weak deactivator that is still an ortho para director."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And this name is just a byproduct of the first observations of quasars. Because all they look like were these kind of point-like sources of electromagnetic radiation, mainly in the radio part of the spectrum. So that's why we call them quasi-stellar radio sources. Now it turns out that they are neither stars or even quasi-stellar. And their main energy isn't even being released in the radio band of the electromagnetic spectrum. They're far more energetic than that. What they really are are the active nucleuses of galaxies."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now it turns out that they are neither stars or even quasi-stellar. And their main energy isn't even being released in the radio band of the electromagnetic spectrum. They're far more energetic than that. What they really are are the active nucleuses of galaxies. So let's think about that a little bit. So if we have a supermassive black hole at the center of a galaxy, so let me draw that right over here. So that's our supermassive black hole."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What they really are are the active nucleuses of galaxies. So let's think about that a little bit. So if we have a supermassive black hole at the center of a galaxy, so let me draw that right over here. So that's our supermassive black hole. Maybe that's the surface of the event horizon of the supermassive black hole. The actual mass of the black hole is in the center of that event horizon. If there's material that's passing by this black hole, it's going to get attracted to it and it's going to form an accretion disk around it."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So that's our supermassive black hole. Maybe that's the surface of the event horizon of the supermassive black hole. The actual mass of the black hole is in the center of that event horizon. If there's material that's passing by this black hole, it's going to get attracted to it and it's going to form an accretion disk around it. This material is going to start rotating around this black hole and some of it, if it doesn't have enough velocity, is going to actually fall into the black hole. So you have all of this material going around the black hole and some of it, if it doesn't have enough angular velocity, not enough to orbit around the black hole, it's actually going to fall in. Now while things, let me label this."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If there's material that's passing by this black hole, it's going to get attracted to it and it's going to form an accretion disk around it. This material is going to start rotating around this black hole and some of it, if it doesn't have enough velocity, is going to actually fall into the black hole. So you have all of this material going around the black hole and some of it, if it doesn't have enough angular velocity, not enough to orbit around the black hole, it's actually going to fall in. Now while things, let me label this. This is the accretion disk. So as things are getting faster and faster, as they fall closer and closer to this black hole and bumping into each other more and more, that gravitational potential energy from things falling into it is being turned into actual energy, actual temperature. And so what you have is things start to get really unbelievably hot near the surface."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now while things, let me label this. This is the accretion disk. So as things are getting faster and faster, as they fall closer and closer to this black hole and bumping into each other more and more, that gravitational potential energy from things falling into it is being turned into actual energy, actual temperature. And so what you have is things start to get really unbelievably hot near the surface. They get hotter and hotter as they fall closer and closer to that event horizon. And so near the event horizon itself, things are so intense that they're actually releasing electromagnetic radiation, high frequency electromagnetic radiation, mainly in the x-ray part of the spectrum. Now I want to be very clear."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so what you have is things start to get really unbelievably hot near the surface. They get hotter and hotter as they fall closer and closer to that event horizon. And so near the event horizon itself, things are so intense that they're actually releasing electromagnetic radiation, high frequency electromagnetic radiation, mainly in the x-ray part of the spectrum. Now I want to be very clear. So there's two things here. One is when you learn about quasars, or when I first was exposed to quasars in like a NOVA special, they make you think that the quasar, that the radiation, is somehow being released by the black hole itself. And I would scratch my head because I was just told that nothing can escape the event horizon of a black hole, including electromagnetic radiation."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now I want to be very clear. So there's two things here. One is when you learn about quasars, or when I first was exposed to quasars in like a NOVA special, they make you think that the quasar, that the radiation, is somehow being released by the black hole itself. And I would scratch my head because I was just told that nothing can escape the event horizon of a black hole, including electromagnetic radiation. So how could that be being emitted by the black hole? And the answer is it's not being emitted by the black hole, it's being emitted by the matter in the accretion disk that hasn't quite gotten to the event horizon yet. Once it's inside of the event horizon, any electromagnetic radiation that it might emit will not be able to escape the black hole anymore, will not be able to escape the actual event horizon."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I would scratch my head because I was just told that nothing can escape the event horizon of a black hole, including electromagnetic radiation. So how could that be being emitted by the black hole? And the answer is it's not being emitted by the black hole, it's being emitted by the matter in the accretion disk that hasn't quite gotten to the event horizon yet. Once it's inside of the event horizon, any electromagnetic radiation that it might emit will not be able to escape the black hole anymore, will not be able to escape the actual event horizon. So all of this is from the accretion disk around the supermassive black hole. And the other question that used to pop in my mind is why does it kind of come out at these kind of perpendicular, orthogonal, to the plane of the actual accretion disk? And they're just, at least my logic tells me, well things aren't going to pop out along the direction of the accretion disk because then they're going to be absorbed by other things."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Once it's inside of the event horizon, any electromagnetic radiation that it might emit will not be able to escape the black hole anymore, will not be able to escape the actual event horizon. So all of this is from the accretion disk around the supermassive black hole. And the other question that used to pop in my mind is why does it kind of come out at these kind of perpendicular, orthogonal, to the plane of the actual accretion disk? And they're just, at least my logic tells me, well things aren't going to pop out along the direction of the accretion disk because then they're going to be absorbed by other things. In fact, that's what's going to cause other things to get heated up. Closer to the actual event horizon. So any energy that's going out in that direction is just going to be absorbed and make other things hotter."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And they're just, at least my logic tells me, well things aren't going to pop out along the direction of the accretion disk because then they're going to be absorbed by other things. In fact, that's what's going to cause other things to get heated up. Closer to the actual event horizon. So any energy that's going out in that direction is just going to be absorbed and make other things hotter. And only when you go roughly perpendicular to the plane of the accretion disk is that energy allowed to kind of go and transmit freely into space. Now I want to be very clear. Quasars are the most luminous things that we know of in the universe."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So any energy that's going out in that direction is just going to be absorbed and make other things hotter. And only when you go roughly perpendicular to the plane of the accretion disk is that energy allowed to kind of go and transmit freely into space. Now I want to be very clear. Quasars are the most luminous things that we know of in the universe. So most luminous things that we know of in the universe. The brightest, or actually many quasars, are on the order of a trillion suns in luminosity. So they can be brighter than an entire galaxy."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Quasars are the most luminous things that we know of in the universe. So most luminous things that we know of in the universe. The brightest, or actually many quasars, are on the order of a trillion suns in luminosity. So they can be brighter than an entire galaxy. And that's just coming from material around a fairly small region of space. Much smaller than an actual galaxy. It's the very center."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they can be brighter than an entire galaxy. And that's just coming from material around a fairly small region of space. Much smaller than an actual galaxy. It's the very center. It's kind of just the galactic core. Now another interesting thing about quasars, and this kind of gives credence to this notion of a constantly changing universe, and even to some degree the Big Bang itself, is you have these supermassive black holes that may be formed shortly after the Big Bang. Now you can imagine, at an early stage in the universe's development, there would have been a lot of mass that would have been near these black holes, that didn't have quite the velocities to be able to escape them or be able to orbit around them."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's the very center. It's kind of just the galactic core. Now another interesting thing about quasars, and this kind of gives credence to this notion of a constantly changing universe, and even to some degree the Big Bang itself, is you have these supermassive black holes that may be formed shortly after the Big Bang. Now you can imagine, at an early stage in the universe's development, there would have been a lot of mass that would have been near these black holes, that didn't have quite the velocities to be able to escape them or be able to orbit around them. And so these would actually start falling into the black hole, and then over time, all of the mass that had to fall into the black hole, into the supermassive black hole, will have fallen into the supermassive black hole. And if you imagine some future period of time, you should still have the supermassive black hole, but all you should see is mostly things orbiting around it. Anything that had to fall into it would have already fallen into it."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now you can imagine, at an early stage in the universe's development, there would have been a lot of mass that would have been near these black holes, that didn't have quite the velocities to be able to escape them or be able to orbit around them. And so these would actually start falling into the black hole, and then over time, all of the mass that had to fall into the black hole, into the supermassive black hole, will have fallen into the supermassive black hole. And if you imagine some future period of time, you should still have the supermassive black hole, but all you should see is mostly things orbiting around it. Anything that had to fall into it would have already fallen into it. So you're just going to see things orbiting around it. And this is actually what we see. If we look around us, we look at our Milky Way galaxy, we don't observe a lot of things falling in."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Anything that had to fall into it would have already fallen into it. So you're just going to see things orbiting around it. And this is actually what we see. If we look around us, we look at our Milky Way galaxy, we don't observe a lot of things falling in. For example, the Milky Way galaxy does not have an active nucleus, an active core. It is not currently a quasar, the center of the Milky Way galaxy. The supermassive black hole there is not, I guess we could say, digesting or consuming material."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "If we look around us, we look at our Milky Way galaxy, we don't observe a lot of things falling in. For example, the Milky Way galaxy does not have an active nucleus, an active core. It is not currently a quasar, the center of the Milky Way galaxy. The supermassive black hole there is not, I guess we could say, digesting or consuming material. But you could imagine, at some point in the Milky Way's past, there might have been a lot of material that didn't have quite the velocity to be able to orbit. And so that was consumed. And as it was consumed, it would emit all of this X-ray radiation and could be observed as a quasar."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The supermassive black hole there is not, I guess we could say, digesting or consuming material. But you could imagine, at some point in the Milky Way's past, there might have been a lot of material that didn't have quite the velocity to be able to orbit. And so that was consumed. And as it was consumed, it would emit all of this X-ray radiation and could be observed as a quasar. And that's actually what we observe. The closest quasars, and we've observed more than 200,000 quasars. The closest quasars are on the order of 780 million light years away."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And as it was consumed, it would emit all of this X-ray radiation and could be observed as a quasar. And that's actually what we observe. The closest quasars, and we've observed more than 200,000 quasars. The closest quasars are on the order of 780 million light years away. So what does that mean? We don't observe quasars closer than 700 million light years. So what that tells us is, at least in our region of the universe, the most recent quasars were 780 million years in the past."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The closest quasars are on the order of 780 million light years away. So what does that mean? We don't observe quasars closer than 700 million light years. So what that tells us is, at least in our region of the universe, the most recent quasars were 780 million years in the past. When we look at closer parts of the universe, let me draw, let's say this is the observable universe, this is us, so we only start to observe quasars at a certain distance away from us. And that distance is actually also a certain time in the past, because it took the light 780 million years to get to us. And actually, most of the quasars are more than 3 billion light years away, which tells us that they only existed more than 3 billion years in the past, at a younger stage of the actual universe, when there was actual material for these supermassive black holes to consume at the center of galaxies."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So what that tells us is, at least in our region of the universe, the most recent quasars were 780 million years in the past. When we look at closer parts of the universe, let me draw, let's say this is the observable universe, this is us, so we only start to observe quasars at a certain distance away from us. And that distance is actually also a certain time in the past, because it took the light 780 million years to get to us. And actually, most of the quasars are more than 3 billion light years away, which tells us that they only existed more than 3 billion years in the past, at a younger stage of the actual universe, when there was actual material for these supermassive black holes to consume at the center of galaxies. You move closer in time to us, and most of that material has actually been consumed. And we just have material orbiting around these supermassive black holes, which we call galaxies. And so we don't observe quasars anymore."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And actually, most of the quasars are more than 3 billion light years away, which tells us that they only existed more than 3 billion years in the past, at a younger stage of the actual universe, when there was actual material for these supermassive black holes to consume at the center of galaxies. You move closer in time to us, and most of that material has actually been consumed. And we just have material orbiting around these supermassive black holes, which we call galaxies. And so we don't observe quasars anymore. And just to give an idea, I mean, everything we learn in cosmology, there's kind of these mind-bending concepts, unbelievable distances, unbelievable masses, unbelievable brightnesses, I guess you could think about it, but just to give a sense, the brightest known quasars devour on the order of 1,000 solar masses per year. So that's on the order of 10 Earths per second, if I did my math right. 10 Earths per second are being devoured by the brightest quasars."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so we don't observe quasars anymore. And just to give an idea, I mean, everything we learn in cosmology, there's kind of these mind-bending concepts, unbelievable distances, unbelievable masses, unbelievable brightnesses, I guess you could think about it, but just to give a sense, the brightest known quasars devour on the order of 1,000 solar masses per year. So that's on the order of 10 Earths per second, if I did my math right. 10 Earths per second are being devoured by the brightest quasars. And it's that energy of that mass that's accreting around it that's generating all of that energy. And actually, I should say, I shouldn't even talk about it in the present tense. These brightest quasars, this happened in the past."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "10 Earths per second are being devoured by the brightest quasars. And it's that energy of that mass that's accreting around it that's generating all of that energy. And actually, I should say, I shouldn't even talk about it in the present tense. These brightest quasars, this happened in the past. We're just observing it now. As far as all we know, the rest of the universe looks fairly similar to the way our universe does. And so there really aren't that many quasars around."}, {"video_title": "Quasars Stars, black holes and galaxies Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These brightest quasars, this happened in the past. We're just observing it now. As far as all we know, the rest of the universe looks fairly similar to the way our universe does. And so there really aren't that many quasars around. Although, the other side of the coin might be, even though most of the material has already been consumed, maybe even by our own supermassive black hole in the center of the Milky Way, at some point in the future, maybe it will be able to consume on some more stellar material, some more, well, any type of material in the future. And that might happen about 4 or 5 billion years in the future when we actually collide with the Andromeda galaxy. So anyway, hopefully that gave you some food for thought."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "And I think it's pretty obvious from this picture that we have a very long carbon cycle here that we can start, or we can just look at this. We have one, two, three, four, five, six, seven, eight carbons in a cycle right over here. And so we can call this backbone. Which is the backbone of our molecule right over here. This is cyclooctane. Cyclooctane. It's an octane in a cycle."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "Which is the backbone of our molecule right over here. This is cyclooctane. Cyclooctane. It's an octane in a cycle. It's an alkane because it's all single bonds right over here. So that is cyclooctane. Now let's think about the different groups that are attached to this backbone right over here."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "It's an octane in a cycle. It's an alkane because it's all single bonds right over here. So that is cyclooctane. Now let's think about the different groups that are attached to this backbone right over here. So first we can look at this one right over here. And we could ask ourselves, well how many carbons are there? How many carbons?"}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "Now let's think about the different groups that are attached to this backbone right over here. So first we can look at this one right over here. And we could ask ourselves, well how many carbons are there? How many carbons? Well there's one, two, three, four carbons that are attached to the backbones. So our prefix for when we're dealing with four carbons is bute, and it is in a cycle. So this is a cyclobutyl group."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "How many carbons? Well there's one, two, three, four carbons that are attached to the backbones. So our prefix for when we're dealing with four carbons is bute, and it is in a cycle. So this is a cyclobutyl group. Cyclo, cyclobutyl, butyl. And we also have another cyclobutyl right over here. So this is another cyclobutyl."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So this is a cyclobutyl group. Cyclo, cyclobutyl, butyl. And we also have another cyclobutyl right over here. So this is another cyclobutyl. Cyclobutyl, I'm spelling it, cyclobutyl group. Now let's think about this one right over here. How many carbons are there?"}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So this is another cyclobutyl. Cyclobutyl, I'm spelling it, cyclobutyl group. Now let's think about this one right over here. How many carbons are there? We have one, two, three carbons. So our prefix when we're dealing with three carbons is prop, and they're in a cycle as well. So this is a cyclopropyl group."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "How many carbons are there? We have one, two, three carbons. So our prefix when we're dealing with three carbons is prop, and they're in a cycle as well. So this is a cyclopropyl group. Cyclopropyl, this is a cyclopropyl group. And now we can jump over to do these two groups right over here. We see that these are clearly the same groups."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So this is a cyclopropyl group. Cyclopropyl, this is a cyclopropyl group. And now we can jump over to do these two groups right over here. We see that these are clearly the same groups. Let's count the number of carbons they have. We have one, two, three, four carbons. So if we're dealing with four carbons, we would think about the prefix bute, but this isn't in a cycle, and it's not in just a straight chain."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "We see that these are clearly the same groups. Let's count the number of carbons they have. We have one, two, three, four carbons. So if we're dealing with four carbons, we would think about the prefix bute, but this isn't in a cycle, and it's not in just a straight chain. It has this Y shape. And so the common name, when we end up with this Y shape, when we're attached down here, and then we're attached to one carbon here, and that's attached to kind of this carbon right over here, this is then attached to these other two, this is an isobutyl group, which we've seen multiple times before. Isobutyl, isobutyl group, and of course this one is an isobutyl group as well."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So if we're dealing with four carbons, we would think about the prefix bute, but this isn't in a cycle, and it's not in just a straight chain. It has this Y shape. And so the common name, when we end up with this Y shape, when we're attached down here, and then we're attached to one carbon here, and that's attached to kind of this carbon right over here, this is then attached to these other two, this is an isobutyl group, which we've seen multiple times before. Isobutyl, isobutyl group, and of course this one is an isobutyl group as well. So these are both isobutyl groups. Now the next thing we need to think about is how we can number it. We wanna go back to our main carbon chain, which in this case is a cycle, and think about how we wanna number."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "Isobutyl, isobutyl group, and of course this one is an isobutyl group as well. So these are both isobutyl groups. Now the next thing we need to think about is how we can number it. We wanna go back to our main carbon chain, which in this case is a cycle, and think about how we wanna number. And the convention is, is to start numbering at the thing that is first in alphabetical order. And when we consider alphabetical order, we think even the C in cyclo, and we think of the I in isobutyl. If this was tert or sec, you wouldn't think about the T or the S in secbutyl or tertbutyl, but for isobutyl you do consider the I, for the cyclos you consider the C. And so when you look at it that way, all of the Cs come before the Is, and then amongst the Cs, these all have cyclos, but the cyclobutyl comes before cyclopropyl, or cyclooctane, B comes before P or O."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "We wanna go back to our main carbon chain, which in this case is a cycle, and think about how we wanna number. And the convention is, is to start numbering at the thing that is first in alphabetical order. And when we consider alphabetical order, we think even the C in cyclo, and we think of the I in isobutyl. If this was tert or sec, you wouldn't think about the T or the S in secbutyl or tertbutyl, but for isobutyl you do consider the I, for the cyclos you consider the C. And so when you look at it that way, all of the Cs come before the Is, and then amongst the Cs, these all have cyclos, but the cyclobutyl comes before cyclopropyl, or cyclooctane, B comes before P or O. So the thing that is first alphabetically are both of these cyclobutyls. So we can start numbering, we can start numbering at where either of them attached to the main carbon backbone. But we have to, or those are, I guess we should say, those are both options, but we want to number from the one that is going to give us the total lowest numbering."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "If this was tert or sec, you wouldn't think about the T or the S in secbutyl or tertbutyl, but for isobutyl you do consider the I, for the cyclos you consider the C. And so when you look at it that way, all of the Cs come before the Is, and then amongst the Cs, these all have cyclos, but the cyclobutyl comes before cyclopropyl, or cyclooctane, B comes before P or O. So the thing that is first alphabetically are both of these cyclobutyls. So we can start numbering, we can start numbering at where either of them attached to the main carbon backbone. But we have to, or those are, I guess we should say, those are both options, but we want to number from the one that is going to give us the total lowest numbering. So let's think about that a little bit. So we could say, let's number from here, let's make this the one carbon, and so that we hit something else as soon as possible, because remember, we're trying to minimize the numbers here, we might want to go clockwise, so we hit a two right over here, three, four, five, six, seven. So in this situation, where do we hit interesting things?"}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "But we have to, or those are, I guess we should say, those are both options, but we want to number from the one that is going to give us the total lowest numbering. So let's think about that a little bit. So we could say, let's number from here, let's make this the one carbon, and so that we hit something else as soon as possible, because remember, we're trying to minimize the numbers here, we might want to go clockwise, so we hit a two right over here, three, four, five, six, seven. So in this situation, where do we hit interesting things? We hit interesting things at the one, at the two, or where do we have groups attached to the chain? One, two, four, four, six, and seven. So that's one option, we'll have things, groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So in this situation, where do we hit interesting things? We hit interesting things at the one, at the two, or where do we have groups attached to the chain? One, two, four, four, six, and seven. So that's one option, we'll have things, groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise. Our other option, one, two, four, six, seven, our other option is to start, and let me erase those, our other option is to start at this other cyclobutyl, at this other cyclobutyl group here on the left, let me erase these so I don't mess up the diagram too much. The other option is to start here, make this the one, number this as one, and then we would also go clockwise to hit those cyclopropyl as soon as quickly, is to hit something else as soon as possible. So two, three, four, five, six, seven."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So that's one option, we'll have things, groups attached at one, two, four, six, and seven if we start right over there, and if we were to go clockwise. Our other option, one, two, four, six, seven, our other option is to start, and let me erase those, our other option is to start at this other cyclobutyl, at this other cyclobutyl group here on the left, let me erase these so I don't mess up the diagram too much. The other option is to start here, make this the one, number this as one, and then we would also go clockwise to hit those cyclopropyl as soon as quickly, is to hit something else as soon as possible. So two, three, four, five, six, seven. So now where are we hitting interesting things? We have groups at the one carbon, the two carbon, the four carbon, the five carbon, and the seven carbon. So actually, this second numbering is preferable."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So two, three, four, five, six, seven. So now where are we hitting interesting things? We have groups at the one carbon, the two carbon, the four carbon, the five carbon, and the seven carbon. So actually, this second numbering is preferable. Both of them have something at the one, the two, the four, and the seven, but the second one has something at a five, while the first one had something at a six. So we would actually want to do this numbering right over here, and this is the numbering that we now have on our, that we have now listed right over here. Let's now write what the name of this molecule actually is."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So actually, this second numbering is preferable. Both of them have something at the one, the two, the four, and the seven, but the second one has something at a five, while the first one had something at a six. So we would actually want to do this numbering right over here, and this is the numbering that we now have on our, that we have now listed right over here. Let's now write what the name of this molecule actually is. So once again, we start first in alphabetical order. So we're gonna start with the cyclobutyl groups, and we have two of them, one at the one carbon and one at the four carbon. So we could say one comma four, and since there are two cyclobutyl groups, we would say di, di-cyclo-butyl, one four di-cyclo-butyl, and then what comes next in alphabetical order is the cyclopropyl that comes before isobutyl, C comes before I."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "Let's now write what the name of this molecule actually is. So once again, we start first in alphabetical order. So we're gonna start with the cyclobutyl groups, and we have two of them, one at the one carbon and one at the four carbon. So we could say one comma four, and since there are two cyclobutyl groups, we would say di, di-cyclo-butyl, one four di-cyclo-butyl, and then what comes next in alphabetical order is the cyclopropyl that comes before isobutyl, C comes before I. That's at the two carbon. So then we can say two, two cyclopropyl, two cyclopropyl, and now we can get to the two isobutyls. So there's an isobutyl at the five carbon and at the seven carbon."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So we could say one comma four, and since there are two cyclobutyl groups, we would say di, di-cyclo-butyl, one four di-cyclo-butyl, and then what comes next in alphabetical order is the cyclopropyl that comes before isobutyl, C comes before I. That's at the two carbon. So then we can say two, two cyclopropyl, two cyclopropyl, and now we can get to the two isobutyls. So there's an isobutyl at the five carbon and at the seven carbon. So we can say five, seven, and since there's two of them, we'd say di-isobutyl, five, seven, di-isobutyl. So di-isobutyl, and then we just give the name of the backbone. We've tackled, we've done the butyls, the di-cyclobutyl right there, the cyclopropyl, the two isobutyls, and now we just write cyclooctane."}, {"video_title": "Naming a cycloalkane Organic chemistry Khan Academy.mp3", "Sentence": "So there's an isobutyl at the five carbon and at the seven carbon. So we can say five, seven, and since there's two of them, we'd say di-isobutyl, five, seven, di-isobutyl. So di-isobutyl, and then we just give the name of the backbone. We've tackled, we've done the butyls, the di-cyclobutyl right there, the cyclopropyl, the two isobutyls, and now we just write cyclooctane. And we are done. We have successfully named the molecule. And this was kind of fun."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "If we look at the boiling points of ethanol and dimethyl ether, we can see there's a large difference between them. Ethanol has a much higher boiling point, 78 degrees Celsius, whereas dimethyl ether is negative 25 degrees. And this explains the state of matter of these molecules. Ethanol, since its boiling point is higher than room temperature, is, of course, a liquid at room temperature and pressure, whereas dimethyl ether, with a much lower boiling point, has already turned into a gas. And so we can explain the states of matter by looking at the intermolecular forces that are present in these molecules. So if I think about one molecule of ethanol, I know that the bond between oxygen and hydrogen is polarized. I know that oxygen is more electronegative, so it will be partially negative."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "Ethanol, since its boiling point is higher than room temperature, is, of course, a liquid at room temperature and pressure, whereas dimethyl ether, with a much lower boiling point, has already turned into a gas. And so we can explain the states of matter by looking at the intermolecular forces that are present in these molecules. So if I think about one molecule of ethanol, I know that the bond between oxygen and hydrogen is polarized. I know that oxygen is more electronegative, so it will be partially negative. And the hydrogen is partially positive as it loses some electron density. If that molecule of ethanol interacts with another molecule of ethanol, the second molecule of ethanol is also polarized. The oxygen is partially negative, and the hydrogen is partially positive."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "I know that oxygen is more electronegative, so it will be partially negative. And the hydrogen is partially positive as it loses some electron density. If that molecule of ethanol interacts with another molecule of ethanol, the second molecule of ethanol is also polarized. The oxygen is partially negative, and the hydrogen is partially positive. And we know that opposite charges attract. So the partially positively charged hydrogen is attracted to the partially negatively charged oxygen, like that. And there's going to be attraction between those two molecules."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "The oxygen is partially negative, and the hydrogen is partially positive. And we know that opposite charges attract. So the partially positively charged hydrogen is attracted to the partially negatively charged oxygen, like that. And there's going to be attraction between those two molecules. And we call this intermolecular force hydrogen bonding, the strongest type of intermolecular force. So hydrogen bonding is present between molecules of ethanol. And this accounts for its large boiling point."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And there's going to be attraction between those two molecules. And we call this intermolecular force hydrogen bonding, the strongest type of intermolecular force. So hydrogen bonding is present between molecules of ethanol. And this accounts for its large boiling point. Let's look at more details about hydrogen bonding here. So hydrogen bonding exists when you have hydrogen bonded to an electronegative atom, like that, this oxygen. But students forget that you also need another electronegative atom over here to give you more of a difference in charge and to make that hydrogen more partially positive."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And this accounts for its large boiling point. Let's look at more details about hydrogen bonding here. So hydrogen bonding exists when you have hydrogen bonded to an electronegative atom, like that, this oxygen. But students forget that you also need another electronegative atom over here to give you more of a difference in charge and to make that hydrogen more partially positive. So it's really three atoms involved in hydrogen bonding there. Let's look at dimethyl ether and see why it does not exhibit hydrogen bonding. So if I were to draw one molecule of dimethyl ether here and think about the polarization between the oxygen and this carbon right here, oxygen's more electronegative, so it would be partially negative."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "But students forget that you also need another electronegative atom over here to give you more of a difference in charge and to make that hydrogen more partially positive. So it's really three atoms involved in hydrogen bonding there. Let's look at dimethyl ether and see why it does not exhibit hydrogen bonding. So if I were to draw one molecule of dimethyl ether here and think about the polarization between the oxygen and this carbon right here, oxygen's more electronegative, so it would be partially negative. This carbon would be partially positive, like that. If I think about the interaction of that molecule of dimethyl ether with another molecule of dimethyl ether, like that, you might be tempted to say, well, there could be some hydrogen bonding, because I know that this carbon right here has some hydrogens attached to it. And so some students will say, oh, there must be hydrogen bonding between this oxygen down here and this hydrogen."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So if I were to draw one molecule of dimethyl ether here and think about the polarization between the oxygen and this carbon right here, oxygen's more electronegative, so it would be partially negative. This carbon would be partially positive, like that. If I think about the interaction of that molecule of dimethyl ether with another molecule of dimethyl ether, like that, you might be tempted to say, well, there could be some hydrogen bonding, because I know that this carbon right here has some hydrogens attached to it. And so some students will say, oh, there must be hydrogen bonding between this oxygen down here and this hydrogen. But that is not the case, because this hydrogen right here, while it is interacting with an oxygen, this hydrogen is bonded to a carbon, which is not very electronegative. And so there's no large differences in electronegativity in the bond between carbon and hydrogen, even though carbon's a little bit more electronegative. There's not enough to make this a true hydrogen bond."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And so some students will say, oh, there must be hydrogen bonding between this oxygen down here and this hydrogen. But that is not the case, because this hydrogen right here, while it is interacting with an oxygen, this hydrogen is bonded to a carbon, which is not very electronegative. And so there's no large differences in electronegativity in the bond between carbon and hydrogen, even though carbon's a little bit more electronegative. There's not enough to make this a true hydrogen bond. And so really, there's only a small amount of dipole-dipole interaction between two molecules of dimethyl ether. So somewhere on this second molecule, there's a partial negative, partial positive. And so there will be a little bit of dipole-dipole interaction here."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "There's not enough to make this a true hydrogen bond. And so really, there's only a small amount of dipole-dipole interaction between two molecules of dimethyl ether. So somewhere on this second molecule, there's a partial negative, partial positive. And so there will be a little bit of dipole-dipole interaction here. But it's not very strong. And it's certainly nowhere near as strong as the hydrogen bonding exhibited on the left, hydrogen bonding being just the super strong form of dipole-dipole interaction. And so dimethyl ether does not have as high of a boiling point as ethanol."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And so there will be a little bit of dipole-dipole interaction here. But it's not very strong. And it's certainly nowhere near as strong as the hydrogen bonding exhibited on the left, hydrogen bonding being just the super strong form of dipole-dipole interaction. And so dimethyl ether does not have as high of a boiling point as ethanol. Again, the answer is hydrogen bonding. Let's see what happens to the boiling point of ethers as we increase the number of carbons in the alkyl groups. So if we're going to look at dimethyl ether again, and let's compare that to an ether that has more carbons in the alkyl group, so diethyl ether."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And so dimethyl ether does not have as high of a boiling point as ethanol. Again, the answer is hydrogen bonding. Let's see what happens to the boiling point of ethers as we increase the number of carbons in the alkyl groups. So if we're going to look at dimethyl ether again, and let's compare that to an ether that has more carbons in the alkyl group, so diethyl ether. We've already seen the boiling point of dimethyl ether is approximately negative 25 degrees Celsius, whereas diethyl ether is about 35 degrees Celsius. And so there's a large difference in boiling points. Diethyl ether's boiling point is just higher than room temperature."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So if we're going to look at dimethyl ether again, and let's compare that to an ether that has more carbons in the alkyl group, so diethyl ether. We've already seen the boiling point of dimethyl ether is approximately negative 25 degrees Celsius, whereas diethyl ether is about 35 degrees Celsius. And so there's a large difference in boiling points. Diethyl ether's boiling point is just higher than room temperature. So it is still a liquid at room temperature and pressure. So let's see if we can look at why diethyl ether has a higher boiling point. We know that ether molecules can't hydrogen bond with each other."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "Diethyl ether's boiling point is just higher than room temperature. So it is still a liquid at room temperature and pressure. So let's see if we can look at why diethyl ether has a higher boiling point. We know that ether molecules can't hydrogen bond with each other. So that cannot be the intermolecular force responsible for this increase in boiling point. So if we look at two molecules of diethyl ether interacting, one of the other intermolecular forces that we discussed was London dispersion forces. So London dispersion forces, you can watch the earlier video for more details."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "We know that ether molecules can't hydrogen bond with each other. So that cannot be the intermolecular force responsible for this increase in boiling point. So if we look at two molecules of diethyl ether interacting, one of the other intermolecular forces that we discussed was London dispersion forces. So London dispersion forces, you can watch the earlier video for more details. But when you have these large alkyl groups, provides more surface area for a form of attraction called London dispersion. And so that increased attraction between alkyl groups means that it's harder to pull those molecules apart. It requires more energy to pull those molecules apart."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So London dispersion forces, you can watch the earlier video for more details. But when you have these large alkyl groups, provides more surface area for a form of attraction called London dispersion. And so that increased attraction between alkyl groups means that it's harder to pull those molecules apart. It requires more energy to pull those molecules apart. It requires more heat in order to do so. And so that's the reason for the increase in boiling point that we see for diethyl ether up to 35 degrees Celsius. And even though London dispersion forces are the weakest intermolecular forces, they're additive."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "It requires more energy to pull those molecules apart. It requires more heat in order to do so. And so that's the reason for the increase in boiling point that we see for diethyl ether up to 35 degrees Celsius. And even though London dispersion forces are the weakest intermolecular forces, they're additive. So the effect is added when you have lots and lots of molecules. And that's the reason for the large difference between dimethyl ether and for diethyl ether. And so the increase of the number of carbons in the alkyl groups increases the boiling point just above room temperature, but not much above room temperature."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And even though London dispersion forces are the weakest intermolecular forces, they're additive. So the effect is added when you have lots and lots of molecules. And that's the reason for the large difference between dimethyl ether and for diethyl ether. And so the increase of the number of carbons in the alkyl groups increases the boiling point just above room temperature, but not much above room temperature. So this makes diethyl ether an excellent solvent for extraction. The other thing the alkyl groups do is they increase the nonpolar part of the molecule. So it's a little bit more nonpolar due to these alkyl groups right here, which means that diethyl ether is very good for dissolving a lot of nonpolar organic compounds."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And so the increase of the number of carbons in the alkyl groups increases the boiling point just above room temperature, but not much above room temperature. So this makes diethyl ether an excellent solvent for extraction. The other thing the alkyl groups do is they increase the nonpolar part of the molecule. So it's a little bit more nonpolar due to these alkyl groups right here, which means that diethyl ether is very good for dissolving a lot of nonpolar organic compounds. And so if you can dissolve a lot of nonpolar organic compounds and the boiling point is just above room temperature, it's an excellent solvent for extraction because you can dissolve your nonpolar organic molecules. And then you can just boil off the ether and you're left with your organic product. So you'll use diethyl ether a lot for extractions."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So it's a little bit more nonpolar due to these alkyl groups right here, which means that diethyl ether is very good for dissolving a lot of nonpolar organic compounds. And so if you can dissolve a lot of nonpolar organic compounds and the boiling point is just above room temperature, it's an excellent solvent for extraction because you can dissolve your nonpolar organic molecules. And then you can just boil off the ether and you're left with your organic product. So you'll use diethyl ether a lot for extractions. Let's look at another type of ether, which is kind of an interesting one. And we call these ethers crown ethers. So if you look at that gigantic ether there, it's called a crown ether."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So you'll use diethyl ether a lot for extractions. Let's look at another type of ether, which is kind of an interesting one. And we call these ethers crown ethers. So if you look at that gigantic ether there, it's called a crown ether. This was discovered by a guy named Charles Peterson, who won the Nobel Prize for this. And the system of nomenclature for crown ethers would be to first count up how many atoms comprise your ring here, your crown. So if we go 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and 18."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at that gigantic ether there, it's called a crown ether. This was discovered by a guy named Charles Peterson, who won the Nobel Prize for this. And the system of nomenclature for crown ethers would be to first count up how many atoms comprise your ring here, your crown. So if we go 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and 18. So there are 18 parts of this crown. So we would write an 18 right here like that, followed by the name crown, followed by the number of oxygens in here. So we have 1 oxygen, 2, 3, 4, 5, and 6."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So if we go 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and 18. So there are 18 parts of this crown. So we would write an 18 right here like that, followed by the name crown, followed by the number of oxygens in here. So we have 1 oxygen, 2, 3, 4, 5, and 6. So the nomenclature would be 18 crown 6 ether. And that just tells you what sort of crown ether that you are dealing with. So why is it called a crown ether?"}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So we have 1 oxygen, 2, 3, 4, 5, and 6. So the nomenclature would be 18 crown 6 ether. And that just tells you what sort of crown ether that you are dealing with. So why is it called a crown ether? Well, the interesting thing about crown ethers are that they can interact with different ions. For example, the size of the potassium ion, so K plus, happens to fit right in the center of this. So the spacing is just right for a potassium ion to fit in there."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So why is it called a crown ether? Well, the interesting thing about crown ethers are that they can interact with different ions. For example, the size of the potassium ion, so K plus, happens to fit right in the center of this. So the spacing is just right for a potassium ion to fit in there. And since all of these oxygens have lone pairs of electrons on them, so negatively charged, there's an attraction between the positively charged potassium ion and the negatively charged electrons, or the partially negative charged oxygen atoms. So there's attraction. Opposite charges attract."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So the spacing is just right for a potassium ion to fit in there. And since all of these oxygens have lone pairs of electrons on them, so negatively charged, there's an attraction between the positively charged potassium ion and the negatively charged electrons, or the partially negative charged oxygen atoms. So there's attraction. Opposite charges attract. And those negative charges are going to hold that potassium ion in here like that. So it looks like a crown if you think about the potassium ion as being someone's head and then is wearing this ether crown on someone's head like that. And crown ethers have proved to be very useful things."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "Opposite charges attract. And those negative charges are going to hold that potassium ion in here like that. So it looks like a crown if you think about the potassium ion as being someone's head and then is wearing this ether crown on someone's head like that. And crown ethers have proved to be very useful things. For example, if you had some potassium fluoride, so some K plus F minus, well, normally potassium fluoride would not dissolve in a nonpolar organic solvent. But if you use a crown ether, the oxygens can take care of the potassium. And the outside of the crown ether is nonpolar."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And crown ethers have proved to be very useful things. For example, if you had some potassium fluoride, so some K plus F minus, well, normally potassium fluoride would not dissolve in a nonpolar organic solvent. But if you use a crown ether, the oxygens can take care of the potassium. And the outside of the crown ether is nonpolar. So this portion and this portion, the outside of the crown ether is nonpolar, which will dissolve in an organic solvent, in a nonpolar organic solvent like benzene, just like that. So like dissolves like. So this portion would dissolve in benzene."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "And the outside of the crown ether is nonpolar. So this portion and this portion, the outside of the crown ether is nonpolar, which will dissolve in an organic solvent, in a nonpolar organic solvent like benzene, just like that. So like dissolves like. So this portion would dissolve in benzene. And then what that would do is that would free up your fluoride anion. That would increase the nucleophilic strength of your fluoride anion, which could participate in an SN2 reaction. So that's one of the uses of crown ethers, is to go ahead and take the cation, leaving the anion to function as a better nucleophile, because the potassium ion is solvated by the crown ether."}, {"video_title": "Properties of ethers and crown ethers Organic chemistry Khan Academy.mp3", "Sentence": "So this portion would dissolve in benzene. And then what that would do is that would free up your fluoride anion. That would increase the nucleophilic strength of your fluoride anion, which could participate in an SN2 reaction. So that's one of the uses of crown ethers, is to go ahead and take the cation, leaving the anion to function as a better nucleophile, because the potassium ion is solvated by the crown ether. And of course, since different ions have different sizes, you can get different sized crown ethers to take care of those ions. So crown ethers, I just think, are very interesting, interesting molecules. And if you can look at a three-dimensional representation of a crown ether, it's much easier to see that the outside is very nonpolar."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Therefore, all six protons are chemically equivalent and should give us only one signal on an NMR spectrum. And so here's the one signal on a spectrum due to the protons on benzene. If we compare benzene to this compound, this is tetramethylsilane, or TMS. And the protons on TMS are all in the same environment. And so therefore, we would expect one signal for TMS. And here's the signal for TMS right here. In an earlier video, we said that as you go to the right on an NMR spectrum, you're talking about a lower frequency signal."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And the protons on TMS are all in the same environment. And so therefore, we would expect one signal for TMS. And here's the signal for TMS right here. In an earlier video, we said that as you go to the right on an NMR spectrum, you're talking about a lower frequency signal. So a lower frequency signal as you move to the right on an NMR spectrum. And so if you move to the left on an NMR spectrum, you're talking about a higher frequency signal. So a higher frequency signal."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "In an earlier video, we said that as you go to the right on an NMR spectrum, you're talking about a lower frequency signal. So a lower frequency signal as you move to the right on an NMR spectrum. And so if you move to the left on an NMR spectrum, you're talking about a higher frequency signal. So a higher frequency signal. And so therefore, the protons on benzene have a higher frequency signal than the protons on TMS. TMS is actually our standard, because the protons on TMS are more shielded than almost all organic compounds. And so therefore, it's our reference."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So a higher frequency signal. And so therefore, the protons on benzene have a higher frequency signal than the protons on TMS. TMS is actually our standard, because the protons on TMS are more shielded than almost all organic compounds. And so therefore, it's our reference. And so instead of talking about frequency, we could talk about chemical shift values here. And the chemical shift would be a similar idea to the frequency. So as you go to the right, you're talking about a lower chemical shift."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so therefore, it's our reference. And so instead of talking about frequency, we could talk about chemical shift values here. And the chemical shift would be a similar idea to the frequency. So as you go to the right, you're talking about a lower chemical shift. And as you move to the left on an NMR spectrum, you're talking about a higher chemical shift. So a higher shift as you move to the left. So the protons on benzene have a higher chemical shift than the protons on TMS."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So as you go to the right, you're talking about a lower chemical shift. And as you move to the left on an NMR spectrum, you're talking about a higher chemical shift. So a higher shift as you move to the left. So the protons on benzene have a higher chemical shift than the protons on TMS. Actually, we set this equal to 0. So this is our standard. So how do we figure out what the chemical shift is?"}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So the protons on benzene have a higher chemical shift than the protons on TMS. Actually, we set this equal to 0. So this is our standard. So how do we figure out what the chemical shift is? For example, for the protons on benzene, it looks like the signal appears a little bit past 7 here. So how do we get this number for a chemical shift? Well, again, everything is compared to TMS."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So how do we figure out what the chemical shift is? For example, for the protons on benzene, it looks like the signal appears a little bit past 7 here. So how do we get this number for a chemical shift? Well, again, everything is compared to TMS. And so let's look at the formula for calculating chemical shift. So if I move down here, we can see the formula for chemical shift. Chemical shift is equal to the observed shift from TMS in hertz times 10 to the 6th divided by the spectrometer frequency in hertz."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Well, again, everything is compared to TMS. And so let's look at the formula for calculating chemical shift. So if I move down here, we can see the formula for chemical shift. Chemical shift is equal to the observed shift from TMS in hertz times 10 to the 6th divided by the spectrometer frequency in hertz. For example, let's say that we are using an NMR spectrometer operating at 300 megahertz. So we're using a 300 megahertz spectrometer here. If you're using a 300 megahertz spectrometer, the protons on benzene absorb a frequency 2,181 hertz more than the protons on TMS."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Chemical shift is equal to the observed shift from TMS in hertz times 10 to the 6th divided by the spectrometer frequency in hertz. For example, let's say that we are using an NMR spectrometer operating at 300 megahertz. So we're using a 300 megahertz spectrometer here. If you're using a 300 megahertz spectrometer, the protons on benzene absorb a frequency 2,181 hertz more than the protons on TMS. And so once again, TMS is our standard, our reference. So this difference, if you're thinking about frequency, this difference between our two signals is 2,181 hertz if we are using a 300 megahertz spectrometer. And so let's go ahead and figure out the chemical shift for the protons on benzene."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "If you're using a 300 megahertz spectrometer, the protons on benzene absorb a frequency 2,181 hertz more than the protons on TMS. And so once again, TMS is our standard, our reference. So this difference, if you're thinking about frequency, this difference between our two signals is 2,181 hertz if we are using a 300 megahertz spectrometer. And so let's go ahead and figure out the chemical shift for the protons on benzene. So let's get some more room down here. And so here's a symbol for chemical shift. So chemical shift is equal to the observed shift from TMS."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and figure out the chemical shift for the protons on benzene. So let's get some more room down here. And so here's a symbol for chemical shift. So chemical shift is equal to the observed shift from TMS. That was 2,181. So that's 2,181 hertz. And we need to multiply that by 10 to the 6th."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So chemical shift is equal to the observed shift from TMS. That was 2,181. So that's 2,181 hertz. And we need to multiply that by 10 to the 6th. And the reason we multiply that by 10 to the 6th is because the spectrometer is in megahertz here. So 300 megahertz is 300 times 10 to the 6th hertz. And so we can cancel out the hertz."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And we need to multiply that by 10 to the 6th. And the reason we multiply that by 10 to the 6th is because the spectrometer is in megahertz here. So 300 megahertz is 300 times 10 to the 6th hertz. And so we can cancel out the hertz. We can cancel out 10 to the 6th. And so we have a simple calculation here to figure out the chemical shift. And so let's go ahead and do that."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so we can cancel out the hertz. We can cancel out 10 to the 6th. And so we have a simple calculation here to figure out the chemical shift. And so let's go ahead and do that. So we turn the calculator on. 2,181 divided by 300 gives us 7.27. So this is equal to 7.27."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so let's go ahead and do that. So we turn the calculator on. 2,181 divided by 300 gives us 7.27. So this is equal to 7.27. Notice how the hertz will cancel out. And we have right here PPM, or parts per million, because these signals are reported as a fraction of the operating frequency of the spectrometer. And so there's a reason why we do that."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So this is equal to 7.27. Notice how the hertz will cancel out. And we have right here PPM, or parts per million, because these signals are reported as a fraction of the operating frequency of the spectrometer. And so there's a reason why we do that. So we got this number 7.27 here. Let's do this calculation again. Let's say we did this."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "And so there's a reason why we do that. So we got this number 7.27 here. Let's do this calculation again. Let's say we did this. We ran the spectrum on a different spectrometer. Let's say we're using a 60 megahertz spectrometer. So let's change it up a little bit."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we did this. We ran the spectrum on a different spectrometer. Let's say we're using a 60 megahertz spectrometer. So let's change it up a little bit. So a 60 megahertz spectrometer. If you use a 60 megahertz spectrometer, the protons on benzene absorb a frequency 436 hertz more than the protons on TMS. So to calculate the chemical shift now, the difference would be 436 hertz times 10 to the 6th divided by, now we're using a 60 megahertz spectrometer, so 60 times 10 to the 6th hertz."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So let's change it up a little bit. So a 60 megahertz spectrometer. If you use a 60 megahertz spectrometer, the protons on benzene absorb a frequency 436 hertz more than the protons on TMS. So to calculate the chemical shift now, the difference would be 436 hertz times 10 to the 6th divided by, now we're using a 60 megahertz spectrometer, so 60 times 10 to the 6th hertz. Once again, the hertz cancels, the 10 to the 6th cancels, and we can do that calculation. So we take 436, we divide that by 60, and we get 7.27 again. So we get 7.27."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So to calculate the chemical shift now, the difference would be 436 hertz times 10 to the 6th divided by, now we're using a 60 megahertz spectrometer, so 60 times 10 to the 6th hertz. Once again, the hertz cancels, the 10 to the 6th cancels, and we can do that calculation. So we take 436, we divide that by 60, and we get 7.27 again. So we get 7.27. Notice we got the same value we did up here. So 7.27 is a constant no matter what kind of spectrometer you're using. So you could be using a 300 megahertz spectrometer or a 60 megahertz spectrometer."}, {"video_title": "Chemical shift Spectroscopy Organic chemistry Khan Academy.mp3", "Sentence": "So we get 7.27. Notice we got the same value we did up here. So 7.27 is a constant no matter what kind of spectrometer you're using. So you could be using a 300 megahertz spectrometer or a 60 megahertz spectrometer. You're going to get the same value for the chemical shift. So that's why we go through this calculation here. So we get a constant value for the chemical shift."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we have some chloroethene here. And you wouldn't have to call this one chloroeth-one-ene, because if you just go with chloroethene, there's only one way to draw this. And the common name for chloroethene is vinyl chloride. So let's say we have a bunch of chloroethene molecules along with, or mixed with, some hydrogen chloride. And I've drawn all of the valence electrons for the chlorine atom, and I've drawn a little magenta electron, the one that the hydrogen atom brought to the table. So we've seen something like this before. What is likely to happen?"}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's say we have a bunch of chloroethene molecules along with, or mixed with, some hydrogen chloride. And I've drawn all of the valence electrons for the chlorine atom, and I've drawn a little magenta electron, the one that the hydrogen atom brought to the table. So we've seen something like this before. What is likely to happen? Well, maybe one of these carbons is willing to give up an electron. That electron goes to the hydrogen, because this electron's already being hogged by the chlorine. So this hydrogen has a partially positive charge."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What is likely to happen? Well, maybe one of these carbons is willing to give up an electron. That electron goes to the hydrogen, because this electron's already being hogged by the chlorine. So this hydrogen has a partially positive charge. Chlorine has a partially negative charge. So that electron would be attracted to the hydrogen. Then this electron can be completely hogged by the chlorine."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this hydrogen has a partially positive charge. Chlorine has a partially negative charge. So that electron would be attracted to the hydrogen. Then this electron can be completely hogged by the chlorine. And if we had to decide which of these carbons is more likely to give up the electron, you just have to say which one is bonded to things that it can share electrons a little bit with. This carbon's only bonded to hydrogens. So it's already hogging their one electron each."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Then this electron can be completely hogged by the chlorine. And if we had to decide which of these carbons is more likely to give up the electron, you just have to say which one is bonded to things that it can share electrons a little bit with. This carbon's only bonded to hydrogens. So it's already hogging their one electron each. There are no more electrons to share with it. This guy's bonded to a chlorine. So the chlorine has a bunch of valence electrons."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's already hogging their one electron each. There are no more electrons to share with it. This guy's bonded to a chlorine. So the chlorine has a bunch of valence electrons. It might be able to share it a little bit with this carbon if this guy became a carbocation. So this guy will lose an electron. This carbon will form the bond with that hydrogen."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So the chlorine has a bunch of valence electrons. It might be able to share it a little bit with this carbon if this guy became a carbocation. So this guy will lose an electron. This carbon will form the bond with that hydrogen. So let's draw it out. So let's say this carbon's electron is that blue thing right there. It will, we could draw it like this, it goes to the hydrogen, and then the hydrogen's magenta electron goes to the chlorine."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This carbon will form the bond with that hydrogen. So let's draw it out. So let's say this carbon's electron is that blue thing right there. It will, we could draw it like this, it goes to the hydrogen, and then the hydrogen's magenta electron goes to the chlorine. This is just a plausible mechanism. Now, once that happens, what will our setup look like? What will it look like?"}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It will, we could draw it like this, it goes to the hydrogen, and then the hydrogen's magenta electron goes to the chlorine. This is just a plausible mechanism. Now, once that happens, what will our setup look like? What will it look like? It will look, we'll have this carbon over here bonded to two hydrogens. It has its single bond to that other carbon that just lost its electron, which is bonded to a hydrogen and a chlorine. And now this carbon on the left, it is now bonded to the hydrogen."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "What will it look like? It will look, we'll have this carbon over here bonded to two hydrogens. It has its single bond to that other carbon that just lost its electron, which is bonded to a hydrogen and a chlorine. And now this carbon on the left, it is now bonded to the hydrogen. That electron went to the hydrogen and it formed a bond with it. So then it forms a bond. So that little blue electron is at this end of the, I want to make it blue, that little blue electron is at this end of the bond, and it is now the hydrogen's electron."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now this carbon on the left, it is now bonded to the hydrogen. That electron went to the hydrogen and it formed a bond with it. So then it forms a bond. So that little blue electron is at this end of the, I want to make it blue, that little blue electron is at this end of the bond, and it is now the hydrogen's electron. And that magenta electron went to the chlorine. So now it is a negative ion. It is a chloride ion."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that little blue electron is at this end of the, I want to make it blue, that little blue electron is at this end of the bond, and it is now the hydrogen's electron. And that magenta electron went to the chlorine. So now it is a negative ion. It is a chloride ion. So we have chloride ion. It has its standard 7 valence electrons that it started off with. But now it took that magenta electron from the hydrogen."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It is a chloride ion. So we have chloride ion. It has its standard 7 valence electrons that it started off with. But now it took that magenta electron from the hydrogen. And so now it has 8 valence electrons. It gained an electron. It now has a negative charge."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But now it took that magenta electron from the hydrogen. And so now it has 8 valence electrons. It gained an electron. It now has a negative charge. This guy over here lost an electron. He now has a positive charge. Now, the next thing that you might expect to happen, if we just followed the pattern of the last several videos, is you would say, hey, this guy will now take an electron from the chlorine, which is pot, or the chloride anion, I should say, which is completely plausible."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It now has a negative charge. This guy over here lost an electron. He now has a positive charge. Now, the next thing that you might expect to happen, if we just followed the pattern of the last several videos, is you would say, hey, this guy will now take an electron from the chlorine, which is pot, or the chloride anion, I should say, which is completely plausible. But there's also a bunch of the chloroethane. This isn't the only molecule of chloroethene. I should say chloroethene, not chloroethane."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now, the next thing that you might expect to happen, if we just followed the pattern of the last several videos, is you would say, hey, this guy will now take an electron from the chlorine, which is pot, or the chloride anion, I should say, which is completely plausible. But there's also a bunch of the chloroethane. This isn't the only molecule of chloroethene. I should say chloroethene, not chloroethane. Chloroethene sitting around. So let us throw another one of those in there. So we have more molecules of this."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I should say chloroethene, not chloroethane. Chloroethene sitting around. So let us throw another one of those in there. So we have more molecules of this. So he could take an electron from the chloride ion, or he could take an electron from this guy over here, remember. This guy, just like this guy, who was this guy, is OK. It doesn't require a super amount of energy to make this guy lose his electron."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have more molecules of this. So he could take an electron from the chloride ion, or he could take an electron from this guy over here, remember. This guy, just like this guy, who was this guy, is OK. It doesn't require a super amount of energy to make this guy lose his electron. He's bonded to other things that are willing to share with him a little bit. Maybe he's willing to lose his electron as opposed to the chloride ion. So this guy has, let me draw it in, so this end of this bond is green."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It doesn't require a super amount of energy to make this guy lose his electron. He's bonded to other things that are willing to share with him a little bit. Maybe he's willing to lose his electron as opposed to the chloride ion. So this guy has, let me draw it in, so this end of this bond is green. And then this goes and bonds with this carbon. So this will be a long bond right here. So this goes and bonds with that carbon, essentially giving that electron to that carbon."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this guy has, let me draw it in, so this end of this bond is green. And then this goes and bonds with this carbon. So this will be a long bond right here. So this goes and bonds with that carbon, essentially giving that electron to that carbon. And then what will our setup look like? So after that happens, we'll look like this. I probably should have copied and pasted this from the get-go."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this goes and bonds with that carbon, essentially giving that electron to that carbon. And then what will our setup look like? So after that happens, we'll look like this. I probably should have copied and pasted this from the get-go. Actually, let me do that. Before I, let me copy and paste this. So now, let me just copy and paste this whole thing."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I probably should have copied and pasted this from the get-go. Actually, let me do that. Before I, let me copy and paste this. So now, let me just copy and paste this whole thing. Nope, that's not what I wanted to do. Let me select it again. All right."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now, let me just copy and paste this whole thing. Nope, that's not what I wanted to do. Let me select it again. All right. Copy and then paste. There you go. So then we have that thing."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "All right. Copy and then paste. There you go. So then we have that thing. And let me redraw what I had erased so that I could copy and paste. And then we have this guy went over to this carbocation. So he's no longer a carbocation."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So then we have that thing. And let me redraw what I had erased so that I could copy and paste. And then we have this guy went over to this carbocation. So he's no longer a carbocation. So let me erase this because now he's gained an electron. He gained that green electron right there. I'll just draw it right there."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So he's no longer a carbocation. So let me erase this because now he's gained an electron. He gained that green electron right there. I'll just draw it right there. And now he's formed a bond with this carbon. And I'll make it blue just so we know which carbon we're talking about. He's formed a bond."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I'll just draw it right there. And now he's formed a bond with this carbon. And I'll make it blue just so we know which carbon we're talking about. He's formed a bond. This bond now moved over to that carbon because the electron went with it. So now that bond is to this carbon right here, that carbon right over there, which is bonded to 2 hydrogens. And now has a single bond to the electron that gave up the carbon."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "He's formed a bond. This bond now moved over to that carbon because the electron went with it. So now that bond is to this carbon right here, that carbon right over there, which is bonded to 2 hydrogens. And now has a single bond to the electron that gave up the carbon. Has a single bond to that character right over here, who is bonded to a hydrogen and a chlorine. And since he now lost an electron, he now has a positive charge. So if you look at this setup right here, it looks very similar to this setup, although we've added one more vinyl chloride to the mix."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now has a single bond to the electron that gave up the carbon. Has a single bond to that character right over here, who is bonded to a hydrogen and a chlorine. And since he now lost an electron, he now has a positive charge. So if you look at this setup right here, it looks very similar to this setup, although we've added one more vinyl chloride to the mix. And the one that we added lost its electron, or this carbon lost its electron. And now it's a carbocation. So what could happen next?"}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if you look at this setup right here, it looks very similar to this setup, although we've added one more vinyl chloride to the mix. And the one that we added lost its electron, or this carbon lost its electron. And now it's a carbocation. So what could happen next? Well, we have more of this vinyl chloride sitting around. Let me draw another vinyl chloride. So I have a carbon, a hydrogen, a hydrogen, and then it is double bonded to a carbon, a hydrogen, and a chlorine."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So what could happen next? Well, we have more of this vinyl chloride sitting around. Let me draw another vinyl chloride. So I have a carbon, a hydrogen, a hydrogen, and then it is double bonded to a carbon, a hydrogen, and a chlorine. And let me copy and paste this. I think you see where this might be going, how this could keep on going and going and going. So copy."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I have a carbon, a hydrogen, a hydrogen, and then it is double bonded to a carbon, a hydrogen, and a chlorine. And let me copy and paste this. I think you see where this might be going, how this could keep on going and going and going. So copy. And then let me just copy that for now. So what's going to happen now? This guy could go and give an electron to this guy and form a bond, or we could have the same process happen over and over and over again."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So copy. And then let me just copy that for now. So what's going to happen now? This guy could go and give an electron to this guy and form a bond, or we could have the same process happen over and over and over again. So this electron could be given to this carbocation right there. So this electron can be given to that carbocation right there, and then what happens? Well, if that happens, then we're going to get, I'll move to the left now, we have our original molecule."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This guy could go and give an electron to this guy and form a bond, or we could have the same process happen over and over and over again. So this electron could be given to this carbocation right there. So this electron can be given to that carbocation right there, and then what happens? Well, if that happens, then we're going to get, I'll move to the left now, we have our original molecule. I'm going to run out of space soon. We have this original molecule, and now this guy is bonded to that. So this carbon right here is going to be this carbon, and now it is bonded to this guy."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Well, if that happens, then we're going to get, I'll move to the left now, we have our original molecule. I'm going to run out of space soon. We have this original molecule, and now this guy is bonded to that. So this carbon right here is going to be this carbon, and now it is bonded to this guy. That orange electron is now given to this guy who was positive. So he now has, let me make it a little bit neater. I can do a better job than that."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon right here is going to be this carbon, and now it is bonded to this guy. That orange electron is now given to this guy who was positive. So he now has, let me make it a little bit neater. I can do a better job than that. So the carbon's here, the bond goes to this guy, he now has the orange electron. He no longer has a positive charge. He's got all of his valence electrons now, and now this guy is bonded to two hydrogens."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I can do a better job than that. So the carbon's here, the bond goes to this guy, he now has the orange electron. He no longer has a positive charge. He's got all of his valence electrons now, and now this guy is bonded to two hydrogens. And he has a single bond, this single bond right here, to the carbon that just lost his electron, who's bonded to a hydrogen and a chlorine. And because he lost his electron, he now is the carbocation. He is now a carbocation."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "He's got all of his valence electrons now, and now this guy is bonded to two hydrogens. And he has a single bond, this single bond right here, to the carbon that just lost his electron, who's bonded to a hydrogen and a chlorine. And because he lost his electron, he now is the carbocation. He is now a carbocation. So I think you see where we're going. We can just keep adding and adding and adding to this chain of vinyl chloride. So if this process just went on and on and on, we could make it like this."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "He is now a carbocation. So I think you see where we're going. We can just keep adding and adding and adding to this chain of vinyl chloride. So if this process just went on and on and on, we could make it like this. It would look something like this. Let me see how well I can draw it. So it would look like this."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if this process just went on and on and on, we could make it like this. It would look something like this. Let me see how well I can draw it. So it would look like this. So this is a CH3, so I'll just draw it as H3C, and then this is bonded to a carbon. Well, maybe I'll call it a CH, which is bonded to a chlorine. So we're that far in the molecule."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it would look like this. So this is a CH3, so I'll just draw it as H3C, and then this is bonded to a carbon. Well, maybe I'll call it a CH, which is bonded to a chlorine. So we're that far in the molecule. And then we have, let's see, the part that repeats. This part right here is going to keep repeating. So that part right there, and I'm going to do it like this."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're that far in the molecule. And then we have, let's see, the part that repeats. This part right here is going to keep repeating. So that part right there, and I'm going to do it like this. So I'm just going to draw one of them. So you have a CH2, that's that right there, and then connected to a CH, which is that right there, which is connected to a chlorine. And so that part right there will keep repeating."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So that part right there, and I'm going to do it like this. So I'm just going to draw one of them. So you have a CH2, that's that right there, and then connected to a CH, which is that right there, which is connected to a chlorine. And so that part right there will keep repeating. And then maybe the very last one, so you have this guy right here, maybe the very last one that joins on. I mean, this could happen millions of times. I just made it happen two or three times."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so that part right there will keep repeating. And then maybe the very last one, so you have this guy right here, maybe the very last one that joins on. I mean, this could happen millions of times. I just made it happen two or three times. It could happen millions of times and form a super long chain or a polymer. And what we are describing in this video is actually a polymer that you have probably dealt with at some point in your life. In fact, I guarantee there's some of it in your house right now."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "I just made it happen two or three times. It could happen millions of times and form a super long chain or a polymer. And what we are describing in this video is actually a polymer that you have probably dealt with at some point in your life. In fact, I guarantee there's some of it in your house right now. So then we'd have that part right there, and we could just make that as CH2CHCl. And now the way we've run it, it's a carbocation, but maybe we've run out of all of the vinyl chloride molecules, or we could also call them chloroethene molecules. And now finally, when everything's said and done, this last guy, since he's run out of vinyl chloride molecules to take their electrons from, he now finally takes it from the chlorine."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "In fact, I guarantee there's some of it in your house right now. So then we'd have that part right there, and we could just make that as CH2CHCl. And now the way we've run it, it's a carbocation, but maybe we've run out of all of the vinyl chloride molecules, or we could also call them chloroethene molecules. And now finally, when everything's said and done, this last guy, since he's run out of vinyl chloride molecules to take their electrons from, he now finally takes it from the chlorine. So you could imagine, after this happens many, many times, then finally, one of these electrons from the chlorine go to that final carbocation, because they've run out of other vinyl chlorides, and then he attaches right over here to the chlorine. Now, when we say that this might happen, this repeats many times, you might write an N here just to show that it repeats many, many times. If you know how many times it means, if you know that there were 1,000 molecules here, you would write 1,000 repetitions, but this is called a polymer."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now finally, when everything's said and done, this last guy, since he's run out of vinyl chloride molecules to take their electrons from, he now finally takes it from the chlorine. So you could imagine, after this happens many, many times, then finally, one of these electrons from the chlorine go to that final carbocation, because they've run out of other vinyl chlorides, and then he attaches right over here to the chlorine. Now, when we say that this might happen, this repeats many times, you might write an N here just to show that it repeats many, many times. If you know how many times it means, if you know that there were 1,000 molecules here, you would write 1,000 repetitions, but this is called a polymer. Polymer. And the name for this molecule right here is, each of these units is vinyl chloride, right? Vinyl chloride."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "If you know how many times it means, if you know that there were 1,000 molecules here, you would write 1,000 repetitions, but this is called a polymer. Polymer. And the name for this molecule right here is, each of these units is vinyl chloride, right? Vinyl chloride. I guess the official name is chloroethene, but the typical name, the one people actually use, is vinyl chloride. That's for each of these units. It's a polymer."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Vinyl chloride. I guess the official name is chloroethene, but the typical name, the one people actually use, is vinyl chloride. That's for each of these units. It's a polymer. We have many of them, so we'll put a poly in front of it. And we put a poly in front of it. So this molecule right here is polyvinyl chloride, or, and now I think it'll ring a bell, or PVC for short."}, {"video_title": "Polymerization of alkenes with acid Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's a polymer. We have many of them, so we'll put a poly in front of it. And we put a poly in front of it. So this molecule right here is polyvinyl chloride, or, and now I think it'll ring a bell, or PVC for short. And you've probably heard of PVC piping. It's what most people have for their plumbing. It's those plastic pipes."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say I'm over here. I'm going to do two scenarios. So I'm an observer over here. This is me. And then maybe even better, I should just draw my eyeball because we're going to be observing light. So I'm just going to draw my eyeball. So this is me in the first scenario, or this is one of my eyeballs, and then this is one of my eyeballs in the second scenario."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is me. And then maybe even better, I should just draw my eyeball because we're going to be observing light. So I'm just going to draw my eyeball. So this is me in the first scenario, or this is one of my eyeballs, and then this is one of my eyeballs in the second scenario. Now in the first scenario, so in both scenarios we're going to have an object. We're going to have some type of source of light. But in the first scenario, relative to me, the source of light will not be moving."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is me in the first scenario, or this is one of my eyeballs, and then this is one of my eyeballs in the second scenario. Now in the first scenario, so in both scenarios we're going to have an object. We're going to have some type of source of light. But in the first scenario, relative to me, the source of light will not be moving. While in the second scenario, the source of light, just for the sake of discussion, just for fun, will be moving at half the speed of light. Unimaginably fast speed, but let's just assuming it is. So it's moving at, it has a velocity of 1 half the speed of light, 1 half light speed away from me, who is the observer."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But in the first scenario, relative to me, the source of light will not be moving. While in the second scenario, the source of light, just for the sake of discussion, just for fun, will be moving at half the speed of light. Unimaginably fast speed, but let's just assuming it is. So it's moving at, it has a velocity of 1 half the speed of light, 1 half light speed away from me, who is the observer. Now let's just imagine what would happen. They're both emitting light. So they're both going to start emitting light at the exact same time."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it's moving at, it has a velocity of 1 half the speed of light, 1 half light speed away from me, who is the observer. Now let's just imagine what would happen. They're both emitting light. So they're both going to start emitting light at the exact same time. And when they start emitting light, they're both at the exact same distance from my eye. The only difference is that this is stationary, relative to me, while this is moving away from me at half the speed of light. So let's say that after some period of time, the light wave from this source reaches my eye."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So they're both going to start emitting light at the exact same time. And when they start emitting light, they're both at the exact same distance from my eye. The only difference is that this is stationary, relative to me, while this is moving away from me at half the speed of light. So let's say that after some period of time, the light wave from this source reaches my eye. And then it looks something like this. I'll try my best to draw it. So let's say I have, I want to draw a couple of wavelengths here."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say that after some period of time, the light wave from this source reaches my eye. And then it looks something like this. I'll try my best to draw it. So let's say I have, I want to draw a couple of wavelengths here. So let's say that's half a wavelength. That's a full wavelength. That's another half a full wavelength."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let's say I have, I want to draw a couple of wavelengths here. So let's say that's half a wavelength. That's a full wavelength. That's another half a full wavelength. Another half full wavelength. And then a half, and then a full wavelength. So let me see if I can draw that."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's another half a full wavelength. Another half full wavelength. And then a half, and then a full wavelength. So let me see if I can draw that. So it would look like full wavelength. This is not easy to do. And then you get another full wavelength."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So let me see if I can draw that. So it would look like full wavelength. This is not easy to do. And then you get another full wavelength. So it would look something like that, the actual waveform. And so this is just the front of the waveform is just getting to my eye. And then as the waveforms keep going past my eye, my eye will perceive some type of a wavelength or frequency and perceive it to be some type of color, assuming that we're in the visible part of the electromagnetic spectrum."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then you get another full wavelength. So it would look something like that, the actual waveform. And so this is just the front of the waveform is just getting to my eye. And then as the waveforms keep going past my eye, my eye will perceive some type of a wavelength or frequency and perceive it to be some type of color, assuming that we're in the visible part of the electromagnetic spectrum. Now let's think about what's going to happen with this source. So the first thing is that the front of the waveform is going to reach me at the exact same time. One of those neat and amazing things about light traveling in general, or especially in a vacuum, it doesn't matter that this is moving away from me at half the speed of light."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And then as the waveforms keep going past my eye, my eye will perceive some type of a wavelength or frequency and perceive it to be some type of color, assuming that we're in the visible part of the electromagnetic spectrum. Now let's think about what's going to happen with this source. So the first thing is that the front of the waveform is going to reach me at the exact same time. One of those neat and amazing things about light traveling in general, or especially in a vacuum, it doesn't matter that this is moving away from me at half the speed of light. The light will still move towards me at the speed of light. It's absolute. Doesn't matter if this is going away at 0.9 the speed of light, the light will still travel to me at the speed of light."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "One of those neat and amazing things about light traveling in general, or especially in a vacuum, it doesn't matter that this is moving away from me at half the speed of light. The light will still move towards me at the speed of light. It's absolute. Doesn't matter if this is going away at 0.9 the speed of light, the light will still travel to me at the speed of light. And it's very unintuitive. Because in our everyday sense, if I'm moving away from you at half the speed of a bullet and I shoot a bullet, the bullet will only move towards you at that half of its velocity will be subtracted, and it'll only move towards me at half of its normal velocity relative to whether it was stationary. But not the case with light."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Doesn't matter if this is going away at 0.9 the speed of light, the light will still travel to me at the speed of light. And it's very unintuitive. Because in our everyday sense, if I'm moving away from you at half the speed of a bullet and I shoot a bullet, the bullet will only move towards you at that half of its velocity will be subtracted, and it'll only move towards me at half of its normal velocity relative to whether it was stationary. But not the case with light. So with that out of the way, let's think about what the waveform would look like. So by the time the light reached here, we need to think, let me actually redraw this over here. Let me redraw this eyeball right over here."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But not the case with light. So with that out of the way, let's think about what the waveform would look like. So by the time the light reached here, we need to think, let me actually redraw this over here. Let me redraw this eyeball right over here. So this is me again. So by the time the light reaches my eye, so they both started emitting the light at the exact same time. This guy has traveled half this distance."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me redraw this eyeball right over here. So this is me again. So by the time the light reaches my eye, so they both started emitting the light at the exact same time. This guy has traveled half this distance. If it took light a certain amount of time to get this far, this guy will get half as far in that same amount of time. So by the time the light reaches my eye, this guy will have traveled about half that distance. So he would have traveled about that far."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This guy has traveled half this distance. If it took light a certain amount of time to get this far, this guy will get half as far in that same amount of time. So by the time the light reaches my eye, this guy will have traveled about half that distance. So he would have traveled about that far. But they started emitting the light at the same time. So that very first photon, if you view light as a particle, will reach my eye at the very same time as the very first photon from this guy. So the waveform is going to essentially be stretched."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So he would have traveled about that far. But they started emitting the light at the same time. So that very first photon, if you view light as a particle, will reach my eye at the very same time as the very first photon from this guy. So the waveform is going to essentially be stretched. So we're still going to have 1, 2, 3, 4 full wavelengths. But they'll now be stretched. Let me see if I can draw 4 full wavelengths."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the waveform is going to essentially be stretched. So we're still going to have 1, 2, 3, 4 full wavelengths. But they'll now be stretched. Let me see if I can draw 4 full wavelengths. So let me cut this in half over here. Let me cut each of those in half. So each of these are going to be a full wavelength."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me see if I can draw 4 full wavelengths. So let me cut this in half over here. Let me cut each of those in half. So each of these are going to be a full wavelength. And then they're going to have a half wavelength in between. And so the waveform is going to look like this. Let me try my best to draw it."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So each of these are going to be a full wavelength. And then they're going to have a half wavelength in between. And so the waveform is going to look like this. Let me try my best to draw it. This is the hardest part, drawing this stretched out waveform. And there you go. It's going to look like this."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me try my best to draw it. This is the hardest part, drawing this stretched out waveform. And there you go. It's going to look like this. And so when it gets to my eye, my eye is going to perceive it as having a longer wavelength, even though from the perspective of each of these objects, if you're traveling with each of them, the frequency and the wavelength of the light emitted is the same. The only difference is this guy is moving away from me, or I'm moving away from it, depending on how you want to view it, while I am stationary, or it is stationary, while in this first case, the observer and the source are both stationary. Now in this situation, what's my eye going to say?"}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's going to look like this. And so when it gets to my eye, my eye is going to perceive it as having a longer wavelength, even though from the perspective of each of these objects, if you're traveling with each of them, the frequency and the wavelength of the light emitted is the same. The only difference is this guy is moving away from me, or I'm moving away from it, depending on how you want to view it, while I am stationary, or it is stationary, while in this first case, the observer and the source are both stationary. Now in this situation, what's my eye going to say? Well, my eye will get each of these successive pulses, or each of these successive wavetrains. And it's going to say, hey, there's a longer wavelength a perceived longer wavelength. Let me write that."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now in this situation, what's my eye going to say? Well, my eye will get each of these successive pulses, or each of these successive wavetrains. And it's going to say, hey, there's a longer wavelength a perceived longer wavelength. Let me write that. And also a perceived lower frequency. So what would that do to the perception of the light? Let's say that this is green light."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let me write that. And also a perceived lower frequency. So what would that do to the perception of the light? Let's say that this is green light. So if we're stationary with the observer, it would be green light. So let's look at the electromagnetic spectrum. I got this off of Wikipedia."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's say that this is green light. So if we're stationary with the observer, it would be green light. So let's look at the electromagnetic spectrum. I got this off of Wikipedia. So if I was stationary with the observer, we'd be in the green light part of the spectrum. So a 500 nanometer wavelength. But if all of a sudden, because the object is moving away from me at this huge velocity, the perceived wavelength becomes wider."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I got this off of Wikipedia. So if I was stationary with the observer, we'd be in the green light part of the spectrum. So a 500 nanometer wavelength. But if all of a sudden, because the object is moving away from me at this huge velocity, the perceived wavelength becomes wider. So from my perception, it's going to have a wider wavelength. And you can see what's happening. It will look redder."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if all of a sudden, because the object is moving away from me at this huge velocity, the perceived wavelength becomes wider. So from my perception, it's going to have a wider wavelength. And you can see what's happening. It will look redder. It will move towards the red part of the spectrum. And this phenomenon is called redshift. And I've done a bunch of videos in the physics playlist on the Doppler effect."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It will look redder. It will move towards the red part of the spectrum. And this phenomenon is called redshift. And I've done a bunch of videos in the physics playlist on the Doppler effect. And over there, I talk about sound waves and the perceived frequency of sound as something travels towards you versus away from you. It's the exact same idea. This is the Doppler effect applied to light."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And I've done a bunch of videos in the physics playlist on the Doppler effect. And over there, I talk about sound waves and the perceived frequency of sound as something travels towards you versus away from you. It's the exact same idea. This is the Doppler effect applied to light. And the reason why the Doppler effect works for light traveling through space and for sound traveling through air is because a sound wave in air, regardless of whether the source is moving away or towards you, the sound wave is going to move at the speed of sound in air at a certain pressure and all of that. And light is the same thing. But in a vacuum, regardless of what the source is doing, the actual light wave itself will always travel at the same velocity."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is the Doppler effect applied to light. And the reason why the Doppler effect works for light traveling through space and for sound traveling through air is because a sound wave in air, regardless of whether the source is moving away or towards you, the sound wave is going to move at the speed of sound in air at a certain pressure and all of that. And light is the same thing. But in a vacuum, regardless of what the source is doing, the actual light wave itself will always travel at the same velocity. The only difference is that its perceived frequency and wavelength will change. And now the whole reason why I'm talking about this is you can use this property of light, that it gets redshift, to see whether things are traveling away or towards you. And people talk about redshift because frankly, most things are traveling away from us."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But in a vacuum, regardless of what the source is doing, the actual light wave itself will always travel at the same velocity. The only difference is that its perceived frequency and wavelength will change. And now the whole reason why I'm talking about this is you can use this property of light, that it gets redshift, to see whether things are traveling away or towards you. And people talk about redshift because frankly, most things are traveling away from us. And that's one of the reasons why we tend to believe in the Big Bang. The opposite, if something is traveling towards me at super high velocities, then we would have something called, and you don't hear the word, it would be violetshift. The frequency would increase."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And people talk about redshift because frankly, most things are traveling away from us. And that's one of the reasons why we tend to believe in the Big Bang. The opposite, if something is traveling towards me at super high velocities, then we would have something called, and you don't hear the word, it would be violetshift. The frequency would increase. So it would look bluer or more purple. Now the other thing I want to highlight is this redshift phenomena, this idea, it doesn't apply only to visible light. So it could even apply to things that we can't even see."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "The frequency would increase. So it would look bluer or more purple. Now the other thing I want to highlight is this redshift phenomena, this idea, it doesn't apply only to visible light. So it could even apply to things that we can't even see. So it would become redder, but it's not like you can even see. It could even apply to things that are even more red than red. So maybe it's a microwave that is being emitted, but because the source is moving away from us so fast, it could be perceived as an actual radio wave."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it could even apply to things that we can't even see. So it would become redder, but it's not like you can even see. It could even apply to things that are even more red than red. So maybe it's a microwave that is being emitted, but because the source is moving away from us so fast, it could be perceived as an actual radio wave. And actually, I should have talked about this in the video on the microwave background radiation, is that we're perceiving it as microwaves. But the sources were moving away from us. They were being redshift."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So maybe it's a microwave that is being emitted, but because the source is moving away from us so fast, it could be perceived as an actual radio wave. And actually, I should have talked about this in the video on the microwave background radiation, is that we're perceiving it as microwaves. But the sources were moving away from us. They were being redshift. So they were not actually emitting microwave radiation. Just what we observe, and this is actually what would be predicted based on the Big Bang, is actually microwave radiation. So anyway, hopefully that gives you a sense of what redshift is, and now we can use this tool to explain why we think many, many things are moving away from us."}, {"video_title": "Red shift Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They were being redshift. So they were not actually emitting microwave radiation. Just what we observe, and this is actually what would be predicted based on the Big Bang, is actually microwave radiation. So anyway, hopefully that gives you a sense of what redshift is, and now we can use this tool to explain why we think many, many things are moving away from us. And now let me just actually make sure you get that idea. If I have two objects, let's say that these are suns, or both galaxies, either way, and because of other properties, and I won't talk about them right now, we know that they are probably emitting light of the same color. Because we know other properties of that star or of that galaxy."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "For us modern humans with our easy access to the local supermarket, it's easy to forget that throughout human history and even today, the amount of humans or the human population has been limited by our ability to get calories or get human consumable food from the land. So what I wanna do in this video is give us a little bit of a framework for thinking about how humans have been getting calories from the land and how that's placed an upper limit on the number of humans that can live in any given area or the population density of humans. So right over here, you have some gentlemen looking for food. They are hunter-gatherers. I'll say HG for short. And the H part, the hunter part, they might actually find some animals. I think these guys right over here are trying to trap some rabbits."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They are hunter-gatherers. I'll say HG for short. And the H part, the hunter part, they might actually find some animals. I think these guys right over here are trying to trap some rabbits. And the gathering part, they're just literally looking for food. Maybe they find fruit of some sort or some nuts or maybe some roots that are edible by humans. So literally, they just walk around, either try to kill things or find things that they can consume."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "I think these guys right over here are trying to trap some rabbits. And the gathering part, they're just literally looking for food. Maybe they find fruit of some sort or some nuts or maybe some roots that are edible by humans. So literally, they just walk around, either try to kill things or find things that they can consume. So I'll call this right over here stage one. And actually, let me write it over here. So this is hunter-gatherers."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So literally, they just walk around, either try to kill things or find things that they can consume. So I'll call this right over here stage one. And actually, let me write it over here. So this is hunter-gatherers. And this is what most humans have done through most of human history. And just to give us a little bit of a framework for maybe how much they could get from the land, and I looked at some of our best sense of studying hunter-gatherer populations. In land like this, maybe they can get about 200 calories."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is hunter-gatherers. And this is what most humans have done through most of human history. And just to give us a little bit of a framework for maybe how much they could get from the land, and I looked at some of our best sense of studying hunter-gatherer populations. In land like this, maybe they can get about 200 calories. And I'll make this whole column right over here. This whole column right over here is the amount that they could get in terms of calories per square kilometer per day. Now, it's obviously going to be hugely dependent on the number of animals that are there, the type of land that's there."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In land like this, maybe they can get about 200 calories. And I'll make this whole column right over here. This whole column right over here is the amount that they could get in terms of calories per square kilometer per day. Now, it's obviously going to be hugely dependent on the number of animals that are there, the type of land that's there. If they're next to a stream where maybe fish are just jumping out of the stream, this number would be much higher. If they were in some type of a desert, this number would be much lower. But this is actually fairly in line with some of the studies of hunter-gatherer cultures."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Now, it's obviously going to be hugely dependent on the number of animals that are there, the type of land that's there. If they're next to a stream where maybe fish are just jumping out of the stream, this number would be much higher. If they were in some type of a desert, this number would be much lower. But this is actually fairly in line with some of the studies of hunter-gatherer cultures. Now, if this is the number of calories that they can get from each square kilometer per day, how many humans can live in a square kilometer per day? Or what is the density of humans? Well, to figure that out, we have to know on average how many calories does a human need to survive."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But this is actually fairly in line with some of the studies of hunter-gatherer cultures. Now, if this is the number of calories that they can get from each square kilometer per day, how many humans can live in a square kilometer per day? Or what is the density of humans? Well, to figure that out, we have to know on average how many calories does a human need to survive. And for the sake of this video, I'm going to make the assumption that a human being needs 2,000 calories per day to survive in a non-malnourished state. And obviously, it's hugely dependent on how active this person is or how large they are. And one other note, in this whole video, I'm going to be using calories with a capital C. And the calories with a capital C are the calories that people are used to referring to when you go to the gym and you run on the treadmill and it says how many calories you've burned or you look at the back of your candy bar and it says 200 calories."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Well, to figure that out, we have to know on average how many calories does a human need to survive. And for the sake of this video, I'm going to make the assumption that a human being needs 2,000 calories per day to survive in a non-malnourished state. And obviously, it's hugely dependent on how active this person is or how large they are. And one other note, in this whole video, I'm going to be using calories with a capital C. And the calories with a capital C are the calories that people are used to referring to when you go to the gym and you run on the treadmill and it says how many calories you've burned or you look at the back of your candy bar and it says 200 calories. These are the calories I'm talking about. They are a slightly different notion than the calories that you encounter in chemistry class. Those calories are calorie with a lowercase c. And just so that you can be optimally confused, it turns out that one calorie with an uppercase C is equal to 1,000 calories with a lowercase c. 1,000 calories."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And one other note, in this whole video, I'm going to be using calories with a capital C. And the calories with a capital C are the calories that people are used to referring to when you go to the gym and you run on the treadmill and it says how many calories you've burned or you look at the back of your candy bar and it says 200 calories. These are the calories I'm talking about. They are a slightly different notion than the calories that you encounter in chemistry class. Those calories are calorie with a lowercase c. And just so that you can be optimally confused, it turns out that one calorie with an uppercase C is equal to 1,000 calories with a lowercase c. 1,000 calories. And the lowercase c calories is the amount of energy needed to heat one gram of water one degrees Celsius. And so this is what you see in your chemistry class, but this is not what we're going to be talking about in this video. We're talking about the capital C, the calories that dietitians are always talking about."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Those calories are calorie with a lowercase c. And just so that you can be optimally confused, it turns out that one calorie with an uppercase C is equal to 1,000 calories with a lowercase c. 1,000 calories. And the lowercase c calories is the amount of energy needed to heat one gram of water one degrees Celsius. And so this is what you see in your chemistry class, but this is not what we're going to be talking about in this video. We're talking about the capital C, the calories that dietitians are always talking about. So with this assumption that the average human eats 2,000 calories a day to not get malnourished, and obviously men would need more, women would need less, children would need even less, but with this assumption, what is the density of humans that could be supported by this culture right over here? Well, 200 calories is 1 10th of the average daily human requirement if you believe this assumption. So the population density, so the density in, let me write it, humans per square kilometer, you can only support 1 10th of a human with this calorie output."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We're talking about the capital C, the calories that dietitians are always talking about. So with this assumption that the average human eats 2,000 calories a day to not get malnourished, and obviously men would need more, women would need less, children would need even less, but with this assumption, what is the density of humans that could be supported by this culture right over here? Well, 200 calories is 1 10th of the average daily human requirement if you believe this assumption. So the population density, so the density in, let me write it, humans per square kilometer, you can only support 1 10th of a human with this calorie output. So you can only support 1 10th of a human per square kilometer. So one human would actually need 10 square kilometers of hunter-gatherer, to hunt from and gather from in order to support just themselves. They would need maybe 30 or 40 square kilometers to support an entire family so that they could wander around and kill the animals and find whatever they need to find on that land."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So the population density, so the density in, let me write it, humans per square kilometer, you can only support 1 10th of a human with this calorie output. So you can only support 1 10th of a human per square kilometer. So one human would actually need 10 square kilometers of hunter-gatherer, to hunt from and gather from in order to support just themselves. They would need maybe 30 or 40 square kilometers to support an entire family so that they could wander around and kill the animals and find whatever they need to find on that land. Now let's go to kind of, you can view it as maybe the next stage, although it's not always the case that herding is going to be more productive than hunter-gatherers, especially in the case where the fish are jumping out of the water, but let's go to this scenario right over here. So this is, we can call this a pastoral lifestyle. So this is two, I'll call it pastoral."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They would need maybe 30 or 40 square kilometers to support an entire family so that they could wander around and kill the animals and find whatever they need to find on that land. Now let's go to kind of, you can view it as maybe the next stage, although it's not always the case that herding is going to be more productive than hunter-gatherers, especially in the case where the fish are jumping out of the water, but let's go to this scenario right over here. So this is, we can call this a pastoral lifestyle. So this is two, I'll call it pastoral. And over here is the realization that, look, you have all of this vegetation that maybe humans can't consume, but there are other animals that can consume this vegetation, and they can turn those calories into calories that can be consumed by a human, and namely, the calories are themselves. So this gentleman right over here, after he gets these sheep to be nice and fat, he can either eat the sheep or he can drink their milk. So one way to think about it is these cattle or these sheep right over here, by herding them and letting them eat the grass, he's turning non-human consumable calories into human consumable calories."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this is two, I'll call it pastoral. And over here is the realization that, look, you have all of this vegetation that maybe humans can't consume, but there are other animals that can consume this vegetation, and they can turn those calories into calories that can be consumed by a human, and namely, the calories are themselves. So this gentleman right over here, after he gets these sheep to be nice and fat, he can either eat the sheep or he can drink their milk. So one way to think about it is these cattle or these sheep right over here, by herding them and letting them eat the grass, he's turning non-human consumable calories into human consumable calories. And so for the sake of our thought experiment, let's say we get a 10 times increase in the human consumable calories per square kilometer. So now instead of 200, we're up to 2,000. And so instead of one human per square kilometer, we have enough calories per square kilometer per day to support one human."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So one way to think about it is these cattle or these sheep right over here, by herding them and letting them eat the grass, he's turning non-human consumable calories into human consumable calories. And so for the sake of our thought experiment, let's say we get a 10 times increase in the human consumable calories per square kilometer. So now instead of 200, we're up to 2,000. And so instead of one human per square kilometer, we have enough calories per square kilometer per day to support one human. So instead of, sorry, instead of 0.1, we can now support one human. So in that 10 square kilometers, we can now support 10 people. In 100 square kilometers, we could now support 100 people."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And so instead of one human per square kilometer, we have enough calories per square kilometer per day to support one human. So instead of, sorry, instead of 0.1, we can now support one human. So in that 10 square kilometers, we can now support 10 people. In 100 square kilometers, we could now support 100 people. Now, the next stage, and I'm skipping a bunch of stages, so because you have things like subsistence agriculture and various forms, and they're not going to be equally productive, and it depends what the land is like, and it depends what the tools are at your disposal. But the next stage that I'll just kind of jump to, we can call traditional agriculture. So this right over here, let's call that traditional agriculture."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In 100 square kilometers, we could now support 100 people. Now, the next stage, and I'm skipping a bunch of stages, so because you have things like subsistence agriculture and various forms, and they're not going to be equally productive, and it depends what the land is like, and it depends what the tools are at your disposal. But the next stage that I'll just kind of jump to, we can call traditional agriculture. So this right over here, let's call that traditional agriculture. And that's this one over here as well. So both of these, I'm gonna call traditional agriculture. And for the purposes of this video, the difference between traditional agriculture and modern agriculture, in traditional agriculture, you didn't have mechanization, so you didn't, or very primitive mechanization."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So this right over here, let's call that traditional agriculture. And that's this one over here as well. So both of these, I'm gonna call traditional agriculture. And for the purposes of this video, the difference between traditional agriculture and modern agriculture, in traditional agriculture, you didn't have mechanization, so you didn't, or very primitive mechanization. You definitely did not have fossil fuel-based engines. You didn't have modern pesticides. You did not have modern genetically engineered crops."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And for the purposes of this video, the difference between traditional agriculture and modern agriculture, in traditional agriculture, you didn't have mechanization, so you didn't, or very primitive mechanization. You definitely did not have fossil fuel-based engines. You didn't have modern pesticides. You did not have modern genetically engineered crops. But you did have some of the basic science of breeding crops and irrigating and using animals as tools. So in this stage right over here, and once again, it completely depends on where you are on the planet, how fertile the land is, how good your tools are, what crops you're actually producing. Let's assume that we got a hundredfold increase in productivity."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You did not have modern genetically engineered crops. But you did have some of the basic science of breeding crops and irrigating and using animals as tools. So in this stage right over here, and once again, it completely depends on where you are on the planet, how fertile the land is, how good your tools are, what crops you're actually producing. Let's assume that we got a hundredfold increase in productivity. And looking at some of the historical records, it looks like, depending on, once again, where you are, that's not out of the realm of possibility. So you have a hundredfold increase. So instead of 2,000 calories per square kilometer per day, you can get 200,000."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Let's assume that we got a hundredfold increase in productivity. And looking at some of the historical records, it looks like, depending on, once again, where you are, that's not out of the realm of possibility. So you have a hundredfold increase. So instead of 2,000 calories per square kilometer per day, you can get 200,000. 200,000 calories per square kilometer per day. And now you could support 100 humans per square kilometer, if you wanted to. So you might not have a hundred humans."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So instead of 2,000 calories per square kilometer per day, you can get 200,000. 200,000 calories per square kilometer per day. And now you could support 100 humans per square kilometer, if you wanted to. So you might not have a hundred humans. One, not all the land you might be able to farm from. Or there are other limits on the population for whatever they might be. But the important thing to think about this upper bound, in this traditional, if you are able to get this type of productivity from your land, and you're able to, in theory, support a hundred people per square kilometer, that means if all of a sudden you have 200 people living, there may be everyone's migrated to this land, because it seems especially fertile, or all the really smart farmers live there, then all of a sudden, not everyone's going to be able to get 2,000 calories a day."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you might not have a hundred humans. One, not all the land you might be able to farm from. Or there are other limits on the population for whatever they might be. But the important thing to think about this upper bound, in this traditional, if you are able to get this type of productivity from your land, and you're able to, in theory, support a hundred people per square kilometer, that means if all of a sudden you have 200 people living, there may be everyone's migrated to this land, because it seems especially fertile, or all the really smart farmers live there, then all of a sudden, not everyone's going to be able to get 2,000 calories a day. Some people might get malnourished. Other people might actually starve. There's this upper bound on the actual number of people that can be there, based on how productive the land actually is."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the important thing to think about this upper bound, in this traditional, if you are able to get this type of productivity from your land, and you're able to, in theory, support a hundred people per square kilometer, that means if all of a sudden you have 200 people living, there may be everyone's migrated to this land, because it seems especially fertile, or all the really smart farmers live there, then all of a sudden, not everyone's going to be able to get 2,000 calories a day. Some people might get malnourished. Other people might actually starve. There's this upper bound on the actual number of people that can be there, based on how productive the land actually is. And now let's move over to modern agriculture. And we've already talked a little bit about what exactly is modern agriculture. You have machines like this combine over here that does a lot of the human labor."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's this upper bound on the actual number of people that can be there, based on how productive the land actually is. And now let's move over to modern agriculture. And we've already talked a little bit about what exactly is modern agriculture. You have machines like this combine over here that does a lot of the human labor. One human can, and I'll talk about the different dimensions because there's actually two dimensions here. How much calories can you get from the land? And how much energy can one human, how much labor can one human input into the land using tools at their disposal?"}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You have machines like this combine over here that does a lot of the human labor. One human can, and I'll talk about the different dimensions because there's actually two dimensions here. How much calories can you get from the land? And how much energy can one human, how much labor can one human input into the land using tools at their disposal? So in this case, cattle, you need this ox pulling this plow, or in this case, this combine that's fueled by fossil fuels. But in modern agriculture, because of all of the things, you have these amazing tools, you have genetically engineered crops, you have modern pesticides, and not everyone is a fan of all of these things, but they have hugely increased our productivity. So you have modern agriculture."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And how much energy can one human, how much labor can one human input into the land using tools at their disposal? So in this case, cattle, you need this ox pulling this plow, or in this case, this combine that's fueled by fossil fuels. But in modern agriculture, because of all of the things, you have these amazing tools, you have genetically engineered crops, you have modern pesticides, and not everyone is a fan of all of these things, but they have hugely increased our productivity. So you have modern agriculture. And let's say that you get another factor of 10 from traditional agriculture. So now you can get 2 million calories per square kilometer per day, or you can support 1,000 humans per square kilometers. And once again, this right over here, this right over here is an upper bound."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So you have modern agriculture. And let's say that you get another factor of 10 from traditional agriculture. So now you can get 2 million calories per square kilometer per day, or you can support 1,000 humans per square kilometers. And once again, this right over here, this right over here is an upper bound. And just to give a sense, and I picked these numbers just so that the numbers would be clean. I looked at some historical records. These aren't completely out of line with what it looks like humans have been able to do in the past."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And once again, this right over here, this right over here is an upper bound. And just to give a sense, and I picked these numbers just so that the numbers would be clean. I looked at some historical records. These aren't completely out of line with what it looks like humans have been able to do in the past. But to give you a sense of what human population densities look like right now and why this upper bound seems to be right about correct, in a place like the United States, so in the United States, the population density is 30 people per square kilometer. So this is 30 humans per square kilometer. In a more dense country or significantly more dense country like India, the population density is 300 humans per square kilometer."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These aren't completely out of line with what it looks like humans have been able to do in the past. But to give you a sense of what human population densities look like right now and why this upper bound seems to be right about correct, in a place like the United States, so in the United States, the population density is 30 people per square kilometer. So this is 30 humans per square kilometer. In a more dense country or significantly more dense country like India, the population density is 300 humans per square kilometer. And in the most population dense country in the world, which is where I come from, or actually I was born in New Orleans, but where some of my ancestors came from, which is Bangladesh. So there's a lot of people like me, I guess. In Bangladesh, you have a population density of 900 humans per square kilometer."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In a more dense country or significantly more dense country like India, the population density is 300 humans per square kilometer. And in the most population dense country in the world, which is where I come from, or actually I was born in New Orleans, but where some of my ancestors came from, which is Bangladesh. So there's a lot of people like me, I guess. In Bangladesh, you have a population density of 900 humans per square kilometer. To some degree, this is a testament to the fertility of the land and whatever else. But this is pretty near the limits, depending on agricultural productivity and whatnot in the land, of modern technology. So it really makes you think if you don't get population under control, you might end up with some of these kind of hitting the wall type of scenarios."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In Bangladesh, you have a population density of 900 humans per square kilometer. To some degree, this is a testament to the fertility of the land and whatever else. But this is pretty near the limits, depending on agricultural productivity and whatnot in the land, of modern technology. So it really makes you think if you don't get population under control, you might end up with some of these kind of hitting the wall type of scenarios. And so the last thing I want you to think about, and this is what I refer to a little bit more, is just think about those two dimensions, because sometimes they get a little bit muddled. One is the kind of the productivity of land, productivity of land, and then the other is the productivity of labor, productivity of labor. So right over here in a hunter-gatherer, they're not getting money calories from their land, so they're right over there."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So it really makes you think if you don't get population under control, you might end up with some of these kind of hitting the wall type of scenarios. And so the last thing I want you to think about, and this is what I refer to a little bit more, is just think about those two dimensions, because sometimes they get a little bit muddled. One is the kind of the productivity of land, productivity of land, and then the other is the productivity of labor, productivity of labor. So right over here in a hunter-gatherer, they're not getting money calories from their land, so they're right over there. And the humans have to do all the labor. They don't have animals helping them in any way. They definitely don't have robots or any type of engines helping them in any way."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So right over here in a hunter-gatherer, they're not getting money calories from their land, so they're right over there. And the humans have to do all the labor. They don't have animals helping them in any way. They definitely don't have robots or any type of engines helping them in any way. And so they have to spend a lot of human time and a lot of human labor doing the work, getting that productivity from the land. But as we progress, so as we progress with things that aid humans, so for example, if all of a sudden you have cattle helping you or you have other tools that help you, you get more human productivity. So less and less of human labor has to be used to get that productivity of the land, and so maybe other humans can go do other things like paint pictures or become blacksmiths or whatever."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They definitely don't have robots or any type of engines helping them in any way. And so they have to spend a lot of human time and a lot of human labor doing the work, getting that productivity from the land. But as we progress, so as we progress with things that aid humans, so for example, if all of a sudden you have cattle helping you or you have other tools that help you, you get more human productivity. So less and less of human labor has to be used to get that productivity of the land, and so maybe other humans can go do other things like paint pictures or become blacksmiths or whatever. And in this direction, you get higher productivity per unit of land. And so that comes from moving from hunter-gatherer to a pastoral lifestyle, to traditional farming with irrigation, to modern farming. And so on this graph right over here, kind of tools for the individuals move us up, getting more productivity of the land move us to the right."}, {"video_title": "Land productivity limiting human population Cosmology & Astronomy Khan Academy.mp3", "Sentence": "So less and less of human labor has to be used to get that productivity of the land, and so maybe other humans can go do other things like paint pictures or become blacksmiths or whatever. And in this direction, you get higher productivity per unit of land. And so that comes from moving from hunter-gatherer to a pastoral lifestyle, to traditional farming with irrigation, to modern farming. And so on this graph right over here, kind of tools for the individuals move us up, getting more productivity of the land move us to the right. Modern agriculture gets us right over here. So we're getting much, much more calories per unit land, and we're getting much, much more calories per unit labor. So you need a much smaller percentage of the human population actually involved in the farming."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Here we have a pair of enantiomers. On the left we have R-Carvone, and on the right we have S-Carvone. Both of these compounds have the same melting point, the same boiling point, and the same density. However, there are a few important differences. R-Carvone is the major component of spearmint oil, so R-Carvone smells like spearmint. S-Carvone is the major component of caraway oil, and this smells like caraway. It's pretty amazing that our noses can tell the difference between these two enantiomers."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "However, there are a few important differences. R-Carvone is the major component of spearmint oil, so R-Carvone smells like spearmint. S-Carvone is the major component of caraway oil, and this smells like caraway. It's pretty amazing that our noses can tell the difference between these two enantiomers. The science of smell is a really, really fascinating topic. Another important difference between these two enantiomers is their optical activity. Enantiomers exhibit different behavior when exposed to plain polarized light."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It's pretty amazing that our noses can tell the difference between these two enantiomers. The science of smell is a really, really fascinating topic. Another important difference between these two enantiomers is their optical activity. Enantiomers exhibit different behavior when exposed to plain polarized light. Let's examine what I mean by that. Here we have our unpolarized light, which is usually from a sodium lamp. Normally we're talking about the D-line of sodium with a wavelength of 589 nanometers."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Enantiomers exhibit different behavior when exposed to plain polarized light. Let's examine what I mean by that. Here we have our unpolarized light, which is usually from a sodium lamp. Normally we're talking about the D-line of sodium with a wavelength of 589 nanometers. This unpolarized light tries to pass through our filter. This filter here, if you look at it, notice we have these slits, these vertical slits. Not all of the unpolarized light can pass through."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Normally we're talking about the D-line of sodium with a wavelength of 589 nanometers. This unpolarized light tries to pass through our filter. This filter here, if you look at it, notice we have these slits, these vertical slits. Not all of the unpolarized light can pass through. Only this vertical plane of light can pass through one of our vertical filters. We now have a plane of polarized light. This plane of polarized light comes to a tube."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Not all of the unpolarized light can pass through. Only this vertical plane of light can pass through one of our vertical filters. We now have a plane of polarized light. This plane of polarized light comes to a tube. This is the tube of the polarimeter. That's what this device is called. Let me go ahead and write this down."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This plane of polarized light comes to a tube. This is the tube of the polarimeter. That's what this device is called. Let me go ahead and write this down. This is our tube. In our tube, we have a solution of an optically active compound. Let me go ahead and draw in some of our compound here."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and write this down. This is our tube. In our tube, we have a solution of an optically active compound. Let me go ahead and draw in some of our compound here. Imagine a solution. This compound is dissolved in something. Our plane of polarized light rotates when it hits our compound."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and draw in some of our compound here. Imagine a solution. This compound is dissolved in something. Our plane of polarized light rotates when it hits our compound. Imagine this plane here, which starts off up and down. It starts to rotate. The more molecules it hits here, the more it rotates."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Our plane of polarized light rotates when it hits our compound. Imagine this plane here, which starts off up and down. It starts to rotate. The more molecules it hits here, the more it rotates. By the time it leaves our tube, it's at a different angle from how it entered. Next we have the analyzer portion. This is our analyzer."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The more molecules it hits here, the more it rotates. By the time it leaves our tube, it's at a different angle from how it entered. Next we have the analyzer portion. This is our analyzer. Imagine that you are right here. Your eye is here looking at the analyzer. The analyzer, let's say, started off with the slits up and down, just like we had on the filter here."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is our analyzer. Imagine that you are right here. Your eye is here looking at the analyzer. The analyzer, let's say, started off with the slits up and down, just like we had on the filter here. That wouldn't allow this plane of light to pass through. We would have to rotate the analyzer to allow our plane of light to pass through. You can see I've already shown that with this drawing."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The analyzer, let's say, started off with the slits up and down, just like we had on the filter here. That wouldn't allow this plane of light to pass through. We would have to rotate the analyzer to allow our plane of light to pass through. You can see I've already shown that with this drawing. The slits are now going in this direction to allow our plane to pass through. We had to rotate our analyzer to the right to allow that plane to get through. This angle here, alpha, is called the observed rotation."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "You can see I've already shown that with this drawing. The slits are now going in this direction to allow our plane to pass through. We had to rotate our analyzer to the right to allow that plane to get through. This angle here, alpha, is called the observed rotation. This is the observed rotation. In this case, we had to rotate the analyzer to the right. Let's start up here."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This angle here, alpha, is called the observed rotation. This is the observed rotation. In this case, we had to rotate the analyzer to the right. Let's start up here. Let's say we started vertically. The plane of light was rotated to the right when it ran into our compound here. That means that we rotated the analyzer to the right."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's start up here. Let's say we started vertically. The plane of light was rotated to the right when it ran into our compound here. That means that we rotated the analyzer to the right. That's said to be an observed rotation that is positive. That's a positive rotation. That's a clockwise rotation."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That means that we rotated the analyzer to the right. That's said to be an observed rotation that is positive. That's a positive rotation. That's a clockwise rotation. This is also called dextrorotatory. Let me write that down here. This is dextrorotatory."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That's a clockwise rotation. This is also called dextrorotatory. Let me write that down here. This is dextrorotatory. What if our plane was rotated to the left? What if the light was rotated to the left? Let's say we started off vertical and our light was rotated in this direction this time."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is dextrorotatory. What if our plane was rotated to the left? What if the light was rotated to the left? Let's say we started off vertical and our light was rotated in this direction this time. We had to rotate the analyzer to the left. The observed rotation is said to be negative. This is a negative rotation, a counterclockwise rotation."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Let's say we started off vertical and our light was rotated in this direction this time. We had to rotate the analyzer to the left. The observed rotation is said to be negative. This is a negative rotation, a counterclockwise rotation. This is called levorotatory. Let me write that in here. This is levorotatory."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a negative rotation, a counterclockwise rotation. This is called levorotatory. Let me write that in here. This is levorotatory. The observed rotation alpha depends on the number of molecules that are hit by our polarized light. Let's say we increase the concentration. I'm going to go ahead and draw some more red dots in here to indicate increasing the concentration of our compound."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is levorotatory. The observed rotation alpha depends on the number of molecules that are hit by our polarized light. Let's say we increase the concentration. I'm going to go ahead and draw some more red dots in here to indicate increasing the concentration of our compound. That means that our light is going to rotate even more. Our light starts off vertical. It runs into more molecules."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to go ahead and draw some more red dots in here to indicate increasing the concentration of our compound. That means that our light is going to rotate even more. Our light starts off vertical. It runs into more molecules. It rotates even more. It exits our tube at a different angle. That changes our observed rotation."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "It runs into more molecules. It rotates even more. It exits our tube at a different angle. That changes our observed rotation. It turns out if you double the concentration, you double the observed rotation. You can also change the observed rotation by changing the path length, by changing the length of this tube here. I'll call that length L. If you hold the concentration constant and you double the path length, you double the observed rotation because that means that your light is running into more molecules because your tube is longer."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That changes our observed rotation. It turns out if you double the concentration, you double the observed rotation. You can also change the observed rotation by changing the path length, by changing the length of this tube here. I'll call that length L. If you hold the concentration constant and you double the path length, you double the observed rotation because that means that your light is running into more molecules because your tube is longer. Let's take these ideas of observed rotation and concentration and path length and let's turn them into an equation here. If we take the observed rotation, which is alpha, and this is measured in degrees. Something could rotate."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'll call that length L. If you hold the concentration constant and you double the path length, you double the observed rotation because that means that your light is running into more molecules because your tube is longer. Let's take these ideas of observed rotation and concentration and path length and let's turn them into an equation here. If we take the observed rotation, which is alpha, and this is measured in degrees. Something could rotate. You could have an angle in here. Think about an angle in degrees for your observed rotation. If you divide your observed rotation by the concentration of what's in your tube, and the concentration is in grams per ml, and then the concentration is multiplied by the path length L, which is in decimeters."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Something could rotate. You could have an angle in here. Think about an angle in degrees for your observed rotation. If you divide your observed rotation by the concentration of what's in your tube, and the concentration is in grams per ml, and then the concentration is multiplied by the path length L, which is in decimeters. This is in decimeters. You'll get something called the specific rotation. That would be alpha in brackets."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "If you divide your observed rotation by the concentration of what's in your tube, and the concentration is in grams per ml, and then the concentration is multiplied by the path length L, which is in decimeters. This is in decimeters. You'll get something called the specific rotation. That would be alpha in brackets. This is the specific rotation. The nice thing about the specific rotation is this is a constant. This is a constant."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "That would be alpha in brackets. This is the specific rotation. The nice thing about the specific rotation is this is a constant. This is a constant. Your observed rotation might change depending on what concentration you're using, depending on what your path length is. If you take the observed rotation and you divide it by the concentration times the path length, you get the specific rotation. Having this as a physical constant is very useful because you can look up the specific rotations for specific compounds."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This is a constant. Your observed rotation might change depending on what concentration you're using, depending on what your path length is. If you take the observed rotation and you divide it by the concentration times the path length, you get the specific rotation. Having this as a physical constant is very useful because you can look up the specific rotations for specific compounds. For example, you could look up the specific rotation for, we talked about S-carvone earlier. That would be the specific rotation. This can also change depending on temperature and wavelength."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Having this as a physical constant is very useful because you can look up the specific rotations for specific compounds. For example, you could look up the specific rotation for, we talked about S-carvone earlier. That would be the specific rotation. This can also change depending on temperature and wavelength. You need to specify the temperature here and the wavelength here. For S-carvone at 20 degrees, let me write this up here. The specific rotation of S-carvone at 20 degrees Celsius and using the D line of sodium, this is equal to positive 61."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This can also change depending on temperature and wavelength. You need to specify the temperature here and the wavelength here. For S-carvone at 20 degrees, let me write this up here. The specific rotation of S-carvone at 20 degrees Celsius and using the D line of sodium, this is equal to positive 61. That's the specific rotation of S-carvone. The four specific rotation is normally unitless. Normally you don't see anything with this number."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "The specific rotation of S-carvone at 20 degrees Celsius and using the D line of sodium, this is equal to positive 61. That's the specific rotation of S-carvone. The four specific rotation is normally unitless. Normally you don't see anything with this number. However, a lot of times you do. I've seen a degree sign here for a lot of things. I'm going to take it off right here because usually the degree sign is left for the observed rotation."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "Normally you don't see anything with this number. However, a lot of times you do. I've seen a degree sign here for a lot of things. I'm going to take it off right here because usually the degree sign is left for the observed rotation. That's how you would see a specific rotation. We just saw that S-carvone has a specific rotation of positive 61. This enantiomer is dextrorotatory."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "I'm going to take it off right here because usually the degree sign is left for the observed rotation. That's how you would see a specific rotation. We just saw that S-carvone has a specific rotation of positive 61. This enantiomer is dextrorotatory. We have a positive rotation. We put a positive sign up here. R-carvone has a specific rotation of negative 61."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This enantiomer is dextrorotatory. We have a positive rotation. We put a positive sign up here. R-carvone has a specific rotation of negative 61. This enantiomer is levorotatory. We have a negative rotation. We put a negative sign up here."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "R-carvone has a specific rotation of negative 61. This enantiomer is levorotatory. We have a negative rotation. We put a negative sign up here. Notice the difference in the specific rotations for our pair of enantiomers. Enantiomers have specific rotations that are equal in magnitude. This one's 61 and this one's 61, but opposite in sign."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "We put a negative sign up here. Notice the difference in the specific rotations for our pair of enantiomers. Enantiomers have specific rotations that are equal in magnitude. This one's 61 and this one's 61, but opposite in sign. This one's negative and this one is positive. Louis Pasteur was the first one to realize this relationship. Pretty amazing that he was able to figure this out."}, {"video_title": "Optical activity Stereochemistry Organic chemistry Khan Academy.mp3", "Sentence": "This one's 61 and this one's 61, but opposite in sign. This one's negative and this one is positive. Louis Pasteur was the first one to realize this relationship. Pretty amazing that he was able to figure this out. Also, I wanted to point out that R and S have nothing to do with negative and positive. The fact that this is S-carvone has nothing to do with the fact that this is positive. R and S are used to assign a configuration to a chiral center."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "In this video, we're going to take a look at two ways to reduce alkynes. The first way is a reaction we've seen before. This is the hydrogenation reaction. And we saw it before when we hydrogenated alkenes to form alkanes. Here we're going to hydrogenate an alkyne to form an alkene. And to do a hydrogenation reaction, we need some hydrogen gas, so some H2 right here, and then a metal catalyst. So we're going to use linalar palladium, which is a special type of catalyst."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we saw it before when we hydrogenated alkenes to form alkanes. Here we're going to hydrogenate an alkyne to form an alkene. And to do a hydrogenation reaction, we need some hydrogen gas, so some H2 right here, and then a metal catalyst. So we're going to use linalar palladium, which is a special type of catalyst. It will catalyze the reduction of the alkyne on the left, the alkene on the right. However, the reduction of the alkene to the alkane down here is slow, so slow that we can stop it if our goal is to just make an alkene. So this reaction will form a cis alkene."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we're going to use linalar palladium, which is a special type of catalyst. It will catalyze the reduction of the alkyne on the left, the alkene on the right. However, the reduction of the alkene to the alkane down here is slow, so slow that we can stop it if our goal is to just make an alkene. So this reaction will form a cis alkene. And it was a syn addition of our hydrogens. So we're going to get the two hydrogens adding on to the same side. And this has to do with the mechanism of a hydrogenation reaction."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this reaction will form a cis alkene. And it was a syn addition of our hydrogens. So we're going to get the two hydrogens adding on to the same side. And this has to do with the mechanism of a hydrogenation reaction. So you can check out the earlier video on hydrogenation of alkenes to see more details. So linalar palladium, a poison catalyst, it will reduce an alkyne to an alkene. It will produce a cis alkene."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And this has to do with the mechanism of a hydrogenation reaction. So you can check out the earlier video on hydrogenation of alkenes to see more details. So linalar palladium, a poison catalyst, it will reduce an alkyne to an alkene. It will produce a cis alkene. So that's how to make a cis alkene. Let's take a look at how to make a trans alkene. So how do we reduce an alkyne to make a trans alkene?"}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It will produce a cis alkene. So that's how to make a cis alkene. Let's take a look at how to make a trans alkene. So how do we reduce an alkyne to make a trans alkene? So here is our alkyne. So we have our triple bond like that. And we're going to add sodium metal."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So how do we reduce an alkyne to make a trans alkene? So here is our alkyne. So we have our triple bond like that. And we're going to add sodium metal. And we're also going to add liquid ammonia like that. So we're going to form a trans alkene. So I'm going to put, this time my two hydrogens are going to be on opposite sides of each other."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we're going to add sodium metal. And we're also going to add liquid ammonia like that. So we're going to form a trans alkene. So I'm going to put, this time my two hydrogens are going to be on opposite sides of each other. So this is formation of a trans alkene like that. And it does this by an anti-addition of hydrogens. So these are adding from opposite sides like that."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I'm going to put, this time my two hydrogens are going to be on opposite sides of each other. So this is formation of a trans alkene like that. And it does this by an anti-addition of hydrogens. So these are adding from opposite sides like that. Let's take a look at the mechanism to form a trans alkene. So I start with my alkyne. So I go ahead and put in my carbons there and put R group on the left side."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these are adding from opposite sides like that. Let's take a look at the mechanism to form a trans alkene. So I start with my alkyne. So I go ahead and put in my carbons there and put R group on the left side. And I'll make this an R prime group to distinguish it from the R group over there. So we start with sodium, which we know, being in group one, has one valence electron like that. And in the first step of the mechanism, the sodium atom is going to donate its valence electron to the alkyne."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So I go ahead and put in my carbons there and put R group on the left side. And I'll make this an R prime group to distinguish it from the R group over there. So we start with sodium, which we know, being in group one, has one valence electron like that. And in the first step of the mechanism, the sodium atom is going to donate its valence electron to the alkyne. So when we're showing the movement of one electron, we use a half-headed arrow. So I'm going to show this electron moving over here. But it's only one electron."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And in the first step of the mechanism, the sodium atom is going to donate its valence electron to the alkyne. So when we're showing the movement of one electron, we use a half-headed arrow. So I'm going to show this electron moving over here. But it's only one electron. So I'm only going to do a half-headed arrow like that, not a full-headed arrow. So one of these bonds here between the carbons is going to break. And one of the electrons is going to move over here to this carbon like that."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But it's only one electron. So I'm only going to do a half-headed arrow like that, not a full-headed arrow. So one of these bonds here between the carbons is going to break. And one of the electrons is going to move over here to this carbon like that. And one of the electrons is going to move over to the carbon on the left. So let's go ahead and draw the result of all those electrons moving around. So we have an R group here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And one of the electrons is going to move over here to this carbon like that. And one of the electrons is going to move over to the carbon on the left. So let's go ahead and draw the result of all those electrons moving around. So we have an R group here. And we had a triple bond. But now we only have a double bond between our two carbons. And then we have R prime over here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have an R group here. And we had a triple bond. But now we only have a double bond between our two carbons. And then we have R prime over here. So the carbon on the right picked up an electron from sodium. And it also picked up an electron from the breaking of that one bond there. So now it has two electrons around it like that, which gives us a negative 1 formal charge on this carbon."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we have R prime over here. So the carbon on the right picked up an electron from sodium. And it also picked up an electron from the breaking of that one bond there. So now it has two electrons around it like that, which gives us a negative 1 formal charge on this carbon. So it's a carbanion. It's an anion here. The carbon on the left picked up one electron for the breaking of that bond like that."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now it has two electrons around it like that, which gives us a negative 1 formal charge on this carbon. So it's a carbanion. It's an anion here. The carbon on the left picked up one electron for the breaking of that bond like that. So that's a radical. That's a thing we haven't talked about before. So we actually form what's called a radical anion here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The carbon on the left picked up one electron for the breaking of that bond like that. So that's a radical. That's a thing we haven't talked about before. So we actually form what's called a radical anion here. So let's go ahead and write that. This is a radical anion. So radical because there's an unpaired electron there."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we actually form what's called a radical anion here. So let's go ahead and write that. This is a radical anion. So radical because there's an unpaired electron there. And then it also has a carbanion in the same molecule like that. So we have these electrons that are pretty close together, at least how I've drawn them. So we know that electrons are all negatively charged."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So radical because there's an unpaired electron there. And then it also has a carbanion in the same molecule like that. So we have these electrons that are pretty close together, at least how I've drawn them. So we know that electrons are all negatively charged. So all these electrons are going to repel each other. So this isn't the most stable way for this molecule to have in terms of a confirmation. These electrons are going to repel."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we know that electrons are all negatively charged. So all these electrons are going to repel each other. So this isn't the most stable way for this molecule to have in terms of a confirmation. These electrons are going to repel. And they're going to want to try to be as far away from each other as they possibly can. So what's going to happen is we have our two carbons right here. And let's say that these two electrons stay over here on this side."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "These electrons are going to repel. And they're going to want to try to be as far away from each other as they possibly can. So what's going to happen is we have our two carbons right here. And let's say that these two electrons stay over here on this side. This one electron is going to go over to the opposite side. They're going to try to get as far away from each other as they possibly can. And same thing with these R groups here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's say that these two electrons stay over here on this side. This one electron is going to go over to the opposite side. They're going to try to get as far away from each other as they possibly can. And same thing with these R groups here. So this R group is going to try to get as far away from this R prime group as it possibly can. So this trans-confirmation is the more stable one. So this is our negatively charged carbanion right here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And same thing with these R groups here. So this R group is going to try to get as far away from this R prime group as it possibly can. So this trans-confirmation is the more stable one. So this is our negatively charged carbanion right here. So in the next step of the mechanism, we remember ammonia is present. So let's go ahead and draw an ammonia molecule floating around like that. So here is our ammonia molecule."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is our negatively charged carbanion right here. So in the next step of the mechanism, we remember ammonia is present. So let's go ahead and draw an ammonia molecule floating around like that. So here is our ammonia molecule. And the carbanion is going to act as a base. And it's going to take a proton from the ammonia molecule. So this lone pair of electrons is going to form a new bond with this proton."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So here is our ammonia molecule. And the carbanion is going to act as a base. And it's going to take a proton from the ammonia molecule. So this lone pair of electrons is going to form a new bond with this proton. And these electrons are going to kick off onto the nitrogen. So let's go ahead and draw the result of that acid-base reaction. So now we have our two carbons with an R group right here, R prime right here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this lone pair of electrons is going to form a new bond with this proton. And these electrons are going to kick off onto the nitrogen. So let's go ahead and draw the result of that acid-base reaction. So now we have our two carbons with an R group right here, R prime right here. And now this carbon on the right is bonded to a proton, bonded to a hydrogen like that. And then we still have our radical down here. So there's one electron on that carbon as well."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now we have our two carbons with an R group right here, R prime right here. And now this carbon on the right is bonded to a proton, bonded to a hydrogen like that. And then we still have our radical down here. So there's one electron on that carbon as well. So the next step of our mechanism, well, there's plenty of sodium present. So here's a sodium atom with one valence electron. The sodium is going to donate this electron to this carbon."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So there's one electron on that carbon as well. So the next step of our mechanism, well, there's plenty of sodium present. So here's a sodium atom with one valence electron. The sodium is going to donate this electron to this carbon. So we'll just use a half-headed arrow to show the movement of one electron. So if that sodium atom donates that one valence electron to that carbon, let's go ahead and draw the results of that. So we have two carbons double bonded, an R group over here, a hydrogen, and an R prime."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The sodium is going to donate this electron to this carbon. So we'll just use a half-headed arrow to show the movement of one electron. So if that sodium atom donates that one valence electron to that carbon, let's go ahead and draw the results of that. So we have two carbons double bonded, an R group over here, a hydrogen, and an R prime. And this carbon had one electron around it. It just picked up one more from a sodium atom. So it's like that, which would give it a negative 1 formal charge."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we have two carbons double bonded, an R group over here, a hydrogen, and an R prime. And this carbon had one electron around it. It just picked up one more from a sodium atom. So it's like that, which would give it a negative 1 formal charge. So this carbon has a negative 1 formal charge. So let's go ahead and draw that negative 1 formal charge. It's a carb anion."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's like that, which would give it a negative 1 formal charge. So this carbon has a negative 1 formal charge. So let's go ahead and draw that negative 1 formal charge. It's a carb anion. And once again, ammonia is floating around. So let's go ahead and draw ammonia right here. So NH3 like that."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's a carb anion. And once again, ammonia is floating around. So let's go ahead and draw ammonia right here. So NH3 like that. And the same thing is going to happen as did before. The negative charge is going to grab a proton. It's going to act as a base."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So NH3 like that. And the same thing is going to happen as did before. The negative charge is going to grab a proton. It's going to act as a base. And these electrons are going to kick off onto the nitrogen here. And so we protonate our carbon ion. And we have completed our mechanism."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's going to act as a base. And these electrons are going to kick off onto the nitrogen here. And so we protonate our carbon ion. And we have completed our mechanism. Because now we have our two R groups across from each other. And we added on two hydrogens across from each other as well like that. So we formed a trans alkene."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And we have completed our mechanism. Because now we have our two R groups across from each other. And we added on two hydrogens across from each other as well like that. So we formed a trans alkene. So that's the mechanism to form a trans alkene. Let's look at a few examples. So let's start with this alkene right here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we formed a trans alkene. So that's the mechanism to form a trans alkene. Let's look at a few examples. So let's start with this alkene right here. So carbon triple bonded to another carbon. And we'll put a methyl group on each side like that. So let's do a few different reactions with the same substrate here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's start with this alkene right here. So carbon triple bonded to another carbon. And we'll put a methyl group on each side like that. So let's do a few different reactions with the same substrate here. So our first reaction will just be a normal hydrogenation with hydrogen gas. And let's use platinum as our catalyst. So this is not a poison catalyst."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's do a few different reactions with the same substrate here. So our first reaction will just be a normal hydrogenation with hydrogen gas. And let's use platinum as our catalyst. So this is not a poison catalyst. This is a normal catalyst. So what's going to happen is first you're going to reduce the alkyne to an alkene. And then, since there's no way of stopping it, it's going to reduce the alkene to an alkane."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is not a poison catalyst. This is a normal catalyst. So what's going to happen is first you're going to reduce the alkyne to an alkene. And then, since there's no way of stopping it, it's going to reduce the alkene to an alkane. So this is going to reduce the alkyne all the way to an alkane. So if we go back up here to the beginning, remember we said that a poison catalyst will stop at the alkene. But if it's not a poison catalyst, it's just going to hydrogenate your alkene to an alkane down here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then, since there's no way of stopping it, it's going to reduce the alkene to an alkane. So this is going to reduce the alkyne all the way to an alkane. So if we go back up here to the beginning, remember we said that a poison catalyst will stop at the alkene. But if it's not a poison catalyst, it's just going to hydrogenate your alkene to an alkane down here. So this reaction is going to produce an alkane. Let's go ahead and draw the product. So we know that there are four carbons in my starting material, so there's going to be four carbons when I'm done here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But if it's not a poison catalyst, it's just going to hydrogenate your alkene to an alkane down here. So this reaction is going to produce an alkane. Let's go ahead and draw the product. So we know that there are four carbons in my starting material, so there's going to be four carbons when I'm done here. So these two carbons in the center here are going to turn into CH2s. And then on either side, we still have our CH3s. So this is going to form."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we know that there are four carbons in my starting material, so there's going to be four carbons when I'm done here. So these two carbons in the center here are going to turn into CH2s. And then on either side, we still have our CH3s. So this is going to form. So it's going to form butane as the product. This time, let's use hydrogen gas. And let's use lindlar palladium here."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this is going to form. So it's going to form butane as the product. This time, let's use hydrogen gas. And let's use lindlar palladium here. So lindlar palladium. This is our poisoned catalyst. So it's going to reduce our alkyne to an alkene, and then it's going to stop."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And let's use lindlar palladium here. So lindlar palladium. This is our poisoned catalyst. So it's going to reduce our alkyne to an alkene, and then it's going to stop. And you have to think, what kind of alkene will you get? You will get a cis-alkene. So if we draw our two hydrogens adding onto the same side, so now we have our methyl groups going like that."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's going to reduce our alkyne to an alkene, and then it's going to stop. And you have to think, what kind of alkene will you get? You will get a cis-alkene. So if we draw our two hydrogens adding onto the same side, so now we have our methyl groups going like that. So our methyl groups will be going like this, and this would be our product, a cis-alkene. Let's do one more. Same starting material."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So if we draw our two hydrogens adding onto the same side, so now we have our methyl groups going like that. So our methyl groups will be going like this, and this would be our product, a cis-alkene. Let's do one more. Same starting material. So this one right here, except this time, we're going to add sodium. And we're going to use ammonia as our solvent. And remember, this will reduce our alkyne to an alkene, but it will form a trans-alkene as your product."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Same starting material. So this one right here, except this time, we're going to add sodium. And we're going to use ammonia as our solvent. And remember, this will reduce our alkyne to an alkene, but it will form a trans-alkene as your product. So when you're drawing your product down here, you want to make sure that your two hydrogens are trans to each other. So they add on the mechanism, and then your two methyl groups would also be on the opposite side like that. So look very closely as to what you are reacting things with."}, {"video_title": "Reduction of alkynes Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And remember, this will reduce our alkyne to an alkene, but it will form a trans-alkene as your product. So when you're drawing your product down here, you want to make sure that your two hydrogens are trans to each other. So they add on the mechanism, and then your two methyl groups would also be on the opposite side like that. So look very closely as to what you are reacting things with. Is it a normal hydrogenation reaction? Is it a hydrogenation reaction with a poison catalyst, which would form a cis-alkene? Or is it reduction with sodium and ammonia, which will give you a trans-alkene?"}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we started with this alkene, and we got this alcohol, where the OH added on to the less substituted carbon. Let's take a look at the mechanism for this reaction. So in step one, we add our borane, BH3. And if we look down here at the dot structure, boron is sp2 hybridized, which means trigonal planar geometry around the boron and also an empty p orbital, which is capable of accepting a pair of electrons. Because, as you can see, boron has only six electrons around it, and the fact that it lacks an octet makes it very reactive. Actually, borane can react with itself, and that's why we have the THF here to stabilize it. So when borane approaches our alkene, right, so here's our alkene, so we have the two methyl groups right here, and then we have the two hydrogens on this carbon."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And if we look down here at the dot structure, boron is sp2 hybridized, which means trigonal planar geometry around the boron and also an empty p orbital, which is capable of accepting a pair of electrons. Because, as you can see, boron has only six electrons around it, and the fact that it lacks an octet makes it very reactive. Actually, borane can react with itself, and that's why we have the THF here to stabilize it. So when borane approaches our alkene, right, so here's our alkene, so we have the two methyl groups right here, and then we have the two hydrogens on this carbon. So here are the two hydrogens. The borane approaches the carbon on the left side of our double bond, so the boron gets closer to the carbon on the left side. And one reason for that is because these methyl groups are relatively bulky, right, so there's some steric hindrance that prevents the boron from getting too close to the carbon on the right side of the double bond."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So when borane approaches our alkene, right, so here's our alkene, so we have the two methyl groups right here, and then we have the two hydrogens on this carbon. So here are the two hydrogens. The borane approaches the carbon on the left side of our double bond, so the boron gets closer to the carbon on the left side. And one reason for that is because these methyl groups are relatively bulky, right, so there's some steric hindrance that prevents the boron from getting too close to the carbon on the right side of the double bond. And the pi electrons are going to attack the boron. Right, so the pi electrons are going to attack the empty orbital of the boron, and we're gonna form a bond between the carbon on the left side of the double bond and the boron. So these pi electrons are going to move into here."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And one reason for that is because these methyl groups are relatively bulky, right, so there's some steric hindrance that prevents the boron from getting too close to the carbon on the right side of the double bond. And the pi electrons are going to attack the boron. Right, so the pi electrons are going to attack the empty orbital of the boron, and we're gonna form a bond between the carbon on the left side of the double bond and the boron. So these pi electrons are going to move into here. And as that bond between this carbon and this boron is forming, we are withdrawing electron density from the carbon on the right. So I'm only gonna put a partial positive charge here because this is pretty much a concerted mechanism, this step is concerted. But as this bond forms, right, we're increasing the partial positive charge on this carbon, and that triggers a hydride shift."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these pi electrons are going to move into here. And as that bond between this carbon and this boron is forming, we are withdrawing electron density from the carbon on the right. So I'm only gonna put a partial positive charge here because this is pretty much a concerted mechanism, this step is concerted. But as this bond forms, right, we're increasing the partial positive charge on this carbon, and that triggers a hydride shift. Remember what hydride is, right, so hydrogen with two electrons, so a negative one formal charge. So we can think about a hydride being right here. So let me mark these electrons in blue, right, so these electrons down here in blue are attracted to the developing partial positive charge."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But as this bond forms, right, we're increasing the partial positive charge on this carbon, and that triggers a hydride shift. Remember what hydride is, right, so hydrogen with two electrons, so a negative one formal charge. So we can think about a hydride being right here. So let me mark these electrons in blue, right, so these electrons down here in blue are attracted to the developing partial positive charge. So we're gonna form a bond right here between this carbon and the hydrogen as the electrons in blue move in here. So the formation of this partial positive charge triggers the hydride shift, and we move those electrons in blue there to form a bond. So let's draw the product."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me mark these electrons in blue, right, so these electrons down here in blue are attracted to the developing partial positive charge. So we're gonna form a bond right here between this carbon and the hydrogen as the electrons in blue move in here. So the formation of this partial positive charge triggers the hydride shift, and we move those electrons in blue there to form a bond. So let's draw the product. So on the left side of where the double bond used to be, we would have BH2, and on the right side we would have H. We would have a methyl group coming out at us in space and a methyl group going away from us in space. On the left side we would have a hydrogen coming out at us and a hydrogen going away from us. So let's follow those electrons along."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw the product. So on the left side of where the double bond used to be, we would have BH2, and on the right side we would have H. We would have a methyl group coming out at us in space and a methyl group going away from us in space. On the left side we would have a hydrogen coming out at us and a hydrogen going away from us. So let's follow those electrons along. The electrons in magenta right here form this bond between the carbon and the boron, and the formation of the carbon-boron bond triggers the hydride shift. So the electrons in blue move into here, and we get that. All right, later in this mechanism we'll see that the OH is gonna go where the boron is, right?"}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons along. The electrons in magenta right here form this bond between the carbon and the boron, and the formation of the carbon-boron bond triggers the hydride shift. So the electrons in blue move into here, and we get that. All right, later in this mechanism we'll see that the OH is gonna go where the boron is, right? So the OH is going to go there. And so now we can talk about the regiochemistry and the stereochemistry for this reaction. So let's first think about regiochemistry."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "All right, later in this mechanism we'll see that the OH is gonna go where the boron is, right? So the OH is going to go there. And so now we can talk about the regiochemistry and the stereochemistry for this reaction. So let's first think about regiochemistry. Why does the OH add on to the less substituted carbon? Well, that's because of two reasons, right? One reason would be the steric hindrance, right?"}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's first think about regiochemistry. Why does the OH add on to the less substituted carbon? Well, that's because of two reasons, right? One reason would be the steric hindrance, right? So these methyl groups prevent the borane from getting on or getting close to the carbon on the right side of the double bond. So steric hindrance is one reason. And also the formation of the developing partial positive charge on this carbon is better stabilized by the presence of these methyl groups, right?"}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "One reason would be the steric hindrance, right? So these methyl groups prevent the borane from getting on or getting close to the carbon on the right side of the double bond. So steric hindrance is one reason. And also the formation of the developing partial positive charge on this carbon is better stabilized by the presence of these methyl groups, right? So we have two reasons, steric hindrance and the stabilizing factor of these methyls, right? Stabilizing the developing partially positive charge. That explains why the boron adds on to the carbon on the left side of the double bond."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And also the formation of the developing partial positive charge on this carbon is better stabilized by the presence of these methyl groups, right? So we have two reasons, steric hindrance and the stabilizing factor of these methyls, right? Stabilizing the developing partially positive charge. That explains why the boron adds on to the carbon on the left side of the double bond. And then eventually that's of course where the OH goes. So that explains the regiochemistry. And the stereochemistry, we see the OH and the H add on to the same side because this is pretty much concerted, right?"}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "That explains why the boron adds on to the carbon on the left side of the double bond. And then eventually that's of course where the OH goes. So that explains the regiochemistry. And the stereochemistry, we see the OH and the H add on to the same side because this is pretty much concerted, right? This pretty much happens at the same time. So your stereochemistry is locked in at this portion of your mechanism. Let's continue on with the mechanism."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And the stereochemistry, we see the OH and the H add on to the same side because this is pretty much concerted, right? This pretty much happens at the same time. So your stereochemistry is locked in at this portion of your mechanism. Let's continue on with the mechanism. Let me redraw this guy right here. So this is a little bit too complicated. Let me make it a little bit easier to look at."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let's continue on with the mechanism. Let me redraw this guy right here. So this is a little bit too complicated. Let me make it a little bit easier to look at. So we would have our methyl groups, right? And then we would have our BH2 right here. This process occurs two more times, all right?"}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me make it a little bit easier to look at. So we would have our methyl groups, right? And then we would have our BH2 right here. This process occurs two more times, all right? Repeat times two because there are two more hydrogens on the boron. And so we actually form a trialkylborane. So let me draw in the boron with three alkyl groups on it."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "This process occurs two more times, all right? Repeat times two because there are two more hydrogens on the boron. And so we actually form a trialkylborane. So let me draw in the boron with three alkyl groups on it. So it should look like this to form a trialkylborane. Some textbooks just leave it as a monoalkylborane and proceed to the oxidation part of this reaction. But most of the time it's a trialkylborane."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw in the boron with three alkyl groups on it. So it should look like this to form a trialkylborane. Some textbooks just leave it as a monoalkylborane and proceed to the oxidation part of this reaction. But most of the time it's a trialkylborane. So that's what I'm gonna show you here. When you move on to the oxidation step, remember this is the second step here, you have hydrogen peroxide and you have hydroxide ions. So let's draw those out."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "But most of the time it's a trialkylborane. So that's what I'm gonna show you here. When you move on to the oxidation step, remember this is the second step here, you have hydrogen peroxide and you have hydroxide ions. So let's draw those out. All right, we have some hydrogen peroxide. So let me draw the dot structure for hydrogen peroxide. Put in lone pairs of electrons on the oxygens."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's draw those out. All right, we have some hydrogen peroxide. So let me draw the dot structure for hydrogen peroxide. Put in lone pairs of electrons on the oxygens. And we have hydroxide, so OH minus. Hydroxide's gonna function as a base. It's gonna take a proton from hydrogen peroxide, leaving these electrons behind on the oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Put in lone pairs of electrons on the oxygens. And we have hydroxide, so OH minus. Hydroxide's gonna function as a base. It's gonna take a proton from hydrogen peroxide, leaving these electrons behind on the oxygen. So we would make the hydroperoxide ions. So I'm sketching that in over here. Three lone pairs of electrons on the oxygen on the right, therefore a negative one formal charge."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "It's gonna take a proton from hydrogen peroxide, leaving these electrons behind on the oxygen. So we would make the hydroperoxide ions. So I'm sketching that in over here. Three lone pairs of electrons on the oxygen on the right, therefore a negative one formal charge. So we can show these electrons in magenta coming off onto the oxygen. So we have the hydroperoxide ion, which is gonna function as a nucleophile. And our nucleophile's going to attack the empty orbital of our boron."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Three lone pairs of electrons on the oxygen on the right, therefore a negative one formal charge. So we can show these electrons in magenta coming off onto the oxygen. So we have the hydroperoxide ion, which is gonna function as a nucleophile. And our nucleophile's going to attack the empty orbital of our boron. So the nucleophile attacks the trialkylborane. And let's show the result of that. So now there's a bond between the boron and the oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And our nucleophile's going to attack the empty orbital of our boron. So the nucleophile attacks the trialkylborane. And let's show the result of that. So now there's a bond between the boron and the oxygen. And then there's an oxygen-oxygen bond, and then a hydrogen here. And let me draw in lone pairs of electrons on the oxygens. So lone pairs of electrons here."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So now there's a bond between the boron and the oxygen. And then there's an oxygen-oxygen bond, and then a hydrogen here. And let me draw in lone pairs of electrons on the oxygens. So lone pairs of electrons here. And then the boron still has our three alkyl groups. So I'm sketching in the alkyl groups here like that. The next step of the mechanism, oh, we have a negative one formal charge on the boron here."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So lone pairs of electrons here. And then the boron still has our three alkyl groups. So I'm sketching in the alkyl groups here like that. The next step of the mechanism, oh, we have a negative one formal charge on the boron here. The next step is the migration of an alkyl group. So this is the weird part. So these electrons right here are actually going to form a bond with the oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "The next step of the mechanism, oh, we have a negative one formal charge on the boron here. The next step is the migration of an alkyl group. So this is the weird part. So these electrons right here are actually going to form a bond with the oxygen. At the same time, these electrons are gonna come off onto this oxygen. So we form hydroxide. Let me go ahead and show that first."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons right here are actually going to form a bond with the oxygen. At the same time, these electrons are gonna come off onto this oxygen. So we form hydroxide. Let me go ahead and show that first. That's maybe the easier part to understand of what's going on here. So we have three lone pairs of electrons on the oxygen, negative one formal charge. Let me make these electrons green."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me go ahead and show that first. That's maybe the easier part to understand of what's going on here. So we have three lone pairs of electrons on the oxygen, negative one formal charge. Let me make these electrons green. So these electrons in green here come off onto the oxygen. So the oxygen-oxygen bond is weak. So it's relatively easy to break this bond."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Let me make these electrons green. So these electrons in green here come off onto the oxygen. So the oxygen-oxygen bond is weak. So it's relatively easy to break this bond. And let me highlight the other electrons here. Well, these electrons in magenta formed this bond between the oxygen and the boron. And now we get our alkyl group migration."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So it's relatively easy to break this bond. And let me highlight the other electrons here. Well, these electrons in magenta formed this bond between the oxygen and the boron. And now we get our alkyl group migration. And I'll show these electrons in blue. So the electrons in blue are forming a bond between this carbon and this oxygen. So let me draw that out."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And now we get our alkyl group migration. And I'll show these electrons in blue. So the electrons in blue are forming a bond between this carbon and this oxygen. So let me draw that out. It's pretty hard to see. Let me go ahead and draw what we have. And then we'll highlight some electrons here."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let me draw that out. It's pretty hard to see. Let me go ahead and draw what we have. And then we'll highlight some electrons here. So now we have our oxygen bonded to this group. And our boron has two alkyl groups still bonded to it like that. So let's follow those electrons."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then we'll highlight some electrons here. So now we have our oxygen bonded to this group. And our boron has two alkyl groups still bonded to it like that. So let's follow those electrons. The electrons in magenta represented the bond between the oxygen and the boron. And the electrons in blue would be these electrons right here. So this carbon is now bonded to this oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So let's follow those electrons. The electrons in magenta represented the bond between the oxygen and the boron. And the electrons in blue would be these electrons right here. So this carbon is now bonded to this oxygen. So the migration of the alkyl group removes the formal charge on the boron, breaks the weak oxygen-oxygen bond, and kicks off hydroxide. So now this process happens two more times because we have these two other alkyl groups. And so when that happens two more times, we get, let me go ahead and draw it in, a trialkoxyborane."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So this carbon is now bonded to this oxygen. So the migration of the alkyl group removes the formal charge on the boron, breaks the weak oxygen-oxygen bond, and kicks off hydroxide. So now this process happens two more times because we have these two other alkyl groups. And so when that happens two more times, we get, let me go ahead and draw it in, a trialkoxyborane. So we would have an oxygen here, and then we have our group coming off of that. And then over here we would have our oxygen, and then we have our group coming off of that. And then also down here."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And so when that happens two more times, we get, let me go ahead and draw it in, a trialkoxyborane. So we would have an oxygen here, and then we have our group coming off of that. And then over here we would have our oxygen, and then we have our group coming off of that. And then also down here. All right, our next step is the hydroxide anion functions as a nucleophile. So the hydroxide attacks the boron. Once again, our boron has an empty orbital."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And then also down here. All right, our next step is the hydroxide anion functions as a nucleophile. So the hydroxide attacks the boron. Once again, our boron has an empty orbital. So nucleophilic attack. Let's draw the results of our nucleophilic attack. Now we have OH bonded to the boron."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Once again, our boron has an empty orbital. So nucleophilic attack. Let's draw the results of our nucleophilic attack. Now we have OH bonded to the boron. And the boron still has all of these groups around it. So let me sketch all of those in. And you can see why this mechanism is getting rather cumbersome at this point."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "Now we have OH bonded to the boron. And the boron still has all of these groups around it. So let me sketch all of those in. And you can see why this mechanism is getting rather cumbersome at this point. Drawing in all these groups is a little bit annoying here. But we still have a negative one, I should say we have a negative one formal charge in the boron, and we still have some lone pairs of electrons on this oxygen. And I'm putting those in because in the next step to get rid of the negative one formal charge, these electrons come off onto the oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And you can see why this mechanism is getting rather cumbersome at this point. Drawing in all these groups is a little bit annoying here. But we still have a negative one, I should say we have a negative one formal charge in the boron, and we still have some lone pairs of electrons on this oxygen. And I'm putting those in because in the next step to get rid of the negative one formal charge, these electrons come off onto the oxygen. And that would form an alkoxide anion. So now we have this oxygen would have, let me do a better job of drawing in our alkoxide anion. So we'd have three lone pairs of electrons on our oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "And I'm putting those in because in the next step to get rid of the negative one formal charge, these electrons come off onto the oxygen. And that would form an alkoxide anion. So now we have this oxygen would have, let me do a better job of drawing in our alkoxide anion. So we'd have three lone pairs of electrons on our oxygen. That's a negative one formal charge. And let me highlight those electrons. So these electrons in here in magenta come off onto the oxygen."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So we'd have three lone pairs of electrons on our oxygen. That's a negative one formal charge. And let me highlight those electrons. So these electrons in here in magenta come off onto the oxygen. And we form our alkoxide anion. We are almost done. So in the last step, we have water present."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So these electrons in here in magenta come off onto the oxygen. And we form our alkoxide anion. We are almost done. So in the last step, we have water present. So water can donate a proton to our alkoxide anion. So the alkoxide anion picks a proton up from water. We protonate the alkoxide anion, we form our alcohol."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "So in the last step, we have water present. So water can donate a proton to our alkoxide anion. So the alkoxide anion picks a proton up from water. We protonate the alkoxide anion, we form our alcohol. And so we're finally done. A very long mechanism, a lot of drawing. But we formed the alcohol that we know, because we know the OH added on to, let me highlight it here, this carbon."}, {"video_title": "Hydroboration-oxidation Mechanism Alkenes and Alkynes Organic chemistry Khan Academy.mp3", "Sentence": "We protonate the alkoxide anion, we form our alcohol. And so we're finally done. A very long mechanism, a lot of drawing. But we formed the alcohol that we know, because we know the OH added on to, let me highlight it here, this carbon. So again, you could probably do this mechanism just doing the monoalkylborane. And I'm sure some professors will let you do that. Most of the time it's the trialkylborane."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Before we start going off into things outside of our solar system, I want to take a few steps back because I found this neat picture of the sun over here. And the reason why, at least in my mind, it's kind of mind-blowing is because at this scale, the sun is obviously still a huge object at this scale, the Earth would be roughly, and this is an approximation, roughly that big. And so for me at least, this is mind-blowing because it's this idea that our whole planet, everything could fit into one of these plasma flares coming off of the sun. And you can only imagine, we can't realistically be there, but if you were in some type of protected capsule, what it would be like to be in this type of an environment. So I just thought this was kind of a fascinating concept. But anyway, with that out of the way, let's just think about what it means to be at the boundary of the solar system. In the last video, we explored the Oort Belt, which was about, it started a little under one light year away from the sun."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And you can only imagine, we can't realistically be there, but if you were in some type of protected capsule, what it would be like to be in this type of an environment. So I just thought this was kind of a fascinating concept. But anyway, with that out of the way, let's just think about what it means to be at the boundary of the solar system. In the last video, we explored the Oort Belt, which was about, it started a little under one light year away from the sun. But depending on what you view as the boundary of the solar system, it could be something way farther in or it could be something as far out as something like the Oort Cloud. So if the sun, we see these things being ejected, but even in unseen ways, or unseen particles, super high energy electrons and protons are also being ejected from the sun. It's super high velocities, 400 kilometers per second."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "In the last video, we explored the Oort Belt, which was about, it started a little under one light year away from the sun. But depending on what you view as the boundary of the solar system, it could be something way farther in or it could be something as far out as something like the Oort Cloud. So if the sun, we see these things being ejected, but even in unseen ways, or unseen particles, super high energy electrons and protons are also being ejected from the sun. It's super high velocities, 400 kilometers per second. Let me write that down. 400 kilometers per second. And on Earth, we're protected from these highly energetic particles because of Earth's magnetic field."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's super high velocities, 400 kilometers per second. Let me write that down. 400 kilometers per second. And on Earth, we're protected from these highly energetic particles because of Earth's magnetic field. But if you're on the surface of the moon, when the sun is on top, and you're not on the dark side of the moon, you'll have direct contact with these. And as you can imagine, not the best thing to hang around in too long. But the whole reason why I'm even talking about these, these charged particles that are coming out at huge velocities from the surface of the sun, these are considered the solar wind."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "And on Earth, we're protected from these highly energetic particles because of Earth's magnetic field. But if you're on the surface of the moon, when the sun is on top, and you're not on the dark side of the moon, you'll have direct contact with these. And as you can imagine, not the best thing to hang around in too long. But the whole reason why I'm even talking about these, these charged particles that are coming out at huge velocities from the surface of the sun, these are considered the solar wind. These are the solar wind. And I'll put wind in quotes because it's really very different than our traditional association of a nice breeze. These are just charged particles that are going out at super high velocities from the sun."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the whole reason why I'm even talking about these, these charged particles that are coming out at huge velocities from the surface of the sun, these are considered the solar wind. These are the solar wind. And I'll put wind in quotes because it's really very different than our traditional association of a nice breeze. These are just charged particles that are going out at super high velocities from the sun. I'm even going into the idea of the solar wind because to some degree, they can help us with one definition of maybe the limits of the solar system. And that's the limits of how far the solar wind is getting before it kind of comes in confrontation with the interstellar medium. This right here shows a depiction of that."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These are just charged particles that are going out at super high velocities from the sun. I'm even going into the idea of the solar wind because to some degree, they can help us with one definition of maybe the limits of the solar system. And that's the limits of how far the solar wind is getting before it kind of comes in confrontation with the interstellar medium. This right here shows a depiction of that. So the Oort cloud, at least the edges of the dense part of it, is way outside of this. As we saw, this is just where Voyager 1, Voyager 2, if we wanted the orbit of Sedna, it would be something like, the close part would be something over here and then it would go out. But the Oort cloud is much, much further out."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This right here shows a depiction of that. So the Oort cloud, at least the edges of the dense part of it, is way outside of this. As we saw, this is just where Voyager 1, Voyager 2, if we wanted the orbit of Sedna, it would be something like, the close part would be something over here and then it would go out. But the Oort cloud is much, much further out. So if you look at this kind of view of the solar system as the extent of the solar wind, it's much smaller than the Oort cloud, but it's still fairly large. So this is right here, this heliopause right here, and I got this from Wikipedia. This is essentially where the velocity and the forces of the solar wind are counteracted."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But the Oort cloud is much, much further out. So if you look at this kind of view of the solar system as the extent of the solar wind, it's much smaller than the Oort cloud, but it's still fairly large. So this is right here, this heliopause right here, and I got this from Wikipedia. This is essentially where the velocity and the forces of the solar wind are counteracted. The pressure is so diluted at this point that it's counteracted by mainly the hydrogen and the helium that's in the interstellar medium that's just kind of out there. After this point, it's not really being injected out anymore. There's this kind of pause, I guess you could say."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "This is essentially where the velocity and the forces of the solar wind are counteracted. The pressure is so diluted at this point that it's counteracted by mainly the hydrogen and the helium that's in the interstellar medium that's just kind of out there. After this point, it's not really being injected out anymore. There's this kind of pause, I guess you could say. Voyager 1 and Voyager 2 have essentially gotten pretty close to, people believe, that pause over there. So that's one view of the edges of the solar system. There's never going to be any hard edge to it."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's this kind of pause, I guess you could say. Voyager 1 and Voyager 2 have essentially gotten pretty close to, people believe, that pause over there. So that's one view of the edges of the solar system. There's never going to be any hard edge to it. Another view would be something like the Oort cloud, the area where you have the still objects out there. This is all actually, we haven't directly observed objects in the Oort cloud. We think that they are out there."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "There's never going to be any hard edge to it. Another view would be something like the Oort cloud, the area where you have the still objects out there. This is all actually, we haven't directly observed objects in the Oort cloud. We think that they are out there. Then maybe the most abstract definition would be a significant influence from the sun's gravitational pull. All of those ways are to imagine the extent of the solar system, but they all kind of leave a gray area for what is and what is not in the solar system. My whole point here, what I want to do is start exploring a little bit outside of the solar system and just give you a sense of the scale as we just go to the closest star."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We think that they are out there. Then maybe the most abstract definition would be a significant influence from the sun's gravitational pull. All of those ways are to imagine the extent of the solar system, but they all kind of leave a gray area for what is and what is not in the solar system. My whole point here, what I want to do is start exploring a little bit outside of the solar system and just give you a sense of the scale as we just go to the closest star. If we go right over here, this shows our local neighborhood from a stellar point of view. Even though these stars look pretty big, if you actually were to draw, this is our solar system right here. You might be saying, oh, maybe that's the sun."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "My whole point here, what I want to do is start exploring a little bit outside of the solar system and just give you a sense of the scale as we just go to the closest star. If we go right over here, this shows our local neighborhood from a stellar point of view. Even though these stars look pretty big, if you actually were to draw, this is our solar system right here. You might be saying, oh, maybe that's the sun. No, the sun, if you were to draw it here, it wouldn't even make up one pixel. In fact, the entire orbit of Pluto, everything inside of it, still would not make up one pixel on the screen right here. What we see right here, which is a radius of about, give or take, a light year, this is roughly maybe the radius of the Oort cloud."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "You might be saying, oh, maybe that's the sun. No, the sun, if you were to draw it here, it wouldn't even make up one pixel. In fact, the entire orbit of Pluto, everything inside of it, still would not make up one pixel on the screen right here. What we see right here, which is a radius of about, give or take, a light year, this is roughly maybe the radius of the Oort cloud. We saw in the last video how huge that was, especially relative to the radius of, say, Pluto's orbit, which is roughly like that. That itself is a huge, huge diameter or a huge distance away from the sun. That wouldn't even make a pixel on this diagram right over here."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "What we see right here, which is a radius of about, give or take, a light year, this is roughly maybe the radius of the Oort cloud. We saw in the last video how huge that was, especially relative to the radius of, say, Pluto's orbit, which is roughly like that. That itself is a huge, huge diameter or a huge distance away from the sun. That wouldn't even make a pixel on this diagram right over here. Just to give you an idea of how far we are, we're a speck of a speck of a speck inside here, of a pixel of a pixel in the center here. We're in the center here to make it from our solar system, or in particular from Earth maybe, to the nearest star, or maybe the nearest cluster of stars, the Alpha Centauri. They're the nearest cluster of stars."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That wouldn't even make a pixel on this diagram right over here. Just to give you an idea of how far we are, we're a speck of a speck of a speck inside here, of a pixel of a pixel in the center here. We're in the center here to make it from our solar system, or in particular from Earth maybe, to the nearest star, or maybe the nearest cluster of stars, the Alpha Centauri. They're the nearest cluster of stars. There's three stars, Alpha Centauri A, which is the largest, Alpha Centauri B, and then there's one that you can't observe with the naked eye, Alpha Proximus, or I think it's Proximus Centauri, I think is what it's called, not Alpha Proximus. Proximus Centauri. That's a much smaller star, but that's the closest star."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "They're the nearest cluster of stars. There's three stars, Alpha Centauri A, which is the largest, Alpha Centauri B, and then there's one that you can't observe with the naked eye, Alpha Proximus, or I think it's Proximus Centauri, I think is what it's called, not Alpha Proximus. Proximus Centauri. That's a much smaller star, but that's the closest star. You can view it as this whole cluster of stars right here. They're the closest. It's about 4.2 light years away."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "That's a much smaller star, but that's the closest star. You can view it as this whole cluster of stars right here. They're the closest. It's about 4.2 light years away. Or another way to think about it, if someone were to shine a light on one of these planets, and assuming that light could get to us, it would take 4.2 years to get to us. If these guys just disappeared or blew up, we wouldn't know it for 4.2 years. We'd say, hey, that's not too bad."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "It's about 4.2 light years away. Or another way to think about it, if someone were to shine a light on one of these planets, and assuming that light could get to us, it would take 4.2 years to get to us. If these guys just disappeared or blew up, we wouldn't know it for 4.2 years. We'd say, hey, that's not too bad. We should take a trip over there and check them out, see if there are any other people there that we can meet and exchange technologies with or whatnot. But this is a huge distance. Just this 4.2 light years is an unbelievably ridiculous distance."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "We'd say, hey, that's not too bad. We should take a trip over there and check them out, see if there are any other people there that we can meet and exchange technologies with or whatnot. But this is a huge distance. Just this 4.2 light years is an unbelievably ridiculous distance. Just to give you a sense, the Voyager 1 and 2, we talked about in the last video, and we can even see how far they've gotten. They've gotten pretty much to the heliopause. These guys are traveling at 60,000 kilometers an hour, which is the same thing as 17 kilometers per second."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "Just this 4.2 light years is an unbelievably ridiculous distance. Just to give you a sense, the Voyager 1 and 2, we talked about in the last video, and we can even see how far they've gotten. They've gotten pretty much to the heliopause. These guys are traveling at 60,000 kilometers an hour, which is the same thing as 17 kilometers per second. If we were able to get up to those type of velocities, and these guys got up to those type of velocities by leveraging the gravitational pull of some of the larger planets to accelerate and keep accelerating. This is a pretty hard velocity to actually reach. But if you were able to reach that velocity and go straight in the direction of the Alpha Centauri system, the closest stars to Earth, it would take you 80,000 years traveling at the same velocity as Voyager 1, which is the fastest of the Voyagers."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "These guys are traveling at 60,000 kilometers an hour, which is the same thing as 17 kilometers per second. If we were able to get up to those type of velocities, and these guys got up to those type of velocities by leveraging the gravitational pull of some of the larger planets to accelerate and keep accelerating. This is a pretty hard velocity to actually reach. But if you were able to reach that velocity and go straight in the direction of the Alpha Centauri system, the closest stars to Earth, it would take you 80,000 years traveling at the same velocity as Voyager 1, which is the fastest of the Voyagers. It's a ridiculously long time. We're going to have to figure out some better way to do that."}, {"video_title": "Scale of distance to closest stars Scale of the universe Cosmology & Astronomy Khan Academy.mp3", "Sentence": "But if you were able to reach that velocity and go straight in the direction of the Alpha Centauri system, the closest stars to Earth, it would take you 80,000 years traveling at the same velocity as Voyager 1, which is the fastest of the Voyagers. It's a ridiculously long time. We're going to have to figure out some better way to do that."}]